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-\Day{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\section{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -\tikzset{every tree node/.style={minimum width=2em,draw,circle}, - blank/.style={draw=none}, - edge from parent/.style= - {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, - level distance=1.5cm} -\begin{align*} -\begin{tikzpicture}[sibling distance=40pt] -\Tree -[.{} - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - ] - \edge[thick]; - [.\node[thick]{+}; - \edge[thick]; [.\node[thick]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[thick]; \node[thick]{+}; - ] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+}; - ] - ] -] -\end{tikzpicture} -\end{align*} -%--------------------------------------------------------------------- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\Day{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\Day{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\section{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\Day{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\NotesBy{Notes by John Suice} + +\Day{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\section{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +\tikzset{every tree node/.style={minimum width=2em,draw,circle}, + blank/.style={draw=none}, + edge from parent/.style= + {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, + level distance=1.5cm} +\begin{align*} +\begin{tikzpicture}[sibling distance=40pt] +\Tree +[.{} + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + ] + \edge[thick]; + [.\node[thick]{+}; + \edge[thick]; [.\node[thick]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[thick]; \node[thick]{+}; + ] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+}; + ] + ] +] +\end{tikzpicture} +\end{align*} +%--------------------------------------------------------------------- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\Day{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\Day{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\section{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\Day{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/all_notes/week_10/December_8_12.tex b/Other/all_notes/week_10/December_8_12.tex index f3120a42..18f65db3 100644 --- a/Other/all_notes/week_10/December_8_12.tex +++ b/Other/all_notes/week_10/December_8_12.tex @@ -1,471 +1,471 @@ -\global\long\def\N{\mathbb{N}} -\global\long\def\Z{\mathbb{Z}} -\global\long\def\Q{\mathbb{Q}} -\global\long\def\R{\mathbb{R}} -\global\long\def\lto{\longrightarrow} -\global\long\def\es{\emptyset} -\global\long\def\F{\mathcal{F}} -\global\long\def\force{\Vdash} -\global\long\def\dom{\textrm{dom}} -\global\long\def\em{\prec} -\global\long\def\cf{\textrm{cf}} -\global\long\def\model{\vDash} -\global\long\def\crit{\mathrm{crit}} -\global\long\def\ult{\mathrm{Ult}} -\global\long\def\inj{\hookrightarrow} -\global\long\def\u{\mathcal{U}} -\global\long\def\dprime{\prime\prime} -\global\long\def\C{\mathbb{C}} -\global\long\def\v{\mathcal{V}} -\global\long\def\w{\mathcal{W}} -\global\long\def\i{\imath} -\global\long\def\P{\mathbb{P}} -\global\long\def\del{\partial} -\global\long\def\an{\mathrm{An}} -\global\long\def\I{\mathrm{I}} -\global\long\def\L{\mathcal{L}} -\global\long\def\D{\mathcal{D}} - -\NotesBy{Notes by Asaaf Sahni} -\Week{Week 10} - -Proving, after all, that $\C\left[\left[X\right]\right]$ is faithfully -flat over $\C\left\{ X\right\} $. - -We do this by induction on the number of variables in $X$. We only -show that $\C\left[\left[X,T\right]\right]$ is flat over $\C\left\{ X,T\right\} $.\\ -Assume that $\C\left[\left[X\right]\right]$ is flat over $\C\left\{ X\right\} $, -and consider $\C\left[\left[X,T\right]\right]$ and $\C\left\{ X,T\right\} $. -\\ -Consider the equation -\[ -\left(\ast\right)_{0}\qquad f_{1}y_{1}+...+f_{n}y_{n}=0,\qquad\textrm{where }f_{i}\in\C\left\{ X,T\right\} . -\] -We can assume that all $f_{i}$'s are non zero, and regular in $T$. -\\ -Apply W.P to $f_{i}$ in $\C\left\{ X,T\right\} $, $f_{i}=u_{i}w_{i}$ -where $u_{i}\in\C\left\{ X,T\right\} ^{\times}$ and $w_{i}\in\C\left\{ X\right\} \left[T\right]$ -is monic of degree $d_{i}$. Note that if $\left(y_{1},...,y_{n}\right)$ -is a solution to $\left(\ast\right)_{0}$, then $\left(u_{1}^{-1}y_{1},...,u_{n}^{-1}y_{n}\right)$ -is a solution to -\[ -\left(\ast\right)\qquad w_{1}y_{1}+...+w_{n}y_{n}=0. -\] -So it is enough to show that each solution to $\left(\ast\right)$ -in $\C\left[\left[X,T\right]\right]$ is a linear combination of solutions -from $\C\left\{ X,T\right\} $.\\ -Consider $z_{2}=\left(w_{2},-w_{1},0,...,0\right)$, $z_{3}=\left(w_{3},0,-w_{1},0,...,0\right)$, -... $z_{n}=\left(w_{n},0,...,0,-w_{1}\right)$. Each $z_{i}$ is a -solution to $\left(\ast\right)$ (and are in $\C\left\{ X,T\right\} $).\\ -Now suppose $y=\left(y_{1},...,y_{n}\right)$ is a solution to $\left(\ast\right)$ -in $\C\left[\left[X,T\right]\right]$. \\ -For each $i$, write $y_{i}=q_{i}w_{1}+r_{i}$ where $q_{i}\in\C\left[\left[X,T\right]\right]$ -and $r_{i}\in\C\left[\left[X\right]\right]\left[T\right]$ of degree -$0}$. -\end{itemize} -Let $\mathrm{An}\left(\u\right)$ be all analytic functions on $\u$. -$\an\left(\u\right)$ has an $\R$-algebra structure. If $f\in\an\left(\u\right)$, -then $\frac{\del f}{\del X_{i}}\in\an\left(\u\right).$ - -For each $a\in\u$ there is a map of $\R$-algebras: -\begin{eqnarray*} -\an\left(\u\right) & \lto & \R\left\{ X\right\} .\\ -f & \mapsto & f_{a} -\end{eqnarray*} - -\begin{prop*} -\label{11.2} (Analytic continuation) If $\u$ is connected, then the map -above is injective. In particular, $\an\left(\u\right)$ is an integral -domain.\end{prop*} -\begin{cor*} -\label{11.3} If $\u$ is connected, $f,g\in\an\left(\u\right)$ agree on -a non empty open subset of $\u$, then $f=g$.\end{cor*} -\begin{prop*} -\label{11.4} Let $f_{1},...,f_{n}\in\an\left(\u\right)$, $\v\subset\R^{n}$ -open s.t $f\left(\u\right)\subset V$, where $f=\left(f_{1},...,f_{n}\right)\colon\u\lto\R^{n}$. - -Then for any $g\in\an\left(\v\right)$ we have $g\circ f\in\an\left(\u\right)$ -and -\[ -\left(g\circ f\right)_{a}=g_{f\left(a\right)}\left(f_{a}-f\left(a\right)\right). -\] - -\end{prop*} -\label{11.5} \uline{Notation}: For $x=\left(x_{1},...,x_{m}\right)\in\R^{m}$, -$\left|x\right|=\max\left\{ \left|x\right|_{1},...,\left|x_{m}\right|\right\} $. -\\ -For $\delta>0$, let $\delta=\left(\delta,...,\delta\right)$ and -$\R\left\{ X\right\} _{\delta^{+}}=\bigcup_{r>\delta}\R\left\{ X\right\} _{r}$.\\ -Each $f\in\R\left\{ X\right\} _{\delta^{+}}$ gives rise to a function -$x\mapsto f\left(x\right)\colon\bar{B}_{\delta}\left(0\right)\lto\R$ -which extendes to an analytic function on an open nbhd of $\bar{B}_{\delta}$. - -Let $Y=\left(Y_{1},...,Y_{n}\right)$, $n\geq1$. Fix $f=f\left(X,Y\right)\in\R\left\{ X,Y\right\} $. -Then for small $x$ we have $f\left(x,Y\right)\in\R\left\{ Y\right\} $.\\ -Consider the following question: How does W.P for $f\left(x,Y\right)$ -depend on $x$, for small $x$?\\ -We will show that for some $\epsilon>0$, $\bar{B}_{\epsilon}\subset\R^{m}$ -can be covered by finitely many ``special sets'', on each of which -W.P is uniform in $x$. -\begin{defn*} -\label{11.6} A special subset of $\bar{B}_{\epsilon}$ is a finite union -of sets of the following form -\[ -\left\{ x\in\bar{B}_{\epsilon};\, f\left(x\right)=0,\, g_{1}\left(x\right)>0,\,...,\, g_{k}\left(x\right)>0\right\} , -\] -where $f,g_{1},...,g_{k}\in\R\left\{ X\right\} _{\epsilon^{+}}$. -\end{defn*} -We first show that $\mathrm{ord}\left(f\left(x,Y\right)\right)$ takes -only finitely many values as $x$ ranges over a neighbourhood of $0\in\R^{m}$. -Write -\[ -f\left(X,Y\right)=\sum_{j}f_{j}\left(X\right)Y^{j}, -\] -where $f_{j}\left(X\right)\in\R\left\{ X\right\} $. Since $\R\left\{ X\right\} $ -is neotherian, the ideal generated by $\left\{ f_{j}\left(X\right)\right\} _{j\in\N^{n}}$ -is finitely generated, hence is generated by $\left\{ f_{j}\left(X\right)\right\} _{\left|j\right|d$, -\[ -f_{j}\left(X\right)=\sum_{\left|i\right|\leq d}g_{ij}f_{i}. -\] -In the following, $i$ ranges over elements of $\N^{n}$ s.t $\left|i\right|\leq d$ -and $j$ over elements of $\N^{n}$ s.t $\left|j\right|>d$. \\ -Substituting the above, we get -\[ -\left(1\right)\qquad f\left(X,Y\right)=\sum_{i}f_{i}\left(X,Y\right)\left(Y^{i}+\sum_{j}g_{ij}\left(X\right)Y^{j}\right), -\] -in $\R\left[\left[X,Y\right]\right]$. Note that for each $i,j$, -$g_{ij}\in\R\left\{ X\right\} $, and so there is some $\delta$ s.t -$g_{ij}\in\R\left\{ X\right\} _{\delta}$. However, in order to have -equation $\left(1\right)$ as an equality in $\R\left\{ X\right\} $, -we need to find a uniform $\delta$ that works for all $i,j$. -\begin{claim*} -$\exists\delta\in\left(0,1\right]$ and there are $f_{i},\, g_{ij}\in\R\left\{ X\right\} _{\delta^{+}}$ -(maybe different than above) such that $\sum_{j}g_{ij}\left(X\right)Y^{j}\in\R\left\{ X,Y\right\} _{\delta^{+}}$ -and $\left(1\right)$ holds in $\R\left\{ X\right\} _{\delta^{+}}$ -with $f_{i},g_{ij}$.\end{claim*} -\begin{proof} -Consider the linear equation -\[ -f=\sum_{i}f_{i}\left(Y^{i}+\sum_{\left|j\right|=d+1}Z_{ij}Y^{j}\right). -\] -By $\left(1\right)$ above, there is a solution $\left\{ Z_{ij};\,\left|i\right|\leq d,\,\left|j\right|=d+1\right\} $ -in $\R\left[\left[X,Y\right]\right]$ (all the higher terms, $Y^{j}$ -for $\left|j\right|>d+1$, are inside the $Z_{ij}$'s). Since $\R\left[\left[X,Y\right]\right]$ -is f.f. over $\R\left\{ X,Y\right\} $, there is a solution $\left\{ Z_{ij}\right\} $ -in $\R\left\{ X,Y\right\} $. Take $\delta$ for all $i,j$, $Z_{ij}\in\R\left\{ X,Y\right\} _{\delta^{+}}$. -Now write $Z_{ij}$ as power series in $Y$ with coefficients in $\R\left\{ X\right\} $ -to get $\left(1\right)$ with $g_{ij}\left(X\right)\in\R\left\{ X\right\} _{\delta^{+}}$ -for all $\left|i\right|\leq d$, $\left|j\right|>d$. -\end{proof} -We work with equation $\left(1\right)$ in $\R\left\{ X,Y\right\} _{\delta^{+}}$ -as given by the claim above. For $\left|x\right|\leq\delta$, -\[ -f\left(x,Y\right)=0\iff\forall i\left(f_{i}\left(x\right)=0\right). -\] -Also, if $f_{i}\left(x\right)\neq0$, then $\mathrm{ord}\left(f\left(x,Y\right)\right)\leq\left|i\right|$. -So -\[ -f\left(x,Y\right)\neq0\implies\mathrm{ord}\left(f\left(x,Y\right)\right)\leq d. -\] -Define -\begin{eqnarray*} -Z_{\delta} & = & \left\{ x\in\bar{B}_{\delta};\,\forall i\, f_{i}\left(x\right)=0\right\} .\\ -S_{i} & = & \left\{ x\in\bar{B}_{\delta};\, f_{i}\left(x\right)\neq0\wedge\forall i'\neq i\left(\left|f_{i}\left(x\right)\right|\geq\left|f_{i'}\left(x\right)\right|\right)\right\} . -\end{eqnarray*} -Note that $\bar{B}_{\delta}=Z_{\delta}\cup\bigcup_{i}S_{i}$.\\ -Fix some $i$, formally divide the expression for $f$ in $\left(1\right)$ -by $f_{i}$, and introduce new variables $V_{i,i'}$ for the quotients -$\nicefrac{f_{i}}{f_{i'}}$, for $i\neq i'$. Let $V_{i}=\left(V_{ii'}\right)_{i'\neq i}$. -Define -\[ -F_{i}=Y^{i}+\sum_{j}g_{ij}Y^{j}+\sum_{i'\neq i}V_{ii'}\left(Y^{i'}+\sum_{j}g_{i'j}Y^{j}\right)\in\R\left\{ X,V_{i},Y\right\} . -\] -For $x\in S_{i}$, let $v_{i}\left(x\right)=\left(\nicefrac{f_{i'}\left(x\right)}{f_{i}\left(x\right)}\right)_{i'\neq i}$. -Then for $x\in S_{i}$, $\left|v_{i}\left(x\right)\right|\leq1$, -and -\[ -\left(2\right)\qquad f\left(x,Y\right)=f_{i}\left(x\right)F_{i}\left(x,v_{i}\left(x\right),Y\right). -\] - - -Idea: apply W.P to $F_{i}$'s locally around every point $X=0$, $V_{i}=c$, -$Y=0$. \\ -For $c=\left(c_{i'}\right)_{i'\neq i}$ with $\left|c\right|\leq1$, -put -\[ -F_{i,c}=F_{i}\left(X,c+V_{i},Y\right). -\] -Then for $x\in S_{i}$, -\[ -f\left(x,Y\right)=f_{i}\left(x\right)F_{i,c}\left(x,v_{i}\left(x\right)-c,Y\right). -\] - -\begin{claim*} -For each $c$ there is $\lambda=\lambda\left(c\right)\in\R^{n-1}$ -with $\left|\lambda\right|\leq1$ such that $F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)$ -is regular in $Y_{n}$ of order$\leq\left|i\right|$, where for $Y=\left(Y_{1},...,Y_{n}\right)$, -$\lambda\left(Y\right)=\left(Y_{1}+\lambda Y_{n},...,Y_{n-1}+\lambda_{n-1}Y_{n},Y_{n}\right)$.\end{claim*} -\begin{proof} -Exercise. -\end{proof} -By W.P., for such $\lambda$, -\[ -\left(3\right)\qquad F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)=\u_{i,c}\w_{i,c},\quad\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} ^{\ast},\:\w_{i,c}\in\R\left\{ X,V_{i},Y_{1},...,Y_{n-1}\right\} \left[Y_{n}\right]. -\] -Take $\epsilon\left(i,c\right)\in\left(0,\delta\right]$ s.t -\begin{itemize} -\item $\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}^{\ast}$. -\item $\w_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}.$ -\end{itemize} -Let $\Gamma\left(i\right)=\left\{ i';\, i'\neq i\right\} $. Note -that our $c$'s vary over $I^{\Gamma\left(i\right)}$, where $\I=\left[-1,1\right]$. -By compactness, there is a finite set $C\left(i\right)$ s.t -\[ -\I^{\Gamma\left(i\right)}\subset\bigcup_{c\in C\left(i\right)}B_{\epsilon\left(i,c\right)}\left(c\right). -\] -Now consider the finite set $\Gamma=\left\{ \left(i,c\right);\,\left|i\right|\leq d,\, d\in C\left(i\right)\right\} $.\\ -Take $\epsilon>0$ s.t $\epsilon\leq\frac{\epsilon\left(i,c\right)}{4}$ -for all $\left(i,c\right)\in\Gamma.$ For $\gamma=\left(i,c\right)\in\Gamma$, -let -\[ -S_{\gamma}=\left\{ x\in S_{i};\,\left|x\right|\leq\epsilon,\,\left|v_{i}\left(x\right)-c\right|<\epsilon\left(i,c\right)\right\} . -\] -Then $S_{i}\cap\bar{B}_{\epsilon}=\bigcup\left\{ S_{\gamma};\,\gamma=\left(i,c\right),\, c\in C\left(i\right)\right\} $. -\\ -So $\bar{B}_{\epsilon}=\left(\underbrace{Z_{\delta}\cap\bar{B}_{\epsilon}}_{\equiv Z}\right)\cup\bigcup_{\gamma}S_{\gamma}$.\\ -For $\gamma=\left(i,c\right)\in\Gamma$, by $\left(2\right)$ and -$\left(3\right)$, -\[ -f\left(x,\lambda\left(Y\right)\right)=f_{i}\left(x\right)\u_{\gamma}\left(x,v_{i}\left(x\right),Y\right)W_{\gamma}\left(x,v_{i}\left(x\right),Y\right). -\] -Note: $\u_{\gamma}$ does not change sign on $\bar{B}_{\epsilon\left(\gamma\right)}$. -Therefore there is $\sigma\left(\gamma\right)\in\left\{ \pm1\right\} $ -s.t -\[ -\mathrm{sign}\left(f\left(x,\lambda\left(y\right)\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),y\right)\right) -\] -for $x\in S_{\gamma}$ and $\left|y\right|\leq2\epsilon$. Note that -$f\left(x,\lambda\left(y\right)\right)$ is defined since $\left|\lambda\right|\leq1$, -so $\left|\lambda\left(y\right)\right|\leq4\epsilon\leq\delta$.\\ -For $\left|y\right|\leq\epsilon$, $\left|\left(-\lambda\right)\left(y\right)\right|\leq2\epsilon$. -Also, $\lambda\left(\left(-\lambda\right)\left(y\right)\right)=y$. -Thus for $\left|y\right|\leq\epsilon$ -\[ -\mathrm{sign}\left(f\left(x,y\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),\left(-\lambda\right)\left(y\right)\right)\right). -\] -Let $Z=\left(Z_{1},...,Z_{n}\right)$ be new variables and -\[ -\hat{\w}_{\gamma}\left(X,V_{i},Z\right)=f_{i}\left(\epsilon X\right)\w_{\gamma}\left(\epsilon X,\epsilon\left(\gamma\right)V_{\gamma},2\epsilon Z\right)\in\R\left\{ X,V_{i},Z\right\} _{1^{+}}. -\] -Then if $\left(x,y\right)\in I^{m+n}$ and $\epsilon x\in S_{\gamma}$, -then -\[ -\mathrm{sign}f\left(x,y\right)=\sigma\left(\gamma\right)\hat{\w}_{\gamma}\left(x,\nicefrac{v_{i}\left(\epsilon x\right)}{\epsilon\left(\gamma\right)},\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right). -\] - -\section{Quantifier elimination for $\R_{an}$.} - -Recall that $\L_{an}$ is the language $\left\{ 0,1,+,-,\cdot,\leq\right\} $ -of ordered rings augmented by a function symbol for every restricted -analytic function $\R^{m}\lto\R$. \\ -Let $\R_{an}$ be the ordered field of reals as an $\L_{an}$-structure -together with the map $^{-1}\colon\R\lto\R$ defined as $x^{-1}=\begin{cases} -\frac{1}{x} & x\neq0\\ -x=0 & 0 -\end{cases}$. Let $\L_{an}\left(^{-1}\right)=\L_{an}\cup\left\{ ^{-1}\right\} $. -\begin{thm*} -\label{12.1} (Denef-v. d. Dries 1988) $\left(\R_{an},\,^{-1}\right)$ has -quantifier elimination. -\end{thm*} -Call $f\colon\I^{m}\lto\R$ analytic if it extends to an analytic -function on an open nbhd of $\I^{m}$. Let $\L_{a}$ be the language -$\left\{ <\right\} $ expanded by $m$-ary function symbols for each -analytic $f\colon\I^{m}\lto\R$ with $f\left(\I^{m}\right)\subset\I$.\\ -We consider $\I$ as an $\L_{a}$-structure.\\ -Define $\D\colon\I^{2}\lto\I$ by -\[ -D\left(x,y\right)=\begin{cases} -\nicefrac{x}{y} & \textrm{if }\left|x\right|\leq\left|y\right|\textrm{ and }y\neq0,\\ -0 & \textrm{otherwise.} -\end{cases} -\] -Let $\L_{a,\D}=\L_{a}\cup\left\{ \D\right\} $. -\begin{thm*} -\label{12.2} The $\L_{a,\D}$-structure $\I$ has $q.e.$ -\end{thm*} -Some general logical considerations:\\ -Let $T$ be an $\L$-theory, and $T'$ be a definitional expansion -of $T$ to an $\L'$-theory. Assume further that $\L'\setminus\L$ -consists only of function symbols, and for each $f\in\L'\setminus\L$ -there is an existential $\L$-formula $\delta_{f}\left(x,y\right)$ -s.t -\[ -T'\vdash f\left(x\right)=y\longleftrightarrow\delta_{f}\left(x,y\right). -\] -(e.g. $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$.) -\begin{lem*} -\label{12.6} Every $\exists\L'$-formula is $T'$-equivalent to an $\exists\L$-formula. -\begin{lem*} -\label{12.7} Suppose that for each q.f. $\L$-formula $\varphi\left(x,y_{1},...,y_{n}\right)$, -$n\geq1$, there is a q.f. $\L'$-formula $\varphi'$$\left(x,z_{1},...,z_{n-1}\right)$ -such that -\begin{enumerate} -\item $T'\vdash\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\varphi'\left(x,z\right)$. -\item The function symbols in $\L'\setminus\L$ are only applied in $\varphi'$ -to terms involving only $x$. -\end{enumerate} -\end{lem*} -Then $T'$ has $q.e$. (So by \eqref{12.6}, $T$ is model -complete). -\end{lem*} -We now verify that the conditions in lemma \eqref{12.7} are -satisfied for $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$. -\begin{lem*} -\label{12.8} Basic Lemma. - -Let $\varphi\left(x,y\right)=\varphi\left(x_{1},...,x_{m},y_{1},...,y_{n}\right)$, -$n\geq1$ be a q.f. $\L_{a}$-formula. Then there is a q.f. $\L_{a,\D}$-formula -$\hat{\varphi}\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$ -such that -\begin{enumerate} -\item $\I\model\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\hat{\varphi}\left(x,z\right)$. -\item In $\hat{\varphi}$, $\D$ is only applied to terms not involving -$z$, and $z_{n}$ only appears polynomially in $\hat{\varphi}$. -\end{enumerate} -\end{lem*} -Given the Basic Lemma, we can use Tarsky's quantifier elimination -for $\R$ (as an ordered ring) to eliminate $\exists z_{n}$ and produce -$\varphi'$ satisfying the hypothesis of \eqref{12.7}. - -For an $\L_{a}$-formula and $\left(a,b\right)\in\I^{m+n}$, $\eta>0$, -consider the formula -\[ -\varphi_{a,b,\eta}=\varphi\wedge``\left|\left(x,y\right)-\left(a,b\right)\right|<\eta", -\] -where $``"$ means the formal sentence for that phrase. By compactness -of $I^{m+n}$, it is enough to show that for any $\left(a,b\right)$, -there is $\eta>0$ such that the Basic Lemma holds for the formula -$\varphi_{a,b,\eta}$. Using an affine transformation, it is enough -to consider only $\left(a,b\right)=\left(0,0\right)$. Let $\varphi_{\eta}=\varphi_{0,0,\eta}$. -Thus the lemma is reduced to the following: -\begin{lem*} -\label{12.9} Local Basic Lemma. - -Let $\varphi\left(x,y\right)$ be a q.f. $\L_{a}$-formula . Then -there is $\epsilon\in\left(0,1\right)$ and a q.f. $\L_{a,\D}$ formula -$\varphi'\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$, s.t -\begin{enumerate} -\item $\I\model\exists y\varphi_{\epsilon}\iff\exists z\left(\varphi_{\epsilon}^{\prime}\right)$. -\item In $\varphi'$, the function symbol $\D$ is applied only to terms -not involving $z$, and $z_{n}$ appears only polynomially. -\end{enumerate} -\end{lem*} -\begin{proof} -We may assume that all atomic subformulae of $\varphi$ are of the -form $f\left(x,y\right)>0$ or $f\left(x,y\right)=0$ for some analytic -$f\colon\I^{m}\lto\I$.\\ -Apply uniform W.P. to the Taylor series at $0$ of all such $f$. -We get \end{proof} -\begin{itemize} -\item $\epsilon\in\left(0,1\right)$, -\item Finite cover $S_{\gamma}$ of $B_{\epsilon}\subset\R^{m}$ by special -sets. -\item For each $\gamma$, a $\lambda=\lambda\left(\gamma\right)\in\R^{n-1}$ -with $\left|\lambda\right|\leq1$. -\item For each $f$ in $\varphi$, an $\L_{a,\D}$-term $t_{\gamma,f}\left(x,z\right)$; -\item $t_{f,\gamma}$ is polynomial in $z_{n}$, -\item $\D$ is only applied to terms involving only $x$ within $t_{f,\gamma}$. -\item $\forall x\in I^{m+n}$ with $\epsilon x\in S_{\gamma}$, -\[ -f\left(\epsilon x,\epsilon y\right)\geq0\iff t_{f,\gamma}\left(x,\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right)\geq0. -\] - -\end{itemize} -Replace $f\left(x,y\right)$ by $t_{f,\gamma}$ in $\varphi$ to get -$\varphi_{\gamma}$. Then -\[ -\I\model\exists y\varphi\left(\epsilon x,\epsilon y\right)\longleftrightarrow\exists z\left(\bigvee_{\gamma}x\in S_{\gamma}\wedge\varphi_{\gamma}\left(x,z\right)\wedge\left|\lambda\left(z\right)\right|\leq\frac{1}{2}\right). -\] -Finally we can convert this to a formula $\varphi'$ satisfying $\left(1\right)$ +\global\long\def\N{\mathbb{N}} +\global\long\def\Z{\mathbb{Z}} +\global\long\def\Q{\mathbb{Q}} +\global\long\def\R{\mathbb{R}} +\global\long\def\lto{\longrightarrow} +\global\long\def\es{\emptyset} +\global\long\def\F{\mathcal{F}} +\global\long\def\force{\Vdash} +\global\long\def\dom{\textrm{dom}} +\global\long\def\em{\prec} +\global\long\def\cf{\textrm{cf}} +\global\long\def\model{\vDash} +\global\long\def\crit{\mathrm{crit}} +\global\long\def\ult{\mathrm{Ult}} +\global\long\def\inj{\hookrightarrow} +\global\long\def\u{\mathcal{U}} +\global\long\def\dprime{\prime\prime} +\global\long\def\C{\mathbb{C}} +\global\long\def\v{\mathcal{V}} +\global\long\def\w{\mathcal{W}} +\global\long\def\i{\imath} +\global\long\def\P{\mathbb{P}} +\global\long\def\del{\partial} +\global\long\def\an{\mathrm{An}} +\global\long\def\I{\mathrm{I}} +\global\long\def\L{\mathcal{L}} +\global\long\def\D{\mathcal{D}} + +\NotesBy{Notes by Asaaf Sahni} +\Week{Week 10} + +Proving, after all, that $\C\left[\left[X\right]\right]$ is faithfully +flat over $\C\left\{ X\right\} $. + +We do this by induction on the number of variables in $X$. We only +show that $\C\left[\left[X,T\right]\right]$ is flat over $\C\left\{ X,T\right\} $.\\ +Assume that $\C\left[\left[X\right]\right]$ is flat over $\C\left\{ X\right\} $, +and consider $\C\left[\left[X,T\right]\right]$ and $\C\left\{ X,T\right\} $. +\\ +Consider the equation +\[ +\left(\ast\right)_{0}\qquad f_{1}y_{1}+...+f_{n}y_{n}=0,\qquad\textrm{where }f_{i}\in\C\left\{ X,T\right\} . +\] +We can assume that all $f_{i}$'s are non zero, and regular in $T$. +\\ +Apply W.P to $f_{i}$ in $\C\left\{ X,T\right\} $, $f_{i}=u_{i}w_{i}$ +where $u_{i}\in\C\left\{ X,T\right\} ^{\times}$ and $w_{i}\in\C\left\{ X\right\} \left[T\right]$ +is monic of degree $d_{i}$. Note that if $\left(y_{1},...,y_{n}\right)$ +is a solution to $\left(\ast\right)_{0}$, then $\left(u_{1}^{-1}y_{1},...,u_{n}^{-1}y_{n}\right)$ +is a solution to +\[ +\left(\ast\right)\qquad w_{1}y_{1}+...+w_{n}y_{n}=0. +\] +So it is enough to show that each solution to $\left(\ast\right)$ +in $\C\left[\left[X,T\right]\right]$ is a linear combination of solutions +from $\C\left\{ X,T\right\} $.\\ +Consider $z_{2}=\left(w_{2},-w_{1},0,...,0\right)$, $z_{3}=\left(w_{3},0,-w_{1},0,...,0\right)$, +... $z_{n}=\left(w_{n},0,...,0,-w_{1}\right)$. Each $z_{i}$ is a +solution to $\left(\ast\right)$ (and are in $\C\left\{ X,T\right\} $).\\ +Now suppose $y=\left(y_{1},...,y_{n}\right)$ is a solution to $\left(\ast\right)$ +in $\C\left[\left[X,T\right]\right]$. \\ +For each $i$, write $y_{i}=q_{i}w_{1}+r_{i}$ where $q_{i}\in\C\left[\left[X,T\right]\right]$ +and $r_{i}\in\C\left[\left[X\right]\right]\left[T\right]$ of degree +$0}$. +\end{itemize} +Let $\mathrm{An}\left(\u\right)$ be all analytic functions on $\u$. +$\an\left(\u\right)$ has an $\R$-algebra structure. If $f\in\an\left(\u\right)$, +then $\frac{\del f}{\del X_{i}}\in\an\left(\u\right).$ + +For each $a\in\u$ there is a map of $\R$-algebras: +\begin{eqnarray*} +\an\left(\u\right) & \lto & \R\left\{ X\right\} .\\ +f & \mapsto & f_{a} +\end{eqnarray*} + +\begin{prop*} +\label{11.2} (Analytic continuation) If $\u$ is connected, then the map +above is injective. In particular, $\an\left(\u\right)$ is an integral +domain.\end{prop*} +\begin{cor*} +\label{11.3} If $\u$ is connected, $f,g\in\an\left(\u\right)$ agree on +a non empty open subset of $\u$, then $f=g$.\end{cor*} +\begin{prop*} +\label{11.4} Let $f_{1},...,f_{n}\in\an\left(\u\right)$, $\v\subset\R^{n}$ +open s.t $f\left(\u\right)\subset V$, where $f=\left(f_{1},...,f_{n}\right)\colon\u\lto\R^{n}$. + +Then for any $g\in\an\left(\v\right)$ we have $g\circ f\in\an\left(\u\right)$ +and +\[ +\left(g\circ f\right)_{a}=g_{f\left(a\right)}\left(f_{a}-f\left(a\right)\right). +\] + +\end{prop*} +\label{11.5} \uline{Notation}: For $x=\left(x_{1},...,x_{m}\right)\in\R^{m}$, +$\left|x\right|=\max\left\{ \left|x\right|_{1},...,\left|x_{m}\right|\right\} $. +\\ +For $\delta>0$, let $\delta=\left(\delta,...,\delta\right)$ and +$\R\left\{ X\right\} _{\delta^{+}}=\bigcup_{r>\delta}\R\left\{ X\right\} _{r}$.\\ +Each $f\in\R\left\{ X\right\} _{\delta^{+}}$ gives rise to a function +$x\mapsto f\left(x\right)\colon\bar{B}_{\delta}\left(0\right)\lto\R$ +which extendes to an analytic function on an open nbhd of $\bar{B}_{\delta}$. + +Let $Y=\left(Y_{1},...,Y_{n}\right)$, $n\geq1$. Fix $f=f\left(X,Y\right)\in\R\left\{ X,Y\right\} $. +Then for small $x$ we have $f\left(x,Y\right)\in\R\left\{ Y\right\} $.\\ +Consider the following question: How does W.P for $f\left(x,Y\right)$ +depend on $x$, for small $x$?\\ +We will show that for some $\epsilon>0$, $\bar{B}_{\epsilon}\subset\R^{m}$ +can be covered by finitely many ``special sets'', on each of which +W.P is uniform in $x$. +\begin{defn*} +\label{11.6} A special subset of $\bar{B}_{\epsilon}$ is a finite union +of sets of the following form +\[ +\left\{ x\in\bar{B}_{\epsilon};\, f\left(x\right)=0,\, g_{1}\left(x\right)>0,\,...,\, g_{k}\left(x\right)>0\right\} , +\] +where $f,g_{1},...,g_{k}\in\R\left\{ X\right\} _{\epsilon^{+}}$. +\end{defn*} +We first show that $\mathrm{ord}\left(f\left(x,Y\right)\right)$ takes +only finitely many values as $x$ ranges over a neighbourhood of $0\in\R^{m}$. +Write +\[ +f\left(X,Y\right)=\sum_{j}f_{j}\left(X\right)Y^{j}, +\] +where $f_{j}\left(X\right)\in\R\left\{ X\right\} $. Since $\R\left\{ X\right\} $ +is neotherian, the ideal generated by $\left\{ f_{j}\left(X\right)\right\} _{j\in\N^{n}}$ +is finitely generated, hence is generated by $\left\{ f_{j}\left(X\right)\right\} _{\left|j\right|d$, +\[ +f_{j}\left(X\right)=\sum_{\left|i\right|\leq d}g_{ij}f_{i}. +\] +In the following, $i$ ranges over elements of $\N^{n}$ s.t $\left|i\right|\leq d$ +and $j$ over elements of $\N^{n}$ s.t $\left|j\right|>d$. \\ +Substituting the above, we get +\[ +\left(1\right)\qquad f\left(X,Y\right)=\sum_{i}f_{i}\left(X,Y\right)\left(Y^{i}+\sum_{j}g_{ij}\left(X\right)Y^{j}\right), +\] +in $\R\left[\left[X,Y\right]\right]$. Note that for each $i,j$, +$g_{ij}\in\R\left\{ X\right\} $, and so there is some $\delta$ s.t +$g_{ij}\in\R\left\{ X\right\} _{\delta}$. However, in order to have +equation $\left(1\right)$ as an equality in $\R\left\{ X\right\} $, +we need to find a uniform $\delta$ that works for all $i,j$. +\begin{claim*} +$\exists\delta\in\left(0,1\right]$ and there are $f_{i},\, g_{ij}\in\R\left\{ X\right\} _{\delta^{+}}$ +(maybe different than above) such that $\sum_{j}g_{ij}\left(X\right)Y^{j}\in\R\left\{ X,Y\right\} _{\delta^{+}}$ +and $\left(1\right)$ holds in $\R\left\{ X\right\} _{\delta^{+}}$ +with $f_{i},g_{ij}$.\end{claim*} +\begin{proof} +Consider the linear equation +\[ +f=\sum_{i}f_{i}\left(Y^{i}+\sum_{\left|j\right|=d+1}Z_{ij}Y^{j}\right). +\] +By $\left(1\right)$ above, there is a solution $\left\{ Z_{ij};\,\left|i\right|\leq d,\,\left|j\right|=d+1\right\} $ +in $\R\left[\left[X,Y\right]\right]$ (all the higher terms, $Y^{j}$ +for $\left|j\right|>d+1$, are inside the $Z_{ij}$'s). Since $\R\left[\left[X,Y\right]\right]$ +is f.f. over $\R\left\{ X,Y\right\} $, there is a solution $\left\{ Z_{ij}\right\} $ +in $\R\left\{ X,Y\right\} $. Take $\delta$ for all $i,j$, $Z_{ij}\in\R\left\{ X,Y\right\} _{\delta^{+}}$. +Now write $Z_{ij}$ as power series in $Y$ with coefficients in $\R\left\{ X\right\} $ +to get $\left(1\right)$ with $g_{ij}\left(X\right)\in\R\left\{ X\right\} _{\delta^{+}}$ +for all $\left|i\right|\leq d$, $\left|j\right|>d$. +\end{proof} +We work with equation $\left(1\right)$ in $\R\left\{ X,Y\right\} _{\delta^{+}}$ +as given by the claim above. For $\left|x\right|\leq\delta$, +\[ +f\left(x,Y\right)=0\iff\forall i\left(f_{i}\left(x\right)=0\right). +\] +Also, if $f_{i}\left(x\right)\neq0$, then $\mathrm{ord}\left(f\left(x,Y\right)\right)\leq\left|i\right|$. +So +\[ +f\left(x,Y\right)\neq0\implies\mathrm{ord}\left(f\left(x,Y\right)\right)\leq d. +\] +Define +\begin{eqnarray*} +Z_{\delta} & = & \left\{ x\in\bar{B}_{\delta};\,\forall i\, f_{i}\left(x\right)=0\right\} .\\ +S_{i} & = & \left\{ x\in\bar{B}_{\delta};\, f_{i}\left(x\right)\neq0\wedge\forall i'\neq i\left(\left|f_{i}\left(x\right)\right|\geq\left|f_{i'}\left(x\right)\right|\right)\right\} . +\end{eqnarray*} +Note that $\bar{B}_{\delta}=Z_{\delta}\cup\bigcup_{i}S_{i}$.\\ +Fix some $i$, formally divide the expression for $f$ in $\left(1\right)$ +by $f_{i}$, and introduce new variables $V_{i,i'}$ for the quotients +$\nicefrac{f_{i}}{f_{i'}}$, for $i\neq i'$. Let $V_{i}=\left(V_{ii'}\right)_{i'\neq i}$. +Define +\[ +F_{i}=Y^{i}+\sum_{j}g_{ij}Y^{j}+\sum_{i'\neq i}V_{ii'}\left(Y^{i'}+\sum_{j}g_{i'j}Y^{j}\right)\in\R\left\{ X,V_{i},Y\right\} . +\] +For $x\in S_{i}$, let $v_{i}\left(x\right)=\left(\nicefrac{f_{i'}\left(x\right)}{f_{i}\left(x\right)}\right)_{i'\neq i}$. +Then for $x\in S_{i}$, $\left|v_{i}\left(x\right)\right|\leq1$, +and +\[ +\left(2\right)\qquad f\left(x,Y\right)=f_{i}\left(x\right)F_{i}\left(x,v_{i}\left(x\right),Y\right). +\] + + +Idea: apply W.P to $F_{i}$'s locally around every point $X=0$, $V_{i}=c$, +$Y=0$. \\ +For $c=\left(c_{i'}\right)_{i'\neq i}$ with $\left|c\right|\leq1$, +put +\[ +F_{i,c}=F_{i}\left(X,c+V_{i},Y\right). +\] +Then for $x\in S_{i}$, +\[ +f\left(x,Y\right)=f_{i}\left(x\right)F_{i,c}\left(x,v_{i}\left(x\right)-c,Y\right). +\] + +\begin{claim*} +For each $c$ there is $\lambda=\lambda\left(c\right)\in\R^{n-1}$ +with $\left|\lambda\right|\leq1$ such that $F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)$ +is regular in $Y_{n}$ of order$\leq\left|i\right|$, where for $Y=\left(Y_{1},...,Y_{n}\right)$, +$\lambda\left(Y\right)=\left(Y_{1}+\lambda Y_{n},...,Y_{n-1}+\lambda_{n-1}Y_{n},Y_{n}\right)$.\end{claim*} +\begin{proof} +Exercise. +\end{proof} +By W.P., for such $\lambda$, +\[ +\left(3\right)\qquad F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)=\u_{i,c}\w_{i,c},\quad\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} ^{\ast},\:\w_{i,c}\in\R\left\{ X,V_{i},Y_{1},...,Y_{n-1}\right\} \left[Y_{n}\right]. +\] +Take $\epsilon\left(i,c\right)\in\left(0,\delta\right]$ s.t +\begin{itemize} +\item $\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}^{\ast}$. +\item $\w_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}.$ +\end{itemize} +Let $\Gamma\left(i\right)=\left\{ i';\, i'\neq i\right\} $. Note +that our $c$'s vary over $I^{\Gamma\left(i\right)}$, where $\I=\left[-1,1\right]$. +By compactness, there is a finite set $C\left(i\right)$ s.t +\[ +\I^{\Gamma\left(i\right)}\subset\bigcup_{c\in C\left(i\right)}B_{\epsilon\left(i,c\right)}\left(c\right). +\] +Now consider the finite set $\Gamma=\left\{ \left(i,c\right);\,\left|i\right|\leq d,\, d\in C\left(i\right)\right\} $.\\ +Take $\epsilon>0$ s.t $\epsilon\leq\frac{\epsilon\left(i,c\right)}{4}$ +for all $\left(i,c\right)\in\Gamma.$ For $\gamma=\left(i,c\right)\in\Gamma$, +let +\[ +S_{\gamma}=\left\{ x\in S_{i};\,\left|x\right|\leq\epsilon,\,\left|v_{i}\left(x\right)-c\right|<\epsilon\left(i,c\right)\right\} . +\] +Then $S_{i}\cap\bar{B}_{\epsilon}=\bigcup\left\{ S_{\gamma};\,\gamma=\left(i,c\right),\, c\in C\left(i\right)\right\} $. +\\ +So $\bar{B}_{\epsilon}=\left(\underbrace{Z_{\delta}\cap\bar{B}_{\epsilon}}_{\equiv Z}\right)\cup\bigcup_{\gamma}S_{\gamma}$.\\ +For $\gamma=\left(i,c\right)\in\Gamma$, by $\left(2\right)$ and +$\left(3\right)$, +\[ +f\left(x,\lambda\left(Y\right)\right)=f_{i}\left(x\right)\u_{\gamma}\left(x,v_{i}\left(x\right),Y\right)W_{\gamma}\left(x,v_{i}\left(x\right),Y\right). +\] +Note: $\u_{\gamma}$ does not change sign on $\bar{B}_{\epsilon\left(\gamma\right)}$. +Therefore there is $\sigma\left(\gamma\right)\in\left\{ \pm1\right\} $ +s.t +\[ +\mathrm{sign}\left(f\left(x,\lambda\left(y\right)\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),y\right)\right) +\] +for $x\in S_{\gamma}$ and $\left|y\right|\leq2\epsilon$. Note that +$f\left(x,\lambda\left(y\right)\right)$ is defined since $\left|\lambda\right|\leq1$, +so $\left|\lambda\left(y\right)\right|\leq4\epsilon\leq\delta$.\\ +For $\left|y\right|\leq\epsilon$, $\left|\left(-\lambda\right)\left(y\right)\right|\leq2\epsilon$. +Also, $\lambda\left(\left(-\lambda\right)\left(y\right)\right)=y$. +Thus for $\left|y\right|\leq\epsilon$ +\[ +\mathrm{sign}\left(f\left(x,y\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),\left(-\lambda\right)\left(y\right)\right)\right). +\] +Let $Z=\left(Z_{1},...,Z_{n}\right)$ be new variables and +\[ +\hat{\w}_{\gamma}\left(X,V_{i},Z\right)=f_{i}\left(\epsilon X\right)\w_{\gamma}\left(\epsilon X,\epsilon\left(\gamma\right)V_{\gamma},2\epsilon Z\right)\in\R\left\{ X,V_{i},Z\right\} _{1^{+}}. +\] +Then if $\left(x,y\right)\in I^{m+n}$ and $\epsilon x\in S_{\gamma}$, +then +\[ +\mathrm{sign}f\left(x,y\right)=\sigma\left(\gamma\right)\hat{\w}_{\gamma}\left(x,\nicefrac{v_{i}\left(\epsilon x\right)}{\epsilon\left(\gamma\right)},\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right). +\] + +\section{Quantifier elimination for $\R_{an}$.} + +Recall that $\L_{an}$ is the language $\left\{ 0,1,+,-,\cdot,\leq\right\} $ +of ordered rings augmented by a function symbol for every restricted +analytic function $\R^{m}\lto\R$. \\ +Let $\R_{an}$ be the ordered field of reals as an $\L_{an}$-structure +together with the map $^{-1}\colon\R\lto\R$ defined as $x^{-1}=\begin{cases} +\frac{1}{x} & x\neq0\\ +x=0 & 0 +\end{cases}$. Let $\L_{an}\left(^{-1}\right)=\L_{an}\cup\left\{ ^{-1}\right\} $. +\begin{thm*} +\label{12.1} (Denef-v. d. Dries 1988) $\left(\R_{an},\,^{-1}\right)$ has +quantifier elimination. +\end{thm*} +Call $f\colon\I^{m}\lto\R$ analytic if it extends to an analytic +function on an open nbhd of $\I^{m}$. Let $\L_{a}$ be the language +$\left\{ <\right\} $ expanded by $m$-ary function symbols for each +analytic $f\colon\I^{m}\lto\R$ with $f\left(\I^{m}\right)\subset\I$.\\ +We consider $\I$ as an $\L_{a}$-structure.\\ +Define $\D\colon\I^{2}\lto\I$ by +\[ +D\left(x,y\right)=\begin{cases} +\nicefrac{x}{y} & \textrm{if }\left|x\right|\leq\left|y\right|\textrm{ and }y\neq0,\\ +0 & \textrm{otherwise.} +\end{cases} +\] +Let $\L_{a,\D}=\L_{a}\cup\left\{ \D\right\} $. +\begin{thm*} +\label{12.2} The $\L_{a,\D}$-structure $\I$ has $q.e.$ +\end{thm*} +Some general logical considerations:\\ +Let $T$ be an $\L$-theory, and $T'$ be a definitional expansion +of $T$ to an $\L'$-theory. Assume further that $\L'\setminus\L$ +consists only of function symbols, and for each $f\in\L'\setminus\L$ +there is an existential $\L$-formula $\delta_{f}\left(x,y\right)$ +s.t +\[ +T'\vdash f\left(x\right)=y\longleftrightarrow\delta_{f}\left(x,y\right). +\] +(e.g. $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$.) +\begin{lem*} +\label{12.6} Every $\exists\L'$-formula is $T'$-equivalent to an $\exists\L$-formula. +\begin{lem*} +\label{12.7} Suppose that for each q.f. $\L$-formula $\varphi\left(x,y_{1},...,y_{n}\right)$, +$n\geq1$, there is a q.f. $\L'$-formula $\varphi'$$\left(x,z_{1},...,z_{n-1}\right)$ +such that +\begin{enumerate} +\item $T'\vdash\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\varphi'\left(x,z\right)$. +\item The function symbols in $\L'\setminus\L$ are only applied in $\varphi'$ +to terms involving only $x$. +\end{enumerate} +\end{lem*} +Then $T'$ has $q.e$. (So by \eqref{12.6}, $T$ is model +complete). +\end{lem*} +We now verify that the conditions in lemma \eqref{12.7} are +satisfied for $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$. +\begin{lem*} +\label{12.8} Basic Lemma. + +Let $\varphi\left(x,y\right)=\varphi\left(x_{1},...,x_{m},y_{1},...,y_{n}\right)$, +$n\geq1$ be a q.f. $\L_{a}$-formula. Then there is a q.f. $\L_{a,\D}$-formula +$\hat{\varphi}\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$ +such that +\begin{enumerate} +\item $\I\model\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\hat{\varphi}\left(x,z\right)$. +\item In $\hat{\varphi}$, $\D$ is only applied to terms not involving +$z$, and $z_{n}$ only appears polynomially in $\hat{\varphi}$. +\end{enumerate} +\end{lem*} +Given the Basic Lemma, we can use Tarsky's quantifier elimination +for $\R$ (as an ordered ring) to eliminate $\exists z_{n}$ and produce +$\varphi'$ satisfying the hypothesis of \eqref{12.7}. + +For an $\L_{a}$-formula and $\left(a,b\right)\in\I^{m+n}$, $\eta>0$, +consider the formula +\[ +\varphi_{a,b,\eta}=\varphi\wedge``\left|\left(x,y\right)-\left(a,b\right)\right|<\eta", +\] +where $``"$ means the formal sentence for that phrase. By compactness +of $I^{m+n}$, it is enough to show that for any $\left(a,b\right)$, +there is $\eta>0$ such that the Basic Lemma holds for the formula +$\varphi_{a,b,\eta}$. Using an affine transformation, it is enough +to consider only $\left(a,b\right)=\left(0,0\right)$. Let $\varphi_{\eta}=\varphi_{0,0,\eta}$. +Thus the lemma is reduced to the following: +\begin{lem*} +\label{12.9} Local Basic Lemma. + +Let $\varphi\left(x,y\right)$ be a q.f. $\L_{a}$-formula . Then +there is $\epsilon\in\left(0,1\right)$ and a q.f. $\L_{a,\D}$ formula +$\varphi'\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$, s.t +\begin{enumerate} +\item $\I\model\exists y\varphi_{\epsilon}\iff\exists z\left(\varphi_{\epsilon}^{\prime}\right)$. +\item In $\varphi'$, the function symbol $\D$ is applied only to terms +not involving $z$, and $z_{n}$ appears only polynomially. +\end{enumerate} +\end{lem*} +\begin{proof} +We may assume that all atomic subformulae of $\varphi$ are of the +form $f\left(x,y\right)>0$ or $f\left(x,y\right)=0$ for some analytic +$f\colon\I^{m}\lto\I$.\\ +Apply uniform W.P. to the Taylor series at $0$ of all such $f$. +We get \end{proof} +\begin{itemize} +\item $\epsilon\in\left(0,1\right)$, +\item Finite cover $S_{\gamma}$ of $B_{\epsilon}\subset\R^{m}$ by special +sets. +\item For each $\gamma$, a $\lambda=\lambda\left(\gamma\right)\in\R^{n-1}$ +with $\left|\lambda\right|\leq1$. +\item For each $f$ in $\varphi$, an $\L_{a,\D}$-term $t_{\gamma,f}\left(x,z\right)$; +\item $t_{f,\gamma}$ is polynomial in $z_{n}$, +\item $\D$ is only applied to terms involving only $x$ within $t_{f,\gamma}$. +\item $\forall x\in I^{m+n}$ with $\epsilon x\in S_{\gamma}$, +\[ +f\left(\epsilon x,\epsilon y\right)\geq0\iff t_{f,\gamma}\left(x,\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right)\geq0. +\] + +\end{itemize} +Replace $f\left(x,y\right)$ by $t_{f,\gamma}$ in $\varphi$ to get +$\varphi_{\gamma}$. Then +\[ +\I\model\exists y\varphi\left(\epsilon x,\epsilon y\right)\longleftrightarrow\exists z\left(\bigvee_{\gamma}x\in S_{\gamma}\wedge\varphi_{\gamma}\left(x,z\right)\wedge\left|\lambda\left(z\right)\right|\leq\frac{1}{2}\right). +\] +Finally we can convert this to a formula $\varphi'$ satisfying $\left(1\right)$ and $\left(2\right)$ of the Local Basic Lemma. \ No newline at end of file diff --git a/Other/all_notes/week_6/g_week_6.tex b/Other/all_notes/week_6/g_week_6.tex index 11fed4bb..655c7331 100644 --- a/Other/all_notes/week_6/g_week_6.tex +++ b/Other/all_notes/week_6/g_week_6.tex @@ -52,8 +52,8 @@ \end{theorem} \textbf{Notation:} for $h(x) = \sum_{i < \alpha} h_i x^{a_i}$ and $\gamma \leq \alpha$ we write -\begin{align*} - h(x)\midr \gamma = \sum_{i < \gamma} h_i x^{a_i} +\begin{align*} + h(x)\midr \gamma = \sum_{i < \gamma} h_i x^{a_i} \end{align*} \begin{proof}[Proof of inequality.] %\WikiBold{Proof of inequality} diff --git a/Other/old/All notes - Copy (2)/Global.aux b/Other/old/All notes - Copy (2)/Global.aux deleted file mode 100644 index 452a9eb6..00000000 --- a/Other/old/All notes - Copy (2)/Global.aux +++ /dev/null @@ -1,11 +0,0 @@ -\relax -\@input{week_1/john_susice_surreal_numbers_notes_fall2014.aux} -\@input{week_2/g_week_2.aux} -\@input{week_3/zach.aux} -\@input{week_4/g_week_4.aux} -\@input{week_5/g_week_5.aux} -\@input{week_6/g_week_6.aux} -\@input{week_7/285D_notes_nov_17_18_21.aux} -\@input{week_8/g_week_8.aux} -\@input{week_9/g_week_9.aux} -\@input{week_11/g_week_11.aux} diff --git a/Other/old/All notes - Copy (2)/Global.bbl b/Other/old/All notes - Copy (2)/Global.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/All notes - Copy (2)/Global.blg b/Other/old/All notes - Copy (2)/Global.blg deleted file mode 100644 index 579f4a7b..00000000 --- a/Other/old/All notes - Copy (2)/Global.blg +++ /dev/null @@ -1,15 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Global.aux -A level-1 auxiliary file: week_1/john_susice_surreal_numbers_notes_fall2014.aux -A level-1 auxiliary file: week_2/week_2.aux -A level-1 auxiliary file: week_3/zach.aux -A level-1 auxiliary file: week_4/week_4.aux -A level-1 auxiliary file: week_5/week_5.aux -A level-1 auxiliary file: week_6/week_6.aux -A level-1 auxiliary file: week_7/285D_notes_nov_17_18_21.aux -A level-1 auxiliary file: week_8/week_8.aux -A level-1 auxiliary file: week_9/week_9.aux -A level-1 auxiliary file: week_11/week_11.aux -I found no \citation commands---while reading file Global.aux -I found no \bibdata command---while reading file Global.aux -I found no \bibstyle command---while reading file Global.aux -(There were 3 error messages) diff --git a/Other/old/All notes - 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Copy (2)/week_1/g_g_john_susice_surreal_numbers_notes_fall2014.tex b/Other/old/All notes - Copy (2)/week_1/g_g_john_susice_surreal_numbers_notes_fall2014.tex index ee5539e1..ccd1e085 100644 --- a/Other/old/All notes - Copy (2)/week_1/g_g_john_susice_surreal_numbers_notes_fall2014.tex +++ b/Other/old/All notes - Copy (2)/week_1/g_g_john_susice_surreal_numbers_notes_fall2014.tex @@ -1,475 +1,475 @@ -\textit{Notes by John Suice} - -\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c diff --git a/Other/old/All notes - Copy (2)/week_1/g_john_susice_surreal_numbers_notes_fall2014.tex b/Other/old/All notes - Copy (2)/week_1/g_john_susice_surreal_numbers_notes_fall2014.tex index 1f758a2c..715fdf72 100644 --- a/Other/old/All notes - Copy (2)/week_1/g_john_susice_surreal_numbers_notes_fall2014.tex +++ b/Other/old/All notes - Copy (2)/week_1/g_john_susice_surreal_numbers_notes_fall2014.tex @@ -1,508 +1,508 @@ -\textit{Notes by John Suice} - -\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. -\end{proof} +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. +\end{proof} diff --git a/Other/old/All notes - Copy (2)/week_1/g_week_1.tex b/Other/old/All notes - Copy (2)/week_1/g_week_1.tex index 1c06ce99..37b7dab3 100644 --- a/Other/old/All notes - Copy (2)/week_1/g_week_1.tex +++ b/Other/old/All notes - Copy (2)/week_1/g_week_1.tex @@ -1,13 +1,13 @@ -\section{Week 1} - -Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. - -Proof - -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - +\section{Week 1} + +Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. + +Proof + +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + diff --git a/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.aux b/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.aux deleted file mode 100644 index 16edaca4..00000000 --- a/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.aux +++ /dev/null @@ -1,35 +0,0 @@ -\relax -\newlabel{}{{1}{2}} -\newlabel{}{{2}{2}} -\newlabel{}{{3}{3}} -\newlabel{lemma_on_length_of_cuts}{{1}{4}} -\newlabel{cofinality_theorem}{{4}{4}} -\newlabel{inverse_cofinality_theorem}{{2}{5}} -\newlabel{}{{5}{5}} -\newlabel{defn_of_surreal_sum}{{1}{6}} -\@setckpt{week_1/john_susice_surreal_numbers_notes_fall2014}{ -\setcounter{page}{7} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.tex b/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.tex index 9347f272..2ccd4c36 100644 --- a/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.tex +++ b/Other/old/All notes - Copy (2)/week_1/john_susice_surreal_numbers_notes_fall2014.tex @@ -1,508 +1,508 @@ -\textit{Notes by John Suice} - -\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/old/All notes - Copy (2)/week_1/week_1.aux b/Other/old/All notes - Copy (2)/week_1/week_1.aux deleted file mode 100644 index 7174f85a..00000000 --- a/Other/old/All notes - Copy (2)/week_1/week_1.aux +++ /dev/null @@ -1,26 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1}Week 1}{2}} -\@setckpt{week_1/week_1}{ -\setcounter{page}{3} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_1/week_1.tex.aux b/Other/old/All notes - Copy (2)/week_1/week_1.tex.aux deleted file mode 100644 index 9b66fc00..00000000 --- a/Other/old/All notes - Copy (2)/week_1/week_1.tex.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_1/week_1.tex}{ -\setcounter{page}{2} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_10/week_10.aux b/Other/old/All notes - Copy (2)/week_10/week_10.aux deleted file mode 100644 index 1ecedcfa..00000000 --- a/Other/old/All notes - Copy (2)/week_10/week_10.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_10/week_10}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_11/g_g_week_11.tex b/Other/old/All notes - Copy (2)/week_11/g_g_week_11.tex index 8a2d286a..2554594e 100644 --- a/Other/old/All notes - Copy (2)/week_11/g_g_week_11.tex +++ b/Other/old/All notes - Copy (2)/week_11/g_g_week_11.tex @@ -1,83 +1,83 @@ -\section{ Siddharth's extra lectures } - -\subsection{Part 1} -''notes by Bill Chen'' - -A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== -* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. -* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. -* $\{0|\emptyset\}$. Here Left has a winning strategy. -* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. - -====Definition 1==== -* $G>0$ if Left has a winning strategy. -* $G<0$ if Right has a winning strategy. -* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) -* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) -* $G\ge 0$ means $G>0$ or $G\sim 0$. - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -'''Proof:''' - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -'''Remark:''' -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== -* $-(G+H)=-G+-H$. -* $--G=G$. -* $G\sim 0$ iff $-G\sim 0$. -* $G>0$ iff $-G<0$. -* $G\parallel 0$ iff $-G\parallel 0$. - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: -* If $G\sim 0$, then $G+H\sim 0$. -* If $G>0$, then $G+H>0$. -* If $G\parallel 0$, then $G+H\parallel 0$. -* If $G+H\sim 0$, then $G\sim 0$. -* If $G+H>0$, then $G>0$. -* If $G+H\parallel 0$, then $G\parallel 0$. - -'''Proof:''' - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== -* $G+ -G\sim 0$. -* If $G>0$ and $H>0$ then $G+H>0$. - -'''Proof:''' - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== -* $G\sim H$ if $G-H\sim 0$. +\section{ Siddharth's extra lectures } + +\subsection{Part 1} +''notes by Bill Chen'' + +A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== +* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. +* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. +* $\{0|\emptyset\}$. Here Left has a winning strategy. +* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. + +====Definition 1==== +* $G>0$ if Left has a winning strategy. +* $G<0$ if Right has a winning strategy. +* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) +* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) +* $G\ge 0$ means $G>0$ or $G\sim 0$. + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +'''Proof:''' + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +'''Remark:''' +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== +* $-(G+H)=-G+-H$. +* $--G=G$. +* $G\sim 0$ iff $-G\sim 0$. +* $G>0$ iff $-G<0$. +* $G\parallel 0$ iff $-G\parallel 0$. + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: +* If $G\sim 0$, then $G+H\sim 0$. +* If $G>0$, then $G+H>0$. +* If $G\parallel 0$, then $G+H\parallel 0$. +* If $G+H\sim 0$, then $G\sim 0$. +* If $G+H>0$, then $G>0$. +* If $G+H\parallel 0$, then $G\parallel 0$. + +'''Proof:''' + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== +* $G+ -G\sim 0$. +* If $G>0$ and $H>0$ then $G+H>0$. + +'''Proof:''' + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== +* $G\sim H$ if $G-H\sim 0$. diff --git a/Other/old/All notes - Copy (2)/week_11/g_week_11.aux b/Other/old/All notes - Copy (2)/week_11/g_week_11.aux deleted file mode 100644 index afc1a001..00000000 --- a/Other/old/All notes - Copy (2)/week_11/g_week_11.aux +++ /dev/null @@ -1,29 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {8} Siddharth's extra lectures }{37}} -\@writefile{toc}{\contentsline {subsection}{\numberline {8.1}Part 1}{37}} -\@setckpt{week_11/g_week_11}{ -\setcounter{page}{39} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{8} -\setcounter{subsection}{1} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_11/g_week_11.tex b/Other/old/All notes - Copy (2)/week_11/g_week_11.tex index a860e0d4..c5778b1b 100644 --- a/Other/old/All notes - Copy (2)/week_11/g_week_11.tex +++ b/Other/old/All notes - Copy (2)/week_11/g_week_11.tex @@ -1,84 +1,84 @@ -\section{ Siddharth's extra lectures } - -\subsection{Part 1} -''notes by Bill Chen'' - -A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== -* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. -* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. -* $\{0|\emptyset\}$. Here Left has a winning strategy. -* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. - -====Definition 1==== -* $G>0$ if Left has a winning strategy. -* $G<0$ if Right has a winning strategy. -* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) -* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) -* $G\ge 0$ means $G>0$ or $G\sim 0$. - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -'''Proof:''' - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -'''Remark:''' -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== -* $-(G+H)=-G+-H$. -* $--G=G$. -* $G\sim 0$ iff $-G\sim 0$. -* $G>0$ iff $-G<0$. -* $G\parallel 0$ iff $-G\parallel 0$. - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: -* If $G\sim 0$, then $G+H\sim 0$. -* If $G>0$, then $G+H>0$. -* If $G\parallel 0$, then $G+H\parallel 0$. -* If $G+H\sim 0$, then $G\sim 0$. -* If $G+H>0$, then $G>0$. -* If $G+H\parallel 0$, then $G\parallel 0$. - -'''Proof:''' - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== -* $G+ -G\sim 0$. -* If $G>0$ and $H>0$ then $G+H>0$. - -'''Proof:''' - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== -* $G\sim H$ if $G-H\sim 0$. -* $G>H$ if $G-H>0$. +\section{ Siddharth's extra lectures } + +\subsection{Part 1} +''notes by Bill Chen'' + +A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== +* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. +* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. +* $\{0|\emptyset\}$. Here Left has a winning strategy. +* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. + +====Definition 1==== +* $G>0$ if Left has a winning strategy. +* $G<0$ if Right has a winning strategy. +* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) +* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) +* $G\ge 0$ means $G>0$ or $G\sim 0$. + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +'''Proof:''' + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +'''Remark:''' +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== +* $-(G+H)=-G+-H$. +* $--G=G$. +* $G\sim 0$ iff $-G\sim 0$. +* $G>0$ iff $-G<0$. +* $G\parallel 0$ iff $-G\parallel 0$. + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: +* If $G\sim 0$, then $G+H\sim 0$. +* If $G>0$, then $G+H>0$. +* If $G\parallel 0$, then $G+H\parallel 0$. +* If $G+H\sim 0$, then $G\sim 0$. +* If $G+H>0$, then $G>0$. +* If $G+H\parallel 0$, then $G\parallel 0$. + +'''Proof:''' + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== +* $G+ -G\sim 0$. +* If $G>0$ and $H>0$ then $G+H>0$. + +'''Proof:''' + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== +* $G\sim H$ if $G-H\sim 0$. +* $G>H$ if $G-H>0$. diff --git a/Other/old/All notes - Copy (2)/week_11/week_11.aux b/Other/old/All notes - Copy (2)/week_11/week_11.aux deleted file mode 100644 index 6cfb559d..00000000 --- a/Other/old/All notes - Copy (2)/week_11/week_11.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_11/week_11}{ -\setcounter{page}{38} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_11/week_11.tex b/Other/old/All notes - Copy (2)/week_11/week_11.tex index 2bf7b1f3..7572257f 100644 --- a/Other/old/All notes - Copy (2)/week_11/week_11.tex +++ b/Other/old/All notes - Copy (2)/week_11/week_11.tex @@ -1,84 +1,84 @@ -== Siddharth's extra lectures == - -===Part 1=== -''notes by Bill Chen'' - -A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== -* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. -* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. -* $\{0|\emptyset\}$. Here Left has a winning strategy. -* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. - -====Definition 1==== -* $G>0$ if Left has a winning strategy. -* $G<0$ if Right has a winning strategy. -* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) -* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) -* $G\ge 0$ means $G>0$ or $G\sim 0$. - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -'''Proof:''' - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -'''Remark:''' -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== -* $-(G+H)=-G+-H$. -* $--G=G$. -* $G\sim 0$ iff $-G\sim 0$. -* $G>0$ iff $-G<0$. -* $G\parallel 0$ iff $-G\parallel 0$. - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: -* If $G\sim 0$, then $G+H\sim 0$. -* If $G>0$, then $G+H>0$. -* If $G\parallel 0$, then $G+H\parallel 0$. -* If $G+H\sim 0$, then $G\sim 0$. -* If $G+H>0$, then $G>0$. -* If $G+H\parallel 0$, then $G\parallel 0$. - -'''Proof:''' - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== -* $G+ -G\sim 0$. -* If $G>0$ and $H>0$ then $G+H>0$. - -'''Proof:''' - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== -* $G\sim H$ if $G-H\sim 0$. -* $G>H$ if $G-H>0$. +== Siddharth's extra lectures == + +===Part 1=== +''notes by Bill Chen'' + +A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== +* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. +* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. +* $\{0|\emptyset\}$. Here Left has a winning strategy. +* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. + +====Definition 1==== +* $G>0$ if Left has a winning strategy. +* $G<0$ if Right has a winning strategy. +* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) +* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) +* $G\ge 0$ means $G>0$ or $G\sim 0$. + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +'''Proof:''' + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +'''Remark:''' +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== +* $-(G+H)=-G+-H$. +* $--G=G$. +* $G\sim 0$ iff $-G\sim 0$. +* $G>0$ iff $-G<0$. +* $G\parallel 0$ iff $-G\parallel 0$. + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: +* If $G\sim 0$, then $G+H\sim 0$. +* If $G>0$, then $G+H>0$. +* If $G\parallel 0$, then $G+H\parallel 0$. +* If $G+H\sim 0$, then $G\sim 0$. +* If $G+H>0$, then $G>0$. +* If $G+H\parallel 0$, then $G\parallel 0$. + +'''Proof:''' + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== +* $G+ -G\sim 0$. +* If $G>0$ and $H>0$ then $G+H>0$. + +'''Proof:''' + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== +* $G\sim H$ if $G-H\sim 0$. +* $G>H$ if $G-H>0$. diff --git a/Other/old/All notes - Copy (2)/week_11/week_6.aux b/Other/old/All notes - Copy (2)/week_11/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/All notes - Copy (2)/week_11/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_2/g_g_week_2.tex b/Other/old/All notes - Copy (2)/week_2/g_g_week_2.tex index 07057784..600a9e15 100644 --- a/Other/old/All notes - Copy (2)/week_2/g_g_week_2.tex +++ b/Other/old/All notes - Copy (2)/week_2/g_g_week_2.tex @@ -1,250 +1,250 @@ -\section{ Week 2 } - -(Notes by John Lensmire) - -\subsection{ Monday 10-13-2014 } - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -'''Proof:''' - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: -* $b <_s a \Rightarrow -b <_s -a$ -* $b < a \Rightarrow -b > -a$ - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: -* $a_L + (-a) < a_L + (-a_L) = 0$ -* $a + (-a_R) < a_R + (-a_R) = 0$ -* $a_R + (-a) > a_R + (-a_R) = 0$ -* $a + (-a_L) > a_L + (-a_L) = 0$ -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -'''Proof:''' - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: -* $f(a_L, b_L) < f(a_L, b_R)$ -* $f(a_R, b_R) < f(a_R, b_L)$ -* $f(a_L, b_L) < f(a_R, b_R)$ -* $f(a_R, b_R) < f(a_L, b_L)$ - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -\subsection{ Wednesday 10-15-2014 } - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -'''Proof:''' - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -\subsection{ Friday 10-17-2014 } - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\i +\section{ Week 2 } + +(Notes by John Lensmire) + +\subsection{ Monday 10-13-2014 } + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +'''Proof:''' + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: +* $b <_s a \Rightarrow -b <_s -a$ +* $b < a \Rightarrow -b > -a$ + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: +* $a_L + (-a) < a_L + (-a_L) = 0$ +* $a + (-a_R) < a_R + (-a_R) = 0$ +* $a_R + (-a) > a_R + (-a_R) = 0$ +* $a + (-a_L) > a_L + (-a_L) = 0$ +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +'''Proof:''' + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: +* $f(a_L, b_L) < f(a_L, b_R)$ +* $f(a_R, b_R) < f(a_R, b_L)$ +* $f(a_L, b_L) < f(a_R, b_R)$ +* $f(a_R, b_R) < f(a_L, b_L)$ + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +\subsection{ Wednesday 10-15-2014 } + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +'''Proof:''' + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +\subsection{ Friday 10-17-2014 } + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\i diff --git a/Other/old/All notes - Copy (2)/week_2/g_week_2.aux b/Other/old/All notes - Copy (2)/week_2/g_week_2.aux deleted file mode 100644 index bb8f38aa..00000000 --- a/Other/old/All notes - Copy (2)/week_2/g_week_2.aux +++ /dev/null @@ -1,31 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1} Week 2 }{7}} -\@writefile{toc}{\contentsline {subsection}{\numberline {1.1} Monday 10-13-2014 }{7}} -\@writefile{toc}{\contentsline {subsection}{\numberline {1.2} Wednesday 10-15-2014 }{8}} -\@writefile{toc}{\contentsline {subsection}{\numberline {1.3} Friday 10-17-2014 }{9}} -\@setckpt{week_2/g_week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{3} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_2/g_week_2.tex b/Other/old/All notes - Copy (2)/week_2/g_week_2.tex index f05b1791..3bb5cc5f 100644 --- a/Other/old/All notes - Copy (2)/week_2/g_week_2.tex +++ b/Other/old/All notes - Copy (2)/week_2/g_week_2.tex @@ -1,271 +1,271 @@ -\section{ Week 2 } - -(Notes by John Lensmire) - -\subsection{ Monday 10-13-2014 } - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -'''Proof:''' - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: -* $b <_s a \Rightarrow -b <_s -a$ -* $b < a \Rightarrow -b > -a$ - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: -* $a_L + (-a) < a_L + (-a_L) = 0$ -* $a + (-a_R) < a_R + (-a_R) = 0$ -* $a_R + (-a) > a_R + (-a_R) = 0$ -* $a + (-a_L) > a_L + (-a_L) = 0$ -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -'''Proof:''' - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: -* $f(a_L, b_L) < f(a_L, b_R)$ -* $f(a_R, b_R) < f(a_R, b_L)$ -* $f(a_L, b_L) < f(a_R, b_R)$ -* $f(a_R, b_R) < f(a_L, b_L)$ - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -\subsection{ Wednesday 10-15-2014 } - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -'''Proof:''' - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -\subsection{ Friday 10-17-2014 } - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). +\section{ Week 2 } + +(Notes by John Lensmire) + +\subsection{ Monday 10-13-2014 } + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +'''Proof:''' + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: +* $b <_s a \Rightarrow -b <_s -a$ +* $b < a \Rightarrow -b > -a$ + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: +* $a_L + (-a) < a_L + (-a_L) = 0$ +* $a + (-a_R) < a_R + (-a_R) = 0$ +* $a_R + (-a) > a_R + (-a_R) = 0$ +* $a + (-a_L) > a_L + (-a_L) = 0$ +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +'''Proof:''' + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: +* $f(a_L, b_L) < f(a_L, b_R)$ +* $f(a_R, b_R) < f(a_R, b_L)$ +* $f(a_L, b_L) < f(a_R, b_R)$ +* $f(a_R, b_R) < f(a_L, b_L)$ + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +\subsection{ Wednesday 10-15-2014 } + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +'''Proof:''' + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +\subsection{ Friday 10-17-2014 } + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/All notes - Copy (2)/week_2/week_2.aux b/Other/old/All notes - Copy (2)/week_2/week_2.aux deleted file mode 100644 index ded0313c..00000000 --- a/Other/old/All notes - Copy (2)/week_2/week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_2/week_2.tex b/Other/old/All notes - Copy (2)/week_2/week_2.tex index 7441881e..20e3641a 100644 --- a/Other/old/All notes - Copy (2)/week_2/week_2.tex +++ b/Other/old/All notes - Copy (2)/week_2/week_2.tex @@ -1,271 +1,271 @@ -== Week 2 == - -(Notes by John Lensmire) - -=== Monday 10-13-2014 === - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -'''Proof:''' - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: -* $b <_s a \Rightarrow -b <_s -a$ -* $b < a \Rightarrow -b > -a$ - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: -* $a_L + (-a) < a_L + (-a_L) = 0$ -* $a + (-a_R) < a_R + (-a_R) = 0$ -* $a_R + (-a) > a_R + (-a_R) = 0$ -* $a + (-a_L) > a_L + (-a_L) = 0$ -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -'''Proof:''' - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: -* $f(a_L, b_L) < f(a_L, b_R)$ -* $f(a_R, b_R) < f(a_R, b_L)$ -* $f(a_L, b_L) < f(a_R, b_R)$ -* $f(a_R, b_R) < f(a_L, b_L)$ - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -=== Wednesday 10-15-2014 === - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -'''Proof:''' - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -=== Friday 10-17-2014 === - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). +== Week 2 == + +(Notes by John Lensmire) + +=== Monday 10-13-2014 === + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +'''Proof:''' + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: +* $b <_s a \Rightarrow -b <_s -a$ +* $b < a \Rightarrow -b > -a$ + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: +* $a_L + (-a) < a_L + (-a_L) = 0$ +* $a + (-a_R) < a_R + (-a_R) = 0$ +* $a_R + (-a) > a_R + (-a_R) = 0$ +* $a + (-a_L) > a_L + (-a_L) = 0$ +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +'''Proof:''' + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: +* $f(a_L, b_L) < f(a_L, b_R)$ +* $f(a_R, b_R) < f(a_R, b_L)$ +* $f(a_L, b_L) < f(a_R, b_R)$ +* $f(a_R, b_R) < f(a_L, b_L)$ + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +=== Wednesday 10-15-2014 === + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +'''Proof:''' + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +=== Friday 10-17-2014 === + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/All notes - Copy (2)/week_3/g_zach.tex b/Other/old/All notes - Copy (2)/week_3/g_zach.tex index 1a414c00..60176a59 100644 --- a/Other/old/All notes - Copy (2)/week_3/g_zach.tex +++ b/Other/old/All notes - Copy (2)/week_3/g_zach.tex @@ -1 +1 @@ -Notes by Zach +Notes by Zach diff --git a/Other/old/All notes - Copy (2)/week_3/zach.aux b/Other/old/All notes - Copy (2)/week_3/zach.aux deleted file mode 100644 index 51500fd3..00000000 --- a/Other/old/All notes - Copy (2)/week_3/zach.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_3/zach}{ -\setcounter{page}{12} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{3} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_4/g_g_week_4.tex b/Other/old/All notes - Copy (2)/week_4/g_g_week_4.tex index 728cd784..ccf06aaf 100644 --- a/Other/old/All notes - Copy (2)/week_4/g_g_week_4.tex +++ b/Other/old/All notes - Copy (2)/week_4/g_g_week_4.tex @@ -1,126 +1,126 @@ -\section{ Week 4 } -Notes by Madeline Barnicle -\subsection{Monday, October 27, 2014} -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. - -\subsection{Wednesday, October 29, 2014} -* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). -* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. -*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. -*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? - -\subsubsection{Section 4: Combinatorics of Ordered Sets} -Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). - -We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. -*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. -*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. -*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. -*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. -*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. -*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. - -*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. - - -\subsection{Friday, October 31, 2014} -Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -'''Proof''' (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -'''Proof''': -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and str +\section{ Week 4 } +Notes by Madeline Barnicle +\subsection{Monday, October 27, 2014} +We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. + +Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. + +\subsection{Wednesday, October 29, 2014} +* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). +* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. +*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. +*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? + +\subsubsection{Section 4: Combinatorics of Ordered Sets} +Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). + +We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. +*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. +*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. +*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. +*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. +*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. +*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. + +*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. + +====Definition 4.1==== +$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. + + +\subsection{Friday, October 31, 2014} +Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +'''Proof''' (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +'''Proof''': +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and str diff --git a/Other/old/All notes - Copy (2)/week_4/g_week_4.aux b/Other/old/All notes - Copy (2)/week_4/g_week_4.aux deleted file mode 100644 index f9542684..00000000 --- a/Other/old/All notes - Copy (2)/week_4/g_week_4.aux +++ /dev/null @@ -1,33 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {2} Week 4 }{12}} -\@writefile{toc}{\contentsline {subsection}{\numberline {2.1}Monday, October 27, 2014}{12}} -\@writefile{toc}{\contentsline {subsection}{\numberline {2.2}Wednesday, October 29, 2014}{12}} -\@writefile{toc}{\contentsline {subsubsection}{\numberline {2.2.1}Section 4: Combinatorics of Ordered Sets}{13}} -\@writefile{toc}{\contentsline {subsection}{\numberline {2.3}Friday, October 31, 2014}{13}} -\@writefile{toc}{\contentsline {subsubsection}{\numberline {2.3.1}Hahn Fields}{14}} -\@setckpt{week_4/g_week_4}{ -\setcounter{page}{15} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{2} -\setcounter{subsection}{3} -\setcounter{subsubsection}{1} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_4/g_week_4.tex b/Other/old/All notes - Copy (2)/week_4/g_week_4.tex index cde139d3..63ae35d1 100644 --- a/Other/old/All notes - Copy (2)/week_4/g_week_4.tex +++ b/Other/old/All notes - Copy (2)/week_4/g_week_4.tex @@ -1,130 +1,130 @@ -\section{ Week 4 } -Notes by Madeline Barnicle -\subsection{Monday, October 27, 2014} -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. - -\subsection{Wednesday, October 29, 2014} -* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). -* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. -*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. -*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? - -\subsubsection{Section 4: Combinatorics of Ordered Sets} -Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). - -We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. -*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. -*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. -*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. -*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. -*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. -*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. - -*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. - - -\subsection{Friday, October 31, 2014} -Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -'''Proof''' (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -'''Proof''': -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -\subsubsection{Hahn Fields} - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. +\section{ Week 4 } +Notes by Madeline Barnicle +\subsection{Monday, October 27, 2014} +We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. + +Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. + +\subsection{Wednesday, October 29, 2014} +* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). +* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. +*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. +*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? + +\subsubsection{Section 4: Combinatorics of Ordered Sets} +Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). + +We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. +*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. +*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. +*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. +*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. +*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. +*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. + +*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. + +====Definition 4.1==== +$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. + + +\subsection{Friday, October 31, 2014} +Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +'''Proof''' (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +'''Proof''': +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +\subsubsection{Hahn Fields} + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/All notes - Copy (2)/week_4/week_2.aux b/Other/old/All notes - Copy (2)/week_4/week_2.aux deleted file mode 100644 index 31eef818..00000000 --- a/Other/old/All notes - Copy (2)/week_4/week_2.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/All notes - Copy (2)/week_4/week_4.aux b/Other/old/All notes - Copy (2)/week_4/week_4.aux deleted file mode 100644 index 48e65042..00000000 --- a/Other/old/All notes - Copy (2)/week_4/week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/week_4}{ -\setcounter{page}{15} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_4/week_4.tex b/Other/old/All notes - Copy (2)/week_4/week_4.tex index 60371053..71ccc79b 100644 --- a/Other/old/All notes - Copy (2)/week_4/week_4.tex +++ b/Other/old/All notes - Copy (2)/week_4/week_4.tex @@ -1,130 +1,130 @@ -== Week 4 == -Notes by Madeline Barnicle -===Monday, October 27, 2014=== -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. - -===Wednesday, October 29, 2014=== -* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). -* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. -*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. -*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? - -====Section 4: Combinatorics of Ordered Sets==== -Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). - -We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. -*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. -*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. -*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. -*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. -*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. -*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. - -*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. - - -===Friday, October 31, 2014=== -Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -'''Proof''' (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -'''Proof''': -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -====Hahn Fields==== - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. +== Week 4 == +Notes by Madeline Barnicle +===Monday, October 27, 2014=== +We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. + +Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. + +===Wednesday, October 29, 2014=== +* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). +* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. +*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. +*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? + +====Section 4: Combinatorics of Ordered Sets==== +Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). + +We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. +*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. +*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. +*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. +*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. +*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. +*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. + +*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. + +====Definition 4.1==== +$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. + + +===Friday, October 31, 2014=== +Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +'''Proof''' (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +'''Proof''': +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +====Hahn Fields==== + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/All notes - Copy (2)/week_5/g_g_week_5.tex b/Other/old/All notes - Copy (2)/week_5/g_g_week_5.tex index a7dba98e..7717ef00 100644 --- a/Other/old/All notes - Copy (2)/week_5/g_g_week_5.tex +++ b/Other/old/All notes - Copy (2)/week_5/g_g_week_5.tex @@ -1,139 +1,139 @@ -\section{ Week 5 } - -\subsection{November 3, 2014 } -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -\subsubsection{ Section 5: The $\omega^-$ map } -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: -* $[-\alpha]=[\alpha]$. -* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. -* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. - - -We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (H�lder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -'''Proof:''' - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: -* $0<\omega^b$, -* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -'''Proof:''' - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -\subsection{November 7, 2014} -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -''Case 1.'' $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -'''Proof:''' - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, -* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). -* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) -* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +\subsubsection{ Section 5: The $\omega^-$ map } +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: +* $[-\alpha]=[\alpha]$. +* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. +* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. + + +We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (H�lder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +'''Proof:''' + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: +* $0<\omega^b$, +* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +'''Proof:''' + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +\subsection{November 7, 2014} +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +''Case 1.'' $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +'''Proof:''' + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, +* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). +* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) +* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -\subsubsection{ Section 5: The $\omega^-$ map } -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: -* $[-\alpha]=[\alpha]$. -* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. -* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. - - -We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (H�lder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -'''Proof:''' - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: -* $0<\omega^b$, -* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -'''Proof:''' - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -\subsection{November 7, 2014} -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -''Case 1.'' $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -'''Proof:''' - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, -* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). -* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) -* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. +\section{ Week 5 } + +\subsection{November 3, 2014 } +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +\subsubsection{ Section 5: The $\omega^-$ map } +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: +* $[-\alpha]=[\alpha]$. +* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. +* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. + + +We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (H�lder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +'''Proof:''' + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: +* $0<\omega^b$, +* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +'''Proof:''' + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +\subsection{November 7, 2014} +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +''Case 1.'' $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +'''Proof:''' + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, +* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). +* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) +* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/All notes - Copy (2)/week_5/week_5.aux b/Other/old/All notes - Copy (2)/week_5/week_5.aux deleted file mode 100644 index 8ab1f07c..00000000 --- a/Other/old/All notes - Copy (2)/week_5/week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/week_5}{ -\setcounter{page}{18} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_5/week_5.tex b/Other/old/All notes - Copy (2)/week_5/week_5.tex index 87a287da..8d4fbb0e 100644 --- a/Other/old/All notes - Copy (2)/week_5/week_5.tex +++ b/Other/old/All notes - Copy (2)/week_5/week_5.tex @@ -1,139 +1,139 @@ -== Week 5 == - -===November 3, 2014 === -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -==== Section 5: The $\omega^-$ map ==== -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: -* $[-\alpha]=[\alpha]$. -* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. -* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. - - -We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (Hölder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -'''Proof:''' - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: -* $0<\omega^b$, -* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -'''Proof:''' - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -===November 7, 2014=== -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -''Case 1.'' $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -'''Proof:''' - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, -* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). -* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) -* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. +== Week 5 == + +===November 3, 2014 === +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +==== Section 5: The $\omega^-$ map ==== +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: +* $[-\alpha]=[\alpha]$. +* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. +* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. + + +We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (Hölder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +'''Proof:''' + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: +* $0<\omega^b$, +* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +'''Proof:''' + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +===November 7, 2014=== +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +''Case 1.'' $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +'''Proof:''' + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, +* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). +* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) +* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/All notes - Copy (2)/week_6/g_g_week_6.tex b/Other/old/All notes - Copy (2)/week_6/g_g_week_6.tex index 212ebbd9..67741569 100644 --- a/Other/old/All notes - Copy (2)/week_6/g_g_week_6.tex +++ b/Other/old/All notes - Copy (2)/week_6/g_g_week_6.tex @@ -1,374 +1,374 @@ -\section{ Week 6 } - -Notes by Anton Bobkov - -\subsection{Monday, November 10, 2014} -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - -* '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ -* '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -'''Proof of inequality''' - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -'''Proof of tail property''' - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -''Case 1'': $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -'''Proof of well-definiteness''' - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -\subsection{Wednesday, November 12, 2014} - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -'''Proof''' -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -'''Proof''' - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ -*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ - -''Case 1'': $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -''Case 2'': $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -'''Proof''' -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -\subsection{Friday, November 14, 2014} - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -'''Proof''' - -''Case 1'': $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^ +\section{ Week 6 } + +Notes by Anton Bobkov + +\subsection{Monday, November 10, 2014} +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + +* '''Inequality:''' For \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ +* '''Tail property:''' if $\gamma < \kappa < \alpha$ +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +'''Proof of inequality''' + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +'''Proof of tail property''' + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +''Case 1'': $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +'''Proof of well-definiteness''' + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +\subsection{Wednesday, November 12, 2014} + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +'''Proof''' +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +'''Proof''' + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ +*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ + +''Case 1'': $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +''Case 2'': $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +'''Proof''' +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +\subsection{Friday, November 14, 2014} + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +'''Proof''' + +''Case 1'': $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^ diff --git a/Other/old/All notes - Copy (2)/week_6/g_week_6.aux b/Other/old/All notes - Copy (2)/week_6/g_week_6.aux deleted file mode 100644 index ec3a2e3c..00000000 --- a/Other/old/All notes - Copy (2)/week_6/g_week_6.aux +++ /dev/null @@ -1,31 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {4} Week 6 }{18}} -\@writefile{toc}{\contentsline {subsection}{\numberline {4.1}Monday, November 10, 2014}{18}} -\@writefile{toc}{\contentsline {subsection}{\numberline {4.2}Wednesday, November 12, 2014}{22}} -\@writefile{toc}{\contentsline {subsection}{\numberline {4.3}Friday, November 14, 2014}{24}} -\@setckpt{week_6/g_week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{4} -\setcounter{subsection}{3} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_6/g_week_6.tex b/Other/old/All notes - Copy (2)/week_6/g_week_6.tex index 5e3190c6..5c462d73 100644 --- a/Other/old/All notes - Copy (2)/week_6/g_week_6.tex +++ b/Other/old/All notes - Copy (2)/week_6/g_week_6.tex @@ -1,390 +1,390 @@ -\section{ Week 6 } - -Notes by Anton Bobkov - -\subsection{Monday, November 10, 2014} -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - -* '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ -* '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -'''Proof of inequality''' - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -'''Proof of tail property''' - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -''Case 1'': $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -'''Proof of well-definiteness''' - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -\subsection{Wednesday, November 12, 2014} - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -'''Proof''' -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -'''Proof''' - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ -*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ - -''Case 1'': $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -''Case 2'': $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -'''Proof''' -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -\subsection{Friday, November 14, 2014} - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -'''Proof''' - -''Case 1'': $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -''Case 2'': $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. +\section{ Week 6 } + +Notes by Anton Bobkov + +\subsection{Monday, November 10, 2014} +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + +* '''Inequality:''' For \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ +* '''Tail property:''' if $\gamma < \kappa < \alpha$ +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +'''Proof of inequality''' + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +'''Proof of tail property''' + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +''Case 1'': $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +'''Proof of well-definiteness''' + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +\subsection{Wednesday, November 12, 2014} + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +'''Proof''' +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +'''Proof''' + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ +*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ + +''Case 1'': $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +''Case 2'': $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +'''Proof''' +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +\subsection{Friday, November 14, 2014} + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +'''Proof''' + +''Case 1'': $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +''Case 2'': $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/All notes - Copy (2)/week_6/week_6.aux b/Other/old/All notes - Copy (2)/week_6/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/All notes - Copy (2)/week_6/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_6/week_6.tex b/Other/old/All notes - Copy (2)/week_6/week_6.tex index 7d2437eb..9c0f6dc8 100644 --- a/Other/old/All notes - Copy (2)/week_6/week_6.tex +++ b/Other/old/All notes - Copy (2)/week_6/week_6.tex @@ -1,390 +1,390 @@ -== Week 6 == - -Notes by Anton Bobkov - -===Monday, November 10, 2014=== -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - -* '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ -* '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -'''Proof of inequality''' - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -'''Proof of tail property''' - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -''Case 1'': $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -'''Proof of well-definiteness''' - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -===Wednesday, November 12, 2014=== - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -'''Proof''' -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -'''Proof''' - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ -*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ - -''Case 1'': $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -''Case 2'': $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -'''Proof''' -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -===Friday, November 14, 2014=== - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -'''Proof''' - -''Case 1'': $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -''Case 2'': $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. +== Week 6 == + +Notes by Anton Bobkov + +===Monday, November 10, 2014=== +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + +* '''Inequality:''' For \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ +* '''Tail property:''' if $\gamma < \kappa < \alpha$ +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +'''Proof of inequality''' + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +'''Proof of tail property''' + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +''Case 1'': $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +'''Proof of well-definiteness''' + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +===Wednesday, November 12, 2014=== + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +'''Proof''' +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +'''Proof''' + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ +*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ + +''Case 1'': $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +''Case 2'': $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +'''Proof''' +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +===Friday, November 14, 2014=== + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +'''Proof''' + +''Case 1'': $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +''Case 2'': $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/All notes - Copy (2)/week_7/285D.aux b/Other/old/All notes - Copy (2)/week_7/285D.aux deleted file mode 100644 index cd380abd..00000000 --- a/Other/old/All notes - Copy (2)/week_7/285D.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_7/285D}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/All notes - Copy (2)/week_7/285D_notes_nov_17_18_21.aux b/Other/old/All notes - Copy (2)/week_7/285D_notes_nov_17_18_21.aux deleted file mode 100644 index 5b2eedb5..00000000 --- a/Other/old/All notes - Copy (2)/week_7/285D_notes_nov_17_18_21.aux +++ /dev/null @@ -1,32 +0,0 @@ -\relax -\newlabel{6.7}{{1}{25}} -\newlabel{6.8}{{2}{26}} -\@writefile{toc}{\contentsline {section}{\numberline {5}The Surreals as a Real Closed Field}{29}} -\newlabel{7.1}{{9}{29}} -\newlabel{7.2}{{10}{29}} -\@setckpt{week_7/285D_notes_nov_17_18_21}{ -\setcounter{page}{32} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{5} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_7/g_285D notes nov 17 18 21.tex b/Other/old/All notes - Copy (2)/week_7/g_285D notes nov 17 18 21.tex index 44104621..47fc2f01 100644 --- a/Other/old/All notes - Copy (2)/week_7/g_285D notes nov 17 18 21.tex +++ b/Other/old/All notes - Copy (2)/week_7/g_285D notes nov 17 18 21.tex @@ -1,273 +1,273 @@ -\documentclass[12pt]{article} -\RequirePackage[left=1.5in,right=1.5in,top=1.5in,bottom=1.5in]{geometry} %margins -\usepackage{amsmath} %general math symbols - -\usepackage{fourier} %Font -\usepackage{amssymb} %\textbf -%\usepackage{mathrsfs} %\mathscr -\usepackage{dsfont} %\mathds -\usepackage{bussproofs} -%\usepackage{latexsym} -\usepackage{multicol} - -% This is the "centered" symbol -\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} - -% Optional to turn on the short abbreviations -\EnableBpAbbreviations - -\usepackage{tikz} %commutative diagrams -\usepackage{enumerate} %lists - -\usepackage{amsthm} -\usepackage{bm} - - -\swapnumbers -\theoremstyle{theorem} %bold title, italicized font -\newtheorem{theorem}{Theorem}[section] - -\theoremstyle{definition} %bold title, regular font -\newtheorem{example}[theorem]{Example} -\newtheorem{definition}[theorem]{Definition} -\newtheorem{proposition}[theorem]{Proposition} -\newtheorem{lemma}[theorem]{Lemma} -\newtheorem{corollary}[theorem]{Corollary} -\newtheorem{exercise}[theorem]{Exercise} -\newtheorem*{problem}{Problem} -\newtheorem{warning}[theorem]{Warning} -\newtheorem*{solution}{Solution} -\newtheorem*{remark}{Remark} - -\theoremstyle{empty} -\newtheorem{namedtheorem}{} - -\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} - -\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} - - -\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} -\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} - -\renewcommand{\restriction}{\mathord{\upharpoonright}} - -\begin{document} -\tikzset{node distance=2cm, auto} - -\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ -\text{} \\ -\begin{large} Tyler Arant \end{large} -\end{center} - - - - -\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, -$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -\end{lemma} - -\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have -\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ - &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ - & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} -where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. - -In the case where $\beta$ is a limit ordinal, we let -$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ -A typical element of this cut is -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ -By inductive hypothesis, this equals -$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ -But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. -\end{proof} - -\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ -\end{proposition} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then -\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ - & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ - & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} -where the third equality uses the induction hypothesis. - -Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form -$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ -or of the form -$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ -We have -\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} - & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ - & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} - where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with - $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ - Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ -\end{proof} - - -\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ -Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then -\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b - &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ - & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ - & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} -where the third equality uses the inductive hypothesis. - -Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, -$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ -and -$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ -Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that -$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ -and -$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ -It follows that -\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ -The verification for (2) is similar. So, by (1), (2) and confinality, -$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ - d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ -We claim that we can further simplify this to -$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ -then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and -$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ -We claim that -$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since -$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ -where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. -\end{proof} - -\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ -\end{proposition} - -\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put -$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ -Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is -\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} -where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, -$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ -and -$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ -where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, -\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} - Now, -\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} -Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. - -\end{proof} - -All together, this completes the proof of the following theorem. - -\begin{theorem} The map -$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ -is an ordered field isomorphism. \end{theorem} - - - -\section{The Surreals as a Real Closed Field} - -Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. - -\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. -\begin{enumerate}[(1)] -\item $K$ is orderable, but has no proper orderable algebraic field extension. -\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. -\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. -\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. -\end{enumerate} -We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} -\end{theorem} - -\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} - -\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. - -Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. -\end{proof} - -The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. - -\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, -\begin{enumerate}[(1)] -\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. -\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} - -\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} - -We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} - -\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} - -\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write -$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ -By replacing $P(X)$ by $P(X-a_{n-1})$, we get -$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for -$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ -we have -$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ -for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that -\begin{enumerate}[$\bullet$] -\item $b_\beta >b_i$ for all $i<\beta$. -\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. -\end{enumerate} -To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then -$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ -for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation -$$S=Q_0H+R_0G,$$ -where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. - -\end{proof} - - - - - -\end{document} - - +\documentclass[12pt]{article} +\RequirePackage[left=1.5in,right=1.5in,top=1.5in,bottom=1.5in]{geometry} %margins +\usepackage{amsmath} %general math symbols + +\usepackage{fourier} %Font +\usepackage{amssymb} %\textbf +%\usepackage{mathrsfs} %\mathscr +\usepackage{dsfont} %\mathds +\usepackage{bussproofs} +%\usepackage{latexsym} +\usepackage{multicol} + +% This is the "centered" symbol +\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} + +% Optional to turn on the short abbreviations +\EnableBpAbbreviations + +\usepackage{tikz} %commutative diagrams +\usepackage{enumerate} %lists + +\usepackage{amsthm} +\usepackage{bm} + + +\swapnumbers +\theoremstyle{theorem} %bold title, italicized font +\newtheorem{theorem}{Theorem}[section] + +\theoremstyle{definition} %bold title, regular font +\newtheorem{example}[theorem]{Example} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{lemma}[theorem]{Lemma} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{exercise}[theorem]{Exercise} +\newtheorem*{problem}{Problem} +\newtheorem{warning}[theorem]{Warning} +\newtheorem*{solution}{Solution} +\newtheorem*{remark}{Remark} + +\theoremstyle{empty} +\newtheorem{namedtheorem}{} + +\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} + +\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} + + +\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} +\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} + +\renewcommand{\restriction}{\mathord{\upharpoonright}} + +\begin{document} +\tikzset{node distance=2cm, auto} + +\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ +\text{} \\ +\begin{large} Tyler Arant \end{large} +\end{center} + + + + +\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, +$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +\end{lemma} + +\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have +\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ + &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ + & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} +where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. + +In the case where $\beta$ is a limit ordinal, we let +$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ +A typical element of this cut is +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ +By inductive hypothesis, this equals +$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ +But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. +\end{proof} + +\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ +\end{proposition} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then +\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ + & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ + & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} +where the third equality uses the induction hypothesis. + +Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form +$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ +or of the form +$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ +We have +\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} + & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ + & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} + where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with + $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ + Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ +\end{proof} + + +\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ +Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then +\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b + &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ + & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ + & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} +where the third equality uses the inductive hypothesis. + +Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, +$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ +and +$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ +Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that +$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ +and +$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ +It follows that +\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ +The verification for (2) is similar. So, by (1), (2) and confinality, +$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ + d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ +We claim that we can further simplify this to +$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ +then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and +$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ +We claim that +$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since +$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ +where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. +\end{proof} + +\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ +\end{proposition} + +\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put +$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ +Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is +\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} +where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, +$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ +and +$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ +where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, +\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} + Now, +\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} +Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. + +\end{proof} + +All together, this completes the proof of the following theorem. + +\begin{theorem} The map +$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ +is an ordered field isomorphism. \end{theorem} + + + +\section{The Surreals as a Real Closed Field} + +Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. + +\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. +\begin{enumerate}[(1)] +\item $K$ is orderable, but has no proper orderable algebraic field extension. +\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. +\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. +\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. +\end{enumerate} +We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} +\end{theorem} + +\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} + +\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. + +Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. +\end{proof} + +The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. + +\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, +\begin{enumerate}[(1)] +\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. +\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} + +\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} + +We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} + +\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} + +\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write +$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ +By replacing $P(X)$ by $P(X-a_{n-1})$, we get +$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for +$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ +we have +$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ +for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that +\begin{enumerate}[$\bullet$] +\item $b_\beta >b_i$ for all $i<\beta$. +\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. +\end{enumerate} +To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then +$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ +for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation +$$S=Q_0H+R_0G,$$ +where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. + +\end{proof} + + + + + +\end{document} + + diff --git a/Other/old/All notes - Copy (2)/week_7/g_285D_notes_nov_17_18_21.tex b/Other/old/All notes - Copy (2)/week_7/g_285D_notes_nov_17_18_21.tex index 14cd6104..834de7c6 100644 --- a/Other/old/All notes - Copy (2)/week_7/g_285D_notes_nov_17_18_21.tex +++ b/Other/old/All notes - Copy (2)/week_7/g_285D_notes_nov_17_18_21.tex @@ -1,205 +1,205 @@ -\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ -\text{} \\ -\begin{large} Tyler Arant \end{large} -\end{center} - -\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, -$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -\end{lemma} - -\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have -\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ - &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ - & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} -where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. - -In the case where $\beta$ is a limit ordinal, we let -$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ -A typical element of this cut is -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ -By inductive hypothesis, this equals -$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ -But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. -\end{proof} - -\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ -\end{proposition} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then -\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ - & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ - & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} -where the third equality uses the induction hypothesis. - -Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form -$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ -or of the form -$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ -We have -\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} - & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ - & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} - where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with - $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ - Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ -\end{proof} - - -\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ -Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then -\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b - &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ - & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ - & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} -where the third equality uses the inductive hypothesis. - -Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, -$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ -and -$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ -Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that -$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ -and -$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ -It follows that -\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ -The verification for (2) is similar. So, by (1), (2) and confinality, -$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ - d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ -We claim that we can further simplify this to -$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ -then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and -$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ -We claim that -$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since -$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ -where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. -\end{proof} - -\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ -\end{proposition} - -\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put -$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ -Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is -\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} -where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, -$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ -and -$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ -where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, -\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} - Now, -\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} -Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. - -\end{proof} - -All together, this completes the proof of the following theorem. - -\begin{theorem} The map -$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ -is an ordered field isomorphism. \end{theorem} - - - -\section{The Surreals as a Real Closed Field} - -Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. - -\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. -\begin{enumerate}[(1)] -\item $K$ is orderable, but has no proper orderable algebraic field extension. -\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. -\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. -\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. -\end{enumerate} -We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} -\end{theorem} - -\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} - -\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. - -Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. -\end{proof} - -The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. - -\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, -\begin{enumerate}[(1)] -\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. -\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} - -\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} - -We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} - -\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} - -\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write -$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ -By replacing $P(X)$ by $P(X-a_{n-1})$, we get -$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for -$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ -we have -$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ -for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that -\begin{enumerate}[$\bullet$] -\item $b_\beta >b_i$ for all $i<\beta$. -\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. -\end{enumerate} -To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then -$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ -for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation -$$S=Q_0H+R_0G,$$ -where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. - -\end{proof} +\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ +\text{} \\ +\begin{large} Tyler Arant \end{large} +\end{center} + +\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, +$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +\end{lemma} + +\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have +\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ + &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ + & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} +where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. + +In the case where $\beta$ is a limit ordinal, we let +$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ +A typical element of this cut is +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ +By inductive hypothesis, this equals +$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ +But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. +\end{proof} + +\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ +\end{proposition} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then +\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ + & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ + & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} +where the third equality uses the induction hypothesis. + +Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form +$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ +or of the form +$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ +We have +\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} + & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ + & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} + where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with + $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ + Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ +\end{proof} + + +\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ +Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then +\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b + &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ + & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ + & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} +where the third equality uses the inductive hypothesis. + +Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, +$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ +and +$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ +Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that +$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ +and +$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ +It follows that +\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ +The verification for (2) is similar. So, by (1), (2) and confinality, +$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ + d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ +We claim that we can further simplify this to +$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ +then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and +$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ +We claim that +$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since +$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ +where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. +\end{proof} + +\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ +\end{proposition} + +\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put +$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ +Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is +\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} +where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, +$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ +and +$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ +where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, +\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} + Now, +\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} +Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. + +\end{proof} + +All together, this completes the proof of the following theorem. + +\begin{theorem} The map +$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ +is an ordered field isomorphism. \end{theorem} + + + +\section{The Surreals as a Real Closed Field} + +Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. + +\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. +\begin{enumerate}[(1)] +\item $K$ is orderable, but has no proper orderable algebraic field extension. +\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. +\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. +\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. +\end{enumerate} +We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} +\end{theorem} + +\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} + +\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. + +Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. +\end{proof} + +The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. + +\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, +\begin{enumerate}[(1)] +\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. +\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} + +\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} + +We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} + +\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} + +\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write +$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ +By replacing $P(X)$ by $P(X-a_{n-1})$, we get +$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for +$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ +we have +$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ +for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that +\begin{enumerate}[$\bullet$] +\item $b_\beta >b_i$ for all $i<\beta$. +\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. +\end{enumerate} +To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then +$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ +for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation +$$S=Q_0H+R_0G,$$ +where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. + +\end{proof} diff --git a/Other/old/All notes - Copy (2)/week_7/g_g_285D notes nov 17 18 21.tex b/Other/old/All notes - Copy (2)/week_7/g_g_285D notes nov 17 18 21.tex index 44104621..47fc2f01 100644 --- a/Other/old/All notes - Copy (2)/week_7/g_g_285D notes nov 17 18 21.tex +++ b/Other/old/All notes - Copy (2)/week_7/g_g_285D notes nov 17 18 21.tex @@ -1,273 +1,273 @@ -\documentclass[12pt]{article} -\RequirePackage[left=1.5in,right=1.5in,top=1.5in,bottom=1.5in]{geometry} %margins -\usepackage{amsmath} %general math symbols - -\usepackage{fourier} %Font -\usepackage{amssymb} %\textbf -%\usepackage{mathrsfs} %\mathscr -\usepackage{dsfont} %\mathds -\usepackage{bussproofs} -%\usepackage{latexsym} -\usepackage{multicol} - -% This is the "centered" symbol -\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} - -% Optional to turn on the short abbreviations -\EnableBpAbbreviations - -\usepackage{tikz} %commutative diagrams -\usepackage{enumerate} %lists - -\usepackage{amsthm} -\usepackage{bm} - - -\swapnumbers -\theoremstyle{theorem} %bold title, italicized font -\newtheorem{theorem}{Theorem}[section] - -\theoremstyle{definition} %bold title, regular font -\newtheorem{example}[theorem]{Example} -\newtheorem{definition}[theorem]{Definition} -\newtheorem{proposition}[theorem]{Proposition} -\newtheorem{lemma}[theorem]{Lemma} -\newtheorem{corollary}[theorem]{Corollary} -\newtheorem{exercise}[theorem]{Exercise} -\newtheorem*{problem}{Problem} -\newtheorem{warning}[theorem]{Warning} -\newtheorem*{solution}{Solution} -\newtheorem*{remark}{Remark} - -\theoremstyle{empty} -\newtheorem{namedtheorem}{} - -\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} - -\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} - - -\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} -\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} - -\renewcommand{\restriction}{\mathord{\upharpoonright}} - -\begin{document} -\tikzset{node distance=2cm, auto} - -\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ -\text{} \\ -\begin{large} Tyler Arant \end{large} -\end{center} - - - - -\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, -$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -\end{lemma} - -\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have -\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ - &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ - & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} -where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. - -In the case where $\beta$ is a limit ordinal, we let -$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ -A typical element of this cut is -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ -By inductive hypothesis, this equals -$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ -But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. -\end{proof} - -\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ -\end{proposition} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then -\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ - & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ - & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} -where the third equality uses the induction hypothesis. - -Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form -$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ -or of the form -$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ -We have -\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} - & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ - & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} - where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with - $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ - Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ -\end{proof} - - -\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ -Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then -\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b - &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ - & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ - & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} -where the third equality uses the inductive hypothesis. - -Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, -$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ -and -$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ -Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that -$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ -and -$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ -It follows that -\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ -The verification for (2) is similar. So, by (1), (2) and confinality, -$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ - d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ -We claim that we can further simplify this to -$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ -then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and -$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ -We claim that -$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since -$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ -where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. -\end{proof} - -\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ -\end{proposition} - -\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put -$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ -Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is -\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} -where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, -$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ -and -$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ -where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, -\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} - Now, -\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} -Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. - -\end{proof} - -All together, this completes the proof of the following theorem. - -\begin{theorem} The map -$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ -is an ordered field isomorphism. \end{theorem} - - - -\section{The Surreals as a Real Closed Field} - -Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. - -\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. -\begin{enumerate}[(1)] -\item $K$ is orderable, but has no proper orderable algebraic field extension. -\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. -\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. -\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. -\end{enumerate} -We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} -\end{theorem} - -\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} - -\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. - -Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. -\end{proof} - -The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. - -\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, -\begin{enumerate}[(1)] -\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. -\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} - -\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} - -We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} - -\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} - -\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write -$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ -By replacing $P(X)$ by $P(X-a_{n-1})$, we get -$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for -$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ -we have -$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ -for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that -\begin{enumerate}[$\bullet$] -\item $b_\beta >b_i$ for all $i<\beta$. -\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. -\end{enumerate} -To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then -$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ -for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation -$$S=Q_0H+R_0G,$$ -where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. - -\end{proof} - - - - - -\end{document} - - +\documentclass[12pt]{article} +\RequirePackage[left=1.5in,right=1.5in,top=1.5in,bottom=1.5in]{geometry} %margins +\usepackage{amsmath} %general math symbols + +\usepackage{fourier} %Font +\usepackage{amssymb} %\textbf +%\usepackage{mathrsfs} %\mathscr +\usepackage{dsfont} %\mathds +\usepackage{bussproofs} +%\usepackage{latexsym} +\usepackage{multicol} + +% This is the "centered" symbol +\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} + +% Optional to turn on the short abbreviations +\EnableBpAbbreviations + +\usepackage{tikz} %commutative diagrams +\usepackage{enumerate} %lists + +\usepackage{amsthm} +\usepackage{bm} + + +\swapnumbers +\theoremstyle{theorem} %bold title, italicized font +\newtheorem{theorem}{Theorem}[section] + +\theoremstyle{definition} %bold title, regular font +\newtheorem{example}[theorem]{Example} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{lemma}[theorem]{Lemma} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{exercise}[theorem]{Exercise} +\newtheorem*{problem}{Problem} +\newtheorem{warning}[theorem]{Warning} +\newtheorem*{solution}{Solution} +\newtheorem*{remark}{Remark} + +\theoremstyle{empty} +\newtheorem{namedtheorem}{} + +\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} + +\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} + + +\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} +\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} + +\renewcommand{\restriction}{\mathord{\upharpoonright}} + +\begin{document} +\tikzset{node distance=2cm, auto} + +\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ +\text{} \\ +\begin{large} Tyler Arant \end{large} +\end{center} + + + + +\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, +$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +\end{lemma} + +\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have +\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ + &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ + & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} +where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. + +In the case where $\beta$ is a limit ordinal, we let +$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ +A typical element of this cut is +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ +By inductive hypothesis, this equals +$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ +But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. +\end{proof} + +\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ +\end{proposition} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then +\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ + & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ + & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} +where the third equality uses the induction hypothesis. + +Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form +$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ +or of the form +$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ +We have +\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} + & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ + & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} + where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with + $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ + Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ +\end{proof} + + +\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ +Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then +\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b + &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ + & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ + & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} +where the third equality uses the inductive hypothesis. + +Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, +$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ +and +$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ +Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that +$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ +and +$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ +It follows that +\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ +The verification for (2) is similar. So, by (1), (2) and confinality, +$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ + d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ +We claim that we can further simplify this to +$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ +then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and +$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ +We claim that +$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since +$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ +where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. +\end{proof} + +\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ +\end{proposition} + +\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put +$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ +Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is +\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} +where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, +$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ +and +$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ +where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, +\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} + Now, +\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} +Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. + +\end{proof} + +All together, this completes the proof of the following theorem. + +\begin{theorem} The map +$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ +is an ordered field isomorphism. \end{theorem} + + + +\section{The Surreals as a Real Closed Field} + +Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. + +\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. +\begin{enumerate}[(1)] +\item $K$ is orderable, but has no proper orderable algebraic field extension. +\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. +\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. +\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. +\end{enumerate} +We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} +\end{theorem} + +\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} + +\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. + +Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. +\end{proof} + +The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. + +\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, +\begin{enumerate}[(1)] +\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. +\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} + +\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} + +We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} + +\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} + +\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write +$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ +By replacing $P(X)$ by $P(X-a_{n-1})$, we get +$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for +$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ +we have +$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ +for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that +\begin{enumerate}[$\bullet$] +\item $b_\beta >b_i$ for all $i<\beta$. +\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. +\end{enumerate} +To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then +$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ +for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation +$$S=Q_0H+R_0G,$$ +where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. + +\end{proof} + + + + + +\end{document} + + diff --git a/Other/old/All notes - Copy (2)/week_7/g_g_285D_notes_nov_17_18_21.tex b/Other/old/All notes - Copy (2)/week_7/g_g_285D_notes_nov_17_18_21.tex index 14cd6104..834de7c6 100644 --- a/Other/old/All notes - Copy (2)/week_7/g_g_285D_notes_nov_17_18_21.tex +++ b/Other/old/All notes - Copy (2)/week_7/g_g_285D_notes_nov_17_18_21.tex @@ -1,205 +1,205 @@ -\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ -\text{} \\ -\begin{large} Tyler Arant \end{large} -\end{center} - -\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, -$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -\end{lemma} - -\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have -\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ - &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ - & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} -where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. - -In the case where $\beta$ is a limit ordinal, we let -$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ -Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ -A typical element of this cut is -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ -By inductive hypothesis, this equals -$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ -But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. -\end{proof} - -\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ -\end{proposition} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then -\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ - & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ - & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} -where the third equality uses the induction hypothesis. - -Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form -$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ -or of the form -$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ -We have -\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} - & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ - & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} - where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with - $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ - Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ -\end{proof} - - -\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ -Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} - -\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then -\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b - &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ - & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ - & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ - & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} -where the third equality uses the inductive hypothesis. - -Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, -$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ -and -$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ -Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that -$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ -and -$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ -It follows that -\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ -The verification for (2) is similar. So, by (1), (2) and confinality, -$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ - d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ -We claim that we can further simplify this to -$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ -then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and -$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ -We claim that -$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since -$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ -where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. -\end{proof} - -\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, -$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ -\end{proposition} - -\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put -$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ -Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is -\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} -where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, -$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ -and -$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ -where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, -\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} - Now, -\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ - &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} -Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. - -\end{proof} - -All together, this completes the proof of the following theorem. - -\begin{theorem} The map -$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ -is an ordered field isomorphism. \end{theorem} - - - -\section{The Surreals as a Real Closed Field} - -Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. - -\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. -\begin{enumerate}[(1)] -\item $K$ is orderable, but has no proper orderable algebraic field extension. -\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. -\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. -\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. -\end{enumerate} -We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} -\end{theorem} - -\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} - -\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. - -Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. -\end{proof} - -The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. - -\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, -\begin{enumerate}[(1)] -\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. -\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} - -\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} - -We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. - -\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} - -\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} - -\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write -$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ -By replacing $P(X)$ by $P(X-a_{n-1})$, we get -$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for -$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ -we have -$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ -for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that -\begin{enumerate}[$\bullet$] -\item $b_\beta >b_i$ for all $i<\beta$. -\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. -\end{enumerate} -To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then -$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ -for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation -$$S=Q_0H+R_0G,$$ -where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. - -\end{proof} +\begin{center} \begin{Large} Math 285D Notes: 11/17, 11/19, 11/21 \end{Large}\\ +\text{} \\ +\begin{large} Tyler Arant \end{large} +\end{center} + +\begin{lemma}[Associativity] \label{6.7} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha+\beta}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i\in \mathds{R}$ for $i<\alpha+\beta$. Then, +$$\sum_{i<\alpha+\beta} f_i\omega^{a_i}=\sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +\end{lemma} + +\begin{proof} We proceed by induction on $\beta$. In the case that $\beta=\gamma+1$ is a successor ordinal, we have +\begin{align*} \sum_{i<\alpha+(\gamma+1)} f_i\omega^{a_i}&= \sum_{i<\alpha+\gamma} f_i\omega^{a_i} + f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}} \\ + &= \sum_{i<\alpha} f_i\omega^{a_i}+ \sum_{j<\gamma} f_{\alpha+j} \omega^{a_{\alpha+j}}+ f_{\alpha+\gamma}\omega^{a_{\alpha+\gamma}}\\ + & = \sum_{i<\alpha} f_i\omega^{a_i} + \sum_{j<\gamma+1}f_{\alpha+j}\omega^{a_{\alpha+j}}, \end{align*} +where the first and third equality use the definition of $\sum$ and the second equality uses the induction hypothesis. + +In the case where $\beta$ is a limit ordinal, we let +$$\{L | R\} = \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}}.$$ +Using the definition of addition between surreal numbers and a simple cofinality argument, we obtain +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j<\beta}f_{\alpha+j}\omega^{a_{\alpha+j}} = \left \{\sum_{i<\alpha}f_i\omega^{a_i} + L \biggl | \sum_{i<\alpha}f_i\omega^{a_i} + R \right \}.$$ +A typical element of this cut is +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{j\leq \gamma} f_{\alpha+j}\omega^{a_{\alpha+j}}-\varepsilon \omega^{a_{\alpha+\gamma}} \qquad (\gamma<\beta, \varepsilon \in \mathds{R}^{>0}).$$ +By inductive hypothesis, this equals +$$\sum_{i<\alpha+\gamma}f_i \omega^{a_i}- \varepsilon \omega^{a_{\alpha+\gamma}}.$$ +But these elements are cofinal in the cut defining $\sum_{i<\alpha+\beta} f_i\omega^{a_i}$; hence, the claim follows by cofinality. +\end{proof} + +\begin{proposition} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$ and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i} = \sum_{i<\alpha}(f_i+g_i)\omega^{a_i}.$$ +\end{proposition} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta+1$ is a successor, then +\begin{align*} \sum_{i<\beta+1}f_i\omega^{a_i} + \sum_{i<\beta+1} g_i \omega^{a_i} & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right ) + \left ( \sum_{i<\beta} g_i \omega^{a_i} + g_\beta \omega^{a_\beta} \right ) \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i} + \sum_{i<\beta} g_i \omega^{a_i} \right ) + ( f_\beta \omega^{a_\beta} + g_\beta \omega^{a_\beta}) \\ + & = \sum_{i<\beta} (f_i+g_i) \omega^{a_i} + (f_\beta+g_\beta)\omega^{a_\beta} \\ + & = \sum_{i<\beta+1}(f_i+g_i)\omega^{a_i}, \end{align*} +where the third equality uses the induction hypothesis. + +Now suppose $\alpha$ is a limit. One type of element from the lef-hand-side of the cut defining $\sum_{i<\alpha}f_i\omega^{a_i} + \sum_{i<\alpha} g_i \omega^{a_i}$ is of the form +$$\sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i}$$ +or of the form +$$\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$$ +We have +\begin{align*} \sum_{i\leq \beta} f_i \omega^{a_i}-\varepsilon \omega^{a_\beta} +\sum_{i<\alpha} g_i \omega^{a_i} + & = \sum_{i\leq \beta} f_i \omega^{a_i} + \sum_{i\leq \beta} g_i \omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta} \\ + & = \sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} + \sum_{\beta< i<\alpha} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}, \end{align*} + where the first equality follows from $(\ref{6.7})$ and the second equality uses the inductive hypothesis. But this is mutually cofinal with + $$\sum_{i\leq \beta}(f_i+g_i)\omega^{a_i} - \varepsilon \omega^{a_\beta}.$$ + Similarly if we star with $\sum_{i<\alpha} f_i \omega^{a_i} +\sum_{i\leq \beta} g_i \omega^{a_i} -\varepsilon \omega^{a_\beta}.$ +\end{proof} + + +\begin{lemma} \label{6.8} Let $\alpha \in \textbf{On}$, $(a_i)_{i<\alpha}$ be a strictly decreasing sequence in $\textbf{No}$, $b\in \textbf{No}$, and $f_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left ( \sum_{i<\alpha} f_i\omega^{a_i} \right ) \omega^b = \sum_{i<\alpha}f_i\omega^{a_i+b}.$$ +Note that the sequence $(a_i+b)_i$ is also strictly decreasing.\end{lemma} + +\begin{proof} We proceed by induction on $\alpha$. If $\alpha=\beta +1$, then +\begin{align*}\left ( \sum_{i<\beta + 1} f_i\omega^{a_i} \right ) \omega^b + &= \left ( \sum_{i<\beta} f_i\omega^{a_i} + f_\beta \omega^{a_\beta} \right )\omega^b \\ + & = \left ( \sum_{i<\beta} f_i\omega^{a_i}\right ) \omega^b + f_\beta \omega^{a_\beta}\cdot \omega^b \\ + & = \sum_{i<\beta}f_i\omega^{a_i+b} + f_\beta\omega^{a_\beta+b} \\ + & = \sum_{i<\beta+1}f_i\omega^{a_i+b}, \end{align*} +where the third equality uses the inductive hypothesis. + +Now suppose $\alpha$ is a limit. Recall that, by their respective definitions, +$$\omega^b=\{0, s\omega^{b'} \ | \ t\omega^{b''}\}$$ +and +$$\sum_{i<\alpha}f_i\omega^{a_i} = \left \{ \sum_{i\leq \beta} f_i\omega^{a_i} - \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0} \ \biggl | \ \sum_{i\leq \beta} f_i\omega^{a_i} + \varepsilon \omega^{a_\beta} \ : \ \beta<\alpha, \varepsilon\in \mathds{R}^{>0}\right \}.$$ +Set $d:=\sum_{i<\alpha}f_i\omega^{a_i}$ and let $d', d''$ be elements from the left and right-hand sides, respectively, of the defining cut determined by the same choice of $\varepsilon$. Note that +$$d-d' = \varepsilon \omega^{a_\beta} + c', \quad \text{where} \ c'\ll \omega^{a_\beta},$$ +and +$$d''-d = \varepsilon \omega^{a_\beta} + c'', \quad \text{where} \ c''\ll \omega^{a_\beta}.$$ +It follows that +\begin{equation}\varepsilon_1\omega^{a_\beta} 2\varepsilon_2 \omega^{a_\beta}\omega^b\geq (d''-d)\omega^b.$$ +The verification for (2) is similar. So, by (1), (2) and confinality, +$$ d\omega^b = \{d'\omega^b, d' \omega^b+(d-d')s\omega^{b'} \ | \ + d''\omega^b, d''\omega^b-(d''-d)s\omega^{b'}\}.$$ +We claim that we can further simplify this to +$$d\omega^b=\{d'\omega^b \ | \ d''\omega^b\},$$ +then we are done by inductive hypothesis. Let now $\varepsilon_{1, 2}\in \mathds{R}^{>0}$ with $\varepsilon_1<\varepsilon<\varepsilon_2$ and +$$d_1'= \sum_{i\geq \beta}f_i\omega^{a_i}-\varepsilon_1\omega^{a_\beta}, \quad d_1''=\sum_{i\geq \beta}f_i\omega^{a_i}+\varepsilon_1\omega^{a_\beta}.$$ +We claim that +$$d_1'\omega^b>d'\omega^b+(d-d')s\omega^{b'}, \quad d_1''\omega^b(d-d')s\omega^{b'}$. But this inequality holds since +$$(d_1'-d)\omega^b= (\varepsilon-\varepsilon_2)\omega^{a_\beta}\omega^b> \varepsilon_2s\omega^{a_\beta}\omega^{b'} \geq (d-d')s\omega^{b'},$$ +where the first inequality holds since $\omega^b\gg \omega^{b'}$ and the second inequality holds by $(*)$. The second part of the claim is proved similarly. +\end{proof} + +\begin{proposition} Let $\alpha, \beta \in \textbf{On}$, $(a_i)_{i<\alpha}$, $(b_j)_{j<\beta}$ be strictly decreasing sequences in $\textbf{No}$, and $f_i, g_i\in \mathds{R}$ for $i<\alpha$. Then, +$$\left (\sum_{i<\alpha}f_i\omega^{a_i} \right ) \left ( \sum_{j<\beta}g_j\omega^{b_j} \right ) = \sum_{i<\alpha, j<\beta} f_ig_j \omega^{a_i+b_j}.$$ +\end{proposition} + +\begin{proof} If either $\alpha$ or $\beta$ are successor ordinals, we verify the proposition by using the inductive hypothesis and lemma $(\ref{6.8})$. Thus, we only need to consider the case where $\alpha$ and $\beta$ are both limits. Put +$$f=\sum_{i<\alpha}f_i X^{a_i}, \quad g=\sum_{j<\beta}g_jX^{a_j}\in K.$$ +Recall that the typical element in the cut of $f(\omega)\cdot g(\omega)$ is +\begin{equation} f(\omega)g(\omega)_{**} + f(\omega)_{*}g(\omega)-f(\omega)_*g(\omega)_{**}, \tag{$\dagger$} \end{equation} +where $*, **$ are either $L$ or $R$. Moreover, this element is $\gamma$. Similarly, $g-g_{**}= \pm \varepsilon_2X^{b_\delta} + h_2$, where $\delta <\beta$ and all the terms in $h_2$ have degree $>\delta$. Thus, +$$(f-f_*)(g-g_{**})= \pm \varepsilon_1\varepsilon_2 X^{a_\gamma + b_\delta} + \text{higher order terms},$$ +and +$$[(f-f_*)(g-g_{**})](\omega) = \pm \varepsilon_1\varepsilon_2\omega^{\alpha_\gamma+b_\delta} + h_3(\omega),$$ +where $h_3(\omega)\ll\omega^{a_\gamma+b_\delta}$. So by cofinality, +\begin{align*}f(\omega)g(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad (f\cdot g)(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta, \varepsilon\in \mathds{R}^{>0} \}. \end{align*} + Now, +\begin{align*}(f\cdot g)(\omega) &=\{ (f\cdot g)(\omega)-\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g), \varepsilon\in \mathds{R}^{>0} \ | \\ + &\qquad f\cdot g(\omega)+\varepsilon\omega^{a_\gamma+b_\delta} \ : \ \gamma<\alpha, \delta < \beta \ \text{s.t} \ a_\alpha+b_\delta \in \text{supp}(f\cdot g),\varepsilon\in \mathds{R}^{>0} \}. \end{align*} +Thus, $(f\cdot g)(\omega)$ satisfies the cut for $f(\omega)\cdot g(\omega)$ and the claim follows by cofinality. + +\end{proof} + +All together, this completes the proof of the following theorem. + +\begin{theorem} The map +$$\mathds{R}((t^{\bf No})) \xrightarrow{\sim} {\bf No}, \quad \sum_{i<\alpha}f_iX^{a_i} \mapsto \sum_{i<\alpha} f_i\omega^{a_i},$$ +is an ordered field isomorphism. \end{theorem} + + + +\section{The Surreals as a Real Closed Field} + +Let $K$ be a field. We call $K$ \textit{orderable} if some ordering on $K$ makes it an ordered field. If $K$ is orderable, then $\text{char}(K)=0$ and $K$ is not algebraically closed. \footnote{To prove that $K$ is not algebraically closed: suppose $K$ is an algebraically closed ordered field and derive a contradiction using $i$, the square root of $-1$.} We call $K$ \textit{euclidean} if $x^2+y^2\neq -1$ for all $x, y \in K$ and $K=\{\pm x^2 \ : \ x\in K\}$. If $K$ is euclidean, then $K$ isa an ordered field for a unique ordering---namely, $a\geq 0 \iff \exists x\in K. x^2=a$. + +\begin{theorem}[Artin $\&$ Schreier, 1927] \label {7.1} For a field $K$, the following are equivalent. +\begin{enumerate}[(1)] +\item $K$ is orderable, but has no proper orderable algebraic field extension. +\item $K$ is euclidean and every polynomial $p\in K[X]$ of odd degree has a zero in $K$. +\item $K$ is not algebraically closed, but $K(i)$, $i^2=-1$, is algebraically closed. +\item $K$ is not algebraically closed, but has an algebraically closed field extension $L$ with $[L:K]<\infty$. +\end{enumerate} +We call $K$ \textit{real closed} if it satisfies one of these equivalent conditions.\footnote{See Lange's \textit{Algebra} for partial proof.} +\end{theorem} + +\begin{corollary}\label{7.2} Let $K'$ be a subfield of a real closed field $K$. Then $K'$ is real closed if and only if $K'$ is algebraically closed in $K$. \end{corollary} + +\begin{proof} Suppose $K'$ is not algebraically closed in $K$. Fix $a\in K\setminus K'$ that is algebraic over $K'$. Then, $K'(a)$ is an proper orderable algebraic field extension of $K'$. Thus, $K'$ is not real closed by $(1)$ of theorem $(\ref{7.1})$. + +Conversely, suppose $K'$ is algebraically closed in $K$. We verify that condition (2) of theorem $(\ref{7.1})$ holds for $K'$. Since $K'$ is algebraically closed in $K$, any zero of a polynomial of the form $X^2-a$ or $-X^2-a$, where $a\in K'$, must be in $K'$. This along with the fact that $K$ is euclidean implies that $K'$ is euclidean. Moreover, if $p\in K'[X]$ has odd degree, then since $K$ satisfies (2), $p$ has a zero $a\in K$. But, $a\in K'$ since $K'$ is algebraically closed in $K$. Thus, $K$ is real closed. +\end{proof} + +The archetypical example of a real closed field is $\mathds{R}$. By corollary $(\ref{7.2})$, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ is also real closed. In fact, the algebraic closure of $\mathds{Q}$ in $\mathds{R}$ can be embedded into any real closed field. + +\begin{proposition} Suppose $K$ is real closed and $p\in K[X]$. Then, +\begin{enumerate}[(1)] +\item $p$ is monic and irreducible if and only if $p=X-a$ for some $a\in K$ or $p=(X-a)^2+b^2$ for some $a, b\in K$, $b\neq 0$. +\item The map $x\mapsto p(x): K \rightarrow K$ has the intermediate value theorem. \end{enumerate}\end{proposition} + +\begin{theorem}[Tarksi] The theory of real closed ordered fields in the language $\mathcal{L}=\{0, 1, +, -, \cdot, \leq\}$ of ordered rings admits quantifier elimination. Hence, for any real closed field $K$, $\mathds{R}\equiv K$ and, if $\mathds{R}$ is a subfield of $K$, then $\mathds{R}\preceq K$.\end{theorem} + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be a real closed field. Then, $K=k((t^\Gamma))$ is real closed. \end{theorem} + +We have $K[i]\cong k[i]((t^\Gamma))$, so it's enough to show the following theorem. + +\begin{theorem} Let $\Gamma$ be a divisible ordered abelian group and let $k$ be an algebraically closed field of characteristic $0$. Then, $K=k((t^\Gamma))$ is algebraically closed. \end{theorem} + +\begin{remark} This theorem is still true if we drop the characteristic $0$ assumption, but it would require a different proof than the one given below. \end{remark} + +\begin{proof} Let $P\in K[X]$ be monic and irreducible, and write +$$P=X^n+a_{n-1}X^{n-1}+\cdots +a_0 \quad (a_i\in K, n\geq n).$$ +By replacing $P(X)$ by $P(X-a_{n-1})$, we get +$$P\left(X-\frac{a_{n-1}}{n}\right) = X^n + \text{terms of degree $0$, and $P_0=\overline{P}$. Suppose we have a strictly increasing sequence $(b_i)_{i<\beta}$ in $\Gamma$ and sequences $(Q_i)_{i<\beta}$, $(R_i)_{i<\beta}$ of polynomials in $k[X]$ of degree $<\deg Q_0$ and $<\deg R_0$, respectively, such that for +$$Q_{<\beta}:= \sum_{i<\beta}Q_it^{b_i}, \quad R_{<\beta}:= \sum_{i<\beta}R_it^{b_i}$$ +we have +$$P\equiv Q_{<\beta}R_{<\beta} \mod{(t^b\mathcal{O})}$$ +for all $b\in \Gamma$ with $b\leq b_i$ for some $i$. Suppose $P\neq Q_{<\beta}R_{<\beta}$; we are going to find $b_\beta\in \Gamma$ and $Q_\beta, R_\beta\in k[X]$ of degrees $< \deg Q_0$ and $< \deg R_0$, respectively, such that +\begin{enumerate}[$\bullet$] +\item $b_\beta >b_i$ for all $i<\beta$. +\item $P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$ for all $b\leq b_\beta$. +\end{enumerate} +To this end, let $\gamma:= v(P-R_{<\beta}Q_{<\beta})\in \Gamma$. Then, $b_\beta:= \gamma>b_i$ for all $i<\beta$. Consider any $G, H\in k[X]$; then +$$P\equiv (Q_{<\beta}+Q_\beta t^{b_\beta})(R_{<\beta} + R_\beta t^{b_\beta}) \mod{(t^b\mathcal{O})}$$ +for all $b\leq b_\beta$. To get this congruence to hold also for $b=b_\beta$, we need $G, H$ to satisfy an equation +$$S=Q_0H+R_0G,$$ +where $S\in k[X]$ has degree $<0$. But we can find such $G, H$ since $Q_0, R_0$ are relatively prime. Then, take $Q_\beta=G$ and $R_\beta= G$ for such $G, H$. + +\end{proof} diff --git a/Other/old/All notes - Copy (2)/week_8/g_g_week_8.tex b/Other/old/All notes - Copy (2)/week_8/g_g_week_8.tex index 3b1ea927..199b1d02 100644 --- a/Other/old/All notes - Copy (2)/week_8/g_g_week_8.tex +++ b/Other/old/All notes - Copy (2)/week_8/g_g_week_8.tex @@ -1,103 +1,103 @@ -\section{ Week 8 } -\subsection{November 24, 2014} -''Notes for today by Madeline Barnicle'' - -Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the ''sign sequence'' (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: -\subsubsection{Section 8: Analytic functions on $\mathbf{No}$} -Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -'''Digression on exp for $\mathbf{No}$''' - -An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -\subsubsection{Section 9: Power series and Weierstrass Preparation} -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -\subsection{ November 26, 2014 } -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -'''Proof:''' - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: -* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, -* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +'''Digression on exp for $\mathbf{No}$''' + +An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +\subsubsection{Section 9: Power series and Weierstrass Preparation} +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +\subsection{ November 26, 2014 } +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +'''Proof:''' + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, +* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. + +The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. + +====Theorem 9.3 (Division with remainder)==== +Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -'''Digression on exp for $\mathbf{No}$''' - -An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -\subsubsection{Section 9: Power series and Weierstrass Preparation} -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -\subsection{ November 26, 2014 } -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -'''Proof:''' - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: -* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, -* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +'''Digression on exp for $\mathbf{No}$''' + +An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +\subsubsection{Section 9: Power series and Weierstrass Preparation} +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +\subsection{ November 26, 2014 } +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +'''Proof:''' + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, +* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. + +The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. + +====Theorem 9.3 (Division with remainder)==== +Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -'''Digression on exp for $\mathbf{No}$''' - -An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -====Section 9: Power series and Weierstrass Preparation==== -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -=== November 26, 2014 === -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -'''Proof:''' - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: -* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, -* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +'''Digression on exp for $\mathbf{No}$''' + +An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +====Section 9: Power series and Weierstrass Preparation==== +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +=== November 26, 2014 === +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +'''Proof:''' + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, +* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. + +The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. + +====Theorem 9.3 (Division with remainder)==== +Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TRpolyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write -* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. -* $r < s \Leftrightarrow r_i < s_i$ for each $i$. -* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: -* $\|f\| = 0 \Leftrightarrow f = 0$. -* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. -* $\|f + g\| \leq \|f\| + \|g\|$. -* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. -* $\|X^i f\| = r^i \|f\|$. -* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - -* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. -* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. - -'''Proof:''' -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -'''Proof:''' - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2 +\section{ Week 9 } + +\subsection{ Monday 12-1-2014 } + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +'''Proof:''' + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +'''Remark:''' + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +'''Proof:''' + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write +* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. +* $r < s \Leftrightarrow r_i < s_i$ for each $i$. +* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: +* $\|f\| = 0 \Leftrightarrow f = 0$. +* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. +* $\|f + g\| \leq \|f\| + \|g\|$. +* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. +* $\|X^i f\| = r^i \|f\|$. +* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + +* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. +* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. + +'''Proof:''' +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +'''Proof:''' + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2 diff --git a/Other/old/All notes - Copy (2)/week_9/g_week_9.aux b/Other/old/All notes - Copy (2)/week_9/g_week_9.aux deleted file mode 100644 index 4e3a6b76..00000000 --- a/Other/old/All notes - Copy (2)/week_9/g_week_9.aux +++ /dev/null @@ -1,32 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {7} Week 9 }{35}} -\@writefile{toc}{\contentsline {subsection}{\numberline {7.1} Monday 12-1-2014 }{35}} -\@writefile{toc}{\contentsline {subsubsection}{\numberline {7.1.1} Section 10: Convergent Power Series }{35}} -\@writefile{toc}{\contentsline {subsection}{\numberline {7.2} Wednesday 12-3-2014 }{36}} -\@writefile{toc}{\contentsline {subsection}{\numberline {7.3} Friday 12-5-2014 }{36}} -\@setckpt{week_9/g_week_9}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{7} -\setcounter{subsection}{3} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_9/g_week_9.tex b/Other/old/All notes - Copy (2)/week_9/g_week_9.tex index 685ab92a..9afcf4cb 100644 --- a/Other/old/All notes - Copy (2)/week_9/g_week_9.tex +++ b/Other/old/All notes - Copy (2)/week_9/g_week_9.tex @@ -1,126 +1,126 @@ -\section{ Week 9 } - -\subsection{ Monday 12-1-2014 } - -==== Corollary 9.4 (Weierstrauss Preparation) ==== - -(Notes today by John Lensmire.) - -Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. -Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. - -'''Proof:''' - -Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. - -Let $x=0$ to get -$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) -+ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ -Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. -This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. - -Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. - -Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) - -'''Remark:''' - -The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial -$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. - -==== Corollary 9.5 ==== - -Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. - -'''Proof:''' - -We proceed by induction on $m$. - -If $m=0$, $K$ is a field, hence noetherian. - -From $m$ to $m+1$: -Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. -Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ -(see last time) we can assume that $f$ is regular in $T$ of some order $d$. -Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write -* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. -* $r < s \Leftrightarrow r_i < s_i$ for each $i$. -* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: -* $\|f\| = 0 \Leftrightarrow f = 0$. -* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. -* $\|f + g\| \leq \|f\| + \|g\|$. -* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. -* $\|X^i f\| = r^i \|f\|$. -* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - -* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. -* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. - -'''Proof:''' -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -'''Proof:''' - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -\subsection{ Wednesday 12-3-2014 } -''notes by Asaaf Shani'' - -'''''Coming soon''''' - -\subsection{ Friday 12-5-2014 } -''notes by Tyler Arant'' - -PDF: [[Media:285D notes 12 5.pdf]] +\section{ Week 9 } + +\subsection{ Monday 12-1-2014 } + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +'''Proof:''' + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +'''Remark:''' + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +'''Proof:''' + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write +* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. +* $r < s \Leftrightarrow r_i < s_i$ for each $i$. +* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: +* $\|f\| = 0 \Leftrightarrow f = 0$. +* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. +* $\|f + g\| \leq \|f\| + \|g\|$. +* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. +* $\|X^i f\| = r^i \|f\|$. +* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + +* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. +* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. + +'''Proof:''' +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +'''Proof:''' + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +\subsection{ Wednesday 12-3-2014 } +''notes by Asaaf Shani'' + +'''''Coming soon''''' + +\subsection{ Friday 12-5-2014 } +''notes by Tyler Arant'' + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/All notes - Copy (2)/week_9/week_6.aux b/Other/old/All notes - Copy (2)/week_9/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/All notes - Copy (2)/week_9/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_9/week_9.aux b/Other/old/All notes - Copy (2)/week_9/week_9.aux deleted file mode 100644 index 113be7d4..00000000 --- a/Other/old/All notes - Copy (2)/week_9/week_9.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_9/week_9}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy (2)/week_9/week_9.tex b/Other/old/All notes - Copy (2)/week_9/week_9.tex index 4ff4a128..e646286d 100644 --- a/Other/old/All notes - Copy (2)/week_9/week_9.tex +++ b/Other/old/All notes - Copy (2)/week_9/week_9.tex @@ -1,126 +1,126 @@ -== Week 9 == - -=== Monday 12-1-2014 === - -==== Corollary 9.4 (Weierstrauss Preparation) ==== - -(Notes today by John Lensmire.) - -Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. -Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. - -'''Proof:''' - -Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. - -Let $x=0$ to get -$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) -+ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ -Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. -This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. - -Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. - -Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) - -'''Remark:''' - -The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial -$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. - -==== Corollary 9.5 ==== - -Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. - -'''Proof:''' - -We proceed by induction on $m$. - -If $m=0$, $K$ is a field, hence noetherian. - -From $m$ to $m+1$: -Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. -Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ -(see last time) we can assume that $f$ is regular in $T$ of some order $d$. -Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write -* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. -* $r < s \Leftrightarrow r_i < s_i$ for each $i$. -* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: -* $\|f\| = 0 \Leftrightarrow f = 0$. -* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. -* $\|f + g\| \leq \|f\| + \|g\|$. -* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. -* $\|X^i f\| = r^i \|f\|$. -* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - -* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. -* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. - -'''Proof:''' -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -'''Proof:''' - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -=== Wednesday 12-3-2014 === -''notes by Asaaf Shani'' - -'''''Coming soon''''' - -=== Friday 12-5-2014 === -''notes by Tyler Arant'' - -PDF: [[Media:285D notes 12 5.pdf]] +== Week 9 == + +=== Monday 12-1-2014 === + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +'''Proof:''' + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +'''Remark:''' + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +'''Proof:''' + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write +* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. +* $r < s \Leftrightarrow r_i < s_i$ for each $i$. +* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: +* $\|f\| = 0 \Leftrightarrow f = 0$. +* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. +* $\|f + g\| \leq \|f\| + \|g\|$. +* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. +* $\|X^i f\| = r^i \|f\|$. +* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + +* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. +* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. + +'''Proof:''' +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +'''Proof:''' + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +=== Wednesday 12-3-2014 === +''notes by Asaaf Shani'' + +'''''Coming soon''''' + +=== Friday 12-5-2014 === +''notes by Tyler Arant'' + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/All notes - Copy/Global.aux b/Other/old/All notes - Copy/Global.aux deleted file mode 100644 index 445536f3..00000000 --- a/Other/old/All notes - Copy/Global.aux +++ /dev/null @@ -1,11 +0,0 @@ -\relax -\@input{week_1/john_susice_surreal_numbers_notes_fall2014.aux} -\@input{week_2/week_2.aux} -\@input{week_3/zach.aux} -\@input{week_4/week_4.aux} -\@input{week_5/week_5.aux} -\@input{week_6/week_6.aux} -\@input{week_7/285D_notes_nov_17_18_21.aux} -\@input{week_8/week_8.aux} -\@input{week_9/week_9.aux} -\@input{week_11/week_11.aux} diff --git a/Other/old/All notes - Copy/Global.bbl b/Other/old/All notes - Copy/Global.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/All notes - 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-\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} - - -\renewcommand{\restriction}{\mathord{\upharpoonright}} - - -\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} -\author{Class Taught by Prof. Aschenbrenner} -\date{\today} -\begin{document} -\maketitle{} - -\include{week_1/john_susice_surreal_numbers_notes_fall2014} -\include{week_2/week_2} -\include{week_3/zach} -\include{week_4/week_4} -\include{week_5/week_5} -\include{week_6/week_6} -\include{week_7/285D_notes_nov_17_18_21} -\include{week_8/week_8} -\include{week_9/week_9} -%\include{week_6/week_6} -\include{week_11/week_11} - +\documentclass{article} + +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{enumerate} +\usepackage{colonequals} +\usepackage{fullpage} + +\usepackage{dsfont} + +\newtheorem{theorem}{Theorem} +\newtheorem{defn}{Definition} +\newtheorem{cor}{Corollary} +\newtheorem{claim}{Claim} +\newtheorem{lem}{Lemma} + +\newtheorem{lemma}{Lemma} + +\newcommand{\R}{\mathbb{R}} +%\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation +\newcommand{\concat}{\frown} %sequence concatenation +\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} +\newcommand{\ZFC}{\mathsf{ZFC}} +\newcommand{\NBG}{\mathsf{NBG}} +\newcommand{\coloneq}{\colonequals} +\newcommand{\N}{\mathbb{N}} + +\newcommand{\No}{\mathbf{No}} +\newcommand{\On}{\mathbf{On}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\brac}[1]{\left[ #1 \right]} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\abs}[1]{\left| #1 \right|} +\newcommand{\rar}{\rightarrow} +\newcommand{\arr}{\rightarrow} + +\DeclareMathOperator{\supp}{supp} +\DeclareMathOperator{\lt}{lt} + +\newcommand{\w}{\omega} +\newcommand{\midr}[1]{\restriction_{#1}} + +% This is the "centered" symbol +\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} + +\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} +\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} + +\swapnumbers +\theoremstyle{theorem} %bold title, italicized font + +\theoremstyle{definition} %bold title, regular font +\newtheorem{example}[theorem]{Example} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{exercise}[theorem]{Exercise} +\newtheorem*{problem}{Problem} +\newtheorem{warning}[theorem]{Warning} +\newtheorem*{solution}{Solution} +\newtheorem*{remark}{Remark} + +\theoremstyle{empty} +\newtheorem{namedtheorem}{} + +\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} + +\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} + + +\renewcommand{\restriction}{\mathord{\upharpoonright}} + + +\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} +\author{Class Taught by Prof. Aschenbrenner} +\date{\today} +\begin{document} +\maketitle{} + +\include{week_1/john_susice_surreal_numbers_notes_fall2014} +\include{week_2/week_2} +\include{week_3/zach} +\include{week_4/week_4} +\include{week_5/week_5} +\include{week_6/week_6} +\include{week_7/285D_notes_nov_17_18_21} +\include{week_8/week_8} +\include{week_9/week_9} +%\include{week_6/week_6} +\include{week_11/week_11} + \end{document} \ No newline at end of file diff --git a/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.aux b/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.aux deleted file mode 100644 index 16edaca4..00000000 --- a/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.aux +++ /dev/null @@ -1,35 +0,0 @@ -\relax -\newlabel{}{{1}{2}} -\newlabel{}{{2}{2}} -\newlabel{}{{3}{3}} -\newlabel{lemma_on_length_of_cuts}{{1}{4}} -\newlabel{cofinality_theorem}{{4}{4}} -\newlabel{inverse_cofinality_theorem}{{2}{5}} -\newlabel{}{{5}{5}} -\newlabel{defn_of_surreal_sum}{{1}{6}} -\@setckpt{week_1/john_susice_surreal_numbers_notes_fall2014}{ -\setcounter{page}{7} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.tex b/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.tex index 9347f272..2ccd4c36 100644 --- a/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.tex +++ b/Other/old/All notes - Copy/week_1/john_susice_surreal_numbers_notes_fall2014.tex @@ -1,508 +1,508 @@ -\textit{Notes by John Suice} - -\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/old/All notes - Copy/week_1/week_1.aux b/Other/old/All notes - Copy/week_1/week_1.aux deleted file mode 100644 index 7174f85a..00000000 --- a/Other/old/All notes - Copy/week_1/week_1.aux +++ /dev/null @@ -1,26 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1}Week 1}{2}} -\@setckpt{week_1/week_1}{ -\setcounter{page}{3} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/All notes - Copy/week_1/week_1.tex b/Other/old/All notes - Copy/week_1/week_1.tex index 1c06ce99..37b7dab3 100644 --- a/Other/old/All notes - Copy/week_1/week_1.tex +++ b/Other/old/All notes - Copy/week_1/week_1.tex @@ -1,13 +1,13 @@ -\section{Week 1} - -Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. - -Proof - -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - +\section{Week 1} + +Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. + +Proof + +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + diff --git a/Other/old/All notes - Copy/week_1/week_1.tex.aux b/Other/old/All notes - Copy/week_1/week_1.tex.aux deleted file mode 100644 index 9b66fc00..00000000 --- a/Other/old/All notes - Copy/week_1/week_1.tex.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_1/week_1.tex}{ -\setcounter{page}{2} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/All notes - Copy/week_10/week_10.aux b/Other/old/All notes - Copy/week_10/week_10.aux deleted file mode 100644 index 1ecedcfa..00000000 --- a/Other/old/All notes - Copy/week_10/week_10.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_10/week_10}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_11/week_11.aux b/Other/old/All notes - Copy/week_11/week_11.aux deleted file mode 100644 index 6cfb559d..00000000 --- a/Other/old/All notes - Copy/week_11/week_11.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_11/week_11}{ -\setcounter{page}{38} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_11/week_11.tex b/Other/old/All notes - Copy/week_11/week_11.tex index 2bf7b1f3..7572257f 100644 --- a/Other/old/All notes - Copy/week_11/week_11.tex +++ b/Other/old/All notes - Copy/week_11/week_11.tex @@ -1,84 +1,84 @@ -== Siddharth's extra lectures == - -===Part 1=== -''notes by Bill Chen'' - -A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== -* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. -* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. -* $\{0|\emptyset\}$. Here Left has a winning strategy. -* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. - -====Definition 1==== -* $G>0$ if Left has a winning strategy. -* $G<0$ if Right has a winning strategy. -* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) -* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) -* $G\ge 0$ means $G>0$ or $G\sim 0$. - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -'''Proof:''' - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -'''Remark:''' -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== -* $-(G+H)=-G+-H$. -* $--G=G$. -* $G\sim 0$ iff $-G\sim 0$. -* $G>0$ iff $-G<0$. -* $G\parallel 0$ iff $-G\parallel 0$. - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: -* If $G\sim 0$, then $G+H\sim 0$. -* If $G>0$, then $G+H>0$. -* If $G\parallel 0$, then $G+H\parallel 0$. -* If $G+H\sim 0$, then $G\sim 0$. -* If $G+H>0$, then $G>0$. -* If $G+H\parallel 0$, then $G\parallel 0$. - -'''Proof:''' - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== -* $G+ -G\sim 0$. -* If $G>0$ and $H>0$ then $G+H>0$. - -'''Proof:''' - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== -* $G\sim H$ if $G-H\sim 0$. -* $G>H$ if $G-H>0$. +== Siddharth's extra lectures == + +===Part 1=== +''notes by Bill Chen'' + +A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== +* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. +* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. +* $\{0|\emptyset\}$. Here Left has a winning strategy. +* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. + +====Definition 1==== +* $G>0$ if Left has a winning strategy. +* $G<0$ if Right has a winning strategy. +* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) +* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) +* $G\ge 0$ means $G>0$ or $G\sim 0$. + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +'''Proof:''' + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +'''Remark:''' +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== +* $-(G+H)=-G+-H$. +* $--G=G$. +* $G\sim 0$ iff $-G\sim 0$. +* $G>0$ iff $-G<0$. +* $G\parallel 0$ iff $-G\parallel 0$. + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: +* If $G\sim 0$, then $G+H\sim 0$. +* If $G>0$, then $G+H>0$. +* If $G\parallel 0$, then $G+H\parallel 0$. +* If $G+H\sim 0$, then $G\sim 0$. +* If $G+H>0$, then $G>0$. +* If $G+H\parallel 0$, then $G\parallel 0$. + +'''Proof:''' + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== +* $G+ -G\sim 0$. +* If $G>0$ and $H>0$ then $G+H>0$. + +'''Proof:''' + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== +* $G\sim H$ if $G-H\sim 0$. +* $G>H$ if $G-H>0$. diff --git a/Other/old/All notes - Copy/week_11/week_6.aux b/Other/old/All notes - Copy/week_11/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/All notes - Copy/week_11/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_2/week_2.aux b/Other/old/All notes - Copy/week_2/week_2.aux deleted file mode 100644 index ded0313c..00000000 --- a/Other/old/All notes - Copy/week_2/week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_2/week_2.tex b/Other/old/All notes - Copy/week_2/week_2.tex index 7441881e..20e3641a 100644 --- a/Other/old/All notes - Copy/week_2/week_2.tex +++ b/Other/old/All notes - Copy/week_2/week_2.tex @@ -1,271 +1,271 @@ -== Week 2 == - -(Notes by John Lensmire) - -=== Monday 10-13-2014 === - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -'''Proof:''' - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: -* $b <_s a \Rightarrow -b <_s -a$ -* $b < a \Rightarrow -b > -a$ - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: -* $a_L + (-a) < a_L + (-a_L) = 0$ -* $a + (-a_R) < a_R + (-a_R) = 0$ -* $a_R + (-a) > a_R + (-a_R) = 0$ -* $a + (-a_L) > a_L + (-a_L) = 0$ -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -'''Proof:''' - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: -* $f(a_L, b_L) < f(a_L, b_R)$ -* $f(a_R, b_R) < f(a_R, b_L)$ -* $f(a_L, b_L) < f(a_R, b_R)$ -* $f(a_R, b_R) < f(a_L, b_L)$ - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -=== Wednesday 10-15-2014 === - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -'''Proof:''' - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -=== Friday 10-17-2014 === - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). +== Week 2 == + +(Notes by John Lensmire) + +=== Monday 10-13-2014 === + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +'''Proof:''' + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: +* $b <_s a \Rightarrow -b <_s -a$ +* $b < a \Rightarrow -b > -a$ + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: +* $a_L + (-a) < a_L + (-a_L) = 0$ +* $a + (-a_R) < a_R + (-a_R) = 0$ +* $a_R + (-a) > a_R + (-a_R) = 0$ +* $a + (-a_L) > a_L + (-a_L) = 0$ +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +'''Proof:''' + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: +* $f(a_L, b_L) < f(a_L, b_R)$ +* $f(a_R, b_R) < f(a_R, b_L)$ +* $f(a_L, b_L) < f(a_R, b_R)$ +* $f(a_R, b_R) < f(a_L, b_L)$ + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +=== Wednesday 10-15-2014 === + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +'''Proof:''' + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +=== Friday 10-17-2014 === + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/All notes - Copy/week_3/zach.aux b/Other/old/All notes - Copy/week_3/zach.aux deleted file mode 100644 index 3d34b19a..00000000 --- a/Other/old/All notes - Copy/week_3/zach.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_3/zach}{ -\setcounter{page}{12} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_4/week_2.aux b/Other/old/All notes - Copy/week_4/week_2.aux deleted file mode 100644 index 31eef818..00000000 --- a/Other/old/All notes - Copy/week_4/week_2.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/All notes - Copy/week_4/week_4.aux b/Other/old/All notes - Copy/week_4/week_4.aux deleted file mode 100644 index 48e65042..00000000 --- a/Other/old/All notes - Copy/week_4/week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/week_4}{ -\setcounter{page}{15} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_4/week_4.tex b/Other/old/All notes - Copy/week_4/week_4.tex index 60371053..71ccc79b 100644 --- a/Other/old/All notes - Copy/week_4/week_4.tex +++ b/Other/old/All notes - Copy/week_4/week_4.tex @@ -1,130 +1,130 @@ -== Week 4 == -Notes by Madeline Barnicle -===Monday, October 27, 2014=== -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. - -===Wednesday, October 29, 2014=== -* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). -* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. -*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. -*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? - -====Section 4: Combinatorics of Ordered Sets==== -Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). - -We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. -*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. -*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. -*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. -*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. -*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. -*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. - -*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. - - -===Friday, October 31, 2014=== -Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -'''Proof''' (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -'''Proof''': -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -====Hahn Fields==== - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. +== Week 4 == +Notes by Madeline Barnicle +===Monday, October 27, 2014=== +We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. + +Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. + +===Wednesday, October 29, 2014=== +* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). +* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. +*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. +*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? + +====Section 4: Combinatorics of Ordered Sets==== +Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). + +We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. +*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. +*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. +*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. +*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. +*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. +*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. + +*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. + +====Definition 4.1==== +$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. + + +===Friday, October 31, 2014=== +Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +'''Proof''' (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +'''Proof''': +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +====Hahn Fields==== + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/All notes - Copy/week_5/week_5.aux b/Other/old/All notes - Copy/week_5/week_5.aux deleted file mode 100644 index 8ab1f07c..00000000 --- a/Other/old/All notes - Copy/week_5/week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/week_5}{ -\setcounter{page}{18} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_5/week_5.tex b/Other/old/All notes - Copy/week_5/week_5.tex index 87a287da..8d4fbb0e 100644 --- a/Other/old/All notes - Copy/week_5/week_5.tex +++ b/Other/old/All notes - Copy/week_5/week_5.tex @@ -1,139 +1,139 @@ -== Week 5 == - -===November 3, 2014 === -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -==== Section 5: The $\omega^-$ map ==== -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: -* $[-\alpha]=[\alpha]$. -* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. -* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. - - -We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (Hölder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -'''Proof:''' - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: -* $0<\omega^b$, -* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -'''Proof:''' - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -===November 7, 2014=== -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -''Case 1.'' $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -'''Proof:''' - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, -* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). -* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) -* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. +== Week 5 == + +===November 3, 2014 === +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +==== Section 5: The $\omega^-$ map ==== +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: +* $[-\alpha]=[\alpha]$. +* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. +* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. + + +We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (Hölder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +'''Proof:''' + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: +* $0<\omega^b$, +* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +'''Proof:''' + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +===November 7, 2014=== +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +''Case 1.'' $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +'''Proof:''' + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, +* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). +* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) +* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/All notes - Copy/week_6/week_6.aux b/Other/old/All notes - Copy/week_6/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/All notes - Copy/week_6/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_6/week_6.tex b/Other/old/All notes - Copy/week_6/week_6.tex index 7d2437eb..9c0f6dc8 100644 --- a/Other/old/All notes - Copy/week_6/week_6.tex +++ b/Other/old/All notes - Copy/week_6/week_6.tex @@ -1,390 +1,390 @@ -== Week 6 == - -Notes by Anton Bobkov - -===Monday, November 10, 2014=== -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - -* '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ -* '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -'''Proof of inequality''' - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -'''Proof of tail property''' - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -''Case 1'': $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -'''Proof of well-definiteness''' - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -===Wednesday, November 12, 2014=== - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -'''Proof''' -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -'''Proof''' - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ -*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ - -''Case 1'': $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -''Case 2'': $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -'''Proof''' -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -===Friday, November 14, 2014=== - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -'''Proof''' - -''Case 1'': $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -''Case 2'': $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. +== Week 6 == + +Notes by Anton Bobkov + +===Monday, November 10, 2014=== +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + +* '''Inequality:''' For \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ +* '''Tail property:''' if $\gamma < \kappa < \alpha$ +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +'''Proof of inequality''' + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +'''Proof of tail property''' + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +''Case 1'': $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +'''Proof of well-definiteness''' + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +===Wednesday, November 12, 2014=== + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +'''Proof''' +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +'''Proof''' + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ +*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ + +''Case 1'': $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +''Case 2'': $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +'''Proof''' +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +===Friday, November 14, 2014=== + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +'''Proof''' + +''Case 1'': $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +''Case 2'': $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/All notes - Copy/week_7/285D.aux b/Other/old/All notes - Copy/week_7/285D.aux deleted file mode 100644 index cd380abd..00000000 --- a/Other/old/All notes - Copy/week_7/285D.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_7/285D}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/All notes - Copy/week_7/285D_notes_nov_17_18_21.aux b/Other/old/All notes - 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Copy/week_8/week_8.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_8/week_8}{ -\setcounter{page}{35} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/All notes - Copy/week_8/week_8.tex b/Other/old/All notes - Copy/week_8/week_8.tex index 4c05097e..7c1fbbc5 100644 --- a/Other/old/All notes - Copy/week_8/week_8.tex +++ b/Other/old/All notes - Copy/week_8/week_8.tex @@ -1,111 +1,111 @@ -== Week 8 == -===November 24, 2014=== -''Notes for today by Madeline Barnicle'' - -Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the ''sign sequence'' (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: -====Section 8: Analytic functions on $\mathbf{No}$==== -Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -'''Digression on exp for $\mathbf{No}$''' - -An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -====Section 9: Power series and Weierstrass Preparation==== -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -=== November 26, 2014 === -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -'''Proof:''' - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: -* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, -* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +'''Digression on exp for $\mathbf{No}$''' + +An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +====Section 9: Power series and Weierstrass Preparation==== +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +=== November 26, 2014 === +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +'''Proof:''' + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, +* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. + +The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. + +====Theorem 9.3 (Division with remainder)==== +Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TRpolyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write -* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. -* $r < s \Leftrightarrow r_i < s_i$ for each $i$. -* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: -* $\|f\| = 0 \Leftrightarrow f = 0$. -* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. -* $\|f + g\| \leq \|f\| + \|g\|$. -* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. -* $\|X^i f\| = r^i \|f\|$. -* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - -* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. -* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. - -'''Proof:''' -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -'''Proof:''' - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -=== Wednesday 12-3-2014 === -''notes by Asaaf Shani'' - -'''''Coming soon''''' - -=== Friday 12-5-2014 === -''notes by Tyler Arant'' - -PDF: [[Media:285D notes 12 5.pdf]] +== Week 9 == + +=== Monday 12-1-2014 === + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +'''Proof:''' + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +'''Remark:''' + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +'''Proof:''' + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write +* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. +* $r < s \Leftrightarrow r_i < s_i$ for each $i$. +* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: +* $\|f\| = 0 \Leftrightarrow f = 0$. +* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. +* $\|f + g\| \leq \|f\| + \|g\|$. +* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. +* $\|X^i f\| = r^i \|f\|$. +* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + +* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. +* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. + +'''Proof:''' +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +'''Proof:''' + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +=== Wednesday 12-3-2014 === +''notes by Asaaf Shani'' + +'''''Coming soon''''' + +=== Friday 12-5-2014 === +''notes by Tyler Arant'' + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/Anton Notes/LaTeX1.aux b/Other/old/Anton Notes/LaTeX1.aux deleted file mode 100644 index 54cd7ee0..00000000 --- a/Other/old/Anton Notes/LaTeX1.aux +++ /dev/null @@ -1 +0,0 @@ -\relax diff --git a/Other/old/Anton Notes/LaTeX1.bbl b/Other/old/Anton Notes/LaTeX1.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/Anton Notes/LaTeX1.blg b/Other/old/Anton Notes/LaTeX1.blg deleted file mode 100644 index 0ec97163..00000000 --- a/Other/old/Anton Notes/LaTeX1.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: LaTeX1.aux -I found no \citation commands---while reading file LaTeX1.aux -I found no \bibdata command---while reading file LaTeX1.aux -I found no \bibstyle command---while reading file LaTeX1.aux -(There were 3 error messages) diff --git a/Other/old/Anton Notes/LaTeX1.log b/Other/old/Anton Notes/LaTeX1.log deleted file mode 100644 index 4609bd7d..00000000 --- a/Other/old/Anton Notes/LaTeX1.log +++ /dev/null @@ -1,170 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 16 DEC 2014 16:03 -entering extended mode -**LaTeX1.tex -("C:\Users\Anton\SparkleShare\Research\Other\1. 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Surreal\LaTeX1.aux") ) -Here is how much of TeX's memory you used: - 1337 strings out of 493922 - 15592 string characters out of 3144898 - 69586 words of memory out of 3000000 - 4649 multiletter control sequences out of 15000+200000 - 11210 words of font info for 44 fonts, out of 3000000 for 9000 - 841 hyphenation exceptions out of 8191 - 27i,10n,26p,222b,201s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on LaTeX1.dvi (3 pages, 13288 bytes). diff --git a/Other/old/Anton Notes/LaTeX1.tex b/Other/old/Anton Notes/LaTeX1.tex index 19ca170d..c706292b 100644 --- a/Other/old/Anton Notes/LaTeX1.tex +++ b/Other/old/Anton Notes/LaTeX1.tex @@ -1,201 +1,201 @@ -\documentclass{article} - -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{amsthm} -\usepackage{enumerate} -\usepackage{colonequals} -\usepackage{fullpage} - -\newtheorem{theorem}{Theorem} -\newtheorem{defn}{Definition} -\newtheorem{cor}{Corollary} -\newtheorem{claim}{Claim} -\newtheorem{lem}{Lemma} -\newcommand{\R}{\mathbb{R}} -\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation -\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} -\newcommand{\ZFC}{\mathsf{ZFC}} -\newcommand{\NBG}{\mathsf{NBG}} -\newcommand{\coloneq}{\colonequals} -\newcommand{\N}{\mathbb{N}} - -\newcommand{\No}{\mathbf{No}} -\newcommand{\On}{\mathbf{On}} -\newcommand{\paren}[1]{\left( #1 \right)} -\newcommand{\brac}[1]{\left[ #1 \right]} -\newcommand{\curly}[1]{\left\{ #1 \right\}} -\newcommand{\abs}[1]{\left| #1 \right|} -\newcommand{\rar}{\rightarrow} -\newcommand{\arr}{\rightarrow} - -\DeclareMathOperator{\supp}{supp} -\DeclareMathOperator{\lt}{lt} - -\newcommand{\w}{\omega} -\newcommand{\midr}[1]{\restriction_{#1}} - - -\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} -\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} -\date{\today} -\begin{document} -\maketitle{} - -Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. - -Proof - -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - - -Lemma The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -Proof - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -\begin{itemize} - \item $a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ - \item $f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ -\end{itemize} - -$\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -$\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -Lemma Let $r \in \R, a \in \No$. Then - -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -Proof -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -Corollary 6.5 -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -Proof - -$\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -If $\alpha$ is arbitrary, so $\alpha = \alpha' + n$ where $\alpha'$ is a limit ordinal and $n \in \N$, we proceed by induction on $n$. -For simplicity assume that $\alpha$ is a limit and we want to show the result for $\alpha + 1$. - -$\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. - -\end{document} - +\documentclass{article} + +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{enumerate} +\usepackage{colonequals} +\usepackage{fullpage} + +\newtheorem{theorem}{Theorem} +\newtheorem{defn}{Definition} +\newtheorem{cor}{Corollary} +\newtheorem{claim}{Claim} +\newtheorem{lem}{Lemma} +\newcommand{\R}{\mathbb{R}} +\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation +\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} +\newcommand{\ZFC}{\mathsf{ZFC}} +\newcommand{\NBG}{\mathsf{NBG}} +\newcommand{\coloneq}{\colonequals} +\newcommand{\N}{\mathbb{N}} + +\newcommand{\No}{\mathbf{No}} +\newcommand{\On}{\mathbf{On}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\brac}[1]{\left[ #1 \right]} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\abs}[1]{\left| #1 \right|} +\newcommand{\rar}{\rightarrow} +\newcommand{\arr}{\rightarrow} + +\DeclareMathOperator{\supp}{supp} +\DeclareMathOperator{\lt}{lt} + +\newcommand{\w}{\omega} +\newcommand{\midr}[1]{\restriction_{#1}} + + +\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} +\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} +\date{\today} +\begin{document} +\maketitle{} + +Lemma Let $f \in K$ be such that $l(f) = \w \cdot \alpha$. Then $l(f(\w)) \geq \alpha$. + +Proof + +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + + +Lemma The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +Proof + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +\begin{itemize} + \item $a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ + \item $f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ +\end{itemize} + +$\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +$\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +Lemma Let $r \in \R, a \in \No$. Then + +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +Proof +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +Corollary 6.5 +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +Proof + +$\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +If $\alpha$ is arbitrary, so $\alpha = \alpha' + n$ where $\alpha'$ is a limit ordinal and $n \in \N$, we proceed by induction on $n$. +For simplicity assume that $\alpha$ is a limit and we want to show the result for $\alpha + 1$. + +$\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. + +\end{document} + diff --git a/Other/old/Anton Notes/Notes.aux b/Other/old/Anton Notes/Notes.aux deleted file mode 100644 index 54cd7ee0..00000000 --- a/Other/old/Anton Notes/Notes.aux +++ /dev/null @@ -1 +0,0 @@ -\relax diff --git a/Other/old/Anton Notes/Notes.bbl b/Other/old/Anton Notes/Notes.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/Anton Notes/Notes.blg b/Other/old/Anton Notes/Notes.blg deleted file mode 100644 index 7a014044..00000000 --- a/Other/old/Anton Notes/Notes.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Notes.aux -I found no \citation commands---while reading file Notes.aux -I found no \bibdata command---while reading file Notes.aux -I found no \bibstyle command---while reading file Notes.aux -(There were 3 error messages) diff --git a/Other/old/Anton Notes/Notes.log b/Other/old/Anton Notes/Notes.log deleted file mode 100644 index 34ee59f5..00000000 --- a/Other/old/Anton Notes/Notes.log +++ /dev/null @@ -1,163 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 3 DEC 2014 21:14 -entering extended mode -**Notes.tex -("C:\Users\Anton\SparkleShare\Research\Other\1. 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Surreal\Notes.aux") ) -Here is how much of TeX's memory you used: - 1328 strings out of 493922 - 15395 string characters out of 3144898 - 70586 words of memory out of 3000000 - 4641 multiletter control sequences out of 15000+200000 - 12161 words of font info for 47 fonts, out of 3000000 for 9000 - 841 hyphenation exceptions out of 8191 - 27i,15n,26p,256b,201s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on Notes.dvi (5 pages, 19972 bytes). diff --git a/Other/old/Anton Notes/Notes.tcp b/Other/old/Anton Notes/Notes.tcp index d686bb0d..b58c0f43 100644 --- a/Other/old/Anton Notes/Notes.tcp +++ b/Other/old/Anton Notes/Notes.tcp @@ -1,12 +1,12 @@ -[FormatInfo] -Type=TeXnicCenterProjectInformation -Version=4 - -[ProjectInfo] -MainFile=Notes.tex -UseBibTeX=0 -UseMakeIndex=0 -ActiveProfile=LaTeX ⇨ DVI -ProjectLanguage=en -ProjectDialect=US - +[FormatInfo] +Type=TeXnicCenterProjectInformation +Version=4 + +[ProjectInfo] +MainFile=Notes.tex +UseBibTeX=0 +UseMakeIndex=0 +ActiveProfile=LaTeX ⇨ DVI +ProjectLanguage=en +ProjectDialect=US + diff --git a/Other/old/Anton Notes/Notes.tex b/Other/old/Anton Notes/Notes.tex index ac662576..bee91704 100644 --- a/Other/old/Anton Notes/Notes.tex +++ b/Other/old/Anton Notes/Notes.tex @@ -1,282 +1,282 @@ -\documentclass{article} - -\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{amsthm} -\usepackage{enumerate} -\usepackage{colonequals} -\usepackage{fullpage} - -\newcommand{\No}{\mathbf{No}} -\newcommand{\On}{\mathbf{On}} -\newcommand{\paren}[1]{\left( #1 \right)} -\newcommand{\brac}[1]{\left[ #1 \right]} -\newcommand{\curly}[1]{\left\{ #1 \right\}} -\newcommand{\abs}[1]{\left| #1 \right|} -\newcommand{\rar}{\rightarrow} - -\newtheorem{theorem}{Theorem} -\newtheorem{question}{Question} -\newtheorem{defn}{Definition} -\newtheorem{cor}{Corollary} -\newtheorem{claim}{Claim} -\newtheorem{lem}{Lemma} - -\newcommand{\R}{\mathbb{R}} -\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation -\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} -\newcommand{\ZFC}{\mathsf{ZFC}} -\newcommand{\NBG}{\mathsf{NBG}} -\newcommand{\coloneq}{\colonequals} -\newcommand{\N}{\mathbb{N}} - -\DeclareMathOperator{\supp}{supp} - -\newcommand{\w}{\omega} -\newcommand{\midr}[1]{\restriction_{#1}} - -\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} -\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} -\date{\today} -\begin{document} -\maketitle{} - -\section*{Day 1: Friday October 3, 2014} - -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub\textbf{set}. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -\begin{question} - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? -\end{question} - -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: -\begin{enumerate} - \item \emph{Inequality:} for - \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} - \end{align*} - we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ - \item \emph{Tail property:} if $\gamma < \kappa < \alpha$ - \begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} - \end{align*} -\end{enumerate} - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -Case 1: $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -%\begin{align*} -% f(x) &= \sum_{i < \gamma} f_i x^{a_i} + f_\gamma x^{a_\gamma} + \sum_{\gamma < i < \beta} f_i x^{a_i} + f_\beta x^{a_\beta} \\ -% g(x) &= \sum_{i < \gamma} g_i x^{a_i} + g_\gamma x^{a_\gamma} + \sum_{\gamma < i < \beta} g_i x^{a_i} + g_\beta x^{a_\beta} -%\end{align*} -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -Case 2: $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -Tail property - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -Case 1: $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -Case 2: $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -Case 1: $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -Case 2: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -Case 3: $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -\end{document} - +\documentclass{article} + +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{enumerate} +\usepackage{colonequals} +\usepackage{fullpage} + +\newcommand{\No}{\mathbf{No}} +\newcommand{\On}{\mathbf{On}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\brac}[1]{\left[ #1 \right]} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\abs}[1]{\left| #1 \right|} +\newcommand{\rar}{\rightarrow} + +\newtheorem{theorem}{Theorem} +\newtheorem{question}{Question} +\newtheorem{defn}{Definition} +\newtheorem{cor}{Corollary} +\newtheorem{claim}{Claim} +\newtheorem{lem}{Lemma} + +\newcommand{\R}{\mathbb{R}} +\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation +\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} +\newcommand{\ZFC}{\mathsf{ZFC}} +\newcommand{\NBG}{\mathsf{NBG}} +\newcommand{\coloneq}{\colonequals} +\newcommand{\N}{\mathbb{N}} + +\DeclareMathOperator{\supp}{supp} + +\newcommand{\w}{\omega} +\newcommand{\midr}[1]{\restriction_{#1}} + +\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} +\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} +\date{\today} +\begin{document} +\maketitle{} + +\section*{Day 1: Friday October 3, 2014} + +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub\textbf{set}. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +\begin{question} + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? +\end{question} + +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: +\begin{enumerate} + \item \emph{Inequality:} for + \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} + \end{align*} + we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ + \item \emph{Tail property:} if $\gamma < \kappa < \alpha$ + \begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} + \end{align*} +\end{enumerate} + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +Case 1: $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +%\begin{align*} +% f(x) &= \sum_{i < \gamma} f_i x^{a_i} + f_\gamma x^{a_\gamma} + \sum_{\gamma < i < \beta} f_i x^{a_i} + f_\beta x^{a_\beta} \\ +% g(x) &= \sum_{i < \gamma} g_i x^{a_i} + g_\gamma x^{a_\gamma} + \sum_{\gamma < i < \beta} g_i x^{a_i} + g_\beta x^{a_\beta} +%\end{align*} +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +Case 2: $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +Tail property + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +Case 1: $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +Case 2: $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +Case 1: $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +Case 2: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +Case 3: $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +\end{document} + diff --git a/Other/old/Anton Notes/Notes.tps b/Other/old/Anton Notes/Notes.tps deleted file mode 100644 index efe4f604..00000000 --- a/Other/old/Anton Notes/Notes.tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - -[Frame0] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=1289 -NormalPos.bottom=728 -Class=LaTeXView -Document=Notes.tex - -[Frame0_View0,0] -TopLine=75 -Cursor=3105 - -[SessionInfo] -FrameCount=1 -ActiveFrame=0 - diff --git a/Other/old/all_notes/Global.aux b/Other/old/all_notes/Global.aux deleted file mode 100644 index 930c2be7..00000000 --- a/Other/old/all_notes/Global.aux +++ /dev/null @@ -1,11 +0,0 @@ -\relax -\@input{week_1/john_susice_surreal_numbers_notes_fall2014.aux} -\@input{week_2/g_week_2.aux} -\@input{week_3/zach.aux} -\@input{week_4/g_week_4.aux} -\@input{week_5/g_week_5.aux} -\@input{week_6/g_week_6.aux} -\@input{week_7/285D_notes_nov_17_18_21.aux} -\@input{week_8/g_week_8.aux} -\@input{week_9/g_week_9.aux} -\@input{week_11/week_11.aux} diff --git a/Other/old/all_notes/Global.bbl b/Other/old/all_notes/Global.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/all_notes/Global.blg b/Other/old/all_notes/Global.blg deleted file mode 100644 index 579f4a7b..00000000 --- a/Other/old/all_notes/Global.blg +++ /dev/null @@ -1,15 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Global.aux -A level-1 auxiliary file: week_1/john_susice_surreal_numbers_notes_fall2014.aux -A level-1 auxiliary file: week_2/week_2.aux -A level-1 auxiliary file: week_3/zach.aux -A level-1 auxiliary file: week_4/week_4.aux -A level-1 auxiliary file: week_5/week_5.aux -A level-1 auxiliary file: week_6/week_6.aux -A level-1 auxiliary file: week_7/285D_notes_nov_17_18_21.aux -A level-1 auxiliary file: week_8/week_8.aux -A level-1 auxiliary file: week_9/week_9.aux -A level-1 auxiliary file: week_11/week_11.aux -I found no \citation commands---while reading file Global.aux -I found no \bibdata command---while reading file Global.aux -I found no \bibstyle command---while reading file Global.aux -(There were 3 error messages) diff --git a/Other/old/all_notes/Global.log b/Other/old/all_notes/Global.log deleted file mode 100644 index caa1c31c..00000000 --- a/Other/old/all_notes/Global.log +++ /dev/null @@ -1,273 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 17 DEC 2014 22:27 -entering extended mode -**Global.tex -("C:\Users\Anton\SparkleShare\Research\Other\All notes\Global.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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+\documentclass{article} + +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{enumerate} +\usepackage{colonequals} +\usepackage{fullpage} + +\usepackage{dsfont} + +\newtheorem{theorem}{Theorem} +\newtheorem{defn}{Definition} +\newtheorem{cor}{Corollary} +\newtheorem{claim}{Claim} +\newtheorem{lem}{Lemma} + +\newtheorem{lemma}{Lemma} + +\newcommand{\R}{\mathbb{R}} +%\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation +\newcommand{\concat}{\frown} %sequence concatenation +\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} +\newcommand{\ZFC}{\mathsf{ZFC}} +\newcommand{\NBG}{\mathsf{NBG}} +\newcommand{\coloneq}{\colonequals} +\newcommand{\N}{\mathbb{N}} + +\newcommand{\No}{\mathbf{No}} +\newcommand{\On}{\mathbf{On}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\brac}[1]{\left[ #1 \right]} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\abs}[1]{\left| #1 \right|} +\newcommand{\rar}{\rightarrow} +\newcommand{\arr}{\rightarrow} + +\DeclareMathOperator{\supp}{supp} +\DeclareMathOperator{\lt}{lt} + +\newcommand{\w}{\omega} +\newcommand{\midr}[1]{\restriction_{#1}} + +% This is the "centered" symbol +\def\fCenter{{\mbox{\Large{$\rightarrow$}}}} + +\newcommand{\bigslant}[2]{{\raisebox{.2em}{$#1$}\left/\raisebox{-.2em}{$#2$}\right.}} +\def\dotminus{\mathbin{\ooalign{\hss\raise1ex\hbox{.}\hss\cr\mathsurround=0pt$-$}}} + +\swapnumbers +\theoremstyle{theorem} %bold title, italicized font + +\theoremstyle{definition} %bold title, regular font +\newtheorem{example}[theorem]{Example} +\newtheorem{definition}[theorem]{Definition} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{corollary}[theorem]{Corollary} +\newtheorem{exercise}[theorem]{Exercise} +\newtheorem*{problem}{Problem} +\newtheorem{warning}[theorem]{Warning} +\newtheorem*{solution}{Solution} +\newtheorem*{remark}{Remark} + +\theoremstyle{empty} +\newtheorem{namedtheorem}{} + +\newcommand{\customtheorem}[3]{\theoremstyle{theorem} \newtheorem{theorem#1}[theorem]{#1} \begin{theorem#1}[#2]#3 \end{theorem#1}} + +\newcommand{\customdefinition}[2]{\theoremstyle{definition} \newtheorem{definition#1}[theorem]{#1} \begin{definition#1}#2 \end{definition#1}} + + +\renewcommand{\restriction}{\mathord{\upharpoonright}} + +\newcommand{\WikiLevelTwo}[1]{\section*{#1}} +\newcommand{\WikiLevelThree}[1]{\subsection*{#1}} +\newcommand{\WikiLevelFour}[1]{\subsubsection*{#1}} + +\newcommand{\WikiSigleStar}{} + +\newcommand{\WikiItalic}[1]{\textit{#1}} +\newcommand{\WikiBold}[1]{\textbf{#1}} +\newcommand{\WikiBoldItalic}[1]{\emph{#1}} + + +\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} +\author{Class Taught by Prof. Aschenbrenner} +\date{\today} +\begin{document} +\maketitle{} + +\include{week_1/john_susice_surreal_numbers_notes_fall2014} +\include{week_2/g_week_2} +\include{week_3/zach} +\include{week_4/g_week_4} +\include{week_5/g_week_5} +\include{week_6/g_week_6} +\include{week_7/285D_notes_nov_17_18_21} +\include{week_8/g_week_8} +\include{week_9/g_week_9} +%\include{week_6/week_6} +\include{week_11/week_11} + \end{document} \ No newline at end of file diff --git a/Other/old/all_notes/Global.tps b/Other/old/all_notes/Global.tps deleted file mode 100644 index db3b9e3e..00000000 --- a/Other/old/all_notes/Global.tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - 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-\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/old/all_notes/week_1/week_1.aux b/Other/old/all_notes/week_1/week_1.aux deleted file mode 100644 index 7174f85a..00000000 --- a/Other/old/all_notes/week_1/week_1.aux +++ /dev/null @@ -1,26 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1}Week 1}{2}} -\@setckpt{week_1/week_1}{ -\setcounter{page}{3} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/all_notes/week_1/week_1.tex.aux b/Other/old/all_notes/week_1/week_1.tex.aux deleted file mode 100644 index 9b66fc00..00000000 --- a/Other/old/all_notes/week_1/week_1.tex.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_1/week_1.tex}{ -\setcounter{page}{2} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/all_notes/week_10/week_10.aux b/Other/old/all_notes/week_10/week_10.aux deleted file mode 100644 index 1ecedcfa..00000000 --- a/Other/old/all_notes/week_10/week_10.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_10/week_10}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_11/g_week_11.aux b/Other/old/all_notes/week_11/g_week_11.aux deleted file mode 100644 index afc1a001..00000000 --- a/Other/old/all_notes/week_11/g_week_11.aux +++ /dev/null @@ -1,29 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {8} Siddharth's extra lectures }{37}} -\@writefile{toc}{\contentsline {subsection}{\numberline {8.1}Part 1}{37}} -\@setckpt{week_11/g_week_11}{ -\setcounter{page}{39} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{8} -\setcounter{subsection}{1} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_11/g_week_11.tex b/Other/old/all_notes/week_11/g_week_11.tex index bd243ace..86d17d97 100644 --- a/Other/old/all_notes/week_11/g_week_11.tex +++ b/Other/old/all_notes/week_11/g_week_11.tex @@ -1,19 +1,19 @@ -\WikiLevelTwo{ Siddharth's extra lectures } - -\WikiLevelThree{Part 1} -\WikiItalic{notes by Bill Chen} - -A crucial concept for these lectures will be \WikiItalic{games}. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== +\WikiLevelTwo{ Siddharth's extra lectures } + +\WikiLevelThree{Part 1} +\WikiItalic{notes by Bill Chen} + +A crucial concept for these lectures will be \WikiItalic{games}. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== \begin{enumerate} \item $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. \item $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. \item $\{0|\emptyset\}$. Here Left has a winning strategy. \item $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. \end{enumerate} - -====Definition 1==== + +====Definition 1==== \begin{enumerate} \item $G>0$ if Left has a winning strategy. \item $G<0$ if Right has a winning strategy. @@ -21,30 +21,30 @@ \item $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) \item $G\ge 0$ means $G>0$ or $G\sim 0$. \end{enumerate} - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -\WikiBold{Proof:} - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the \WikiItalic{disjunctive sum} $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -\WikiBold{Remark:} -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the \WikiItalic{negation} $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +\WikiBold{Proof:} + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the \WikiItalic{disjunctive sum} $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +\WikiBold{Remark:} +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the \WikiItalic{negation} $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== \begin{enumerate} \item $-(G+H)=-G+-H$. \item $--G=G$. @@ -52,11 +52,11 @@ \item $G>0$ iff $-G<0$. \item $G\parallel 0$ iff $-G\parallel 0$. \end{enumerate} - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: \begin{enumerate} \item If $G\sim 0$, then $G+H\sim 0$. \item If $G>0$, then $G+H>0$. @@ -65,28 +65,28 @@ \item If $G+H>0$, then $G>0$. \item If $G+H\parallel 0$, then $G\parallel 0$. \end{enumerate} - -\WikiBold{Proof:} - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== + +\WikiBold{Proof:} + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== \begin{enumerate} \item $G+ -G\sim 0$. \item If $G>0$ and $H>0$ then $G+H>0$. \end{enumerate} - -\WikiBold{Proof:} - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== + +\WikiBold{Proof:} + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== diff --git a/Other/old/all_notes/week_11/week_11.aux b/Other/old/all_notes/week_11/week_11.aux deleted file mode 100644 index 4018c7a3..00000000 --- a/Other/old/all_notes/week_11/week_11.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_11/week_11}{ -\setcounter{page}{39} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_11/week_11.tex b/Other/old/all_notes/week_11/week_11.tex index 2bf7b1f3..7572257f 100644 --- a/Other/old/all_notes/week_11/week_11.tex +++ b/Other/old/all_notes/week_11/week_11.tex @@ -1,84 +1,84 @@ -== Siddharth's extra lectures == - -===Part 1=== -''notes by Bill Chen'' - -A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== -* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. -* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. -* $\{0|\emptyset\}$. Here Left has a winning strategy. -* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. - -====Definition 1==== -* $G>0$ if Left has a winning strategy. -* $G<0$ if Right has a winning strategy. -* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) -* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) -* $G\ge 0$ means $G>0$ or $G\sim 0$. - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -'''Proof:''' - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -'''Remark:''' -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== -* $-(G+H)=-G+-H$. -* $--G=G$. -* $G\sim 0$ iff $-G\sim 0$. -* $G>0$ iff $-G<0$. -* $G\parallel 0$ iff $-G\parallel 0$. - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: -* If $G\sim 0$, then $G+H\sim 0$. -* If $G>0$, then $G+H>0$. -* If $G\parallel 0$, then $G+H\parallel 0$. -* If $G+H\sim 0$, then $G\sim 0$. -* If $G+H>0$, then $G>0$. -* If $G+H\parallel 0$, then $G\parallel 0$. - -'''Proof:''' - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== -* $G+ -G\sim 0$. -* If $G>0$ and $H>0$ then $G+H>0$. - -'''Proof:''' - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== -* $G\sim H$ if $G-H\sim 0$. -* $G>H$ if $G-H>0$. +== Siddharth's extra lectures == + +===Part 1=== +''notes by Bill Chen'' + +A crucial concept for these lectures will be ''games''. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== +* $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. +* $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. +* $\{0|\emptyset\}$. Here Left has a winning strategy. +* $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. + +====Definition 1==== +* $G>0$ if Left has a winning strategy. +* $G<0$ if Right has a winning strategy. +* $G\sim 0$ if the second player has a winning strategy. ($G$ is ''similar'' to $0$.) +* $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) +* $G\ge 0$ means $G>0$ or $G\sim 0$. + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +'''Proof:''' + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the ''disjunctive sum'' $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +'''Remark:''' +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the ''negation'' $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== +* $-(G+H)=-G+-H$. +* $--G=G$. +* $G\sim 0$ iff $-G\sim 0$. +* $G>0$ iff $-G<0$. +* $G\parallel 0$ iff $-G\parallel 0$. + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: +* If $G\sim 0$, then $G+H\sim 0$. +* If $G>0$, then $G+H>0$. +* If $G\parallel 0$, then $G+H\parallel 0$. +* If $G+H\sim 0$, then $G\sim 0$. +* If $G+H>0$, then $G>0$. +* If $G+H\parallel 0$, then $G\parallel 0$. + +'''Proof:''' + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== +* $G+ -G\sim 0$. +* If $G>0$ and $H>0$ then $G+H>0$. + +'''Proof:''' + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== +* $G\sim H$ if $G-H\sim 0$. +* $G>H$ if $G-H>0$. diff --git a/Other/old/all_notes/week_11/week_6.aux b/Other/old/all_notes/week_11/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/all_notes/week_11/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_2/g_week_2.aux b/Other/old/all_notes/week_2/g_week_2.aux deleted file mode 100644 index 3539c1e2..00000000 --- a/Other/old/all_notes/week_2/g_week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/g_week_2}{ -\setcounter{page}{12} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_2/g_week_2.tex b/Other/old/all_notes/week_2/g_week_2.tex index 12cdfb3a..989d98cf 100644 --- a/Other/old/all_notes/week_2/g_week_2.tex +++ b/Other/old/all_notes/week_2/g_week_2.tex @@ -1,277 +1,277 @@ -\WikiLevelTwo{ Week 2 } - -(Notes by John Lensmire) - -\WikiLevelThree{ Monday 10-13-2014 } - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -\WikiBold{Proof:} - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: +\WikiLevelTwo{ Week 2 } + +(Notes by John Lensmire) + +\WikiLevelThree{ Monday 10-13-2014 } + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +\WikiBold{Proof:} + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: \begin{enumerate} \item $b <_s a \Rightarrow -b <_s -a$ \item $b < a \Rightarrow -b > -a$ \end{enumerate} - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: \begin{enumerate} \item $a_L + (-a) < a_L + (-a_L) = 0$ \item $a + (-a_R) < a_R + (-a_R) = 0$ \item $a_R + (-a) > a_R + (-a_R) = 0$ \item $a + (-a_L) > a_L + (-a_L) = 0$ \end{enumerate} -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -\WikiBold{Proof:} - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +\WikiBold{Proof:} + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: \begin{enumerate} \item $f(a_L, b_L) < f(a_L, b_R)$ \item $f(a_R, b_R) < f(a_R, b_L)$ \item $f(a_L, b_L) < f(a_R, b_R)$ \item $f(a_R, b_R) < f(a_L, b_L)$ \end{enumerate} - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -\WikiLevelThree{ Wednesday 10-15-2014 } - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -\WikiBold{Proof:} - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -\WikiLevelThree{ Friday 10-17-2014 } - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +\WikiLevelThree{ Wednesday 10-15-2014 } + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +\WikiBold{Proof:} + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +\WikiLevelThree{ Friday 10-17-2014 } + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/all_notes/week_2/week_2.aux b/Other/old/all_notes/week_2/week_2.aux deleted file mode 100644 index ded0313c..00000000 --- a/Other/old/all_notes/week_2/week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_2/week_2.tex b/Other/old/all_notes/week_2/week_2.tex index db58d236..df2497be 100644 --- a/Other/old/all_notes/week_2/week_2.tex +++ b/Other/old/all_notes/week_2/week_2.tex @@ -1,271 +1,271 @@ -\WikiLevelTwo{ Week 2 } - -(Notes by John Lensmire) - -=== Monday 10-13-2014 === - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -'''Proof:''' - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: -* $b <_s a \Rightarrow -b <_s -a$ -* $b < a \Rightarrow -b > -a$ - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: -* $a_L + (-a) < a_L + (-a_L) = 0$ -* $a + (-a_R) < a_R + (-a_R) = 0$ -* $a_R + (-a) > a_R + (-a_R) = 0$ -* $a + (-a_L) > a_L + (-a_L) = 0$ -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -'''Proof:''' - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: -* $f(a_L, b_L) < f(a_L, b_R)$ -* $f(a_R, b_R) < f(a_R, b_L)$ -* $f(a_L, b_L) < f(a_R, b_R)$ -* $f(a_R, b_R) < f(a_L, b_L)$ - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -=== Wednesday 10-15-2014 === - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -'''Proof:''' - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -=== Friday 10-17-2014 === - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). +\WikiLevelTwo{ Week 2 } + +(Notes by John Lensmire) + +=== Monday 10-13-2014 === + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +'''Proof:''' + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: +* $b <_s a \Rightarrow -b <_s -a$ +* $b < a \Rightarrow -b > -a$ + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: +* $a_L + (-a) < a_L + (-a_L) = 0$ +* $a + (-a_R) < a_R + (-a_R) = 0$ +* $a_R + (-a) > a_R + (-a_R) = 0$ +* $a + (-a_L) > a_L + (-a_L) = 0$ +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +'''Proof:''' + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: +* $f(a_L, b_L) < f(a_L, b_R)$ +* $f(a_R, b_R) < f(a_R, b_L)$ +* $f(a_L, b_L) < f(a_R, b_R)$ +* $f(a_R, b_R) < f(a_L, b_L)$ + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +=== Wednesday 10-15-2014 === + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +'''Proof:''' + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +=== Friday 10-17-2014 === + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/all_notes/week_3/zach.aux b/Other/old/all_notes/week_3/zach.aux deleted file mode 100644 index 6e8b1dae..00000000 --- a/Other/old/all_notes/week_3/zach.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_3/zach}{ -\setcounter{page}{13} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_4/g_week_4.aux b/Other/old/all_notes/week_4/g_week_4.aux deleted file mode 100644 index a0eae918..00000000 --- a/Other/old/all_notes/week_4/g_week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/g_week_4}{ -\setcounter{page}{16} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_4/g_week_4.tex b/Other/old/all_notes/week_4/g_week_4.tex index 50f66b95..097f9dae 100644 --- a/Other/old/all_notes/week_4/g_week_4.tex +++ b/Other/old/all_notes/week_4/g_week_4.tex @@ -1,71 +1,71 @@ -\WikiLevelTwo{ Week 4 } -Notes by Madeline Barnicle -\WikiLevelThree{Monday, October 27, 2014} -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. \end{enumerate} - -\WikiLevelThree{Wednesday, October 29, 2014} + +\WikiLevelThree{Wednesday, October 29, 2014} \begin{enumerate} \item $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). \item $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. \end{enumerate} - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. \begin{enumerate} \item $r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. \item What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? \end{enumerate} - -\WikiLevelFour{Section 4: Combinatorics of Ordered Sets} -Let $S$ be a set. An \WikiItalic{ordering} $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an \WikiItalic{ordered set} (partial or total). - -We say $\leq$ is \WikiItalic{total} if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is \WikiItalic{increasing} if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, \WikiItalic{strictly increasing} if $x < y \rightarrow \phi(x) < \phi(y)$, a \WikiItalic{quasi-embedding} if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. + +\WikiLevelFour{Section 4: Combinatorics of Ordered Sets} +Let $S$ be a set. An \WikiItalic{ordering} $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an \WikiItalic{ordered set} (partial or total). + +We say $\leq$ is \WikiItalic{total} if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is \WikiItalic{increasing} if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, \WikiItalic{strictly increasing} if $x < y \rightarrow \phi(x) < \phi(y)$, a \WikiItalic{quasi-embedding} if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. \begin{enumerate} \item $S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. \item $S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. @@ -74,65 +74,65 @@ \item The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. \item $\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. \end{enumerate} - + \WikiSigleStar Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is \WikiItalic{noetherian} if it is well-founded and has no infinite antichains. - - -\WikiLevelThree{Friday, October 31, 2014} -Call an infinite sequence $(x_n)$ in $S$ \WikiItalic{good} if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -\WikiBold{Proof} (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -\WikiBold{Proof}: -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -\WikiLevelFour{Hahn Fields} - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose \WikiItalic{support} supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is \WikiItalic{well-ordered}. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. + +====Definition 4.1==== +$(S, \leq)$ is \WikiItalic{noetherian} if it is well-founded and has no infinite antichains. + + +\WikiLevelThree{Friday, October 31, 2014} +Call an infinite sequence $(x_n)$ in $S$ \WikiItalic{good} if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +\WikiBold{Proof} (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +\WikiBold{Proof}: +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +\WikiLevelFour{Hahn Fields} + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose \WikiItalic{support} supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is \WikiItalic{well-ordered}. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/all_notes/week_4/week_2.aux b/Other/old/all_notes/week_4/week_2.aux deleted file mode 100644 index 31eef818..00000000 --- a/Other/old/all_notes/week_4/week_2.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/all_notes/week_4/week_4.aux b/Other/old/all_notes/week_4/week_4.aux deleted file mode 100644 index 48e65042..00000000 --- a/Other/old/all_notes/week_4/week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/week_4}{ -\setcounter{page}{15} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_4/week_4.tex b/Other/old/all_notes/week_4/week_4.tex index 60371053..71ccc79b 100644 --- a/Other/old/all_notes/week_4/week_4.tex +++ b/Other/old/all_notes/week_4/week_4.tex @@ -1,130 +1,130 @@ -== Week 4 == -Notes by Madeline Barnicle -===Monday, October 27, 2014=== -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. - -===Wednesday, October 29, 2014=== -* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). -* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. -*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. -*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? - -====Section 4: Combinatorics of Ordered Sets==== -Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). - -We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. -*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. -*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. -*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. -*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. -*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. -*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. - -*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. - - -===Friday, October 31, 2014=== -Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -'''Proof''' (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -'''Proof''': -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -====Hahn Fields==== - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. +== Week 4 == +Notes by Madeline Barnicle +===Monday, October 27, 2014=== +We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. + +Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. + +===Wednesday, October 29, 2014=== +* $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). +* $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. +*$r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. +*What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? + +====Section 4: Combinatorics of Ordered Sets==== +Let $S$ be a set. An ''ordering'' $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an ''ordered set'' (partial or total). + +We say $\leq$ is ''total'' if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is ''increasing'' if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, ''strictly increasing'' if $x < y \rightarrow \phi(x) < \phi(y)$, a ''quasi-embedding'' if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. +*$S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. +*$S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. +*Let $S^*$ be the set of finite words on $S$. Define $x_{1}...x_{m} \leq^{*} y_{1}...y_{m}$ if there exists a strictly increasing $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for every $i=1...m$. +*Let $S^\diamond$ be the set of "commutative" finite words on $S=S*/ \sim$, where $x_{1}...x_{m} \sim y_{1}...y_{m}$ if $m=n$ and there exists a permutation of $\{1...m\}$ such that $x_i = y_{\phi(i)}$ for all $i$. Define $x_{1}...x_{m} \leq^{\diamond} y_{1}...y_{m} \leftrightarrow$ there exists an injective $\phi: \{1...m\} \rightarrow \{1...n\}$ such that $x_i \leq_S y_{\phi(i)}$ for $i=1...m$. +*The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. +*$\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. + +*Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. + +====Definition 4.1==== +$(S, \leq)$ is ''noetherian'' if it is well-founded and has no infinite antichains. + + +===Friday, October 31, 2014=== +Call an infinite sequence $(x_n)$ in $S$ ''good'' if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +'''Proof''' (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +'''Proof''': +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +====Hahn Fields==== + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose ''support'' supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is ''well-ordered''. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/all_notes/week_5/g_week_5.aux b/Other/old/all_notes/week_5/g_week_5.aux deleted file mode 100644 index e9136496..00000000 --- a/Other/old/all_notes/week_5/g_week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/g_week_5}{ -\setcounter{page}{19} -\setcounter{equation}{1} -\setcounter{enumi}{4} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_5/g_week_5.tex b/Other/old/all_notes/week_5/g_week_5.tex index 24370214..75d4ea56 100644 --- a/Other/old/all_notes/week_5/g_week_5.tex +++ b/Other/old/all_notes/week_5/g_week_5.tex @@ -1,145 +1,145 @@ -\WikiLevelTwo{ Week 5 } - -\WikiLevelThree{November 3, 2014 } -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -\WikiLevelFour{ Section 5: The $\omega^-$ map } -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are \WikiItalic{archimedean equivalent} if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: +\WikiLevelTwo{ Week 5 } + +\WikiLevelThree{November 3, 2014 } +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +\WikiLevelFour{ Section 5: The $\omega^-$ map } +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are \WikiItalic{archimedean equivalent} if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: \begin{enumerate} \item $[-\alpha]=[\alpha]$. \item $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. \item If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. \end{enumerate} - - -We say that $\Gamma$ is \WikiItalic{archimedean} if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (H�lder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -\WikiBold{Proof:} - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: + + +We say that $\Gamma$ is \WikiItalic{archimedean} if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (H�lder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +\WikiBold{Proof:} + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: \begin{enumerate} \item $0<\omega^b$, \item $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -\WikiBold{Proof:} - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -\WikiLevelThree{November 7, 2014} -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -\WikiItalic{Case 1.} $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -\WikiBold{Proof:} - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, +Then (5.3) makes sense for $b\in \mathbf{No}$ with $l(b)\le \gamma$ and $(1),(2)$ above hold for all $b,c\in\mathbf{No}$ with $\ell(b),\ell(c)\le \gamma$. + +\WikiBold{Proof} +Suppose $\ell(b)=\gamma$. Then $\ell(b_L),\ell(b_R)<\gamma$, so $\omega^{b_L} \ll \omega^{b_R}$. Therefore $r\omega^{b_L}0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +\WikiBold{Proof:} + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +\WikiLevelThree{November 7, 2014} +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +\WikiItalic{Case 1.} $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +\WikiBold{Proof:} + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, \begin{enumerate} \item $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). \item $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) \item $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. + +That was messy, but nice because it all fell out of the definition of the $\omega^-$ map. + +We now check that ordinal exponentiation agrees with the $\omega^-$ map. + +====Lemma 5.8:==== +For $a\in\mathbf{On}$, $\omega^a\in\mathbf{No}$ is the same as the ordinal $\omega^a$ (ordinal exponentiation). + +\WikiBold{Proof:} + +Write $\omega\uparrow a$ for ordinal exponentiation. By induction on $a\in \mathbf{On}$, we show that $\omega^a=\omega\uparrow a$. The base case was already done. Let $a=\{a_L\mid \emptyset\}$. Then using the induction hypothesis, +$$\omega^a=\{0,r\omega^{a_L}\mid \emptyset\}=\{0,r\omega\uparrow a_L\mid\emptyset\}$$ +$$=\omega\uparrow a,$$ +using the definition of ordinal exponentiation for the last equality. + +\WikiLevelFour{Section 6. The Normal Form} + +Let $K=\mathbb{R}((t^\mathbb{No}))$ be the Hahn field and set $x:=\frac{1}{t}$. We think of the elements of $K$ as formal series in $x$: +$$f(x)=\sum_{i<\alpha}f_ix^{a_i}$$ +where $\alpha\in \mathbf{On}$, $(a_i)$ is a strictly decreasing sequence in $\mathbf{No}$, and $f_i\in \mathbb{R}\setminus \{0\}$. So $\alpha$ is the order-type of $\mathrm{supp}(f)$, which we will denote by $\ell(f)$ (the agreement of the choice of this notation with the length of a surreal number is not a coincidence). We turn $K$ into an ordered field such that $f>0$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/all_notes/week_5/week_5.aux b/Other/old/all_notes/week_5/week_5.aux deleted file mode 100644 index 8ab1f07c..00000000 --- a/Other/old/all_notes/week_5/week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/week_5}{ -\setcounter{page}{18} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_5/week_5.tex b/Other/old/all_notes/week_5/week_5.tex index 87a287da..8d4fbb0e 100644 --- a/Other/old/all_notes/week_5/week_5.tex +++ b/Other/old/all_notes/week_5/week_5.tex @@ -1,139 +1,139 @@ -== Week 5 == - -===November 3, 2014 === -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -==== Section 5: The $\omega^-$ map ==== -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: -* $[-\alpha]=[\alpha]$. -* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. -* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. - - -We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (Hölder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -'''Proof:''' - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: -* $0<\omega^b$, -* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -'''Proof:''' - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -===November 7, 2014=== -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -''Case 1.'' $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -'''Proof:''' - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, -* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). -* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) -* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. +== Week 5 == + +===November 3, 2014 === +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +==== Section 5: The $\omega^-$ map ==== +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are ''archimedean equivalent'' if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: +* $[-\alpha]=[\alpha]$. +* $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. +* If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. + + +We say that $\Gamma$ is ''archimedean'' if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (Hölder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +'''Proof:''' + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: +* $0<\omega^b$, +* $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +'''Proof:''' + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +===November 7, 2014=== +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +''Case 1.'' $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +'''Proof:''' + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, +* $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). +* $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) +* $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/all_notes/week_6/g_week_6.aux b/Other/old/all_notes/week_6/g_week_6.aux deleted file mode 100644 index 3032f66a..00000000 --- a/Other/old/all_notes/week_6/g_week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/g_week_6}{ -\setcounter{page}{26} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_6/g_week_6.tex b/Other/old/all_notes/week_6/g_week_6.tex index 2536982f..0e43ec6a 100644 --- a/Other/old/all_notes/week_6/g_week_6.tex +++ b/Other/old/all_notes/week_6/g_week_6.tex @@ -1,392 +1,392 @@ -\WikiLevelTwo{ Week 6 } - -Notes by Anton Bobkov - -\WikiLevelThree{Monday, November 10, 2014} -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub\WikiItalic{set}. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - +\WikiLevelTwo{ Week 6 } + +Notes by Anton Bobkov + +\WikiLevelThree{Monday, November 10, 2014} +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub\WikiItalic{set}. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + \WikiSigleStar '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ \WikiSigleStar '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -\WikiBold{Proof of inequality} - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -\WikiBold{Proof of tail property} - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -\WikiItalic{Case 1}: $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -\WikiBold{Proof of well-definiteness} - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -\WikiItalic{Case 2}: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -\WikiItalic{Case 3}: $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -\WikiLevelThree{Wednesday, November 12, 2014} - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -\WikiBold{Proof} -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -\WikiBold{Proof} - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +\WikiBold{Proof of inequality} + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +\WikiBold{Proof of tail property} + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +\WikiItalic{Case 1}: $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +\WikiBold{Proof of well-definiteness} + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +\WikiItalic{Case 2}: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +\WikiItalic{Case 3}: $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +\WikiLevelThree{Wednesday, November 12, 2014} + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +\WikiBold{Proof} +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +\WikiBold{Proof} + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + \begin{enumerate} \item $a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ \item $f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ \end{enumerate} - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -\WikiItalic{Case 2}: $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -\WikiBold{Proof} -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -\WikiLevelThree{Friday, November 14, 2014} - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -\WikiBold{Proof} - -\WikiItalic{Case 1}: $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -\WikiItalic{Case 2}: $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +\WikiItalic{Case 2}: $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +\WikiBold{Proof} +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +\WikiLevelThree{Friday, November 14, 2014} + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +\WikiBold{Proof} + +\WikiItalic{Case 1}: $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +\WikiItalic{Case 2}: $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/all_notes/week_6/week_6.aux b/Other/old/all_notes/week_6/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/all_notes/week_6/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_6/week_6.tex b/Other/old/all_notes/week_6/week_6.tex index 7d2437eb..9c0f6dc8 100644 --- a/Other/old/all_notes/week_6/week_6.tex +++ b/Other/old/all_notes/week_6/week_6.tex @@ -1,390 +1,390 @@ -== Week 6 == - -Notes by Anton Bobkov - -===Monday, November 10, 2014=== -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - -* '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ -* '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -'''Proof of inequality''' - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -'''Proof of tail property''' - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -''Case 1'': $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -''Case 2'': $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -'''Proof of well-definiteness''' - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -''Case 1'': $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -===Wednesday, November 12, 2014=== - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -'''Proof''' -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -'''Proof''' - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - -*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ -*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ - -''Case 1'': $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -''Case 2'': $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -'''Proof''' -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -===Friday, November 14, 2014=== - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -'''Proof''' - -''Case 1'': $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -''Case 2'': $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. +== Week 6 == + +Notes by Anton Bobkov + +===Monday, November 10, 2014=== +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub''set''. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + +* '''Inequality:''' For \begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ +* '''Tail property:''' if $\gamma < \kappa < \alpha$ +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +'''Proof of inequality''' + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +'''Proof of tail property''' + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +''Case 1'': $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +''Case 2'': $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +'''Proof of well-definiteness''' + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +''Case 1'': $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +''Case 2'': $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +''Case 3'': $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +===Wednesday, November 12, 2014=== + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +'''Proof''' +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +'''Proof''' + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + +*$a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ +*$f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ + +''Case 1'': $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +''Case 2'': $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +'''Proof''' +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +===Friday, November 14, 2014=== + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +'''Proof''' + +''Case 1'': $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +''Case 2'': $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/all_notes/week_7/285D.aux b/Other/old/all_notes/week_7/285D.aux deleted file mode 100644 index cd380abd..00000000 --- a/Other/old/all_notes/week_7/285D.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_7/285D}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/all_notes/week_7/285D_notes_nov_17_18_21.aux b/Other/old/all_notes/week_7/285D_notes_nov_17_18_21.aux deleted file mode 100644 index f5313b87..00000000 --- a/Other/old/all_notes/week_7/285D_notes_nov_17_18_21.aux +++ /dev/null @@ -1,32 +0,0 @@ -\relax -\newlabel{6.7}{{1}{26}} -\newlabel{6.8}{{2}{27}} -\@writefile{toc}{\contentsline {section}{\numberline {1}The Surreals as a Real Closed Field}{30}} -\newlabel{7.1}{{9}{30}} -\newlabel{7.2}{{10}{30}} -\@setckpt{week_7/285D_notes_nov_17_18_21}{ -\setcounter{page}{33} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_8/g_week_8.aux b/Other/old/all_notes/week_8/g_week_8.aux deleted file mode 100644 index 8849a921..00000000 --- a/Other/old/all_notes/week_8/g_week_8.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_8/g_week_8}{ -\setcounter{page}{36} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_8/g_week_8.tex b/Other/old/all_notes/week_8/g_week_8.tex index 2a1db579..b7aa0b43 100644 --- a/Other/old/all_notes/week_8/g_week_8.tex +++ b/Other/old/all_notes/week_8/g_week_8.tex @@ -1,115 +1,115 @@ -\WikiLevelTwo{ Week 8 } -\WikiLevelThree{November 24, 2014} -\WikiItalic{Notes for today by Madeline Barnicle} - -Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the \WikiItalic{sign sequence} (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: -\WikiLevelFour{Section 8: Analytic functions on $\mathbf{No}$} -Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -\WikiBold{Digression on exp for $\mathbf{No}$} - -An \WikiItalic{exponential function} on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -\WikiBold{Theorem} (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a \WikiItalic{set}, and $\Gamma \neq \{0\}$, then there is \WikiItalic{no} exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -\WikiBold{Theorem} (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -\WikiLevelFour{Section 9: Power series and Weierstrass Preparation} -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, \WikiItalic{the ring of formal power series in $X$ over $A$.} Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -\WikiLevelThree{ November 26, 2014 } -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its \WikiItalic{constant term}. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -\WikiBold{Proof:} - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define \WikiItalic{substitution}. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ \WikiItalic{regular in $T$ of order $d$} if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +\WikiLevelTwo{ Week 8 } +\WikiLevelThree{November 24, 2014} +\WikiItalic{Notes for today by Madeline Barnicle} + +Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the \WikiItalic{sign sequence} (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: +\WikiLevelFour{Section 8: Analytic functions on $\mathbf{No}$} +Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +\WikiBold{Digression on exp for $\mathbf{No}$} + +An \WikiItalic{exponential function} on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +\WikiBold{Theorem} (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a \WikiItalic{set}, and $\Gamma \neq \{0\}$, then there is \WikiItalic{no} exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +\WikiBold{Theorem} (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +\WikiLevelFour{Section 9: Power series and Weierstrass Preparation} +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, \WikiItalic{the ring of formal power series in $X$ over $A$.} Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +\WikiLevelThree{ November 26, 2014 } +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its \WikiItalic{constant term}. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +\WikiBold{Proof:} + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define \WikiItalic{substitution}. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ \WikiItalic{regular in $T$ of order $d$} if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: \begin{enumerate} \item $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, \item $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. \end{enumerate} - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a \WikiItalic{polynomial} in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -'''Digression on exp for $\mathbf{No}$''' - -An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -====Section 9: Power series and Weierstrass Preparation==== -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -=== November 26, 2014 === -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -'''Proof:''' - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: -* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, -* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TR 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +'''Digression on exp for $\mathbf{No}$''' + +An ''exponential function'' on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +'''Theorem''' (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a ''set'', and $\Gamma \neq \{0\}$, then there is ''no'' exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +'''Theorem''' (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +====Section 9: Power series and Weierstrass Preparation==== +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, ''the ring of formal power series in $X$ over $A$.'' Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +=== November 26, 2014 === +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its ''constant term''. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +'''Proof:''' + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define ''substitution''. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ ''regular in $T$ of order $d$'' if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +* $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, +* $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. + +The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. + +====Theorem 9.3 (Division with remainder)==== +Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a ''polynomial'' in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TRpolyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write +\WikiLevelTwo{ Week 9 } + +\WikiLevelThree{ Monday 12-1-2014 } + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +\WikiBold{Proof:} + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +\WikiBold{Remark:} + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +\WikiBold{Proof:} + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write \begin{enumerate} \item $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. \item $r < s \Leftrightarrow r_i < s_i$ for each $i$. \item $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. \end{enumerate} - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: \begin{enumerate} \item $\|f\| = 0 \Leftrightarrow f = 0$. \item $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. @@ -71,62 +71,62 @@ \item $\|X^i f\| = r^i \|f\|$. \item $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. \end{enumerate} - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + \begin{enumerate} \item $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. \item $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. \end{enumerate} - -\WikiBold{Proof:} -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -\WikiBold{Proof:} - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -\WikiLevelThree{ Wednesday 12-3-2014 } -\WikiItalic{notes by Asaaf Shani} - -\WikiBoldItalic{Coming soon} - -\WikiLevelThree{ Friday 12-5-2014 } -\WikiItalic{notes by Tyler Arant} - -PDF: [[Media:285D notes 12 5.pdf]] + +\WikiBold{Proof:} +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +\WikiBold{Proof:} + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +\WikiLevelThree{ Wednesday 12-3-2014 } +\WikiItalic{notes by Asaaf Shani} + +\WikiBoldItalic{Coming soon} + +\WikiLevelThree{ Friday 12-5-2014 } +\WikiItalic{notes by Tyler Arant} + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/all_notes/week_9/week_6.aux b/Other/old/all_notes/week_9/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/all_notes/week_9/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_9/week_9.aux b/Other/old/all_notes/week_9/week_9.aux deleted file mode 100644 index 113be7d4..00000000 --- a/Other/old/all_notes/week_9/week_9.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_9/week_9}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/all_notes/week_9/week_9.tex b/Other/old/all_notes/week_9/week_9.tex index 4ff4a128..e646286d 100644 --- a/Other/old/all_notes/week_9/week_9.tex +++ b/Other/old/all_notes/week_9/week_9.tex @@ -1,126 +1,126 @@ -== Week 9 == - -=== Monday 12-1-2014 === - -==== Corollary 9.4 (Weierstrauss Preparation) ==== - -(Notes today by John Lensmire.) - -Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. -Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. - -'''Proof:''' - -Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. - -Let $x=0$ to get -$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) -+ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ -Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. -This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. - -Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. - -Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) - -'''Remark:''' - -The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial -$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. - -==== Corollary 9.5 ==== - -Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. - -'''Proof:''' - -We proceed by induction on $m$. - -If $m=0$, $K$ is a field, hence noetherian. - -From $m$ to $m+1$: -Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. -Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ -(see last time) we can assume that $f$ is regular in $T$ of some order $d$. -Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write -* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. -* $r < s \Leftrightarrow r_i < s_i$ for each $i$. -* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: -* $\|f\| = 0 \Leftrightarrow f = 0$. -* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. -* $\|f + g\| \leq \|f\| + \|g\|$. -* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. -* $\|X^i f\| = r^i \|f\|$. -* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - -* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. -* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. - -'''Proof:''' -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -'''Proof:''' - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -=== Wednesday 12-3-2014 === -''notes by Asaaf Shani'' - -'''''Coming soon''''' - -=== Friday 12-5-2014 === -''notes by Tyler Arant'' - -PDF: [[Media:285D notes 12 5.pdf]] +== Week 9 == + +=== Monday 12-1-2014 === + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +'''Proof:''' + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +'''Remark:''' + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +'''Proof:''' + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write +* $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. +* $r < s \Leftrightarrow r_i < s_i$ for each $i$. +* $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: +* $\|f\| = 0 \Leftrightarrow f = 0$. +* $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. +* $\|f + g\| \leq \|f\| + \|g\|$. +* $\|f\cdot g\| \leq \|f\| \cdot \|g\|$. +* $\|X^i f\| = r^i \|f\|$. +* $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + +* $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. +* $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. + +'''Proof:''' +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +'''Proof:''' + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +=== Wednesday 12-3-2014 === +''notes by Asaaf Shani'' + +'''''Coming soon''''' + +=== Friday 12-5-2014 === +''notes by Tyler Arant'' + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/final/Global.aux b/Other/old/final/Global.aux deleted file mode 100644 index b2173348..00000000 --- a/Other/old/final/Global.aux +++ /dev/null @@ -1,33 +0,0 @@ -\relax -\providecommand\hyper@newdestlabel[2]{} -\providecommand\HyperFirstAtBeginDocument{\AtBeginDocument} -\HyperFirstAtBeginDocument{\ifx\hyper@anchor\@undefined -\global\let\oldcontentsline\contentsline -\gdef\contentsline#1#2#3#4{\oldcontentsline{#1}{#2}{#3}} -\global\let\oldnewlabel\newlabel -\gdef\newlabel#1#2{\newlabelxx{#1}#2} -\gdef\newlabelxx#1#2#3#4#5#6{\oldnewlabel{#1}{{#2}{#3}}} -\AtEndDocument{\ifx\hyper@anchor\@undefined -\let\contentsline\oldcontentsline -\let\newlabel\oldnewlabel -\fi} -\fi} -\global\let\hyper@last\relax -\gdef\HyperFirstAtBeginDocument#1{#1} -\providecommand\HyField@AuxAddToFields[1]{} -\providecommand\HyField@AuxAddToCoFields[2]{} -\select@language{english} -\@writefile{toc}{\select@language{english}} -\@writefile{lof}{\select@language{english}} -\@writefile{lot}{\select@language{english}} -\@input{week_1/john_susice_surreal_numbers_notes_fall2014.aux} -\@input{week_2/g_week_2.aux} -\@input{week_3/zach.aux} -\@input{week_4/g_week_4.aux} -\@input{week_5/g_week_5.aux} -\@input{week_6/g_week_6.aux} -\@input{week_7/285D_notes_nov_17_18_21.aux} -\@input{week_8/g_week_8.aux} -\@input{week_9/g_week_9.aux} -\@input{week_10/December_8_12.aux} -\@input{week_11/g_week_11.aux} diff --git a/Other/old/final/Global.bbl b/Other/old/final/Global.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/final/Global.blg b/Other/old/final/Global.blg deleted file mode 100644 index 1f003bed..00000000 --- a/Other/old/final/Global.blg +++ /dev/null @@ -1,16 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Global.aux -A level-1 auxiliary file: week_1/john_susice_surreal_numbers_notes_fall2014.aux -A level-1 auxiliary file: week_2/g_week_2.aux -A level-1 auxiliary file: week_3/zach.aux -A level-1 auxiliary file: week_4/g_week_4.aux -A level-1 auxiliary file: week_5/g_week_5.aux -A level-1 auxiliary file: week_6/g_week_6.aux -A level-1 auxiliary file: week_7/285D_notes_nov_17_18_21.aux -A level-1 auxiliary file: week_8/g_week_8.aux -A level-1 auxiliary file: week_9/g_week_9.aux -A level-1 auxiliary file: week_10/December_8_12.aux -A level-1 auxiliary file: week_11/g_week_11.aux -I found no \citation commands---while reading file Global.aux -I found no \bibdata command---while reading file Global.aux -I found no \bibstyle command---while reading file Global.aux -(There were 3 error messages) diff --git a/Other/old/final/Global.log b/Other/old/final/Global.log deleted file mode 100644 index f0d51209..00000000 --- a/Other/old/final/Global.log +++ /dev/null @@ -1,998 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.5-1.40.14 (MiKTeX 2.9) (preloaded format=pdflatex 2014.9.19) 23 DEC 2014 22:03 -entering extended mode -**Global.tex -(C:\Users\Anton\SparkleShare\Research\Other\all_notes\Global.tex -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2013-05-26, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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map }{}% 5 -\BOOKMARK [1][-]{section.6}{The Normal Form}{}% 6 -\BOOKMARK [1][-]{section.7}{The Surreals as a Real Closed Field}{}% 7 -\BOOKMARK [1][-]{section.8}{Analytic functions on No}{}% 8 -\BOOKMARK [1][-]{section.9}{Power series and Weierstrass Preparation}{}% 9 -\BOOKMARK [1][-]{section.10}{Convergent Power Series }{}% 10 -\BOOKMARK [1][-]{section.11}{Restricted Analytic Functions}{}% 11 -\BOOKMARK [1][-]{section.12}{Quantifier elimination for Ran.}{}% 12 -\BOOKMARK [1][-]{section.13}{ Siddharth's extra lectures }{}% 13 diff --git a/Other/old/final/Global.tcp b/Other/old/final/Global.tcp index 57c3829e..cdf7f258 100644 --- a/Other/old/final/Global.tcp +++ b/Other/old/final/Global.tcp @@ -1,12 +1,12 @@ -[FormatInfo] -Type=TeXnicCenterProjectInformation -Version=4 - -[ProjectInfo] -MainFile=Global.tex -UseBibTeX=0 -UseMakeIndex=0 -ActiveProfile=LaTeX ⇨ PDF -ProjectLanguage=en -ProjectDialect=US - +[FormatInfo] +Type=TeXnicCenterProjectInformation +Version=4 + +[ProjectInfo] +MainFile=Global.tex +UseBibTeX=0 +UseMakeIndex=0 +ActiveProfile=LaTeX ⇨ PDF +ProjectLanguage=en +ProjectDialect=US + diff --git a/Other/old/final/Global.toc b/Other/old/final/Global.toc deleted file mode 100644 index 340b1bf6..00000000 --- a/Other/old/final/Global.toc +++ /dev/null @@ -1,14 +0,0 @@ -\select@language {english} -\contentsline {section}{\numberline {1}Basic Definitions and Existence Theorem}{2}{section.1} -\contentsline {section}{\numberline {2}Arithmetic Operators}{6}{section.2} -\contentsline {section}{\numberline {3}Real Numbers}{13}{section.3} -\contentsline {section}{\numberline {4}Combinatorics of Ordered Sets}{19}{section.4} -\contentsline {section}{\numberline {5} The $\omega ^-$ map }{22}{section.5} -\contentsline {section}{\numberline {6}The Normal Form}{24}{section.6} -\contentsline {section}{\numberline {7}The Surreals as a Real Closed Field}{37}{section.7} -\contentsline {section}{\numberline {8}Analytic functions on $\mathbf {No}$}{40}{section.8} -\contentsline {section}{\numberline {9}Power series and Weierstrass Preparation}{41}{section.9} -\contentsline {section}{\numberline {10}Convergent Power Series }{44}{section.10} -\contentsline {section}{\numberline {11}Restricted Analytic Functions}{50}{section.11} -\contentsline {section}{\numberline {12}Quantifier elimination for $\mathbb {R}_{an}$.}{55}{section.12} -\contentsline {section}{\numberline {13} Siddharth's extra lectures }{57}{section.13} diff --git a/Other/old/final/Global.tps b/Other/old/final/Global.tps deleted file mode 100644 index 6ac22248..00000000 --- a/Other/old/final/Global.tps +++ /dev/null @@ -1,62 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - 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-\usepackage{amsmath} -\usepackage{amssymb} -\usepackage{amsthm} -\usepackage{enumerate} -\usepackage{colonequals} -\usepackage{tikz-qtree} -\usepackage{fullpage} - -\newcommand{\No}{\mathbf{No}} -\newcommand{\On}{\mathbf{On}} -\newcommand{\paren}[1]{\left( #1 \right)} -\newcommand{\brac}[1]{\left[ #1 \right]} -\newcommand{\curly}[1]{\left\{ #1 \right\}} -\newcommand{\rar}{\rightarrow} -\newtheorem{theorem}{Theorem} -\newtheorem{defn}{Definition} -\newtheorem{cor}{Corollary} -\newtheorem{claim}{Claim} -\newtheorem{lem}{Lemma} -\newcommand{\R}{\mathbb{R}} -\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation -\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} -\newcommand{\ZFC}{\mathsf{ZFC}} -\newcommand{\NBG}{\mathsf{NBG}} -\newcommand{\coloneq}{\colonequals} -\newcommand{\N}{\mathbb{N}} - - -\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} -\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} -\date{\today} -\begin{document} -\maketitle{} - -\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -\tikzset{every tree node/.style={minimum width=2em,draw,circle}, - blank/.style={draw=none}, - edge from parent/.style= - {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, - level distance=1.5cm} -\begin{align*} -\begin{tikzpicture}[sibling distance=40pt] -\Tree -[.{} - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - ] - \edge[thick]; - [.\node[thick]{+}; - \edge[thick]; [.\node[thick]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[thick]; \node[thick]{+}; - ] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+}; - ] - ] -] -\end{tikzpicture} -\end{align*} -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. -\end{proof} - - - - - - - - - - - - - - - - - - - - - - - - - - - -\end{document} - +\documentclass{article} + +\usepackage{amsmath} +\usepackage{amssymb} +\usepackage{amsthm} +\usepackage{enumerate} +\usepackage{colonequals} +\usepackage{tikz-qtree} +\usepackage{fullpage} + +\newcommand{\No}{\mathbf{No}} +\newcommand{\On}{\mathbf{On}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\brac}[1]{\left[ #1 \right]} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\rar}{\rightarrow} +\newtheorem{theorem}{Theorem} +\newtheorem{defn}{Definition} +\newtheorem{cor}{Corollary} +\newtheorem{claim}{Claim} +\newtheorem{lem}{Lemma} +\newcommand{\R}{\mathbb{R}} +\newcommand{\concat}{\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}} %sequence concatenation +\newcommand{\dom}[1]{\operatorname{dom}\paren{#1}} +\newcommand{\ZFC}{\mathsf{ZFC}} +\newcommand{\NBG}{\mathsf{NBG}} +\newcommand{\coloneq}{\colonequals} +\newcommand{\N}{\mathbb{N}} + + +\title{Notes on Surreal Numbers \\ Math 285: Fall 2014} +\author{Class Taught by Prof. Aschenbrenner \\ Notes by John Susice} +\date{\today} +\begin{document} +\maketitle{} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +\tikzset{every tree node/.style={minimum width=2em,draw,circle}, + blank/.style={draw=none}, + edge from parent/.style= + {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, + level distance=1.5cm} +\begin{align*} +\begin{tikzpicture}[sibling distance=40pt] +\Tree +[.{} + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + ] + \edge[thick]; + [.\node[thick]{+}; + \edge[thick]; [.\node[thick]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[thick]; \node[thick]{+}; + ] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+}; + ] + ] +] +\end{tikzpicture} +\end{align*} +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. +\end{proof} + + + + + + + + + + + + + + + + + + + + + + + + + + + +\end{document} + diff --git a/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.aux b/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.aux deleted file mode 100644 index 8c5d0c07..00000000 --- a/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.aux +++ /dev/null @@ -1,37 +0,0 @@ -\relax -\providecommand\hyper@newdestlabel[2]{} -\newlabel{}{{0.1}{2}{Kruskal, 1980s}{theorem.0.1}{}} -\newlabel{}{{0.2}{2}{van den Dries-Ehrlich, c. 2000}{theorem.0.2}{}} -\@writefile{toc}{\contentsline {section}{\numberline {1}Basic Definitions and Existence Theorem}{2}{section.1}} -\newlabel{}{{1.4}{3}{Existence Theorem}{theorem.1.4}{}} -\newlabel{lemma_on_length_of_cuts}{{1.7}{4}{}{theorem.1.7}{}} -\newlabel{cofinality_theorem}{{1.9}{5}{The ``Cofinality Theorem''}{theorem.1.9}{}} -\newlabel{inverse_cofinality_theorem}{{1.11}{5}{``Inverse Cofinality Theorem''}{theorem.1.11}{}} -\@writefile{toc}{\contentsline {section}{\numberline {2}Arithmetic Operators}{6}{section.2}} -\newlabel{}{{2.1}{6}{Cantor's Normal Form Theorem}{theorem.2.1}{}} -\newlabel{defn_of_surreal_sum}{{1}{6}{}{equation.2.1}{}} -\@setckpt{week_1/john_susice_surreal_numbers_notes_fall2014}{ -\setcounter{page}{8} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{2} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{Item}{9} -\setcounter{Hfootnote}{0} -\setcounter{bookmark@seq@number}{2} -\setcounter{theorem}{6} -\setcounter{namedtheorem}{0} -\setcounter{section@level}{1} -} diff --git a/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.tex b/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.tex index 8530c585..c4d434be 100644 --- a/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.tex +++ b/Other/old/final/week_1/john_susice_surreal_numbers_notes_fall2014.tex @@ -1,543 +1,543 @@ -\NotesBy{Notes by John Suice} - -\Day{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\section{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -\tikzset{every tree node/.style={minimum width=2em,draw,circle}, - blank/.style={draw=none}, - edge from parent/.style= - {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, - level distance=1.5cm} -\begin{align*} -\begin{tikzpicture}[sibling distance=40pt] -\Tree -[.{} - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; - [.\node[dashed]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+};] - ] - \edge[thick]; - [.\node[thick]{+}; - \edge[thick]; [.\node[thick]{-}; - \edge[dashed]; \node[dashed]{-}; - \edge[thick]; \node[thick]{+}; - ] - \edge[dashed]; [.\node[dashed]{+}; - \edge[dashed]; \node[dashed]{-}; - \edge[dashed]; \node[dashed]{+}; - ] - ] -] -\end{tikzpicture} -\end{align*} -%--------------------------------------------------------------------- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\Day{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\Day{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\section{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\Day{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\NotesBy{Notes by John Suice} + +\Day{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\section{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +\tikzset{every tree node/.style={minimum width=2em,draw,circle}, + blank/.style={draw=none}, + edge from parent/.style= + {draw, edge from parent path={(\tikzparentnode) -- (\tikzchildnode)}}, + level distance=1.5cm} +\begin{align*} +\begin{tikzpicture}[sibling distance=40pt] +\Tree +[.{} + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; + [.\node[dashed]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + %[.\node[dashed]{+}; \node[dashed]{-}; \node[dashed]{+};] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+};] + ] + \edge[thick]; + [.\node[thick]{+}; + \edge[thick]; [.\node[thick]{-}; + \edge[dashed]; \node[dashed]{-}; + \edge[thick]; \node[thick]{+}; + ] + \edge[dashed]; [.\node[dashed]{+}; + \edge[dashed]; \node[dashed]{-}; + \edge[dashed]; \node[dashed]{+}; + ] + ] +] +\end{tikzpicture} +\end{align*} +%--------------------------------------------------------------------- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\Day{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\Day{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\section{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\Day{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/old/final/week_1/week_1.aux b/Other/old/final/week_1/week_1.aux deleted file mode 100644 index 7174f85a..00000000 --- a/Other/old/final/week_1/week_1.aux +++ /dev/null @@ -1,26 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1}Week 1}{2}} -\@setckpt{week_1/week_1}{ -\setcounter{page}{3} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} 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-\newlabel{12.8}{{12.5}{55}{}{theorem.12.5}{}} -\newlabel{12.9}{{12.6}{56}{}{theorem.12.6}{}} -\@setckpt{week_10/December_8_12}{ -\setcounter{page}{57} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{12} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{Item}{112} -\setcounter{Hfootnote}{2} -\setcounter{bookmark@seq@number}{12} -\setcounter{theorem}{6} -\setcounter{namedtheorem}{0} -\setcounter{section@level}{1} -} diff --git a/Other/old/final/week_10/December_8_12.bbl b/Other/old/final/week_10/December_8_12.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/final/week_10/December_8_12.blg b/Other/old/final/week_10/December_8_12.blg deleted file mode 100644 index 881ed8c1..00000000 --- a/Other/old/final/week_10/December_8_12.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: December_8_12.aux -I found no \citation commands---while reading file December_8_12.aux -I found no \bibdata command---while reading file December_8_12.aux -I found no \bibstyle command---while reading file December_8_12.aux -(There were 3 error messages) diff --git a/Other/old/final/week_10/December_8_12.log b/Other/old/final/week_10/December_8_12.log deleted file mode 100644 index 41abf18c..00000000 --- a/Other/old/final/week_10/December_8_12.log +++ /dev/null @@ -1,10271 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.5-1.40.14 (MiKTeX 2.9) (preloaded format=latex 2014.9.14) 23 DEC 2014 21:43 -entering extended mode -**December_8_12.tex -(C:\Users\Anton\SparkleShare\Research\Other\all_notes\week_10\December_8_12.tex -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, 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Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 35--41 -[][] - [] - - -Overfull \hbox (9.06943pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 X$ - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (29.70969pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (28.59857pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 35--41 -\OML/cmm/m/it/10 C []$ - [] - -! Undefined control sequence. -\C ->\mathbb - {C} -l.42 ...{n}y_{n}=0,\qquad\textrm{where }f_{i}\in\C - \left\{ X,T\right\} . -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (198.58188pt too wide) detected at line 43 -[][] \OML/cmm/m/it/10 f[]y[] \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 ::: \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 f[]y[] \OT1/cmr/ -m/n/10 = 0\OML/cmm/m/it/10 ; []f[] \OMS/cmsy/m/n/10 2 \OML/cmm/m/it/10 C [] : - [] - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no z in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no A in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.46 Apply W.P to $f_{i}$ in $\C - \left\{ X,T\right\} $, $f_{i}=u_{i}w_{i}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.47 where $u_{i}\in\C - \left\{ X,T\right\} ^{\times}$ and $w_{i}\in\C\left\{ ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.47 ...t\{ X,T\right\} ^{\times}$ and $w_{i}\in\C - \left\{ X\right\} \left[T\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! - -Overfull \hbox (8.22514pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 f[]$ - [] - - -Overfull \hbox (7.23265pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 T$ - [] - - -Overfull \hbox (8.22514pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 f[]$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (18.78065pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 f[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.54231pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 u[]w[]$ - [] - - -Overfull \hbox (18.49826pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 u[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (46.4701pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 C [][]$ - [] - - -Overfull \hbox (19.93285pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 w[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (43.05342pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (8.53415pt too wide) in paragraph at lines 43--51 -\OML/cmm/m/it/10 d[]$ - [] - - -Overfull \hbox (44.73508pt too wide) in paragraph at lines 43--51 -[]$ - [] - - -Overfull \hbox (17.26395pt too wide) in paragraph at lines 43--51 -[][]$ - [] - - -Overfull \hbox (79.04544pt too wide) in paragraph at lines 43--51 -[]$ - [] - - -Overfull \hbox (103.4273pt too wide) detected at line 53 -[] \OML/cmm/m/it/10 w[]y[] \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 ::: \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 w[]y[] \OT1/cmr/m/ -n/10 = 0\OML/cmm/m/it/10 : - [] - -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.55 in $\C - \left[\left[X,T\right]\right]$ is a linear combination of solutions -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.56 from $\C - \left\{ X,T\right\} $.\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no E in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.59 ...ion to $\left(\ast\right)$ (and are in $\C - \left\{ X,T\right\} $).\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.61 in $\C - \left[\left[X,T\right]\right]$. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.62 ...$y_{i}=q_{i}w_{1}+r_{i}$ where $q_{i}\in\C - \left[\left[X,T\right]\rig... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.63 and $r_{i}\in\C - \left[\left[X\right]\right]\left[T\right]$ of degree -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (19.69214pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 z[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (74.95723pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (19.69214pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 z[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (84.40166pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (20.64934pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 z[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (75.91443pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (7.97978pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 z[]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (15.81711pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 y \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (44.73508pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 53--65 -[]$ - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (3.44513pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 i$ - [] - - -Overfull \hbox (18.78761pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 y[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (27.2165pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 q[]w[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (7.84087pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 r[]$ - [] - - -Overfull \hbox (17.23781pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 q[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (17.28526pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 r[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 < - [] - - -Overfull \hbox (9.691pt too wide) in paragraph at lines 53--65 -\OML/cmm/m/it/10 d[]$ - [] - - -Overfull \hbox (116.52554pt too wide) detected at line 67 -\OML/cmm/m/it/10 y \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 q[]z[] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 : - [] - -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (220.76764pt too wide) detected at line 71 -\OML/cmm/m/it/10 y \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 q[]z[] \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 ::: \OT1/cmr/m/n/10 + \O -ML/cmm/m/it/10 q[]z[] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 ; []r[]: - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.74 from $\C - \left\{ X,T\right\} $.\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.75 First we claim that $r_{1}$ is also in $\C - \left[\left[X\right]\right]\l... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.77 so $w_{1}r_{1}\in\C - \left[\left[X\right]\right]\left[T\right]$. By -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.78 division in the ring of polynomials $\C - \left[\left[X\right]\right]\left... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.79 there are some $h,r\in\C - \left[\left[X\right]\right]\left[T\right]$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (15.34488pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (43.9526pt too wide) in paragraph at lines 71--81 -[]$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 71--81 -[]$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (8.99771pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 r[]$ - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 71--81 -[]$ - [] - - -Overfull \hbox (31.19852pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 w[]r[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (93.20027pt too wide) in paragraph at lines 71--81 -\OMS/cmsy/m/n/10 []$ - [] - - -Overfull \hbox (30.0874pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 w[]r[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (24.43976pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 h; r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (32.1504pt too wide) in paragraph at lines 71--81 -[] \OML/cmm/m/it/10 r < - [] - - -Overfull \hbox (9.691pt too wide) in paragraph at lines 71--81 -\OML/cmm/m/it/10 d[]$ - [] - - -Overfull \hbox (66.1725pt too wide) detected at line 83 -\OML/cmm/m/it/10 w[]r[] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 w[]h \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 r: - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.84 In $\C - \left[\left[X,T\right]\right]$, we have -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 83--85 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (70.17484pt too wide) detected at line 87 -\OML/cmm/m/it/10 w[]r[] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 w[]r[] \OT1/cmr/m/n/10 + 0\OML/cmm/m/it/10 : - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.88 ... division, we have $r=0$ and $r_{1}=h\in\C - \left[\left[X\right]\right... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.34488pt too wide) in paragraph at lines 87--89 -\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 87--89 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (19.55322pt too wide) in paragraph at lines 87--89 -\OML/cmm/m/it/10 r[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (15.20598pt too wide) in paragraph at lines 87--89 -\OML/cmm/m/it/10 h \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 87--89 -\OML/cmm/m/it/10 C [] []$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.90 S - o $r\in\left(\C\left[\left[X\right]\right]\left[T\right]\right)^{n}$ -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.90 So $r\in\left(\C - \left[\left[X\right]\right]\left[T\right]\right)^{n}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.92 is a linear combination (over $\C - \left[\left[X,T\right]\right]$) -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ) in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.93 of solutions from $\C - \left\{ X,T\right\} $. We claim that in fact -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.94 $r$ is a linear combination in $\C - \left[\left[X\right]\right]\left[T\ri... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.95 of solutions from $\C - \left\{ X\right\} \left[T\right]$. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no R in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.96 Recall that, by the inductive hypothesis, $\C - \left[\left[X\right]\right]$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\C ->\mathbb - {C} -l.97 is flat over $\C - \left\{ X\right\} $. Thus we have reduced the theorem -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no : in font nullfont! - -! LaTeX Error: Environment prop* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.99 \begin{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no S in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{prop*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.101 $S\left[T\right]$.\end{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.102 \begin{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\N ->\mathbb - {N} -l.106 Take some $d\in\N - $ s.t $\deg f_{i},\deg x_{i} for immediate help. - ... - -l.119 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 90--121 -[][] - [] - - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (57.38567pt too wide) in paragraph at lines 90--121 -[][]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 90--121 -[]$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (40.83119pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (39.72008pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (44.16454pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (43.05342pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C [] []$ - [] - - -Overfull \hbox (29.70969pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (28.59857pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (22.125pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (21.16316pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S []$ - [] - - -Overfull \hbox (27.36119pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 f[]x[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (16.11116pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 ::: \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (32.0533pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 f[]x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 90--121 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (17.66954pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 f[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (21.16316pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S []$ - [] - - -Overfull \hbox (55.80438pt too wide) in paragraph at lines 90--121 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (22.125pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (14.64926pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 d \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 N$ - [] - - -Overfull \hbox (65.88068pt too wide) in paragraph at lines 90--121 -[] \OML/cmm/m/it/10 f[]; [] x[] < - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (3.44513pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 i$ - [] - - -Overfull \hbox (19.60007pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (51.32556pt too wide) in paragraph at lines 90--121 -[][] \OML/cmm/m/it/10 x[]T[]$ - [] - - -Overfull \hbox (18.78065pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 f[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (50.50615pt too wide) in paragraph at lines 90--121 -[][] \OML/cmm/m/it/10 f[]T[]$ - [] - - -Overfull \hbox (7.23265pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 T$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 90--121 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (17.4688pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 d[] \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 90--121 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (12.75812pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 x[]$ - [] - - -Overfull \hbox (11.9387pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 f[]$ - [] - - -Overfull \hbox (22.75815pt too wide) in paragraph at lines 90--121 -[]$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (22.75815pt too wide) in paragraph at lines 90--121 -[]$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (23.61238pt too wide) in paragraph at lines 90--121 -[]$ - [] - - -Overfull \hbox (20.11574pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (15.24301pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 s \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (27.74304pt too wide) in paragraph at lines 90--121 -\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 ; :::; k$ - [] - - -Overfull \hbox (23.31363pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (39.11354pt too wide) in paragraph at lines 90--121 -[][] \OML/cmm/m/it/10 []y[]$ - [] - - -Overfull \hbox (21.39006pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 y[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (20.09143pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 y[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (50.5131pt too wide) in paragraph at lines 90--121 -[][] \OML/cmm/m/it/10 y[]T[]$ - [] - - -Overfull \hbox (19.60007pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (36.7038pt too wide) in paragraph at lines 90--121 -[][] \OML/cmm/m/it/10 []y[]$ - [] - - -Overfull \hbox (44.88205pt too wide) in paragraph at lines 90--121 -[]$ - [] - - -Overfull \hbox (21.16316pt too wide) in paragraph at lines 90--121 -\OML/cmm/m/it/10 S []$ - [] - - -! LaTeX Error: The font size command \normalsize is not defined: - there is probably something wrong with the class file. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.121 - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -[1] - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.124 A - s a consequence of (11.1) , -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no A in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 124--125 -[][] - [] - -! Undefined control sequence. -\C ->\mathbb - {C} -l.126 ...re }e_{k}=\left(0,...,1,...,0\right)\in\C - ^{m}. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (294.71077pt too wide) detected at line 127 -[] [] \OT1/cmr/m/n/10 = [] []\OML/cmm/m/it/10 ; []e[] \OT1/cmr/m/n/10 = [] \OMS/cmsy/m/n/10 2 \OML/cmm/m/it/10 C[]: - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.128 Let $\u\subset\R - ^{m}$ be open. A function $f\colon\u\lto\R$ is analytic -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no A in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.128 ...{m}$ be open. A function $f\colon\u\lto\R - $ is analytic -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.129 ...\in\u$, there is some $r$ and $f_{a}\in\R - \left\{ X\right\} _{r}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (17.80688pt too wide) in paragraph at lines 127--131 -\OMS/cmsy/m/n/10 U  - [] - - -Overfull \hbox (15.26627pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (36.55695pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 f\OT1/cmr/m/n/10 : \OMS/cmsy/m/n/10 U [][]! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (14.73029pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 a \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 127--131 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (19.1779pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 f[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (32.83449pt too wide) in paragraph at lines 127--131 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (88.01913pt too wide) detected at line 133 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 f[] [] ; - [] - -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! - -Overfull \hbox (9.06943pt too wide) in paragraph at lines 133--136 -\OML/cmm/m/it/10 X$ - [] - - -Overfull \hbox (5.28589pt too wide) in paragraph at lines 133--136 -\OML/cmm/m/it/10 a$ - [] - - -Overfull \hbox (5.97226pt too wide) in paragraph at lines 133--136 -\OML/cmm/m/it/10 f$ - [] - - -Overfull \hbox (16.33473pt too wide) in paragraph at lines 133--136 -\OML/cmm/m/it/10 C[]$ - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 133--136 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (134.94023pt too wide) detected at line 138 -\OML/cmm/m/it/10 f[] [] \OT1/cmr/m/n/10 = [] [] [] [] \OML/cmm/m/it/10 X[]: - [] - -Missing character: There is no A in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (52.54965pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (28.24736pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 f[] []$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 x$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.141 E - xamples: -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no E in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no : in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 141--142 -[][] - [] - -Missing character: There is no P in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.144 \item $\exp$, $\sin$, $\cos$, on $\R - $. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.27782pt too wide) in paragraph at lines 144--145 -[]$[]$ - [] - - -Overfull \hbox (12.2778pt too wide) in paragraph at lines 144--145 -[]$ - [] - - -Overfull \hbox (13.3889pt too wide) in paragraph at lines 144--145 -[]$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 144--145 -\OML/cmm/m/it/10 R$ - [] - -Missing character: There is no , in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.145 \item $\log$, $x^{a}$ (for $a\in\R - $), on $\R^{>0}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ) in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.145 \item $\log$, $x^{a}$ (for $a\in\R$), on $\R - ^{>0}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (12.91669pt too wide) in paragraph at lines 145--146 -[]$[]$ - [] - - -Overfull \hbox (10.55292pt too wide) in paragraph at lines 145--146 -\OML/cmm/m/it/10 x[]$ - [] - - -Overfull \hbox (14.73029pt too wide) in paragraph at lines 145--146 -\OML/cmm/m/it/10 a \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 145--146 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (18.4063pt too wide) in paragraph at lines 145--146 -\OML/cmm/m/it/10 R[]$ - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.148 $\an\left(\u\right)$ has an $\R - $-algebra structure. If $f\in\an\left(\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no - in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 147--150 -[] []$ - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 147--150 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 147--150 -[] []$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 147--150 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (15.41666pt too wide) in paragraph at lines 147--150 -\OML/cmm/m/it/10 f \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 147--150 -[] []$ - [] - - -Overfull \hbox (26.27129pt too wide) in paragraph at lines 147--150 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (34.1958pt too wide) in paragraph at lines 147--150 -[] [] \OML/cmm/m/it/10 :$ - [] - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.151 For each $a\in\u$ there is a map of $\R - $-algebras: -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no - in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no : in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 151--152 -[][] - [] - - -Overfull \hbox (14.73029pt too wide) in paragraph at lines 151--152 -\OML/cmm/m/it/10 a \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 151--152 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 151--152 -\OML/cmm/m/it/10 R$ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.153 \an\left(\u\right) & \lto & \R - \left\{ X\right\} .\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (78.71323pt too wide) in alignment at lines 152--155 - [][][] [] - [] - - -! LaTeX Error: Environment prop* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.157 \begin{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no ( in font nullfont! -Missing character: There is no A in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{prop*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.160 domain.\end{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment cor* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.161 \begin{cor*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{cor*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.163 ...pen subset of $\u$, then $f=g$.\end{cor*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment prop* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.164 \begin{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.165 ...f_{n}\in\an\left(\u\right)$, $\v\subset\R - ^{n}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.166 ...left(f_{1},...,f_{n}\right)\colon\u\lto\R - ^{n}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 158--167 -[][] - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 158--167 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 158--167 -[] []$ - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 158--167 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (24.43398pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 f; g \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 158--167 -[] []$ - [] - - -Overfull \hbox (7.25137pt too wide) in paragraph at lines 158--167 -\OMS/cmsy/m/n/10 U$ - [] - - -Overfull \hbox (16.52777pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 f \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.12846pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 g$ - [] - - -Overfull \hbox (46.38776pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 f[]; :::; f[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 158--167 -[] []$ - [] - - -Overfull \hbox (17.50555pt too wide) in paragraph at lines 158--167 -\OMS/cmsy/m/n/10 V  - [] - - -Overfull \hbox (13.11348pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (33.22357pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 f [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (8.05556pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 V$ - [] - - -Overfull \hbox (16.52777pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 f \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (76.97247pt too wide) in paragraph at lines 158--167 -[] \OT1/cmr/m/n/10 : \OMS/cmsy/m/n/10 U [][]! - [] - - -Overfull \hbox (13.11348pt too wide) in paragraph at lines 158--167 -\OML/cmm/m/it/10 R[]$ - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 168--170 -[][] - [] - - -Overfull \hbox (14.57286pt too wide) in paragraph at lines 168--170 -\OML/cmm/m/it/10 g \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (29.45006pt too wide) in paragraph at lines 168--170 -[] []$ - [] - - -Overfull \hbox (10.12848pt too wide) in paragraph at lines 168--170 -\OML/cmm/m/it/10 g \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (15.41666pt too wide) in paragraph at lines 168--170 -\OML/cmm/m/it/10 f \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (29.75139pt too wide) in paragraph at lines 168--170 -[] []$ - [] - - -Overfull \hbox (123.58224pt too wide) detected at line 172 -[][] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 g[] [] : - [] - - -! LaTeX Error: \begin{document} ended by \end{prop*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.174 \end{prop*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -! Undefined control sequence. -l.175 \label{11.5} \uline - {Notation}: For $x=\left(x_{1},...,x_{m}\right)\in\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.175 ... For $x=\left(x_{1},...,x_{m}\right)\in\R - ^{m}$, -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.179 $\R - \left\{ X\right\} _{\delta^{+}}=\bigcup_{r>\delta}\R\left\{ X\right... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -\R ->\mathbb - {R} -l.179 ...ight\} _{\delta^{+}}=\bigcup_{r>\delta}\R - \left\{ X\right\} _{r}$.\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no E in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.180 Each $f\in\R - \left\{ X\right\} _{\delta^{+}}$ gives rise to a function -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.181 ...colon\bar{B}_{\delta}\left(0\right)\lto\R - $ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 175--183 -[][] - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 x \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (57.95717pt too wide) in paragraph at lines 175--183 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.26627pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (21.82635pt too wide) in paragraph at lines 175--183 -[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (83.79057pt too wide) in paragraph at lines 175--183 -[] []$ - [] - - -Overfull \hbox (15.37842pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10  > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 175--183 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.37842pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (34.09024pt too wide) in paragraph at lines 175--183 -[]$ - [] - - -Overfull \hbox (48.98927pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 R [][] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (57.40129pt too wide) in paragraph at lines 175--183 -[][] \OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (15.41666pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 f \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (38.43376pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (18.493pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 7! - [] - - -Overfull \hbox (72.54997pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 : [] [] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 175--183 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (11.97363pt too wide) in paragraph at lines 175--183 -[]$ - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no x in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.184 ..., $n\geq1$. Fix $f=f\left(X,Y\right)\in\R - \left\{ X,Y\right\} $. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.185 ...small $x$ we have $f\left(x,Y\right)\in\R - \left\{ Y\right\} $.\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no H in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no ? in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.188 ...\epsilon>0$, $\bar{B}_{\epsilon}\subset\R - ^{m}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ` in font nullfont! -Missing character: There is no ` in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment defn* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.191 \begin{defn*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no A in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 184--194 -[][] - [] - - -Overfull \hbox (18.5833pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 Y \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (46.54056pt too wide) in paragraph at lines 184--194 -[]$ - [] - - -Overfull \hbox (16.55786pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 184--194 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (16.52777pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 f \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (45.84715pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 f [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (40.32288pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (43.04855pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 f [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (27.3646pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (33.60416pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 f []$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10  > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 184--194 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (21.97359pt too wide) in paragraph at lines 184--194 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (15.26627pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 184--194 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (11.41808pt too wide) in paragraph at lines 184--194 -[]$ - [] - - -Overfull \hbox (203.64082pt too wide) detected at line 196 -[] \OML/cmm/m/it/10 ; - [] - -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.197 where $f,g_{1},...,g_{k}\in\R - \left\{ X\right\} _{\epsilon^{+}}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{defn*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.198 \end{defn*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.200 ...x$ ranges over a neighbourhood of $0\in\R - ^{m}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! - -Overfull \hbox (55.47336pt too wide) in paragraph at lines 196--202 -\OML/cmm/m/it/10 f; g[]; :::; g[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (37.8782pt too wide) in paragraph at lines 196--202 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (57.52084pt too wide) in paragraph at lines 196--202 -[] []$ - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 196--202 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (14.44441pt too wide) in paragraph at lines 196--202 -\OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.26627pt too wide) in paragraph at lines 196--202 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (110.15611pt too wide) detected at line 204 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 f[] [] Y[]; - [] - -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.205 where $f_{j}\left(X\right)\in\R - \left\{ X\right\} $. Since $\R\left\{ X... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.205 ...right)\in\R\left\{ X\right\} $. Since $\R - \left\{ X\right\} $ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -! Undefined control sequence. -\N ->\mathbb - {N} -l.206 ...ft\{ f_{j}\left(X\right)\right\} _{j\in\N - ^{n}}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\N ->\mathbb - {N} -l.208 for some $d\in\N - $. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.209 Thus there are $g_{ij}\in\R - \left\{ X\right\} $ such that for every -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (37.06767pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 f[] [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (59.2145pt too wide) in paragraph at lines 204--211 -[][]$ - [] - - -Overfull \hbox (56.99976pt too wide) in paragraph at lines 204--211 -[][]$ - [] - - -Overfull \hbox (14.64926pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 d \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 N$ - [] - - -Overfull \hbox (21.25693pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 g[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (20.80157pt too wide) in paragraph at lines 204--211 -[] \OML/cmm/m/it/10 > - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 204--211 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (83.43076pt too wide) detected at line 213 -\OML/cmm/m/it/10 f[] [] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 g[]f[]: - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -\N ->\mathbb - {N} -l.214 ...ollowing, $i$ ranges over elements of $\N - ^{n}$ s.t $\left|i\right|\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -\N ->\mathbb - {N} -l.215 and $j$ over elements of $\N - ^{n}$ s.t $\left|j\right|>d$. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (3.44513pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 i$ - [] - - -Overfull \hbox (14.5683pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 N[]$ - [] - - -Overfull \hbox (19.55621pt too wide) in paragraph at lines 213--217 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (4.69049pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 j$ - [] - - -Overfull \hbox (14.5683pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 N[]$ - [] - - -Overfull \hbox (20.80157pt too wide) in paragraph at lines 213--217 -[] \OML/cmm/m/it/10 > - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 213--217 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (226.4829pt too wide) detected at line 219 -[] \OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 f[] [] [] ; - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.220 in $\R - \left[\left[X,Y\right]\right]$. Note that for each $i,j$, -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.221 $g_{ij}\in\R - \left\{ X\right\} $, and so there is some $\delta$ s.t -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.222 $g_{ij}\in\R - \left\{ X\right\} _{\delta}$. However, in order to have -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no H in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.223 ...on $\left(1\right)$ as an equality in $\R - \left\{ X\right\} $, -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment claim* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.225 \begin{claim*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.226 ...ght]$ and there are $f_{i},\, g_{ij}\in\R - \left\{ X\right\} _{\delta... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ( in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.227 ...t $\sum_{j}g_{ij}\left(X\right)Y^{j}\in\R - \left\{ X,Y\right\} _{\del... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.228 and $\left(1\right)$ holds in $\R - \left\{ X\right\} _{\delta^{+}}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{claim*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.229 with $f_{i},g_{ij}$.\end{claim*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.230 \begin{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! - -Overfull \hbox (41.43399pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (12.58003pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 i; j$ - [] - - -Overfull \hbox (21.25693pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 g[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (4.8229pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (21.25693pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 g[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (32.79478pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 219--232 -[]$ - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (4.8229pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (12.58003pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 i; j$ - [] - - -Overfull \hbox (19.82288pt too wide) in paragraph at lines 219--232 -\OMS/cmsy/m/n/10 9\OML/cmm/m/it/10  \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (21.11113pt too wide) in paragraph at lines 219--232 -[]$ - [] - - -Overfull \hbox (35.59311pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 f[]; g[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (38.43376pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (70.11453pt too wide) in paragraph at lines 219--232 -[][] \OML/cmm/m/it/10 g[] [] Y[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (50.3504pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 219--232 -[]$ - [] - - -Overfull \hbox (38.43376pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (24.48209pt too wide) in paragraph at lines 219--232 -\OML/cmm/m/it/10 f[]; g[]$ - [] - - -Overfull \hbox (147.49962pt too wide) detected at line 234 -\OML/cmm/m/it/10 f \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 f[] [] : - [] - -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.236 in $\R - \left[\left[X,Y\right]\right]$ (all the higher terms, $Y^{j}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ( in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.237 ...1$, are inside the $Z_{ij}$'s). Since $\R - \left[\left[X,Y\right]\rig... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.238 is f.f. over $\R - \left\{ X,Y\right\} $, there is a solution $\left\{ Z_... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.239 in $\R - \left\{ X,Y\right\} $. Take $\delta$ for all $i,j$, $Z_{ij}\in\R... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.239 ...Take $\delta$ for all $i,j$, $Z_{ij}\in\R - \left\{ X,Y\right\} _{\del... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.240 ...er series in $Y$ with coefficients in $\R - \left\{ X\right\} $ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.241 ...1\right)$ with $g_{ij}\left(X\right)\in\R - \left\{ X\right\} _{\delta... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.243 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.244 ...ork with equation $\left(1\right)$ in $\R - \left\{ X,Y\right\} _{\del... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 234--246 -[]$ - [] - - -Overfull \hbox (109.63644pt too wide) in paragraph at lines 234--246 -[]$ - [] - - -Overfull \hbox (41.43399pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (12.24135pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 Y[]$ - [] - - -Overfull \hbox (20.80157pt too wide) in paragraph at lines 234--246 -[] \OML/cmm/m/it/10 > - [] - - -Overfull \hbox (12.98267pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 d \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 234--246 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (13.86925pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 Z[]$ - [] - - -Overfull \hbox (41.43399pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (40.32288pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (23.86928pt too wide) in paragraph at lines 234--246 -[]$ - [] - - -Overfull \hbox (40.32288pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (4.8229pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (12.58003pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 i; j$ - [] - - -Overfull \hbox (23.31364pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 Z[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (50.3504pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (13.86925pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 Z[]$ - [] - - -Overfull \hbox (8.02779pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 Y$ - [] - - -Overfull \hbox (28.40623pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R []$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 234--246 -[]$ - [] - - -Overfull \hbox (39.77078pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 g[] [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (38.43376pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (19.55621pt too wide) in paragraph at lines 234--246 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (20.80157pt too wide) in paragraph at lines 234--246 -[] \OML/cmm/m/it/10 > - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 234--246 -[]$ - [] - - -Overfull \hbox (50.3504pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 R [][]$ - [] - - -Overfull \hbox (21.82635pt too wide) in paragraph at lines 234--246 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (4.8229pt too wide) in paragraph at lines 234--246 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (145.98926pt too wide) detected at line 248 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = 0 \OMS/cmsy/m/n/10 ([]) 8\OML/cmm/m/it/10 i [] : - [] - -Missing character: There is no A in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! - -Overfull \hbox (33.94035pt too wide) in paragraph at lines 248--251 -\OML/cmm/m/it/10 f[] [] \OMS/cmsy/m/n/10 6\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 248--251 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (68.07635pt too wide) in paragraph at lines 248--251 -[] [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.0007pt too wide) in paragraph at lines 248--251 -[]$ - [] - -! Undefined control sequence. -l.252 f\left(x,Y\right)\neq0\implies - \mathrm{ord}\left(f\left(x,Y\right)\righ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (130.77411pt too wide) detected at line 253 -\OML/cmm/m/it/10 f [] \OMS/cmsy/m/n/10 6\OT1/cmr/m/n/10 = 0[] [] \OMS/cmsy/m/n/10  \OML/cmm/m/it/10 d: - [] - -Missing character: There is no D in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! - -Overfull \hbox (235.903pt too wide) in alignment at lines 255--258 - [][][] [] - [] - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. - \nicefrac - -l.262 $\nicefrac - {f_{i}}{f_{i'}}$, for $i\neq i'$. Let $V_{i}=\left(V_{ii'}\r... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! - -Overfull \hbox (22.52914pt too wide) in paragraph at lines 258--264 -[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (17.88164pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 Z[] \OMS/cmsy/m/n/10 [ - [] - - -Overfull \hbox (22.7905pt too wide) in paragraph at lines 258--264 -[][] \OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (3.44513pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 i$ - [] - - -Overfull \hbox (5.97226pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 f$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 258--264 -[]$ - [] - - -Overfull \hbox (8.22514pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 f[]$ - [] - - -Overfull \hbox (17.07178pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 V[]$ - [] - - -Overfull \hbox (19.15517pt too wide) in paragraph at lines 258--264 -[][]$ - [] - - -Overfull \hbox (14.00064pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 i \OMS/cmsy/m/n/10 6\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (6.25069pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 i[]$ - [] - - -Overfull \hbox (19.71812pt too wide) in paragraph at lines 258--264 -\OML/cmm/m/it/10 V[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (37.47694pt too wide) in paragraph at lines 258--264 -[][]$ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.265 ...t(Y^{i'}+\sum_{j}g_{i'j}Y^{j}\right)\in\R - \left\{ X,V_{i},Y\right\} . -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (279.9535pt too wide) detected at line 266 -\OML/cmm/m/it/10 F[] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 Y[] \OT1/cmr/m/n/10 + [] \OML/cmm/m/it/10 g[]Y[] \OT1/cmr/m/n/10 - + [] \OML/cmm/m/it/10 V[] [] \OMS/cmsy/m/n/10 2 \OML/cmm/m/it/10 R [] : - [] - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -l.267 ... let $v_{i}\left(x\right)=\left(\nicefrac - {f_{i'}\left(x\right)}{f_{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! - -Overfull \hbox (15.15967pt too wide) in paragraph at lines 266--270 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.46123pt too wide) in paragraph at lines 266--270 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (33.89172pt too wide) in paragraph at lines 266--270 -\OML/cmm/m/it/10 v[] [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (72.25473pt too wide) in paragraph at lines 266--270 -[][]$ - [] - - -Overfull \hbox (15.15967pt too wide) in paragraph at lines 266--270 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.46123pt too wide) in paragraph at lines 266--270 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (39.4473pt too wide) in paragraph at lines 266--270 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 266--270 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (157.71669pt too wide) detected at line 272 -[] \OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 f[] [] F[] [] : - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 275--279 -[][] - [] - - -Overfull \hbox (9.75984pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 F[]$ - [] - - -Overfull \hbox (19.62494pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 X \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 275--279 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (19.71812pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 V[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (4.32756pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 c$ - [] - - -Overfull \hbox (18.5833pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 Y \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 275--279 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (14.88307pt too wide) in paragraph at lines 275--279 -\OML/cmm/m/it/10 c \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (33.14189pt too wide) in paragraph at lines 275--279 -[][]$ - [] - - -Overfull \hbox (20.43864pt too wide) in paragraph at lines 275--279 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 275--279 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (103.83328pt too wide) detected at line 281 -\OML/cmm/m/it/10 F[] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 F[] [] : - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (15.15967pt too wide) in paragraph at lines 281--283 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.46123pt too wide) in paragraph at lines 281--283 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (164.10406pt too wide) detected at line 285 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 f[] [] F[] [] : - [] - - -! LaTeX Error: Environment claim* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.287 \begin{claim*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.288 ...re is $\lambda=\lambda\left(c\right)\in\R - ^{n-1}$ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{claim*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.291 ...bda_{n-1}Y_{n},Y_{n}\right)$.\end{claim*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.292 \begin{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no E in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.294 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 288--296 -[][] - [] - - -Overfull \hbox (4.32756pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 c$ - [] - - -Overfull \hbox (16.38887pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (29.04974pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10  [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (23.34962pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (21.94444pt too wide) in paragraph at lines 288--296 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 288--296 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (75.0239pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 F[] []$ - [] - - -Overfull \hbox (11.24889pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 Y[]$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 288--296 -\OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.0007pt too wide) in paragraph at lines 288--296 -[]$ - [] - - -Overfull \hbox (18.5833pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 Y \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (46.54056pt too wide) in paragraph at lines 288--296 -[]$ - [] - - -Overfull \hbox (33.86108pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10  [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (146.75824pt too wide) in paragraph at lines 288--296 -[]$ - [] - - -Overfull \hbox (5.83336pt too wide) in paragraph at lines 288--296 -\OML/cmm/m/it/10 $ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.297 ...ight)=\u_{i,c}\w_{i,c},\quad\u_{i,c}\in\R - \left\{ X,V_{i},Y\right\} ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -\R ->\mathbb - {R} -l.297 ...X,V_{i},Y\right\} ^{\ast},\:\w_{i,c}\in\R - \left\{ X,V_{i},Y_{1},...,... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (383.52927pt too wide) detected at line 298 -[] \OML/cmm/m/it/10 F[] [] \OT1/cmr/m/n/10 = \OMS/cmsy/m/n/10 U[]W[]\OML/cmm/m/it/10 ; \OMS/cmsy/m/n/10 U[] 2 \OML/cmm -/m/it/10 R [][] ; \OMS/cmsy/m/n/10 W[] 2 \OML/cmm/m/it/10 R [] [] : - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (35.16496pt too wide) in paragraph at lines 298--300 -\OML/cmm/m/it/10  [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (20.93402pt too wide) in paragraph at lines 298--300 -[]$ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.301 \item $\u_{i,c}\in\R - \left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\righ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (24.98074pt too wide) in paragraph at lines 301--302 -[]$\OMS/cmsy/m/n/10 U[] 2 - [] - - -Overfull \hbox (78.42993pt too wide) in paragraph at lines 301--302 -\OML/cmm/m/it/10 R [][]$ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.302 \item $\w_{i,c}\in\R - \left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\righ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (28.60025pt too wide) in paragraph at lines 302--303 -[]$\OMS/cmsy/m/n/10 W[] 2 - [] - - -Overfull \hbox (82.87434pt too wide) in paragraph at lines 302--303 -\OML/cmm/m/it/10 R [][] :$ - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (29.69508pt too wide) in paragraph at lines 304--307 -\OT1/cmr/m/n/10 [] = - [] - - -Overfull \hbox (45.3908pt too wide) in paragraph at lines 304--307 -[]$ - [] - - -Overfull \hbox (4.32756pt too wide) in paragraph at lines 304--307 -\OML/cmm/m/it/10 c$ - [] - - -Overfull \hbox (19.70432pt too wide) in paragraph at lines 304--307 -\OML/cmm/m/it/10 I[]$ - [] - - -Overfull \hbox (14.16663pt too wide) in paragraph at lines 304--307 -\OT1/cmr/m/n/10 I = - [] - - -Overfull \hbox (27.77782pt too wide) in paragraph at lines 304--307 -[]$ - [] - - -Overfull \hbox (20.75204pt too wide) in paragraph at lines 304--307 -\OML/cmm/m/it/10 C []$ - [] - - -Overfull \hbox (102.06522pt too wide) detected at line 309 -\OT1/cmr/m/n/10 I[] \OMS/cmsy/m/n/10  [] \OML/cmm/m/it/10 B[] [] : - [] - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (16.80553pt too wide) in paragraph at lines 309--314 -\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (109.60144pt too wide) in paragraph at lines 309--314 -[]$ - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 309--314 -\OML/cmm/m/it/10  > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 309--314 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 309--314 -\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (20.76105pt too wide) in paragraph at lines 309--314 -[]$ - [] - - -Overfull \hbox (29.4393pt too wide) in paragraph at lines 309--314 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.0278pt too wide) in paragraph at lines 309--314 -\OT1/cmr/m/n/10 \OML/cmm/m/it/10 :$ - [] - - -Overfull \hbox (16.28836pt too wide) in paragraph at lines 309--314 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (29.4393pt too wide) in paragraph at lines 309--314 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (6.25002pt too wide) in paragraph at lines 309--314 -\OT1/cmr/m/n/10 $ - [] - - -Overfull \hbox (191.81973pt too wide) detected at line 316 -\OML/cmm/m/it/10 S[] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 : - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (16.12791pt too wide) in paragraph at lines 316--322 -\OML/cmm/m/it/10 S[] \OMS/cmsy/m/n/10 \ - [] - - -Overfull \hbox (21.97359pt too wide) in paragraph at lines 316--322 -[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (121.51442pt too wide) in paragraph at lines 316--322 -[] []$ - [] - - -Overfull \hbox (21.97359pt too wide) in paragraph at lines 316--322 -[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (57.91077pt too wide) in paragraph at lines 316--322 -[] \OMS/cmsy/m/n/10 [ - [] - - -Overfull \hbox (26.39407pt too wide) in paragraph at lines 316--322 -[][] \OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (16.28836pt too wide) in paragraph at lines 316--322 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (29.4393pt too wide) in paragraph at lines 316--322 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (6.25002pt too wide) in paragraph at lines 316--322 -\OT1/cmr/m/n/10 $ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 316--322 -[]$ - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 316--322 -[]$ - [] - - -Overfull \hbox (233.50087pt too wide) detected at line 324 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 f[] [] \OMS/cmsy/m/n/10 U[] [] \OML/cmm/m/it/10 W[] [] : - [] - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no t in font nullfont! - -Overfull \hbox (11.38939pt too wide) in paragraph at lines 324--328 -\OMS/cmsy/m/n/10 U[]$ - [] - - -Overfull \hbox (22.29918pt too wide) in paragraph at lines 324--328 -[]$ - [] - - -Overfull \hbox (30.69458pt too wide) in paragraph at lines 324--328 -\OML/cmm/m/it/10  [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (22.77785pt too wide) in paragraph at lines 324--328 -[]$ - [] - - -Overfull \hbox (230.1836pt too wide) detected at line 330 -[] [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10  [] [] [] - [] - -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no A in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! - -Overfull \hbox (15.15967pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (11.26302pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (21.37268pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.05905pt too wide) in paragraph at lines 330--337 -\OT1/cmr/m/n/10 2\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (46.11575pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 f []$ - [] - - -Overfull \hbox (21.94444pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 330--337 -\OT1/cmr/m/n/10 1$ - [] - - -Overfull \hbox (36.65047pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (19.61456pt too wide) in paragraph at lines 330--337 -\OT1/cmr/m/n/10 4\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (4.8229pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (21.37268pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (4.05904pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (52.20607pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.05905pt too wide) in paragraph at lines 330--337 -\OT1/cmr/m/n/10 2\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (61.92828pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10  [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.2616pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 y$ - [] - - -Overfull \hbox (21.37268pt too wide) in paragraph at lines 330--337 -[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (4.05904pt too wide) in paragraph at lines 330--337 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (250.18361pt too wide) detected at line 339 -[] [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10  [] [] [] : - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! - -Overfull \hbox (18.09718pt too wide) in paragraph at lines 339--341 -\OML/cmm/m/it/10 Z \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (48.58224pt too wide) in paragraph at lines 339--341 -[]$ - [] - -! Undefined control sequence. -\R ->\mathbb - {R} -l.342 ...\right)V_{\gamma},2\epsilon Z\right)\in\R - \left\{ X,V_{i},Z\right\} ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (278.97672pt too wide) detected at line 343 -[] [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10 f[] [] \OMS/cmsy/m/n/10 W[] [] 2 \OML/cmm/m/it/10 R [][] : - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! - -Overfull \hbox (32.64348pt too wide) in paragraph at lines 343--346 -[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (23.8589pt too wide) in paragraph at lines 343--346 -\OML/cmm/m/it/10 I[]$ - [] - - -Overfull \hbox (19.2187pt too wide) in paragraph at lines 343--346 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (11.26302pt too wide) in paragraph at lines 343--346 -\OML/cmm/m/it/10 S[]$ - [] - -! Undefined control sequence. -l.347 ...\right)\hat{\w}_{\gamma}\left(x,\nicefrac - {v_{i}\left(\epsilon x\rig... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -l.347 ...)}{\epsilon\left(\gamma\right)},\nicefrac - {\left(-\lambda\right)\lef... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (215.59421pt too wide) detected at line 348 -[]\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = \OML/cmm/m/it/10  [] [] [] : - [] - -! Undefined control sequence. -l.350 \section - {Quantifier elimination for $\R_{an}$.} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no Q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.350 \section{Quantifier elimination for $\R - _{an}$.} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 350--351 -[][] - [] - - -Overfull \hbox (17.37387pt too wide) in paragraph at lines 350--351 -\OML/cmm/m/it/10 R[]$ - [] - -Missing character: There is no R in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.354 analytic function $\R - ^{m}\lto\R$. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -\R ->\mathbb - {R} -l.354 analytic function $\R^{m}\lto\R - $. \\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.355 Let $\R - _{an}$ be the ordered field of reals as an $\L_{an}$-structure -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.356 together with the map $^{-1}\colon\R - \lto\R$ defined as $x^{-1}=\begin{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -\R ->\mathbb - {R} -l.356 together with the map $^{-1}\colon\R\lto\R - $ defined as $x^{-1}=\begin{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -! Argument of \frac has an extra }. - - \par -l.357 \frac - {1}{x} & x\neq0\\ -I've run across a `}' that doesn't seem to match anything. -For example, `\def\a#1{...}' and `\a}' would produce -this error. If you simply proceed now, the `\par' that -I've just inserted will cause me to report a runaway -argument that might be the root of the problem. But if -your `}' was spurious, just type `2' and it will go away. - -Runaway argument? -! Paragraph ended before \frac was complete. - - \par -l.357 \frac - {1}{x} & x\neq0\\ -I suspect you've forgotten a `}', causing me to apply this -control sequence to too much text. How can we recover? -My plan is to forget the whole thing and hope for the best. - -! Missing $ inserted. - - $ -l.357 \frac - {1}{x} & x\neq0\\ -I've inserted a begin-math/end-math symbol since I think -you left one out. Proceed, with fingers crossed. - -! Missing \cr inserted. - - \cr -l.357 \frac - {1}{x} & x\neq0\\ -I'm guessing that you meant to end an alignment here. - -! Missing $ inserted. - - $ -l.357 \frac - {1}{x} & x\neq0\\ -I've inserted a begin-math/end-math symbol since I think -you left one out. Proceed, with fingers crossed. - -! Misplaced alignment tab character &. -l.357 \frac{1}{x} & - x\neq0\\ -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - -! Misplaced alignment tab character &. -l.358 x=0 & - 0 -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - -Missing character: There is no . in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment thm* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.360 \begin{thm*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no ( in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 9 in font nullfont! -Missing character: There is no 8 in font nullfont! -Missing character: There is no 8 in font nullfont! -Missing character: There is no ) in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.361 ...{12.1} (Denef-v. d. Dries 1988) $\left(\R - _{an},\,^{-1}\right)$ has -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{thm*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.363 \end{thm*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.364 Call $f\colon\I^{m}\lto\R - $ analytic if it extends to an analytic -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.367 analytic $f\colon\I^{m}\lto\R - $ with $f\left(\I^{m}\right)\subset\I$.\\ -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 352--370 -[][] - [] - - -Overfull \hbox (16.67822pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[]$ - [] - - -Overfull \hbox (68.33331pt too wide) in paragraph at lines 352--370 -[]$ - [] - - -Overfull \hbox (34.15517pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R[] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (17.37387pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (16.67822pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[]$ - [] - - -Overfull \hbox (41.73961pt too wide) in paragraph at lines 352--370 -[]\OT1/cmr/m/n/10 : \OML/cmm/m/it/10 R [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (27.00693pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (36.51933pt too wide) in paragraph at lines 352--370 -[] \OT1/cmr/m/n/10 1\OML/cmm/m/it/10 xx \OMS/cmsy/m/n/10 6\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 0[] - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 x \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (10.00003pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 00$ - [] - - -Overfull \hbox (48.80322pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[] [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (23.34491pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[] [ - [] - - -Overfull \hbox (22.40286pt too wide) in paragraph at lines 352--370 -[]$ - [] - - -Overfull \hbox (43.38777pt too wide) in paragraph at lines 352--370 -[]$ - [] - - -Overfull \hbox (40.5128pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 f\OT1/cmr/m/n/10 : I[] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (11.20723pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 I[]$ - [] - - -Overfull \hbox (11.7349pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[]$ - [] - - -Overfull \hbox (17.77783pt too wide) in paragraph at lines 352--370 -[]$ - [] - - -Overfull \hbox (8.78014pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 m$ - [] - - -Overfull \hbox (40.5128pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 f\OT1/cmr/m/n/10 : I[] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 R$ - [] - - -Overfull \hbox (37.17943pt too wide) in paragraph at lines 352--370 -\OML/cmm/m/it/10 f [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (3.61111pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 I$ - [] - - -Overfull \hbox (3.61111pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 I$ - [] - - -Overfull \hbox (11.7349pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 L[]$ - [] - - -Overfull \hbox (39.42227pt too wide) in paragraph at lines 352--370 -\OMS/cmsy/m/n/10 D\OT1/cmr/m/n/10 : I[] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (3.61111pt too wide) in paragraph at lines 352--370 -\OT1/cmr/m/n/10 I$ - [] - -! Undefined control sequence. - \nicefrac - -l.372 \nicefrac - {x}{y} & \textrm{if }\left|x\right|\leq\left|y\right|\textrm{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Misplaced alignment tab character &. -l.372 \nicefrac{x}{y} & - \textrm{if }\left|x\right|\leq\left|y\right|\textrm{... -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - -! Misplaced alignment tab character &. -l.373 0 & - \textrm{otherwise.} -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - - -Overfull \hbox (214.97655pt too wide) detected at line 375 -\OML/cmm/m/it/10 D [] \OT1/cmr/m/n/10 = [] \OML/cmm/m/it/10 xy[] [] \OMS/cmsy/m/n/10  [] []\OML/cmm/m/it/10 y \OMS/cmsy -/m/n/10 6\OT1/cmr/m/n/10 = 0\OML/cmm/m/it/10 ; [] \OT1/cmr/m/n/10 0[] - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment thm* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.377 \begin{thm*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{thm*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.379 \end{thm*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no d in font nullfont! 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LaTeX Error: \begin{document} ended by \end{lem*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.401 \end{lem*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{lem*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.404 \end{lem*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment lem* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.407 \begin{lem*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no B in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (10.03821pt too wide) in paragraph at lines 402--409 -\OML/cmm/m/it/10 T[]$ - [] - - -Overfull \hbox (12.25696pt too wide) in paragraph at lines 402--409 -\OML/cmm/m/it/10 q:e$ - [] - - -Overfull \hbox (25.55563pt too wide) in paragraph at lines 402--409 -[]$ - [] - - -Overfull \hbox (7.23265pt too wide) in paragraph at lines 402--409 -\OML/cmm/m/it/10 T$ - [] - - -Overfull \hbox (25.55563pt too wide) in paragraph at lines 402--409 -[]$ - [] - - -Overfull \hbox (17.78816pt too wide) in paragraph at lines 402--409 -\OML/cmm/m/it/10 T \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (36.24747pt too wide) in paragraph at lines 402--409 -[][] []$ - [] - - -Overfull \hbox (20.59372pt too wide) in paragraph at lines 402--409 -\OML/cmm/m/it/10 T[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (44.13231pt too wide) in paragraph at lines 402--409 -[][] []$ - [] - -Missing character: There is no L in font nullfont! 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Undefined control sequence. -\model ->\vDash - -l.415 \item $\I\model - \exists y\varphi\left(x,y\right)\longleftrightarrow\exi... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -Overfull \hbox (66.94678pt too wide) in paragraph at lines 415--416 -[]$\OT1/cmr/m/n/10 I\OMS/cmsy/m/n/10 9\OML/cmm/m/it/10 y' [] \OMS/cmsy/m/n/10 []! - [] - - -Overfull \hbox (41.88196pt too wide) in paragraph at lines 415--416 -\OMS/cmsy/m/n/10 9\OML/cmm/m/it/10 z[] []$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (6.54167pt too wide) in paragraph at lines 416--418 -[]$ - [] - - -Overfull \hbox (7.99171pt too wide) in paragraph at lines 416--418 -\OMS/cmsy/m/n/10 D$ - [] - - -Overfull \hbox (5.0903pt too wide) in paragraph at lines 416--418 -\OML/cmm/m/it/10 z$ - [] - - -Overfull \hbox (10.09383pt too wide) in paragraph at lines 416--418 -\OML/cmm/m/it/10 z[]$ - [] - - -Overfull \hbox (6.54167pt too wide) in paragraph at lines 416--418 -[]$ - [] - - -! LaTeX Error: \begin{document} ended by \end{lem*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.419 \end{lem*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no G in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -\R ->\mathbb - {R} -l.421 for $\R - $ (as an ordered ring) to eliminate $\exists z_{n}$ and produce -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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Undefined control sequence. -\model ->\vDash - -l.442 \item $\I\model - \exists y\varphi_{\epsilon}\iff\exists z\left(\varphi_{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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LaTeX Error: \begin{document} ended by \end{lem*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.446 \end{lem*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! 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LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.452 We get \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (6.54167pt too wide) in paragraph at lines 448--453 -\OML/cmm/m/it/10 '$ - [] - - -Overfull \hbox (41.39348pt too wide) in paragraph at lines 448--453 -\OML/cmm/m/it/10 f [] > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 448--453 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (41.39348pt too wide) in paragraph at lines 448--453 -\OML/cmm/m/it/10 f [] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 448--453 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (40.5128pt too wide) in paragraph at lines 448--453 -\OML/cmm/m/it/10 f\OT1/cmr/m/n/10 : I[] [][]\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (3.61111pt too wide) in paragraph at lines 448--453 -\OT1/cmr/m/n/10 I$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 448--453 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (5.97226pt too wide) in paragraph at lines 448--453 -\OML/cmm/m/it/10 f$ - [] - -Missing character: There is no , in font nullfont! - -Overfull \hbox (13.50343pt too wide) in paragraph at lines 454--455 -[]$\OML/cmm/m/it/10  \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (22.22224pt too wide) in paragraph at lines 454--455 -[]$ - [] - -Missing character: There is no F in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -! 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Undefined control sequence. -\model ->\vDash - -l.471 \I\model - \exists y\varphi\left(\epsilon x,\epsilon y\right)\longleftrig... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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Emergency stop. -<*> December_8_12.tex - -*** (job aborted, no legal \end found) - - -Here is how much of TeX's memory you used: - 45 strings out of 493922 - 535 string characters out of 3144899 - 96620 words of memory out of 3000000 - 3437 multiletter control sequences out of 15000+200000 - 3640 words of font info for 14 fonts, out of 3000000 for 9000 - 841 hyphenation exceptions out of 8191 - 18i,10n,12p,250b,113s stack positions out of 5000i,500n,10000p,200000b,50000s -Output written on December_8_12.dvi (1 page, 5616 bytes). diff --git a/Other/old/final/week_10/December_8_12.tex b/Other/old/final/week_10/December_8_12.tex index f3120a42..18f65db3 100644 --- a/Other/old/final/week_10/December_8_12.tex +++ b/Other/old/final/week_10/December_8_12.tex @@ -1,471 +1,471 @@ -\global\long\def\N{\mathbb{N}} -\global\long\def\Z{\mathbb{Z}} -\global\long\def\Q{\mathbb{Q}} -\global\long\def\R{\mathbb{R}} -\global\long\def\lto{\longrightarrow} -\global\long\def\es{\emptyset} -\global\long\def\F{\mathcal{F}} -\global\long\def\force{\Vdash} -\global\long\def\dom{\textrm{dom}} -\global\long\def\em{\prec} -\global\long\def\cf{\textrm{cf}} -\global\long\def\model{\vDash} -\global\long\def\crit{\mathrm{crit}} -\global\long\def\ult{\mathrm{Ult}} -\global\long\def\inj{\hookrightarrow} -\global\long\def\u{\mathcal{U}} -\global\long\def\dprime{\prime\prime} -\global\long\def\C{\mathbb{C}} -\global\long\def\v{\mathcal{V}} -\global\long\def\w{\mathcal{W}} -\global\long\def\i{\imath} -\global\long\def\P{\mathbb{P}} -\global\long\def\del{\partial} -\global\long\def\an{\mathrm{An}} -\global\long\def\I{\mathrm{I}} -\global\long\def\L{\mathcal{L}} -\global\long\def\D{\mathcal{D}} - -\NotesBy{Notes by Asaaf Sahni} -\Week{Week 10} - -Proving, after all, that $\C\left[\left[X\right]\right]$ is faithfully -flat over $\C\left\{ X\right\} $. - -We do this by induction on the number of variables in $X$. We only -show that $\C\left[\left[X,T\right]\right]$ is flat over $\C\left\{ X,T\right\} $.\\ -Assume that $\C\left[\left[X\right]\right]$ is flat over $\C\left\{ X\right\} $, -and consider $\C\left[\left[X,T\right]\right]$ and $\C\left\{ X,T\right\} $. -\\ -Consider the equation -\[ -\left(\ast\right)_{0}\qquad f_{1}y_{1}+...+f_{n}y_{n}=0,\qquad\textrm{where }f_{i}\in\C\left\{ X,T\right\} . -\] -We can assume that all $f_{i}$'s are non zero, and regular in $T$. -\\ -Apply W.P to $f_{i}$ in $\C\left\{ X,T\right\} $, $f_{i}=u_{i}w_{i}$ -where $u_{i}\in\C\left\{ X,T\right\} ^{\times}$ and $w_{i}\in\C\left\{ X\right\} \left[T\right]$ -is monic of degree $d_{i}$. Note that if $\left(y_{1},...,y_{n}\right)$ -is a solution to $\left(\ast\right)_{0}$, then $\left(u_{1}^{-1}y_{1},...,u_{n}^{-1}y_{n}\right)$ -is a solution to -\[ -\left(\ast\right)\qquad w_{1}y_{1}+...+w_{n}y_{n}=0. -\] -So it is enough to show that each solution to $\left(\ast\right)$ -in $\C\left[\left[X,T\right]\right]$ is a linear combination of solutions -from $\C\left\{ X,T\right\} $.\\ -Consider $z_{2}=\left(w_{2},-w_{1},0,...,0\right)$, $z_{3}=\left(w_{3},0,-w_{1},0,...,0\right)$, -... $z_{n}=\left(w_{n},0,...,0,-w_{1}\right)$. Each $z_{i}$ is a -solution to $\left(\ast\right)$ (and are in $\C\left\{ X,T\right\} $).\\ -Now suppose $y=\left(y_{1},...,y_{n}\right)$ is a solution to $\left(\ast\right)$ -in $\C\left[\left[X,T\right]\right]$. \\ -For each $i$, write $y_{i}=q_{i}w_{1}+r_{i}$ where $q_{i}\in\C\left[\left[X,T\right]\right]$ -and $r_{i}\in\C\left[\left[X\right]\right]\left[T\right]$ of degree -$0}$. -\end{itemize} -Let $\mathrm{An}\left(\u\right)$ be all analytic functions on $\u$. -$\an\left(\u\right)$ has an $\R$-algebra structure. If $f\in\an\left(\u\right)$, -then $\frac{\del f}{\del X_{i}}\in\an\left(\u\right).$ - -For each $a\in\u$ there is a map of $\R$-algebras: -\begin{eqnarray*} -\an\left(\u\right) & \lto & \R\left\{ X\right\} .\\ -f & \mapsto & f_{a} -\end{eqnarray*} - -\begin{prop*} -\label{11.2} (Analytic continuation) If $\u$ is connected, then the map -above is injective. In particular, $\an\left(\u\right)$ is an integral -domain.\end{prop*} -\begin{cor*} -\label{11.3} If $\u$ is connected, $f,g\in\an\left(\u\right)$ agree on -a non empty open subset of $\u$, then $f=g$.\end{cor*} -\begin{prop*} -\label{11.4} Let $f_{1},...,f_{n}\in\an\left(\u\right)$, $\v\subset\R^{n}$ -open s.t $f\left(\u\right)\subset V$, where $f=\left(f_{1},...,f_{n}\right)\colon\u\lto\R^{n}$. - -Then for any $g\in\an\left(\v\right)$ we have $g\circ f\in\an\left(\u\right)$ -and -\[ -\left(g\circ f\right)_{a}=g_{f\left(a\right)}\left(f_{a}-f\left(a\right)\right). -\] - -\end{prop*} -\label{11.5} \uline{Notation}: For $x=\left(x_{1},...,x_{m}\right)\in\R^{m}$, -$\left|x\right|=\max\left\{ \left|x\right|_{1},...,\left|x_{m}\right|\right\} $. -\\ -For $\delta>0$, let $\delta=\left(\delta,...,\delta\right)$ and -$\R\left\{ X\right\} _{\delta^{+}}=\bigcup_{r>\delta}\R\left\{ X\right\} _{r}$.\\ -Each $f\in\R\left\{ X\right\} _{\delta^{+}}$ gives rise to a function -$x\mapsto f\left(x\right)\colon\bar{B}_{\delta}\left(0\right)\lto\R$ -which extendes to an analytic function on an open nbhd of $\bar{B}_{\delta}$. - -Let $Y=\left(Y_{1},...,Y_{n}\right)$, $n\geq1$. Fix $f=f\left(X,Y\right)\in\R\left\{ X,Y\right\} $. -Then for small $x$ we have $f\left(x,Y\right)\in\R\left\{ Y\right\} $.\\ -Consider the following question: How does W.P for $f\left(x,Y\right)$ -depend on $x$, for small $x$?\\ -We will show that for some $\epsilon>0$, $\bar{B}_{\epsilon}\subset\R^{m}$ -can be covered by finitely many ``special sets'', on each of which -W.P is uniform in $x$. -\begin{defn*} -\label{11.6} A special subset of $\bar{B}_{\epsilon}$ is a finite union -of sets of the following form -\[ -\left\{ x\in\bar{B}_{\epsilon};\, f\left(x\right)=0,\, g_{1}\left(x\right)>0,\,...,\, g_{k}\left(x\right)>0\right\} , -\] -where $f,g_{1},...,g_{k}\in\R\left\{ X\right\} _{\epsilon^{+}}$. -\end{defn*} -We first show that $\mathrm{ord}\left(f\left(x,Y\right)\right)$ takes -only finitely many values as $x$ ranges over a neighbourhood of $0\in\R^{m}$. -Write -\[ -f\left(X,Y\right)=\sum_{j}f_{j}\left(X\right)Y^{j}, -\] -where $f_{j}\left(X\right)\in\R\left\{ X\right\} $. Since $\R\left\{ X\right\} $ -is neotherian, the ideal generated by $\left\{ f_{j}\left(X\right)\right\} _{j\in\N^{n}}$ -is finitely generated, hence is generated by $\left\{ f_{j}\left(X\right)\right\} _{\left|j\right|d$, -\[ -f_{j}\left(X\right)=\sum_{\left|i\right|\leq d}g_{ij}f_{i}. -\] -In the following, $i$ ranges over elements of $\N^{n}$ s.t $\left|i\right|\leq d$ -and $j$ over elements of $\N^{n}$ s.t $\left|j\right|>d$. \\ -Substituting the above, we get -\[ -\left(1\right)\qquad f\left(X,Y\right)=\sum_{i}f_{i}\left(X,Y\right)\left(Y^{i}+\sum_{j}g_{ij}\left(X\right)Y^{j}\right), -\] -in $\R\left[\left[X,Y\right]\right]$. Note that for each $i,j$, -$g_{ij}\in\R\left\{ X\right\} $, and so there is some $\delta$ s.t -$g_{ij}\in\R\left\{ X\right\} _{\delta}$. However, in order to have -equation $\left(1\right)$ as an equality in $\R\left\{ X\right\} $, -we need to find a uniform $\delta$ that works for all $i,j$. -\begin{claim*} -$\exists\delta\in\left(0,1\right]$ and there are $f_{i},\, g_{ij}\in\R\left\{ X\right\} _{\delta^{+}}$ -(maybe different than above) such that $\sum_{j}g_{ij}\left(X\right)Y^{j}\in\R\left\{ X,Y\right\} _{\delta^{+}}$ -and $\left(1\right)$ holds in $\R\left\{ X\right\} _{\delta^{+}}$ -with $f_{i},g_{ij}$.\end{claim*} -\begin{proof} -Consider the linear equation -\[ -f=\sum_{i}f_{i}\left(Y^{i}+\sum_{\left|j\right|=d+1}Z_{ij}Y^{j}\right). -\] -By $\left(1\right)$ above, there is a solution $\left\{ Z_{ij};\,\left|i\right|\leq d,\,\left|j\right|=d+1\right\} $ -in $\R\left[\left[X,Y\right]\right]$ (all the higher terms, $Y^{j}$ -for $\left|j\right|>d+1$, are inside the $Z_{ij}$'s). Since $\R\left[\left[X,Y\right]\right]$ -is f.f. over $\R\left\{ X,Y\right\} $, there is a solution $\left\{ Z_{ij}\right\} $ -in $\R\left\{ X,Y\right\} $. Take $\delta$ for all $i,j$, $Z_{ij}\in\R\left\{ X,Y\right\} _{\delta^{+}}$. -Now write $Z_{ij}$ as power series in $Y$ with coefficients in $\R\left\{ X\right\} $ -to get $\left(1\right)$ with $g_{ij}\left(X\right)\in\R\left\{ X\right\} _{\delta^{+}}$ -for all $\left|i\right|\leq d$, $\left|j\right|>d$. -\end{proof} -We work with equation $\left(1\right)$ in $\R\left\{ X,Y\right\} _{\delta^{+}}$ -as given by the claim above. For $\left|x\right|\leq\delta$, -\[ -f\left(x,Y\right)=0\iff\forall i\left(f_{i}\left(x\right)=0\right). -\] -Also, if $f_{i}\left(x\right)\neq0$, then $\mathrm{ord}\left(f\left(x,Y\right)\right)\leq\left|i\right|$. -So -\[ -f\left(x,Y\right)\neq0\implies\mathrm{ord}\left(f\left(x,Y\right)\right)\leq d. -\] -Define -\begin{eqnarray*} -Z_{\delta} & = & \left\{ x\in\bar{B}_{\delta};\,\forall i\, f_{i}\left(x\right)=0\right\} .\\ -S_{i} & = & \left\{ x\in\bar{B}_{\delta};\, f_{i}\left(x\right)\neq0\wedge\forall i'\neq i\left(\left|f_{i}\left(x\right)\right|\geq\left|f_{i'}\left(x\right)\right|\right)\right\} . -\end{eqnarray*} -Note that $\bar{B}_{\delta}=Z_{\delta}\cup\bigcup_{i}S_{i}$.\\ -Fix some $i$, formally divide the expression for $f$ in $\left(1\right)$ -by $f_{i}$, and introduce new variables $V_{i,i'}$ for the quotients -$\nicefrac{f_{i}}{f_{i'}}$, for $i\neq i'$. Let $V_{i}=\left(V_{ii'}\right)_{i'\neq i}$. -Define -\[ -F_{i}=Y^{i}+\sum_{j}g_{ij}Y^{j}+\sum_{i'\neq i}V_{ii'}\left(Y^{i'}+\sum_{j}g_{i'j}Y^{j}\right)\in\R\left\{ X,V_{i},Y\right\} . -\] -For $x\in S_{i}$, let $v_{i}\left(x\right)=\left(\nicefrac{f_{i'}\left(x\right)}{f_{i}\left(x\right)}\right)_{i'\neq i}$. -Then for $x\in S_{i}$, $\left|v_{i}\left(x\right)\right|\leq1$, -and -\[ -\left(2\right)\qquad f\left(x,Y\right)=f_{i}\left(x\right)F_{i}\left(x,v_{i}\left(x\right),Y\right). -\] - - -Idea: apply W.P to $F_{i}$'s locally around every point $X=0$, $V_{i}=c$, -$Y=0$. \\ -For $c=\left(c_{i'}\right)_{i'\neq i}$ with $\left|c\right|\leq1$, -put -\[ -F_{i,c}=F_{i}\left(X,c+V_{i},Y\right). -\] -Then for $x\in S_{i}$, -\[ -f\left(x,Y\right)=f_{i}\left(x\right)F_{i,c}\left(x,v_{i}\left(x\right)-c,Y\right). -\] - -\begin{claim*} -For each $c$ there is $\lambda=\lambda\left(c\right)\in\R^{n-1}$ -with $\left|\lambda\right|\leq1$ such that $F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)$ -is regular in $Y_{n}$ of order$\leq\left|i\right|$, where for $Y=\left(Y_{1},...,Y_{n}\right)$, -$\lambda\left(Y\right)=\left(Y_{1}+\lambda Y_{n},...,Y_{n-1}+\lambda_{n-1}Y_{n},Y_{n}\right)$.\end{claim*} -\begin{proof} -Exercise. -\end{proof} -By W.P., for such $\lambda$, -\[ -\left(3\right)\qquad F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)=\u_{i,c}\w_{i,c},\quad\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} ^{\ast},\:\w_{i,c}\in\R\left\{ X,V_{i},Y_{1},...,Y_{n-1}\right\} \left[Y_{n}\right]. -\] -Take $\epsilon\left(i,c\right)\in\left(0,\delta\right]$ s.t -\begin{itemize} -\item $\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}^{\ast}$. -\item $\w_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}.$ -\end{itemize} -Let $\Gamma\left(i\right)=\left\{ i';\, i'\neq i\right\} $. Note -that our $c$'s vary over $I^{\Gamma\left(i\right)}$, where $\I=\left[-1,1\right]$. -By compactness, there is a finite set $C\left(i\right)$ s.t -\[ -\I^{\Gamma\left(i\right)}\subset\bigcup_{c\in C\left(i\right)}B_{\epsilon\left(i,c\right)}\left(c\right). -\] -Now consider the finite set $\Gamma=\left\{ \left(i,c\right);\,\left|i\right|\leq d,\, d\in C\left(i\right)\right\} $.\\ -Take $\epsilon>0$ s.t $\epsilon\leq\frac{\epsilon\left(i,c\right)}{4}$ -for all $\left(i,c\right)\in\Gamma.$ For $\gamma=\left(i,c\right)\in\Gamma$, -let -\[ -S_{\gamma}=\left\{ x\in S_{i};\,\left|x\right|\leq\epsilon,\,\left|v_{i}\left(x\right)-c\right|<\epsilon\left(i,c\right)\right\} . -\] -Then $S_{i}\cap\bar{B}_{\epsilon}=\bigcup\left\{ S_{\gamma};\,\gamma=\left(i,c\right),\, c\in C\left(i\right)\right\} $. -\\ -So $\bar{B}_{\epsilon}=\left(\underbrace{Z_{\delta}\cap\bar{B}_{\epsilon}}_{\equiv Z}\right)\cup\bigcup_{\gamma}S_{\gamma}$.\\ -For $\gamma=\left(i,c\right)\in\Gamma$, by $\left(2\right)$ and -$\left(3\right)$, -\[ -f\left(x,\lambda\left(Y\right)\right)=f_{i}\left(x\right)\u_{\gamma}\left(x,v_{i}\left(x\right),Y\right)W_{\gamma}\left(x,v_{i}\left(x\right),Y\right). -\] -Note: $\u_{\gamma}$ does not change sign on $\bar{B}_{\epsilon\left(\gamma\right)}$. -Therefore there is $\sigma\left(\gamma\right)\in\left\{ \pm1\right\} $ -s.t -\[ -\mathrm{sign}\left(f\left(x,\lambda\left(y\right)\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),y\right)\right) -\] -for $x\in S_{\gamma}$ and $\left|y\right|\leq2\epsilon$. Note that -$f\left(x,\lambda\left(y\right)\right)$ is defined since $\left|\lambda\right|\leq1$, -so $\left|\lambda\left(y\right)\right|\leq4\epsilon\leq\delta$.\\ -For $\left|y\right|\leq\epsilon$, $\left|\left(-\lambda\right)\left(y\right)\right|\leq2\epsilon$. -Also, $\lambda\left(\left(-\lambda\right)\left(y\right)\right)=y$. -Thus for $\left|y\right|\leq\epsilon$ -\[ -\mathrm{sign}\left(f\left(x,y\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),\left(-\lambda\right)\left(y\right)\right)\right). -\] -Let $Z=\left(Z_{1},...,Z_{n}\right)$ be new variables and -\[ -\hat{\w}_{\gamma}\left(X,V_{i},Z\right)=f_{i}\left(\epsilon X\right)\w_{\gamma}\left(\epsilon X,\epsilon\left(\gamma\right)V_{\gamma},2\epsilon Z\right)\in\R\left\{ X,V_{i},Z\right\} _{1^{+}}. -\] -Then if $\left(x,y\right)\in I^{m+n}$ and $\epsilon x\in S_{\gamma}$, -then -\[ -\mathrm{sign}f\left(x,y\right)=\sigma\left(\gamma\right)\hat{\w}_{\gamma}\left(x,\nicefrac{v_{i}\left(\epsilon x\right)}{\epsilon\left(\gamma\right)},\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right). -\] - -\section{Quantifier elimination for $\R_{an}$.} - -Recall that $\L_{an}$ is the language $\left\{ 0,1,+,-,\cdot,\leq\right\} $ -of ordered rings augmented by a function symbol for every restricted -analytic function $\R^{m}\lto\R$. \\ -Let $\R_{an}$ be the ordered field of reals as an $\L_{an}$-structure -together with the map $^{-1}\colon\R\lto\R$ defined as $x^{-1}=\begin{cases} -\frac{1}{x} & x\neq0\\ -x=0 & 0 -\end{cases}$. Let $\L_{an}\left(^{-1}\right)=\L_{an}\cup\left\{ ^{-1}\right\} $. -\begin{thm*} -\label{12.1} (Denef-v. d. Dries 1988) $\left(\R_{an},\,^{-1}\right)$ has -quantifier elimination. -\end{thm*} -Call $f\colon\I^{m}\lto\R$ analytic if it extends to an analytic -function on an open nbhd of $\I^{m}$. Let $\L_{a}$ be the language -$\left\{ <\right\} $ expanded by $m$-ary function symbols for each -analytic $f\colon\I^{m}\lto\R$ with $f\left(\I^{m}\right)\subset\I$.\\ -We consider $\I$ as an $\L_{a}$-structure.\\ -Define $\D\colon\I^{2}\lto\I$ by -\[ -D\left(x,y\right)=\begin{cases} -\nicefrac{x}{y} & \textrm{if }\left|x\right|\leq\left|y\right|\textrm{ and }y\neq0,\\ -0 & \textrm{otherwise.} -\end{cases} -\] -Let $\L_{a,\D}=\L_{a}\cup\left\{ \D\right\} $. -\begin{thm*} -\label{12.2} The $\L_{a,\D}$-structure $\I$ has $q.e.$ -\end{thm*} -Some general logical considerations:\\ -Let $T$ be an $\L$-theory, and $T'$ be a definitional expansion -of $T$ to an $\L'$-theory. Assume further that $\L'\setminus\L$ -consists only of function symbols, and for each $f\in\L'\setminus\L$ -there is an existential $\L$-formula $\delta_{f}\left(x,y\right)$ -s.t -\[ -T'\vdash f\left(x\right)=y\longleftrightarrow\delta_{f}\left(x,y\right). -\] -(e.g. $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$.) -\begin{lem*} -\label{12.6} Every $\exists\L'$-formula is $T'$-equivalent to an $\exists\L$-formula. -\begin{lem*} -\label{12.7} Suppose that for each q.f. $\L$-formula $\varphi\left(x,y_{1},...,y_{n}\right)$, -$n\geq1$, there is a q.f. $\L'$-formula $\varphi'$$\left(x,z_{1},...,z_{n-1}\right)$ -such that -\begin{enumerate} -\item $T'\vdash\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\varphi'\left(x,z\right)$. -\item The function symbols in $\L'\setminus\L$ are only applied in $\varphi'$ -to terms involving only $x$. -\end{enumerate} -\end{lem*} -Then $T'$ has $q.e$. (So by \eqref{12.6}, $T$ is model -complete). -\end{lem*} -We now verify that the conditions in lemma \eqref{12.7} are -satisfied for $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$. -\begin{lem*} -\label{12.8} Basic Lemma. - -Let $\varphi\left(x,y\right)=\varphi\left(x_{1},...,x_{m},y_{1},...,y_{n}\right)$, -$n\geq1$ be a q.f. $\L_{a}$-formula. Then there is a q.f. $\L_{a,\D}$-formula -$\hat{\varphi}\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$ -such that -\begin{enumerate} -\item $\I\model\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\hat{\varphi}\left(x,z\right)$. -\item In $\hat{\varphi}$, $\D$ is only applied to terms not involving -$z$, and $z_{n}$ only appears polynomially in $\hat{\varphi}$. -\end{enumerate} -\end{lem*} -Given the Basic Lemma, we can use Tarsky's quantifier elimination -for $\R$ (as an ordered ring) to eliminate $\exists z_{n}$ and produce -$\varphi'$ satisfying the hypothesis of \eqref{12.7}. - -For an $\L_{a}$-formula and $\left(a,b\right)\in\I^{m+n}$, $\eta>0$, -consider the formula -\[ -\varphi_{a,b,\eta}=\varphi\wedge``\left|\left(x,y\right)-\left(a,b\right)\right|<\eta", -\] -where $``"$ means the formal sentence for that phrase. By compactness -of $I^{m+n}$, it is enough to show that for any $\left(a,b\right)$, -there is $\eta>0$ such that the Basic Lemma holds for the formula -$\varphi_{a,b,\eta}$. Using an affine transformation, it is enough -to consider only $\left(a,b\right)=\left(0,0\right)$. Let $\varphi_{\eta}=\varphi_{0,0,\eta}$. -Thus the lemma is reduced to the following: -\begin{lem*} -\label{12.9} Local Basic Lemma. - -Let $\varphi\left(x,y\right)$ be a q.f. $\L_{a}$-formula . Then -there is $\epsilon\in\left(0,1\right)$ and a q.f. $\L_{a,\D}$ formula -$\varphi'\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$, s.t -\begin{enumerate} -\item $\I\model\exists y\varphi_{\epsilon}\iff\exists z\left(\varphi_{\epsilon}^{\prime}\right)$. -\item In $\varphi'$, the function symbol $\D$ is applied only to terms -not involving $z$, and $z_{n}$ appears only polynomially. -\end{enumerate} -\end{lem*} -\begin{proof} -We may assume that all atomic subformulae of $\varphi$ are of the -form $f\left(x,y\right)>0$ or $f\left(x,y\right)=0$ for some analytic -$f\colon\I^{m}\lto\I$.\\ -Apply uniform W.P. to the Taylor series at $0$ of all such $f$. -We get \end{proof} -\begin{itemize} -\item $\epsilon\in\left(0,1\right)$, -\item Finite cover $S_{\gamma}$ of $B_{\epsilon}\subset\R^{m}$ by special -sets. -\item For each $\gamma$, a $\lambda=\lambda\left(\gamma\right)\in\R^{n-1}$ -with $\left|\lambda\right|\leq1$. -\item For each $f$ in $\varphi$, an $\L_{a,\D}$-term $t_{\gamma,f}\left(x,z\right)$; -\item $t_{f,\gamma}$ is polynomial in $z_{n}$, -\item $\D$ is only applied to terms involving only $x$ within $t_{f,\gamma}$. -\item $\forall x\in I^{m+n}$ with $\epsilon x\in S_{\gamma}$, -\[ -f\left(\epsilon x,\epsilon y\right)\geq0\iff t_{f,\gamma}\left(x,\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right)\geq0. -\] - -\end{itemize} -Replace $f\left(x,y\right)$ by $t_{f,\gamma}$ in $\varphi$ to get -$\varphi_{\gamma}$. Then -\[ -\I\model\exists y\varphi\left(\epsilon x,\epsilon y\right)\longleftrightarrow\exists z\left(\bigvee_{\gamma}x\in S_{\gamma}\wedge\varphi_{\gamma}\left(x,z\right)\wedge\left|\lambda\left(z\right)\right|\leq\frac{1}{2}\right). -\] -Finally we can convert this to a formula $\varphi'$ satisfying $\left(1\right)$ +\global\long\def\N{\mathbb{N}} +\global\long\def\Z{\mathbb{Z}} +\global\long\def\Q{\mathbb{Q}} +\global\long\def\R{\mathbb{R}} +\global\long\def\lto{\longrightarrow} +\global\long\def\es{\emptyset} +\global\long\def\F{\mathcal{F}} +\global\long\def\force{\Vdash} +\global\long\def\dom{\textrm{dom}} +\global\long\def\em{\prec} +\global\long\def\cf{\textrm{cf}} +\global\long\def\model{\vDash} +\global\long\def\crit{\mathrm{crit}} +\global\long\def\ult{\mathrm{Ult}} +\global\long\def\inj{\hookrightarrow} +\global\long\def\u{\mathcal{U}} +\global\long\def\dprime{\prime\prime} +\global\long\def\C{\mathbb{C}} +\global\long\def\v{\mathcal{V}} +\global\long\def\w{\mathcal{W}} +\global\long\def\i{\imath} +\global\long\def\P{\mathbb{P}} +\global\long\def\del{\partial} +\global\long\def\an{\mathrm{An}} +\global\long\def\I{\mathrm{I}} +\global\long\def\L{\mathcal{L}} +\global\long\def\D{\mathcal{D}} + +\NotesBy{Notes by Asaaf Sahni} +\Week{Week 10} + +Proving, after all, that $\C\left[\left[X\right]\right]$ is faithfully +flat over $\C\left\{ X\right\} $. + +We do this by induction on the number of variables in $X$. We only +show that $\C\left[\left[X,T\right]\right]$ is flat over $\C\left\{ X,T\right\} $.\\ +Assume that $\C\left[\left[X\right]\right]$ is flat over $\C\left\{ X\right\} $, +and consider $\C\left[\left[X,T\right]\right]$ and $\C\left\{ X,T\right\} $. +\\ +Consider the equation +\[ +\left(\ast\right)_{0}\qquad f_{1}y_{1}+...+f_{n}y_{n}=0,\qquad\textrm{where }f_{i}\in\C\left\{ X,T\right\} . +\] +We can assume that all $f_{i}$'s are non zero, and regular in $T$. +\\ +Apply W.P to $f_{i}$ in $\C\left\{ X,T\right\} $, $f_{i}=u_{i}w_{i}$ +where $u_{i}\in\C\left\{ X,T\right\} ^{\times}$ and $w_{i}\in\C\left\{ X\right\} \left[T\right]$ +is monic of degree $d_{i}$. Note that if $\left(y_{1},...,y_{n}\right)$ +is a solution to $\left(\ast\right)_{0}$, then $\left(u_{1}^{-1}y_{1},...,u_{n}^{-1}y_{n}\right)$ +is a solution to +\[ +\left(\ast\right)\qquad w_{1}y_{1}+...+w_{n}y_{n}=0. +\] +So it is enough to show that each solution to $\left(\ast\right)$ +in $\C\left[\left[X,T\right]\right]$ is a linear combination of solutions +from $\C\left\{ X,T\right\} $.\\ +Consider $z_{2}=\left(w_{2},-w_{1},0,...,0\right)$, $z_{3}=\left(w_{3},0,-w_{1},0,...,0\right)$, +... $z_{n}=\left(w_{n},0,...,0,-w_{1}\right)$. Each $z_{i}$ is a +solution to $\left(\ast\right)$ (and are in $\C\left\{ X,T\right\} $).\\ +Now suppose $y=\left(y_{1},...,y_{n}\right)$ is a solution to $\left(\ast\right)$ +in $\C\left[\left[X,T\right]\right]$. \\ +For each $i$, write $y_{i}=q_{i}w_{1}+r_{i}$ where $q_{i}\in\C\left[\left[X,T\right]\right]$ +and $r_{i}\in\C\left[\left[X\right]\right]\left[T\right]$ of degree +$0}$. +\end{itemize} +Let $\mathrm{An}\left(\u\right)$ be all analytic functions on $\u$. +$\an\left(\u\right)$ has an $\R$-algebra structure. If $f\in\an\left(\u\right)$, +then $\frac{\del f}{\del X_{i}}\in\an\left(\u\right).$ + +For each $a\in\u$ there is a map of $\R$-algebras: +\begin{eqnarray*} +\an\left(\u\right) & \lto & \R\left\{ X\right\} .\\ +f & \mapsto & f_{a} +\end{eqnarray*} + +\begin{prop*} +\label{11.2} (Analytic continuation) If $\u$ is connected, then the map +above is injective. In particular, $\an\left(\u\right)$ is an integral +domain.\end{prop*} +\begin{cor*} +\label{11.3} If $\u$ is connected, $f,g\in\an\left(\u\right)$ agree on +a non empty open subset of $\u$, then $f=g$.\end{cor*} +\begin{prop*} +\label{11.4} Let $f_{1},...,f_{n}\in\an\left(\u\right)$, $\v\subset\R^{n}$ +open s.t $f\left(\u\right)\subset V$, where $f=\left(f_{1},...,f_{n}\right)\colon\u\lto\R^{n}$. + +Then for any $g\in\an\left(\v\right)$ we have $g\circ f\in\an\left(\u\right)$ +and +\[ +\left(g\circ f\right)_{a}=g_{f\left(a\right)}\left(f_{a}-f\left(a\right)\right). +\] + +\end{prop*} +\label{11.5} \uline{Notation}: For $x=\left(x_{1},...,x_{m}\right)\in\R^{m}$, +$\left|x\right|=\max\left\{ \left|x\right|_{1},...,\left|x_{m}\right|\right\} $. +\\ +For $\delta>0$, let $\delta=\left(\delta,...,\delta\right)$ and +$\R\left\{ X\right\} _{\delta^{+}}=\bigcup_{r>\delta}\R\left\{ X\right\} _{r}$.\\ +Each $f\in\R\left\{ X\right\} _{\delta^{+}}$ gives rise to a function +$x\mapsto f\left(x\right)\colon\bar{B}_{\delta}\left(0\right)\lto\R$ +which extendes to an analytic function on an open nbhd of $\bar{B}_{\delta}$. + +Let $Y=\left(Y_{1},...,Y_{n}\right)$, $n\geq1$. Fix $f=f\left(X,Y\right)\in\R\left\{ X,Y\right\} $. +Then for small $x$ we have $f\left(x,Y\right)\in\R\left\{ Y\right\} $.\\ +Consider the following question: How does W.P for $f\left(x,Y\right)$ +depend on $x$, for small $x$?\\ +We will show that for some $\epsilon>0$, $\bar{B}_{\epsilon}\subset\R^{m}$ +can be covered by finitely many ``special sets'', on each of which +W.P is uniform in $x$. +\begin{defn*} +\label{11.6} A special subset of $\bar{B}_{\epsilon}$ is a finite union +of sets of the following form +\[ +\left\{ x\in\bar{B}_{\epsilon};\, f\left(x\right)=0,\, g_{1}\left(x\right)>0,\,...,\, g_{k}\left(x\right)>0\right\} , +\] +where $f,g_{1},...,g_{k}\in\R\left\{ X\right\} _{\epsilon^{+}}$. +\end{defn*} +We first show that $\mathrm{ord}\left(f\left(x,Y\right)\right)$ takes +only finitely many values as $x$ ranges over a neighbourhood of $0\in\R^{m}$. +Write +\[ +f\left(X,Y\right)=\sum_{j}f_{j}\left(X\right)Y^{j}, +\] +where $f_{j}\left(X\right)\in\R\left\{ X\right\} $. Since $\R\left\{ X\right\} $ +is neotherian, the ideal generated by $\left\{ f_{j}\left(X\right)\right\} _{j\in\N^{n}}$ +is finitely generated, hence is generated by $\left\{ f_{j}\left(X\right)\right\} _{\left|j\right|d$, +\[ +f_{j}\left(X\right)=\sum_{\left|i\right|\leq d}g_{ij}f_{i}. +\] +In the following, $i$ ranges over elements of $\N^{n}$ s.t $\left|i\right|\leq d$ +and $j$ over elements of $\N^{n}$ s.t $\left|j\right|>d$. \\ +Substituting the above, we get +\[ +\left(1\right)\qquad f\left(X,Y\right)=\sum_{i}f_{i}\left(X,Y\right)\left(Y^{i}+\sum_{j}g_{ij}\left(X\right)Y^{j}\right), +\] +in $\R\left[\left[X,Y\right]\right]$. Note that for each $i,j$, +$g_{ij}\in\R\left\{ X\right\} $, and so there is some $\delta$ s.t +$g_{ij}\in\R\left\{ X\right\} _{\delta}$. However, in order to have +equation $\left(1\right)$ as an equality in $\R\left\{ X\right\} $, +we need to find a uniform $\delta$ that works for all $i,j$. +\begin{claim*} +$\exists\delta\in\left(0,1\right]$ and there are $f_{i},\, g_{ij}\in\R\left\{ X\right\} _{\delta^{+}}$ +(maybe different than above) such that $\sum_{j}g_{ij}\left(X\right)Y^{j}\in\R\left\{ X,Y\right\} _{\delta^{+}}$ +and $\left(1\right)$ holds in $\R\left\{ X\right\} _{\delta^{+}}$ +with $f_{i},g_{ij}$.\end{claim*} +\begin{proof} +Consider the linear equation +\[ +f=\sum_{i}f_{i}\left(Y^{i}+\sum_{\left|j\right|=d+1}Z_{ij}Y^{j}\right). +\] +By $\left(1\right)$ above, there is a solution $\left\{ Z_{ij};\,\left|i\right|\leq d,\,\left|j\right|=d+1\right\} $ +in $\R\left[\left[X,Y\right]\right]$ (all the higher terms, $Y^{j}$ +for $\left|j\right|>d+1$, are inside the $Z_{ij}$'s). Since $\R\left[\left[X,Y\right]\right]$ +is f.f. over $\R\left\{ X,Y\right\} $, there is a solution $\left\{ Z_{ij}\right\} $ +in $\R\left\{ X,Y\right\} $. Take $\delta$ for all $i,j$, $Z_{ij}\in\R\left\{ X,Y\right\} _{\delta^{+}}$. +Now write $Z_{ij}$ as power series in $Y$ with coefficients in $\R\left\{ X\right\} $ +to get $\left(1\right)$ with $g_{ij}\left(X\right)\in\R\left\{ X\right\} _{\delta^{+}}$ +for all $\left|i\right|\leq d$, $\left|j\right|>d$. +\end{proof} +We work with equation $\left(1\right)$ in $\R\left\{ X,Y\right\} _{\delta^{+}}$ +as given by the claim above. For $\left|x\right|\leq\delta$, +\[ +f\left(x,Y\right)=0\iff\forall i\left(f_{i}\left(x\right)=0\right). +\] +Also, if $f_{i}\left(x\right)\neq0$, then $\mathrm{ord}\left(f\left(x,Y\right)\right)\leq\left|i\right|$. +So +\[ +f\left(x,Y\right)\neq0\implies\mathrm{ord}\left(f\left(x,Y\right)\right)\leq d. +\] +Define +\begin{eqnarray*} +Z_{\delta} & = & \left\{ x\in\bar{B}_{\delta};\,\forall i\, f_{i}\left(x\right)=0\right\} .\\ +S_{i} & = & \left\{ x\in\bar{B}_{\delta};\, f_{i}\left(x\right)\neq0\wedge\forall i'\neq i\left(\left|f_{i}\left(x\right)\right|\geq\left|f_{i'}\left(x\right)\right|\right)\right\} . +\end{eqnarray*} +Note that $\bar{B}_{\delta}=Z_{\delta}\cup\bigcup_{i}S_{i}$.\\ +Fix some $i$, formally divide the expression for $f$ in $\left(1\right)$ +by $f_{i}$, and introduce new variables $V_{i,i'}$ for the quotients +$\nicefrac{f_{i}}{f_{i'}}$, for $i\neq i'$. Let $V_{i}=\left(V_{ii'}\right)_{i'\neq i}$. +Define +\[ +F_{i}=Y^{i}+\sum_{j}g_{ij}Y^{j}+\sum_{i'\neq i}V_{ii'}\left(Y^{i'}+\sum_{j}g_{i'j}Y^{j}\right)\in\R\left\{ X,V_{i},Y\right\} . +\] +For $x\in S_{i}$, let $v_{i}\left(x\right)=\left(\nicefrac{f_{i'}\left(x\right)}{f_{i}\left(x\right)}\right)_{i'\neq i}$. +Then for $x\in S_{i}$, $\left|v_{i}\left(x\right)\right|\leq1$, +and +\[ +\left(2\right)\qquad f\left(x,Y\right)=f_{i}\left(x\right)F_{i}\left(x,v_{i}\left(x\right),Y\right). +\] + + +Idea: apply W.P to $F_{i}$'s locally around every point $X=0$, $V_{i}=c$, +$Y=0$. \\ +For $c=\left(c_{i'}\right)_{i'\neq i}$ with $\left|c\right|\leq1$, +put +\[ +F_{i,c}=F_{i}\left(X,c+V_{i},Y\right). +\] +Then for $x\in S_{i}$, +\[ +f\left(x,Y\right)=f_{i}\left(x\right)F_{i,c}\left(x,v_{i}\left(x\right)-c,Y\right). +\] + +\begin{claim*} +For each $c$ there is $\lambda=\lambda\left(c\right)\in\R^{n-1}$ +with $\left|\lambda\right|\leq1$ such that $F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)$ +is regular in $Y_{n}$ of order$\leq\left|i\right|$, where for $Y=\left(Y_{1},...,Y_{n}\right)$, +$\lambda\left(Y\right)=\left(Y_{1}+\lambda Y_{n},...,Y_{n-1}+\lambda_{n-1}Y_{n},Y_{n}\right)$.\end{claim*} +\begin{proof} +Exercise. +\end{proof} +By W.P., for such $\lambda$, +\[ +\left(3\right)\qquad F_{i,c}\left(X,V_{i},\lambda\left(Y\right)\right)=\u_{i,c}\w_{i,c},\quad\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} ^{\ast},\:\w_{i,c}\in\R\left\{ X,V_{i},Y_{1},...,Y_{n-1}\right\} \left[Y_{n}\right]. +\] +Take $\epsilon\left(i,c\right)\in\left(0,\delta\right]$ s.t +\begin{itemize} +\item $\u_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}^{\ast}$. +\item $\w_{i,c}\in\R\left\{ X,V_{i},Y\right\} _{\epsilon\left(i,c\right)^{+}}.$ +\end{itemize} +Let $\Gamma\left(i\right)=\left\{ i';\, i'\neq i\right\} $. Note +that our $c$'s vary over $I^{\Gamma\left(i\right)}$, where $\I=\left[-1,1\right]$. +By compactness, there is a finite set $C\left(i\right)$ s.t +\[ +\I^{\Gamma\left(i\right)}\subset\bigcup_{c\in C\left(i\right)}B_{\epsilon\left(i,c\right)}\left(c\right). +\] +Now consider the finite set $\Gamma=\left\{ \left(i,c\right);\,\left|i\right|\leq d,\, d\in C\left(i\right)\right\} $.\\ +Take $\epsilon>0$ s.t $\epsilon\leq\frac{\epsilon\left(i,c\right)}{4}$ +for all $\left(i,c\right)\in\Gamma.$ For $\gamma=\left(i,c\right)\in\Gamma$, +let +\[ +S_{\gamma}=\left\{ x\in S_{i};\,\left|x\right|\leq\epsilon,\,\left|v_{i}\left(x\right)-c\right|<\epsilon\left(i,c\right)\right\} . +\] +Then $S_{i}\cap\bar{B}_{\epsilon}=\bigcup\left\{ S_{\gamma};\,\gamma=\left(i,c\right),\, c\in C\left(i\right)\right\} $. +\\ +So $\bar{B}_{\epsilon}=\left(\underbrace{Z_{\delta}\cap\bar{B}_{\epsilon}}_{\equiv Z}\right)\cup\bigcup_{\gamma}S_{\gamma}$.\\ +For $\gamma=\left(i,c\right)\in\Gamma$, by $\left(2\right)$ and +$\left(3\right)$, +\[ +f\left(x,\lambda\left(Y\right)\right)=f_{i}\left(x\right)\u_{\gamma}\left(x,v_{i}\left(x\right),Y\right)W_{\gamma}\left(x,v_{i}\left(x\right),Y\right). +\] +Note: $\u_{\gamma}$ does not change sign on $\bar{B}_{\epsilon\left(\gamma\right)}$. +Therefore there is $\sigma\left(\gamma\right)\in\left\{ \pm1\right\} $ +s.t +\[ +\mathrm{sign}\left(f\left(x,\lambda\left(y\right)\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),y\right)\right) +\] +for $x\in S_{\gamma}$ and $\left|y\right|\leq2\epsilon$. Note that +$f\left(x,\lambda\left(y\right)\right)$ is defined since $\left|\lambda\right|\leq1$, +so $\left|\lambda\left(y\right)\right|\leq4\epsilon\leq\delta$.\\ +For $\left|y\right|\leq\epsilon$, $\left|\left(-\lambda\right)\left(y\right)\right|\leq2\epsilon$. +Also, $\lambda\left(\left(-\lambda\right)\left(y\right)\right)=y$. +Thus for $\left|y\right|\leq\epsilon$ +\[ +\mathrm{sign}\left(f\left(x,y\right)\right)=\sigma\left(\gamma\right)\mathrm{sign}\left(f_{i}\left(x\right)\w_{\gamma}\left(x,v_{i}\left(x\right),\left(-\lambda\right)\left(y\right)\right)\right). +\] +Let $Z=\left(Z_{1},...,Z_{n}\right)$ be new variables and +\[ +\hat{\w}_{\gamma}\left(X,V_{i},Z\right)=f_{i}\left(\epsilon X\right)\w_{\gamma}\left(\epsilon X,\epsilon\left(\gamma\right)V_{\gamma},2\epsilon Z\right)\in\R\left\{ X,V_{i},Z\right\} _{1^{+}}. +\] +Then if $\left(x,y\right)\in I^{m+n}$ and $\epsilon x\in S_{\gamma}$, +then +\[ +\mathrm{sign}f\left(x,y\right)=\sigma\left(\gamma\right)\hat{\w}_{\gamma}\left(x,\nicefrac{v_{i}\left(\epsilon x\right)}{\epsilon\left(\gamma\right)},\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right). +\] + +\section{Quantifier elimination for $\R_{an}$.} + +Recall that $\L_{an}$ is the language $\left\{ 0,1,+,-,\cdot,\leq\right\} $ +of ordered rings augmented by a function symbol for every restricted +analytic function $\R^{m}\lto\R$. \\ +Let $\R_{an}$ be the ordered field of reals as an $\L_{an}$-structure +together with the map $^{-1}\colon\R\lto\R$ defined as $x^{-1}=\begin{cases} +\frac{1}{x} & x\neq0\\ +x=0 & 0 +\end{cases}$. Let $\L_{an}\left(^{-1}\right)=\L_{an}\cup\left\{ ^{-1}\right\} $. +\begin{thm*} +\label{12.1} (Denef-v. d. Dries 1988) $\left(\R_{an},\,^{-1}\right)$ has +quantifier elimination. +\end{thm*} +Call $f\colon\I^{m}\lto\R$ analytic if it extends to an analytic +function on an open nbhd of $\I^{m}$. Let $\L_{a}$ be the language +$\left\{ <\right\} $ expanded by $m$-ary function symbols for each +analytic $f\colon\I^{m}\lto\R$ with $f\left(\I^{m}\right)\subset\I$.\\ +We consider $\I$ as an $\L_{a}$-structure.\\ +Define $\D\colon\I^{2}\lto\I$ by +\[ +D\left(x,y\right)=\begin{cases} +\nicefrac{x}{y} & \textrm{if }\left|x\right|\leq\left|y\right|\textrm{ and }y\neq0,\\ +0 & \textrm{otherwise.} +\end{cases} +\] +Let $\L_{a,\D}=\L_{a}\cup\left\{ \D\right\} $. +\begin{thm*} +\label{12.2} The $\L_{a,\D}$-structure $\I$ has $q.e.$ +\end{thm*} +Some general logical considerations:\\ +Let $T$ be an $\L$-theory, and $T'$ be a definitional expansion +of $T$ to an $\L'$-theory. Assume further that $\L'\setminus\L$ +consists only of function symbols, and for each $f\in\L'\setminus\L$ +there is an existential $\L$-formula $\delta_{f}\left(x,y\right)$ +s.t +\[ +T'\vdash f\left(x\right)=y\longleftrightarrow\delta_{f}\left(x,y\right). +\] +(e.g. $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$.) +\begin{lem*} +\label{12.6} Every $\exists\L'$-formula is $T'$-equivalent to an $\exists\L$-formula. +\begin{lem*} +\label{12.7} Suppose that for each q.f. $\L$-formula $\varphi\left(x,y_{1},...,y_{n}\right)$, +$n\geq1$, there is a q.f. $\L'$-formula $\varphi'$$\left(x,z_{1},...,z_{n-1}\right)$ +such that +\begin{enumerate} +\item $T'\vdash\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\varphi'\left(x,z\right)$. +\item The function symbols in $\L'\setminus\L$ are only applied in $\varphi'$ +to terms involving only $x$. +\end{enumerate} +\end{lem*} +Then $T'$ has $q.e$. (So by \eqref{12.6}, $T$ is model +complete). +\end{lem*} +We now verify that the conditions in lemma \eqref{12.7} are +satisfied for $T=\mathrm{Th}_{\L_{a}}\left(\I\right)$ and $T'=\mathrm{Th}_{\L_{a,\D}}\left(\I\right)$. +\begin{lem*} +\label{12.8} Basic Lemma. + +Let $\varphi\left(x,y\right)=\varphi\left(x_{1},...,x_{m},y_{1},...,y_{n}\right)$, +$n\geq1$ be a q.f. $\L_{a}$-formula. Then there is a q.f. $\L_{a,\D}$-formula +$\hat{\varphi}\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$ +such that +\begin{enumerate} +\item $\I\model\exists y\varphi\left(x,y\right)\longleftrightarrow\exists z\hat{\varphi}\left(x,z\right)$. +\item In $\hat{\varphi}$, $\D$ is only applied to terms not involving +$z$, and $z_{n}$ only appears polynomially in $\hat{\varphi}$. +\end{enumerate} +\end{lem*} +Given the Basic Lemma, we can use Tarsky's quantifier elimination +for $\R$ (as an ordered ring) to eliminate $\exists z_{n}$ and produce +$\varphi'$ satisfying the hypothesis of \eqref{12.7}. + +For an $\L_{a}$-formula and $\left(a,b\right)\in\I^{m+n}$, $\eta>0$, +consider the formula +\[ +\varphi_{a,b,\eta}=\varphi\wedge``\left|\left(x,y\right)-\left(a,b\right)\right|<\eta", +\] +where $``"$ means the formal sentence for that phrase. By compactness +of $I^{m+n}$, it is enough to show that for any $\left(a,b\right)$, +there is $\eta>0$ such that the Basic Lemma holds for the formula +$\varphi_{a,b,\eta}$. Using an affine transformation, it is enough +to consider only $\left(a,b\right)=\left(0,0\right)$. Let $\varphi_{\eta}=\varphi_{0,0,\eta}$. +Thus the lemma is reduced to the following: +\begin{lem*} +\label{12.9} Local Basic Lemma. + +Let $\varphi\left(x,y\right)$ be a q.f. $\L_{a}$-formula . Then +there is $\epsilon\in\left(0,1\right)$ and a q.f. $\L_{a,\D}$ formula +$\varphi'\left(x,z\right)$, $z=\left(z_{1},...,z_{n}\right)$, s.t +\begin{enumerate} +\item $\I\model\exists y\varphi_{\epsilon}\iff\exists z\left(\varphi_{\epsilon}^{\prime}\right)$. +\item In $\varphi'$, the function symbol $\D$ is applied only to terms +not involving $z$, and $z_{n}$ appears only polynomially. +\end{enumerate} +\end{lem*} +\begin{proof} +We may assume that all atomic subformulae of $\varphi$ are of the +form $f\left(x,y\right)>0$ or $f\left(x,y\right)=0$ for some analytic +$f\colon\I^{m}\lto\I$.\\ +Apply uniform W.P. to the Taylor series at $0$ of all such $f$. +We get \end{proof} +\begin{itemize} +\item $\epsilon\in\left(0,1\right)$, +\item Finite cover $S_{\gamma}$ of $B_{\epsilon}\subset\R^{m}$ by special +sets. +\item For each $\gamma$, a $\lambda=\lambda\left(\gamma\right)\in\R^{n-1}$ +with $\left|\lambda\right|\leq1$. +\item For each $f$ in $\varphi$, an $\L_{a,\D}$-term $t_{\gamma,f}\left(x,z\right)$; +\item $t_{f,\gamma}$ is polynomial in $z_{n}$, +\item $\D$ is only applied to terms involving only $x$ within $t_{f,\gamma}$. +\item $\forall x\in I^{m+n}$ with $\epsilon x\in S_{\gamma}$, +\[ +f\left(\epsilon x,\epsilon y\right)\geq0\iff t_{f,\gamma}\left(x,\nicefrac{\left(-\lambda\right)\left(y\right)}{2}\right)\geq0. +\] + +\end{itemize} +Replace $f\left(x,y\right)$ by $t_{f,\gamma}$ in $\varphi$ to get +$\varphi_{\gamma}$. Then +\[ +\I\model\exists y\varphi\left(\epsilon x,\epsilon y\right)\longleftrightarrow\exists z\left(\bigvee_{\gamma}x\in S_{\gamma}\wedge\varphi_{\gamma}\left(x,z\right)\wedge\left|\lambda\left(z\right)\right|\leq\frac{1}{2}\right). +\] +Finally we can convert this to a formula $\varphi'$ satisfying $\left(1\right)$ and $\left(2\right)$ of the Local Basic Lemma. \ No newline at end of file diff --git a/Other/old/final/week_10/g_week_10.aux b/Other/old/final/week_10/g_week_10.aux deleted file mode 100644 index 2feb6e1d..00000000 --- a/Other/old/final/week_10/g_week_10.aux +++ /dev/null @@ -1,22 +0,0 @@ -\relax -\@setckpt{week_10/g_week_10}{ -\setcounter{page}{47} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{11} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{1} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/final/week_10/week_10.aux b/Other/old/final/week_10/week_10.aux deleted file mode 100644 index 1ecedcfa..00000000 --- a/Other/old/final/week_10/week_10.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_10/week_10}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/final/week_11/g_week_11.aux b/Other/old/final/week_11/g_week_11.aux deleted file mode 100644 index b9c2502e..00000000 --- a/Other/old/final/week_11/g_week_11.aux +++ /dev/null @@ -1,28 +0,0 @@ -\relax -\providecommand\hyper@newdestlabel[2]{} -\@writefile{toc}{\contentsline {section}{\numberline {13} Siddharth's extra lectures }{57}{section.13}} -\@setckpt{week_11/g_week_11}{ -\setcounter{page}{60} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{13} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{Item}{137} -\setcounter{Hfootnote}{2} -\setcounter{bookmark@seq@number}{13} -\setcounter{theorem}{12} -\setcounter{namedtheorem}{0} -\setcounter{section@level}{1} -} diff --git a/Other/old/final/week_11/g_week_11.bbl b/Other/old/final/week_11/g_week_11.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/Other/old/final/week_11/g_week_11.blg b/Other/old/final/week_11/g_week_11.blg deleted file mode 100644 index 5808d9ba..00000000 --- a/Other/old/final/week_11/g_week_11.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: g_week_11.aux -I found no \citation commands---while reading file g_week_11.aux -I found no \bibdata command---while reading file g_week_11.aux -I found no \bibstyle command---while reading file g_week_11.aux -(There were 3 error messages) diff --git a/Other/old/final/week_11/g_week_11.log b/Other/old/final/week_11/g_week_11.log deleted file mode 100644 index 432d0120..00000000 --- a/Other/old/final/week_11/g_week_11.log +++ /dev/null @@ -1,4979 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 19 DEC 2014 21:47 -entering extended mode -**g_week_11.tex -(C:\Users\Anton\SparkleShare\Research\Other\all_notes\week_11\g_week_11.tex -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - kannada, kurmanji, latin, latvian, lithuanian, malayalam, marathi, mongolian, mongolianlmc, monogreek, ngerman, ngerman --x-2012-05-30, nynorsk, oriya, panjabi, pinyin, polish, portuguese, romanian, russian, sanskrit, serbian, slovak, sloven -ian, spanish, swedish, swissgerman, tamil, telugu, turkish, turkmen, ukenglish, ukrainian, uppersorbian, usenglishmax, w -elsh, loaded. -! 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Undefined control sequence. -l.5 ...cept for these lectures will be \WikiItalic - {games}. These games where... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no R in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -LaTeX Font Info: External font `cmex10' loaded for size -(Font) <7> on input line 5. -LaTeX Font Info: External font `cmex10' loaded for size -(Font) <5> on input line 5. -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no " in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no " in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ) in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 5--6 -[][] - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 5--6 -\OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 5--6 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (41.04903pt too wide) in paragraph at lines 5--6 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 G[]\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 G[]\OMS/cmsy/m/n/10 g$ - [] - - -! LaTeX Error: Environment example undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.7 \begin{example} - \ %====Example==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.7 \begin{example}\ - %====Example==== -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 7--8 -[][] - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no z in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (22.77785pt too wide) in paragraph at lines 9--10 -[]$\OMS/cmsy/m/n/10 f;j;g$ - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no R in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no R in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! 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-Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.34 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 33--35 -[][] - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (30.03131pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 A\OT1/cmr/m/n/10 &\OMS/cmsy/m/n/10 :\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 G < - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (30.03131pt too wide) in paragraph at lines 33--35 -\OMS/cmsy/m/n/10 :\OML/cmm/m/it/10 A\OT1/cmr/m/n/10 &\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (36.698pt too wide) in paragraph at lines 33--35 -\OMS/cmsy/m/n/10 :\OML/cmm/m/it/10 A\OT1/cmr/m/n/10 &\OMS/cmsy/m/n/10 :\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (15.64021pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (23.36462pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 A\OT1/cmr/m/n/10 &\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (23.36462pt too wide) in paragraph at lines 33--35 -\OML/cmm/m/it/10 A\OT1/cmr/m/n/10 &\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 33--35 -\OT1/cmr/m/n/10 0$ - [] - - -! LaTeX Error: Environment definition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.36 \begin{definition} - % ====Definition 3==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: \begin{document} ended by \end{definition}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.37 \end{definition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -l.38 If $G,H$ are games, the \WikiItalic - {disjunctive sum} $G+H$ is the game ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no " in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no " in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 38--39 -[][] - [] - - -Overfull \hbox (21.43187pt too wide) in paragraph at lines 38--39 -\OML/cmm/m/it/10 G; H$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 38--39 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 38--39 -\OML/cmm/m/it/10 H$ - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 38--39 -\OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 38--39 -\OML/cmm/m/it/10 H$ - [] - - -Overfull \hbox (183.96036pt too wide) detected at line 39 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 H \OT1/cmr/m/n/10 = \OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 G[] \OT1/cm -r/m/n/10 + \OML/cmm/m/it/10 H; G \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 H[]\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 G[] \OT1/cmr/m -/n/10 + \OML/cmm/m/it/10 H; G \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 H[]\OMS/cmsy/m/n/10 g\OML/cmm/m/it/10 : - [] - - -! LaTeX Error: Environment remark undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.41 \begin{remark} - %\WikiBold{Remark:} -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{remark}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.43 \end{remark} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 42--44 -[][] - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 42--44 -\OT1/cmr/m/n/10 +$ - [] - - -! LaTeX Error: Environment definition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.45 \begin{definition} - % ====Definition 4==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -l.46 If $G$ is a game, the \WikiItalic - {negation} $-G$ is the game obtained b... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no R in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 46--47 -[][] - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 46--47 -\OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 46--47 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (88.35593pt too wide) detected at line 47 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OT1/cmr/m/n/10 = \OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 G[]\OMS/cmsy/m/n/10 j \OML/ -cmm/m/it/10 G[]\OMS/cmsy/m/n/10 g\OML/cmm/m/it/10 : - [] - - -! LaTeX Error: \begin{document} ended by \end{definition}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.48 \end{definition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 50--51 -[][] - [] - - -! LaTeX Error: Environment lemma undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.52 \begin{lemma} - % ====Lemma 5 (Basic properties of $+$ and $-$)==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 53--53 -[][] - [] - -Missing character: There is no . in font nullfont! - -Overfull \hbox (27.30699pt too wide) in paragraph at lines 54--55 -[]$\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (23.56938pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 H\OT1/cmr/m/n/10 ) = - [] - - -Overfull \hbox (23.41809pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (16.90277pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H$ - [] - -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 55--56 -[]$\OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 G$ - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 56--57 -[]$\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 56--57 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (26.1958pt too wide) in paragraph at lines 56--57 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 56--57 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 57--58 -[]$\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 57--58 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (26.1958pt too wide) in paragraph at lines 57--58 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G < - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 57--58 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64021pt too wide) in paragraph at lines 58--59 -[]$\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (23.41801pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 0$ - [] - - -! LaTeX Error: \begin{document} ended by \end{lemma}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.60 \end{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 62--63 -[][] - [] - - -! LaTeX Error: Environment lemma undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.64 \begin{lemma} - % ====Lemma 6==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no : in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 65--66 -[][] - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 65--66 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 65--66 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 67--68 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 67--68 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 67--68 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 67--68 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 67--68 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 68--69 -\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 68--69 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 68--69 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 68--69 -\OML/cmm/m/it/10 H > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 68--69 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64021pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 69--70 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (16.9027pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 69--70 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 70--71 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 70--71 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 70--71 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 70--71 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 70--71 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 H > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 71--72 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 71--72 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 72--73 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (16.9027pt too wide) in paragraph at lines 72--73 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 72--73 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.64021pt too wide) in paragraph at lines 72--73 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 k - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 72--73 -\OT1/cmr/m/n/10 0$ - [] - - -! LaTeX Error: \begin{document} ended by \end{lemma}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.74 \end{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.76 \begin{proof} - %\WikiBold{Proof:} -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! 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LaTeX Error: Environment lemma undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.89 \begin{lemma} - \ % ====Lemma 7==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 89--90 -[][] - [] - -Missing character: There is no . in font nullfont! - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 91--92 -[]$\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (26.1958pt too wide) in paragraph at lines 91--92 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 91--92 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 92--93 -\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 92--93 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 92--93 -\OML/cmm/m/it/10 H > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 92--93 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 92--93 -\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 92--93 -\OML/cmm/m/it/10 H > - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 92--93 -\OT1/cmr/m/n/10 0$ - [] - - -! LaTeX Error: \begin{document} ended by \end{lemma}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.94 \end{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.96 \begin{proof} - %\WikiBold{Proof:} -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no " in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no G in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no " in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.99 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 98--100 -[][] - [] - -! Undefined control sequence. -l.101 \NotesBy - {notes by Zach Norwood} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no Z in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! - -! LaTeX Error: Environment definition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.102 \begin{definition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 101--103 -[][] - [] - - -Overfull \hbox (105.6412pt too wide) detected at line 103 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 ([]) \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 \ -OML/cmm/m/it/10 H \OT1/cmr/m/n/10 = 0\OML/cmm/m/it/10 : - [] - - -! LaTeX Error: The font size command \normalsize is not defined: - there is probably something wrong with the class file. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.103 \[ G \sim H \iff G - H = 0.\] - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -[1] -Overfull \hbox (89.53018pt too wide) detected at line 104 -\OML/cmm/m/it/10 G > H \OMS/cmsy/m/n/10  \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H > \OT1/cmr/m/n/10 0\O -ML/cmm/m/it/10 : - [] - - -! LaTeX Error: \begin{document} ended by \end{definition}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.105 \end{definition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment lemma undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.107 \begin{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{lemma}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.109 \end{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.110 \begin{proof} - \ -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (27.7778pt too wide) in paragraph at lines 108--111 -[][]$[]$ - [] - - -Overfull \hbox (286.8379pt too wide) detected at line 112 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  \OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 ([]) \ -OT1/cmr/m/n/10 (\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10  \OT1/cmr/m/ -n/10 0 \OMS/cmsy/m/n/10 ([]) \OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  \OT1/cmr/m/n/ -10 0 \OMS/cmsy/m/n/10 ([]) \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  \OT1/cmr/m/n/10 - 0\OML/cmm/m/it/10 ; - [] - -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 112--114 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (33.0138pt too wide) in paragraph at lines 112--114 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 ([]) - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 112--114 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 112--114 -\OML/cmm/m/it/10 G$ - [] - -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 114--115 -[]$\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 114--115 -\OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 114--115 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 114--115 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 114--115 -\OT1/cmr/m/n/10 0$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no : in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (10.66805pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 G[]$ - [] - - -Overfull \hbox (21.22356pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 G[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 H$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 115--116 -\OML/cmm/m/it/10 H$ - [] - -! Undefined control sequence. -l.116 \[ G\sim G'\sim 0 \text - { and } G'-H\sim 0 \Rightarrow G-G'+G'-H\sim 0 ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - - -Overfull \hbox (269.57149pt too wide) detected at line 116 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  \OML/cmm/m/it/10 G[] \OMS/cmsy/m/n/10  \OT1/cmr/m/n/10 0[]\OML/cmm/m/it/10 G[] \O -MS/cmsy/m/n/10 \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  \OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 ) \OML/cmm/m/it/10 G \OMS/cmsy -/m/n/10 \OML/cmm/m/it/10 G[] \OT1/cmr/m/n/10 + \OML/cmm/m/it/10 G[] \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H \OMS/cmsy/m/ -n/10  \OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 ) \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10 \OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  -\OT1/cmr/m/n/10 0\OML/cmm/m/it/10 : - [] - - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.118 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment lemma undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.120 \begin{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{lemma}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.122 \end{lemma} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.123 \begin{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.125 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (27.7778pt too wide) in paragraph at lines 121--126 -[][]$[]$ - [] - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 127--129 -[][] - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 127--129 -[]$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 127--129 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 127--129 -[]$ - [] - - -Overfull \hbox (22.77785pt too wide) in paragraph at lines 127--129 -\OMS/cmsy/m/n/10 f;j;g$ - [] - - -Overfull \hbox (18.418pt too wide) in paragraph at lines 127--129 -\OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (10.66805pt too wide) in paragraph at lines 127--129 -\OML/cmm/m/it/10 G[]$ - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 127--129 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (11.93053pt too wide) in paragraph at lines 127--129 -\OML/cmm/m/it/10 H[]$ - [] - - -Overfull \hbox (15.64029pt too wide) in paragraph at lines 130--131 -[]$\OML/cmm/m/it/10 G \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.68048pt too wide) in paragraph at lines 130--131 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.44585pt too wide) in paragraph at lines 130--131 -\OML/cmm/m/it/10 G[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (11.93053pt too wide) in paragraph at lines 130--131 -\OML/cmm/m/it/10 H[]$ - [] - - -Overfull \hbox (26.1958pt too wide) in paragraph at lines 131--132 -[]$\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.44585pt too wide) in paragraph at lines 131--132 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 G[]$ - [] - -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.418pt too wide) in paragraph at lines 132--133 -[]$\OML/cmm/m/it/10 G > - [] - - -Overfull \hbox (33.0138pt too wide) in paragraph at lines 132--133 -\OML/cmm/m/it/10 H \OMS/cmsy/m/n/10 ([]) - [] - - -Overfull \hbox (21.22356pt too wide) in paragraph at lines 132--133 -\OML/cmm/m/it/10 G[] > - [] - - -Overfull \hbox (11.93053pt too wide) in paragraph at lines 132--133 -\OML/cmm/m/it/10 H[]$ - [] - - -! LaTeX Error: Environment theorem undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.135 \begin{theorem} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -! Undefined control sequence. - \No - -l.136 $\No - $ forms an ordered abelian subgroup of the partially ordered abeli... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{theorem}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.137 \end{theorem} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 136--138 -[][]$$ - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 136--138 -\OML/cmm/m/it/10 G$ - [] - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.139 \begin{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -! Undefined control sequence. -l.140 We define a map $\pi\colon\No - \to G$ and prove that it's an injective g... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no : in font nullfont! - -! LaTeX Error: Environment align* undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.141 \begin{align*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -! Missing $ inserted. - - $ -l.142 \pi - \colon& 0\mapsto 0 \\ -I've inserted a begin-math/end-math symbol since I think -you left one out. Proceed, with fingers crossed. - -! Misplaced alignment tab character &. -l.142 \pi \colon& - 0\mapsto 0 \\ -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - -! Misplaced alignment tab character &. -l.143 \pi\colon& - \{L | R\} \mapsto \{\pi(L) | \pi(R) \} -I can't figure out why you would want to use a tab mark -here. If you just want an ampersand, the remedy is -simple: Just type `I\&' now. But if some right brace -up above has ended a previous alignment prematurely, -you're probably due for more error messages, and you -might try typing `S' now just to see what is salvageable. - - -! LaTeX Error: \begin{document} ended by \end{align*}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.144 \end{align*} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -! Missing $ inserted. - - $ -l.144 \end{align*} - -I've inserted something that you may have forgotten. -(See the above.) -With luck, this will get me unwedged. But if you -really didn't forget anything, try typing `2' now; then -my insertion and my current dilemma will both disappear. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 140--146 -[][] - [] - - -Overfull \hbox (20.50348pt too wide) in paragraph at lines 140--146 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 : \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.86249pt too wide) in paragraph at lines 140--146 -\OML/cmm/m/it/10 G$ - [] - - -Overfull \hbox (28.2812pt too wide) in paragraph at lines 140--146 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 : 0 \OMS/cmsy/m/n/10 7! - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 140--146 -\OT1/cmr/m/n/10 0[] - [] - - -Overfull \hbox (50.53473pt too wide) in paragraph at lines 140--146 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 : \OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 L\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 R\OMS/cmsy/m/ -n/10 g 7! - [] - - -Overfull \hbox (54.92725pt too wide) in paragraph at lines 140--146 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 L\OT1/cmr/m/n/10 )\OMS/cmsy/m/n/10 j\OML/cmm/m/it/ -10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 R\OT1/cmr/m/n/10 )\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (6.05905pt too wide) in paragraph at lines 140--146 -\OML/cmm/m/it/10 $ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 147--148 -[][] - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 147--148 -\OT1/cmr/m/n/10 +$ - [] - - -Overfull \hbox (6.05905pt too wide) in paragraph at lines 147--148 -\OML/cmm/m/it/10 $ - [] - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! 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g_week_4.aux -(There were 3 error messages) diff --git a/Other/old/final/week_4/g_week_4.log b/Other/old/final/week_4/g_week_4.log deleted file mode 100644 index 8624c2c3..00000000 --- a/Other/old/final/week_4/g_week_4.log +++ /dev/null @@ -1,8903 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 19 DEC 2014 20:27 -entering extended mode -**g_week_4.tex -(C:\Users\Anton\SparkleShare\Research\Other\all_notes\week_4\g_week_4.tex -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - kannada, kurmanji, latin, latvian, lithuanian, malayalam, marathi, mongolian, mongolianlmc, monogreek, ngerman, ngerman 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LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.9 C - laim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mat... -You're in trouble here. 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Undefined control sequence. -l.9 ...\in S$ satisfies $s>r$, take $d \in \mathbb - {D}$ with $s>d>r$. Then $d... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.11 S - o $r$ is an upper bound. If $L to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no S in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 11--12 -[][] - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (17.36108pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 L < - [] - - -Overfull \hbox (18.15044pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r[] < - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (7.59492pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r[]$ - [] - - -Overfull \hbox (5.20486pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 d$ - [] - - -Overfull \hbox (18.15044pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r[] < - [] - - -Overfull \hbox (15.76038pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 d < - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (14.64926pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 d \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 11--12 -\OML/cmm/m/it/10 R$ - [] - - -! LaTeX Error: Environment corollary undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.13 \begin{corollary} - % ====Corollary 3.13==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.14 C - orollary to \eqref{3.8} and \eqref{3.12}: Let $a \in \mathbb{R} \setmi... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -! Undefined control sequence. -l.14 Corollary to \eqref - {3.8} and \eqref{3.12}: Let $a \in \mathbb{R} \setmi... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no 3 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no 8 in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -l.14 Corollary to \eqref{3.8} and \eqref - {3.12}: Let $a \in \mathbb{R} \setmi... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no 3 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no : in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -! Undefined control sequence. -l.14 ...{3.8} and \eqref{3.12}: Let $a \in \mathbb - {R} \setminus \mathbb{D}$.... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -l.14 ...}: Let $a \in \mathbb{R} \setminus \mathbb - {D}$. Then $\lim_{n \to +\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -l.14 ... Then $\lim_{n \to +\infty} a \restriction - n =a$ in $\mathbb{R}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. - \mathbb - -l.14 ... +\infty} a \restriction n =a$ in $\mathbb - {R}$. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{corollary}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.15 \end{corollary} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 14--16 -[][] - [] - - -Overfull \hbox (14.73029pt too wide) in paragraph at lines 14--16 -\OML/cmm/m/it/10 a \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (12.67017pt too wide) in paragraph at lines 14--16 -\OML/cmm/m/it/10 R \OMS/cmsy/m/n/10 n - [] - - -Overfull \hbox (8.55695pt too wide) in paragraph at lines 14--16 -\OML/cmm/m/it/10 D$ - [] - - -Overfull \hbox (64.92604pt too wide) in paragraph at lines 14--16 -[][] \OML/cmm/m/it/10 an \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.28589pt too wide) in paragraph at lines 14--16 -\OML/cmm/m/it/10 a$ - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 14--16 -\OML/cmm/m/it/10 R$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.17 \textbf{Ordinals} - -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - - -Overfull \hbox (62.6678pt too wide) in paragraph at lines 17--18 -[][]\OT1/cmr/bx/n/10 Ordinals - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.19 W - e identify each $n \in \mathbf{On}$ wih $a_\alpha \in \mathbf{No}$, wh... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no ? in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 19--20 -[][] - [] - - -Overfull \hbox (15.44675pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.0277pt too wide) in paragraph at lines 19--20 -[]$ - [] - - -Overfull \hbox (20.42905pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (14.74992pt too wide) in paragraph at lines 19--20 -[]$ - [] - - -Overfull \hbox (32.49854pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 l\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 a[]\OT1/cmr/m/n/10 ) = - [] - - -Overfull \hbox (6.43404pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (35.50198pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 a[]\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 \OT1/cmr/m/n/10 ) = - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 19--20 -\OT1/cmr/m/n/10 +$ - [] - - -Overfull \hbox (16.73953pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (6.43404pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (26.50687pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 ; \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (36.46165pt too wide) in paragraph at lines 19--20 -[]\OML/cmm/m/it/10 ; \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.96175pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (29.31796pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 a[] <\OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (10.68002pt too wide) in paragraph at lines 19--20 -\OML/cmm/m/it/10 a[]$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.21 T - he canonical representation of $a_\alpha$ is $a_\alpha = \{a_{\beta}, ... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 21--22 -[][] - [] - - -Overfull \hbox (10.98465pt too wide) in paragraph at lines 21--22 -\OML/cmm/m/it/10 a[]$ - [] - - -Overfull \hbox (21.54016pt too wide) in paragraph at lines 21--22 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (36.86398pt too wide) in paragraph at lines 21--22 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[]; \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (19.21185pt too wide) in paragraph at lines 21--22 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 j;g$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.23 I - f $H$ is any nonempty set of ordinals and $\alpha =$ sup$ H$, then $a_... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 23--24 -[][] - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 23--24 -\OML/cmm/m/it/10 H$ - [] - - -Overfull \hbox (16.98955pt too wide) in paragraph at lines 23--24 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 =$ - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 23--24 -\OML/cmm/m/it/10 H$ - [] - - -Overfull \hbox (21.54016pt too wide) in paragraph at lines 23--24 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (35.75287pt too wide) in paragraph at lines 23--24 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[]; \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (21.90279pt too wide) in paragraph at lines 23--24 -\OML/cmm/m/it/10 H\OMS/cmsy/m/n/10 j;g$ - [] - - -! LaTeX Error: Environment proposition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.25 \begin{proposition} - % ====Proposition 3.14==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.26 F - or all $\alpha, \beta \in \mathbf{On}, a_\alpha + a_\beta = a_{\alpha ... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{proposition}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.27 \end{proposition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 26--28 -[][] - [] - - -Overfull \hbox (26.50687pt too wide) in paragraph at lines 26--28 -\OML/cmm/m/it/10 ; \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (38.23456pt too wide) in paragraph at lines 26--28 -[]\OML/cmm/m/it/10 ; a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 26--28 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (42.55788pt too wide) in paragraph at lines 26--28 -\OML/cmm/m/it/10 a[]; a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 26--28 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (22.1288pt too wide) in paragraph at lines 26--28 -\OML/cmm/m/it/10 a[]$ - [] - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.30 \begin{proof} - %\WikiBold{Proof:} -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.32 B - y induction, $a_\alpha + a_\beta =$ sup $\{a_\alpha + a_{\beta'}, a_{\... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 32--33 -[][] - [] - - -Overfull \hbox (18.76245pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 =$ - [] - - -Overfull \hbox (23.76247pt too wide) in paragraph at lines 32--33 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (39.29665pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 a[]; a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (33.25291pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 a[]\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (30.42354pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 ; [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.73955pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 =$ - [] - - -Overfull \hbox (81.68468pt too wide) in paragraph at lines 32--33 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[]; a[]\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (30.42354pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 ; [] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.73955pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (22.1288pt too wide) in paragraph at lines 32--33 -\OML/cmm/m/it/10 a[]$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.34 F - or multiplication, using the result for addition, and the distributive... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no F in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -l.34 ...d the distributive laws for $+, \centerdot - ,$ on $\mathbf{No}$ and $\... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no 8 in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 34--35 -[][] - [] - - -Overfull \hbox (15.0pt too wide) in paragraph at lines 34--35 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 ; ;$ - [] - - -Overfull \hbox (14.74992pt too wide) in paragraph at lines 34--35 -[]$ - [] - - -Overfull \hbox (20.00002pt too wide) in paragraph at lines 34--35 -\OMS/cmsy/m/n/10 \OML/cmm/m/it/10 ; \OMS/cmsy/m/n/10 -$ - [] - - -Overfull \hbox (15.0277pt too wide) in paragraph at lines 34--35 -[]$ - [] - - -Overfull \hbox (16.98955pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (32.19551pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 ![]; \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (38.11217pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 ![]; r; s \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.0277pt too wide) in paragraph at lines 34--35 -[]$ - [] - - -Overfull \hbox (15.98466pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (25.10498pt too wide) in paragraph at lines 34--35 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (17.45224pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (15.41698pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (22.70334pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (14.67445pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (15.97246pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (15.37842pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (31.59041pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 ![]; \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (26.56717pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 ![]\OMS/cmsy/m/n/10 j;g\OML/cmm/m/it/10 :$ - [] - - -Overfull \hbox (15.98466pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (20.12442pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.0277pt too wide) in paragraph at lines 34--35 -[]$ - [] - - -Overfull \hbox (20.10497pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (17.45224pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (15.41698pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (22.70334pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (14.67445pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (20.97247pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (73.82732pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[] < - [] - - -Overfull \hbox (26.65019pt too wide) in paragraph at lines 34--35 -\OML/cmm/m/it/10 a[]:$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.36 A - nd $l(a_\alpha * a_\beta) \leq l(\omega^{r \oplus s}) = \omega^{r \opl... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no A in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 36--37 -[][] - [] - - -Overfull \hbox (23.05414pt too wide) in paragraph at lines 36--37 -\OML/cmm/m/it/10 l\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (25.12444pt too wide) in paragraph at lines 36--37 -\OML/cmm/m/it/10 a[]\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (42.54979pt too wide) in paragraph at lines 36--37 -\OML/cmm/m/it/10 l\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 ![]\OT1/cmr/m/n/10 ) = - [] - - -Overfull \hbox (21.0359pt too wide) in paragraph at lines 36--37 -\OML/cmm/m/it/10 ![]$ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.38 N - ow we show the $\geq$ direction. -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 38--39 -[][] - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 38--39 -\OMS/cmsy/m/n/10 $ - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.40 L - et $r' a_{\omega... -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -! Undefined control sequence. -l.40 Let $r' a_{\omega... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -l.40 ... s}_n}$. Similarly if $s' for immediate help. - ... - -l.41 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 40--42 -[][] - [] - - -Overfull \hbox (18.15044pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 r[] < - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (15.44675pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 N$ - [] - - -Overfull \hbox (15.98466pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] > - [] - - -Overfull \hbox (22.80984pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (25.58763pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 - - [] - - -Overfull \hbox (6.18402pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 40--42 -\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (26.57729pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[]$ - [] - - -Overfull \hbox (18.04857pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 s[] < - [] - - -Overfull \hbox (24.57866pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 s; n \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (9.12497pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 N$ - [] - - -Overfull \hbox (15.98466pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] > - [] - - -Overfull \hbox (26.57729pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[]$ - [] - - -Overfull \hbox (15.98466pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (21.23553pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[] \OMS/cmsy/m/n/10 $ - [] - - -Overfull \hbox (68.1545pt too wide) in paragraph at lines 40--42 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 a[]; a[] \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (18.15044pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 r[] < - [] - - -Overfull \hbox (26.72679pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 r; s[] < - [] - - -Overfull \hbox (24.57866pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 s; n \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (24.6805pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 N\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (23.59465pt too wide) in paragraph at lines 40--42 -\OML/cmm/m/it/10 a[]$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 40--42 -\OMS/cmsy/m/n/10 $ - [] - - -! LaTeX Error: The font size command \normalsize is not defined: - there is probably something wrong with the class file. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.42 - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -[1] - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.43 = - ===Examples==== -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no E in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no = in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 43--44 -[][] - [] - - -! LaTeX Error: Missing \begin{document}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.45 \begin{enumerate} - -You're in trouble here. Try typing to proceed. -If that doesn't work, type X to quit. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 45--45 -[][] - [] - -Missing character: There is no + in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 46--47 -[]$\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (15.55553pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 1 = - [] - - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 46--47 -\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (31.11113pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 (\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 1) = - [] - - -Overfull \hbox (31.55798pt too wide) in paragraph at lines 46--47 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (33.33336pt too wide) in paragraph at lines 46--47 -\OMS/cmsy/m/n/10 f;j\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (18.78017pt too wide) in paragraph at lines 46--47 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (22.13892pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (20.55554pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (35.91899pt too wide) in paragraph at lines 46--47 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 !\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 (+ + - [] - - -Overfull \hbox (41.02786pt too wide) in paragraph at lines 46--47 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 :::\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 )\OML/cmm/m/it/10 :\OT1/cmr/m/n/10 (\OML/cmm/m/it -/10 !$ - [] - -Missing character: There is no ? in font nullfont! -Missing character: There is no B in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! 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-Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no + in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 47--48 -[]$\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (20.44684pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (8.88892pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 1)$ - [] - - -Overfull \hbox (15.20605pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 w \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (19.33565pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 (+ + - [] - - -Overfull \hbox (23.88896pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 ::: \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (20.00006pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 :::\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 !$ - [] - - -Overfull \hbox (8.78014pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m$ - [] - - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (20.44684pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.44443pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 1) = - [] - - -Overfull \hbox (31.55798pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (20.55562pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 f;j - [] - - -Overfull \hbox (24.33566pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (18.78017pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (22.13892pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (24.33566pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (28.14127pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (24.33566pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 (+ + - [] - - -Overfull \hbox (23.88896pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 ::: \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (16.11116pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 ::: \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (11.6667pt too wide) in paragraph at lines 47--48 -\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 !$ - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 47--48 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 47--48 -\OT1/cmr/m/n/10 1$ - [] - -Missing character: There is no ( in font nullfont! -Missing character: There is no + in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ) in font nullfont! - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 48--49 -[]$\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (16.94164pt too wide) in paragraph at lines 48--49 -[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (31.55798pt too wide) in paragraph at lines 48--49 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (33.33336pt too wide) in paragraph at lines 48--49 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (18.78017pt too wide) in paragraph at lines 48--49 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (25.19165pt too wide) in paragraph at lines 48--49 -[]\OML/cmm/m/it/10 ; ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (22.13892pt too wide) in paragraph at lines 48--49 -\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (20.55554pt too wide) in paragraph at lines 48--49 -\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (28.72223pt too wide) in paragraph at lines 48--49 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 !\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (20.55554pt too wide) in paragraph at lines 48--49 -\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 48--49 -\OT1/cmr/m/n/10 (+ + - [] - - -Overfull \hbox (23.88896pt too wide) in paragraph at lines 48--49 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 ::: \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (11.6667pt too wide) in paragraph at lines 48--49 -\OMS/cmsy/m/n/10 \OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 48--49 -\OML/cmm/m/it/10 !$ - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -! Undefined control sequence. -l.49 \item In general, if $r \in \mathbb - {R}^{>0}, \omega + r = \{n|\empty... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ( in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -! Undefined control sequence. -l.49 ...--the full result holds for $r \in \mathbb - {Z}^{<0}$ as well. -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (37.21182pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 R[]; ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (15.34488pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (31.55798pt too wide) in paragraph at lines 49--50 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (27.25354pt too wide) in paragraph at lines 49--50 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 L\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 R\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (25.03014pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 L; n \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.92827pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (23.22568pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 R\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.36113pt too wide) in paragraph at lines 49--50 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (23.94447pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 L\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (12.67017pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 R\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (19.36104pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 ! _ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (6.80557pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 L$ - [] - - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (18.27782pt too wide) in paragraph at lines 49--50 -\OML/cmm/m/it/10 Z[]$ - [] - -! Undefined control sequence. -l.52 \WikiLevelThree - {Wednesday, October 29, 2014} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no O in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 9 in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 4 in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 52--53 -[][] - [] - -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no + in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (23.52495pt too wide) in paragraph at lines 54--55 -[]$[]\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (27.77786pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 g  - [] - - -Overfull \hbox (34.3357pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 =$ - [] - - -Overfull \hbox (25.16629pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 f[]\OML/cmm/m/it/10 n \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (11.58333pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (11.00237pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (27.94408pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j[]\OML/cmm/m/it/10 n \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (11.58333pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 1 \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (11.00237pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (20.55554pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 1\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (34.52739pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 f[]\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (27.944pt too wide) in paragraph at lines 54--55 -[]\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 =$ - [] - - -Overfull \hbox (28.14127pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (21.55788pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 n\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 (+ + - [] - - -Overfull \hbox (23.88896pt too wide) in paragraph at lines 54--55 -\OT1/cmr/m/n/10 +\OML/cmm/m/it/10 ::: \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 54--55 -\OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (12.22226pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 :::\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 !$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 54--55 -\OML/cmm/m/it/10 !$ - [] - -Missing character: There is no ( in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no ? in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -! Undefined control sequence. -l.55 ...n$. $0<\epsilon 0}$, i.e. $\epsilon$... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no I in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no G in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (18.24443pt too wide) in paragraph at lines 55--56 -[]$[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.4445pt too wide) in paragraph at lines 55--56 -\OT1/cmr/m/n/10 (+ \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 55--56 -\OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (12.22226pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 :::\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 !$ - [] - - -Overfull \hbox (4.05904pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (15.55553pt too wide) in paragraph at lines 55--56 -\OT1/cmr/m/n/10 0 \OML/cmm/m/it/10 < - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10  < - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (18.4063pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 R[]$ - [] - - -Overfull \hbox (4.05904pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (6.58331pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10 !$ - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (7.68892pt too wide) in paragraph at lines 55--56 -[]$ - [] - - -Overfull \hbox (9.61453pt too wide) in paragraph at lines 55--56 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (29.06795pt too wide) in paragraph at lines 55--56 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j[]g$ - [] - -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! - -Overfull \hbox (41.19786pt too wide) in paragraph at lines 58--59 -[][]$\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (34.06796pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j[]g  - [] - - -Overfull \hbox (37.11348pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 j;g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (30.12117pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j[]g  - [] - - -Overfull \hbox (26.55797pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 j;g$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (14.05907pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 58--59 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (10.00003pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (12.77782pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (10.00003pt too wide) in paragraph at lines 58--59 -\OT1/cmr/m/n/10 0 \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.42015pt too wide) in paragraph at lines 58--59 -\OML/cmm/m/it/10 !\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 58--59 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.70444pt too wide) in paragraph at lines 58--59 -[]\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (31.67899pt too wide) in paragraph at lines 58--59 -[]\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (14.05907pt too wide) in paragraph at lines 58--59 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (20.61697pt too wide) in paragraph at lines 58--59 -\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 58--59 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.70444pt too wide) in paragraph at lines 58--59 -[]\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (23.90126pt too wide) in paragraph at lines 58--59 -[]\OML/cmm/m/it/10 m\OMS/cmsy/m/n/10 g\OML/cmm/m/it/10 :$ - [] - -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 60--61 -[][] - [] - - -Overfull \hbox (9.05905pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (8.78014pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10 m$ - [] - - -Overfull \hbox (23.39468pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10 m < - [] - - -Overfull \hbox (23.33333pt too wide) in paragraph at lines 60--61 -\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :\OT1/cmr/m/n/10 1 \OML/cmm/m/it/10 < - [] - - -Overfull \hbox (9.05905pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10  \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (16.55794pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (21.70444pt too wide) in paragraph at lines 60--61 -[]\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (16.12346pt too wide) in paragraph at lines 60--61 -[]\OML/cmm/m/it/10 m$ - [] - - -Overfull \hbox (25.59334pt too wide) in paragraph at lines 60--61 -[]\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 ! \OMS/cmsy/m/n/10 - [] - - -Overfull \hbox (9.89125pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10 n\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 60--61 -\OT1/cmr/m/n/10 0$ - [] - - -Overfull \hbox (21.19786pt too wide) in paragraph at lines 60--61 -\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 60--61 -\OT1/cmr/m/n/10 1$ - [] - -! Undefined control sequence. -l.62 \item $r + \epsilon (r \in \mathbb - {R})$. First assume $r \in \mathbb{... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no F in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -! Undefined control sequence. -l.62 ... \mathbb{R})$. First assume $r \in \mathbb - {D}$, so $r=\{r_L|r_R\}, r... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -! Undefined control sequence. -l.62 ...$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb - {D}$. $r + \epsilon = \{r_... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no . in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no = in font nullfont! -! Undefined control sequence. -l.62 ...frac{1}{n}\}$ by cofinality = $\{r|\mathbb - {D}^{>r}\}=r\frown(+)$, of... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no , in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (12.56717pt too wide) in paragraph at lines 62--63 -[]$\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (22.1817pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (11.55905pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 R\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (8.55695pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 D$ - [] - - -Overfull \hbox (15.34488pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (74.24983pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 r[]\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 r[]\OMS/cmsy/m/n/10 g\OML/cmm/m/it/10 ; r[]; r[ -] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (8.55695pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 D$ - [] - - -Overfull \hbox (12.56717pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (14.61455pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (23.27554pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 r[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (23.84949pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 r[]\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (35.67667pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 0\OMS/cmsy/m/n/10 j[]g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (17.56718pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (27.71996pt too wide) in paragraph at lines 62--63 -\OT1/cmr/m/n/10 0\OML/cmm/m/it/10 ; r[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.40399pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (30.6372pt too wide) in paragraph at lines 62--63 -[]\OML/cmm/m/it/10 ; r[] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (19.61456pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (25.13434pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 r\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 r \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (12.34334pt too wide) in paragraph at lines 62--63 -[]\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (47.35791pt too wide) in paragraph at lines 62--63 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 r\OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 D[]\OMS/cmsy/m/n/10 g \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (17.5671pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 r _ - [] - - -Overfull \hbox (15.5556pt too wide) in paragraph at lines 62--63 -\OT1/cmr/m/n/10 (+)$ - [] - - -Overfull \hbox (14.36111pt too wide) in paragraph at lines 62--63 -\OML/cmm/m/it/10 ! \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 62--63 -\OT1/cmr/m/n/10 1$ - [] - -Missing character: There is no W in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -l.63 ... $\lambda$ a limit ordinal, $r \in \mathbb - {R}$? -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no ? in font nullfont! - -Overfull \hbox (13.61116pt too wide) in paragraph at lines 63--64 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (4.78937pt too wide) in paragraph at lines 63--64 -\OML/cmm/m/it/10 r$ - [] - - -Overfull \hbox (5.83336pt too wide) in paragraph at lines 63--64 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (14.23376pt too wide) in paragraph at lines 63--64 -\OML/cmm/m/it/10 r \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.67015pt too wide) in paragraph at lines 63--64 -\OML/cmm/m/it/10 R$ - [] - -! Undefined control sequence. -l.66 \section - {Combinatorics of Ordered Sets} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no C in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no O in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no S in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no A in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -l.67 Let $S$ be a set. An \WikiItalic - {ordering} $\leq$ on $S$ is an reflexiv... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -l.67 ...ion on $S$. Call $(S,\leq)$ an \WikiItalic - {ordered set} (partial or ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 66--68 -[][] - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 66--68 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 66--68 -\OMS/cmsy/m/n/10 $ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 66--68 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 66--68 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (22.26387pt too wide) in paragraph at lines 66--68 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 S; \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (3.8889pt too wide) in paragraph at lines 66--68 -\OT1/cmr/m/n/10 )$ - [] - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. -l.69 We say $\leq$ is \WikiItalic - {total} if $x \leq y$ or $y \leq x$ for eac... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 69--70 -[][] - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 69--70 -\OMS/cmsy/m/n/10 $ - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.2616pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 y$ - [] - - -Overfull \hbox (15.81711pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 y \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (24.86568pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 x; y \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 69--70 -\OML/cmm/m/it/10 S$ - [] - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. -l.71 ...arrow T$ a map. Then $\phi$ is \WikiItalic - {increasing} if $x \leq y ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. -l.71 ...htarrow \phi(x) \leq \phi(y)$, \WikiItalic - {strictly increasing} if $... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -! Undefined control sequence. -l.71 ...ghtarrow \phi(x) < \phi(y)$, a \WikiItalic - {quasi-embedding} if $\phi... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (20.0pt too wide) in paragraph at lines 71--72 -[][] - [] - - -Overfull \hbox (7.23265pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 T$ - [] - - -Overfull \hbox (11.51384pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (19.48604pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 S \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (7.23265pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 T$ - [] - - -Overfull \hbox (5.95834pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 $ - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.03932pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 y \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (30.00693pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 x\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (18.99774pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 y\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 x < - [] - - -Overfull \hbox (18.03932pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 y \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (30.00693pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 x\OT1/cmr/m/n/10 ) \OML/cmm/m/it/10 < - [] - - -Overfull \hbox (18.99774pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 y\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (30.00693pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 x\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (31.77547pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 y\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (5.2616pt too wide) in paragraph at lines 71--72 -\OML/cmm/m/it/10 y$ - [] - -Missing character: There is no E in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no a in font nullfont! 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-Missing character: There is no . in font nullfont! - -Overfull \hbox (11.30554pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (46.99889pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 x[]:::x[] \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (30.22124pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 y[]:::y[]$ - [] - - -Overfull \hbox (11.51384pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (44.89127pt too wide) in paragraph at lines 77--78 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::m\OMS/cmsy/m/n/10 g ! - [] - - -Overfull \hbox (29.33575pt too wide) in paragraph at lines 77--78 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::n\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (25.39314pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 x[] \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (19.29047pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 y[]$ - [] - - -Overfull \hbox (14.00064pt too wide) in paragraph at lines 77--78 -\OML/cmm/m/it/10 i \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (22.11351pt too wide) in paragraph at lines 77--78 -\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::m$ - [] - -Missing character: There is no L in font nullfont! 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-Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (11.30554pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (17.26382pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 S \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (11.70833pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 S \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (15.55553pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 = \OMS/cmsy/m/n/10 $ - [] - - -Overfull \hbox (42.40166pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 x[]:::x[] \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (30.22124pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 y[]:::y[]$ - [] - - -Overfull \hbox (19.33565pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 m \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (6.00235pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 n$ - [] - - -Overfull \hbox (32.11354pt too wide) in paragraph at lines 78--79 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::m\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (19.60007pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 x[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.29047pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 y[]$ - [] - - -Overfull \hbox (3.44513pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 i$ - [] - - -Overfull \hbox (46.99889pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 x[]:::x[] \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (42.99896pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 y[]:::y[] \OMS/cmsy/m/n/10 $$ - [] - - -Overfull \hbox (11.51384pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10  \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (44.89127pt too wide) in paragraph at lines 78--79 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::m\OMS/cmsy/m/n/10 g ! - [] - - -Overfull \hbox (29.33575pt too wide) in paragraph at lines 78--79 -\OMS/cmsy/m/n/10 f\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::n\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (25.39314pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 x[] \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (19.29047pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 y[]$ - [] - - -Overfull \hbox (14.00064pt too wide) in paragraph at lines 78--79 -\OML/cmm/m/it/10 i \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (22.11351pt too wide) in paragraph at lines 78--79 -\OT1/cmr/m/n/10 1\OML/cmm/m/it/10 :::m$ - [] - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (32.01389pt too wide) in paragraph at lines 79--80 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 S[]; \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (16.66663pt too wide) in paragraph at lines 79--80 -\OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (32.01389pt too wide) in paragraph at lines 79--80 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 S[]; \OMS/cmsy/m/n/10 [] - [] - - -Overfull \hbox (3.8889pt too wide) in paragraph at lines 79--80 -\OT1/cmr/m/n/10 )$ - [] - -! Undefined control sequence. - \mathbb - -l.80 \item $\mathbb - {N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -l.80 \item $\mathbb{N}^m=\mathbb - {N} * \mathbb{N} * ...\mathbb{N}$ with the... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -l.80 \item $\mathbb{N}^m=\mathbb{N} * \mathbb - {N} * ...\mathbb{N}$ with the... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -! Undefined control sequence. -l.80 ...{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb - {N}$ with the product orde... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -! Undefined control sequence. - \mathbb - -l.80 ...es with trivial ordering. The map $\mathbb - {N}^m= \rightarrow X^\diam... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -Overfull \hbox (27.2766pt too wide) in paragraph at lines 80--81 -[]$\OML/cmm/m/it/10 N[] \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (14.12498pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 N \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (14.12498pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 N \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (17.45833pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 :::N$ - [] - - -Overfull \hbox (19.62494pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 X \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (41.84618pt too wide) in paragraph at lines 80--81 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 x[]:::x[]\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (37.27661pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 N[] \OT1/cmr/m/n/10 =\OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (69.97804pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 X[]; \OT1/cmr/m/n/10 (\OML/cmm/m/it/10 v[]:::v[]\OT1/cmr/m/n/10 ) = - [] - - -Overfull \hbox (46.01524pt too wide) in paragraph at lines 80--81 -\OML/cmm/m/it/10 X[]:::X[]$ - [] - - -Overfull \hbox (12.37503pt too wide) in paragraph at lines 80--81 -\OMS/cmsy/m/n/10 []$ - [] - -! Undefined control sequence. -l.83 \WikiSigleStar - Let $S$ be an ordered set. Call $F \subset S$ a ''final ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no G in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no P in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! 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- [] - - -Overfull \hbox (16.27078pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 x \OMS/cmsy/m/n/10 6 - [] - - -Overfull \hbox (5.2616pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 y$ - [] - - -Overfull \hbox (15.81711pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 y \OMS/cmsy/m/n/10 6 - [] - - -Overfull \hbox (5.71527pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 x$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (20.75691pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 x[] > - [] - - -Overfull \hbox (18.53476pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 x[]:::$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 83--84 -\OML/cmm/m/it/10 S$ - [] - - -! LaTeX Error: Environment definition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.85 \begin{definition} - % ====Definition 4.1==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -! Undefined control sequence. -l.86 $(S, \leq)$ is \WikiItalic - {noetherian} if it is well-founded and has no... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{definition}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.87 \end{definition} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (42.26387pt too wide) in paragraph at lines 86--88 -[][]$\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 S; \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (3.8889pt too wide) in paragraph at lines 86--88 -\OT1/cmr/m/n/10 )$ - [] - -! Undefined control sequence. -l.90 \WikiLevelThree - {Friday, October 31, 2014} -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no F in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no O in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no 3 in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 4 in font nullfont! -Missing character: There is no C in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no q in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -l.91 ...finite sequence $(x_n)$ in $S$ \WikiItalic - {good} if $x_i \leq x_j$ f... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no g in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment proposition undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.92 \begin{proposition} - % ====Proposition 4.2==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! 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LaTeX Error: \begin{document} ended by \end{proof}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.111 \end{proof} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (37.77774pt too wide) in paragraph at lines 110--112 -[][]$\OT1/cmr/m/n/10 4 \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 110--112 -\OT1/cmr/m/n/10 5$ - [] - - -Overfull \hbox (17.77774pt too wide) in paragraph at lines 110--112 -\OT1/cmr/m/n/10 5 \OMS/cmsy/m/n/10 ! - [] - - -Overfull \hbox (5.00002pt too wide) in paragraph at lines 110--112 -\OT1/cmr/m/n/10 1$ - [] - - -! LaTeX Error: Environment corollary undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.113 \begin{corollary} - % ====Corollary 4.3==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 114--114 -[][] - [] - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no x in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no j in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -! Undefined control sequence. -l.115 ...creasing surjection $S \twoheadrightarrow - T$ and $S$ is noetherian,... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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LaTeX Error: \begin{document} ended by \end{corollary}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.119 \end{corollary} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no I in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no , in font nullfont! -! Undefined control sequence. - \mathbb - -l.121 In particular, $\mathbb - {N}^m$ is noetherian (``Dickson's Lemma''). -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no ` in font nullfont! -Missing character: There is no ` in font nullfont! -Missing character: There is no D in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no k in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: Environment theorem undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.122 \begin{theorem} - % ====Theorem 4.4 (Higman)==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no I in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no 4 in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no 3 in font nullfont! -Missing character: There is no ( in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no ) in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{theorem}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.124 \end{theorem} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 121--125 -[][] - [] - - -Overfull \hbox (16.72108pt too wide) in paragraph at lines 121--125 -\OML/cmm/m/it/10 N[]$ - [] - - -Overfull \hbox (6.70831pt too wide) in paragraph at lines 121--125 -\OML/cmm/m/it/10 S$ - [] - - -Overfull \hbox (11.30554pt too wide) in paragraph at lines 121--125 -\OML/cmm/m/it/10 S[]$ - [] - - -Overfull \hbox (11.30554pt too wide) in paragraph at lines 121--125 -\OML/cmm/m/it/10 S[]$ - [] - - -! 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Undefined control sequence. -l.133 ... $\phi: S \rightarrow T$ is a \WikiItalic - {strictly} increasing map ... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). 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LaTeX Error: \begin{document} ended by \end{corollary}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.139 \end{corollary} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 138--140 -[][] - [] - - -Overfull \hbox (20.03123pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A; B$ - [] - - -Overfull \hbox (6.25002pt too wide) in paragraph at lines 138--140 -\OT1/cmr/m/n/10 $ - [] - - -Overfull \hbox (14.1667pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A \OMS/cmsy/m/n/10 [ - [] - - -Overfull \hbox (8.0868pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (15.27782pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (18.64232pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 B \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (19.21185pt too wide) in paragraph at lines 138--140 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (24.84024pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 j\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (27.57285pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A; \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (13.08682pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 B\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (15.17725pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.27782pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (8.0868pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (34.28467pt too wide) in paragraph at lines 138--140 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 ; \OT1/cmr/m/n/10 ) \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (15.27782pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 A \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (8.0868pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 B$ - [] - - -Overfull \hbox (16.28836pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (14.21184pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (6.18402pt too wide) in paragraph at lines 138--140 -\OML/cmm/m/it/10 $ - [] - -Missing character: There is no S in font nullfont! 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- [] - - -Overfull \hbox (19.16672pt too wide) in paragraph at lines 141--142 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 A \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (11.97571pt too wide) in paragraph at lines 141--142 -\OML/cmm/m/it/10 B\OT1/cmr/m/n/10 )$ - [] - - -! LaTeX Error: Environment corollary undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.143 \begin{corollary} - % ====Corollary 4.6==== -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no L in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no f in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -! Undefined control sequence. -l.144 ...\alpha_1...\alpha_n)$ with $n \in \mathbb - {N}, \alpha_1...\alpha_n \... -The control sequence at the end of the top line -of your error message was never \def'ed. If you have -misspelled it (e.g., `\hobx'), type `I' and the correct -spelling (e.g., `I\hbox'). Otherwise just continue, -and I'll forget about whatever was undefined. - -Missing character: There is no s in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no . in font nullfont! - -! LaTeX Error: \begin{document} ended by \end{corollary}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.145 \end{corollary} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - - -Overfull \hbox (20.0pt too wide) in paragraph at lines 144--146 -[][] - [] - - -Overfull \hbox (18.05553pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 A \OMS/cmsy/m/n/10  - [] - - -Overfull \hbox (16.98616pt too wide) in paragraph at lines 144--146 -\OT1/cmr/m/n/10 []$ - [] - - -Overfull \hbox (7.7778pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 < - [] - - -Overfull \hbox (25.83333pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 A >\OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (23.66096pt too wide) in paragraph at lines 144--146 -\OMS/cmsy/m/n/10 f\OML/cmm/m/it/10 [] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (25.7292pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 ::: [] \OT1/cmr/m/n/10 : - [] - - -Overfull \hbox (19.1707pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (12.50003pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 A\OMS/cmsy/m/n/10 g$ - [] - - -Overfull \hbox (22.95505pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 \OMS/cmsy/m/n/10 2\OML/cmm/m/it/10 < - [] - - -Overfull \hbox (18.05553pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 A >$ - [] - - -Overfull \hbox (49.28142pt too wide) in paragraph at lines 144--146 -\OT1/cmr/m/n/10 (\OML/cmm/m/it/10 n; []::: []\OT1/cmr/m/n/10 )$ - [] - - -Overfull \hbox (15.44675pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 n \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (53.51508pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 N; []::: [] \OMS/cmsy/m/n/10 2 - [] - - -Overfull \hbox (7.50002pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 A$ - [] - - -Overfull \hbox (16.28836pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 \OT1/cmr/m/n/10 = - [] - - -Overfull \hbox (18.66095pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 [] \OT1/cmr/m/n/10 + - [] - - -Overfull \hbox (20.1737pt too wide) in paragraph at lines 144--146 -\OML/cmm/m/it/10 ::: []$ - [] - - -! LaTeX Error: Environment proof undefined. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.147 \begin{proof} - %\WikiBold{Proof}: -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -Missing character: There is no W in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no y in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no . in font nullfont! -Missing character: There is no N in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no H in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ' in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no T in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no . in font nullfont! - -! 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-\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{Item}{65} -\setcounter{Hfootnote}{0} -\setcounter{bookmark@seq@number}{6} -\setcounter{theorem}{5} -\setcounter{namedtheorem}{0} -\setcounter{section@level}{1} -} diff --git a/Other/old/final/week_6/g_week_6.tex b/Other/old/final/week_6/g_week_6.tex index 11fed4bb..655c7331 100644 --- a/Other/old/final/week_6/g_week_6.tex +++ b/Other/old/final/week_6/g_week_6.tex @@ -52,8 +52,8 @@ \end{theorem} \textbf{Notation:} for $h(x) = \sum_{i < \alpha} h_i x^{a_i}$ and $\gamma \leq \alpha$ we write -\begin{align*} - h(x)\midr \gamma = \sum_{i < \gamma} h_i x^{a_i} +\begin{align*} + h(x)\midr \gamma = \sum_{i < \gamma} h_i x^{a_i} \end{align*} \begin{proof}[Proof of inequality.] %\WikiBold{Proof of inequality} diff --git a/Other/old/final/week_6/week_6.aux b/Other/old/final/week_6/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/final/week_6/week_6.aux +++ /dev/null @@ -1,27 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-\begin{defn*} -\label{10.4} $\C\left\{ X\right\} =\bigcup_{r}\C\left\{ X\right\} _{r}$.\\ -$\C\left\{ X\right\} $ is a subring of $\C\left[\left[X\right]\right]$ -containing $\C\left[X\right]$, called the ring of convergent power -series in $X$.\end{defn*} -\begin{xca*} -\label{10.5} Let $\left\langle f_{j}\right\rangle _{j\in J}$ be a family -in $\C\left\{ X\right\} _{r}$ such that $\mathrm{ord}\left(f_{j}\right)\lto\infty$, -so $\sum_{j\in J}\in\C\left[\left[X\right]\right]$ exists. Then $\left\Vert \sum_{j}f_{j}\right\Vert _{r}\leq\sum_{j}\left\Vert f_{j}\right\Vert _{r}$. -Assume $\sum_{j}\left\Vert f_{j}\right\Vert <\infty$, then $\sum_{j}f_{j}\in\C\left\{ X\right\} _{r}$ -and\end{xca*} -\begin{enumerate} -\item $\forall\epsilon>0\ \exists\textrm{ finite }I_{\epsilon}\subset J$ s.t -$\forall\textrm{ finite }I_{\epsilon}\subset I\subset J$, $\left\Vert \sum_{j\in J}f_{j}-\sum_{j\in I}f_{j}\right\Vert _{r}<\epsilon$. -\item $\forall x\in D_{r}\left(0\right)$, $\left(\sum_{j\in J}f_{j}\right)\left(x\right)=\sum_{j\in J}f_{j}\left(x\right)$.\end{enumerate} -\begin{lem*} -\label{10.6} (Abel) Let $f\in\C\left[\left[X\right]\right]$, $s\in\left(\R^{>0}\right)^{m}$, -$L\in\R^{>0}$, s.t $\left|f_{i}\right|s^{i}\leq L$ for all $i$.\\ -Then $f\in\C\left\{ X\right\} _{r}$ for all $r0}$ small -enough such that for $r=\left(r_{1},...,r_{m+1}\right)$, $\norm[g]_{r}$, -$\norm[u^{-1}]_{r}$, \\ -$\norm[f_{0}]_{r'}$,...,$\norm[f_{d-1}]_{r'}<\infty$. Thus -\[ -\norm[F]_{r}\leq\norm[u^{-1}]_{r}\sum_{i0\ \exists\textrm{ finite }I_{\epsilon}\subset J$ s.t +$\forall\textrm{ finite }I_{\epsilon}\subset I\subset J$, $\left\Vert \sum_{j\in J}f_{j}-\sum_{j\in I}f_{j}\right\Vert _{r}<\epsilon$. +\item $\forall x\in D_{r}\left(0\right)$, $\left(\sum_{j\in J}f_{j}\right)\left(x\right)=\sum_{j\in J}f_{j}\left(x\right)$.\end{enumerate} +\begin{lem*} +\label{10.6} (Abel) Let $f\in\C\left[\left[X\right]\right]$, $s\in\left(\R^{>0}\right)^{m}$, +$L\in\R^{>0}$, s.t $\left|f_{i}\right|s^{i}\leq L$ for all $i$.\\ +Then $f\in\C\left\{ X\right\} _{r}$ for all $r0}$ small +enough such that for $r=\left(r_{1},...,r_{m+1}\right)$, $\norm[g]_{r}$, +$\norm[u^{-1}]_{r}$, \\ +$\norm[f_{0}]_{r'}$,...,$\norm[f_{d-1}]_{r'}<\infty$. 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-\section*{Day 1: Friday October 3, 2014} -Surreal numbers were discovered by John Conway. -The class of all surreal numbers is denoted $\No$ and -this class comes equipped with a natural linear ordering and -arithmetic operations making $\No$ a real closed ordered field. - -For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, -where $\omega$ denotes the first infinite ordinal. - -\begin{theorem}[Kruskal, 1980s] - There is an exponential function $\exp \colon \No \rar \No$ - exteding the usual exponential $x \mapsto e^x$ on $\R$. - \label{} -\end{theorem} - -\begin{theorem}[van den Dries-Ehrlich, c. 2000] - $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq - (\No, 0, 1, +, \cdot, \leq, \exp)$. - \label{} -\end{theorem} - -\subsection*{Basic Definitions and Existence Theorem} -Throughout this class, we will work in von Neumann-Bernays-G\"odel -set theory with global choice ($\NBG$). This is conservative over -$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). - -An example of a surreal number is the following: -\begin{align*} - f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ - 0 &\longmapsto + \\ - 1 &\longmapsto - \\ - 2 &\longmapsto + -\end{align*} -This may be depicted in tree form as follows: -%------------------------Beautiful Tree Diagram------------------------------------- -%------------------------DO NOT ALTER IN ANY WAY------------------------------------ -%----------------------Violators WILL be prosecuted--------------------------------- -%----The above is not meant to exclude the possibility of extrajudical punishment--- -%--------------------------------------------------------------------- -We will denote such a surreal number by $f=(+-+)$ -Another example is: -\begin{align*} - f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ - n &\longmapsto + \\ - \omega + n &\longmapsto - -\end{align*} -We write $\No$ for the class of surreal numbers. We often view -$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by -setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. - -\begin{defn} - Let $a, b \in \No$. - \begin{enumerate} - \item We say that $a$ is an \emph{initial segment} of - $b$ if $l(a) \leq l(b)$ and $b \restriction - \dom{a} = a$. We denote this by $a \leq_s b$ - (read: ``$a$ is simpler than $b$''). - \item We say that $a$ is a \emph{proper initial segment} - of $b$ if $a \leq_s b$ and $a \neq b$. We denote - this by $a <_s b$. - \item If $a \leq_s b$, then the \emph{tail} of $a$ in - $b$ is the surreal number $c$ of length - $l(b) - l(a)$ satisfying $c(\alpha) = - a(l(b) + \alpha)$ for all $\alpha$. - \item We define $a \concat b$ to be the surreal number - satisfying: - \begin{align*} - (a \concat b)(\alpha) = - \begin{cases} - a(\alpha) & \alpha < l(a) \\ - b(\alpha - l(a)) & \alpha \geq l(a) - \end{cases} - \end{align*} - (so in particular if $a \leq_s b$ and $c$ is the tail - of $a$ in $b$, then $b = a \concat c$). - \item Suppose $a \neq b$. Then the \emph{common initial - segment} of $a$ and $b$ is the element - $c \in \No$ with minimal length such that - $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) - = a \restriction - l(c) = b \restriction l(c)$. We write - $c = a \wedge b$, and also set $a \wedge a = a$. - \end{enumerate} -\end{defn} -Note that -\begin{align*} - a \leq_s b \iff a \wedge b = a -\end{align*} - -\section*{Day 2: Monday October 6, 2014} -\begin{defn} - We order $\left\{ +, -, 0 \right\}$ by setting - $- < 0 < +$ and for $a, b \in \No$ we define - \begin{align*} - a < b &\iff a < b \text{ lexicographically} \\ - &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) - \text{ where } \alpha_0 = l(a \wedge b) - \end{align*} - As usual we also set $a \leq b \iff a < b \lor a = b$. -\end{defn} -Clearly $\leq$ is a linear ordering on $\No$. - -Examples: -\begin{align*} - (+-+) < (+++ \cdots --- \cdots) \\ - (-+) < () < (+-) < (+) < (++) -\end{align*} -Remark: if $a \leq_s b$ then $a \wedge b = a$ and if -$b \leq_s a$ then $a \wedge b = b$. Suppose that neither -$a \leq_s b$ or $b \leq_s a$. Put: -\begin{align*} - \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} -\end{align*} -Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which -case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, -in which case $a < (a \wedge b) < b$. In either case: -\begin{align*} - a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} -\end{align*} - -\begin{defn} - Let $L, R$ be subsets (or subclasses) of $\No$. We say - $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define - $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. -\end{defn} -Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by -vacuous satisfaction. - -\begin{theorem}[Existence Theorem] - Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. - Then there exists a unique $c \in \No$ of minimal length - such that $L < c < R$. - \label{} -\end{theorem} -\begin{proof} -%--------------Redundant Section (Covered at beginning of next day)------------------ -% First assume that there exists $c \in \No$ with $L < c < R$. -% By minimizing over the lengths of all such $c$ (using the fact that -% the ordinals are well-ordered), we may assume without loss of -% generality that $c$ has minimal length. But then it is immediate -% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied -% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by -% the note at the beginning of this section we would have: -% \begin{align*} -% L < \min{ \curly{c, \tilde{c}}} -% < (c \land \tilde{c}) < \max{ \curly{c, -% \tilde{c}}} < R -% \end{align*} -% contradicting minimality of $l(c)$. -% -% Now for existence: let -%------------------------------------------------------------------------------------ - We first prove existence. Let - \begin{align*} - \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} - \end{align*} - be the least strict upper bound of lengths of elements of - $L \cup R$ (it is here that we use that $L$ and $R$ are sets - rather than proper classes). For each ordinal $\alpha$, - denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha - \colon l \in L}$, and similarly for $R$. Note that - $L \restriction \gamma = L$ and $R \restriction \gamma = R$. - We construct $c$ of length $\gamma$ by defining the - values $c(\alpha)$ by induction on - $\alpha \leq \gamma$ as follows: - \begin{align*} - c(\alpha) = - \begin{cases} - - & \text{ if } - (c \restriction \alpha \concat (-) ) \geq - L \restriction (\alpha + 1) \\ - + & \text{ otherwise} - \end{cases} - \end{align*} - \begin{claim} - $c \geq L$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there is $l \in L$ such that - $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ - where $\alpha_0 = l(c \wedge l)$. Since - $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. - is nonzero) this implies $c(\alpha_0) = -$ even though - $(c \restriction \alpha_0 \concat (-)) \not \geq - l \restriction (\alpha_0 + 1)$, a contradiction. - \end{proof} - \begin{claim} - $c \leq R$ - \end{claim} - \begin{proof}[Proof of Claim] - Otherwise there exists $r \in R$ such that - $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ - where $\alpha_0 = l(r \land c)$. - %We may assume - %that $\alpha_0$ is least possible, i.e. that - %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ - %for all $r' \in R$. - Since $c(\alpha_0) > r(\alpha_0)$, - we must be in the ``$c(\alpha_0) = +$'' case, and so - there is some $l \in L$ such that - $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) - \concat (-) = (r \restriction \alpha_0) \concat (-)$. - In particular $l(\alpha_0) \in \curly{0, +}$. - So if $r(\alpha_0) = -$ then $r < l$, and if - $r(\alpha_0) = 0$ then $r \leq l$, in either - case contradicting $L < R$. - \end{proof} - At this point we have shown $L \leq c \leq R$. - But by construction $c$ has length $\gamma$, and so - in particular cannot be an element of $L \cup R$. - Thus - \begin{align*} - L < c < R - \end{align*} - as desired. -\end{proof} - -\section*{Day 3: Wednesday October 8, 2014} -Last time we showed that there is $c \in \No$ with $L < c < R$. -The well-ordering principle on $\On$ gives us such a $c$ of minimal -length. Let now $d \in \No$ satisfy $L < d < R$. Then -$L < (c \wedge d) < R$. By minimality of $l(c)$ and since -$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. -Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. - -Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ -of minimal length with $L < c < R$. Some remarks: -\begin{enumerate}[(1)] - \item $\left\{ L \vert \emptyset \right\}$ consists only of - $+$'s. - \item $\left\{ \emptyset \vert R \right\}$ consists only of - $-$'s. -\end{enumerate} -\begin{lem} - If $L < R$ are subsets of $\No$, then - \begin{align*} - l( \curly{L \vert R}) \leq - \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all - $b \in L \cup R$} }} - \end{align*} - Conversely, every $a \in \No$ is of the form - $a = \curly{L \vert R}$ where $L < R$ are subsets of - $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. - \label{lemma_on_length_of_cuts} -\end{lem} -\begin{proof} - Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > - \alpha > l(b)$ for all $b \in L \cup R$. Then - $c \coloneq \curly{L \vert R} \restriction \alpha$ also - satsfies $L < c < R$, contradicting the minimality of - $l(\left\{ L \vert R \right\})$. For the second part, let - $a \in \No$ and set $\alpha \coloneq l(a)$. Put: - \begin{align*} - L &\coloneq \curly{b \in \No \colon b < a - \text{ and } l(b) < \alpha} \\ - R &\coloneq \curly{b \in \No \colon - b > a \text{ and } l(b) < \alpha} - \end{align*} - Then $L < a < R$ and $L \cup R$ contains all surreals of - length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. -\end{proof} -\begin{defn} - Let $L, L', R, R'$ be subsets of $\No$. We say that - $(L', R')$ is \emph{cofinal} in $(L, R)$ if: - \begin{itemize} - \item $(\forall a \in L)(\exists a' \in L')$ - such that $a \leq a'$, and - \item $(\forall b \in R)(\exists b' \in R')$ - such that $b \geq b'$. - \end{itemize} -\end{defn} -Some trivial observations: -\begin{itemize} - \item If $L' \supseteq L$ and $R' \supseteq R$, then - $(L', R')$ is cofinal in $(L, R)$ and in - particular $(L, R)$ is cofinal in $(L, R)$. - \item Cofinality is transitive. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < R'$, then $L < R$. - \item If $(L', R')$ is cofinal in $(L, R)$ and - $L' < a < R'$, then $L < a < R$. -\end{itemize} -\begin{theorem}[The ``Cofinality Theorem''] - Let $L, L', R, R'$ be subsets of $\No$ with - $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and - $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert - R\right\} = \curly{L' \vert R'}$. - \label{cofinality_theorem} -\end{theorem} -\begin{proof} - Suppose that $L' < a < R'$. Then $L < a < R$ since - $(L', R')$ is cofinal in $(L, R)$. Hence - $l(a) \geq l( \curly{L \vert R})$. Thus - $\left\{ L \vert R \right\} = \curly{L \vert R'}$. -\end{proof} -\begin{cor}[Canonical Representation] - Let $a \in \No$ and set - \begin{align*} - L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ - R' &= \curly{b \colon b > a \text{ and } b <_s a} - \end{align*} - Then $a = \curly{L' \vert R'}$. -\end{cor} -\begin{proof} - By Lemma \ref{lemma_on_length_of_cuts} take - $L < R$ such that $a = \curly{L \vert R}$ and - $l(b) < l(a)$ for all $b \in L \cup R$. Then - $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is - cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} - it remains to show that $(L', R')$ is cofinal in - $(L, R)$. - - For this let $b \in L$ be arbitrary. Then - $l(a \wedge b) \leq l(b) < l(a)$ and - thus $b \leq (a \wedge b) < a$. Therefore - $a \wedge b \in L'$. Similarly for $R$. -\end{proof} -Exercise: let $a = \curly{L' \vert R'}$ be the canonical -representation of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = -} -\end{align*} - -Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation -of $a \in \No$. Then -\begin{align*} - L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ - R' &= \curly{a \restriction \beta \colon a(\beta) = 1} -\end{align*} -For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ -and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of -$L'$ decrease in the ordering as their length increases, whereas those -of $R'$ do the opposite. Also note that the canonical representation -is not minimal, as $a$ may also be realized as the cut -$a = \curly{(++-+--) \vert (++-+-)}$. -\begin{cor}[``Inverse Cofinality Theorem''] - Let $a = \curly{L \vert R}$ be the canonical representation - of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary - representation. Then $(L', R')$ is cofinal in $(L, R)$. - \label{inverse_cofinality_theorem} -\end{cor} -\begin{proof} - Let $b \in L$ and suppose that for a contradiction that - $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, - contradicting $a = \curly{L' \vert R'}$. -\end{proof} -\subsection*{Arithmetic Operators} -We will define addition and multiplication on $\No$ and we will -show that they, together with the ordering, make $\No$ into -an ordered field. -\section*{Day 4: Friday, October 10, 2014} -We begin by recalling some facts about ordinal arithmetic: -\begin{theorem}[Cantor's Normal Form Theorem] - Every ordinal $\alpha$ can be uniquely represented as - \begin{align*} - \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} - a_2 + \cdots + \omega^{\alpha_n} a_n - \end{align*} - where $\alpha_1 > \cdots > \alpha_n$ are ordinals and - $a_1, \cdots, a_n \in \N \setminus \curly{0}$. - \label{} -\end{theorem} -\begin{defn} - The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of - two ordinals - \begin{align*} - \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} - a_n \\ - \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} - b_n - \end{align*} - where $\gamma_1 > \cdots > \gamma_n$ are ordinals and - $a_i, b_j \in \N$, is defined by: - \begin{align*} - a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots - + \omega^{\gamma_n}(a_n + b_n) - \end{align*} -\end{defn} -The operation $\oplus$ is associative, commutative, and strictly increasing -in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus -\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence -$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus -\gamma \implies \alpha = \beta$. There is also a notion of -\emph{natural product} of ordinals: -\begin{defn} - For $\alpha, \beta$ as above, set - \begin{align*} - \alpha \otimes \beta \coloneq - \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i - b_j} - \end{align*} -\end{defn} -The natural product is also associative, commutative, and strictly -increasing in each argument. The distributive law also holds for -$\oplus$, $\otimes$: -\begin{align*} - \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) - \oplus (\alpha \otimes \gamma) -\end{align*} -In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover -strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = -1 + \omega$. - -%In the following, if $a = \curly{L \vert R}$ is the canonical -%representation of $a \in \No$ then we let $a_L$ range over -%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). -In the following, if $a = \curly{L \vert R}$ is the canonical -representation of $a \in \No$, we set $L(a) = L$ and -$R(a) = R$. We will use the shorthand $X + a = -\left\{ x + a \colon x \in X \right\}$ (and its obvious -variations) for $X$ a subset of -$\No$ and $a \in \No$. - -\begin{defn} - Let $a, b \in \No$. Set - \begin{align} - a + b \coloneq - \left\{ (L(a) + b) \cup (L(b) + a) \vert - (R(a) + b) \cup (R(b) + a) \right\} - \label{defn_of_surreal_sum} - \end{align} -\end{defn} -Some remarks: -\begin{enumerate}[(1)] - \item This is an inductive definition on $l(a) \oplus l(b)$. - There is no special treatment needed for the base - case: $\left\{ \emptyset \vert \emptyset \right\} = - + \curly{\emptyset \vert \emptyset} = - \left\{ \emptyset \vert \emptyset \right\}$. - \item To justify the definition we need to check that - the sets $L, R$ used in defining $a + b = - \left\{ L \vert R \right\}$ satisfy $L < R$. -\end{enumerate} -\begin{lem} - Suppose that for all $a, b \in \No$ with $l(a) \oplus - l(b) < \gamma$ we have defined $a + b$ so that - Equation \ref{defn_of_surreal_sum} holds and - \begin{align*} - b > c \implies a + b > a + c - \text{ and } b + a > c + a - \tag{$*$} - \end{align*} - holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then - for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have - \begin{align*} - (L(a) + b) \cup (L(b) + a) < - (R(a) + b) \cup (R(b) + a) - \end{align*} - and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, - $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus - l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. -\end{lem} -\begin{proof} - The first part is immediate from $(*)$ in conjunction with the - fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ - for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and - $b_R \in R(b)$. -Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq -\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose -$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq -\gamma$, and $b > c$. Then by definition we have -\begin{align*} - a + b_L < \;& a + b \\ - & a + c < a + c_R -\end{align*} -for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then -we can take $b_L = c$ and get $a + b > a + c$. Similarly, if -$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. -Suppose neither $c <_s b$ nor $b <_s c$ and put -$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and -$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. - -We may show $b + a > c + a$ similarly. -\end{proof} -\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] - Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. - Then - \begin{align*} - a + a' = - \left\{ (L + a') \cup (a' + L) \vert - (R + a') \cup (a + R') \right\} - \end{align*} -\end{lem} -\begin{proof} - Let $a = \curly{L_a \vert R_a}$ be the canonical - representation. By Corollary \ref{inverse_cofinality_theorem} - $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is - cofinal in $(L_{a'}, R_{a'})$. Hence - \begin{align*} - \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} - \end{align*} - is cofinal in - \begin{align*} - \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup - (a + R_{a'})} - \end{align*} - Moreover, - \begin{align*} - (L + a') \cup (a + L') < a + a' < - (R + a') \cup (a + R') - \end{align*} - Now use Theorem \ref{cofinality_theorem} to conclude the - proof. +\textit{Notes by John Suice} + +\section*{Day 1: Friday October 3, 2014} +Surreal numbers were discovered by John Conway. +The class of all surreal numbers is denoted $\No$ and +this class comes equipped with a natural linear ordering and +arithmetic operations making $\No$ a real closed ordered field. + +For example, $1/ \omega, \omega - \pi, \sqrt{\omega} \in \No$, +where $\omega$ denotes the first infinite ordinal. + +\begin{theorem}[Kruskal, 1980s] + There is an exponential function $\exp \colon \No \rar \No$ + exteding the usual exponential $x \mapsto e^x$ on $\R$. + \label{} +\end{theorem} + +\begin{theorem}[van den Dries-Ehrlich, c. 2000] + $(\R, 0, 1, +, \cdot, \leq, e^x) \preccurlyeq + (\No, 0, 1, +, \cdot, \leq, \exp)$. + \label{} +\end{theorem} + +\subsection*{Basic Definitions and Existence Theorem} +Throughout this class, we will work in von Neumann-Bernays-G\"odel +set theory with global choice ($\NBG$). This is conservative over +$\ZFC$ (see Ehrlich, \emph{Absolutely Saturated Models}). + +An example of a surreal number is the following: +\begin{align*} + f \colon \curly{0, 1, 2} &\longrightarrow \curly{+, -} \\ + 0 &\longmapsto + \\ + 1 &\longmapsto - \\ + 2 &\longmapsto + +\end{align*} +This may be depicted in tree form as follows: +%------------------------Beautiful Tree Diagram------------------------------------- +%------------------------DO NOT ALTER IN ANY WAY------------------------------------ +%----------------------Violators WILL be prosecuted--------------------------------- +%----The above is not meant to exclude the possibility of extrajudical punishment--- +%--------------------------------------------------------------------- +We will denote such a surreal number by $f=(+-+)$ +Another example is: +\begin{align*} + f \colon \omega + \omega &\longrightarrow \curly{+, -} \\ + n &\longmapsto + \\ + \omega + n &\longmapsto - +\end{align*} +We write $\No$ for the class of surreal numbers. We often view +$f \in \No$ as a function $f \colon \On \rar \curly{+, -, 0}$ by +setting $f(\alpha) = 0$ for $\alpha \notin \dom{f}$. + +\begin{defn} + Let $a, b \in \No$. + \begin{enumerate} + \item We say that $a$ is an \emph{initial segment} of + $b$ if $l(a) \leq l(b)$ and $b \restriction + \dom{a} = a$. We denote this by $a \leq_s b$ + (read: ``$a$ is simpler than $b$''). + \item We say that $a$ is a \emph{proper initial segment} + of $b$ if $a \leq_s b$ and $a \neq b$. We denote + this by $a <_s b$. + \item If $a \leq_s b$, then the \emph{tail} of $a$ in + $b$ is the surreal number $c$ of length + $l(b) - l(a)$ satisfying $c(\alpha) = + a(l(b) + \alpha)$ for all $\alpha$. + \item We define $a \concat b$ to be the surreal number + satisfying: + \begin{align*} + (a \concat b)(\alpha) = + \begin{cases} + a(\alpha) & \alpha < l(a) \\ + b(\alpha - l(a)) & \alpha \geq l(a) + \end{cases} + \end{align*} + (so in particular if $a \leq_s b$ and $c$ is the tail + of $a$ in $b$, then $b = a \concat c$). + \item Suppose $a \neq b$. Then the \emph{common initial + segment} of $a$ and $b$ is the element + $c \in \No$ with minimal length such that + $a(l(c)) \neq b(l(c))$ and $c \restriction l(c) + = a \restriction + l(c) = b \restriction l(c)$. We write + $c = a \wedge b$, and also set $a \wedge a = a$. + \end{enumerate} +\end{defn} +Note that +\begin{align*} + a \leq_s b \iff a \wedge b = a +\end{align*} + +\section*{Day 2: Monday October 6, 2014} +\begin{defn} + We order $\left\{ +, -, 0 \right\}$ by setting + $- < 0 < +$ and for $a, b \in \No$ we define + \begin{align*} + a < b &\iff a < b \text{ lexicographically} \\ + &\iff a \neq b \land a(\alpha_0) < b(\alpha_0) + \text{ where } \alpha_0 = l(a \wedge b) + \end{align*} + As usual we also set $a \leq b \iff a < b \lor a = b$. +\end{defn} +Clearly $\leq$ is a linear ordering on $\No$. + +Examples: +\begin{align*} + (+-+) < (+++ \cdots --- \cdots) \\ + (-+) < () < (+-) < (+) < (++) +\end{align*} +Remark: if $a \leq_s b$ then $a \wedge b = a$ and if +$b \leq_s a$ then $a \wedge b = b$. Suppose that neither +$a \leq_s b$ or $b \leq_s a$. Put: +\begin{align*} + \alpha_0 = \min{ \curly{\alpha \colon a(\alpha) \neq b(\alpha)}} +\end{align*} +Then either $a(\alpha_0) = +$ and $b(\alpha_0) = -$, in which +case $b < (a \wedge b) < a$, or $a(\alpha_0) = -$ and $b(\alpha_0) = +$, +in which case $a < (a \wedge b) < b$. In either case: +\begin{align*} + a \wedge b \in \brac{ \min{ \curly{a, b}, \max{ \curly{a, b}}}} +\end{align*} + +\begin{defn} + Let $L, R$ be subsets (or subclasses) of $\No$. We say + $L < R$ if $l < r$ for all $l \in L$ and $r \in R$. We define + $A < c$ for $A \cup \curly{c} \subseteq \No$ in the obvious manner. +\end{defn} +Note that $\emptyset < A$ and $A < \emptyset$ for all $A \subseteq \No$ by +vacuous satisfaction. + +\begin{theorem}[Existence Theorem] + Let $L, R$ be sub\emph{sets} of $\No$ with $L < R$. + Then there exists a unique $c \in \No$ of minimal length + such that $L < c < R$. + \label{} +\end{theorem} +\begin{proof} +%--------------Redundant Section (Covered at beginning of next day)------------------ +% First assume that there exists $c \in \No$ with $L < c < R$. +% By minimizing over the lengths of all such $c$ (using the fact that +% the ordinals are well-ordered), we may assume without loss of +% generality that $c$ has minimal length. But then it is immediate +% that $c$ is unique; for if $\tilde{c} \neq c$ satisfied +% $L < \tilde{c} < R$ and $l(c) = l(\tilde{c})$, then by +% the note at the beginning of this section we would have: +% \begin{align*} +% L < \min{ \curly{c, \tilde{c}}} +% < (c \land \tilde{c}) < \max{ \curly{c, +% \tilde{c}}} < R +% \end{align*} +% contradicting minimality of $l(c)$. +% +% Now for existence: let +%------------------------------------------------------------------------------------ + We first prove existence. Let + \begin{align*} + \gamma = \sup_{a \in L \cup R}{(l(a) + 1)} + \end{align*} + be the least strict upper bound of lengths of elements of + $L \cup R$ (it is here that we use that $L$ and $R$ are sets + rather than proper classes). For each ordinal $\alpha$, + denote by $L\restriction \alpha$ the set $\curly{l \restriction \alpha + \colon l \in L}$, and similarly for $R$. Note that + $L \restriction \gamma = L$ and $R \restriction \gamma = R$. + We construct $c$ of length $\gamma$ by defining the + values $c(\alpha)$ by induction on + $\alpha \leq \gamma$ as follows: + \begin{align*} + c(\alpha) = + \begin{cases} + - & \text{ if } + (c \restriction \alpha \concat (-) ) \geq + L \restriction (\alpha + 1) \\ + + & \text{ otherwise} + \end{cases} + \end{align*} + \begin{claim} + $c \geq L$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there is $l \in L$ such that + $c < l$. This means $c(\alpha_0) < l(\alpha_0)$ + where $\alpha_0 = l(c \wedge l)$. Since + $c(\alpha_0)$ must be in $\left\{ -, + \right\}$ (i.e. + is nonzero) this implies $c(\alpha_0) = -$ even though + $(c \restriction \alpha_0 \concat (-)) \not \geq + l \restriction (\alpha_0 + 1)$, a contradiction. + \end{proof} + \begin{claim} + $c \leq R$ + \end{claim} + \begin{proof}[Proof of Claim] + Otherwise there exists $r \in R$ such that + $r < c$. This means $r(\alpha_0) < c(\alpha_0)$ + where $\alpha_0 = l(r \land c)$. + %We may assume + %that $\alpha_0$ is least possible, i.e. that + %$c \restriction \alpha_0 \leq r' \restriction \alpha_0$ + %for all $r' \in R$. + Since $c(\alpha_0) > r(\alpha_0)$, + we must be in the ``$c(\alpha_0) = +$'' case, and so + there is some $l \in L$ such that + $l \restriction (\alpha_0 + 1) > (c \restriction \alpha_0) + \concat (-) = (r \restriction \alpha_0) \concat (-)$. + In particular $l(\alpha_0) \in \curly{0, +}$. + So if $r(\alpha_0) = -$ then $r < l$, and if + $r(\alpha_0) = 0$ then $r \leq l$, in either + case contradicting $L < R$. + \end{proof} + At this point we have shown $L \leq c \leq R$. + But by construction $c$ has length $\gamma$, and so + in particular cannot be an element of $L \cup R$. + Thus + \begin{align*} + L < c < R + \end{align*} + as desired. +\end{proof} + +\section*{Day 3: Wednesday October 8, 2014} +Last time we showed that there is $c \in \No$ with $L < c < R$. +The well-ordering principle on $\On$ gives us such a $c$ of minimal +length. Let now $d \in \No$ satisfy $L < d < R$. Then +$L < (c \wedge d) < R$. By minimality of $l(c)$ and since +$(c \wedge d) \leq_s c$ we have $l(c \wedge d) = l(c)$. +Therefore $(c \wedge d) = c$, or in other words $c \leq_s d$. + +Notation: $\left\{ L \vert R \right\}$ denotes the $c \in \No$ +of minimal length with $L < c < R$. Some remarks: +\begin{enumerate}[(1)] + \item $\left\{ L \vert \emptyset \right\}$ consists only of + $+$'s. + \item $\left\{ \emptyset \vert R \right\}$ consists only of + $-$'s. +\end{enumerate} +\begin{lem} + If $L < R$ are subsets of $\No$, then + \begin{align*} + l( \curly{L \vert R}) \leq + \min{ \curly{\alpha \colon l(b) < \alpha \text{ for all + $b \in L \cup R$} }} + \end{align*} + Conversely, every $a \in \No$ is of the form + $a = \curly{L \vert R}$ where $L < R$ are subsets of + $\No$ such that $l(b) < l(a)$ for all $b \in L \cup R$. + \label{lemma_on_length_of_cuts} +\end{lem} +\begin{proof} + Suppose that $\alpha$ satisfies $l(\left\{ L \vert R \right\}) > + \alpha > l(b)$ for all $b \in L \cup R$. Then + $c \coloneq \curly{L \vert R} \restriction \alpha$ also + satsfies $L < c < R$, contradicting the minimality of + $l(\left\{ L \vert R \right\})$. For the second part, let + $a \in \No$ and set $\alpha \coloneq l(a)$. Put: + \begin{align*} + L &\coloneq \curly{b \in \No \colon b < a + \text{ and } l(b) < \alpha} \\ + R &\coloneq \curly{b \in \No \colon + b > a \text{ and } l(b) < \alpha} + \end{align*} + Then $L < a < R$ and $L \cup R$ contains all surreals of + length $< \alpha = l(a)$. So $a = \curly{L \vert R}$. +\end{proof} +\begin{defn} + Let $L, L', R, R'$ be subsets of $\No$. We say that + $(L', R')$ is \emph{cofinal} in $(L, R)$ if: + \begin{itemize} + \item $(\forall a \in L)(\exists a' \in L')$ + such that $a \leq a'$, and + \item $(\forall b \in R)(\exists b' \in R')$ + such that $b \geq b'$. + \end{itemize} +\end{defn} +Some trivial observations: +\begin{itemize} + \item If $L' \supseteq L$ and $R' \supseteq R$, then + $(L', R')$ is cofinal in $(L, R)$ and in + particular $(L, R)$ is cofinal in $(L, R)$. + \item Cofinality is transitive. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < R'$, then $L < R$. + \item If $(L', R')$ is cofinal in $(L, R)$ and + $L' < a < R'$, then $L < a < R$. +\end{itemize} +\begin{theorem}[The ``Cofinality Theorem''] + Let $L, L', R, R'$ be subsets of $\No$ with + $L < R$. Suppose $L' < \curly{L \vert R} < R'$ and + $(L', R')$ is cofinal in $(L, R)$. Then $\left\{ L \vert + R\right\} = \curly{L' \vert R'}$. + \label{cofinality_theorem} +\end{theorem} +\begin{proof} + Suppose that $L' < a < R'$. Then $L < a < R$ since + $(L', R')$ is cofinal in $(L, R)$. Hence + $l(a) \geq l( \curly{L \vert R})$. Thus + $\left\{ L \vert R \right\} = \curly{L \vert R'}$. +\end{proof} +\begin{cor}[Canonical Representation] + Let $a \in \No$ and set + \begin{align*} + L' &= \curly{b \colon b < a \text{ and } b <_s a} \\ + R' &= \curly{b \colon b > a \text{ and } b <_s a} + \end{align*} + Then $a = \curly{L' \vert R'}$. +\end{cor} +\begin{proof} + By Lemma \ref{lemma_on_length_of_cuts} take + $L < R$ such that $a = \curly{L \vert R}$ and + $l(b) < l(a)$ for all $b \in L \cup R$. Then + $L' \subseteq L$ and $R' \subseteq R$, so $(L, R)$ is + cofinal in $(L', R')$. By Theorem \ref{cofinality_theorem} + it remains to show that $(L', R')$ is cofinal in + $(L, R)$. + + For this let $b \in L$ be arbitrary. Then + $l(a \wedge b) \leq l(b) < l(a)$ and + thus $b \leq (a \wedge b) < a$. Therefore + $a \wedge b \in L'$. Similarly for $R$. +\end{proof} +Exercise: let $a = \curly{L' \vert R'}$ be the canonical +representation of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = -} +\end{align*} + +Exercise: Let $a = \curly{L' \vert R'}$ be the canonical representation +of $a \in \No$. Then +\begin{align*} + L' &= \curly{a \restriction \beta \colon a(\beta) = +} \\ + R' &= \curly{a \restriction \beta \colon a(\beta) = 1} +\end{align*} +For example, if $a = (++-+--+)$, then $L' = \{(), (+), (++-), (++-+--)\}$ +and $R' = \{(++), (++-+), (++-+-)\}$. Note that the elements of +$L'$ decrease in the ordering as their length increases, whereas those +of $R'$ do the opposite. Also note that the canonical representation +is not minimal, as $a$ may also be realized as the cut +$a = \curly{(++-+--) \vert (++-+-)}$. +\begin{cor}[``Inverse Cofinality Theorem''] + Let $a = \curly{L \vert R}$ be the canonical representation + of $a$ and let $a = \curly{L' \vert R'}$ be an arbitrary + representation. Then $(L', R')$ is cofinal in $(L, R)$. + \label{inverse_cofinality_theorem} +\end{cor} +\begin{proof} + Let $b \in L$ and suppose that for a contradiction that + $L' < b$. Then $L' < b < a < R'$, and $l(b) < l(a)$, + contradicting $a = \curly{L' \vert R'}$. +\end{proof} +\subsection*{Arithmetic Operators} +We will define addition and multiplication on $\No$ and we will +show that they, together with the ordering, make $\No$ into +an ordered field. +\section*{Day 4: Friday, October 10, 2014} +We begin by recalling some facts about ordinal arithmetic: +\begin{theorem}[Cantor's Normal Form Theorem] + Every ordinal $\alpha$ can be uniquely represented as + \begin{align*} + \alpha = \omega^{\alpha_1} a_1 + \omega^{\alpha_2} + a_2 + \cdots + \omega^{\alpha_n} a_n + \end{align*} + where $\alpha_1 > \cdots > \alpha_n$ are ordinals and + $a_1, \cdots, a_n \in \N \setminus \curly{0}$. + \label{} +\end{theorem} +\begin{defn} + The (Hessenberg) \emph{natural sum} $\alpha \oplus \beta$ of + two ordinals + \begin{align*} + \alpha &= \omega^{\gamma_1} a_1 + \cdots \omega^{\gamma_n} + a_n \\ + \beta &= \omega^{\gamma_1} b_1 + \cdots \omega^{\gamma_n} + b_n + \end{align*} + where $\gamma_1 > \cdots > \gamma_n$ are ordinals and + $a_i, b_j \in \N$, is defined by: + \begin{align*} + a \oplus \beta = \omega^{\gamma_1}(a_1 + b_1) + \cdots + + \omega^{\gamma_n}(a_n + b_n) + \end{align*} +\end{defn} +The operation $\oplus$ is associative, commutative, and strictly increasing +in each argument, i.e. $\alpha < \beta \implies a \oplus \gamma < \beta \oplus +\gamma$ for all $\alpha, \beta, \gamma \in \On$. Hence +$\oplus$ is \emph{cancellative}: $\alpha \oplus \gamma = \beta \oplus +\gamma \implies \alpha = \beta$. There is also a notion of +\emph{natural product} of ordinals: +\begin{defn} + For $\alpha, \beta$ as above, set + \begin{align*} + \alpha \otimes \beta \coloneq + \bigoplus_{i, j}{\omega^{\gamma_i \oplus \gamma_j}a_i + b_j} + \end{align*} +\end{defn} +The natural product is also associative, commutative, and strictly +increasing in each argument. The distributive law also holds for +$\oplus$, $\otimes$: +\begin{align*} + \alpha \otimes (\beta \oplus \gamma) = (\alpha \otimes \beta) + \oplus (\alpha \otimes \gamma) +\end{align*} +In general $\alpha \oplus \beta \geq \alpha + \beta$. Moreover +strict inequality may occur: $1 \oplus \omega = \omega + 1 > \omega = +1 + \omega$. + +%In the following, if $a = \curly{L \vert R}$ is the canonical +%representation of $a \in \No$ then we let $a_L$ range over +%$L$ and $a_R$ range over $R$ (so in particular $a_L < a < a_R$). +In the following, if $a = \curly{L \vert R}$ is the canonical +representation of $a \in \No$, we set $L(a) = L$ and +$R(a) = R$. We will use the shorthand $X + a = +\left\{ x + a \colon x \in X \right\}$ (and its obvious +variations) for $X$ a subset of +$\No$ and $a \in \No$. + +\begin{defn} + Let $a, b \in \No$. Set + \begin{align} + a + b \coloneq + \left\{ (L(a) + b) \cup (L(b) + a) \vert + (R(a) + b) \cup (R(b) + a) \right\} + \label{defn_of_surreal_sum} + \end{align} +\end{defn} +Some remarks: +\begin{enumerate}[(1)] + \item This is an inductive definition on $l(a) \oplus l(b)$. + There is no special treatment needed for the base + case: $\left\{ \emptyset \vert \emptyset \right\} = + + \curly{\emptyset \vert \emptyset} = + \left\{ \emptyset \vert \emptyset \right\}$. + \item To justify the definition we need to check that + the sets $L, R$ used in defining $a + b = + \left\{ L \vert R \right\}$ satisfy $L < R$. +\end{enumerate} +\begin{lem} + Suppose that for all $a, b \in \No$ with $l(a) \oplus + l(b) < \gamma$ we have defined $a + b$ so that + Equation \ref{defn_of_surreal_sum} holds and + \begin{align*} + b > c \implies a + b > a + c + \text{ and } b + a > c + a + \tag{$*$} + \end{align*} + holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) < \gamma$ and $l(a) \oplus l(c) < \gamma$. Then + for all $a, b \in \No$ with $l(a) \oplus l(b) \leq \gamma$ we have + \begin{align*} + (L(a) + b) \cup (L(b) + a) < + (R(a) + b) \cup (R(b) + a) + \end{align*} + and defining $a + b$ as in Equation \ref{defn_of_surreal_sum}, + $(*)$ holds for all $a, b, c \in \No$ with $l(a) \oplus + l(b) \leq \gamma$ and $l(a) \oplus l(c) \leq \gamma$. +\end{lem} +\begin{proof} + The first part is immediate from $(*)$ in conjunction with the + fact that $l(a_L), l(a_R) < l(a)$, $l(b_L), l(b_R) < l(b)$ + for all $a_L \in L(a), a_R \in R(a)$, $b_L \in L(b)$, and + $b_R \in R(b)$. +Define $a + b$ for $a, b \in \No$ with $l(a) \oplus l(b) \leq +\gamma$ as in Equation \ref{defn_of_surreal_sum}. Suppose +$a, b, c \in \No$ with $l(a) \oplus l(b), l(a) \oplus l(c) \leq +\gamma$, and $b > c$. Then by definition we have +\begin{align*} + a + b_L < \;& a + b \\ + & a + c < a + c_R +\end{align*} +for all $b_L \in L(b)$ and $c_R \in R(c)$. If $c <_s b$ then +we can take $b_L = c$ and get $a + b > a + c$. Similarly, if +$b <_s c$, then we can take $c_R = b$ and also get $a + b > a + c$. +Suppose neither $c <_s b$ nor $b <_s c$ and put +$d \coloneq b \wedge c$. Then $l(d) < l(b), l(c)$ and +$b > d > c$. Hence by $(*)$, $a + b > a + d > a + c$. + +We may show $b + a > c + a$ similarly. +\end{proof} +\begin{lem}[``Uniformity'' of the Definition of $a$ and $b$] + Let $a = \curly{L \vert R}$ and $a' = \curly{L' \vert R'}$. + Then + \begin{align*} + a + a' = + \left\{ (L + a') \cup (a' + L) \vert + (R + a') \cup (a + R') \right\} + \end{align*} +\end{lem} +\begin{proof} + Let $a = \curly{L_a \vert R_a}$ be the canonical + representation. By Corollary \ref{inverse_cofinality_theorem} + $(L, R)$ is cofinal in $(L_a, R_a)$ and $(L', R')$ is + cofinal in $(L_{a'}, R_{a'})$. Hence + \begin{align*} + \paren{(L + a') \cup (a + L'), (R+a') \cup (a + R')} + \end{align*} + is cofinal in + \begin{align*} + \paren{(L_a + a') \cup (a + L_{a'}), (R_a + a') \cup + (a + R_{a'})} + \end{align*} + Moreover, + \begin{align*} + (L + a') \cup (a + L') < a + a' < + (R + a') \cup (a + R') + \end{align*} + Now use Theorem \ref{cofinality_theorem} to conclude the + proof. \end{proof} \ No newline at end of file diff --git a/Other/old/zach_copy/week_1/week_1.aux b/Other/old/zach_copy/week_1/week_1.aux deleted file mode 100644 index 7174f85a..00000000 --- a/Other/old/zach_copy/week_1/week_1.aux +++ /dev/null @@ -1,26 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\numberline {1}Week 1}{2}} -\@setckpt{week_1/week_1}{ -\setcounter{page}{3} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/zach_copy/week_1/week_1.tex.aux b/Other/old/zach_copy/week_1/week_1.tex.aux deleted file mode 100644 index 9b66fc00..00000000 --- a/Other/old/zach_copy/week_1/week_1.tex.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_1/week_1.tex}{ -\setcounter{page}{2} -\setcounter{equation}{0} -\setcounter{enumi}{0} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{0} -\setcounter{defn}{0} -\setcounter{cor}{0} -\setcounter{claim}{0} -\setcounter{lem}{0} -} diff --git a/Other/old/zach_copy/week_10/week_10.aux b/Other/old/zach_copy/week_10/week_10.aux deleted file mode 100644 index 1ecedcfa..00000000 --- a/Other/old/zach_copy/week_10/week_10.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_10/week_10}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_11/g_week_11.aux b/Other/old/zach_copy/week_11/g_week_11.aux deleted file mode 100644 index 1f5577d4..00000000 --- a/Other/old/zach_copy/week_11/g_week_11.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_11/g_week_11}{ -\setcounter{page}{40} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_11/g_week_11.tex b/Other/old/zach_copy/week_11/g_week_11.tex index bd243ace..86d17d97 100644 --- a/Other/old/zach_copy/week_11/g_week_11.tex +++ b/Other/old/zach_copy/week_11/g_week_11.tex @@ -1,19 +1,19 @@ -\WikiLevelTwo{ Siddharth's extra lectures } - -\WikiLevelThree{Part 1} -\WikiItalic{notes by Bill Chen} - -A crucial concept for these lectures will be \WikiItalic{games}. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) - -====Example==== +\WikiLevelTwo{ Siddharth's extra lectures } + +\WikiLevelThree{Part 1} +\WikiItalic{notes by Bill Chen} + +A crucial concept for these lectures will be \WikiItalic{games}. These games where two players Left and Right alternate moves, and a player loses if she has no moves. The information of who goes first is not encoded into the game. Formally, a game $G$ consists of two sets of games, $G=\{G_L|G_R\}$, where the left side consists of the valid games which Left can move to, and similarly for the right side. (We use the "typical element" notation for sets, which carries over from the notation for surreal numbers.) + +====Example==== \begin{enumerate} \item $\{\emptyset|\emptyset\}$ is called the zero game (abbreviated 0). There are no legal moves for either player, and the first player to move loses. \item $\{\emptyset|0\}$ is the game 1. In this game, Left has no legal moves, and Right can move to the 0 game, so Right has a winning strategy no matter who moves first. \item $\{0|\emptyset\}$. Here Left has a winning strategy. \item $\{0|0\}$ First player to move wins. This is a valid game which is not a surreal number as we defined in Week 2. \end{enumerate} - -====Definition 1==== + +====Definition 1==== \begin{enumerate} \item $G>0$ if Left has a winning strategy. \item $G<0$ if Right has a winning strategy. @@ -21,30 +21,30 @@ \item $G\parallel 0$ if the first player has a winning strategy. ($G$ is ''fuzzy''.) \item $G\ge 0$ means $G>0$ or $G\sim 0$. \end{enumerate} - -====Lemma 2 (Determinacy)==== -For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. - -\WikiBold{Proof:} - -Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. - -Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. - -====Definition 3==== -If $G,H$ are games, the \WikiItalic{disjunctive sum} $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, -$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ - -\WikiBold{Remark:} -By induction, can prove that $+$ is associative and commutative. - -====Definition 4==== -If $G$ is a game, the \WikiItalic{negation} $-G$ is the game obtained by switching the roles of Left and Right. Formally, -$$-G=\{-G_R|-G_L\}.$$ - -Notice that these are the same definitions as for surreal numbers. - -====Lemma 5 (Basic properties of $+$ and $-$)==== + +====Lemma 2 (Determinacy)==== +For any game $G$, one of $G>0$, $G<0$, $G\sim 0$, or $G\parallel 0$ holds. + +\WikiBold{Proof:} + +Let $A$ be the assertion that there is a $G_L$ with $G_L\ge 0$, and $B$ be the assertion that there is a $G_R$ with $G_R\le 0$. + +Then one can check that $G>0$ iff $A\& \neg B$, $G<0$ iff $\neg A \& B$, $G\sim 0$ iff $\neg A \& \neg B$, and $G\parallel 0$ iff $A\& B$. For example, if $A \& B$ holds, then the first player can move to a game that is positive or similar $0$. In the first case, the first player clearly wins. In the second case, the first player becomes the second player of the new game similar to $0$, and hence wins. + +====Definition 3==== +If $G,H$ are games, the \WikiItalic{disjunctive sum} $G+H$ is the game in which $G$ and $H$ are "played in parallel." Formally, +$$G+H=\{G_L+H,G+H_L|G_R+H, G+H_R\}.$$ + +\WikiBold{Remark:} +By induction, can prove that $+$ is associative and commutative. + +====Definition 4==== +If $G$ is a game, the \WikiItalic{negation} $-G$ is the game obtained by switching the roles of Left and Right. Formally, +$$-G=\{-G_R|-G_L\}.$$ + +Notice that these are the same definitions as for surreal numbers. + +====Lemma 5 (Basic properties of $+$ and $-$)==== \begin{enumerate} \item $-(G+H)=-G+-H$. \item $--G=G$. @@ -52,11 +52,11 @@ \item $G>0$ iff $-G<0$. \item $G\parallel 0$ iff $-G\parallel 0$. \end{enumerate} - -We won't prove this lemma, but it is not difficult. - -====Lemma 6==== -Let $H\sim 0$. Then: + +We won't prove this lemma, but it is not difficult. + +====Lemma 6==== +Let $H\sim 0$. Then: \begin{enumerate} \item If $G\sim 0$, then $G+H\sim 0$. \item If $G>0$, then $G+H>0$. @@ -65,28 +65,28 @@ \item If $G+H>0$, then $G>0$. \item If $G+H\parallel 0$, then $G\parallel 0$. \end{enumerate} - -\WikiBold{Proof:} - -Formally, this is proved by induction. We just describe the strategies in words. - -For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. - -For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. - -(3), is a similar analysis. - -The next three follow from the first three by using cases based on determinacy. - -====Lemma 7==== + +\WikiBold{Proof:} + +Formally, this is proved by induction. We just describe the strategies in words. + +For (1), if the second player has a winning strategy in $G$ and $H$, then the second player can use the winning strategy corresponding to the game in which the first player plays in. + +For (2), the proof splits into cases. If Right moves first, either Right moves in $H$, so Left can play according to the second player's strategy in the $H$ game, or Right moves in $G$, so Left can play according to his strategy in $G$. If Left moves first, he plays according to his strategy in $G$ and then according to the previous sentence against the subsequent moves of Right. + +(3), is a similar analysis. + +The next three follow from the first three by using cases based on determinacy. + +====Lemma 7==== \begin{enumerate} \item $G+ -G\sim 0$. \item If $G>0$ and $H>0$ then $G+H>0$. \end{enumerate} - -\WikiBold{Proof:} - -The first assertion follows from the strategy of "playing Go on two boards against the same person." - - -====Definition 8==== + +\WikiBold{Proof:} + +The first assertion follows from the strategy of "playing Go on two boards against the same person." + + +====Definition 8==== diff --git a/Other/old/zach_copy/week_11/week_11.aux b/Other/old/zach_copy/week_11/week_11.aux deleted file mode 100644 index 4018c7a3..00000000 --- a/Other/old/zach_copy/week_11/week_11.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_11/week_11}{ -\setcounter{page}{39} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_11/week_6.aux b/Other/old/zach_copy/week_11/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/zach_copy/week_11/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_2/g_week_2.aux b/Other/old/zach_copy/week_2/g_week_2.aux deleted file mode 100644 index 3539c1e2..00000000 --- a/Other/old/zach_copy/week_2/g_week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/g_week_2}{ -\setcounter{page}{12} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_2/g_week_2.tex b/Other/old/zach_copy/week_2/g_week_2.tex index 12cdfb3a..989d98cf 100644 --- a/Other/old/zach_copy/week_2/g_week_2.tex +++ b/Other/old/zach_copy/week_2/g_week_2.tex @@ -1,277 +1,277 @@ -\WikiLevelTwo{ Week 2 } - -(Notes by John Lensmire) - -\WikiLevelThree{ Monday 10-13-2014 } - -Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. - -==== Theorem 2.5 ==== - -$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. - -\WikiBold{Proof:} - -We have already proven that $\leq$ is translation invariant. - -Commutativity is clear from the symmetric nature of the definition. - -We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and -\begin{align*} -a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ -&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ -&= \{a_L | a_R\} \ (\text{by induction}) \\ -&= a -\end{align*} - -We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. -We have -\begin{align*} -(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ -&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} -\end{align*} -where the second equality holds because of uniformity. -An identical calculation shows: -\[ -a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} -\] -and hence $(a+b)+c = a+(b+c)$ holds by induction. - -To show $a + (-a) = 0$ first note: +\WikiLevelTwo{ Week 2 } + +(Notes by John Lensmire) + +\WikiLevelThree{ Monday 10-13-2014 } + +Let $a,b\in \mathbf{No}$. Recall that $a + b = \{a_L + b, a + b_L | a_R + b, a + b_R \}$. + +==== Theorem 2.5 ==== + +$(\mathbf{No},+,<)$ is an ordered abelian group with $0 = () = \{\emptyset, \emptyset \}$ and $-a$ is obtained by reversing all signs in $a$. + +\WikiBold{Proof:} + +We have already proven that $\leq$ is translation invariant. + +Commutativity is clear from the symmetric nature of the definition. + +We show by induction on $l(a)$ that $a+0 = a$. The base case is clear, and +\begin{align*} +a + 0 &= \{a_L + 0, a + 0_L | a_R + 0, a + 0_R \} \\ +&= \{a_L + 0 | a_R + 0\} \ (\text{as } 0_L = 0_R = \emptyset) \\ +&= \{a_L | a_R\} \ (\text{by induction}) \\ +&= a +\end{align*} + +We next show the associative law by induction on $l(a)\oplus l(b)\oplus l(c)$. +We have +\begin{align*} +(a+b)+c &= \{(a+b)_L + c, (a+b) + c_L | (a+b)_R + c, (a+b) + c_R \} \\ +&= \{(a_L+b) + c, (a+b_L) + c, (a+b) + c_L | (a_R+b) + c, (a+b_R) + c, (a+b) + c_R\} +\end{align*} +where the second equality holds because of uniformity. +An identical calculation shows: +\[ +a+(b+c) = \{a_L+ (b + c), a+ (b_L + c), a+ (b + c_L) | a_R+ (b + c), a+ (b_R + c), a+ (b + c_R)\} +\] +and hence $(a+b)+c = a+(b+c)$ holds by induction. + +To show $a + (-a) = 0$ first note: \begin{enumerate} \item $b <_s a \Rightarrow -b <_s -a$ \item $b < a \Rightarrow -b > -a$ \end{enumerate} - -Hence, $-a = \{-a_R | -a_L\}$. Thus, -\[ -a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} -\] -By the induction hypothesis and the fact that $+$ is increasing we have the following: + +Hence, $-a = \{-a_R | -a_L\}$. Thus, +\[ +a + (-a) = \{a_L + (-a), a + (-a_R) | a_R + (-a), a + (-a_L) \} +\] +By the induction hypothesis and the fact that $+$ is increasing we have the following: \begin{enumerate} \item $a_L + (-a) < a_L + (-a_L) = 0$ \item $a + (-a_R) < a_R + (-a_R) = 0$ \item $a_R + (-a) > a_R + (-a_R) = 0$ \item $a + (-a_L) > a_L + (-a_L) = 0$ \end{enumerate} -Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. - -==== Definition 2.6 ==== - -For $a,b\in \mathbf{No}$ set -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. - -==== Lemma 2.7 ==== - -Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where -$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ -Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. - -\WikiBold{Proof:} - -Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. - -Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. - -Claim: -\begin{enumerate} - \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function - \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function - \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function - \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function -\end{enumerate} - -We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then -\begin{align*} -f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ -&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ -&\Leftrightarrow P(a,a',b'_2,b'_1) -\end{align*} -and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. - -1-4 in the claim give us respectively: +Thus, $0$ is a realization of the cut. Since $0$ has minimal length, we have $a+(-a) = 0$ as needed. + +==== Definition 2.6 ==== + +For $a,b\in \mathbf{No}$ set +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +As motivation for this definition, note that in any ordered field: if $a'0$ so in particular $a'b + ab' - a'b' < ab$. + +==== Lemma 2.7 ==== + +Suppose for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) < \gamma$ we have defined $a\cdot b$ so that (2.6) holds, and for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $<\gamma$ $(*)$ holds, where +$(*): a>b, c>d \Rightarrow ac-bc > ad-bd.$ +Then: (2.6) holds for all $a,b\in \mathbf{No}$ with $l(a)\oplus l(b) \leq \gamma$ and $(*)$ holds for all $a,b,c,d\in \mathbf{No}$ with the natural sum of the lengths of each factor is $\leq \gamma$. + +\WikiBold{Proof:} + +Let $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd.$ Because the surreal numbers form an ordered abelian group, $P$ is "transitive in the last two variables" (i.e. $P(a,b,c,d) \ \&\ P(a,b,d,e) \Rightarrow P(a,b,c,e)$) and similarly in the first two variables. + +Fix $a,b\in \mathbf{No}$. For $a' <_s a, b' <_s b$ we define $f(a',b') = a'b + ab' - a'b'$. + +Claim: +\begin{enumerate} + \item $a' < a \Rightarrow b' \mapsto f(a',b')$ is an increasing function + \item $a' > a \Rightarrow b' \mapsto f(a',b')$ is a decreasing function + \item $b' < b \Rightarrow a' \mapsto f(a',b')$ is an increasing function + \item $b' > b \Rightarrow a' \mapsto f(a',b')$ is a decreasing function +\end{enumerate} + +We prove 1 (the rest are left as an exercise). Let $b'_1,b'_2 <_s b$ and $b'_1 < b'_2$. Then +\begin{align*} +f(a',b'_2) > f(a',b'_1) &\Leftrightarrow (a'b + ab'_2 - a'b'_2) > (a'b + ab'_1 - a'b'_1) \\ +&\Leftrightarrow (ab'_2 - a'b'_2) > (ab'_1 - a'b'_1) \\ +&\Leftrightarrow P(a,a',b'_2,b'_1) +\end{align*} +and $P(a,a',b'_2,b'_1)$ holds by induction, proving 1. + +1-4 in the claim give us respectively: \begin{enumerate} \item $f(a_L, b_L) < f(a_L, b_R)$ \item $f(a_R, b_R) < f(a_R, b_L)$ \item $f(a_L, b_L) < f(a_R, b_R)$ \item $f(a_R, b_R) < f(a_L, b_L)$ \end{enumerate} - -These facts exactly give us that $a\cdot b$ is well-defined. - -We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. - -\WikiLevelThree{ Wednesday 10-15-2014 } - -Recall the definition of multiplication from last time: -\[ -a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} -\] -and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. - -Note we can rephrase the defining inequalities for $a\cdot b$ as -\[ -(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) -\] - -To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. - -Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. - -Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. - -Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. - -==== Lemma 2.8 ==== - -The uniformity property holds for multiplication. - -\WikiBold{Proof:} - -Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: - -Claim: -\begin{enumerate} -\item $a' < a \Rightarrow f(a',-)$ is an increasing function -\item $a' > a \Rightarrow f(a',-)$ is a decreasing function -\item $b' < b \Rightarrow f(-,b')$ is an increasing function -\item $b' > b \Rightarrow f(-,b')$ is a decreasing function -\end{enumerate} - -Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. - -Let $a_l, b_l$ range over $L,L'$. As an example, note -\[ -f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) -\Leftrightarrow P(a,a_l,b,b_l) -\] -which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. -Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. - -\WikiLevelThree{ Friday 10-17-2014 } - -==== Definition of Inverses ==== - -Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). - -For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. -Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. -\[ -b\circ a' = [1-(a-a')b]/a'. -\] -This works as inductively we'll have already defined $1/a'$. - -For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. - -Now set (as candidates for defining $1/a$) -\begin{align*} -L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ -R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} -\end{align*} -Note that this definition is an expansion of the naive idea presented at the end of last lecture. - -We first show -\[ -(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. -\] -by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. - -The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. - -For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. - -Claim: -\begin{enumerate} -\item $x > b \Leftrightarrow 1 > ab$ -\item $ax = 1 + (a-a')(x-b)$. -\end{enumerate} -By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. - -Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. -$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. - -The other cases are similar, so $(*)$ holds in general. - -Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. - -The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. - -We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. - -Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). - -If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, -\begin{align*} -a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ -&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ -&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 -\end{align*} -where the last equivalence holds by $(\Delta)$ and $a'>0$. - -Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. - -We have thus shown: - -==== Theorem 2.10 (Conway) ==== - -$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. - -We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. - -We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. - -We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). - -==== Lemma 3.1 ==== - -For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). + +These facts exactly give us that $a\cdot b$ is well-defined. + +We are left to show that $(*)$ continues to hold. We'll continue this on Wednesday. + +\WikiLevelThree{ Wednesday 10-15-2014 } + +Recall the definition of multiplication from last time: +\[ +a\cdot b = \{a_L\cdot b + a\cdot b_L - a_L\cdot b_L, a_R\cdot b + a\cdot b_R - a_R\cdot b_R | a_L\cdot b + a\cdot b_R - a_L\cdot b_R, a_R\cdot b + a\cdot b_L - a_R\cdot b_L \} +\] +and the statement $(*): a>b, c>d \Rightarrow ac-bc > ad-bd$. We'll also continue write $P(a,b,c,d)\Leftrightarrow ac - bc > ad - bd$. + +Note we can rephrase the defining inequalities for $a\cdot b$ as +\[ +(\Delta): P(a,a_L,b,b_L), P(a_R,a,b_R,b), P(a,a_L,b_R,b), P(a_R,a,b,b_L) +\] + +To finish the proof of Lemma 2.7, suppose $a>b>c>d$ (of suitable lengths). We want to show $P(a,b,c,d)$. + +Case 1: Suppose in each pair $\{a,b\}, \{c,d\}$ one of the elements is an initial segment of the other. Then note we are done by $(\Delta)$. + +Case 2: Suppose $a \not<_s b, b\not<_s a$ but $c <_s d$ or $d <_s c$. Then we have that $a > a\wedge b > b$ and by Case 1, $P(a,a\wedge b, c, d)$ and $P(a\wedge b, b, c, d)$. This implies (by transitivity of the first two variables) $P(a,b,c,d)$. + +Case 3: Suppose $c \not<_s d, d\not<_s c$. Then by Case 1 and 2, $P(a,b,c,c\wedge d)$ and $P(a,b,c\wedge d, d)$, so (transitivity of the last two variables) $P(a,b,c,d)$ as needed. This completes the proof of Lemma 2.7. + +==== Lemma 2.8 ==== + +The uniformity property holds for multiplication. + +\WikiBold{Proof:} + +Fix $a,b\in \mathbf{No}$. For any $a',b'\in \mathbf{No}$ we define (as last time) $f(a',b') = a'b + ab' - a'b'$. Using Lemma 2.7, we can extend the Claim from last time to hold in general: + +Claim: +\begin{enumerate} +\item $a' < a \Rightarrow f(a',-)$ is an increasing function +\item $a' > a \Rightarrow f(a',-)$ is a decreasing function +\item $b' < b \Rightarrow f(-,b')$ is an increasing function +\item $b' > b \Rightarrow f(-,b')$ is a decreasing function +\end{enumerate} + +Let $a = \{L|R\}, b = \{L'|R'\}$ be any representations of $a,b$. We want to verify the hypothesis of the Cofinality Theorem 1.10. + +Let $a_l, b_l$ range over $L,L'$. As an example, note +\[ +f(a_l, b_l) < ab \Leftrightarrow 0 < ab - (a_lb + ab_l - a_lb_l) = (ab - a_lb) - (ab_l - a_lb_l) +\Leftrightarrow P(a,a_l,b,b_l) +\] +which holds as $a_l0$, i.e. find a solution to $a\cdot x = 1$. +Note, the naive idea is to set $x = \{1/a_R | 1/a_L, (a_L\neq 0)\}$ but this does not work in general. + +\WikiLevelThree{ Friday 10-17-2014 } + +==== Definition of Inverses ==== + +Let $a\in \mathbf{No}$ with $a>0$. Our aim is to define $1/a$. Let $a = \{L|R\}$ the canonical representation of $a$. Observe that $a'\geq 0$ for all $a'\in L$ (as $a' <_s a$). + +For every finite sequence $(a_1,\ldots,a_n)\in (L\cup R)\setminus \{0\}$ we define $\langle a_1,\ldots, a_n\rangle \in \mathbf{No}$ by induction on $n$. Set $\langle \ \rangle = 0$ and inductively set $\langle a_1,\ldots, a_n, a_{n+1}\rangle = \langle a_1,\ldots, a_n \rangle \circ a_{n+1}$. +Here, for arbitrary $b\in \mathbf{No}$ and $a'\in (L\cup R)\setminus \{0\}$ let $b\circ a'$ be the unique solution to $(a-a')b + a'x = 1$, i.e. +\[ +b\circ a' = [1-(a-a')b]/a'. +\] +This works as inductively we'll have already defined $1/a'$. + +For example, $\langle a_1 \rangle = \langle \ \rangle \circ a_1 = 0 \circ a_1 = 1/a_1$. + +Now set (as candidates for defining $1/a$) +\begin{align*} +L^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is even} \} \\ +R^{-1} &= \{ \langle a_1,\ldots, a_n \rangle | \text{ the number of } a_i \text{ in }L \text{ is odd} \} +\end{align*} +Note that this definition is an expansion of the naive idea presented at the end of last lecture. + +We first show +\[ +(*) x \in L^{-1} \Rightarrow ax < 1 \text{ and } x\in R^{-1} \Rightarrow ax > 1. +\] +by induction on $n$. In particular, this yields that $L^{-1} < R^{-1}$. + +The base case is clear, as $\langle \ \rangle = 0\in L^{-1}$ and $a\cdot 0 = 0 < 1$. + +For the induction, suppose $b\in L^{-1}\cup R^{-1}$ satisfying $(*)$ and $0\neq a' <_s a$. We show that $x = b\circ a'$ also satisfies $(*)$. + +Claim: +\begin{enumerate} +\item $x > b \Leftrightarrow 1 > ab$ +\item $ax = 1 + (a-a')(x-b)$. +\end{enumerate} +By definition, $x$ is the solution to $(a-a')b + a'x = 1$ and $ab = ab - a'b + a'b = (a-a')b + a'b$. These two equations yield $ab = 1 + a'(b-x)$. Both parts of the claim follow. + +Now suppose, $b\in L^{-1}, a'\in L$. Then $x\in R^{-1}$ so want to check $ax > 1$. +$b\in L^{-1}$ and $(*)$ implies that $ab < 1$ and hence Claim 1 tells us that $x > b$. Now by Claim 2, $ax = 1 + (a-a')(x-b)$ hence $ax > 1$ (because $a>a', x>b$) as needed. + +The other cases are similar, so $(*)$ holds in general. + +Thus, we can set $c = \{L^{-1} | R^{-1} \}$. We claim that $1/a = c$, that is, $ac = 1 = \{0 | \emptyset\}$. + +The typical element used to define $ac$ is $a'c + ac' - a'c'$ with $a'\in L\cup R, c'\in L^{-1}\cup R^{-1}$. + +We first show $\{\text{ lower elements for }ac\} < 1 < \{(\text{ upper elements for }ac \}$. + +Suppose that $a' = 0$, then we get an element $a'c + ac' - a'c' = ac'$ which is an upper element for $ac$ if and only if $c'\in R^{-1}$ by definition. However, by $(*)$ if $c'\in R^{-1}$ then $ac' = a'c + ac' - a'c' > 1$ (as needed since an upper element), else $c'\in L^{-1}$ and then $ac' = a'c + ac' - a'c' < 1$ (as needed since a lower element). + +If $a'\neq 0$, then $x = c'\circ a'$ is defined, lies in $L^{-1}\cup R^{-1}$, and obeys $(\Delta): (a-a')c' + a'x = 1$. Hence, +\begin{align*} +a'c + ac' - a'c' \text{ is a lower element for } ac &\Leftrightarrow a' \ \&\ c' \text{ are on the same side of } a \ \&\ \text{(respectively) } c \\ +&\Leftrightarrow x\in R^{-1} \Leftrightarrow x > c \\ +&\Leftrightarrow \text{a typical element } a'c + ac' - a'c' = (a-a')c' + a'c < 1 +\end{align*} +where the last equivalence holds by $(\Delta)$ and $a'>0$. + +Since $1$ satisfies the cut for $ac$ (and $0$ does not) it is the minimial realization, so $ac = 1$ as needed. + +We have thus shown: + +==== Theorem 2.10 (Conway) ==== + +$(\mathbf{No}, +, \cdot, \leq)$ is an ordered field. + +We'll now begin to focus on how to view real numbers and ordinals as surreal numbers. + +We have $0 = ()$ the additive identity of $\mathbf{No}$ and $1 = (+)$ the multiplicative identity of $\mathbf{No}$. + +We also have an embedding of ordered rings $\mathbb{Z} \hookrightarrow \mathbf{No}$ where $k\mapsto k\cdot 1$ (where $k\cdot 1$ is $k$ additions of $+1$ or $-1$). + +==== Lemma 3.1 ==== + +For $n\in \mathbb{N}$, $n\cdot 1 = (+\cdots +)$ (i.e. $n$ $+$'s). diff --git a/Other/old/zach_copy/week_2/week_2.aux b/Other/old/zach_copy/week_2/week_2.aux deleted file mode 100644 index ded0313c..00000000 --- a/Other/old/zach_copy/week_2/week_2.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_3/zach.aux b/Other/old/zach_copy/week_3/zach.aux deleted file mode 100644 index 6e8b1dae..00000000 --- a/Other/old/zach_copy/week_3/zach.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_3/zach}{ -\setcounter{page}{13} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_4/g_week_4.aux b/Other/old/zach_copy/week_4/g_week_4.aux deleted file mode 100644 index a0eae918..00000000 --- a/Other/old/zach_copy/week_4/g_week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/g_week_4}{ -\setcounter{page}{16} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_4/g_week_4.tex b/Other/old/zach_copy/week_4/g_week_4.tex index 50f66b95..097f9dae 100644 --- a/Other/old/zach_copy/week_4/g_week_4.tex +++ b/Other/old/zach_copy/week_4/g_week_4.tex @@ -1,71 +1,71 @@ -\WikiLevelTwo{ Week 4 } -Notes by Madeline Barnicle -\WikiLevelThree{Monday, October 27, 2014} -We need to show that the ordered field of reals in $\mathbf{No}$ is Dedekind-complete. - -Let $S \neq \emptyset$ be a set of reals which is bounded from above. We need to show sup $S$ exists. We can assume $S$ has no maximum. Put $R=\{a \in \mathbb{D}: a>S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. - -Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. - -So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' S\}, L=\mathbb{D} \setminus R$. Using (3.11), the density of $\mathbb{D}$, $L$ has no maximum. If $R$ has a minimum, then this value is the sup of $S$ and we are done. So suppose $R$ has a minimum, then $r=\{L|R\}$ is real. + +Claim: $r=$ sup $S$. Suppose $s \in S$ satisfies $s>r$, take $d \in \mathbb{D}$ with $s>d>r$. Then $d \in L$, contradiction since $d>r$. + +So $r$ is an upper bound. If $L a_{\omega^{r'}_n} * a_{\beta}=a_{\omega^{r'}_n} \otimes \beta$ by hypothesis $=a_{\omega^{r' \oplus s}_n}$. Similarly if $s' a_{\omega^{r \oplus s'}_n}$. Therefore, $a_\alpha * a_\beta \geq$ sup $\{a_{\omega^{r' \oplus s}_n}, a_{\omega^{r \oplus s'}_n} : r' 0}, \omega + r = \{n|\emptyset \} + \{L|R\}$ (the canonical representation for $r$) = $\omega + L, n+r | \omega +R\} = \{\omega + L | \omega + R \}$. Inducting on the length of $r$, this equals $\omega \frown r$. This also works for negative, non-integer $r$ (we need $L$ to be nonempty for the argument to go through)--the full result holds for $r \in \mathbb{Z}^{<0}$ as well. \end{enumerate} - -\WikiLevelThree{Wednesday, October 29, 2014} + +\WikiLevelThree{Wednesday, October 29, 2014} \begin{enumerate} \item $\frac{1}{2} \omega = \{0|1\}*\{n|\emptyset \}=$ by the definition of multiplication, $\{\frac{1}{2} n + \omega * 0 - n*0 | \frac{1}{2} n + \omega * 1 - n*1\}=\{\frac{1}{2} n|\omega -\frac{1}{2} n\}=$ by cofinality $\{n|\omega -n\}= (+++...---...)$ ($\omega$ +s, $\omega$ -s). \item $\frac{1}{\omega}= (+---...)$ ($\omega$ -s)? Call this number $\epsilon$. $0<\epsilon 0}$, i.e. $\epsilon$ is ''infinitesimal''. (It is the unique infinitesimal of length $\omega$.) Guess that $\epsilon = \frac{1}{\omega}$. Canonical representation of $\epsilon: \{0|\frac{1}{2^n}\}$. \end{enumerate} - -$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ - -Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. + +$\epsilon \omega = \{0|\frac{1}{2^n}\}*\{m|\emptyset \}=\{0|\frac{1}{n}\}*\{m|\emptyset \}$ by cofinality $=\{\epsilon *m +0*0-0*\omega|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}= \{\epsilon *m|\epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m\}.$ + +Now $\epsilon *m$ is still infinitesimal, $\epsilon m <1. 1< \epsilon *m+\frac{1}{n}\omega - \frac{1}{n}m$, as $\frac{1}{n}(\omega -n)$ is infinite. Since $0$ does not satisfy the cut, $\epsilon \omega =1$. \begin{enumerate} \item $r + \epsilon (r \in \mathbb{R})$. First assume $r \in \mathbb{D}$, so $r=\{r_L|r_R\}, r_L, r_R \in \mathbb{D}$. $r + \epsilon = \{r_L + r_R\} + \{0|\frac{1}{n}\}=\{r+0, r_L + \epsilon|r+\frac{1}{n}, r_R + \epsilon\}= \{r|r+\frac{1}{n}\}$ by cofinality = $\{r|\mathbb{D}^{>r}\}=r\frown(+)$, of length $\omega +1$. \item What is $\lambda +r$, $\lambda$ a limit ordinal, $r \in \mathbb{R}$? \end{enumerate} - -\WikiLevelFour{Section 4: Combinatorics of Ordered Sets} -Let $S$ be a set. An \WikiItalic{ordering} $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an \WikiItalic{ordered set} (partial or total). - -We say $\leq$ is \WikiItalic{total} if $x \leq y$ or $y \leq x$ for each $x, y \in S$. - -Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is \WikiItalic{increasing} if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, \WikiItalic{strictly increasing} if $x < y \rightarrow \phi(x) < \phi(y)$, a \WikiItalic{quasi-embedding} if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. - -Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. + +\WikiLevelFour{Section 4: Combinatorics of Ordered Sets} +Let $S$ be a set. An \WikiItalic{ordering} $\leq$ on $S$ is an reflexive, transitive, antisymmetric, binary relation on $S$. Call $(S,\leq)$ an \WikiItalic{ordered set} (partial or total). + +We say $\leq$ is \WikiItalic{total} if $x \leq y$ or $y \leq x$ for each $x, y \in S$. + +Let $T$ be an ordered set, and $\phi: S \rightarrow T$ a map. Then $\phi$ is \WikiItalic{increasing} if $x \leq y \rightarrow \phi(x) \leq \phi(y)$, \WikiItalic{strictly increasing} if $x < y \rightarrow \phi(x) < \phi(y)$, a \WikiItalic{quasi-embedding} if $\phi(x) \leq \phi(y) \rightarrow x \leq y$. + +Examples: let $(S, \leq_S), (T, \leq_T)$ be ordered sets. \begin{enumerate} \item $S \coprod T$= disjoint union of $S$ and $T$ with the ordering $\leq_S \cup \leq_T$. \item $S \times T$ can be equipped with the ''product ordering'' $(x,y)\leq(x',y') \leftrightarrow x \leq_S x'$ and $y \leq_T y'$, or the ''lexicographic ordering'' $(x,y) \leq_{lex} (x',y') \leftrightarrow x <_S x'$ or $(x=x'$ and $y \leq_T y')$. The lexicographic ordering extends the product ordering. @@ -74,65 +74,65 @@ \item The natural surjective map $(S^*, \leq^*) \rightarrow (S^\diamond, \leq^\diamond)$ is increasing. \item $\mathbb{N}^m=\mathbb{N} * \mathbb{N} * ...\mathbb{N}$ with the product ordering. $X=\{x_1...x_m\}$ distinct indeterminates with trivial ordering. The map $\mathbb{N}^m= \rightarrow X^\diamond, \nu(v_1...v_n)=X_{1}^{v_1}...X_{m}^{v_m}$ is an isomorphism of ordered sets. $\leq^\diamond$ is divisibility of monomials. \end{enumerate} - + \WikiSigleStar Let $S$ be an ordered set. Call $F \subset S$ a ''final segment'' of $S$ if $x \leq y$ and $x \in F \rightarrow y \in F$ ($F$ is upward closed). Given $X \subset S$, define $(X) \subset F = \{y \in S|\exists x \in X, x \leq y\},$ the final segment of $S$ ''generated'' by $X$ (the notation corresponds to the ideal generated by monomials). Put $\mathcal{F}(S)=$ the set of all finite segments of $S$. Call $A \subset S$ an ''antichain'' if for $x, y \in A, x \neq y \rightarrow x \not\leq y$ and $y \not\leq x$. We say that $S$ is ''well-founded'' if there is no infinite sequence $x_{1}>x_{2}...$ in $S$. - -====Definition 4.1==== -$(S, \leq)$ is \WikiItalic{noetherian} if it is well-founded and has no infinite antichains. - - -\WikiLevelThree{Friday, October 31, 2014} -Call an infinite sequence $(x_n)$ in $S$ \WikiItalic{good} if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. - -$2 \leftrightarrow 3$ is a standard argument. - -$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. - -$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. - -====Corollary 4.3==== -\begin{enumerate} - \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) - \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. - \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) -\end{enumerate} - -In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). -====Theorem 4.4 (Higman)==== -If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). - -\WikiBold{Proof} (Nash-Williams) - -Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. - -\WikiBold{Proof}: -We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. - -\WikiLevelFour{Hahn Fields} - -Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose \WikiItalic{support} supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is \WikiItalic{well-ordered}. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. + +====Definition 4.1==== +$(S, \leq)$ is \WikiItalic{noetherian} if it is well-founded and has no infinite antichains. + + +\WikiLevelThree{Friday, October 31, 2014} +Call an infinite sequence $(x_n)$ in $S$ \WikiItalic{good} if $x_i \leq x_j$ for some $i x_2, x_2 \ni (G)$. This yields an infinite sequence $x_1 > x_2...$, a contradiction. + +$2 \leftrightarrow 3$ is a standard argument. + +$3 \rightarrow 4$: Let $(x_i)$ be a sequence in $S$. We define a subsequence $x_{n_k}$ such that $x_{n_k} \leq x_{n_{k+1}} \forall k$, and there are infinitely many $n > n_k$ such that $x_n > x_{n_k}$. We let $n_1 = 1$. Inductively, suppose we have already defined $n_1...n_k$. Then the final segment $(x_n: n > n_k, x_n \geq x_{n_k})$ is finitely generated, so there is some $x_{n_{k+1}}$ such that for infinitely many $n>k$ with $x_n > x_{n_k}$ we have $x_n > x_{n_{k+1}}$. + +$4 \rightarrow 5$ is obvious, as is $5 \rightarrow 1$. + +====Corollary 4.3==== +\begin{enumerate} + \item If there exists an increasing surjection $S \twoheadrightarrow T$ and $S$ is noetherian, then $T$ is noetherian. (Use condition 5.) + \item If there exists a quasi-embedding $S \rightarrow T$ and $T$ is noetherian, then $S$ is noetherian. + \item If $S, T$ are noetherian, then so are $S \coprod T$ and $S \times T$. (For the product case, take an infinite sequence and apply condition 4 to each component.) +\end{enumerate} + +In particular, $\mathbb{N}^m$ is noetherian ("Dickson's Lemma"). +====Theorem 4.4 (Higman)==== +If $S$ is noetherian, then $S^*$ is noetherian, (and then so is $S^\diamond$ by (4.3 (1))). + +\WikiBold{Proof} (Nash-Williams) + +Suppose $w_1, w_2...$ is a bad sequence for $\leq^*$. We may assume that each $w_n$ is of minimal length under the condition $w_n \ni (w_1...w_{n-1})$. Then $w_n \neq \epsilon$ (the empty word) for any $n$. So $w_n=s_{n}*u_{n}, s_n \in S, u_n \in S^*$ (splitting off the first letter). Since S is noetherian, we can take an infinite subsequence $s_{n_1} \leq s_{n_2}...$ of $s_n$. By minimality, $w_1...w_{n_{1}-1}, u_{n_1}, u_{n_2}...$ is good. So $\exists i0}$ be well-ordered. Then $=\{\alpha_1+...\alpha_n: \alpha_i \in A\}$ is also well-ordered, and for each $\gamma \in $ there are only finitely many $(n, \alpha_1...\alpha_n)$ with $n \in \mathbb{N}, \alpha_1...\alpha_n \in A$ such that $\gamma=\alpha_1+...\alpha_n$. + +\WikiBold{Proof}: +We have the map $A^\diamond \rightarrow , \alpha_1...\alpha_n \rightarrow \alpha_1+...\alpha_n$, onto and strictly increasing since all $\alpha_i >0$. Now use Higman's Theorem. + +\WikiLevelFour{Hahn Fields} + +Let $k$ be a field, $\Gamma$ an ordered abelian group. Define $K=k((t^\Gamma))$ to be the set of all formal series $f(t)=\sum_{\gamma \in \Gamma} f_\gamma t^\gamma (f_\gamma \in k)$, whose \WikiItalic{support} supp$f=\{\gamma \in \Gamma: f_\gamma \neq 0\}$ is \WikiItalic{well-ordered}. Using (4.5), we can now define $f+g=\sum_{\gamma} (f_\gamma + g_\gamma)t^\gamma,$ and $f*g= \sum_{\gamma} (\sum_{\alpha+\beta=\gamma} f_\alpha * g_\beta)t^\gamma$. $K$ is an integral domain, and $k$ includes into $K$ by $a \rightarrow a*t^0$. We will show $K$ is actually a field. diff --git a/Other/old/zach_copy/week_4/week_2.aux b/Other/old/zach_copy/week_4/week_2.aux deleted file mode 100644 index 31eef818..00000000 --- a/Other/old/zach_copy/week_4/week_2.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_2/week_2}{ -\setcounter{page}{11} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/zach_copy/week_4/week_4.aux b/Other/old/zach_copy/week_4/week_4.aux deleted file mode 100644 index 48e65042..00000000 --- a/Other/old/zach_copy/week_4/week_4.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_4/week_4}{ -\setcounter{page}{15} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_5/g_week_5.aux b/Other/old/zach_copy/week_5/g_week_5.aux deleted file mode 100644 index e9136496..00000000 --- a/Other/old/zach_copy/week_5/g_week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/g_week_5}{ -\setcounter{page}{19} -\setcounter{equation}{1} -\setcounter{enumi}{4} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_5/g_week_5.tex b/Other/old/zach_copy/week_5/g_week_5.tex index 24370214..75d4ea56 100644 --- a/Other/old/zach_copy/week_5/g_week_5.tex +++ b/Other/old/zach_copy/week_5/g_week_5.tex @@ -1,145 +1,145 @@ -\WikiLevelTwo{ Week 5 } - -\WikiLevelThree{November 3, 2014 } -We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. - -Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. - -\WikiLevelFour{ Section 5: The $\omega^-$ map } -Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are \WikiItalic{archimedean equivalent} if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by -$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ -We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. - -This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. - -Some properties: +\WikiLevelTwo{ Week 5 } + +\WikiLevelThree{November 3, 2014 } +We will now show that the Hahn field $K$ defined last time is in fact a field. Define $v:K\setminus \{0\}\rightarrow \Gamma$ by $$vF:=\min \mathrm{supp}(f).$$ We will use a profound but simple remark about the structure of $K$ due (probably?) to Neumann: if $vF>0$ (i.e., $\mathrm{supp}(f)\subseteq \Gamma^{>0}$), then $\sum_{n=0}^\infty f^n$ makes sense as an element of $K$. Why? By (4.6), for any $\gamma\in \Gamma$ there are finitely many $n$ such that $\gamma\in\mathrm{supp}(f^n)$. By definition of multiplication in $K$, any $\gamma\in \mathrm{supp}(f^n)$ is an element of $\langle \mathrm{supp}(f)\rangle$. By the second part of (4.6), there are only finitely many such $n$. + +Now write an arbitrary $g\in K\setminus \{0\}$ as $g=ct^\gamma(1-f)$ where $c\in K\setminus \{0\}$, $\gamma\in \Gamma$, and $vF=0$ (this is achieved by taking $\gamma$ to be $\min\mathrm{supp}(g)$). Then it can be checked that $g^{-1}=c^{-1}t^{-\gamma}\sum_{n=0}^\infty f^n$ works. + +\WikiLevelFour{ Section 5: The $\omega^-$ map } +Let $(\Gamma, \le, +)$ be an ordered abelian group. Set $|\gamma|:=\max\{\gamma, -\gamma\}$ for $\gamma\in \Gamma$. We say $\alpha,\beta$ are \WikiItalic{archimedean equivalent} if there is some $n\ge 1$ such that $|\alpha|\le n|\beta|$ and $|\beta|\le n|\alpha|$. This is an equivalence relation on $\Gamma$, we write $\alpha\sim\beta$. Denote the equivalence classes $[\alpha]=\{\beta\in\Gamma:\beta\sim\alpha\}$. These equivalence classes $[\Gamma]=\{[\gamma]:\gamma\in\Gamma\}$ can be ordered by +$$[\alpha]<[\beta]\quad \textrm{if and only if} \quad \textrm{for all }n\ge 1, n|\alpha|<|\beta|.$$ +We also write $\alpha\ll\beta$ or $\alpha=o(\beta)$ instead of $[\alpha]<[\beta]$. This defines a total order on $[\Gamma]$ with smallest element $[0]$. + +This equivalence relation gives a coarse view of the ordered abelian group $\Gamma$. + +Some properties: \begin{enumerate} \item $[-\alpha]=[\alpha]$. \item $[\alpha+\beta]\le \max\{[\alpha],[\beta]\}$ and equality holds if $[\alpha]\neq [\beta]$. \item If $0\le \alpha\le \beta$, then $[\alpha]\le [\beta]$. \end{enumerate} - - -We say that $\Gamma$ is \WikiItalic{archimedean} if $[\Gamma]\setminus \{[0]\}$ is a singleton. -====Lemma 5.1 (H�lder): ==== -If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. - -The proof is easy, using Dedekind cuts. - -====Lemma 5.2:==== - -Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. - -\WikiBold{Proof:} - -There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. - - -==== Lemma 5.4: ==== -Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: + + +We say that $\Gamma$ is \WikiItalic{archimedean} if $[\Gamma]\setminus \{[0]\}$ is a singleton. +====Lemma 5.1 (H�lder): ==== +If $\Gamma$ is archimedean and $\epsilon\in \Gamma^{>0}$, then there is a unique embedding $(\Gamma,\le,+)\rightarrow (\mathbb{R},\le,+)$ with $\epsilon\mapsto 1$. + +The proof is easy, using Dedekind cuts. + +====Lemma 5.2:==== + +Let $a\in \mathbf{No}$. Then there is a unique $x\in \mathbf{No}$ of minimal length with $a\sim x$. + +\WikiBold{Proof:} + +There is certainly an $x\in [a]$ of smallest length. Let $x\neq y$ with $x\sim a\sim y$. Put $z:=x\wedge y$. Then $z\sim x\sim y$ and $l(z)0}$ and $b=\{b_L\mid b_R\}$. + + +==== Lemma 5.4: ==== +Suppose $\gamma\in \mathbf{On}$ and $\omega^b$ has been defined for all $b\in \mathbf{No}$ with $\ell(b)<\gamma$ such that: \begin{enumerate} \item $0<\omega^b$, \item $b0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. - -Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. - -The proof is an exercise using cofinality and inverse cofinality theorems. - -====Lemma 5.6: ==== - -Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. - -\WikiBold{Proof:} - -Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. - -\WikiLevelThree{November 7, 2014} -No class on Wednesday. - -To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put -$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ -$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ - -\WikiItalic{Case 1.} $F\cap G\neq \emptyset$. - -Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ - -====Lemma 5.7:==== -The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. - -\WikiBold{Proof:} - -By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. - -Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then -$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ -Using the definition of multiplication and induction hypothesis, -$\omega^a\cdot\omega^b$ has left part -$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ -and right part -$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ - -We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. - -Also, +Then (5.3) makes sense for $b\in \mathbf{No}$ with $l(b)\le \gamma$ and $(1),(2)$ above hold for all $b,c\in\mathbf{No}$ with $\ell(b),\ell(c)\le \gamma$. + +\WikiBold{Proof} +Suppose $\ell(b)=\gamma$. Then $\ell(b_L),\ell(b_R)<\gamma$, so $\omega^{b_L} \ll \omega^{b_R}$. Therefore $r\omega^{b_L}0}$. So $00}$. Hence $\omega^b$ is defined (i.e., this is actually a cut) and $\omega^b>0$. + +Suppose $b0}\cdot \omega^L\mid \mathbb{R}^{>0}\cdot \omega^R\}$. + +The proof is an exercise using cofinality and inverse cofinality theorems. + +====Lemma 5.6: ==== + +Let $a\in \mathbf{No}$. Then $a=\omega^b$ for some $b\in\mathbf{No}$ if and only if $a$ is the element of minimal length of $[a]$. + +\WikiBold{Proof:} + +Suppose $a=\omega^b=\{0,r\omega^{b_L}\mid s\omega^{b_R}\}$. If $a\sim x$, then $r\omega^{b_L}0}$ so $a\le_s x$. This finishes one direction of the proof, the other will be done next time. + +\WikiLevelThree{November 7, 2014} +No class on Wednesday. + +To finish the other direction of Lemma 5.6, we show that each $a\in\mathbf{No}$, $a>0$, is $\sim$ to some element of the form $\omega^b$ by induction on $\ell(a)$. Suppose $a=\{L\mid R\}$, with $0\in L$. By induction hypothesis, every element of $L\cup R$ is $\sim$ to some $\omega^b$. Put +$$F:=\{y\in\mathbf{No}:\exists a_L\in L:a_L\sim \omega^y\}$$ +$$G:=\{y\in\mathbf{No}:\exists a_R\in R:a_R\sim \omega^y\}.$$ + +\WikiItalic{Case 1.} $F\cap G\neq \emptyset$. + +Let $y\in F\cap G$. Then there are $a_L\in L$, $a_R\in R$ with $a_L\sim\omega^y\sim a_R$. Since $0 \omega^F, \mathbb{R}^>\omega^G)$ is cofinal in $(L,R)$. Let $a_L \in L\setminus \{0\}$. Then by induction hypothesis, there is $x\in F$ with $a_L\sim \omega^x$. So there is some $r\in \mathbb{R}^>$ such that $r\omega^x\ge a_L$. Similarly for each $a_R\in R$ there is some $y\in G$ and $r\in \mathbb{R}^>$ s.t. $r\omega^y\le a_R$. $\Box$ + +====Lemma 5.7:==== +The map $a\mapsto \omega^a$ is an embedding of ordered groups $(\mathbf{No},+,\le)\hookrightarrow (\mathbf{No}^>,\cdot,\le)$. + +\WikiBold{Proof:} + +By definition, $\omega^0=\{0\mid \emptyset\}=1$. By induction on $\ell(a)\oplus \ell(b)$ we show that $\omega^a\omega^b=\omega^{a+b}$. + +Let $a=\{a_L\mid a_R\}$, $b=\{b_L\mid b_R\}$, so $\omega^a=\{0,r\omega^{a_L}\mid s\omega^{a_R}\}$ and $\omega^b=\{0,r'\omega^{b_L}\mid s'\omega^{b_R}\}$, where $r,s,r',s'$ range over $\mathbb{R}^>$. Then +$$\omega^{a+b}=\{0,r\omega^{a_L+b},r'\omega^{a+b_L}\mid s\omega^{a_R+b}, s'\omega^{a+b_R}\}.$$ +Using the definition of multiplication and induction hypothesis, +$\omega^a\cdot\omega^b$ has left part +$$\{0,r\omega^{a_L+b}, r'\omega^{b_L+a},r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L},s\omega^{a_R+b}+s'\omega^{b_R+a}-ss'\omega^{a_R+b_R}\}$$ +and right part +$$s\omega^{a_R+b},s'\omega^{b_R+a},r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R},s\omega^{a_R+b}+r'\omega^{b_L+a}-r's\omega^{a_R+b_L}\}$$ + +We will show that these two cuts are mutually cofinal, proving the lemma. Note that the elements defining $\omega^{a+b}$ also appear in the cut for $\omega^a\omega^b$. + +Also, \begin{enumerate} \item $r\omega^{a_L+b}+r'\omega^{b_L+a}-rr'\omega^{a_L+b_L}\le (r+r')\omega^{\max(a_L+b,a+b_L)}$ (a term appearing in the cut for $\omega^{a+b}$). \item $s\omega^{a_R+b}+s'\omega^{a+b_R}-ss'\omega^{a_R+b_R}<0$. (The third term of the LHS has the highest archimedean class.) \item $r\omega^{a_L+b}+s'\omega^{b_R+a}-rs'\omega^{a_L+b_R}>s''\omega^{b_R+a}$ for any $s''\in\mathbb{R}$ with $0s''\omega^{a_R+b}$ for any $00$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. + +That was messy, but nice because it all fell out of the definition of the $\omega^-$ map. + +We now check that ordinal exponentiation agrees with the $\omega^-$ map. + +====Lemma 5.8:==== +For $a\in\mathbf{On}$, $\omega^a\in\mathbf{No}$ is the same as the ordinal $\omega^a$ (ordinal exponentiation). + +\WikiBold{Proof:} + +Write $\omega\uparrow a$ for ordinal exponentiation. By induction on $a\in \mathbf{On}$, we show that $\omega^a=\omega\uparrow a$. The base case was already done. Let $a=\{a_L\mid \emptyset\}$. Then using the induction hypothesis, +$$\omega^a=\{0,r\omega^{a_L}\mid \emptyset\}=\{0,r\omega\uparrow a_L\mid\emptyset\}$$ +$$=\omega\uparrow a,$$ +using the definition of ordinal exponentiation for the last equality. + +\WikiLevelFour{Section 6. The Normal Form} + +Let $K=\mathbb{R}((t^\mathbb{No}))$ be the Hahn field and set $x:=\frac{1}{t}$. We think of the elements of $K$ as formal series in $x$: +$$f(x)=\sum_{i<\alpha}f_ix^{a_i}$$ +where $\alpha\in \mathbf{On}$, $(a_i)$ is a strictly decreasing sequence in $\mathbf{No}$, and $f_i\in \mathbb{R}\setminus \{0\}$. So $\alpha$ is the order-type of $\mathrm{supp}(f)$, which we will denote by $\ell(f)$ (the agreement of the choice of this notation with the length of a surreal number is not a coincidence). We turn $K$ into an ordered field such that $f>0$ iff $f\neq 0$ and $f_0>0$. We will define an ordered field isomorphism between $K$ and $\mathbf{No}$, which will give a normal form for elements of $\mathbf{No}$ generalizing the Cantor normal form. diff --git a/Other/old/zach_copy/week_5/week_5.aux b/Other/old/zach_copy/week_5/week_5.aux deleted file mode 100644 index 8ab1f07c..00000000 --- a/Other/old/zach_copy/week_5/week_5.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_5/week_5}{ -\setcounter{page}{18} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_6/g_week_6.aux b/Other/old/zach_copy/week_6/g_week_6.aux deleted file mode 100644 index 3032f66a..00000000 --- a/Other/old/zach_copy/week_6/g_week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/g_week_6}{ -\setcounter{page}{26} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_6/g_week_6.tex b/Other/old/zach_copy/week_6/g_week_6.tex index 2536982f..0e43ec6a 100644 --- a/Other/old/zach_copy/week_6/g_week_6.tex +++ b/Other/old/zach_copy/week_6/g_week_6.tex @@ -1,392 +1,392 @@ -\WikiLevelTwo{ Week 6 } - -Notes by Anton Bobkov - -\WikiLevelThree{Monday, November 10, 2014} -We define a map which will eventually be proven to be an ordered field isomorphism. - -\begin{align*} - K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No -\end{align*} - -We have an element written as -\begin{align*} - &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ - &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} -\end{align*} -where $\supp(f)$ is a well-ordered sub\WikiItalic{set}. Now let $x = t^{-1}$ and write -\begin{align*} - f(x) = \sum_{i < \alpha} f_i x^{a_i} -\end{align*} -where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). - -==== Question ==== - What is the relationship of what we are going to do with Kaplansky's results from valuation theory? - -==== Definition/Theorem ==== -For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ -When $\alpha = \beta + 1$ is a successor: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} -\end{align*} -When $\alpha$ is a limit ordinal: -\begin{align*} - \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} -Simultaneously with this definition we prove the following statements by induction: - +\WikiLevelTwo{ Week 6 } + +Notes by Anton Bobkov + +\WikiLevelThree{Monday, November 10, 2014} +We define a map which will eventually be proven to be an ordered field isomorphism. + +\begin{align*} + K = \R((t^\No)) \overset{\sim}{\longrightarrow} \No +\end{align*} + +We have an element written as +\begin{align*} + &f = \sum_{\gamma \in \No} f_\gamma t^\gamma \\ + &\supp(f) = \{\gamma \colon f_\gamma \neq 0\} +\end{align*} +where $\supp(f)$ is a well-ordered sub\WikiItalic{set}. Now let $x = t^{-1}$ and write +\begin{align*} + f(x) = \sum_{i < \alpha} f_i x^{a_i} +\end{align*} +where $(a_i)_{i<\alpha}$ is strictly decreasing in $\No$, $\alpha$ ordinal and $f_i \in \R$ for $i < \alpha$. Also define $l(f(x))$ to be the order type of $\supp(f)$ (which may be smaller than $\alpha$ as we allow zero coefficients). + +==== Question ==== + What is the relationship of what we are going to do with Kaplansky's results from valuation theory? + +==== Definition/Theorem ==== +For $f(x) = \sum_{i < \alpha} f_i x^{a_i}$ define $\sum_{i < \alpha} f_i \w^{a_i} = f(\omega)$ recursively on $\alpha$: \\ +When $\alpha = \beta + 1$ is a successor: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} = \paren{\sum_{i < \beta} f_i \omega^{a_i}} + f_\beta \w^{a_\beta} +\end{align*} +When $\alpha$ is a limit ordinal: +\begin{align*} + \sum_{i < \alpha} f_i \w^{a_i} &= \curly{L \mid R} \\ + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} +Simultaneously with this definition we prove the following statements by induction: + \WikiSigleStar '''Inequality:''' For \begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} we have $f(x) > g(x) \Rightarrow f(\w) > g(\w)$ \WikiSigleStar '''Tail property:''' if $\gamma < \kappa < \alpha$ -\begin{align*} - \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} -\end{align*} - -\WikiBold{Proof of inequality} - -Suppose we have -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ - g(x) &= \sum_{i < \alpha} g_i x^{a_i} -\end{align*} - -with $f(x) < g(x)$ - -Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. -It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -\begin{align*} - f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ - g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} -\end{align*} - -Suppose $\gamma = \beta$. -Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute -\begin{align*} - f(\w) - g(\w) &= \\ - &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ - &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ - &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 -\end{align*} - -Now suppose $\gamma < \beta$. - -Group the terms -\begin{align*} - f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ - g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} -\end{align*} -where -\begin{align*} - h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ - f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ - g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} -\end{align*} - -Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute - -\begin{align*} - f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} -\end{align*} - -We have $f_\gamma > g_\gamma$. -All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. -Thus $f(\w) - g(\w) > 0$ as needed. - -\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. - -$f(\w)$ and $g(\w)$ are defined as - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - g(\w) &= \curly{L_g \mid R_g} -\end{align*} - -Recall that - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. - -Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. -Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. -As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. - -\WikiBold{Proof of tail property} - -It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. - -\WikiItalic{Case 1}: $\alpha = \beta + 1$. - -Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. - -\begin{align*} - &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ - &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} -\end{align*} - -Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. - -\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. - -Write definitions of $f(\w)$ using $\kappa$ - -\begin{align*} - f(\w) &= \curly{L_f \mid R_f} \\ - F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} -\end{align*} - -\begin{align*} - L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ -\end{align*} - -Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, -and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. - -By induction hypothesis we have -\begin{align*} - \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ - \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} -\end{align*} - -\begin{align*} - l(\w) \leq f(\w) \leq r(\w) \\ -\end{align*} -\begin{align*} - l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) -\end{align*} - -Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. - -\WikiBold{Proof of well-definiteness} - -We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. - -\begin{align*} - f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ - \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} -\end{align*} - -We need to check that $f(\w) = \bar f(\w)$ - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -\begin{align*} - f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ - g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ - g(\w) &= \bar g(\w) -\end{align*} - -If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. - -Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. -\begin{align*} - f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) -\end{align*} - -\WikiItalic{Case 2}: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. - -In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover -\begin{align*} - L_{\bar f} &\subseteq L_f \\ - R_{\bar f} &\subseteq R_f -\end{align*} -and are cofinal. Thus $f(\w) = \bar f(\w)$. - -\WikiItalic{Case 3}: $\alpha$ is a limit and for some $\gamma < \alpha$ -\begin{align*} - \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 -\end{align*} - -\begin{align*} - g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ - L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) - \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ - &= \curly{g(\w) + \epsilon \w^{a_\beta} - \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} -\end{align*} - -We have -\begin{align*} - L_f^* &\subseteq L_f \\ - R_f^* &\subseteq R_f -\end{align*} -and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. -Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. - -\WikiLevelThree{Wednesday, November 12, 2014} - -==== Lemma 6.1 ==== -$l(f(\w)) \geq l(f(x))$ - -\WikiBold{Proof} -Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. -\begin{align*} - l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) -\end{align*} -So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. -Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. - -==== Lemma 6.2 ==== -The map - -\begin{align*} - K &\arr \No \\ - f(x) &\mapsto f(\w) -\end{align*} - -is onto. - -\WikiBold{Proof} - -Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. -Put -\begin{align*} - S = \curly{s \in \R \colon s\w^b \leq a} -\end{align*} -Then $S \neq \emptyset$ and bounded from above. -Put $r = \sup S \in \R$. -Then -\begin{align*} - (r + \epsilon)\w^b > a > (r - \epsilon)\w^b -\end{align*} -for all $\epsilon \in \R^{>0}$ -thus -\begin{align*} - \abs{a - r\w^b} << \w^b \tag{*} -\end{align*} - -Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. - -We set $\lt(a) = r\w^b$ - -Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. -We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where - +\begin{align*} + \left| \sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} \right| << \w^{a_\gamma} +\end{align*} + +\WikiBold{Proof of inequality} + +Suppose we have +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i x^{a_i} \\ + g(x) &= \sum_{i < \alpha} g_i x^{a_i} +\end{align*} + +with $f(x) < g(x)$ + +Choose $\gamma$ smallest such that $f_\gamma \neq g_\gamma$. +It has to be that $f_\gamma > g_\gamma$. Also $f(x)\midr\gamma = g(x)\midr\gamma$ + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +\begin{align*} + f(x) &= f(x)\midr\beta + f_\beta x^{a_\beta}\\ + g(x) &= g(x)\midr\beta + g_\beta x^{a_\beta} +\end{align*} + +Suppose $\gamma = \beta$. +Then $\bar f(x) = \bar g(x)$, $\bar f(\w) = \bar g(\w)$, so compute +\begin{align*} + f(\w) - g(\w) &= \\ + &= f(\w)\midr\beta + f_\beta \w^{a_\beta} - g(\w)\midr\beta - g_\beta \w^{a_\beta} \\ + &= f_\beta \w^{a_\beta} - g_\beta \w^{a_\beta} \\ + &= \paren{f_\beta - g_\beta} \w^{a_\beta} > 0 +\end{align*} + +Now suppose $\gamma < \beta$. + +Group the terms +\begin{align*} + f(\w) &= h(\w) + f_\gamma \w^{a_\gamma} + f^* + f_\beta \w^{a_\beta} \\ + g(\w) &= h(\w) + g_\gamma \w^{a_\gamma} + g^* + g_\beta \w^{a_\beta} +\end{align*} +where +\begin{align*} + h(\w) &= f(\w)\midr\gamma = g(\w)\midr\gamma \\ + f^* &= f(\w)\midr\beta - f(\w)\midr{\gamma + 1} \\ + g^* &= g(\w)\midr\beta - g(\w)\midr{\gamma + 1} +\end{align*} + +Then we have by tail property $f^* << x^{a_\gamma}$ and $g^* << x^{a_\gamma}$. Compute + +\begin{align*} + f(\w) - g(\w) &= (f_\gamma - g_\gamma) x^{a_\gamma} + (f* - g*) + (f_\beta - g_\beta) x^{a_\beta} +\end{align*} + +We have $f_\gamma > g_\gamma$. +All $f*$, $g*$ and $(f_\beta - g_\beta) x^{a_\beta}$ are $<< x^{a_\gamma}$. +Thus $f(\w) - g(\w) > 0$ as needed. + +\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. + +$f(\w)$ and $g(\w)$ are defined as + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + g(\w) &= \curly{L_g \mid R_g} +\end{align*} + +Recall that + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_g &= \curly{\sum_{i < \beta} g_i \w^{a_i} + (g_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +Pick any $\beta$ with $\gamma < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar f(\w) \in L_f$ and $\bar g(\w) \in R_g$ corresponding to $\beta, \epsilon$. + +Then $\bar f(x) < \bar g(x)$ as first coefficient where they differ is $x^{a_\gamma}$ and $f_\gamma > g_\gamma$. +Thus by inductive hypothesis $\bar f(\w) < \bar g(\w)$. +As choice of those was arbitrary we have $L_f < R_g$ so $f(\w) > g(\w)$. + +\WikiBold{Proof of tail property} + +It is easy to see that statement holds for all $\gamma < \kappa < \alpha$ iff it holds for all $\gamma < \kappa \leq \alpha$. + +\WikiItalic{Case 1}: $\alpha = \beta + 1$. + +Suppose we have $\gamma < \kappa < \alpha$, then $\gamma < \kappa \leq \beta$ and induction hypothesis applies. + +\begin{align*} + &\sum_{i < \alpha} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i} = \\ + &\brac{\sum_{i < \beta} f_i \w^{a_i} - \sum_{i < \kappa} f_i \w^{a_i}} + f_\alpha \w^{a_\alpha} +\end{align*} + +Expression $\brac{\ldots}$ is $<< \w^{a_\gamma}$ by induction hypothesis. $f_\alpha \w^{a_\alpha} << \w^{a_\gamma}$ as $a_\alpha < a_\gamma$. Thus the entire sum is $<< \w^{a_\gamma}$ as needed. + +\WikiItalic{Case 2}: $\alpha$ is a limit ordinal. + +Write definitions of $f(\w)$ using $\kappa$ + +\begin{align*} + f(\w) &= \curly{L_f \mid R_f} \\ + F(\w) &= f(\w)\midr\kappa = \sum_{i < \kappa} f_i \w^{a_i} +\end{align*} + +\begin{align*} + L_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \beta < \alpha, \epsilon \in \R^{>0}} \\ +\end{align*} + +Pick any $\beta$ with $\kappa < \beta < \alpha$ and $\epsilon \in \R^{>0}$, +and pick limit elements $\bar l(\w) \in L_f$ and $\bar r(\w) \in R_f$ corresponding to $\beta, \epsilon$. + +By induction hypothesis we have +\begin{align*} + \bar l(\w) - F(\w) &= \bar l(\w) - \bar l(\w)\midr\kappa \ << \w^{a_\kappa} \\ + \bar r(\w) - F(\w) &= \bar r(\w) - \bar r(\w)\midr\kappa \ << \w^{a_\kappa} +\end{align*} + +\begin{align*} + l(\w) \leq f(\w) \leq r(\w) \\ +\end{align*} +\begin{align*} + l(\w) - F(\w) \leq f(\w) - F(\w) \leq r(\w) - F(\w) +\end{align*} + +Thus $f(\w) - F(\w) << \w^{a_\kappa}$ as it is between two elements that are $<< \w^{a_\kappa}$. + +\WikiBold{Proof of well-definiteness} + +We also need to check that the function is well-defined. For $f(x)$ define its reduced form, where we only keep non-zero coefficients. + +\begin{align*} + f(x) &= \sum_{i < \alpha} f_i \w^{a_i} \\ + \bar f(x) &= \sum_{j < \alpha'} f_j' \w^{a_j'} +\end{align*} + +We need to check that $f(\w) = \bar f(\w)$ + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +\begin{align*} + f(\w) &= g(\w) + f_\beta \w^{a_\beta} \\ + g(x) &= \sum_{i < \beta} f_i \w^{a_i} \\ + g(\w) &= \bar g(\w) +\end{align*} + +If $f_\beta = 0$ then $\bar f(x) = \bar g(x)$ and $f(\w) = g(\w)$ so $f(\w) = g(\w) = \bar g(\w) = \bar f(\w)$ as needed. + +Suppose $f_\beta \neq 0$. Then $\bar f(x) = \bar g(x) + f_\beta x^{a_\beta}$. +\begin{align*} + f(\w) = g(\w) + f_\beta \w^{a_\beta} = \bar g(\w) + f_\beta \w^{a_\beta} = \bar f(\w) +\end{align*} + +\WikiItalic{Case 2}: $\alpha$ is a limit and non-zero coefficients are cofinal in $\alpha$. + +In this case both $f(x)$ and $\bar f(x)$ have limit ordinals in their definitions. Moreover +\begin{align*} + L_{\bar f} &\subseteq L_f \\ + R_{\bar f} &\subseteq R_f +\end{align*} +and are cofinal. Thus $f(\w) = \bar f(\w)$. + +\WikiItalic{Case 3}: $\alpha$ is a limit and for some $\gamma < \alpha$ +\begin{align*} + \gamma \leq \beta < \alpha \Rightarrow f_\beta = 0 +\end{align*} + +\begin{align*} + g(x) &= \sum_{i < \gamma} f_i x^{a_i} \\ + L_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) - \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + R_f^* &= \curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon) \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} \\ + &= \curly{g(\w) + \epsilon \w^{a_\beta} + \colon \gamma < \beta < \alpha, \epsilon \in \R^{>0}} +\end{align*} + +We have +\begin{align*} + L_f^* &\subseteq L_f \\ + R_f^* &\subseteq R_f +\end{align*} +and are cofinal. $\curly{L_f^* - g(\w) \mid R_f^* - g(\w)} = 0$ as left side is negative and right side is positive. +Thus $f(\w) = \curly{L_f^* \mid R_f^*} = g(\w)$. As $\gamma < \alpha$ this case is covered by induction. + +\WikiLevelThree{Wednesday, November 12, 2014} + +==== Lemma 6.1 ==== +$l(f(\w)) \geq l(f(x))$ + +\WikiBold{Proof} +Suppose $\beta < \alpha$. Then the elements used to define $\sum_{i < \w\cdot\beta} (\ldots)$ also appear in the cut for $\sum_{i < \w\cdot\alpha} (\ldots) = f(\w)$. Thus by uniqueness of the normal form. +\begin{align*} + l\paren{\sum_{i < \w\cdot\beta} (\ldots)} < l(f(\w)) +\end{align*} +So the map $\phi(\beta) = l(\sum_{i < \w\cdot\beta} (\ldots))$ is strictly increasing. +Any strictly increasing map $\phi$ on an initial segment of $\On$ satisfies $\phi(\beta) \geq \beta$. + +==== Lemma 6.2 ==== +The map + +\begin{align*} + K &\arr \No \\ + f(x) &\mapsto f(\w) +\end{align*} + +is onto. + +\WikiBold{Proof} + +Let $a \in \No, a \neq 0$. By (5.6) there is a unique $b \in \No$ such that $\brac{a} = \brac{\w^b}$. +Put +\begin{align*} + S = \curly{s \in \R \colon s\w^b \leq a} +\end{align*} +Then $S \neq \emptyset$ and bounded from above. +Put $r = \sup S \in \R$. +Then +\begin{align*} + (r + \epsilon)\w^b > a > (r - \epsilon)\w^b +\end{align*} +for all $\epsilon \in \R^{>0}$ +thus +\begin{align*} + \abs{a - r\w^b} << \w^b \tag{*} +\end{align*} + +Note $r \neq 0$; $r,b$ subject to $(*)$ are unique. + +We set $\lt(a) = r\w^b$ + +Towards a contradiction assume that $a$ is not in the image of $f(x) \mapsto f(\w)$. +We shall inductively define a sequence $(a_i, f_i)_{i \in \On}$ where + \begin{enumerate} \item $a_i \in \No$ is strictly decreasing; $f_i \in \R - \{0\}$ \item $f_\alpha \w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha} f_i\w^{a_i}}$ for all $\alpha \in \On$ \end{enumerate} - -\WikiItalic{Case 1}: $\alpha = \beta + 1$ - -Take $(a_\alpha, f_\alpha)$ so that -\begin{align*} - f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} -\end{align*} - -By inductive hypothesis, if $\beta < \alpha$ - -\begin{align*} - f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ - \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ - &<< \w^{a_\beta} \text{ by (*)} \\ - \Rightarrow a_\alpha &< a_\beta -\end{align*} - -\WikiItalic{Case 2}: $\alpha$ limit - -Take $(a_\alpha, f_\alpha)$ as above. -Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. -We have - -\begin{align*} - a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} -\end{align*} - -By the tail property -\begin{align*} - &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ - \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ - \Rightarrow &a_\alpha < a_\beta -\end{align*} - -This completes the induction. -Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. - -So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. -Let $\alpha = \w \cdot \alpha'$. -Hence -\begin{align*} - l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' -\end{align*} -by (6.1). -So $l(a)$ is bigger than all limits - contradiction. - -==== Lemma 6.4 ==== -Let $r \in \R, a \in \No$. Then -\begin{align*} - r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} -\end{align*} -where $\epsilon$ ranges over $\R^{>0}$. - -\WikiBold{Proof} -\begin{align*} - r &= \{r - \epsilon \mid r + \epsilon\} \\ - \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} -\end{align*} -\begin{align*} - r\w^a = \{ - &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ - &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ - &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ - &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} -\end{align*} -Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. - -\WikiLevelThree{Friday, November 14, 2014} - -==== Corollary 6.5 ==== -\begin{align*} - \sum_{i \leq \alpha} f_i\w^{a_i} = - \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid - \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } -\end{align*} - -\WikiBold{Proof} - -\WikiItalic{Case 1}: $\alpha$ is a limit -\begin{align*} - &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} - &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid - \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} - + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = - &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, - \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} - &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } -\end{align*} - -\WikiItalic{Case 2}: $\alpha + 1$ - -\begin{align*} - &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ - &\text{(by (6.4) and induction hypothesis)} \\ - = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + - \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ - = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, - \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} -\end{align*} - -and again we are done by cofinality. + +\WikiItalic{Case 1}: $\alpha = \beta + 1$ + +Take $(a_\alpha, f_\alpha)$ so that +\begin{align*} + f_\alpha\w^{a_\alpha} = \lt\paren{a - \sum_{i < \alpha}(\ldots)} +\end{align*} + +By inductive hypothesis, if $\beta < \alpha$ + +\begin{align*} + f_\beta\w^{a_\beta} &= \lt\paren{a - \sum_{i < \beta}(\ldots)} \\ + \Rightarrow f_\alpha\w^{a_\alpha} &= \lt\paren{\paren{a - \sum_{i < \beta}(\ldots)} - f_\beta\w^{a_\beta}} \\ + &<< \w^{a_\beta} \text{ by (*)} \\ + \Rightarrow a_\alpha &< a_\beta +\end{align*} + +\WikiItalic{Case 2}: $\alpha$ limit + +Take $(a_\alpha, f_\alpha)$ as above. +Let $\beta < \alpha$; to show $a_\alpha < a_\beta$. +We have + +\begin{align*} + a - \sum_{i \leq \beta} f_i \w^{a_i} = \paren{a - \sum_{i < \beta} f_i\w^{a_i}} - f_\beta\w^{a_\beta} << \w^{a_\beta} +\end{align*} + +By the tail property +\begin{align*} + &\sum_{i < \alpha} (\ldots) - \sum_{i \leq \beta} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta} \\ + \Rightarrow &\lt\paren{a - \sum_{i < \alpha} (\ldots) } << \w^{a_\beta} \\ + \Rightarrow &a_\alpha < a_\beta +\end{align*} + +This completes the induction. +Note that we showed that if $\alpha$ is a limit then $a - \sum_{i < \alpha} (\ldots) << \w^{a_\beta}$ for all $\beta < \alpha$. + +So if $\curly{L \mid R}$ is the cut used to define $\sum_{i < \alpha} (\ldots)$ then $L < a < R$. +Let $\alpha = \w \cdot \alpha'$. +Hence +\begin{align*} + l(a) > l\paren{\sum_{i < \w\cdot\alpha'} (\ldots)} \geq \alpha' +\end{align*} +by (6.1). +So $l(a)$ is bigger than all limits - contradiction. + +==== Lemma 6.4 ==== +Let $r \in \R, a \in \No$. Then +\begin{align*} + r\w^a = \{(r - \epsilon) \w^a \mid (r+\epsilon)\w^a\} +\end{align*} +where $\epsilon$ ranges over $\R^{>0}$. + +\WikiBold{Proof} +\begin{align*} + r &= \{r - \epsilon \mid r + \epsilon\} \\ + \w^a &= \curly{0, s\w^{a_L} \mid t\w^{a_R}} \text{ where } s,t \in \R^{>0} +\end{align*} +\begin{align*} + r\w^a = \{ + &(r - \epsilon) \w^a, (r - \epsilon)\w^a + \epsilon s \w^{a_L}, \\ + &(r + \epsilon) \w^a - \epsilon t \w^{a_R} \mid \\ + &(r + \epsilon) \w^a, (r + \epsilon)\w^a - \epsilon s \w^{a_R}, \\ + &(r - \epsilon) \w^a + \epsilon t \w^{a_L} \} +\end{align*} +Now use $\w^{a_L} << \w^a << \w^{a_R}$ and cofinality. + +\WikiLevelThree{Friday, November 14, 2014} + +==== Corollary 6.5 ==== +\begin{align*} + \sum_{i \leq \alpha} f_i\w^{a_i} = + \curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid + \sum_{i < \alpha} f_i\w^{a_i} + (f_i + \epsilon) \w^{a_\alpha} } +\end{align*} + +\WikiBold{Proof} + +\WikiItalic{Case 1}: $\alpha$ is a limit +\begin{align*} + &\sum_{i \leq \alpha} f_i\w^{a_i} = \sum_{i < \alpha} f_i\w^{a_i} + f_\alpha\w^{a_\alpha} = \\ \underset{(6.4)}{=} + &\curly{\sum_{i < \beta} f_i \w^{a_i} + (f_\beta - \epsilon)\w^{a_\beta} \mid + \sum_{i < \beta} f_i \w^{a_i} + (f_\beta + \epsilon)\w^{a_\beta}}_{\beta < \alpha, \epsilon \in \R^{>0}} + + \curly{(r - \epsilon) \w^\alpha \mid (r+\epsilon)\w^\alpha} = \\ = + &\curly{\sum_{i \leq \beta} f_i \w^{a_i} - \epsilon\w^{a_\beta} + f_\alpha\w^{a_\alpha}, + \sum_{i < \alpha} f_i \w^{a_i} + (f_\alpha - \epsilon)\w^{a_\alpha} \mid \ldots} = \\ \underset{cofinality}{=} + &\curly{ \sum_{i < \alpha} f_i\w^{a_i} + (f_i - \epsilon) \w^{a_\alpha} \mid \ldots } +\end{align*} + +\WikiItalic{Case 2}: $\alpha + 1$ + +\begin{align*} + &\sum_{i \leq \alpha + 1} f_i \w^{a_i} = \sum_{i \leq \alpha} f_i \w^{a_i} + f_{\alpha + 1} \w^{a_{\alpha + 1}} = \\ + &\text{(by (6.4) and induction hypothesis)} \\ + = &\curly{\sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} \mid \ldots} + + \curly{(f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} = \\ + = &\curly{\sum_{i < \alpha} f_i \w^{a_i} + (f_{\alpha} - \epsilon) \w^{a_\alpha} + f_{\alpha + 1} \w^{a_{\alpha + 1}}, + \sum_{i \leq \alpha} f_i \w^{a_i} + (f_{\alpha + 1} - \epsilon) \w^{a_{\alpha + 1}} \mid \ldots} +\end{align*} + +and again we are done by cofinality. diff --git a/Other/old/zach_copy/week_6/week_6.aux b/Other/old/zach_copy/week_6/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/zach_copy/week_6/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_7/285D.aux b/Other/old/zach_copy/week_7/285D.aux deleted file mode 100644 index cd380abd..00000000 --- a/Other/old/zach_copy/week_7/285D.aux +++ /dev/null @@ -1,25 +0,0 @@ -\relax -\@setckpt{week_7/285D}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -} diff --git a/Other/old/zach_copy/week_7/285D_notes_nov_17_18_21.aux b/Other/old/zach_copy/week_7/285D_notes_nov_17_18_21.aux deleted file mode 100644 index f5313b87..00000000 --- a/Other/old/zach_copy/week_7/285D_notes_nov_17_18_21.aux +++ /dev/null @@ -1,32 +0,0 @@ -\relax -\newlabel{6.7}{{1}{26}} -\newlabel{6.8}{{2}{27}} -\@writefile{toc}{\contentsline {section}{\numberline {1}The Surreals as a Real Closed Field}{30}} -\newlabel{7.1}{{9}{30}} -\newlabel{7.2}{{10}{30}} -\@setckpt{week_7/285D_notes_nov_17_18_21}{ -\setcounter{page}{33} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_8/g_week_8.aux b/Other/old/zach_copy/week_8/g_week_8.aux deleted file mode 100644 index 8849a921..00000000 --- a/Other/old/zach_copy/week_8/g_week_8.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_8/g_week_8}{ -\setcounter{page}{36} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_8/g_week_8.tex b/Other/old/zach_copy/week_8/g_week_8.tex index 2a1db579..b7aa0b43 100644 --- a/Other/old/zach_copy/week_8/g_week_8.tex +++ b/Other/old/zach_copy/week_8/g_week_8.tex @@ -1,115 +1,115 @@ -\WikiLevelTwo{ Week 8 } -\WikiLevelThree{November 24, 2014} -\WikiItalic{Notes for today by Madeline Barnicle} - -Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the \WikiItalic{sign sequence} (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: -\WikiLevelFour{Section 8: Analytic functions on $\mathbf{No}$} -Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) - -$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) - -So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. - -Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. - -\WikiBold{Digression on exp for $\mathbf{No}$} - -An \WikiItalic{exponential function} on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ - -A negative result: -\WikiBold{Theorem} (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a \WikiItalic{set}, and $\Gamma \neq \{0\}$, then there is \WikiItalic{no} exponential function on $\mathbb{R}((t^{\Gamma}))$. - -Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). - -For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. - -Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. - -Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. - -Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. - -Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. - -\WikiBold{Theorem} (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. - -vd Dries and Ehrlich expanded this by adding the exponential function to both sides. - -\WikiLevelFour{Section 9: Power series and Weierstrass Preparation} -Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, \WikiItalic{the ring of formal power series in $X$ over $A$.} Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. - -$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. - -order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. - -Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. - -\WikiLevelThree{ November 26, 2014 } -Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its \WikiItalic{constant term}. - -====Lemma 9.1==== -Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. - -\WikiBold{Proof:} - -The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. - - -Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. - -Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. - -We will also need to define \WikiItalic{substitution}. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. - -Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write -$$f= \sum_{i,j} f_{ij} X^iY^j.$$ -This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. - -The previous result can be sharpened somewhat. -====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that -$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ -where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. - -The proof is an exercise. - -From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ \WikiItalic{regular in $T$ of order $d$} if -$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ -where $c\in K-\{0\}$. - -Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: +\WikiLevelTwo{ Week 8 } +\WikiLevelThree{November 24, 2014} +\WikiItalic{Notes for today by Madeline Barnicle} + +Write $x \in \mathbf{No}$ in normal form. Say all powers of $\omega$ are positive. Take an initial segment of $x$; the segment also has this property. The proof requires the \WikiItalic{sign sequence} (chapter 5 of Gonshor), and we need this to delve into the exponential function. Instead, we will cover: +\WikiLevelFour{Section 8: Analytic functions on $\mathbf{No}$} +Let $\Gamma$ be an ordered abelian group, $K = \mathbb{R}((t^{\Gamma})), x=t^{-1}$. Let $F: I \rightarrow \mathbb{R}$ ($I=(a, b), a 0} \}$ (infinitesimals of $K$) + +$K^{\uparrow}=\{f \in K:$ supp$(f) \subset \Gamma^{< 0} \}$ (the "purely infinite" elements of $K$) + +So $K=K^{\downarrow} \oplus \mathbb{R} \oplus K^{\uparrow}$. $O=K^{\downarrow} \oplus \mathbb{R}$. Then $I_K=\{c+\epsilon | c \in I \subset \mathbb{R}, \epsilon \in K^{\downarrow}\}$. Set $F_K (c+\epsilon)=\sum_{n=0}^{\infty} \frac{F^{(n)}(c)}{n!}\epsilon^{n} \in K$, which converges in $K$ by Neumann's lemma. For example, $c \rightarrow e^c: \mathbb{R} \rightarrow \mathbb{R}^{> 0}$ extends this way to $c+\epsilon \rightarrow e^{c+\epsilon}=e^c \sum_{n=0}^{\infty} \frac{\epsilon^n}{n!}, O \rightarrow K^{> 0}$. + +Likewise, every analytic function $G: U \rightarrow \mathbb{R}$ where $U \subset \mathbb{R}^n$ is open extends to a $K$-valued function whose domain $U_k$ is the set of all points in $K^n$ of infinitesimal distance to a point in $U$. + +\WikiBold{Digression on exp for $\mathbf{No}$} + +An \WikiItalic{exponential function} on $K$ is an isomorphism $(K, \leq, +) \rightarrow (K^{\geq 0}, \geq, \cdot).$ + +A negative result: +\WikiBold{Theorem} (F.-V. and S. Kuhlmann, Shelah): If the underlying class of $\Gamma$ is a \WikiItalic{set}, and $\Gamma \neq \{0\}$, then there is \WikiItalic{no} exponential function on $\mathbb{R}((t^{\Gamma}))$. + +Nevertheless, there is an exponential function on $\mathbf{No} \cong \mathbb{R}((t^{\mathbf{No}}))$ (Gonshor/Kruskal). + +For the rest of the course we will focus on restricted analytic functions on $\mathbf{No}$. + +Let $I=[-1,1] \subset \mathbb{R}$. A restricted analytic function is a function $F: \mathbb{R}^m \rightarrow \mathbb{R}$ such that $F(x)=0$ for $x \in \mathbb{R}^{m} \setminus I^m$, and $F \restriction I^m$ extends to an analytic function $U \rightarrow \mathbb{R}$ for some neighborhood $U$ of $I^m$. + +Example: $F: \mathbb{R} \rightarrow \mathbb{R}$ given by $F(x)=0$ if $|x|>1, F(x)=e^x$ if $|x| \leq 1$. For each such $F$, we have a function $F_K: K^M \rightarrow K$ such that $F_K(x)=0$ if $x \in K^m \setminus {I_K}^M, F_K \restriction {I_K}^m$ extends to a function $G_K: U_K \rightarrow K$ where $G: U \rightarrow \mathbb{R}$ is an analytic extension of $F \restriction I^m$ to an open neighborhood $U$ of $I^m$. + +Example: for $F$ as before, $F_K(x)=0$ if $|x|>1$. $F_K(c+\epsilon)=e^{c}\sum_{n} \frac{\epsilon^n}{n!}$, for $x=c+\epsilon, |x| \leq 1$. + +Let $L_{an}$ be the language $\{0,+, -, \cdot, \leq\}$ of ordered rings, augmented by a function symbol for each restricted analytic $\mathbb{R}^m \rightarrow \mathbb{R}$ (as $m$ varies). Let $\mathbb{R}_{an} = \mathbb{R}$ with the natural $L_{an}$ structure, $K_{an}=\mathbb{R}((t^{\Gamma}))$ with the natural structure (extensions as above). $\mathbb{R}_{an} \leq K_{an}$. + +\WikiBold{Theorem} (van den Dries, Macintyre, Marker, extending Denef-van den Dries): If $\Gamma$ is divisible, then $R_{an} \prec K_{an}$ (elementary substructure). In particular, $\mathbb{R} \prec \mathbf{No}$. + +vd Dries and Ehrlich expanded this by adding the exponential function to both sides. + +\WikiLevelFour{Section 9: Power series and Weierstrass Preparation} +Let $A$ be a commutative ring with $1, X=(x_1...x_m)$ indeterminates. $A[|x|]=A[|x_1, ...x_m)|]=\{f=\sum_{i \in \mathbb{N}^m}f_i x^i, f_i \in A\}$, \WikiItalic{the ring of formal power series in $X$ over $A$.} Here, $X^i= x_{1}^{i_1}...x_{m}^{i_m}$. These terms can be added or multiplied in the obvious way. + +$A \subset A[x] \subset A[|x|]$. For $i=(i_1...i_m ) \in \mathbb{N}^m$, put $|i|=i_1+...i_m$. For $f \in A[|x|]$, order $(f)=$min$\{|i|: f_i \neq 0\}$ if $f \neq 0$, or $\infty$ if $f=0$. + +order $(f+g) \geq$ min (ord($f$), ord($g$)). ord($fg$) $\geq$ ord ($f$) $+$ ord($g$), with equality if and only if $A$ is an integral domain. $A[|x|]$ is an integral domain if and only if $A$ is. + +Let $(f_j)_{j \in J}$ be a family in $A[|x|]$. If for all $d \in \mathbb{N}$ there are only finitely many $j \in J$ with ord($f_j$) $\leq d$, then we can make sense of $\sum_{j \in J}f_j \in A[|x|]$. We often write $f \in A[|x|]$ as $f=\sum_{d \in \mathbb{N}} f_d$ where $f_d=\sum_{|i|=d}f_i x^i$ is the degree-$d$ homogeneous part of $f$. + +\WikiLevelThree{ November 26, 2014 } +Let $A$ be a commutative ring, $X$ a set of $m$ indeterminates $X_1,\ldots,X_m$. For $f=\sum f_i X^i \in A[[X]]$, the map $f: A[[X]]\rightarrow A$ given by $f\mapsto f_0$ (here $0$ is the multi-index $(0,0,\ldots,0)$) is a ring morphism sending $f$ to its \WikiItalic{constant term}. + +====Lemma 9.1==== +Let $f\in A[[X]]$. Then $f$ is a unit in $A[[X]]$ if and only $f_0$ is a unit in $A$. + +\WikiBold{Proof:} + +The "if" direction is clear. For the converse, suppose $f_0g_0=1$, where $g_0\in A$. Then $fg_0=1-h$, where $\rm{ord}(h)\ge 1$. Now we can apply the usual geometric series trick: take $\sum_n h_n\in A[[X]]$, which is defined using the notion of infinite sum defined last time, and check that $g_0\cdot \sum_n h^n$ is an inverse for $f$. + + +Define $\mathfrak{o}=\{f\in A[[X]]:\rm{ord}(f)\ge 1\}=\{f:f_0=0\}$. This is an ideal of $A[[X]]$, and $A[[X]]=A\oplus \mathfrak{o}$ as additive groups. + +Every $f\in \mathfrak{o}$ is of the form $f=x_1g_1+\cdots +x_mg_m$, where $g_i\in A[[X]]$ (of course, this representation is not unique). More generally, set $\mathfrak{o}^d:=$ the ideal of $A[[X]]$ generated by products $f_1,\ldots,f_d$ with $f_i\in \mathfrak{o}$. Equivalently, this is the ideal $\{f:\rm{ord}(f)\ge d\}$ or the ideal generated by monomials of the form $X^i$, where $|i|=d$. That these are indeed equivalent is a straightforward exercise. + +We will also need to define \WikiItalic{substitution}. Let $Y=(y_1,\ldots y_n)$ be another tuple of distinct indeterminates, and let $g_1\ldots, g_m\in A[[Y]]$ with constant term $0$. Define a ring morphism $A[[X]]\rightarrow A[[Y]]$ by $f\mapsto f(g_1,\ldots, g_m)=\sum_i f_i g^i$. In the usual applications of this definition, we'll have $X=Y$. + +Let's introduce some more basic definitions for working with several sets of indeterminates. Suppose that $X$ and $Y$ are sets of indeterminates $\{X_1,\ldots,X_m\}$ and $\{Y_1,\ldots,Y_n\}$ respectively, and that none of the indeterminates in $Y$ appears in $X$. Put $(X,Y):=(X_1,\ldots, X_m, Y_1,\ldots, Y_n)$. Then for $f\in A[[X,Y]]$, we can write +$$f= \sum_{i,j} f_{ij} X^iY^j.$$ +This can be rewritten as $\sum_j(\sum_i f_{ij}X^i)Y^j$ (the sums above are actually the infinite sums defined last time). This gives an identification of $A[[X,Y]]$ with $A[[X]][[Y]]$. + +The previous result can be sharpened somewhat. +====Lemma 9.2==== Let $f\in A[[X,Y]]$. Then there are unique $g_1,\ldots, g_n\in A[[X,Y]]$ such that +$$f(X,Y)=f(X)+X_1 g_1(X, Y_1)+\ldots + Y_n g_n(X, Y_1,\ldots Y_n)$$ +where $g_i\in A[[X,Y_1,\ldots, Y_i]]$. + +The proof is an exercise. + +From now on, we will take $A$ to be a field $K$. Let $T$ be an indeterminate not among $X_1,\ldots, X_m$. We call $f(X,T)\in A[[X,T]]$ \WikiItalic{regular in $T$ of order $d$} if +$$f(0,T)=cT^d+\textrm{ terms of order larger than }d$$ +where $c\in K-\{0\}$. + +Writing $f=\sum_{i\in\mathbb{N}} f_i(X)T^i$, this is also equivalent to either of the following: \begin{enumerate} \item $f_0(0)=\cdots =f_{d-1}(0)=0$ and $f_d(0)\neq 0$, \item $f_0,\ldots, f_{d-1}\in \mathfrak{o}$, $f_d\in \mathfrak{o}$, $f_d\in K[[X]]^\times$. \end{enumerate} - -The reason for this definition is that there is a "Euclidean division" result which holds when dividing by regular power series. - -====Theorem 9.3 (Division with remainder)==== -Let $f\in K[[X,T]]$ be regular in $T$ of order $d$. Then for each $g\in K[[X,T]]$, there is a unique pair $(Q,R)$ where $Q\in K[[X,T]]$ and $R\in K[[X]][T]$ (so it is just a \WikiItalic{polynomial} in $T$) such that $g=Qf+R$ and $\mathrm{deg}_TRpolyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. -Given polyradii $r,s$ we write +\WikiLevelTwo{ Week 9 } + +\WikiLevelThree{ Monday 12-1-2014 } + +==== Corollary 9.4 (Weierstrauss Preparation) ==== + +(Notes today by John Lensmire.) + +Let $f\in K[[X,T]]$ be regular of order $d$ in $T$. +Then $f\in K[[X,T]]^\times$ and $W\in K[[X]][T]$ is monic of degree $d$ in $T$. + +\WikiBold{Proof:} + +Using Theorem 9.3, we can write $T^d = Qf+R$ where $Q\in K[[X,T]], R\in K[[X]][T], \mathrm{deg}_TR < d$. + +Let $x=0$ to get +$$T^d = \left( \sum_i Q_i(0) T^i \right) (f_d(0) + \textrm{ terms of higher order} ) ++ R_0(0) + R_1(0) T + \cdots + R_{d-1}(0) T^{d-1}.$$ +Looking at the coefficient of $T^d$, we have $1 = Q_0(0) f_d(0)$. +This implies $Q_0 \in K[[X]]^\times$ and thus $Q\in K[[X,T]]^\times$. + +Hence, we have $f = uW$ where $u = Q^{-1}$ and $W = T^d - R$. + +Uniqueness follows from the uniqueness in Theorem 9.3 (details are left as an exercise.) + +\WikiBold{Remark:} + +The above proof shows that we can take $W$ to be a Weierstrauss polynomial, i.e. a monic polynomial +$W = T^d + W_{d-1} T^{d-1} + \cdots + W_0$ where $W_0,\ldots, W_{d-1}\in \mathfrak{o}[[X]]$. + +==== Corollary 9.5 ==== + +Suppose $K$ is infinite. Then the ring $K[[X]]$ is noetherian. + +\WikiBold{Proof:} + +We proceed by induction on $m$. + +If $m=0$, $K$ is a field, hence noetherian. + +From $m$ to $m+1$: +Let $\{0\} \neq I \subset K[[X,T]]$ be an ideal. +Take $f\in I\setminus \{0\}$, after replacing $I,f$ by images under a suitable automorphism of $K[[X,T]]$ +(see last time) we can assume that $f$ is regular in $T$ of some order $d$. +Then each $g\in I$ can be written as $g = qf + r$, where $q\in K[[X,T]]$ and $r\in A[T]$ is of degree $polyradius is a vector $r = (r_1,\ldots, r_m)\in (R^{\geq 0})^m$. +Given polyradii $r,s$ we write \begin{enumerate} \item $r\leq s \Leftrightarrow r_i \leq s_i$ for each $i$. \item $r < s \Leftrightarrow r_i < s_i$ for each $i$. \item $r^i = r_1^{i_1}\cdots r_m^{i_m}$ for $i = (i_1,\ldots, i_m)\in \mathbb{N}^m$. \end{enumerate} - -Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i -\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. -Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. - -For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. -Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: + +Given a polyradius $r$ and $a\in \mathbb{C}^m$, $D_r(a):= \{x\in \mathbb{C}^m |\ |x_i - a_i| < r_i +\textrm{ for } i = 1,\ldots,m$, called the open polydisk centered at $a$ with polyradius $r$. +Its closure, the closed polydisk centered at $a$ with polyradius $r$, is $\overline{D_r}(a) = \{x | \ |x_i - a_i| \leq r_i\}$. + +For $f\in \mathbb{C}[[X]]$, define $\|f\|_r := \sum_i |f_i| r^i \in \mathbb{R}^{\leq 0} \cup \{+\infty\}$. +Writing $\|\cdot \| = \|\cdot \|_r$, it is easy to verify: \begin{enumerate} \item $\|f\| = 0 \Leftrightarrow f = 0$. \item $\|c f\| = |c| \cdot \|f\|$ for $c\in \mathbb{C}$. @@ -71,62 +71,62 @@ \item $\|X^i f\| = r^i \|f\|$. \item $r\leq s \Rightarrow \|f\|_r \leq \|f\|_s$. \end{enumerate} - -==== Definition 10.1 ==== - -$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ - -==== Lemma 10.2 ==== - + +==== Definition 10.1 ==== + +$\mathbb{C}\{X\}_r := \{f\in \mathbb{C}[[X]] |\ \|f\|_r < +\infty \}$ + +==== Lemma 10.2 ==== + \begin{enumerate} \item $\mathbb{C} \{X\}_r$ is a subalgebra of $\mathbb{C}[[X]]$ containing $C[X]$. \item $\mathbb{C} \{X\}_r$ is complete with respect to the norm $\|\cdot\|_r$. \end{enumerate} - -\WikiBold{Proof:} -1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, -and is left as an exercise. - -Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: -For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. -This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). - -For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, -and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) -$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. -Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. -Further, we have, -for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, -and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. - -==== Lemma 10.3 ==== - -The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: -$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ -Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. - -\WikiBold{Proof:} - -All claims follow from definitions directly except injectivity. By induction on $m$, we show: -$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. - -If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. - -For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. -This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. - -For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ -with $r' = (r_1,\ldots, r_{m-1})$. -Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. -Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ -such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. - -\WikiLevelThree{ Wednesday 12-3-2014 } -\WikiItalic{notes by Asaaf Shani} - -\WikiBoldItalic{Coming soon} - -\WikiLevelThree{ Friday 12-5-2014 } -\WikiItalic{notes by Tyler Arant} - -PDF: [[Media:285D notes 12 5.pdf]] + +\WikiBold{Proof:} +1. is clear. 2. is routine using a "Cauchy Estimate": for every $i\in \mathbb{N}^m$, $|f_i|\leq \|f\|_r/r^i$, +and is left as an exercise. + +Each $f\in\mathbb{C}\{X\}_r$ gives rise to a function $\overline{D_r}(0)\rightarrow\mathbb{C}$ as follows: +For $x\in\overline{D_r}(0)$, the series $\sum_i f_i x^i$ converges absolutely to a complex number $f(x)$. +This function $x\mapsto f(x)$ is continuous (as it is the limit of uniformly continuous functions). + +For $f\in \mathbb{C}[[X,Y]]$, $f = \sum_{j\in \mathbb{N}^n} f_j(X) Y^j$, +and $(r,s)\in (\mathbb{R}^{\geq 0})^{m+n}$ a polyradius, we have (from the definitions) +$\|f\|_{(r,s)} = \sum_j \|f_j\|_r s^j$. +Hence, $f\in \mathbb{C}\{X,Y\}_{(r,s)}$ and in particular $f_j\in \mathbb{C}\{X\}_r$ for all $j$. +Further, we have, +for $x\in \overline{D_r}(0)$, $f(x,y):= \sum_j f_j(X)Y^j\in \mathbb{C}\{Y\}_s$, +and for $(x,y)\in \overline{D_{(r,s)}}(0)$, $f(x,y) = \left( \sum_j f_j(X)Y^j\right)(y) = \sum_j f_j(x)y^j$. + +==== Lemma 10.3 ==== + +The map that sends $f\in \mathbb{C}\{X\}_r$ to $f_r$ given by $x\mapsto f(x)$ is an injective ring morphism: +$$\mathbb{C}\{X\}_r \rightarrow \{\textrm{ ring of continuous functions } \overline{D_r}(0) \rightarrow \mathbb{C}\}$$ +Further, $\|f_r\|_{\mathrm{sup}} \leq \|f\|_r$. + +\WikiBold{Proof:} + +All claims follow from definitions directly except injectivity. By induction on $m$, we show: +$f\in \mathbb{C}\{X\}_r\setminus \{0\}$ implies $f_r$ does not vanish identically on any open neighborhood of $0\in \mathbb{C}^m$. + +If $m=1$, write $f = X^d (f_d + f_{d+1} X + \cdots )$, where $f_i\in \mathbb{C}, f_d\neq 0$. + +For $|x|\leq r$, the series $f_d + f_{d+1}X + \cdots $ converges to a continuous function of $X$. +This function takes value $f_d\neq 0$ at $x=0$, hence is non-zero in a neighborhood around $0$. + +For $m\geq 2$, write $f = \sum_i f_i(X') X_m^i$, where $X' = (X_1,\ldots, X_{m-1})$, $f_i \in \mathbb{C}\{X'\}_{r'}$ +with $r' = (r_1,\ldots, r_{m-1})$. +Then $\|f\|_r = \sum_i \|f_i\|_{r'} r_m^i$ and $f(x) = \sum_i f_i(X')X_m^i$ for $X = (X',X_m)\in \overline{D_r}(0)$. +Fix $j$ such that $f_j(X') = 0$. By the induction hypothesis, there are $X'\in \mathbb{C}^{m-1}$ as close as we want to $0$ +such that $f_j(X')\neq 0$. For such an $X'$ (as in the $m=1$ case), we have $f(X',X_m)\neq 0$ for all sufficiently small $X_m$. + +\WikiLevelThree{ Wednesday 12-3-2014 } +\WikiItalic{notes by Asaaf Shani} + +\WikiBoldItalic{Coming soon} + +\WikiLevelThree{ Friday 12-5-2014 } +\WikiItalic{notes by Tyler Arant} + +PDF: [[Media:285D notes 12 5.pdf]] diff --git a/Other/old/zach_copy/week_9/week_6.aux b/Other/old/zach_copy/week_9/week_6.aux deleted file mode 100644 index f272cbc6..00000000 --- a/Other/old/zach_copy/week_9/week_6.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_6/week_6}{ -\setcounter{page}{25} -\setcounter{equation}{1} -\setcounter{enumi}{3} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{0} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{5} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{0} -\setcounter{namedtheorem}{0} -} diff --git a/Other/old/zach_copy/week_9/week_9.aux b/Other/old/zach_copy/week_9/week_9.aux deleted file mode 100644 index 113be7d4..00000000 --- a/Other/old/zach_copy/week_9/week_9.aux +++ /dev/null @@ -1,27 +0,0 @@ -\relax -\@setckpt{week_9/week_9}{ -\setcounter{page}{37} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{2} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{1} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{14} -\setcounter{defn}{7} -\setcounter{cor}{2} -\setcounter{claim}{2} -\setcounter{lem}{3} -\setcounter{lemma}{3} -\setcounter{namedtheorem}{0} -} diff --git a/Shelah-Spencer VC (Anton, Nov 6 10h56).tps b/Shelah-Spencer VC (Anton, Nov 6 10h56).tps deleted file mode 100644 index 9a93be45..00000000 --- a/Shelah-Spencer VC (Anton, Nov 6 10h56).tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - 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-\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{diagrams} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\begin{document} - -\title{Some vc-density computations in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We compute vc-densities of minimal extension formulas in Shelah-Spencer random graphs. -\end{abstract} - -\maketitle - -We fix the density of the graph $\alpha$. - -\begin{Lemma} - For any $\A \in \K$ and $\epsilon > 0$ there exists an $\B$ such that $(\A, \B)$ is minimal and $\delta(\B/\A) < \epsilon$. -\end{Lemma} - -\begin{proof} - Let $m$ be an integer such that $m\alpha < 1 < (m+1)\alpha$. Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. -\end{proof} - -\begin{Lemma} - Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair with $\epsilon = \delta (\B_2/\A_2)$. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Assume that it is not that case that $\A_2 \subset \B_2$ and $\A_1$ is disjoint from $\A_2$. Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \epsilon$ -\end{Lemma} - -Fix an embedding $f$. It induces the following substructure diagram in $M$. Denote -\begin{align*} - \A &= \A_1 \cup \A_2 \\ - \B_f^* &= \B_1 \cup f(\B_2) \\ - \B_1^* &= \B_1 \cup \A \\ - \B_2^* &= f(\B_2) \cup \A \\ - \B^* &= (\B_1 \cap f(\B_2)) \cup \A -\end{align*} - -\begin{diagram} - & &\B_f \\ - &\ruLine & &\luLine \\ - \B_1^* & & & &\B_2^* \\ - &\luLine & &\ruLine \\ - & &\B^* \\ - & &\uLine \\ - & &\A\\ -\end{diagram} - -From the diagram we see that -\begin{align*} - \delta(\B_f/\A) \leq \delta(\B_1^*/\A) + \delta(\B_2^*/\B^*) -\end{align*} -Thus all we need to do is to verify that -\begin{align*} - \delta(\B_2^*/\B^*) \leq \epsilon -\end{align*} -Let $\B'$ denote graph induced on all the vertices in $(f(B_2) / B_1) \cup A_2$. -Then $\B'$ is a substructure of $\B_2$ over $\A_2$. By minimality we get that $\delta(\B'/\A_2) \leq \epsilon$. Moreover $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$ thus proving the claimed statement. -Then $\delta (\B^*/A_2)$ has to be less than $\delta (\B_2/\A_2)$ by minimality of $(\B_2, \A_2)$. Relative dimension of the whole construction has to be even smaller. -It is easy to show that this construction induces a proper subpair in $(\A_2, \B_2)$ which has to have smaller dimension. - - - -Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists a minimal extension $M$ over $\{x,y\}$ of relative dimension $\epsilon$. Let $n$ be such that $n\epsilon < 1 < (n+1)\epsilon$. Then we argue that $vc(\phi) = n$. - -Fix a $m$-strong (for any $m > |M|$) set of non-connected vertices $B$. Fix some $a$. We invesitgate the trace of $\phi(x, a)$ on $B$. Suppose we have $b_1, \ldots, b_k$ satisfying $\phi(b_i, a)$ as witnessed by $M_j$. Relative dimension of $M_1 \cup M_2 \cup \ldots \cup M_j \cup {a}$ is minimized when all $M_j$ are disjoint (by minimality). Thus for that dimension to be positive we can have at most $n$ extensions. - -\begin{thebibliography}{9} - -\bibitem{Laskowski} - Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} - Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{diagrams} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\begin{document} + +\title{Some vc-density computations in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We compute vc-densities of minimal extension formulas in Shelah-Spencer random graphs. +\end{abstract} + +\maketitle + +We fix the density of the graph $\alpha$. + +\begin{Lemma} + For any $\A \in \K$ and $\epsilon > 0$ there exists an $\B$ such that $(\A, \B)$ is minimal and $\delta(\B/\A) < \epsilon$. +\end{Lemma} + +\begin{proof} + Let $m$ be an integer such that $m\alpha < 1 < (m+1)\alpha$. Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. +\end{proof} + +\begin{Lemma} + Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair with $\epsilon = \delta (\B_2/\A_2)$. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Assume that it is not that case that $\A_2 \subset \B_2$ and $\A_1$ is disjoint from $\A_2$. Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \epsilon$ +\end{Lemma} + +Fix an embedding $f$. It induces the following substructure diagram in $M$. Denote +\begin{align*} + \A &= \A_1 \cup \A_2 \\ + \B_f^* &= \B_1 \cup f(\B_2) \\ + \B_1^* &= \B_1 \cup \A \\ + \B_2^* &= f(\B_2) \cup \A \\ + \B^* &= (\B_1 \cap f(\B_2)) \cup \A +\end{align*} + +\begin{diagram} + & &\B_f \\ + &\ruLine & &\luLine \\ + \B_1^* & & & &\B_2^* \\ + &\luLine & &\ruLine \\ + & &\B^* \\ + & &\uLine \\ + & &\A\\ +\end{diagram} + +From the diagram we see that +\begin{align*} + \delta(\B_f/\A) \leq \delta(\B_1^*/\A) + \delta(\B_2^*/\B^*) +\end{align*} +Thus all we need to do is to verify that +\begin{align*} + \delta(\B_2^*/\B^*) \leq \epsilon +\end{align*} +Let $\B'$ denote graph induced on all the vertices in $(f(B_2) / B_1) \cup A_2$. +Then $\B'$ is a substructure of $\B_2$ over $\A_2$. By minimality we get that $\delta(\B'/\A_2) \leq \epsilon$. Moreover $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$ thus proving the claimed statement. +Then $\delta (\B^*/A_2)$ has to be less than $\delta (\B_2/\A_2)$ by minimality of $(\B_2, \A_2)$. Relative dimension of the whole construction has to be even smaller. +It is easy to show that this construction induces a proper subpair in $(\A_2, \B_2)$ which has to have smaller dimension. + + + +Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists a minimal extension $M$ over $\{x,y\}$ of relative dimension $\epsilon$. Let $n$ be such that $n\epsilon < 1 < (n+1)\epsilon$. Then we argue that $vc(\phi) = n$. + +Fix a $m$-strong (for any $m > |M|$) set of non-connected vertices $B$. Fix some $a$. We invesitgate the trace of $\phi(x, a)$ on $B$. Suppose we have $b_1, \ldots, b_k$ satisfying $\phi(b_i, a)$ as witnessed by $M_j$. Relative dimension of $M_1 \cup M_2 \cup \ldots \cup M_j \cup {a}$ is minimized when all $M_j$ are disjoint (by minimality). Thus for that dimension to be positive we can have at most $n$ extensions. + +\begin{thebibliography}{9} + +\bibitem{Laskowski} + Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} + Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 + +\end{thebibliography} + \end{document} diff --git a/Shelah-Spencer VC (Anton, Sep 24 18h21).tex b/Shelah-Spencer VC (Anton, Sep 24 18h21).tex index e0f461cb..730cbd2d 100644 --- a/Shelah-Spencer VC (Anton, Sep 24 18h21).tex +++ b/Shelah-Spencer VC (Anton, Sep 24 18h21).tex @@ -1,71 +1,71 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{diagrams} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\begin{document} - -\title{Some vc-density computations in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We compute vc-densities of minimal extension formulas in Shelah-Spencer random graphs. -\end{abstract} - -\maketitle - -We fix the density of the graph $\alpha$. - -\begin{Lemma} - For any $\A \in \K$ and $\epsilon > 0$ there exists an $\B$ such that $(\A, \B)$ is minimal and $\delta(\B/\A) < \epsilon$. -\end{Lemma} - -\begin{proof} - Let $m$ be an integer such that $m\alpha < 1 < (m+1)\alpha$. Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. -\end{proof} - -\begin{Lemma} - Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \delta (\B_2/\A_2)$ -\end{Lemma} - -\begin{diagram} - & &\B_1 \cup \B_2 = \B \\ - &\ruLine & &\luLine \\ - \B_1 \cup \A & & & &\B_2 \cup \A \\ - &\luLine & &\ruLine \\ - & &(\B_1 \cap \B_2) \cup \A \\ - & &\uLine \\ - & &\A_1 \cup \A_2 = \A\\ -\end{diagram} - -From the diagram we see that $\delta(\B/\A) \leq \delta(\B_1 \cup \A/\A) + \delta\left((\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)\right)$. Thus all we need to do is to verify that $\delta\left((\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)\right) \leq \delta (\B_2/\A_2)$. Look at the structure $\B^*$ induced inside of $B_2$ by vertices of $(B_1 \cap B_2) \cup A$. It contains $\A$. Suppose it is not the whole $B_2$. Then we have by minimality that $\delta(\B^*/\A_2) > 0$ and thus $\delta(\B_2/\B^*) < \delta (\B_2/\A_2)$. Now $(\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)$ has the same number of vertices as $\B_2/\B^*$ and at least as many edges. Claim follows. - - - -Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists a minimal extension $M$ over $\{x,y\}$ of relative dimension $\epsilon$. Let $n$ be such that $n\epsilon < 1 < (n+1)\epsilon$. Then we argue that $vc(\phi) = n$. - -Fix a $m$-strong (for any $m > |M|$) set of non-connected vertices $B$. Fix some $a$. We invesitgate the trace of $\phi(x, a)$ on $B$. Suppose we have $b_1, \ldots, b_k$ satisfying $\phi(b_i, a)$ as witnessed by $M_j$. Relative dimension of $M_1 \cup M_2 \cup \ldots \cup M_j \cup {a}$ is minimized when all $M_j$ are disjoint (by minimality). Thus for that dimension to be positive we can have at most $n$ extensions. - -\begin{thebibliography}{9} - -\bibitem{Laskowski} - Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} - Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{diagrams} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\begin{document} + +\title{Some vc-density computations in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We compute vc-densities of minimal extension formulas in Shelah-Spencer random graphs. +\end{abstract} + +\maketitle + +We fix the density of the graph $\alpha$. + +\begin{Lemma} + For any $\A \in \K$ and $\epsilon > 0$ there exists an $\B$ such that $(\A, \B)$ is minimal and $\delta(\B/\A) < \epsilon$. +\end{Lemma} + +\begin{proof} + Let $m$ be an integer such that $m\alpha < 1 < (m+1)\alpha$. Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. +\end{proof} + +\begin{Lemma} + Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \delta (\B_2/\A_2)$ +\end{Lemma} + +\begin{diagram} + & &\B_1 \cup \B_2 = \B \\ + &\ruLine & &\luLine \\ + \B_1 \cup \A & & & &\B_2 \cup \A \\ + &\luLine & &\ruLine \\ + & &(\B_1 \cap \B_2) \cup \A \\ + & &\uLine \\ + & &\A_1 \cup \A_2 = \A\\ +\end{diagram} + +From the diagram we see that $\delta(\B/\A) \leq \delta(\B_1 \cup \A/\A) + \delta\left((\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)\right)$. Thus all we need to do is to verify that $\delta\left((\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)\right) \leq \delta (\B_2/\A_2)$. Look at the structure $\B^*$ induced inside of $B_2$ by vertices of $(B_1 \cap B_2) \cup A$. It contains $\A$. Suppose it is not the whole $B_2$. Then we have by minimality that $\delta(\B^*/\A_2) > 0$ and thus $\delta(\B_2/\B^*) < \delta (\B_2/\A_2)$. Now $(\B_2 \cup \A)/((\B_1 \cap \B_2) \cup \A)$ has the same number of vertices as $\B_2/\B^*$ and at least as many edges. Claim follows. + + + +Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists a minimal extension $M$ over $\{x,y\}$ of relative dimension $\epsilon$. Let $n$ be such that $n\epsilon < 1 < (n+1)\epsilon$. Then we argue that $vc(\phi) = n$. + +Fix a $m$-strong (for any $m > |M|$) set of non-connected vertices $B$. Fix some $a$. We invesitgate the trace of $\phi(x, a)$ on $B$. Suppose we have $b_1, \ldots, b_k$ satisfying $\phi(b_i, a)$ as witnessed by $M_j$. Relative dimension of $M_1 \cup M_2 \cup \ldots \cup M_j \cup {a}$ is minimized when all $M_j$ are disjoint (by minimality). Thus for that dimension to be positive we can have at most $n$ extensions. + +\begin{thebibliography}{9} + +\bibitem{Laskowski} + Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} + Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 + +\end{thebibliography} + \end{document} diff --git a/Shelah-Spencer VC (Anton, Sep 24 18h21).tps b/Shelah-Spencer VC (Anton, Sep 24 18h21).tps deleted file mode 100644 index f01292b7..00000000 --- a/Shelah-Spencer VC (Anton, Sep 24 18h21).tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - -[Frame0] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=745 -NormalPos.bottom=392 -Class=LaTeXView -Document=Shelah-Spencer VC.tex - -[Frame0_View0,0] -TopLine=47 -Cursor=1560 - -[SessionInfo] -FrameCount=1 -ActiveFrame=0 - diff --git a/g_week_2 (Anton, Dec 19 20h17).aux b/g_week_2 (Anton, Dec 19 20h17).aux deleted file mode 100644 index 0bf482bf..00000000 --- a/g_week_2 (Anton, Dec 19 20h17).aux +++ /dev/null @@ -1,24 +0,0 @@ -\relax -\newlabel{d26}{{2.8}{8}} -\newlabel{l27}{{2.9}{9}} -\@setckpt{week_2/g_week_2}{ -\setcounter{page}{13} -\setcounter{equation}{1} -\setcounter{enumi}{2} -\setcounter{enumii}{0} -\setcounter{enumiii}{0} -\setcounter{enumiv}{0} -\setcounter{footnote}{0} -\setcounter{mpfootnote}{0} -\setcounter{part}{0} -\setcounter{section}{2} -\setcounter{subsection}{0} -\setcounter{subsubsection}{0} -\setcounter{paragraph}{0} -\setcounter{subparagraph}{0} -\setcounter{figure}{0} -\setcounter{table}{0} -\setcounter{parentequation}{0} -\setcounter{theorem}{13} -\setcounter{namedtheorem}{0} -} diff --git a/mycommands.xml b/mycommands.xml index 9f7832e3..6cccf3b0 100644 --- a/mycommands.xml +++ b/mycommands.xml @@ -1,158 +1,158 @@ - - - - - - - - - - - - - - - - - - - - - - - - - - - expbefore="\begin{"/> - - - - - - - - - - - - - - - - - - - - - + + + + + + + + + + + + + + + + + + + + + + + + + + + expbefore="\begin{"/> + + + + + + + + + + + + + + + + + + + + + diff --git a/research/01 VCd alternative/VCd alternative.tex b/research/01 VCd alternative/VCd alternative.tex index a18033c8..007060fc 100644 --- a/research/01 VCd alternative/VCd alternative.tex +++ b/research/01 VCd alternative/VCd alternative.tex @@ -1,65 +1,65 @@ -\documentclass{amsart} - -\usepackage{AMC_style} -\usepackage{Research} - -\newcommand{\D}{\Delta} -\newcommand{\F}{\mathcal F} - - -\begin{document} -The following is the proof of the Theorem \textbf{7.1} in \textit{Vapnik-Chervonenkis density in some theories without the independence property, I} without using a stong $\VC d$ property. -\ \\ -\begin{Theorem} - Assume that $\vc(m) \leq r$ and the theory has the $\VC d$ property. Then $\vc(m+1) \leq r + d$. -\end{Theorem} -\begin{proof} - Write $x = (x_0, x')$ with $x' = (x_1, \ldots, x_m)$ so that $|x_0| = 1$ and $|x'| = m$. Let $\D(x;y)$ be given. Define - \begin{align*} - \D_0(x_0;x', y) = \{\phi(x_0; x', y) \mid \phi(x;y) \in \D\} - \end{align*} - Applying VC $d$ property to $\D_0$ we have finitely many families - \begin{align*} - &\F_i = (\phi_i(x', y; y_1, \ldots, y_d))_{\phi \in \D} &(i \in I) - \end{align*} - of $\LL$-formulas with the following property: for any $a_1 \in M^{|x_0|}$, $a_2 \in M^{|x'|}$, any finite $B \subset M^{|y|}$ there are $\vec b \in B^d$ and $i \in I$ such that $\F_i(a_2, y; \vec b)$ defines $\tp_{\D_0}(a_1/a_2B)$, i.e. for all $\phi \in \D, b \in B$ - \begin{align*} - \models \phi(a_1, a_2, b) \iff \models \phi_i(a_2, b, \vec b) - \end{align*} - For each $i \in I$ let - \begin{align*} - \D_i(x';y, y_1 \ldots y_d) = \{\phi_i(x'; y, y_1 \ldots y_d) \mid \phi(x;y) \in \D\} - \end{align*} - As $|x'| = m$ the assumption that $\vc(m) \leq r$ applies to each $\D_i$. Thus there is a constant $K$ such that for any finite $C \subset (M^{|y|})^{(d+1)}$ there is a set of representatives for $S^{\D_i}(C)$ of size at most $K|C|^r$. (More precisely, for each $\D_i$ there is going to be such a constant $K_i$ and we can take $K$ to be the maximum of these). Now fix a finite set $B \subset M^{|y|}$. Let $N = K|B|^r$. For every element $\vec b \in B^d$ fix a set of representatives of $S^{\D_i}(B\vec b)$ (Note that $|B| = |B\vec b|$) - \begin{align*} - \alpha^{i, \vec b}_1, \alpha^{i, \vec b}_2, \ldots, \alpha^{i, \vec b}_{N} - \end{align*} - (Some of the representatives may be repeated). Also fix a set of representatives of every type in $S^\D(B)$. That is we pick some functions - \begin{align*} - F_1 &\colon S^\D(B) \arr M^{|x_0|} \\ - F_2 &\colon S^\D(B) \arr M^{|x'|} - \end{align*} - such that for all $\pp(x_0, x') \in S^\D(B)$ we have $\pp = \tp^\D(F_1(\pp)F_2(\pp) / B)$, i.e. $(F_1(\pp), F_2(\pp))$ is a realization of $\pp$. Now to every type in $S^\D(B)$ we assign a triple of elements $\langle i, \vec b, \alpha \rangle$ where $i \in I, \vec b \in B^d$ and $\alpha$ is one of the chosen representatives. This is done as follows. Given a type $\boldsymbol p \in S^\D(B)$ we pick its realization $(a_1, a_2) = (F_1(\pp), F_2(\pp))$ . By definability of $\D_0$ there is $j \in I$ and $\vec b \in B^d$ such that for all $\phi \in \D, b \in B$ - \begin{align*} - \models \phi(a_1, a_2, b) \iff \models \phi_j(a_2, b, \vec b) - \end{align*} - Pick $\alpha$, a representative of $S^{\D_j}(B\vec b)$ that has the same $\D_j$-type as $a_2$. ($\alpha = \alpha^{i, \vec b}_k$ for some $k \in [N]$). That is for all $b \in B, \phi \in \D$ we have - \begin{align*} - \models \phi_j(a_2, b, \vec b) \iff \models \phi_j(\alpha, b, \vec b) - \end{align*} - To the type $\pp$ we associate triple $\langle j, \vec b, \alpha\rangle$. (In general there might be more than one choice for the triple. To ensure uniqueness pick the smallest triple after fixing some appropriate ordering) This defines a map - \begin{align*} - F \colon S^\D(B) \arr T - \end{align*} - where $T$ is the space of all possible resulting tuples. We have $|T| = I \cdot |B^d| \cdot N = |I||B|^dK|B|^r= K|I||B|^{d + r}$. Once we show that $F$ is injective we will have $S^\D(B) \leq K|I||B|^{d + r}$. As choice of $\D$ and $B$ was arbitrary we will be done. Thus, all that remains is to show injectivity of $F$. We claim that $\pp$ is uniquely determined by its triple. Fix $\pp \in S^\D(B)$ and let $F(\pp) = \langle j, \vec b, \alpha\rangle$. Now for all $\phi \in \D, b \in B$ we have - \begin{align*} - \phi(x_0, x', b) \in \pp &\iff \models \phi(F_1(\pp), F_2(\pp), b) \\ - &\iff \models \phi_j(F_2(\pp), b, \vec b) \iff \models \phi_j(\alpha, b, \vec b) - \end{align*} - This shows that two different types would have different tuples associated to them. Thus $F$ is injective as needed. -\end{proof} -\begin{Corollary} - If $vc(1) = r$ and the theory has $\VC d$ property then $\vc(m) \leq r + d \cdot (m - 1)$ -\end{Corollary} - +\documentclass{amsart} + +\usepackage{AMC_style} +\usepackage{Research} + +\newcommand{\D}{\Delta} +\newcommand{\F}{\mathcal F} + + +\begin{document} +The following is the proof of the Theorem \textbf{7.1} in \textit{Vapnik-Chervonenkis density in some theories without the independence property, I} without using a stong $\VC d$ property. +\ \\ +\begin{Theorem} + Assume that $\vc(m) \leq r$ and the theory has the $\VC d$ property. Then $\vc(m+1) \leq r + d$. +\end{Theorem} +\begin{proof} + Write $x = (x_0, x')$ with $x' = (x_1, \ldots, x_m)$ so that $|x_0| = 1$ and $|x'| = m$. Let $\D(x;y)$ be given. Define + \begin{align*} + \D_0(x_0;x', y) = \{\phi(x_0; x', y) \mid \phi(x;y) \in \D\} + \end{align*} + Applying VC $d$ property to $\D_0$ we have finitely many families + \begin{align*} + &\F_i = (\phi_i(x', y; y_1, \ldots, y_d))_{\phi \in \D} &(i \in I) + \end{align*} + of $\LL$-formulas with the following property: for any $a_1 \in M^{|x_0|}$, $a_2 \in M^{|x'|}$, any finite $B \subset M^{|y|}$ there are $\vec b \in B^d$ and $i \in I$ such that $\F_i(a_2, y; \vec b)$ defines $\tp_{\D_0}(a_1/a_2B)$, i.e. for all $\phi \in \D, b \in B$ + \begin{align*} + \models \phi(a_1, a_2, b) \iff \models \phi_i(a_2, b, \vec b) + \end{align*} + For each $i \in I$ let + \begin{align*} + \D_i(x';y, y_1 \ldots y_d) = \{\phi_i(x'; y, y_1 \ldots y_d) \mid \phi(x;y) \in \D\} + \end{align*} + As $|x'| = m$ the assumption that $\vc(m) \leq r$ applies to each $\D_i$. Thus there is a constant $K$ such that for any finite $C \subset (M^{|y|})^{(d+1)}$ there is a set of representatives for $S^{\D_i}(C)$ of size at most $K|C|^r$. (More precisely, for each $\D_i$ there is going to be such a constant $K_i$ and we can take $K$ to be the maximum of these). Now fix a finite set $B \subset M^{|y|}$. Let $N = K|B|^r$. For every element $\vec b \in B^d$ fix a set of representatives of $S^{\D_i}(B\vec b)$ (Note that $|B| = |B\vec b|$) + \begin{align*} + \alpha^{i, \vec b}_1, \alpha^{i, \vec b}_2, \ldots, \alpha^{i, \vec b}_{N} + \end{align*} + (Some of the representatives may be repeated). Also fix a set of representatives of every type in $S^\D(B)$. That is we pick some functions + \begin{align*} + F_1 &\colon S^\D(B) \arr M^{|x_0|} \\ + F_2 &\colon S^\D(B) \arr M^{|x'|} + \end{align*} + such that for all $\pp(x_0, x') \in S^\D(B)$ we have $\pp = \tp^\D(F_1(\pp)F_2(\pp) / B)$, i.e. $(F_1(\pp), F_2(\pp))$ is a realization of $\pp$. Now to every type in $S^\D(B)$ we assign a triple of elements $\langle i, \vec b, \alpha \rangle$ where $i \in I, \vec b \in B^d$ and $\alpha$ is one of the chosen representatives. This is done as follows. Given a type $\boldsymbol p \in S^\D(B)$ we pick its realization $(a_1, a_2) = (F_1(\pp), F_2(\pp))$ . By definability of $\D_0$ there is $j \in I$ and $\vec b \in B^d$ such that for all $\phi \in \D, b \in B$ + \begin{align*} + \models \phi(a_1, a_2, b) \iff \models \phi_j(a_2, b, \vec b) + \end{align*} + Pick $\alpha$, a representative of $S^{\D_j}(B\vec b)$ that has the same $\D_j$-type as $a_2$. ($\alpha = \alpha^{i, \vec b}_k$ for some $k \in [N]$). That is for all $b \in B, \phi \in \D$ we have + \begin{align*} + \models \phi_j(a_2, b, \vec b) \iff \models \phi_j(\alpha, b, \vec b) + \end{align*} + To the type $\pp$ we associate triple $\langle j, \vec b, \alpha\rangle$. (In general there might be more than one choice for the triple. To ensure uniqueness pick the smallest triple after fixing some appropriate ordering) This defines a map + \begin{align*} + F \colon S^\D(B) \arr T + \end{align*} + where $T$ is the space of all possible resulting tuples. We have $|T| = I \cdot |B^d| \cdot N = |I||B|^dK|B|^r= K|I||B|^{d + r}$. Once we show that $F$ is injective we will have $S^\D(B) \leq K|I||B|^{d + r}$. As choice of $\D$ and $B$ was arbitrary we will be done. Thus, all that remains is to show injectivity of $F$. We claim that $\pp$ is uniquely determined by its triple. Fix $\pp \in S^\D(B)$ and let $F(\pp) = \langle j, \vec b, \alpha\rangle$. Now for all $\phi \in \D, b \in B$ we have + \begin{align*} + \phi(x_0, x', b) \in \pp &\iff \models \phi(F_1(\pp), F_2(\pp), b) \\ + &\iff \models \phi_j(F_2(\pp), b, \vec b) \iff \models \phi_j(\alpha, b, \vec b) + \end{align*} + This shows that two different types would have different tuples associated to them. Thus $F$ is injective as needed. +\end{proof} +\begin{Corollary} + If $vc(1) = r$ and the theory has $\VC d$ property then $\vc(m) \leq r + d \cdot (m - 1)$ +\end{Corollary} + \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.aux b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.aux deleted file mode 100644 index 0b599ef3..00000000 --- a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.aux +++ /dev/null @@ -1,36 +0,0 @@ -\relax -\citation{vc_density} -\citation{parigot_trees} -\citation{simon_dp_min} -\citation{vc_density} -\citation{vc_density} -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Preliminaries}}{1}} -\citation{vc_density} -\citation{vc_density} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Proper Subdivisions}}{2}} -\newlabel{lm_subdivision}{{2.4}{3}} -\newlabel{ex_cone}{{2.6}{3}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Intervals}}{4}} -\newlabel{lm_interval}{{3.2}{4}} -\newlabel{lm_ui}{{3}{4}} -\newlabel{lm_li}{{3.4}{4}} -\newlabel{cor_disj_int}{{3.6}{5}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{Partition into intervals}}{6}} -\newlabel{lm_meet}{{4.1}{6}} -\newlabel{lm_partition}{{4.2}{6}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{5}{Type counting}}{6}} -\newlabel{def_type_count}{{5.3}{7}} -\newlabel{lm_easy_bound}{{5.4}{7}} -\newlabel{lm_hard_bound}{{5.5}{7}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{6}{Proof for one-dimensional case}}{7}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{7}{Main proof}}{8}} -\citation{vc_density} -\bibcite{vc_density}{1} -\bibcite{simon_dp_min}{2} -\bibcite{parigot_trees}{3} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{9}} diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.bbl b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.blg b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.blg deleted file mode 100644 index 2dbd29bb..00000000 --- a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Trees vc-density_2.aux -I found no \bibdata command---while reading file Trees vc-density_2.aux -I found no \bibstyle command---while reading file Trees vc-density_2.aux -(There were 2 error messages) diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.log b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.log deleted file mode 100644 index aa4a334d..00000000 --- a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.log +++ /dev/null @@ -1,460 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 24 JUL 2014 15:23 -entering extended mode -**Trees*vc-density_2.tex -("C:\Users\Anton\SparkleShare\Research\research\02 Trees vc-density\Intervals\Trees\Trees vc-density_2.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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10852 strings out of 493921 - 198805 string characters out of 3144875 - 242242 words of memory out of 3000000 - 13780 multiletter control sequences out of 15000+200000 - 13362 words of font info for 50 fonts, out of 3000000 for 9000 - 1002 hyphenation exceptions out of 8191 - 56i,14n,55p,654b,288s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on "Trees vc-density_2.pdf" (9 pages, 279608 bytes). -PDF statistics: - 119 PDF objects out of 1000 (max. 8388607) - 0 named destinations out of 1000 (max. 500000) - 13 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.tex b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.tex index c3f5bc28..44698f24 100644 --- a/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.tex +++ b/research/02 Trees vc-density/Intervals/Trees/Trees vc-density_2.tex @@ -53,28 +53,28 @@ \section{Preliminaries} We work with finite relational languages. Given a formula we can define its complexity $n$ as the depth of quantifiers used to build up the formula. More precisely %See for example \cite{ynm_notes} Definition 2D.4 pg.72. -\begin{Definition} +\begin{Definition} Define complexity of a formula by induction: -\begin{align*} +\begin{align*} &\cx(\text{q.f. formula}) = 0 \\ &\cx(\exists x \phi(x)) = \cx(\phi(x)) + 1 \\ &\cx(\phi \cap \psi) = \max(\cx(\phi), \cx(\psi)) \\ &\cx(\neg \phi) = \cx(\phi) -\end{align*} +\end{align*} \end{Definition} A simple inductive argument verifies that there are (up to equivalence) only finitely many formulas with fixed complexity and the number of free variables. We will use the following notation for types and Stone Space: -\begin{Definition} Let $n,m$ be naturals, $\B$ a structure, $A$ a parameter set and $a,b$ tuples in $\B$. - \begin{itemize} +\begin{Definition} Let $n,m$ be naturals, $\B$ a structure, $A$ a parameter set and $a,b$ tuples in $\B$. + \begin{itemize} \item $\tp^n_{\B}(a/A)$ will stand for all the $A$-formulas of complexity $\leq n$ that are true of $a$ in $\B$. Note that if $A$ is finite there are finitely many of them (up tp equivalence). Conjunction of those formulas would still have complexity $\leq n$ so we can just associate a single formula describing this type. If $A = \empty$ we may write $\tp^n_{\B}(a)$. $\B$ will be omitted as well if it is clear from context. \item $\B \models a \equiv^n_A b$ means that $a,b$ have the same type of complexity $n$ over $A$ in $\B$, i.e. $\tp^n_{\B}(a/A) = \tp^n_{\B}(b/A)$ \item $S^n_{\B, m}(A)$ will stand for all $m$-types of complexity $n$ over $A$. - \end{itemize} + \end{itemize} \end{Definition} Language for the trees consists of a single binary predicate $\{\leq\}$. Theory of trees states that $\leq$ defines a partial order and for every element $a$ we have $\{x \mid x < a\}$ a linear order. For visualization purposes we assume trees grow upwards, with smaller elements on the bottom and larger elements on the top. If $a \leq b$ we will say that $a$ is below $b$ and $b$ is above $a$. -\begin{Definition} - Work in a tree $\TT$. For $x \in T$ let $I(x) = \{t \in T \mid t \leq x\}$ denote all elements below $x$. \emph{Meet} of two tree elements $a,b$ is the greatest element of $I(a) \cap I(b)$ (if one exists). +\begin{Definition} + Work in a tree $\TT$. For $x \in T$ let $I(x) = \{t \in T \mid t \leq x\}$ denote all elements below $x$. \emph{Meet} of two tree elements $a,b$ is the greatest element of $I(a) \cap I(b)$ (if one exists). \end{Definition} Theory of meet trees requires that any two elements in the same connected component have a meet. Colored trees are trees with a finite number of colors added via unary predicates. @@ -165,7 +165,7 @@ \section{Proper Subdivisions} \begin{Example} \label{ex_cone} Work with a (colored) tree $\TT$ in language $\LL = \{\leq, C_1, \ldots, C_n\}$ with colors interpreted by sets $S_1, \ldots S_n$. Fix $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Consider structures - \begin{align*} + \begin{align*} \A = (A, \leq, S_1 \cap A, \ldots, S_n \cap A, S) \text{ in language with an extra color} \LL_A = \LL \cup \{C\} \\ \B = (B, \leq, S_1 \cap B, \ldots, S_n \cap B) \text{ in language } \LL \end{align*} @@ -325,8 +325,8 @@ \section{Type counting} \end{Definition} That is $\phi(A, B)$ is a collection of subsets of $A$ definable by $\phi$ with parameters from $B$. -\begin{Note} - For $A_1 \subseteq A_2$ we have $|\phi(A_1, B)| \leq |\phi(A_2, B)|$. +\begin{Note} + For $A_1 \subseteq A_2$ we have $|\phi(A_1, B)| \leq |\phi(A_2, B)|$. \end{Note} \begin{Definition} \label{def_type_count} @@ -343,16 +343,16 @@ \section{Type counting} \begin{Lemma} \label{lm_easy_bound} Suppose $\phi(x,y)$ is a formula of complexity $n$ in some language $\LL$ of a (colored) tree. Let $I = (c_1, c_2)$ an interval and let its compliment be $D$. Then - \begin{align*} - |\phi(D^{|x|}, I^{|y|})| < N(n, |y|, \LL') + \begin{align*} + |\phi(D^{|x|}, I^{|y|})| < N(n, |y|, \LL') \end{align*} where $\LL'$ is $\LL$ with two extra constant symbols. \end{Lemma} Note that the bound provided is uniform - it doesn't depend on the model or the interval. -\begin{proof} - Fix $a, b \in I^{|y|}$. Then if $a \equiv^n_{c_1c_2} b$ then $\phi(D^{|x|}, a) = \phi(D^{|x|}, b)$ by lemma \ref{lm_interval}. Thus $|\phi(D^{|x|}, I^{|y|})|$ is bounded by $|S^n(y/c_1c_2)|$. +\begin{proof} + Fix $a, b \in I^{|y|}$. Then if $a \equiv^n_{c_1c_2} b$ then $\phi(D^{|x|}, a) = \phi(D^{|x|}, b)$ by lemma \ref{lm_interval}. Thus $|\phi(D^{|x|}, I^{|y|})|$ is bounded by $|S^n(y/c_1c_2)|$. \end{proof} We extract a similar result from Corollary \ref{cor_disj_int}. @@ -366,7 +366,7 @@ \section{Type counting} \end{Lemma} \begin{proof} - Fix $a_i, b_i \in I_i^{k_i}$ for each $i$ such that $a_i \equiv^n_{E_i} b_i$ where $E_i$ is endpoints of $I_i$. Then by Corollary \ref{cor_disj_int} $\phi(E^{|x|}, a_1, a_2, \ldots, a_m) = \phi(E^{|x|}, b_1, b_2, \ldots, b_m)$. This implies that + Fix $a_i, b_i \in I_i^{k_i}$ for each $i$ such that $a_i \equiv^n_{E_i} b_i$ where $E_i$ is endpoints of $I_i$. Then by Corollary \ref{cor_disj_int} $\phi(E^{|x|}, a_1, a_2, \ldots, a_m) = \phi(E^{|x|}, b_1, b_2, \ldots, b_m)$. This implies that \begin{align*} |\phi(E^{|x|}, I_1^{k_1} \times I_2^{k_2} \times \ldots \times I_m^{k_m})| &\leq |S^n_{k_1}(E_1)| \cdot |S^n_{k_2}(E_2)| \cdot \ldots \cdot |S^n_{k_m}(E_m)| \leq \\ @@ -436,10 +436,10 @@ \section{Main proof} \begin{Corollary} In the theory of infinite (colored) meet trees we have $vc(n) = n$ for all $n$. \end{Corollary} - \begin{Corollary} - In the theory of infinite (colored) trees we have $vc(n) = n$ for all $n$. + \begin{Corollary} + In the theory of infinite (colored) trees we have $vc(n) = n$ for all $n$. \end{Corollary} - \begin{proof} + \begin{proof} Let $\TT'$ be a tree. We can embed it in a larger tree that is closed under meets $\TT' \subset \TT$. Expand $\TT$ by an extra color and interpret it by coloring the subset $\TT'$. Thus we can interpret $\TT'$ in $T^1$. By Corollary 3.17 in \cite{vc_density} we get that $\vc^{\TT'}(n) \leq \vc^T(1 \cdot n) = n$ thus $\vc^{\TT'}(n) = n$ as well. \end{proof} diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees.tcp b/research/02 Trees vc-density/Intervals/Trees/Trees.tcp index bc115402..ba210776 100644 --- a/research/02 Trees vc-density/Intervals/Trees/Trees.tcp +++ b/research/02 Trees vc-density/Intervals/Trees/Trees.tcp @@ -1,12 +1,12 @@ -[FormatInfo] -Type=TeXnicCenterProjectInformation -Version=4 - -[ProjectInfo] -MainFile=Trees vc-density_2.tex -UseBibTeX=0 -UseMakeIndex=0 -ActiveProfile=LaTeX ⇨ PDF -ProjectLanguage=en -ProjectDialect=US - +[FormatInfo] +Type=TeXnicCenterProjectInformation +Version=4 + +[ProjectInfo] +MainFile=Trees vc-density_2.tex +UseBibTeX=0 +UseMakeIndex=0 +ActiveProfile=LaTeX ⇨ PDF +ProjectLanguage=en +ProjectDialect=US + diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees.tex b/research/02 Trees vc-density/Intervals/Trees/Trees.tex index d3f5a12f..8b137891 100644 --- a/research/02 Trees vc-density/Intervals/Trees/Trees.tex +++ b/research/02 Trees vc-density/Intervals/Trees/Trees.tex @@ -1 +1 @@ - + diff --git a/research/02 Trees vc-density/Intervals/Trees/Trees.tps b/research/02 Trees vc-density/Intervals/Trees/Trees.tps deleted file mode 100644 index 15f74082..00000000 --- a/research/02 Trees vc-density/Intervals/Trees/Trees.tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - -[Frame0] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=1476 -NormalPos.bottom=728 -Class=LaTeXView -Document=Trees vc-density_2.tex - -[Frame0_View0,0] -TopLine=10 -Cursor=533 - -[SessionInfo] -FrameCount=1 -ActiveFrame=0 - diff --git a/research/02 Trees vc-density/Trees vc-density.aux b/research/02 Trees vc-density/Trees vc-density.aux deleted file mode 100644 index 0e5bc455..00000000 --- a/research/02 Trees vc-density/Trees vc-density.aux +++ /dev/null @@ -1,34 +0,0 @@ -\relax -\citation{vc_density} -\citation{parigot_trees} -\citation{simon_dp_min} -\citation{vc_density} -\citation{vc_density} -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Preliminaries}}{1}} -\citation{vc_density} -\citation{vc_density} -\citation{vc_density} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Proper Subdivisions: Definition and Properties}}{3}} -\newlabel{lm_subdivision}{{2.4}{3}} -\newlabel{ex_disc}{{2.5}{3}} -\newlabel{ex_cone}{{2.6}{4}} -\newlabel{cor_type_count}{{2.8}{4}} -\newlabel{def_type_count}{{2.9}{4}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Proper Subdivisions: Constructions}}{4}} -\@writefile{lof}{\contentsline {figure}{\numberline {1}{\ignorespaces Proper subdivision for $(A, B) = (A^{c_1}_{c_2}, B^{c_1}_{c_2})$}}{5}} -\@writefile{lof}{\contentsline {figure}{\numberline {2}{\ignorespaces Proper subdivision for $(A, B) = (A_{c_1}, B_{c_1})$}}{5}} -\@writefile{lof}{\contentsline {figure}{\numberline {3}{\ignorespaces Proper subdivision for $(A, B) = (A^{c_1}_G, B^{c_1}_G)$ for $G = \{c_2\}$}}{6}} -\@writefile{lof}{\contentsline {figure}{\numberline {4}{\ignorespaces Proper subdivision for $(A, B) = (A_G, B_G)$ for $G = \{c_1, c_2\}$}}{6}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{4}{Main proof}}{8}} -\newlabel{lm_partition_bound}{{4.1}{8}} -\newlabel{lm_meet}{{4.2}{9}} -\citation{vc_density} -\bibcite{vc_density}{1} -\bibcite{simon_dp_min}{2} -\bibcite{parigot_trees}{3} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{12}} diff --git a/research/02 Trees vc-density/Trees vc-density.backup.tex b/research/02 Trees vc-density/Trees vc-density.backup.tex index dca6d967..5d952a35 100644 --- a/research/02 Trees vc-density/Trees vc-density.backup.tex +++ b/research/02 Trees vc-density/Trees vc-density.backup.tex @@ -1,424 +1,424 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{tikz} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} -\newcommand{\CS}{\mathcal S} - -\begin{document} - -\title{vc-density for trees} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} -%more info - -\begin{abstract} - We show that for the theory of infinite trees we have $\vc(n) = n$ for all $n$. -\end{abstract} - -\maketitle - -VC density was introduced in \cite{vc_density} by Aschenbrenner, Dolich, Haskell, MacPherson, and Starchenko as a natural notion of dimension for NIP theories. In a NIP theory we can define a VC function - -\begin{align*} - \vc : \N \arr \N -\end{align*} - -Where $vc(n)$ measures complexity of definable sets $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. - -In this paper we work with infinite trees viewed as posets. Parigot in \cite{parigot_trees} showed that such theory is NIP. This result was strengthened by Simon in \cite{simon_dp_min} showing that trees are dp-minimal. \cite{vc_density} has the following problem - -\begin{Problem} (\cite{vc_density} p. 47) - Determine the VC density function of each (infinite) tree. -\end{Problem} - -Here we settle this question by showing that theory of trees has $\vc(n) = n$. - -\section{Preliminaries} -%tree finite? meet? disconnected? -%Work in connected trees for simplicity? -%Work in non-meet trees for generality? -%Hodges M.T. for reference -%inequality directions -We use notation $a \in T^n$ for tuples of size $n$. For variable $x$ or tuple $a$ we denote their arity by $|x|$ and $|a|$ respectively. - -We work with finite relational languages. Given a formula we can define its complexity $n$ as the number of quantifiers in its normal form. $S^n_{\A}(x)$ stands for all the types made up of formulas of complexity at most $n$ in a structure $\A$. $\tp^n_{\B}(a)$ stands for such a type. For two structures $\A, \B$ we say $\A \equiv_n \B$ if two structures agree on all sentences of complexity at most $n$. - -\begin{Note} - Saying that $(\A, a_1) \equiv_n (\A, a_2)$ is the same as saying that $a_1$ and $a_2$ have the same $n$-complexity type in $\A$. -\end{Note} -%Define meet more precisely? - -Language for the trees consists of a single binary predicate $\{\leq\}$. Theory of trees states that $\leq$ defines a partial order and for every element $a$ we have $\{x \mid x < a\}$ a linear order. Theory of meet trees requires that in addition tree is closed under meet operation, i.e. for any $a, b$ in the same connected component there exists the greatest upper bound for elements both $\leq$ than $a$ and $b$. Note that we allow our trees to be disconnected or finite unless otherwise stated. - -For completeness we also present definition of VC function. -One should refer to \cite{vc_density} for more details. -Suppose we have a collection $\CS$ of subsets of $X$. We define a \emph{shatter function} $\pi_\CS(n)$ - -\begin{align*} - \pi_\CS(n) = \max \{|A \cap \CS| : A \subset X \text{ and } |A| = n\} -\end{align*} - -Sauer's Lemma asserts that asymptotically $\pi_\CS$ is either $2^n$ or polynomial. -In the polynomial case we define VC density of $\CS$ to be power of polynomial that bounds $\pi_\CS$. -More formally -\begin{align*} - \vc(\CS) = \limsup_{n \to \infty}\frac{\log \pi_\CS}{\log n} -\end{align*} - -Given a model $M \models T$ and a fomula $\phi(x, y)$ we define - -\begin{align*} - \CS_\phi &= \{\phi(M^{|n|}, b) : b \in M^{|y|}\} \\ - \vc(\phi) &= \vc(\CS_\phi) -\end{align*} - -One has to check that this definition is independet of realization of $T$, see \cite{vc_density}, Lemma 3.2. For a theory $T$ we define the VC function - -\begin{align*} - \vc(n) = \sup \{\vc(\phi(x, y)) : |x| = n\} -\end{align*} - -\section{Proper Subdivisions: Definition and Properties} - -% empty ok? -\begin{Definition} - Let $\A$, $\B$, $\TT$ be models in (possibly different) finite relational languages. If $A$, $B$ partition $T$ (i.e. $T = A \sqcup B$) we say that $(\A, \B)$ is a \emph{subdivision} of $\TT$. -\end{Definition} - -\begin{Definition} - $(\A, \B)$ subdivision of $\TT$ is called \emph{$n$-proper} if for all $p,q \in \N$, for all $a_1, a_2 \in A^p$ and $b_1, b_2 \in B^q$ we have - \begin{align*} - (\A, a_1) &\equiv_n (\A, a_2) \\ - (\B, b_1) &\equiv_n (\B, b_2) - \end{align*} - then - \begin{align*} - (\TT, a_1, b_1) \equiv_n (\TT, a_2, b_2) - \end{align*} -\end{Definition} - -\begin{Definition} - $(\A, \B)$ subdivision of $\TT$ is called \emph{proper} if it is $n$-proper for all $n \in \N$. -\end{Definition} - -\begin{Lemma} \label{lm_subdivision} - Consider a subdivision $(\A, \B)$ of $\TT$. If it is $0$-proper then it is proper. -\end{Lemma} - -%only L_A L_B required to be relational -\begin{proof} - % why is one existential quantifier enough? - Prove the subdivision is $n$-proper for all $n$ by induction. Case $n = 0$ is given by the assumption. Suppose $n = k + 1$ and we have $\TT \models \exists x \, \phi^k(x, a_1, b_1)$ where $\phi^k$ is some formula of complexity $k$. Let $a \in T$ witness the existential claim i.e. $\TT \models \phi^k(a, a_1, b_1)$. $a \in A$ or $a \in B$. Without loss of generality assume $a \in A$. Let $\pp = \tp^k_{\A} (a, a_1)$. Then we have - \begin{align*} - \A \models \exists x \, \tp^k_{\A}(x, a_1) = \pp - \end{align*} - Formula $\tp^k_{\A}(x, a_1) = \pp$ is of complexity $k$ so $\exists x \, \tp^k_{\A}(x, a_1) = \pp$ is of complexity $k+1$ by inductive hypothesis we have - \begin{align*} - \A \models \exists x \, \tp^k_{\A}(x, a_2) = \pp - \end{align*} - Let $a'$ witness this existential claim so that - \begin{align*} - \tp^k_{\A}(a', a_2) &= \pp \\ - \tp^k_{\A}(a', a_2) &= \tp^k_{\A}(a, a_1) \\ - (\A, a', a_2) &\equiv_k (\A, a, a_1) - \end{align*} - % careful with induction! - by inductive assumption we have - \begin{align*} - (\TT, a, a_1, b_1) &\equiv_k (\TT, a', a_2, b_2) \\ - \TT &\models \phi^k(a', a_2, b_2) & \text {as } \TT &\models \phi^k(a, a_1, b_1)\\ - \TT &\models \exists x \phi^k(x, a_2, b_2) - \end{align*} -\end{proof} - -We don't require this lemma in full generality. From now on in this paper we'll have $\TT$ to be a model of a tree in the language $\LL = \{\leq\}$ and $\A, \B$ be in some languages $\LL_A, \LL_B$ which will be expands of $\LL$, with $\A, \B$ substructures of $\TT$ as reducts to $\LL$. We'll refer to $(\A, \B)$ as a \emph{proper subdivision} ($\TT$ will be dropped if it is implied from context). - -\begin{Example} - Suppose the tree consists of two connected components $C_1, C_2$. Then $(C_1, \leq)$ and $(C_2, \leq)$ form a proper subdivision. -\end{Example} - -\begin{Example} \label{ex_cone} - Fix $\TT$ and $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Then $(A, \leq, S)$ and $(B, \leq)$ form a proper subdivision, where $\LL_A$ has a unary predicate interpreted by $S$. -\end{Example} - -\begin{Definition} For $\phi(x, y)$, $A \subseteq T^{|x|}$ and $B \subseteq T^{|y|}$ -\begin{itemize} - \item Let $\phi(A, b) = \{a \in A \mid \phi(a, b)\} \subseteq A$ - \item Let $\phi(A, B) = \{\phi(A, b) \mid b \in B\} \subseteq \PP(A)$ -\end{itemize} -\end{Definition} -$\phi(A, B)$ is a collection of subsets of $A$ definable by $\phi$ with parameters from $B$. We notice the following bound when $A, B$ are parts of a proper subdivision. - -\begin{Corollary} \label{cor_type_count} - Suppose $\phi(x,y)$ is a formula of complexity $n$. Let $\A, \B$ be a proper subdivision of $\TT$ and $b_1, b_2 \in B^{|y|}$. Then if $\tp^n_{\B}(b_1) = \tp^n_{\B}(b_2)$ then $\phi(A^{|x|}, b_1) = \phi(A, b_2)$. Thus $|\phi(A^{|x|}, B^{|y|})|$ is bounded by $|S^n_{\B}(y)|$ -\end{Corollary} - -\begin{proof} - Take some $a \in A^{|x|}$. We have $(\B, b_1) \equiv_n (\B, b_2)$ and (trivially) $(\A, a) \equiv_n (\A, a)$. Thus by the Lemma \ref{lm_subdivision} we have $(\TT, a, b_1) \equiv_n (\TT, a, b_2)$ so $\phi(a, b_1) \iff \phi(a, b_2)$. Since $a$ was arbitrary we have $\phi(A^{|x|}, b_1) = \phi(A^{|x|}, b_2)$. -\end{proof} - -Now we note that the number of such types can be bounded uniformly. - -\begin{Note} \label{nt_type_count} - Fix a (finite relational) language $\LL_B$, and $n$, $|y|$. Then there is some $N = N(n, |y|, \LL_B)$ such that for any structure $\B$ in $\LL_B$ we have $|S^n_{\B}(y)| \leq N$ -\end{Note} - -%relations with this number - -\section{Proper Subdivisions: Constructions} - -First, we describe several constructions of proper subdivisions that are needed for the proof. - -%relate to cones -\begin{Definition} - We say that $E(b, c)$ if $b$ and $c$ are connected - \begin{align*} - E(b, c) \ifff \exists x \, (b \geq x) \wedge (c \geq x) - \end{align*} - Similarly $E_a(b, c)$ means that $b$ and $c$ are connected through an element above $a$. More precisely - \begin{align*} - E_a(b, c) \ifff \exists x \, (x > a) \wedge (b \geq x) \wedge (c \geq x) - \end{align*} -\end{Definition} - -In the following four definitions $\B$-structures are going to be in the same language $\LL_B = \{\leq, U\}$ with $U$ a unary predicate. It is not always necessary to have this predicate but for the sake of uniformity we keep it. $\A$-structures will have different $\LL_A$ languages (those are not as important in later applications). - -\input {vc-trees-all_figures} - -% Simon's cone, parigot's notation? -\begin{Definition} - Fix $c_1 < c_2$ in $T$. Let - \begin{align*} - B &= \{b \in T \mid E_{c_1}(c_2, b) \wedge \neg(b \geq c_2)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t < c_1\} \\ - S_2 &= \{t \in T \mid t < c_2\} \\ - S_B &= S_2 - S_1 \\ - T_A &= \{t \in T \mid c_2 \leq t\} - \end{align*} - Define structures $\A^{c_1}_{c_2} = (A, \leq, S_1, T_A)$ and $\B^{c_1}_{c_2} = (B, \leq, S_B)$ where $\LL_A$ is expansion of $\{\leq\}$ by two unary predicates (and $\LL_B$ as defined above). Note that $c_1, c_2 \notin B$. -\end{Definition} - - -\begin{Definition} - Fix $c$ in $T$. Let - \begin{align*} - B &= \{b \in T \mid \neg(b \geq c) \wedge E(b,c)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t < c\} - \end{align*} - Define structures $\A_{c} = (A, \leq)$ and $\B_{c} = (B, \leq, S_1)$ where $\LL_A = \{\leq\}$ (and $\LL_B$ as defined above). Note that $c \notin B$. (cf example \ref{ex_cone}). -\end{Definition} - -\begin{Definition} - Fix $c$ in $T$ and $S \subseteq T$ a finite subset. Let - \begin{align*} - B &= \{b \in T \mid (b > c) \text{ and for all $s \in S$ we have } \neg E_c(s, b)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t \leq c\} - \end{align*} - Define structures $\A^{c}_{S} = (A, \leq, S_1)$ and $\B^{c}_{S} = (B, \leq, B)$ where $L_A$ is expansion of $\{\leq\}$ by a single unary predicate (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $c \notin B$ and $S \cap B = \emptyset$. -\end{Definition} - -\begin{Definition} - Fix $S \subseteq T$ a finite subset. Let - \begin{align*} - B &= \{b \in T \mid \text{ for all $s \in S$ we have } \neg E(s, b)\} \\ - A &= T - B - \end{align*} - Define structures $\A_{S} = (A, \leq)$ and $\B_{S} = (B, \leq, B)$ where $\LL_A = \{\leq\}$ (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $S \cap B = \emptyset$. -\end{Definition} - -\begin{Lemma} - Pairs of structures defined above are all proper subdivisions. -\end{Lemma} - -\begin{proof} - We only show this holds for the first definition $\A = \A^{c_1}_{c_2}$ and $\B = \B^{c_1}_{c_2}$. Other cases follow by a similar argument. $A,B$ partition $T$ by definition, so it is a subdivision. To show that it is proper by Lemma \ref{lm_subdivision} we only need to check that it is $0$-proper. Suppose we have - \begin{align*} - a &= (a_1, a_2, \ldots, a_p) \in A^p \\ - a' &= (a_1', a_2', \ldots, a_p') \in A^p \\ - b &= (b_1, b_2, \ldots, b_q) \in B^q \\ - b' &= (b_1', b_2', \ldots, b_q') \in B^q - \end{align*} - with $(\A, a) \equiv_0 (\A, a')$ and $(\B, b) \equiv_0 (\B, b')$. We need to make sure that $ab$ has the same quantifier free type as $a'b'$. Any two elements in $T$ can be related in the four following ways - \begin{align*} - x &= y \\ - x &< y \\ - x &> y \\ - x&,y \text{ are incomparable} - \end{align*} - We need to check that the same relations hold for pairs of $(a_i, b_j), (a_i', b_j')$ for all $i,j$. - - \begin{itemize} - \item It is impossible that $a_i = b_j$ as they come from disjoint sets. - \item Suppose $a_i < b_j$. This forces $a_i \in S_1$ thus $a_i' \in S_1$ and $a_i' < b_j'$ - \item Suppose $a_i > b_j$ This forces $b_j \in S_B$ and $a \in T_A$, thus $b_j' \in S_B$ and $a_i' \in T_A$ so $a_i' > b_j'$ - \item Suppose $a_i$ and $b_j$ are incomparable. Two cases are possible: - \begin{itemize} - \item $b_j \notin S_B$ and $a_i \in T_A$. Then $b_j' \notin S_B$ and $a_i' \in T_A$ making $a_i', b_j'$ incomparable - \item $b_j \in S_B$, $a_i \notin T_A$, $a_i \notin S_1$. Similarly this forces $a_i', b_j'$ incomparable - \end{itemize} - \end{itemize} -\end{proof} - -\section{Main proof} - -Basic idea for the proof is that we are able to divide our parameter space into $O(n)$ many pieces. Each of $q$ parameters can come from any of those $O(n)$ partitions giving us $O(n)^q$ many choices for parameter configuration. When every parameter coming from a fixed partition the number of definable sets is constant and in fact is uniformly bounded by some $N$. This gives us $N O(n)^q = O(n^q)$ possibilities for different definable sets. - -First, we generalize Corollary \ref{cor_type_count}. (This is only required for computing vc-density for formulas $\phi(x, y)$ with $|y| > 1$) - -\begin{Lemma} \label{lm_partition_bound} - Consider a finite collection $(\A_i, \B_i)_{i \leq n}$ where each $(\A_i, \B_i)$ is a proper subdivision or a singleton: $B_i = \{b_i\}$ with $A_i = T$. Also assume that all $\B_i$ have the same language $\LL_B$. Let $A = \bigcap_{i \in I} A_i$. Fix a formula $\phi(x, y)$ of complexity $m$ . Let $N = N(m, |y|, \LL_B)$ as in Note \ref{nt_type_count}. Consider any $B \subseteq T^{|y|}$ of the form - \begin{align*} - B = B_1^{i_1} \times B_2^{i_2} \times \ldots \times B_n^{i_n} \text { with } i_1 + i_2 + \ldots + i_n = |y| - \end{align*} - (some of the indexes can be zero). Then we have the following bound - \begin{align*} - \phi(A^{|x|}, B) \leq N^{|y|} - \end{align*} -\end{Lemma} - -\begin{proof} - We show this result by counting types. Suppose we have - \begin{align*} - b_1, b_1' &\in B_1^{i_1} \text{ with } b_1 \equiv_m b_1' \text { in } B_1 \\ - b_2, b_2' &\in B_2^{i_2} \text{ with } b_2 \equiv_m b_2' \text { in } B_2 \\ - &\cdots \\ - b_n, b_n' &\in B_n^{i_n} \text{ with } b_n \equiv_m b_n' \text { in } B_n - \end{align*} - Then we have - \begin{align*} - \phi(A^{|x|}, b_1, b_2, \ldots b_n) \ifff \phi(A^{|x|}, b_1', b_2', \ldots b_n') - \end{align*} - This is easy to see by applying Corollary \ref{cor_type_count} one by one for each tuple. This works if $B_i$ is part of a proper subdivision; if it is a singleton then the implication is trivial as $b_i = b_i'$. - This shows that $\phi(A^{|x|}, B)$ only depends on the choice of types for the tuples - \begin{align*} - |\phi(A^{|x|}, B)| \leq |\tp^m_{B_1}(i_1)| \cdot |\tp^m_{B_2}(i_2)| \cdot \ldots \cdot |\tp^m_{B_n}(i_n)| - \end{align*} - Now for each type space we have inequality - \begin{align*} - |\tp^m_{B_1}(i_1)| \leq N(m, i_1, \LL_B) \leq N(m, |y|, \LL_B) \leq N - \end{align*} - (For singletons $|\tp^m_{B_j}(i_j)| = 1 \leq N$). Only non-zero indexes contribute to the product and there are at most $|y|$ of those (by equality $i_1 + i_2 + \ldots + i_n = |y|$). Thus we have - \begin{align*} - |\phi(A^{|x|}, B)| \leq N^{|y|} - \end{align*} - as needed. -\end{proof} - -For subdivisions to work out properly we will need to work with subsets closed under meets. We observe that closure under meets doesn't add too many new elements. - -% write an actual proof! -\begin{Lemma} \label{lm_meet} - Suppose $S \subseteq T$ is a non-empty finite subset of a meet tree of size $n$ and $S'$ its closure under meets. Then $|S'| \leq 2n - 1$. -\end{Lemma} -\begin{proof} - We prove by induction on $n$. Base case $n = 1$ is clear. Suppose we have $S$ of size $k$ with closure of size at most $2k - 1$. Take a new point and look at its meets with all the elements of $S$. Pick the largest one. That is the only element we need to add to $S'$ to make sure the set is closed under meets. -\end{proof} - -Putting all of those results together we are able to compute $\vc$-density of formulas in meet trees. - -\begin{Theorem} - Let $\TT$ be an infinite meet tree and $\phi(x, y)$ a formula with $|x| = p$ and $|y| = q$. Then $\vc(\phi) \leq q$. -\end{Theorem} - -\begin{proof} - Pick a finite subset of $S_0 \subset T^p$ of size $n$. Let $S_1 \subset T$ consist of coordinates of $S_0$. Let $S \subset T$ be a closure of $S_1$ under meets. Using Lemma \ref{lm_meet} we have $|S_2| \leq 2|S_1| \leq 2p|S_0| = 2pn = O(n)$. We have $S_0 \subseteq S^p$, so $|\phi(S_0, T^q)| \leq |\phi(S^p, T^q)|$. Thus it is enough to show $|\phi(S^p, T^q)| = O(n^q)$. - - Label $S = \{c_i\}_{i \in I}$ with $|I| \leq 2pn$. For every $c_i$ we construct two partitions in the following way. We have $c_i$ is either minimal in $S$ or it has a predecessor in $S$ (greatest element less than $c$). If it is minimal construct $(\A_{c_i}, \B_{c_i})$. If there is a predecessor $p$ construct $(\A^p_{c_i}, \B^p_{c_i})$. For the second subdivision let $G$ be all elements in $S$ greater than $c_i$ and construct $(\A^c_G, \B^c_G)$. So far we have constructed two subdivisions for every $i \in I$. Additionally construct $(\A_S, \B_S)$. We end up with a finite collection of proper subdivisons $(\A_j, \B_j)_{j \in J}$ with $|J| = 2|I| + 1$. Before we proceed we note the following two lemmas describing our partitions. - - \begin{Lemma} - For all $j \in J$ we have $S \subseteq A_j$. Thus $S \subseteq \bigcap_{j \in J} A_j$ and $S^p \subseteq \bigcap_{j \in J} (A_j)^p$ - \end{Lemma} - - \begin{proof} - Check this for each possible choices of partition. Cases for partitions of the type $\A_S, \A^c_G, \A_c$ are easy. Suppose we have partition $(\A, \B) = (\A^{c_1}_{c_2}, \B^{c_1}_{c_2})$. We need to show that $B \cap S = \emptyset$. By construction we have $c_1, c_2 \notin B$. Suppose we have some other $c \in S$ with $c \in B$. We have $E_{c_1}(c_2, c)$ i.e. there is some $b$ such that $(b > c_1)$, $(b \leq c_2)$ and $(b \leq c)$. Consider the meet $(c \wedge c_2)$. We have $(c \wedge c_2) \geq b > c_1$. Also as $\neg (c \geq c_2)$ we have $(c \wedge c_2) < c_2$. To summarize $c_2 > (c \wedge c_2) > c_1$. But this contradicts our construction as $S$ is closed under meets, so $(c \wedge c_2) \in S$ and $c_1$ is supposed to be a predecessor of $c_2$ in $S$. - \end{proof} - - \begin{Lemma} - $\{B_j\}_{j \in J}$ partition $T - S$ i.e. $T = \bigsqcup_{j \in J} B_j \sqcup S$ - \end{Lemma} - - \begin{proof} - This more or less follows from the choice of partitions. Pick any $b \in S - T$. Take all elements in $S$ greater than $b$ and take the minimal one $a$. Take all elements in $S$ less than $b$ and take the maximal one $c$ (possible as $S$ is closed under meets). Also take all elements in $S$ incomparable to $b$ and denote them $G$. If both $a$ and $c$ exist we have $b \in \B^a_c$. If only upper bound exists we have $b \in \B^a_G$. If only lower bound exists we have $b \in \B_c$. If neither exists we have $b \in \B_G$. - \end{proof} - - \begin{Note} - Those two lemmas imply $S = \bigcap_{j \in J} A_j$ - \end{Note} - - \begin{Note} - %careful application of note - have different languages and has to be > 1 - For one-dimensional case $q = 1$ we don't need to do any more work. We have partitioned parameter space into $|J| = O(n)$ many pieces and over each piece the number of definable sets is uniformly bounded. By Note \ref{nt_type_count} we have that $|\phi((A_j)^p, B_j)| \leq N$ for any $j \in J$ (letting $N = N(n_\phi, q, \{\leq, S\})$ where $n_\phi$ is complexity of $\phi$ and $S$ is a unary predicate). Compute - %describe steps - \begin{align*} - |\phi(S^p, T)| - &= \left|\bigcup_{j \in J} \phi(S^p, B_j) \cup \phi(S^p, S)\right| \leq \\ - &\leq \sum_{j \in J} |\phi(S^p, B_j)| + |\phi(S^p, S)| \leq \\ - &\leq \sum_{j \in J} |\phi((A_j)^p, B_j)| + |S| \leq \\ - &\leq \sum_{j \in J}N + |I| \leq \\ - &\leq (4pn + 1)N + 2pn = (4pN + 2p)n + N = O(n) - \end{align*} - \end{Note} - Basic idea for the general case $q \geq 1$ is that we have $q$ parameters and $|J| = O(n)$ partitions to pick each parameter from giving us $|J|^q = O(n^q)$ choices for parameter configuration, each giving uniformly constant number of definable subsets of $S$. (If every parameter is picked from a fixed partition, Lemma \ref{lm_partition_bound} provides a uniform bound). This yields $\vc(\phi) \leq q$ as needed. The rest of the proof is stating this idea formally. - - First, we extend our collection of subdivisions $(\A_j, \B_j)_{j \in J}$ by the following singleton sets. For each $c_i \in S$ let $B_i = \{c_i\}$ and $A_i = T$ and add $(\A_i, \B_i)$ to our collection with $\LL_B$ the language of $B_i$ interpreted arbitrarily. We end up with a new collection $(\A_k, \B_k)_{k \in K}$ indexed by some $K$ with $|K| = |J| + |I|$ (we added $|S|$ new pairs). Now we have that $B_k$ partition $T$, so $T = \bigsqcup_{k \in K} B_k$ and $S = \bigcap_{j \in J} A_j = \bigcap_{k \in K} A_k$. For $(k_1, k_2, \ldots k_q) = \vec k \in K^q$ denote - \begin{align*} - B_{\vec k} = B_{k_1} \times B_{k_2} \times \ldots \times B_{k_q} - \end{align*} - Then we have the following identity - \begin{align*} - T^q = (\bigsqcup_{k \in K} B_k)^q = \bigsqcup_{\vec k \in K^q} B_{\vec k} - \end{align*} - Thus we have that $\{B_{\vec k}\}_{\vec k \in K^q}$ partition $T^q$. Compute - \begin{align*} - |\phi(S^p, T^q)| - &= \left|\bigcup_{\vec k \in K^q} \phi(S^p, B_{\vec k}) \right| \leq \\ - &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| - \end{align*} - We can bound $|\phi(S^p, B_{\vec k})|$ uniformly using Lemma \ref{lm_partition_bound}. $(\A_k, \B_k)_{k \in K}$ satisfies the requirements of the lemma and $B_{\vec k}$ looks like $B$ in the lemma after possibly permuting some variables in $\phi$. Applying the lemma we get - \begin{align*} - |\phi(S^p, B_{\vec k})| \leq N^q - \end{align*} - with $N$ only depending on $q$ and complexity of $\phi$. We complete our computation - \begin{align*} - |\phi(S^p, T^q)| - &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| \leq \\ - &\leq \sum_{\vec k \in K^q} N^q \leq \\ - &\leq |K^q| N^q \leq \\ - &\leq (|J| + |I|)^q N^q \leq \\ - &\leq (4pn + 1 + 2pn)^q N^q = N^q (6p + 1/n)^q n^q = O(n^q) - \end{align*} - \end{proof} - \begin{Corollary} - In the theory of infinite meet trees we have $vc(n) = n$ for all $n \in \N^{+}$. - \end{Corollary} - -\begin{thebibliography}{9} - -\bibitem{vc_density} - M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, - \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) - -\bibitem{simon_dp_min} - P. Simon, - \textit{On dp-minimal ordered structures}, - J. Symbolic Logic 76 (2011), no. 2, 448-460 - -\bibitem{parigot_trees} - Michel Parigot. - Th\'eories d'arbres. - \textit{Journal of Symbolic Logic}, 47, 1982. - - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{tikz} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} +\newcommand{\CS}{\mathcal S} + +\begin{document} + +\title{vc-density for trees} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} +%more info + +\begin{abstract} + We show that for the theory of infinite trees we have $\vc(n) = n$ for all $n$. +\end{abstract} + +\maketitle + +VC density was introduced in \cite{vc_density} by Aschenbrenner, Dolich, Haskell, MacPherson, and Starchenko as a natural notion of dimension for NIP theories. In a NIP theory we can define a VC function + +\begin{align*} + \vc : \N \arr \N +\end{align*} + +Where $vc(n)$ measures complexity of definable sets $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. + +In this paper we work with infinite trees viewed as posets. Parigot in \cite{parigot_trees} showed that such theory is NIP. This result was strengthened by Simon in \cite{simon_dp_min} showing that trees are dp-minimal. \cite{vc_density} has the following problem + +\begin{Problem} (\cite{vc_density} p. 47) + Determine the VC density function of each (infinite) tree. +\end{Problem} + +Here we settle this question by showing that theory of trees has $\vc(n) = n$. + +\section{Preliminaries} +%tree finite? meet? disconnected? +%Work in connected trees for simplicity? +%Work in non-meet trees for generality? +%Hodges M.T. for reference +%inequality directions +We use notation $a \in T^n$ for tuples of size $n$. For variable $x$ or tuple $a$ we denote their arity by $|x|$ and $|a|$ respectively. + +We work with finite relational languages. Given a formula we can define its complexity $n$ as the number of quantifiers in its normal form. $S^n_{\A}(x)$ stands for all the types made up of formulas of complexity at most $n$ in a structure $\A$. $\tp^n_{\B}(a)$ stands for such a type. For two structures $\A, \B$ we say $\A \equiv_n \B$ if two structures agree on all sentences of complexity at most $n$. + +\begin{Note} + Saying that $(\A, a_1) \equiv_n (\A, a_2)$ is the same as saying that $a_1$ and $a_2$ have the same $n$-complexity type in $\A$. +\end{Note} +%Define meet more precisely? + +Language for the trees consists of a single binary predicate $\{\leq\}$. Theory of trees states that $\leq$ defines a partial order and for every element $a$ we have $\{x \mid x < a\}$ a linear order. Theory of meet trees requires that in addition tree is closed under meet operation, i.e. for any $a, b$ in the same connected component there exists the greatest upper bound for elements both $\leq$ than $a$ and $b$. Note that we allow our trees to be disconnected or finite unless otherwise stated. + +For completeness we also present definition of VC function. +One should refer to \cite{vc_density} for more details. +Suppose we have a collection $\CS$ of subsets of $X$. We define a \emph{shatter function} $\pi_\CS(n)$ + +\begin{align*} + \pi_\CS(n) = \max \{|A \cap \CS| : A \subset X \text{ and } |A| = n\} +\end{align*} + +Sauer's Lemma asserts that asymptotically $\pi_\CS$ is either $2^n$ or polynomial. +In the polynomial case we define VC density of $\CS$ to be power of polynomial that bounds $\pi_\CS$. +More formally +\begin{align*} + \vc(\CS) = \limsup_{n \to \infty}\frac{\log \pi_\CS}{\log n} +\end{align*} + +Given a model $M \models T$ and a fomula $\phi(x, y)$ we define + +\begin{align*} + \CS_\phi &= \{\phi(M^{|n|}, b) : b \in M^{|y|}\} \\ + \vc(\phi) &= \vc(\CS_\phi) +\end{align*} + +One has to check that this definition is independet of realization of $T$, see \cite{vc_density}, Lemma 3.2. For a theory $T$ we define the VC function + +\begin{align*} + \vc(n) = \sup \{\vc(\phi(x, y)) : |x| = n\} +\end{align*} + +\section{Proper Subdivisions: Definition and Properties} + +% empty ok? +\begin{Definition} + Let $\A$, $\B$, $\TT$ be models in (possibly different) finite relational languages. If $A$, $B$ partition $T$ (i.e. $T = A \sqcup B$) we say that $(\A, \B)$ is a \emph{subdivision} of $\TT$. +\end{Definition} + +\begin{Definition} + $(\A, \B)$ subdivision of $\TT$ is called \emph{$n$-proper} if for all $p,q \in \N$, for all $a_1, a_2 \in A^p$ and $b_1, b_2 \in B^q$ we have + \begin{align*} + (\A, a_1) &\equiv_n (\A, a_2) \\ + (\B, b_1) &\equiv_n (\B, b_2) + \end{align*} + then + \begin{align*} + (\TT, a_1, b_1) \equiv_n (\TT, a_2, b_2) + \end{align*} +\end{Definition} + +\begin{Definition} + $(\A, \B)$ subdivision of $\TT$ is called \emph{proper} if it is $n$-proper for all $n \in \N$. +\end{Definition} + +\begin{Lemma} \label{lm_subdivision} + Consider a subdivision $(\A, \B)$ of $\TT$. If it is $0$-proper then it is proper. +\end{Lemma} + +%only L_A L_B required to be relational +\begin{proof} + % why is one existential quantifier enough? + Prove the subdivision is $n$-proper for all $n$ by induction. Case $n = 0$ is given by the assumption. Suppose $n = k + 1$ and we have $\TT \models \exists x \, \phi^k(x, a_1, b_1)$ where $\phi^k$ is some formula of complexity $k$. Let $a \in T$ witness the existential claim i.e. $\TT \models \phi^k(a, a_1, b_1)$. $a \in A$ or $a \in B$. Without loss of generality assume $a \in A$. Let $\pp = \tp^k_{\A} (a, a_1)$. Then we have + \begin{align*} + \A \models \exists x \, \tp^k_{\A}(x, a_1) = \pp + \end{align*} + Formula $\tp^k_{\A}(x, a_1) = \pp$ is of complexity $k$ so $\exists x \, \tp^k_{\A}(x, a_1) = \pp$ is of complexity $k+1$ by inductive hypothesis we have + \begin{align*} + \A \models \exists x \, \tp^k_{\A}(x, a_2) = \pp + \end{align*} + Let $a'$ witness this existential claim so that + \begin{align*} + \tp^k_{\A}(a', a_2) &= \pp \\ + \tp^k_{\A}(a', a_2) &= \tp^k_{\A}(a, a_1) \\ + (\A, a', a_2) &\equiv_k (\A, a, a_1) + \end{align*} + % careful with induction! + by inductive assumption we have + \begin{align*} + (\TT, a, a_1, b_1) &\equiv_k (\TT, a', a_2, b_2) \\ + \TT &\models \phi^k(a', a_2, b_2) & \text {as } \TT &\models \phi^k(a, a_1, b_1)\\ + \TT &\models \exists x \phi^k(x, a_2, b_2) + \end{align*} +\end{proof} + +We don't require this lemma in full generality. From now on in this paper we'll have $\TT$ to be a model of a tree in the language $\LL = \{\leq\}$ and $\A, \B$ be in some languages $\LL_A, \LL_B$ which will be expands of $\LL$, with $\A, \B$ substructures of $\TT$ as reducts to $\LL$. We'll refer to $(\A, \B)$ as a \emph{proper subdivision} ($\TT$ will be dropped if it is implied from context). + +\begin{Example} + Suppose the tree consists of two connected components $C_1, C_2$. Then $(C_1, \leq)$ and $(C_2, \leq)$ form a proper subdivision. +\end{Example} + +\begin{Example} \label{ex_cone} + Fix $\TT$ and $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Then $(A, \leq, S)$ and $(B, \leq)$ form a proper subdivision, where $\LL_A$ has a unary predicate interpreted by $S$. +\end{Example} + +\begin{Definition} For $\phi(x, y)$, $A \subseteq T^{|x|}$ and $B \subseteq T^{|y|}$ +\begin{itemize} + \item Let $\phi(A, b) = \{a \in A \mid \phi(a, b)\} \subseteq A$ + \item Let $\phi(A, B) = \{\phi(A, b) \mid b \in B\} \subseteq \PP(A)$ +\end{itemize} +\end{Definition} +$\phi(A, B)$ is a collection of subsets of $A$ definable by $\phi$ with parameters from $B$. We notice the following bound when $A, B$ are parts of a proper subdivision. + +\begin{Corollary} \label{cor_type_count} + Suppose $\phi(x,y)$ is a formula of complexity $n$. Let $\A, \B$ be a proper subdivision of $\TT$ and $b_1, b_2 \in B^{|y|}$. Then if $\tp^n_{\B}(b_1) = \tp^n_{\B}(b_2)$ then $\phi(A^{|x|}, b_1) = \phi(A, b_2)$. Thus $|\phi(A^{|x|}, B^{|y|})|$ is bounded by $|S^n_{\B}(y)|$ +\end{Corollary} + +\begin{proof} + Take some $a \in A^{|x|}$. We have $(\B, b_1) \equiv_n (\B, b_2)$ and (trivially) $(\A, a) \equiv_n (\A, a)$. Thus by the Lemma \ref{lm_subdivision} we have $(\TT, a, b_1) \equiv_n (\TT, a, b_2)$ so $\phi(a, b_1) \iff \phi(a, b_2)$. Since $a$ was arbitrary we have $\phi(A^{|x|}, b_1) = \phi(A^{|x|}, b_2)$. +\end{proof} + +Now we note that the number of such types can be bounded uniformly. + +\begin{Note} \label{nt_type_count} + Fix a (finite relational) language $\LL_B$, and $n$, $|y|$. Then there is some $N = N(n, |y|, \LL_B)$ such that for any structure $\B$ in $\LL_B$ we have $|S^n_{\B}(y)| \leq N$ +\end{Note} + +%relations with this number + +\section{Proper Subdivisions: Constructions} + +First, we describe several constructions of proper subdivisions that are needed for the proof. + +%relate to cones +\begin{Definition} + We say that $E(b, c)$ if $b$ and $c$ are connected + \begin{align*} + E(b, c) \ifff \exists x \, (b \geq x) \wedge (c \geq x) + \end{align*} + Similarly $E_a(b, c)$ means that $b$ and $c$ are connected through an element above $a$. More precisely + \begin{align*} + E_a(b, c) \ifff \exists x \, (x > a) \wedge (b \geq x) \wedge (c \geq x) + \end{align*} +\end{Definition} + +In the following four definitions $\B$-structures are going to be in the same language $\LL_B = \{\leq, U\}$ with $U$ a unary predicate. It is not always necessary to have this predicate but for the sake of uniformity we keep it. $\A$-structures will have different $\LL_A$ languages (those are not as important in later applications). + +\input {vc-trees-all_figures} + +% Simon's cone, parigot's notation? +\begin{Definition} + Fix $c_1 < c_2$ in $T$. Let + \begin{align*} + B &= \{b \in T \mid E_{c_1}(c_2, b) \wedge \neg(b \geq c_2)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t < c_1\} \\ + S_2 &= \{t \in T \mid t < c_2\} \\ + S_B &= S_2 - S_1 \\ + T_A &= \{t \in T \mid c_2 \leq t\} + \end{align*} + Define structures $\A^{c_1}_{c_2} = (A, \leq, S_1, T_A)$ and $\B^{c_1}_{c_2} = (B, \leq, S_B)$ where $\LL_A$ is expansion of $\{\leq\}$ by two unary predicates (and $\LL_B$ as defined above). Note that $c_1, c_2 \notin B$. +\end{Definition} + + +\begin{Definition} + Fix $c$ in $T$. Let + \begin{align*} + B &= \{b \in T \mid \neg(b \geq c) \wedge E(b,c)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t < c\} + \end{align*} + Define structures $\A_{c} = (A, \leq)$ and $\B_{c} = (B, \leq, S_1)$ where $\LL_A = \{\leq\}$ (and $\LL_B$ as defined above). Note that $c \notin B$. (cf example \ref{ex_cone}). +\end{Definition} + +\begin{Definition} + Fix $c$ in $T$ and $S \subseteq T$ a finite subset. Let + \begin{align*} + B &= \{b \in T \mid (b > c) \text{ and for all $s \in S$ we have } \neg E_c(s, b)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t \leq c\} + \end{align*} + Define structures $\A^{c}_{S} = (A, \leq, S_1)$ and $\B^{c}_{S} = (B, \leq, B)$ where $L_A$ is expansion of $\{\leq\}$ by a single unary predicate (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $c \notin B$ and $S \cap B = \emptyset$. +\end{Definition} + +\begin{Definition} + Fix $S \subseteq T$ a finite subset. Let + \begin{align*} + B &= \{b \in T \mid \text{ for all $s \in S$ we have } \neg E(s, b)\} \\ + A &= T - B + \end{align*} + Define structures $\A_{S} = (A, \leq)$ and $\B_{S} = (B, \leq, B)$ where $\LL_A = \{\leq\}$ (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $S \cap B = \emptyset$. +\end{Definition} + +\begin{Lemma} + Pairs of structures defined above are all proper subdivisions. +\end{Lemma} + +\begin{proof} + We only show this holds for the first definition $\A = \A^{c_1}_{c_2}$ and $\B = \B^{c_1}_{c_2}$. Other cases follow by a similar argument. $A,B$ partition $T$ by definition, so it is a subdivision. To show that it is proper by Lemma \ref{lm_subdivision} we only need to check that it is $0$-proper. Suppose we have + \begin{align*} + a &= (a_1, a_2, \ldots, a_p) \in A^p \\ + a' &= (a_1', a_2', \ldots, a_p') \in A^p \\ + b &= (b_1, b_2, \ldots, b_q) \in B^q \\ + b' &= (b_1', b_2', \ldots, b_q') \in B^q + \end{align*} + with $(\A, a) \equiv_0 (\A, a')$ and $(\B, b) \equiv_0 (\B, b')$. We need to make sure that $ab$ has the same quantifier free type as $a'b'$. Any two elements in $T$ can be related in the four following ways + \begin{align*} + x &= y \\ + x &< y \\ + x &> y \\ + x&,y \text{ are incomparable} + \end{align*} + We need to check that the same relations hold for pairs of $(a_i, b_j), (a_i', b_j')$ for all $i,j$. + + \begin{itemize} + \item It is impossible that $a_i = b_j$ as they come from disjoint sets. + \item Suppose $a_i < b_j$. This forces $a_i \in S_1$ thus $a_i' \in S_1$ and $a_i' < b_j'$ + \item Suppose $a_i > b_j$ This forces $b_j \in S_B$ and $a \in T_A$, thus $b_j' \in S_B$ and $a_i' \in T_A$ so $a_i' > b_j'$ + \item Suppose $a_i$ and $b_j$ are incomparable. Two cases are possible: + \begin{itemize} + \item $b_j \notin S_B$ and $a_i \in T_A$. Then $b_j' \notin S_B$ and $a_i' \in T_A$ making $a_i', b_j'$ incomparable + \item $b_j \in S_B$, $a_i \notin T_A$, $a_i \notin S_1$. Similarly this forces $a_i', b_j'$ incomparable + \end{itemize} + \end{itemize} +\end{proof} + +\section{Main proof} + +Basic idea for the proof is that we are able to divide our parameter space into $O(n)$ many pieces. Each of $q$ parameters can come from any of those $O(n)$ partitions giving us $O(n)^q$ many choices for parameter configuration. When every parameter coming from a fixed partition the number of definable sets is constant and in fact is uniformly bounded by some $N$. This gives us $N O(n)^q = O(n^q)$ possibilities for different definable sets. + +First, we generalize Corollary \ref{cor_type_count}. (This is only required for computing vc-density for formulas $\phi(x, y)$ with $|y| > 1$) + +\begin{Lemma} \label{lm_partition_bound} + Consider a finite collection $(\A_i, \B_i)_{i \leq n}$ where each $(\A_i, \B_i)$ is a proper subdivision or a singleton: $B_i = \{b_i\}$ with $A_i = T$. Also assume that all $\B_i$ have the same language $\LL_B$. Let $A = \bigcap_{i \in I} A_i$. Fix a formula $\phi(x, y)$ of complexity $m$ . Let $N = N(m, |y|, \LL_B)$ as in Note \ref{nt_type_count}. Consider any $B \subseteq T^{|y|}$ of the form + \begin{align*} + B = B_1^{i_1} \times B_2^{i_2} \times \ldots \times B_n^{i_n} \text { with } i_1 + i_2 + \ldots + i_n = |y| + \end{align*} + (some of the indexes can be zero). Then we have the following bound + \begin{align*} + \phi(A^{|x|}, B) \leq N^{|y|} + \end{align*} +\end{Lemma} + +\begin{proof} + We show this result by counting types. Suppose we have + \begin{align*} + b_1, b_1' &\in B_1^{i_1} \text{ with } b_1 \equiv_m b_1' \text { in } B_1 \\ + b_2, b_2' &\in B_2^{i_2} \text{ with } b_2 \equiv_m b_2' \text { in } B_2 \\ + &\cdots \\ + b_n, b_n' &\in B_n^{i_n} \text{ with } b_n \equiv_m b_n' \text { in } B_n + \end{align*} + Then we have + \begin{align*} + \phi(A^{|x|}, b_1, b_2, \ldots b_n) \ifff \phi(A^{|x|}, b_1', b_2', \ldots b_n') + \end{align*} + This is easy to see by applying Corollary \ref{cor_type_count} one by one for each tuple. This works if $B_i$ is part of a proper subdivision; if it is a singleton then the implication is trivial as $b_i = b_i'$. + This shows that $\phi(A^{|x|}, B)$ only depends on the choice of types for the tuples + \begin{align*} + |\phi(A^{|x|}, B)| \leq |\tp^m_{B_1}(i_1)| \cdot |\tp^m_{B_2}(i_2)| \cdot \ldots \cdot |\tp^m_{B_n}(i_n)| + \end{align*} + Now for each type space we have inequality + \begin{align*} + |\tp^m_{B_1}(i_1)| \leq N(m, i_1, \LL_B) \leq N(m, |y|, \LL_B) \leq N + \end{align*} + (For singletons $|\tp^m_{B_j}(i_j)| = 1 \leq N$). Only non-zero indexes contribute to the product and there are at most $|y|$ of those (by equality $i_1 + i_2 + \ldots + i_n = |y|$). Thus we have + \begin{align*} + |\phi(A^{|x|}, B)| \leq N^{|y|} + \end{align*} + as needed. +\end{proof} + +For subdivisions to work out properly we will need to work with subsets closed under meets. We observe that closure under meets doesn't add too many new elements. + +% write an actual proof! +\begin{Lemma} \label{lm_meet} + Suppose $S \subseteq T$ is a non-empty finite subset of a meet tree of size $n$ and $S'$ its closure under meets. Then $|S'| \leq 2n - 1$. +\end{Lemma} +\begin{proof} + We prove by induction on $n$. Base case $n = 1$ is clear. Suppose we have $S$ of size $k$ with closure of size at most $2k - 1$. Take a new point and look at its meets with all the elements of $S$. Pick the largest one. That is the only element we need to add to $S'$ to make sure the set is closed under meets. +\end{proof} + +Putting all of those results together we are able to compute $\vc$-density of formulas in meet trees. + +\begin{Theorem} + Let $\TT$ be an infinite meet tree and $\phi(x, y)$ a formula with $|x| = p$ and $|y| = q$. Then $\vc(\phi) \leq q$. +\end{Theorem} + +\begin{proof} + Pick a finite subset of $S_0 \subset T^p$ of size $n$. Let $S_1 \subset T$ consist of coordinates of $S_0$. Let $S \subset T$ be a closure of $S_1$ under meets. Using Lemma \ref{lm_meet} we have $|S_2| \leq 2|S_1| \leq 2p|S_0| = 2pn = O(n)$. We have $S_0 \subseteq S^p$, so $|\phi(S_0, T^q)| \leq |\phi(S^p, T^q)|$. Thus it is enough to show $|\phi(S^p, T^q)| = O(n^q)$. + + Label $S = \{c_i\}_{i \in I}$ with $|I| \leq 2pn$. For every $c_i$ we construct two partitions in the following way. We have $c_i$ is either minimal in $S$ or it has a predecessor in $S$ (greatest element less than $c$). If it is minimal construct $(\A_{c_i}, \B_{c_i})$. If there is a predecessor $p$ construct $(\A^p_{c_i}, \B^p_{c_i})$. For the second subdivision let $G$ be all elements in $S$ greater than $c_i$ and construct $(\A^c_G, \B^c_G)$. So far we have constructed two subdivisions for every $i \in I$. Additionally construct $(\A_S, \B_S)$. We end up with a finite collection of proper subdivisons $(\A_j, \B_j)_{j \in J}$ with $|J| = 2|I| + 1$. Before we proceed we note the following two lemmas describing our partitions. + + \begin{Lemma} + For all $j \in J$ we have $S \subseteq A_j$. Thus $S \subseteq \bigcap_{j \in J} A_j$ and $S^p \subseteq \bigcap_{j \in J} (A_j)^p$ + \end{Lemma} + + \begin{proof} + Check this for each possible choices of partition. Cases for partitions of the type $\A_S, \A^c_G, \A_c$ are easy. Suppose we have partition $(\A, \B) = (\A^{c_1}_{c_2}, \B^{c_1}_{c_2})$. We need to show that $B \cap S = \emptyset$. By construction we have $c_1, c_2 \notin B$. Suppose we have some other $c \in S$ with $c \in B$. We have $E_{c_1}(c_2, c)$ i.e. there is some $b$ such that $(b > c_1)$, $(b \leq c_2)$ and $(b \leq c)$. Consider the meet $(c \wedge c_2)$. We have $(c \wedge c_2) \geq b > c_1$. Also as $\neg (c \geq c_2)$ we have $(c \wedge c_2) < c_2$. To summarize $c_2 > (c \wedge c_2) > c_1$. But this contradicts our construction as $S$ is closed under meets, so $(c \wedge c_2) \in S$ and $c_1$ is supposed to be a predecessor of $c_2$ in $S$. + \end{proof} + + \begin{Lemma} + $\{B_j\}_{j \in J}$ partition $T - S$ i.e. $T = \bigsqcup_{j \in J} B_j \sqcup S$ + \end{Lemma} + + \begin{proof} + This more or less follows from the choice of partitions. Pick any $b \in S - T$. Take all elements in $S$ greater than $b$ and take the minimal one $a$. Take all elements in $S$ less than $b$ and take the maximal one $c$ (possible as $S$ is closed under meets). Also take all elements in $S$ incomparable to $b$ and denote them $G$. If both $a$ and $c$ exist we have $b \in \B^a_c$. If only upper bound exists we have $b \in \B^a_G$. If only lower bound exists we have $b \in \B_c$. If neither exists we have $b \in \B_G$. + \end{proof} + + \begin{Note} + Those two lemmas imply $S = \bigcap_{j \in J} A_j$ + \end{Note} + + \begin{Note} + %careful application of note - have different languages and has to be > 1 + For one-dimensional case $q = 1$ we don't need to do any more work. We have partitioned parameter space into $|J| = O(n)$ many pieces and over each piece the number of definable sets is uniformly bounded. By Note \ref{nt_type_count} we have that $|\phi((A_j)^p, B_j)| \leq N$ for any $j \in J$ (letting $N = N(n_\phi, q, \{\leq, S\})$ where $n_\phi$ is complexity of $\phi$ and $S$ is a unary predicate). Compute + %describe steps + \begin{align*} + |\phi(S^p, T)| + &= \left|\bigcup_{j \in J} \phi(S^p, B_j) \cup \phi(S^p, S)\right| \leq \\ + &\leq \sum_{j \in J} |\phi(S^p, B_j)| + |\phi(S^p, S)| \leq \\ + &\leq \sum_{j \in J} |\phi((A_j)^p, B_j)| + |S| \leq \\ + &\leq \sum_{j \in J}N + |I| \leq \\ + &\leq (4pn + 1)N + 2pn = (4pN + 2p)n + N = O(n) + \end{align*} + \end{Note} + Basic idea for the general case $q \geq 1$ is that we have $q$ parameters and $|J| = O(n)$ partitions to pick each parameter from giving us $|J|^q = O(n^q)$ choices for parameter configuration, each giving uniformly constant number of definable subsets of $S$. (If every parameter is picked from a fixed partition, Lemma \ref{lm_partition_bound} provides a uniform bound). This yields $\vc(\phi) \leq q$ as needed. The rest of the proof is stating this idea formally. + + First, we extend our collection of subdivisions $(\A_j, \B_j)_{j \in J}$ by the following singleton sets. For each $c_i \in S$ let $B_i = \{c_i\}$ and $A_i = T$ and add $(\A_i, \B_i)$ to our collection with $\LL_B$ the language of $B_i$ interpreted arbitrarily. We end up with a new collection $(\A_k, \B_k)_{k \in K}$ indexed by some $K$ with $|K| = |J| + |I|$ (we added $|S|$ new pairs). Now we have that $B_k$ partition $T$, so $T = \bigsqcup_{k \in K} B_k$ and $S = \bigcap_{j \in J} A_j = \bigcap_{k \in K} A_k$. For $(k_1, k_2, \ldots k_q) = \vec k \in K^q$ denote + \begin{align*} + B_{\vec k} = B_{k_1} \times B_{k_2} \times \ldots \times B_{k_q} + \end{align*} + Then we have the following identity + \begin{align*} + T^q = (\bigsqcup_{k \in K} B_k)^q = \bigsqcup_{\vec k \in K^q} B_{\vec k} + \end{align*} + Thus we have that $\{B_{\vec k}\}_{\vec k \in K^q}$ partition $T^q$. Compute + \begin{align*} + |\phi(S^p, T^q)| + &= \left|\bigcup_{\vec k \in K^q} \phi(S^p, B_{\vec k}) \right| \leq \\ + &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| + \end{align*} + We can bound $|\phi(S^p, B_{\vec k})|$ uniformly using Lemma \ref{lm_partition_bound}. $(\A_k, \B_k)_{k \in K}$ satisfies the requirements of the lemma and $B_{\vec k}$ looks like $B$ in the lemma after possibly permuting some variables in $\phi$. Applying the lemma we get + \begin{align*} + |\phi(S^p, B_{\vec k})| \leq N^q + \end{align*} + with $N$ only depending on $q$ and complexity of $\phi$. We complete our computation + \begin{align*} + |\phi(S^p, T^q)| + &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| \leq \\ + &\leq \sum_{\vec k \in K^q} N^q \leq \\ + &\leq |K^q| N^q \leq \\ + &\leq (|J| + |I|)^q N^q \leq \\ + &\leq (4pn + 1 + 2pn)^q N^q = N^q (6p + 1/n)^q n^q = O(n^q) + \end{align*} + \end{proof} + \begin{Corollary} + In the theory of infinite meet trees we have $vc(n) = n$ for all $n \in \N^{+}$. + \end{Corollary} + +\begin{thebibliography}{9} + +\bibitem{vc_density} + M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, + \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) + +\bibitem{simon_dp_min} + P. Simon, + \textit{On dp-minimal ordered structures}, + J. Symbolic Logic 76 (2011), no. 2, 448-460 + +\bibitem{parigot_trees} + Michel Parigot. + Th\'eories d'arbres. + \textit{Journal of Symbolic Logic}, 47, 1982. + + +\end{thebibliography} + \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/Trees vc-density.bbl b/research/02 Trees vc-density/Trees vc-density.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/02 Trees vc-density/Trees vc-density.blg b/research/02 Trees vc-density/Trees vc-density.blg deleted file mode 100644 index 7f30fd14..00000000 --- a/research/02 Trees vc-density/Trees vc-density.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Trees vc-density.aux -I found no \bibdata command---while reading file Trees vc-density.aux -I found no \bibstyle command---while reading file Trees vc-density.aux -(There were 2 error messages) diff --git a/research/02 Trees vc-density/Trees vc-density.log b/research/02 Trees vc-density/Trees vc-density.log deleted file mode 100644 index 6e73bd10..00000000 --- a/research/02 Trees vc-density/Trees vc-density.log +++ /dev/null @@ -1,440 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 8 SEP 2015 20:38 -entering extended mode -**Trees*vc-density.tex -("C:\Users\Anton\SparkleShare\Research\research\02 Trees vc-density\Trees vc-density.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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In a NIP theory we can define a VC function - -\begin{align*} - \vc : \N \arr \N -\end{align*} - -Where $vc(n)$ measures complexity of definable sets in an $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. - -In this paper we work with trees viewed as posets. Parigot in \cite{parigot_trees} showed that such models have NIP. This result was strengthened by Simon in \cite{simon_dp_min} showing that trees are dp-minimal. \cite{vc_density} has the following problem - -\begin{Problem} (\cite{vc_density} p. 47) - Determine the VC density function of each (infinite) tree. -\end{Problem} - -Here we settle this question by showing that theory of trees has $\vc(n) = n$. - -\section{Preliminaries} -%tree finite? meet? disconnected? -%MAYBE: Work in connected trees for simplicity! -%inequality directions -We use notation $a \in T^n$ for tuples of size $n$. For variable $x$ or tuple $a$ we denote their arity by $|x|$ and $|a|$ respectively. - -We work with finite relational languages. Given a formula we can define its complexity $n$ as the depth of quantifiers used to build up the formula. More precisely -%See for example \cite{ynm_notes} Definition 2D.4 pg.72. -\begin{Definition} -Define complexity of a formula by induction: -\begin{align*} - &\cx(\text{q.f. formula}) = 0 \\ - &\cx(\exists x \phi(x)) = \cx(\phi(x)) + 1 \\ - &\cx(\phi \wedge \psi) = \max(\cx(\phi), \cx(\psi)) \\ - &\cx(\neg \phi) = \cx(\phi) -\end{align*} -\end{Definition} -A simple inductive argument verifies that there are (up to equivalence) only finitely many formulas with fixed complexity and the number of free variables. We will use the following notation for types: -\begin{Definition} Let $n,m$ be naturals, $\B$ a structure, $A$ a parameter set and $a,b$ tuples in $\B$. - \begin{itemize} - \item $\tp^n_{\B}(a/A)$ will stand for all the $A$-formulas of complexity $\leq n$ that are true of $a$ in $\B$. If $A = \empty$ we may write $\tp^n_{\B}(a)$. $\B$ will be omitted as well if it is clear from context. Note that if $A$ is finite there are finitely many of such formulas (up to an equivalence). Conjunction of those formulas would still have complexity $\leq n$ so we can just associate a single formula describing such a type. - \item $\B \models a \equiv^n_A b$ means that $a,b$ have the same type of complexity $n$ over $A$ in $\B$, i.e. $\tp^n_{\B}(a/A) = \tp^n_{\B}(b/A)$ - \item $S^n_{\B, m}(A)$ will stand for all $m$-types of complexity $n$ over $A$, for example $\tp^n_{\B}(a/A) \in S^n_{\B, m}(A)$ assuming $|a| = m$. - \end{itemize} -\end{Definition} - -Language for the trees consists of a single binary predicate $\{\leq\}$. Theory of trees states that $\leq$ defines a partial order and for every element $a$ we have $\{x \mid x < a\}$ a linear order. For visualization purposes we assume trees grow upwards, with smaller elements on the bottom and larger elements on the top. If $a \leq b$ we will say that $a$ is below $b$ and $b$ is above $a$. - -\begin{Definition} - Work in a tree $\TT$. For $x \in T$ let $I(x) = \{t \in T \mid t \leq x\}$ denote all the elements below $x$. \emph{Meet} of two tree elements $a,b$ is the greatest element of $I(a) \cap I(b)$ (if one exists) and is denoted by $a \wedge b$. -\end{Definition} - -Theory of meet trees requires that any two elements in the same connected component have a meet. Colored trees are trees with a finite number of colors added via unary predicates. - -From now on assume that all trees are colored. We allow our trees to be disconnected or finite unless otherwise stated. - -For completeness we also present definition of VC function. -One should refer to \cite{vc_density} for more details. -Suppose we have a collection $\CS$ of subsets of $X$. We define a \emph{shatter function} $\pi_\CS(n)$ - -\begin{align*} - \pi_\CS(n) = \max \{|A \cap \CS| : A \subset X \text{ and } |A| = n\} -\end{align*} - -Sauer's Lemma asserts that asymptotically $\pi_\CS$ is either $2^n$ or polynomial. -In the polynomial case we define VC density of $\CS$ to be power of polynomial that bounds $\pi_\CS$. -More formally -\begin{align*} - \vc(\CS) = \limsup_{n \to \infty}\frac{\log \pi_\CS}{\log n} -\end{align*} - -Given a model $M \models T$ and a formula $\phi(x, y)$ we define - -\begin{align*} - \CS_\phi &= \{\phi(M^{|n|}, b) : b \in M^{|y|}\} \\ - \vc(\phi) &= \vc(\CS_\phi) -\end{align*} - -One has to check that this definition is independent of realization of $T$, see \cite{vc_density}, Lemma 3.2. For a theory $T$ we define the VC function - -\begin{align*} - \vc(n) = \sup \{\vc(\phi(x, y)) : |x| = n\} -\end{align*} - -Also, \cite{vc_density}, 3.2 tells us that $\vc(m) \geq m$. - -\section{Proper Subdivisions: Definition and Properties} -\begin{Definition} - Let $\A$, $\B$, $\TT$ be models in some (possibly different) finite relational languages. If $A$, $B$ partition $T$ (i.e. $T = A \sqcup B$) we say that $(\A, \B)$ is a \emph{subdivision} of $\TT$. -\end{Definition} - -\begin{Definition} - $(\A, \B)$ subdivision of $\TT$ is called \emph{$n$-proper} if for all $p,q \in \N$, for all $a_1, a_2 \in A^p$ and $b_1, b_2 \in B^q$ we have - \begin{align*} - \A \models a_1 &\equiv_n a_2 \\ - \B \models b_1 &\equiv_n b_2 - \end{align*} - then - \begin{align*} - \TT \models a_1b_1 \equiv_n a_2b_2 - \end{align*} -\end{Definition} - -\begin{Definition} - $(\A, \B)$ subdivision of $\TT$ is called \emph{proper} if it is $n$-proper for all $n \in \N$. -\end{Definition} - -\begin{Lemma} \label{lm_subdivision} - Consider a subdivision $(\A, \B)$ of $\TT$. If it is $0$-proper then it is proper. -\end{Lemma} - -\begin{proof} - Prove the subdivision is $n$-proper for all $n$ by induction. Case $n = 0$ is given by the assumption. Suppose $n = k + 1$ and we have $\TT \models \exists x \, \phi^k(x, a_1, b_1)$ where $\phi^k$ is some formula of complexity $k$. Let $a \in T$ witness the existential claim i.e. $\TT \models \phi^k(a, a_1, b_1)$. We can have $a \in A$ or $a \in B$. Without loss of generality assume $a \in A$. Let $\pp = \tp^k_{\A} (a, a_1)$. Then we have - \begin{align*} - \A \models \exists x \, \tp^k_{\A}(x, a_1) = \pp - \end{align*} - Formula $\tp^k_{\A}(x, a_1) = \pp$ is of complexity $k$ so $\exists x \, \tp^k_{\A}(x, a_1) = \pp$ is of complexity $k+1$. By inductive hypothesis we have - \begin{align*} - \A \models \exists x \, \tp^k_{\A}(x, a_2) = \pp - \end{align*} - Let $a'$ witness this existential claim so that - \begin{align*} - \tp^k_{\A}(a', a_2) &= \pp \\ - \tp^k_{\A}(a', a_2) &= \tp^k_{\A}(a, a_1) \\ - \A \models a'a_2 &\equiv_k aa_1 - \end{align*} - by inductive hypothesis we have - \begin{align*} - \TT \models aa_1b_1 &\equiv_k a'a_2b_2 \\ - \TT &\models \phi^k(a', a_2, b_2) & \text {as } \TT &\models \phi^k(a, a_1, b_1)\\ - \TT &\models \exists x \phi^k(x, a_2, b_2) - \end{align*} -\end{proof} - -We use this lemma for (colored) trees. Suppose we have $\TT$ to be a model of a (colored) tree in some language $\LL = \{\leq, \ldots\}$ and $\A, \B$ be in some languages $\LL_A, \LL_B$ which will be expands of $\LL$, with $\A, \B$ substructures of $\TT$ as reducts to $\LL$. In this case we'll refer to $(\A, \B)$ as a \emph{proper subdivision} (of $\TT$). - -\begin{Example} \label{ex_disc} - Suppose a tree consists of two connected components $C_1, C_2$. Then those components $(C_1, \leq, \ldots)$, $(C_2, \leq, \ldots)$ interpreted as substructures form a proper subdivision. -\end{Example} - -%\begin{Example} \label{ex_cone} -% Work with a (colored) tree $\TT$ in a language $\LL = \{\leq, C_1, \ldots, C_n\}$ with colors interpreted by sets $S_1, \ldots S_n$. Fix $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Let $\LL_A$ be just $\LL$ expanded by an extra unary predicate $\LL_A = \LL \cup \{C\}$. Consider structure $\A$ in the language $\LL_A$ with universe $A$, colors $C_1 \ldots C_n$ interpreted by restriction and $C$ interpreted by $S$. Also let $\B$ be a structure in language $\LL$ and universe $B$, considered as substructure of $\TT$. Then $(\A, \B)$ is a proper subdivision of $\TT$. -%\end{Example} - -\begin{Example} \label{ex_cone} - Fix a tree $\TT$ in the language $\{\leq\}$ and $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Then $(A, \leq, S)$ and $(B, \leq)$ form a proper subdivision, where $\LL_A$ has a unary predicate interpreted by $S$. -\end{Example} - -\begin{Definition} For $\phi(x, y)$, $A \subseteq T^{|x|}$ and $B \subseteq T^{|y|}$ -\begin{itemize} - \item Let $\phi(A, b) = \{a \in A \mid \phi(a, b)\} \subseteq A$ - \item Let $\phi(A, B) = \{\phi(A, b) \mid b \in B\} \subseteq \PP(A)$ -\end{itemize} -\end{Definition} -$\phi(A, B)$ is a collection of subsets of $A$ definable by $\phi$ with parameters from $B$. We notice the following bound when $A, B$ are parts of a proper subdivision. - -\begin{Corollary} \label{cor_type_count} - Let $\A, \B$ be a proper subdivision of $\TT$ and $\phi(x,y)$ is a formula of complexity $n$. Then $|\phi(A^{|x|}, B^{|y|})|$ is bounded by $|S^n_{\B, |y|}|$. -\end{Corollary} - -\begin{proof} - Take some $a \in A^{|x|}$ and $b_1, b_2 \in B^{|y|}$ with $\tp^n_{\B}(b_1) = \tp^n_{\B}(b_2)$. We have $\B \models b_1 \equiv_n b_2$ and (trivially) $\A \models a \equiv_n a$. Thus by the Lemma \ref{lm_subdivision} we have $\TT \models ab_1 \equiv_n ab_2$ so $\phi(a, b_1) \leftrightarrow \phi(a, b_2)$. Since $a$ was arbitrary we have $\phi(A^{|x|}, b_1) = \phi(A^{|x|}, b_2)$. That is different traces can only come from parameters of different types. Thus $|\phi(A^{|x|}, B^{|y|}) \leq |S^n_{\B, |y|}|$. -\end{proof} - -We note that the size of that type space can be bounded uniformly. - -\begin{Definition} \label{def_type_count} - Fix a (finite relational) language $\LL_B$, and $n$, $|y|$. Let $N = N(n, |y|, \LL_B)$ be smallest number such that for any structure $\B$ in $\LL_B$ we have $|S^n_{\B, |y|}| \leq N$. This number is well-defined as there are a finite number (up to equivalence) of possible formulas of complexity $\leq n$ with $|y|$ free variables. Note the following easy inequalities - \begin{align*} - n \leq m &\Rightarrow N(n, |y|, \LL_B) \leq N(m, |y|, \LL_B) \\ - |y| \leq |z| &\Rightarrow N(n, |y|, \LL_B) \leq N(n, |z|, \LL_B) \\ - \LL_A \subseteq \LL_B &\Rightarrow N(n, |y|, \LL_A) \leq N(n, |y|, \LL_B) - \end{align*} -\end{Definition} - -\section{Proper Subdivisions: Constructions} - -From now on work in meet trees unless mentioned otherwise. First, we describe several constructions of proper subdivisions that are needed for the proof. - -\begin{Definition} - We say that $E(b, c)$ if $b$ and $c$ are in the same connected component - \begin{align*} - E(b, c) \ifff \exists x \, (b \geq x) \wedge (c \geq x) - \end{align*} -\end{Definition} -\begin{Definition} - Given a tree element $a$ we can look at all elements above $a$, i.e. $\{x \mid x \geq a\}$. We can think about it as a \emph{closed cone} above $a$. Connected components of that cone can be thought of an \emph{open cones} over $a$. With that interpretation in mind, notation $E_a(b, c)$ means that $b$ and $c$ are in the same open cone over $a$. A more compact way to write it down is through meets: - \begin{align*} - E_a(b, c) \ifff E(b,c) \text{ and } (b \wedge c) > a - \end{align*} -\end{Definition} - -Fix a language for colored trees $\LL = \{\leq, C_1, \ldots C_n\} = \{\leq, \vec C\}$. In the following four definitions $\B$-structures are going to be in the same language $\LL_B = \LL \cup \{U\}$ with $U$ a unary predicate. It is not always necessary to have this predicate but for the sake of uniformity we keep it. $\A$-structures will have different $\LL_A$ languages (those are not as important in later applications). All the colors $\vec C$ are interpreted by colors in $\TT$ by restriction. - -\input {vc-trees-all_figures} - -\begin{Definition} - Fix $c_1 < c_2$ in $T$. Let - \begin{align*} - B &= \{b \in T \mid E_{c_1}(c_2, b) \wedge \neg(b \geq c_2)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t < c_1\} \\ - S_2 &= \{t \in T \mid t < c_2\} \\ - S_B &= S_2 - S_1 \\ - T_A &= \{t \in T \mid c_2 \leq t\} - \end{align*} - Define structures $\A^{c_1}_{c_2} = (A, \leq, \vec C, S_1, T_A)$ and $\B^{c_1}_{c_2} = (B, \leq, \vec C, S_B)$ where $\LL_A$ is expansion of $\LL$ by two unary predicates (and $\LL_B$ as defined above). Note that $c_1, c_2 \notin B$. -\end{Definition} - - -\begin{Definition} - Fix $c$ in $T$. Let - \begin{align*} - B &= \{b \in T \mid \neg(b \geq c) \wedge E(b,c)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t < c\} - \end{align*} - Define structures $\A_{c} = (A, \leq, \vec C)$ and $\B_{c} = (B, \leq, \vec C, S_1)$ where $\LL_A = \LL$ (and $\LL_B$ as defined above). Note that $c \notin B$. (cf example \ref{ex_cone}). -\end{Definition} - -\begin{Definition} - Fix $c$ in $T$ and $S \subseteq T$ a finite subset. Let - \begin{align*} - B &= \{b \in T \mid (b > c) \text{ and for all $s \in S$ we have } \neg E_c(s, b)\} \\ - A &= T - B \\ - S_1 &= \{t \in T \mid t \leq c\} - \end{align*} - Define structures $\A^{c}_{S} = (A, \leq, \vec C, S_1)$ and $\B^{c}_{S} = (B, \leq, \vec C, B)$ where $L_A$ is expansion of $\LL$ by a single unary predicate (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $c \notin B$ and $S \cap B = \emptyset$. -\end{Definition} - -\begin{Definition} - Fix $S \subseteq T$ a finite subset. Let - \begin{align*} - B &= \{b \in T \mid \text{ for all $s \in S$ we have } \neg E(s, b)\} \\ - A &= T - B - \end{align*} - Define structures $\A_{S} = (A, \leq)$ and $\B_{S} = (B, \leq, \vec C, B)$ where $\LL_A = \LL$ (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $S \cap B = \emptyset$. (cf example \ref{ex_disc}) -\end{Definition} - -\begin{Lemma} - Pairs of structures defined above are all proper subdivisions. -\end{Lemma} - -\begin{proof} - We only show this holds for the first definition $\A = \A^{c_1}_{c_2}$ and $\B = \B^{c_1}_{c_2}$. Other cases follow by a similar argument. $A,B$ partition $T$ by definition, so it is a subdivision. To show that it is proper by Lemma \ref{lm_subdivision} we only need to check that it is $0$-proper. Suppose we have - \begin{align*} - a &= (a_1, a_2, \ldots, a_p) \in A^p \\ - a' &= (a_1', a_2', \ldots, a_p') \in A^p \\ - b &= (b_1, b_2, \ldots, b_q) \in B^q \\ - b' &= (b_1', b_2', \ldots, b_q') \in B^q - \end{align*} - with $(\A, a) \equiv_0 (\A, a')$ and $(\B, b) \equiv_0 (\B, b')$. We need to make sure that $ab$ has the same quantifier free type as $a'b'$. Any two elements in $T$ can be related in the four following ways - \begin{align*} - x &= y \\ - x &< y \\ - x &> y \\ - x&,y \text{ are incomparable} - \end{align*} - We need to check that the same relations hold for pairs of $(a_i, b_j), (a_i', b_j')$ for all $i,j$. - - \begin{itemize} - \item It is impossible that $a_i = b_j$ as they come from disjoint sets. - \item Suppose $a_i < b_j$. This forces $a_i \in S_1$ thus $a_i' \in S_1$ and $a_i' < b_j'$ - \item Suppose $a_i > b_j$ This forces $b_j \in S_B$ and $a \in T_A$, thus $b_j' \in S_B$ and $a_i' \in T_A$ so $a_i' > b_j'$ - \item Suppose $a_i$ and $b_j$ are incomparable. Two cases are possible: - \begin{itemize} - \item $b_j \notin S_B$ and $a_i \in T_A$. Then $b_j' \notin S_B$ and $a_i' \in T_A$ making $a_i', b_j'$ incomparable - \item $b_j \in S_B$, $a_i \notin T_A$, $a_i \notin S_1$. Similarly this forces $a_i', b_j'$ incomparable - \end{itemize} - \end{itemize} - Also we need to check that $ab$ has the same colors as $a'b'$. But that is immediate as having the same color in the substructure means having the same color in the whole tree. -\end{proof} - -\section{Main proof} - -Basic idea for the proof is as follows. Suppose we have a formula with $q$ parameters. We are able to split our parameter space into $O(n)$ many partitions. Each of $q$ parameters can come from any of those $O(n)$ partitions giving us $O(n)^q$ many choices for parameter configuration. When every parameter is coming from a fixed partition the number of definable sets is constant and in fact is uniformly bounded above by some $N$. This gives us at most $N \cdot O(n)^q$ possibilities for different definable sets. - -First, we generalize Corollary \ref{cor_type_count}. (This is required for computing vc-density for formulas $\phi(x, y)$ with $|y| > 1$). - -\begin{Lemma} \label{lm_partition_bound} - Consider a finite collection $(\A_i, \B_i)_{i \leq n}$ where each $(\A_i, \B_i)$ is either a proper subdivision or a singleton: $B_i = \{b_i\}$ with $A_i = T$. Also assume that all $\B_i$ have the same language $\LL_B$. Let $A = \bigcap_{i \in I} A_i$. Fix a formula $\phi(x, y)$ of complexity $m$ . Let $N = N(m, |y|, \LL_B)$ as in Definition \ref{def_type_count}. Consider any $B \subseteq T^{|y|}$ of the form - \begin{align*} - B = B_1^{i_1} \times B_2^{i_2} \times \ldots \times B_n^{i_n} \text { with } i_1 + i_2 + \ldots + i_n = |y| - \end{align*} - (some of the indexes can be zero). Then we have the following bound - \begin{align*} - \phi(A^{|x|}, B) \leq N^{|y|} - \end{align*} -\end{Lemma} - -\begin{proof} - We show this result by counting types. Suppose we have - \begin{align*} - b_1, b_1' &\in B_1^{i_1} \text{ with } b_1 \equiv_m b_1' \text { in } \B_1 \\ - b_2, b_2' &\in B_2^{i_2} \text{ with } b_2 \equiv_m b_2' \text { in } \B_2 \\ - &\cdots \\ - b_n, b_n' &\in B_n^{i_n} \text{ with } b_n \equiv_m b_n' \text { in } \B_n - \end{align*} - Then we have - \begin{align*} - \phi(A^{|x|}, b_1, b_2, \ldots b_n) \ifff \phi(A^{|x|}, b_1', b_2', \ldots b_n') - \end{align*} - This is easy to see by applying Corollary \ref{cor_type_count} one by one for each tuple. This works if $\B_i$ is part of a proper subdivision; if it is a singleton then the implication is trivial as $b_i = b_i'$. - This shows that $\phi(A^{|x|}, B)$ only depends on the choice of types for the tuples - \begin{align*} - |\phi(A^{|x|}, B)| \leq |S^m_{\B_1, i_1}| \cdot |S^m_{\B_2, i_2}| \cdot \ldots \cdot |S^m_{\B_n, i_n}| - \end{align*} - Now for each type space we have inequality - \begin{align*} - |S^m_{\B_j, i_j}| \leq N(m, i_j, \LL_B) \leq N(m, |y|, \LL_B) \leq N - \end{align*} - (For singletons $|S^m_{\B_j, i_j}| = 1 \leq N$). Only non-zero indexes contribute to the product and there are at most $|y|$ of those (by equality $i_1 + i_2 + \ldots + i_n = |y|$). Thus we have - \begin{align*} - |\phi(A^{|x|}, B)| \leq N^{|y|} - \end{align*} - as needed. -\end{proof} - -For subdivisions to work out properly we will need to work with subsets closed under meets. We observe that closure under meets doesn't add too many new elements. - -%MAYBE: write a more detailed proof -\begin{Lemma} \label{lm_meet} - Suppose $S \subseteq T$ is a finite subset of size $n$ in a meet tree and $S'$ its closure under meets. Then $|S'| \leq 2n$. -\end{Lemma} -\begin{proof} - We can partition $S$ into connected components and prove the result separately for each component. Thus we may assume elements of $S$ lie in the same connected component. We prove the claim by induction on $n$. Base case $n = 1$ is clear. Suppose we have $S$ of size $k$ with closure of size at most $2k - 1$. Take a new point $s$, and look at its meets with all the elements of $S$. Pick the smallest one, $s'$. Then $S \cup \{s, s'\}$ is closed under meets. -\end{proof} - -Putting all of those results together we are able to compute $\vc$-density of formulas in meet trees. - -\begin{Theorem} - Let $\TT$ be an infinite (colored) meet tree and $\phi(x, y)$ a formula with $|x| = p$ and $|y| = q$. Then $\vc(\phi) \leq q$. -\end{Theorem} - -\begin{proof} - Pick a finite subset of $S_0 \subset T^p$ of size $n$. Let $S_1 \subset T$ consist of coordinates of $S_0$. Let $S \subset T$ be a closure of $S_1$ under meets. Using Lemma \ref{lm_meet} we have $|S| \leq 2|S_1| \leq 2p|S_0| = 2pn = O(n)$. We have $S_0 \subseteq S^p$, so $|\phi(S_0, T^q)| \leq |\phi(S^p, T^q)|$. Thus it is enough to show $|\phi(S^p, T^q)| = O(n^q)$. - - Label $S = \{c_i\}_{i \in I}$ with $|I| \leq 2pn$. For every $c_i$ we construct two partitions in the following way. We have $c_i$ is either minimal in $S$ or it has a predecessor in $S$ (greatest element less than $c$). If it is minimal construct $(\A_{c_i}, \B_{c_i})$. If there is a predecessor $p$ construct $(\A^p_{c_i}, \B^p_{c_i})$. For the second subdivision let $G$ be all elements in $S$ greater than $c_i$ and construct $(\A^c_G, \B^c_G)$. So far we have constructed two subdivisions for every $i \in I$. Additionally construct $(\A_S, \B_S)$. We end up with a finite collection of proper subdivisons $(\A_j, \B_j)_{j \in J}$ with $|J| = 2|I| + 1$. Before we proceed we note the following two lemmas describing our partitions. - - \begin{Lemma} - For all $j \in J$ we have $S \subseteq A_j$. Thus $S \subseteq \bigcap_{j \in J} A_j$ and $S^p \subseteq \bigcap_{j \in J} (A_j)^p$ - \end{Lemma} - - \begin{proof} - Check this for each possible choices of partition. Cases for partitions of the type $\A_S, \A^c_G, \A_c$ are easy. Suppose we have partition $(\A, \B) = (\A^{c_1}_{c_2}, \B^{c_1}_{c_2})$. We need to show that $B \cap S = \emptyset$. By construction we have $c_1, c_2 \notin B$. Suppose we have some other $c \in S$ with $c \in B$. We have $E_{c_1}(c_2, c)$ i.e. there is some $b$ such that $(b > c_1)$, $(b \leq c_2)$ and $(b \leq c)$. Consider the meet $(c \wedge c_2)$. We have $(c \wedge c_2) \geq b > c_1$. Also as $\neg (c \geq c_2)$ we have $(c \wedge c_2) < c_2$. To summarize $c_2 > (c \wedge c_2) > c_1$. But this contradicts our construction as $S$ is closed under meets, so $(c \wedge c_2) \in S$ and $c_1$ is supposed to be a predecessor of $c_2$ in $S$. - \end{proof} - - \begin{Lemma} - $\{B_j\}_{j \in J}$ partition $T - S$ i.e. $T = \bigsqcup_{j \in J} B_j \sqcup S$ - \end{Lemma} - - \begin{proof} - This more or less follows from the choice of partitions. Pick any $b \in S - T$. Take all elements in $S$ greater than $b$ and take the minimal one $a$. Take all elements in $S$ less than $b$ and take the maximal one $c$ (possible as $S$ is closed under meets). Also take all elements in $S$ incomparable to $b$ and denote them $G$. If both $a$ and $c$ exist we have $b \in \B^a_c$. If only upper bound exists we have $b \in \B^a_G$. If only lower bound exists we have $b \in \B_c$. If neither exists we have $b \in \B_G$. - \end{proof} - - \begin{Note} - Those two lemmas imply $S = \bigcap_{j \in J} A_j$ - \end{Note} - - \begin{Note} - %careful application of note - have different languages and has to be > 1 - For one-dimensional case $q = 1$ we don't need to do any more work. We have partitioned parameter space into $|J| = O(n)$ many pieces and over each piece the number of definable sets is uniformly bounded. By Corollary \ref{cor_type_count} we have that $|\phi((A_j)^p, B_j)| \leq N$ for any $j \in J$ (letting $N = N(n_\phi, q, \LL \cup \{S\})$ where $n_\phi$ is complexity of $\phi$ and $S$ is a unary predicate). Compute - %describe steps - \begin{align*} - |\phi(S^p, T)| - &= \left|\bigcup_{j \in J} \phi(S^p, B_j) \cup \phi(S^p, S)\right| \leq \\ - &\leq \sum_{j \in J} |\phi(S^p, B_j)| + |\phi(S^p, S)| \leq \\ - &\leq \sum_{j \in J} |\phi((A_j)^p, B_j)| + |S| \leq \\ - &\leq \sum_{j \in J}N + |I| \leq \\ - &\leq (4pn + 1)N + 2pn = (4pN + 2p)n + N = O(n) - \end{align*} - \end{Note} - Basic idea for the general case $q \geq 1$ is that we have $q$ parameters and $|J| = O(n)$ partitions to pick each parameter from giving us $|J|^q = O(n^q)$ choices for parameter configuration, each giving uniformly constant number of definable subsets of $S$. (If every parameter is picked from a fixed partition, Lemma \ref{lm_partition_bound} provides a uniform bound). This yields $\vc(\phi) \leq q$ as needed. The rest of the proof is stating this idea formally. - - First, we extend our collection of subdivisions $(\A_j, \B_j)_{j \in J}$ by the following singleton sets. For each $c_i \in S$ let $B_i = \{c_i\}$ and $A_i = T$ and add $(\A_i, \B_i)$ to our collection with $\LL_B$ the language of $B_i$ interpreted arbitrarily. We end up with a new collection $(\A_k, \B_k)_{k \in K}$ indexed by some $K$ with $|K| = |J| + |I|$ (we added $|S|$ new pairs). Now we have that $B_k$ partition $T$, so $T = \bigsqcup_{k \in K} B_k$ and $S = \bigcap_{j \in J} A_j = \bigcap_{k \in K} A_k$. For $(k_1, k_2, \ldots k_q) = \vec k \in K^q$ denote - \begin{align*} - B_{\vec k} = B_{k_1} \times B_{k_2} \times \ldots \times B_{k_q} - \end{align*} - Then we have the following identity - \begin{align*} - T^q = (\bigsqcup_{k \in K} B_k)^q = \bigsqcup_{\vec k \in K^q} B_{\vec k} - \end{align*} - Thus we have that $\{B_{\vec k}\}_{\vec k \in K^q}$ partition $T^q$. Compute - \begin{align*} - |\phi(S^p, T^q)| - &= \left|\bigcup_{\vec k \in K^q} \phi(S^p, B_{\vec k}) \right| \leq \\ - &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| - \end{align*} - We can bound $|\phi(S^p, B_{\vec k})|$ uniformly using Lemma \ref{lm_partition_bound}. $(\A_k, \B_k)_{k \in K}$ satisfies the requirements of the lemma and $B_{\vec k}$ looks like $B$ in the lemma after possibly permuting some variables in $\phi$. Applying the lemma we get - \begin{align*} - |\phi(S^p, B_{\vec k})| \leq N^q - \end{align*} - with $N$ only depending on $q$ and complexity of $\phi$. We complete our computation - \begin{align*} - |\phi(S^p, T^q)| - &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| \leq \\ - &\leq \sum_{\vec k \in K^q} N^q \leq \\ - &\leq |K^q| N^q \leq \\ - &\leq (|J| + |I|)^q N^q \leq \\ - &\leq (4pn + 1 + 2pn)^q N^q = N^q (6p + 1/n)^q n^q = O(n^q) - \end{align*} - \end{proof} - \begin{Corollary} - In the theory of infinite (colored) meet trees we have $vc(n) = n$ for all $n$. - \end{Corollary} - We get the general result for trees that aren't necessarily meet trees via an easy application of interpretability. - \begin{Corollary} - In the theory of infinite (colored) trees we have $vc(n) = n$ for all $n$. - \end{Corollary} - \begin{proof} - Let $\TT'$ be a tree. We can embed it in a larger tree that is closed under meets $\TT' \subset \TT$. Expand $\TT$ by an extra color and interpret it by coloring the subset $\TT'$. Thus we can interpret $\TT'$ in $T^1$. By Corollary 3.17 in \cite{vc_density} we get that $\vc^{\TT'}(n) \leq \vc^T(1 \cdot n) = n$ thus $\vc^{\TT'}(n) = n$ as well. - \end{proof} - - \begin{thebibliography}{9} - -\bibitem{vc_density} - M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, - \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) - -\bibitem{simon_dp_min} - P. Simon, - \textit{On dp-minimal ordered structures}, - J. Symbolic Logic 76 (2011), no. 2, 448-460 - -\bibitem{parigot_trees} - Michel Parigot. - Th\'eories d'arbres. - \textit{Journal of Symbolic Logic}, 47, 1982. - - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{tikz} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} +\DeclareMathOperator{\cl}{cl} +\newcommand{\CS}{\mathcal S} + +\DeclareMathOperator{\cx}{Complexity} + +\begin{document} + +\title{vc-density for trees} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} +%more info + +\begin{abstract} + We show that for the theory of infinite trees we have $\vc(n) = n$ for all $n$. +\end{abstract} + +\maketitle + +VC density was introduced in \cite{vc_density} by Aschenbrenner, Dolich, Haskell, MacPherson, and Starchenko as a natural notion of dimension for NIP theories. In a NIP theory we can define a VC function + +\begin{align*} + \vc : \N \arr \N +\end{align*} + +Where $vc(n)$ measures complexity of definable sets in an $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. + +In this paper we work with trees viewed as posets. Parigot in \cite{parigot_trees} showed that such models have NIP. This result was strengthened by Simon in \cite{simon_dp_min} showing that trees are dp-minimal. \cite{vc_density} has the following problem + +\begin{Problem} (\cite{vc_density} p. 47) + Determine the VC density function of each (infinite) tree. +\end{Problem} + +Here we settle this question by showing that theory of trees has $\vc(n) = n$. + +\section{Preliminaries} +%tree finite? meet? disconnected? +%MAYBE: Work in connected trees for simplicity! +%inequality directions +We use notation $a \in T^n$ for tuples of size $n$. For variable $x$ or tuple $a$ we denote their arity by $|x|$ and $|a|$ respectively. + +We work with finite relational languages. Given a formula we can define its complexity $n$ as the depth of quantifiers used to build up the formula. More precisely +%See for example \cite{ynm_notes} Definition 2D.4 pg.72. +\begin{Definition} +Define complexity of a formula by induction: +\begin{align*} + &\cx(\text{q.f. formula}) = 0 \\ + &\cx(\exists x \phi(x)) = \cx(\phi(x)) + 1 \\ + &\cx(\phi \wedge \psi) = \max(\cx(\phi), \cx(\psi)) \\ + &\cx(\neg \phi) = \cx(\phi) +\end{align*} +\end{Definition} +A simple inductive argument verifies that there are (up to equivalence) only finitely many formulas with fixed complexity and the number of free variables. We will use the following notation for types: +\begin{Definition} Let $n,m$ be naturals, $\B$ a structure, $A$ a parameter set and $a,b$ tuples in $\B$. + \begin{itemize} + \item $\tp^n_{\B}(a/A)$ will stand for all the $A$-formulas of complexity $\leq n$ that are true of $a$ in $\B$. If $A = \empty$ we may write $\tp^n_{\B}(a)$. $\B$ will be omitted as well if it is clear from context. Note that if $A$ is finite there are finitely many of such formulas (up to an equivalence). Conjunction of those formulas would still have complexity $\leq n$ so we can just associate a single formula describing such a type. + \item $\B \models a \equiv^n_A b$ means that $a,b$ have the same type of complexity $n$ over $A$ in $\B$, i.e. $\tp^n_{\B}(a/A) = \tp^n_{\B}(b/A)$ + \item $S^n_{\B, m}(A)$ will stand for all $m$-types of complexity $n$ over $A$, for example $\tp^n_{\B}(a/A) \in S^n_{\B, m}(A)$ assuming $|a| = m$. + \end{itemize} +\end{Definition} + +Language for the trees consists of a single binary predicate $\{\leq\}$. Theory of trees states that $\leq$ defines a partial order and for every element $a$ we have $\{x \mid x < a\}$ a linear order. For visualization purposes we assume trees grow upwards, with smaller elements on the bottom and larger elements on the top. If $a \leq b$ we will say that $a$ is below $b$ and $b$ is above $a$. + +\begin{Definition} + Work in a tree $\TT$. For $x \in T$ let $I(x) = \{t \in T \mid t \leq x\}$ denote all the elements below $x$. \emph{Meet} of two tree elements $a,b$ is the greatest element of $I(a) \cap I(b)$ (if one exists) and is denoted by $a \wedge b$. +\end{Definition} + +Theory of meet trees requires that any two elements in the same connected component have a meet. Colored trees are trees with a finite number of colors added via unary predicates. + +From now on assume that all trees are colored. We allow our trees to be disconnected or finite unless otherwise stated. + +For completeness we also present definition of VC function. +One should refer to \cite{vc_density} for more details. +Suppose we have a collection $\CS$ of subsets of $X$. We define a \emph{shatter function} $\pi_\CS(n)$ + +\begin{align*} + \pi_\CS(n) = \max \{|A \cap \CS| : A \subset X \text{ and } |A| = n\} +\end{align*} + +Sauer's Lemma asserts that asymptotically $\pi_\CS$ is either $2^n$ or polynomial. +In the polynomial case we define VC density of $\CS$ to be power of polynomial that bounds $\pi_\CS$. +More formally +\begin{align*} + \vc(\CS) = \limsup_{n \to \infty}\frac{\log \pi_\CS}{\log n} +\end{align*} + +Given a model $M \models T$ and a formula $\phi(x, y)$ we define + +\begin{align*} + \CS_\phi &= \{\phi(M^{|n|}, b) : b \in M^{|y|}\} \\ + \vc(\phi) &= \vc(\CS_\phi) +\end{align*} + +One has to check that this definition is independent of realization of $T$, see \cite{vc_density}, Lemma 3.2. For a theory $T$ we define the VC function + +\begin{align*} + \vc(n) = \sup \{\vc(\phi(x, y)) : |x| = n\} +\end{align*} + +Also, \cite{vc_density}, 3.2 tells us that $\vc(m) \geq m$. + +\section{Proper Subdivisions: Definition and Properties} +\begin{Definition} + Let $\A$, $\B$, $\TT$ be models in some (possibly different) finite relational languages. If $A$, $B$ partition $T$ (i.e. $T = A \sqcup B$) we say that $(\A, \B)$ is a \emph{subdivision} of $\TT$. +\end{Definition} + +\begin{Definition} + $(\A, \B)$ subdivision of $\TT$ is called \emph{$n$-proper} if for all $p,q \in \N$, for all $a_1, a_2 \in A^p$ and $b_1, b_2 \in B^q$ we have + \begin{align*} + \A \models a_1 &\equiv_n a_2 \\ + \B \models b_1 &\equiv_n b_2 + \end{align*} + then + \begin{align*} + \TT \models a_1b_1 \equiv_n a_2b_2 + \end{align*} +\end{Definition} + +\begin{Definition} + $(\A, \B)$ subdivision of $\TT$ is called \emph{proper} if it is $n$-proper for all $n \in \N$. +\end{Definition} + +\begin{Lemma} \label{lm_subdivision} + Consider a subdivision $(\A, \B)$ of $\TT$. If it is $0$-proper then it is proper. +\end{Lemma} + +\begin{proof} + Prove the subdivision is $n$-proper for all $n$ by induction. Case $n = 0$ is given by the assumption. Suppose $n = k + 1$ and we have $\TT \models \exists x \, \phi^k(x, a_1, b_1)$ where $\phi^k$ is some formula of complexity $k$. Let $a \in T$ witness the existential claim i.e. $\TT \models \phi^k(a, a_1, b_1)$. We can have $a \in A$ or $a \in B$. Without loss of generality assume $a \in A$. Let $\pp = \tp^k_{\A} (a, a_1)$. Then we have + \begin{align*} + \A \models \exists x \, \tp^k_{\A}(x, a_1) = \pp + \end{align*} + Formula $\tp^k_{\A}(x, a_1) = \pp$ is of complexity $k$ so $\exists x \, \tp^k_{\A}(x, a_1) = \pp$ is of complexity $k+1$. By inductive hypothesis we have + \begin{align*} + \A \models \exists x \, \tp^k_{\A}(x, a_2) = \pp + \end{align*} + Let $a'$ witness this existential claim so that + \begin{align*} + \tp^k_{\A}(a', a_2) &= \pp \\ + \tp^k_{\A}(a', a_2) &= \tp^k_{\A}(a, a_1) \\ + \A \models a'a_2 &\equiv_k aa_1 + \end{align*} + by inductive hypothesis we have + \begin{align*} + \TT \models aa_1b_1 &\equiv_k a'a_2b_2 \\ + \TT &\models \phi^k(a', a_2, b_2) & \text {as } \TT &\models \phi^k(a, a_1, b_1)\\ + \TT &\models \exists x \phi^k(x, a_2, b_2) + \end{align*} +\end{proof} + +We use this lemma for (colored) trees. Suppose we have $\TT$ to be a model of a (colored) tree in some language $\LL = \{\leq, \ldots\}$ and $\A, \B$ be in some languages $\LL_A, \LL_B$ which will be expands of $\LL$, with $\A, \B$ substructures of $\TT$ as reducts to $\LL$. In this case we'll refer to $(\A, \B)$ as a \emph{proper subdivision} (of $\TT$). + +\begin{Example} \label{ex_disc} + Suppose a tree consists of two connected components $C_1, C_2$. Then those components $(C_1, \leq, \ldots)$, $(C_2, \leq, \ldots)$ interpreted as substructures form a proper subdivision. +\end{Example} + +%\begin{Example} \label{ex_cone} +% Work with a (colored) tree $\TT$ in a language $\LL = \{\leq, C_1, \ldots, C_n\}$ with colors interpreted by sets $S_1, \ldots S_n$. Fix $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Let $\LL_A$ be just $\LL$ expanded by an extra unary predicate $\LL_A = \LL \cup \{C\}$. Consider structure $\A$ in the language $\LL_A$ with universe $A$, colors $C_1 \ldots C_n$ interpreted by restriction and $C$ interpreted by $S$. Also let $\B$ be a structure in language $\LL$ and universe $B$, considered as substructure of $\TT$. Then $(\A, \B)$ is a proper subdivision of $\TT$. +%\end{Example} + +\begin{Example} \label{ex_cone} + Fix a tree $\TT$ in the language $\{\leq\}$ and $a \in T$. Let $B = \{t \in T \mid a < t\}$, $S = \{t \in T \mid t \leq a\}$, $A = T - B$. Then $(A, \leq, S)$ and $(B, \leq)$ form a proper subdivision, where $\LL_A$ has a unary predicate interpreted by $S$. +\end{Example} + +\begin{Definition} For $\phi(x, y)$, $A \subseteq T^{|x|}$ and $B \subseteq T^{|y|}$ +\begin{itemize} + \item Let $\phi(A, b) = \{a \in A \mid \phi(a, b)\} \subseteq A$ + \item Let $\phi(A, B) = \{\phi(A, b) \mid b \in B\} \subseteq \PP(A)$ +\end{itemize} +\end{Definition} +$\phi(A, B)$ is a collection of subsets of $A$ definable by $\phi$ with parameters from $B$. We notice the following bound when $A, B$ are parts of a proper subdivision. + +\begin{Corollary} \label{cor_type_count} + Let $\A, \B$ be a proper subdivision of $\TT$ and $\phi(x,y)$ is a formula of complexity $n$. Then $|\phi(A^{|x|}, B^{|y|})|$ is bounded by $|S^n_{\B, |y|}|$. +\end{Corollary} + +\begin{proof} + Take some $a \in A^{|x|}$ and $b_1, b_2 \in B^{|y|}$ with $\tp^n_{\B}(b_1) = \tp^n_{\B}(b_2)$. We have $\B \models b_1 \equiv_n b_2$ and (trivially) $\A \models a \equiv_n a$. Thus by the Lemma \ref{lm_subdivision} we have $\TT \models ab_1 \equiv_n ab_2$ so $\phi(a, b_1) \leftrightarrow \phi(a, b_2)$. Since $a$ was arbitrary we have $\phi(A^{|x|}, b_1) = \phi(A^{|x|}, b_2)$. That is different traces can only come from parameters of different types. Thus $|\phi(A^{|x|}, B^{|y|}) \leq |S^n_{\B, |y|}|$. +\end{proof} + +We note that the size of that type space can be bounded uniformly. + +\begin{Definition} \label{def_type_count} + Fix a (finite relational) language $\LL_B$, and $n$, $|y|$. Let $N = N(n, |y|, \LL_B)$ be smallest number such that for any structure $\B$ in $\LL_B$ we have $|S^n_{\B, |y|}| \leq N$. This number is well-defined as there are a finite number (up to equivalence) of possible formulas of complexity $\leq n$ with $|y|$ free variables. Note the following easy inequalities + \begin{align*} + n \leq m &\Rightarrow N(n, |y|, \LL_B) \leq N(m, |y|, \LL_B) \\ + |y| \leq |z| &\Rightarrow N(n, |y|, \LL_B) \leq N(n, |z|, \LL_B) \\ + \LL_A \subseteq \LL_B &\Rightarrow N(n, |y|, \LL_A) \leq N(n, |y|, \LL_B) + \end{align*} +\end{Definition} + +\section{Proper Subdivisions: Constructions} + +From now on work in meet trees unless mentioned otherwise. First, we describe several constructions of proper subdivisions that are needed for the proof. + +\begin{Definition} + We say that $E(b, c)$ if $b$ and $c$ are in the same connected component + \begin{align*} + E(b, c) \ifff \exists x \, (b \geq x) \wedge (c \geq x) + \end{align*} +\end{Definition} +\begin{Definition} + Given a tree element $a$ we can look at all elements above $a$, i.e. $\{x \mid x \geq a\}$. We can think about it as a \emph{closed cone} above $a$. Connected components of that cone can be thought of an \emph{open cones} over $a$. With that interpretation in mind, notation $E_a(b, c)$ means that $b$ and $c$ are in the same open cone over $a$. A more compact way to write it down is through meets: + \begin{align*} + E_a(b, c) \ifff E(b,c) \text{ and } (b \wedge c) > a + \end{align*} +\end{Definition} + +Fix a language for colored trees $\LL = \{\leq, C_1, \ldots C_n\} = \{\leq, \vec C\}$. In the following four definitions $\B$-structures are going to be in the same language $\LL_B = \LL \cup \{U\}$ with $U$ a unary predicate. It is not always necessary to have this predicate but for the sake of uniformity we keep it. $\A$-structures will have different $\LL_A$ languages (those are not as important in later applications). All the colors $\vec C$ are interpreted by colors in $\TT$ by restriction. + +\input {vc-trees-all_figures} + +\begin{Definition} + Fix $c_1 < c_2$ in $T$. Let + \begin{align*} + B &= \{b \in T \mid E_{c_1}(c_2, b) \wedge \neg(b \geq c_2)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t < c_1\} \\ + S_2 &= \{t \in T \mid t < c_2\} \\ + S_B &= S_2 - S_1 \\ + T_A &= \{t \in T \mid c_2 \leq t\} + \end{align*} + Define structures $\A^{c_1}_{c_2} = (A, \leq, \vec C, S_1, T_A)$ and $\B^{c_1}_{c_2} = (B, \leq, \vec C, S_B)$ where $\LL_A$ is expansion of $\LL$ by two unary predicates (and $\LL_B$ as defined above). Note that $c_1, c_2 \notin B$. +\end{Definition} + + +\begin{Definition} + Fix $c$ in $T$. Let + \begin{align*} + B &= \{b \in T \mid \neg(b \geq c) \wedge E(b,c)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t < c\} + \end{align*} + Define structures $\A_{c} = (A, \leq, \vec C)$ and $\B_{c} = (B, \leq, \vec C, S_1)$ where $\LL_A = \LL$ (and $\LL_B$ as defined above). Note that $c \notin B$. (cf example \ref{ex_cone}). +\end{Definition} + +\begin{Definition} + Fix $c$ in $T$ and $S \subseteq T$ a finite subset. Let + \begin{align*} + B &= \{b \in T \mid (b > c) \text{ and for all $s \in S$ we have } \neg E_c(s, b)\} \\ + A &= T - B \\ + S_1 &= \{t \in T \mid t \leq c\} + \end{align*} + Define structures $\A^{c}_{S} = (A, \leq, \vec C, S_1)$ and $\B^{c}_{S} = (B, \leq, \vec C, B)$ where $L_A$ is expansion of $\LL$ by a single unary predicate (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $c \notin B$ and $S \cap B = \emptyset$. +\end{Definition} + +\begin{Definition} + Fix $S \subseteq T$ a finite subset. Let + \begin{align*} + B &= \{b \in T \mid \text{ for all $s \in S$ we have } \neg E(s, b)\} \\ + A &= T - B + \end{align*} + Define structures $\A_{S} = (A, \leq)$ and $\B_{S} = (B, \leq, \vec C, B)$ where $\LL_A = \LL$ (and $U \in \LL_B$ vacuously interpreted by $B$). Note that $S \cap B = \emptyset$. (cf example \ref{ex_disc}) +\end{Definition} + +\begin{Lemma} + Pairs of structures defined above are all proper subdivisions. +\end{Lemma} + +\begin{proof} + We only show this holds for the first definition $\A = \A^{c_1}_{c_2}$ and $\B = \B^{c_1}_{c_2}$. Other cases follow by a similar argument. $A,B$ partition $T$ by definition, so it is a subdivision. To show that it is proper by Lemma \ref{lm_subdivision} we only need to check that it is $0$-proper. Suppose we have + \begin{align*} + a &= (a_1, a_2, \ldots, a_p) \in A^p \\ + a' &= (a_1', a_2', \ldots, a_p') \in A^p \\ + b &= (b_1, b_2, \ldots, b_q) \in B^q \\ + b' &= (b_1', b_2', \ldots, b_q') \in B^q + \end{align*} + with $(\A, a) \equiv_0 (\A, a')$ and $(\B, b) \equiv_0 (\B, b')$. We need to make sure that $ab$ has the same quantifier free type as $a'b'$. Any two elements in $T$ can be related in the four following ways + \begin{align*} + x &= y \\ + x &< y \\ + x &> y \\ + x&,y \text{ are incomparable} + \end{align*} + We need to check that the same relations hold for pairs of $(a_i, b_j), (a_i', b_j')$ for all $i,j$. + + \begin{itemize} + \item It is impossible that $a_i = b_j$ as they come from disjoint sets. + \item Suppose $a_i < b_j$. This forces $a_i \in S_1$ thus $a_i' \in S_1$ and $a_i' < b_j'$ + \item Suppose $a_i > b_j$ This forces $b_j \in S_B$ and $a \in T_A$, thus $b_j' \in S_B$ and $a_i' \in T_A$ so $a_i' > b_j'$ + \item Suppose $a_i$ and $b_j$ are incomparable. Two cases are possible: + \begin{itemize} + \item $b_j \notin S_B$ and $a_i \in T_A$. Then $b_j' \notin S_B$ and $a_i' \in T_A$ making $a_i', b_j'$ incomparable + \item $b_j \in S_B$, $a_i \notin T_A$, $a_i \notin S_1$. Similarly this forces $a_i', b_j'$ incomparable + \end{itemize} + \end{itemize} + Also we need to check that $ab$ has the same colors as $a'b'$. But that is immediate as having the same color in the substructure means having the same color in the whole tree. +\end{proof} + +\section{Main proof} + +Basic idea for the proof is as follows. Suppose we have a formula with $q$ parameters. We are able to split our parameter space into $O(n)$ many partitions. Each of $q$ parameters can come from any of those $O(n)$ partitions giving us $O(n)^q$ many choices for parameter configuration. When every parameter is coming from a fixed partition the number of definable sets is constant and in fact is uniformly bounded above by some $N$. This gives us at most $N \cdot O(n)^q$ possibilities for different definable sets. + +First, we generalize Corollary \ref{cor_type_count}. (This is required for computing vc-density for formulas $\phi(x, y)$ with $|y| > 1$). + +\begin{Lemma} \label{lm_partition_bound} + Consider a finite collection $(\A_i, \B_i)_{i \leq n}$ where each $(\A_i, \B_i)$ is either a proper subdivision or a singleton: $B_i = \{b_i\}$ with $A_i = T$. Also assume that all $\B_i$ have the same language $\LL_B$. Let $A = \bigcap_{i \in I} A_i$. Fix a formula $\phi(x, y)$ of complexity $m$ . Let $N = N(m, |y|, \LL_B)$ as in Definition \ref{def_type_count}. Consider any $B \subseteq T^{|y|}$ of the form + \begin{align*} + B = B_1^{i_1} \times B_2^{i_2} \times \ldots \times B_n^{i_n} \text { with } i_1 + i_2 + \ldots + i_n = |y| + \end{align*} + (some of the indexes can be zero). Then we have the following bound + \begin{align*} + \phi(A^{|x|}, B) \leq N^{|y|} + \end{align*} +\end{Lemma} + +\begin{proof} + We show this result by counting types. Suppose we have + \begin{align*} + b_1, b_1' &\in B_1^{i_1} \text{ with } b_1 \equiv_m b_1' \text { in } \B_1 \\ + b_2, b_2' &\in B_2^{i_2} \text{ with } b_2 \equiv_m b_2' \text { in } \B_2 \\ + &\cdots \\ + b_n, b_n' &\in B_n^{i_n} \text{ with } b_n \equiv_m b_n' \text { in } \B_n + \end{align*} + Then we have + \begin{align*} + \phi(A^{|x|}, b_1, b_2, \ldots b_n) \ifff \phi(A^{|x|}, b_1', b_2', \ldots b_n') + \end{align*} + This is easy to see by applying Corollary \ref{cor_type_count} one by one for each tuple. This works if $\B_i$ is part of a proper subdivision; if it is a singleton then the implication is trivial as $b_i = b_i'$. + This shows that $\phi(A^{|x|}, B)$ only depends on the choice of types for the tuples + \begin{align*} + |\phi(A^{|x|}, B)| \leq |S^m_{\B_1, i_1}| \cdot |S^m_{\B_2, i_2}| \cdot \ldots \cdot |S^m_{\B_n, i_n}| + \end{align*} + Now for each type space we have inequality + \begin{align*} + |S^m_{\B_j, i_j}| \leq N(m, i_j, \LL_B) \leq N(m, |y|, \LL_B) \leq N + \end{align*} + (For singletons $|S^m_{\B_j, i_j}| = 1 \leq N$). Only non-zero indexes contribute to the product and there are at most $|y|$ of those (by equality $i_1 + i_2 + \ldots + i_n = |y|$). Thus we have + \begin{align*} + |\phi(A^{|x|}, B)| \leq N^{|y|} + \end{align*} + as needed. +\end{proof} + +For subdivisions to work out properly we will need to work with subsets closed under meets. We observe that closure under meets doesn't add too many new elements. + +%MAYBE: write a more detailed proof +\begin{Lemma} \label{lm_meet} + Suppose $S \subseteq T$ is a finite subset of size $n$ in a meet tree and $S'$ its closure under meets. Then $|S'| \leq 2n$. +\end{Lemma} +\begin{proof} + We can partition $S$ into connected components and prove the result separately for each component. Thus we may assume elements of $S$ lie in the same connected component. We prove the claim by induction on $n$. Base case $n = 1$ is clear. Suppose we have $S$ of size $k$ with closure of size at most $2k - 1$. Take a new point $s$, and look at its meets with all the elements of $S$. Pick the smallest one, $s'$. Then $S \cup \{s, s'\}$ is closed under meets. +\end{proof} + +Putting all of those results together we are able to compute $\vc$-density of formulas in meet trees. + +\begin{Theorem} + Let $\TT$ be an infinite (colored) meet tree and $\phi(x, y)$ a formula with $|x| = p$ and $|y| = q$. Then $\vc(\phi) \leq q$. +\end{Theorem} + +\begin{proof} + Pick a finite subset of $S_0 \subset T^p$ of size $n$. Let $S_1 \subset T$ consist of coordinates of $S_0$. Let $S \subset T$ be a closure of $S_1$ under meets. Using Lemma \ref{lm_meet} we have $|S| \leq 2|S_1| \leq 2p|S_0| = 2pn = O(n)$. We have $S_0 \subseteq S^p$, so $|\phi(S_0, T^q)| \leq |\phi(S^p, T^q)|$. Thus it is enough to show $|\phi(S^p, T^q)| = O(n^q)$. + + Label $S = \{c_i\}_{i \in I}$ with $|I| \leq 2pn$. For every $c_i$ we construct two partitions in the following way. We have $c_i$ is either minimal in $S$ or it has a predecessor in $S$ (greatest element less than $c$). If it is minimal construct $(\A_{c_i}, \B_{c_i})$. If there is a predecessor $p$ construct $(\A^p_{c_i}, \B^p_{c_i})$. For the second subdivision let $G$ be all elements in $S$ greater than $c_i$ and construct $(\A^c_G, \B^c_G)$. So far we have constructed two subdivisions for every $i \in I$. Additionally construct $(\A_S, \B_S)$. We end up with a finite collection of proper subdivisons $(\A_j, \B_j)_{j \in J}$ with $|J| = 2|I| + 1$. Before we proceed we note the following two lemmas describing our partitions. + + \begin{Lemma} + For all $j \in J$ we have $S \subseteq A_j$. Thus $S \subseteq \bigcap_{j \in J} A_j$ and $S^p \subseteq \bigcap_{j \in J} (A_j)^p$ + \end{Lemma} + + \begin{proof} + Check this for each possible choices of partition. Cases for partitions of the type $\A_S, \A^c_G, \A_c$ are easy. Suppose we have partition $(\A, \B) = (\A^{c_1}_{c_2}, \B^{c_1}_{c_2})$. We need to show that $B \cap S = \emptyset$. By construction we have $c_1, c_2 \notin B$. Suppose we have some other $c \in S$ with $c \in B$. We have $E_{c_1}(c_2, c)$ i.e. there is some $b$ such that $(b > c_1)$, $(b \leq c_2)$ and $(b \leq c)$. Consider the meet $(c \wedge c_2)$. We have $(c \wedge c_2) \geq b > c_1$. Also as $\neg (c \geq c_2)$ we have $(c \wedge c_2) < c_2$. To summarize $c_2 > (c \wedge c_2) > c_1$. But this contradicts our construction as $S$ is closed under meets, so $(c \wedge c_2) \in S$ and $c_1$ is supposed to be a predecessor of $c_2$ in $S$. + \end{proof} + + \begin{Lemma} + $\{B_j\}_{j \in J}$ partition $T - S$ i.e. $T = \bigsqcup_{j \in J} B_j \sqcup S$ + \end{Lemma} + + \begin{proof} + This more or less follows from the choice of partitions. Pick any $b \in S - T$. Take all elements in $S$ greater than $b$ and take the minimal one $a$. Take all elements in $S$ less than $b$ and take the maximal one $c$ (possible as $S$ is closed under meets). Also take all elements in $S$ incomparable to $b$ and denote them $G$. If both $a$ and $c$ exist we have $b \in \B^a_c$. If only upper bound exists we have $b \in \B^a_G$. If only lower bound exists we have $b \in \B_c$. If neither exists we have $b \in \B_G$. + \end{proof} + + \begin{Note} + Those two lemmas imply $S = \bigcap_{j \in J} A_j$ + \end{Note} + + \begin{Note} + %careful application of note - have different languages and has to be > 1 + For one-dimensional case $q = 1$ we don't need to do any more work. We have partitioned parameter space into $|J| = O(n)$ many pieces and over each piece the number of definable sets is uniformly bounded. By Corollary \ref{cor_type_count} we have that $|\phi((A_j)^p, B_j)| \leq N$ for any $j \in J$ (letting $N = N(n_\phi, q, \LL \cup \{S\})$ where $n_\phi$ is complexity of $\phi$ and $S$ is a unary predicate). Compute + %describe steps + \begin{align*} + |\phi(S^p, T)| + &= \left|\bigcup_{j \in J} \phi(S^p, B_j) \cup \phi(S^p, S)\right| \leq \\ + &\leq \sum_{j \in J} |\phi(S^p, B_j)| + |\phi(S^p, S)| \leq \\ + &\leq \sum_{j \in J} |\phi((A_j)^p, B_j)| + |S| \leq \\ + &\leq \sum_{j \in J}N + |I| \leq \\ + &\leq (4pn + 1)N + 2pn = (4pN + 2p)n + N = O(n) + \end{align*} + \end{Note} + Basic idea for the general case $q \geq 1$ is that we have $q$ parameters and $|J| = O(n)$ partitions to pick each parameter from giving us $|J|^q = O(n^q)$ choices for parameter configuration, each giving uniformly constant number of definable subsets of $S$. (If every parameter is picked from a fixed partition, Lemma \ref{lm_partition_bound} provides a uniform bound). This yields $\vc(\phi) \leq q$ as needed. The rest of the proof is stating this idea formally. + + First, we extend our collection of subdivisions $(\A_j, \B_j)_{j \in J}$ by the following singleton sets. For each $c_i \in S$ let $B_i = \{c_i\}$ and $A_i = T$ and add $(\A_i, \B_i)$ to our collection with $\LL_B$ the language of $B_i$ interpreted arbitrarily. We end up with a new collection $(\A_k, \B_k)_{k \in K}$ indexed by some $K$ with $|K| = |J| + |I|$ (we added $|S|$ new pairs). Now we have that $B_k$ partition $T$, so $T = \bigsqcup_{k \in K} B_k$ and $S = \bigcap_{j \in J} A_j = \bigcap_{k \in K} A_k$. For $(k_1, k_2, \ldots k_q) = \vec k \in K^q$ denote + \begin{align*} + B_{\vec k} = B_{k_1} \times B_{k_2} \times \ldots \times B_{k_q} + \end{align*} + Then we have the following identity + \begin{align*} + T^q = (\bigsqcup_{k \in K} B_k)^q = \bigsqcup_{\vec k \in K^q} B_{\vec k} + \end{align*} + Thus we have that $\{B_{\vec k}\}_{\vec k \in K^q}$ partition $T^q$. Compute + \begin{align*} + |\phi(S^p, T^q)| + &= \left|\bigcup_{\vec k \in K^q} \phi(S^p, B_{\vec k}) \right| \leq \\ + &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| + \end{align*} + We can bound $|\phi(S^p, B_{\vec k})|$ uniformly using Lemma \ref{lm_partition_bound}. $(\A_k, \B_k)_{k \in K}$ satisfies the requirements of the lemma and $B_{\vec k}$ looks like $B$ in the lemma after possibly permuting some variables in $\phi$. Applying the lemma we get + \begin{align*} + |\phi(S^p, B_{\vec k})| \leq N^q + \end{align*} + with $N$ only depending on $q$ and complexity of $\phi$. We complete our computation + \begin{align*} + |\phi(S^p, T^q)| + &\leq \sum_{\vec k \in K^q} |\phi(S^p, B_{\vec k})| \leq \\ + &\leq \sum_{\vec k \in K^q} N^q \leq \\ + &\leq |K^q| N^q \leq \\ + &\leq (|J| + |I|)^q N^q \leq \\ + &\leq (4pn + 1 + 2pn)^q N^q = N^q (6p + 1/n)^q n^q = O(n^q) + \end{align*} + \end{proof} + \begin{Corollary} + In the theory of infinite (colored) meet trees we have $vc(n) = n$ for all $n$. + \end{Corollary} + We get the general result for trees that aren't necessarily meet trees via an easy application of interpretability. + \begin{Corollary} + In the theory of infinite (colored) trees we have $vc(n) = n$ for all $n$. + \end{Corollary} + \begin{proof} + Let $\TT'$ be a tree. We can embed it in a larger tree that is closed under meets $\TT' \subset \TT$. Expand $\TT$ by an extra color and interpret it by coloring the subset $\TT'$. Thus we can interpret $\TT'$ in $T^1$. By Corollary 3.17 in \cite{vc_density} we get that $\vc^{\TT'}(n) \leq \vc^T(1 \cdot n) = n$ thus $\vc^{\TT'}(n) = n$ as well. + \end{proof} + + \begin{thebibliography}{9} + +\bibitem{vc_density} + M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, + \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) + +\bibitem{simon_dp_min} + P. Simon, + \textit{On dp-minimal ordered structures}, + J. Symbolic Logic 76 (2011), no. 2, 448-460 + +\bibitem{parigot_trees} + Michel Parigot. + Th\'eories d'arbres. + \textit{Journal of Symbolic Logic}, 47, 1982. + + +\end{thebibliography} + \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/Trees vc-density.tps b/research/02 Trees vc-density/Trees vc-density.tps deleted file mode 100644 index 25591a1a..00000000 --- a/research/02 Trees vc-density/Trees vc-density.tps +++ /dev/null @@ -1,62 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - -[Frame0] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=1071 -NormalPos.bottom=621 -Class=LaTeXView -Document=Trees vc-density.tex - -[Frame0_View0,0] -TopLine=231 -Cursor=11750 - -[Frame1] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=1071 -NormalPos.bottom=621 -Class=LaTeXView -Document=vc-trees-all_figures.tex - 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-("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsfonts\umsb.fd" -File: umsb.fd 2013/01/14 v3.01 AMS symbols B -) ABD: EveryShipout initializing macros [1] [2] [3] -[4] [5] - -LaTeX Warning: Reference `pew' on page 6 undefined on input line 346. - -[6] -Overfull \hbox (1.91988pt too wide) in paragraph at lines 353--353 -[] - [] - -[7] [8] [9] [10] [11] [12] ("C:\Users\Anton\SparkleShare\Research\research\02 Trees vc-density\Trees vc-density_2.aux") - -LaTeX Warning: There were undefined references. - - -LaTeX Warning: There were multiply-defined labels. - - ) -Here is how much of TeX's memory you used: - 10262 strings out of 493922 - 189250 string characters out of 3144899 - 239452 words of memory out of 3000000 - 13204 multiletter control sequences out of 15000+200000 - 13362 words of font info for 50 fonts, out of 3000000 for 9000 - 1002 hyphenation exceptions out of 8191 - 56i,14n,55p,809b,313s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on "Trees vc-density_2.dvi" (12 pages, 75920 bytes). diff --git a/research/02 Trees vc-density/tikz_2.tex b/research/02 Trees vc-density/tikz_2.tex index 350e0e6d..9a78529a 100644 --- a/research/02 Trees vc-density/tikz_2.tex +++ b/research/02 Trees vc-density/tikz_2.tex @@ -1,88 +1,88 @@ -\documentclass{amsart} - -\usepackage{tikz} - -\tikzstyle{node}=[circle, draw] - -\tikzstyle{up}=[node, fill = red] -\tikzstyle{c1}=[node, fill = black] -\tikzstyle{md}=[node, fill = orange] -\tikzstyle{c2}=[node, fill = black] -\tikzstyle{dn}=[node, fill = yellow] -\tikzstyle{ds}=[node, fill = blue] -\tikzstyle{ex}=[node, fill = green] -\tikzstyle{nd}=[rectangle, draw] - -\begin{document} - \begin{tikzpicture} - \node[up] {} - [level distance=20mm] - child[grow = south] {node[up] {} - child[grow = south west, level distance=10mm]{node[up]{} %left up - [sibling distance=5mm] - child{node[up]{}} - child{node[up]{}} - } - child{node[c1]{} %main up - [sibling distance=5mm] - child [grow = -120, level distance=10mm] {node[md]{} %1 - child[grow = -120]{node[md]{} %2 - child[grow = -150]{node[md]{} - child{node[md]{}} - child{node[md]{}} - } - child[grow = -120]{node[c2]{} %3 - [grow = south] - child{node[dn]{} - child{node[dn]{}} - } - child{node[dn]{} - child{node[dn]{} - child{node[dn]{}} - child{node[dn]{}} - } - } - } - } - child[grow = south]{node[md]{} - child[grow = -60]{node[md]{}} - } - } - child[grow = -30, level distance=10mm]{node[ds]{} %aux1 middle - [sibling distance=5mm] - child{node[ds]{}} - child{node[ds]{}} - } - child [grow = -60, level distance=10mm] {node[ds]{} %aux2 middle - [sibling distance=5mm] - child{node[ds]{}} - child{node[ds]{}} - } - } - child [grow = south east, level distance=10mm] {node[up]{} %right up - [sibling distance=5mm] - child{node[up]{}} - child{node[up]{}} - } - } - child[grow = east, level distance=40mm, white]{node[black, ex]{} - [grow = south] - [level distance=10mm] - child[black]{node[ex]{} - [sibling distance=5mm] - child{node[ex]{}} - child{node[ex]{}} - } - child[black] {node[ex]{} - [sibling distance=5mm] - child{node[ex]{}} - child{node[ex]{}} - } - } - ; - \draw [black, fill=cyan] (70mm,0) circle [radius=2mm]; - \node [right] at (72mm,0) {$A$}; - \draw [black, fill=cyan] (70mm,-10mm) circle [radius=2mm]; - \node [right] at (72mm,-10mm) {$B$}; - \end{tikzpicture} +\documentclass{amsart} + +\usepackage{tikz} + +\tikzstyle{node}=[circle, draw] + +\tikzstyle{up}=[node, fill = red] +\tikzstyle{c1}=[node, fill = black] +\tikzstyle{md}=[node, fill = orange] +\tikzstyle{c2}=[node, fill = black] +\tikzstyle{dn}=[node, fill = yellow] +\tikzstyle{ds}=[node, fill = blue] +\tikzstyle{ex}=[node, fill = green] +\tikzstyle{nd}=[rectangle, draw] + +\begin{document} + \begin{tikzpicture} + \node[up] {} + [level distance=20mm] + child[grow = south] {node[up] {} + child[grow = south west, level distance=10mm]{node[up]{} %left up + [sibling distance=5mm] + child{node[up]{}} + child{node[up]{}} + } + child{node[c1]{} %main up + [sibling distance=5mm] + child [grow = -120, level distance=10mm] {node[md]{} %1 + child[grow = -120]{node[md]{} %2 + child[grow = -150]{node[md]{} + child{node[md]{}} + child{node[md]{}} + } + child[grow = -120]{node[c2]{} %3 + [grow = south] + child{node[dn]{} + child{node[dn]{}} + } + child{node[dn]{} + child{node[dn]{} + child{node[dn]{}} + child{node[dn]{}} + } + } + } + } + child[grow = south]{node[md]{} + child[grow = -60]{node[md]{}} + } + } + child[grow = -30, level distance=10mm]{node[ds]{} %aux1 middle + [sibling distance=5mm] + child{node[ds]{}} + child{node[ds]{}} + } + child [grow = -60, level distance=10mm] {node[ds]{} %aux2 middle + [sibling distance=5mm] + child{node[ds]{}} + child{node[ds]{}} + } + } + child [grow = south east, level distance=10mm] {node[up]{} %right up + [sibling distance=5mm] + child{node[up]{}} + child{node[up]{}} + } + } + child[grow = east, level distance=40mm, white]{node[black, ex]{} + [grow = south] + [level distance=10mm] + child[black]{node[ex]{} + [sibling distance=5mm] + child{node[ex]{}} + child{node[ex]{}} + } + child[black] {node[ex]{} + [sibling distance=5mm] + child{node[ex]{}} + child{node[ex]{}} + } + } + ; + \draw [black, fill=cyan] (70mm,0) circle [radius=2mm]; + \node [right] at (72mm,0) {$A$}; + \draw [black, fill=cyan] (70mm,-10mm) circle [radius=2mm]; + \node [right] at (72mm,-10mm) {$B$}; + \end{tikzpicture} \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/tikz_figure 1.tex b/research/02 Trees vc-density/tikz_figure 1.tex index 2f215c7b..276ec784 100644 --- a/research/02 Trees vc-density/tikz_figure 1.tex +++ b/research/02 Trees vc-density/tikz_figure 1.tex @@ -1,78 +1,78 @@ -\documentclass{amsart} - -\usepackage{tikz} - -%\tikzstyle{every node}=[circle, draw, fill=black, -% inner sep=0pt, minimum width=2pt] - -%\tikzstyle{every node}=[circle, draw, fill=black] - -\tikzstyle{cr}=[radius=0.1] -\tikzstyle{ln}=[fill, dashed] - -\begin{document} - \begin{tikzpicture}[scale=.5] - %\path [fill=blue] (-10,-15) -- (0,0) -- (10,-15); - %\path [fill=orange] (3.3,-4) -- (6,0) -- (8.6,-4); - %\path [fill=brown] (9.3,-4) -- (12,0) -- (14.6,-4); - - \path [fill=lightgray] (-3.3,-15) -- (0,-5) -- (3.3, -15); - %\path [fill=red] (3.3,-15) -- (0,-5) -- (6.6,-15); - %\path [fill=cyan] (-3.3,-15) -- (0,-5) -- (-6.6,-15); - \path [fill=white] (0,-9) -- (-2,-15) -- (2, -15); - - \draw [gray] (0,0) -- (-10,-15); - \draw [gray] (0,0) -- (10, -15); - - %\draw [gray] (0,-5) -- (-6.6,-15); - %\draw [gray] (0,-5) -- (6.6, -15); - - %\draw [gray] (0,-5) -- (-3.3,-15); - %\draw [gray] (0,-5) -- (3.3, -15); - - %\draw [gray] (0,-9) -- (-2,-15); - %\draw [gray] (0,-9) -- (2, -15); - - \draw [gray] (3.3,-4) -- (6,0) -- (8.6,-4); - \draw [gray] (9.3,-4) -- (12,0) -- (14.6,-4); - - \draw[ln] (0,0) -- (0,-3) circle [cr] node{}; - \draw[ln] (0,-3) --(1,-4) circle [cr] node{}; - \draw[ln] (0,-3) --(-2,-5) circle [cr] node{}; - - \draw[ln] (0,-3) --(0,-5) circle [cr] node[right]{$c_1$}; - - \draw[ln] (0,-5) --(0,-9) circle [cr] node[right]{$c_2$}; - - \draw[ln] (0,-9) --(0,-11) circle [cr] node{}; - \draw[ln] (0,-11) --(-1,-14) circle [cr] node{}; - \draw[ln] (0,-11) --(1,-14) circle [cr] node{}; - - \draw[ln] (0,-5) --(2,-9) circle [cr] node{}; - \draw[ln] (2,-9) --(2,-10) circle [cr] node{}; - \draw[ln] (2,-9) --(3,-11) circle [cr] node{}; - - - \draw[ln] (6,0) -- (6,-2) circle [cr] node{}; - \draw[ln] (6,-2) --(7,-3.5) circle [cr] node{}; - \draw[ln] (6,-2) --(5,-3) circle [cr] node{}; - - \draw [very thick] (0,0) -- (0,-5); - \draw [thick] (0,0) -- (0,-9); - \node [left] at (0, -2) {$S_1$}; - \node [left] at (0, -7) {$S_B$}; - - \node at (1.5, -11) {$B$}; - - \node at (4.5, -11) {$A$}; - \node at (-4.5, -11) {$A$}; - %\node at (-.5, -2) {$A$}; - - \node at (0, -13) {$T_A$}; - - \node at (6,-3) {$A$}; - \node at (12,-2) {$A$}; - - - \end{tikzpicture} +\documentclass{amsart} + +\usepackage{tikz} + +%\tikzstyle{every node}=[circle, draw, fill=black, +% inner sep=0pt, minimum width=2pt] + +%\tikzstyle{every node}=[circle, draw, fill=black] + +\tikzstyle{cr}=[radius=0.1] +\tikzstyle{ln}=[fill, dashed] + +\begin{document} + \begin{tikzpicture}[scale=.5] + %\path [fill=blue] (-10,-15) -- (0,0) -- (10,-15); + %\path [fill=orange] (3.3,-4) -- (6,0) -- (8.6,-4); + %\path [fill=brown] (9.3,-4) -- (12,0) -- (14.6,-4); + + \path [fill=lightgray] (-3.3,-15) -- (0,-5) -- (3.3, -15); + %\path [fill=red] (3.3,-15) -- (0,-5) -- (6.6,-15); + %\path [fill=cyan] (-3.3,-15) -- (0,-5) -- (-6.6,-15); + \path [fill=white] (0,-9) -- (-2,-15) -- (2, -15); + + \draw [gray] (0,0) -- (-10,-15); + \draw [gray] (0,0) -- (10, -15); + + %\draw [gray] (0,-5) -- (-6.6,-15); + %\draw [gray] (0,-5) -- (6.6, -15); + + %\draw [gray] (0,-5) -- (-3.3,-15); + %\draw [gray] (0,-5) -- (3.3, -15); + + %\draw [gray] (0,-9) -- (-2,-15); + %\draw [gray] (0,-9) -- (2, -15); + + \draw [gray] (3.3,-4) -- (6,0) -- (8.6,-4); + \draw [gray] (9.3,-4) -- (12,0) -- (14.6,-4); + + \draw[ln] (0,0) -- (0,-3) circle [cr] node{}; + \draw[ln] (0,-3) --(1,-4) circle [cr] node{}; + \draw[ln] (0,-3) --(-2,-5) circle [cr] node{}; + + \draw[ln] (0,-3) --(0,-5) circle [cr] node[right]{$c_1$}; + + \draw[ln] (0,-5) --(0,-9) circle [cr] node[right]{$c_2$}; + + \draw[ln] (0,-9) --(0,-11) circle [cr] node{}; + \draw[ln] (0,-11) --(-1,-14) circle [cr] node{}; + \draw[ln] (0,-11) --(1,-14) circle [cr] node{}; + + \draw[ln] (0,-5) --(2,-9) circle [cr] node{}; + \draw[ln] (2,-9) --(2,-10) circle [cr] node{}; + \draw[ln] (2,-9) --(3,-11) circle [cr] node{}; + + + \draw[ln] (6,0) -- (6,-2) circle [cr] node{}; + \draw[ln] (6,-2) --(7,-3.5) circle [cr] node{}; + \draw[ln] (6,-2) --(5,-3) circle [cr] node{}; + + \draw [very thick] (0,0) -- (0,-5); + \draw [thick] (0,0) -- (0,-9); + \node [left] at (0, -2) {$S_1$}; + \node [left] at (0, -7) {$S_B$}; + + \node at (1.5, -11) {$B$}; + + \node at (4.5, -11) {$A$}; + \node at (-4.5, -11) {$A$}; + %\node at (-.5, -2) {$A$}; + + \node at (0, -13) {$T_A$}; + + \node at (6,-3) {$A$}; + \node at (12,-2) {$A$}; + + + \end{tikzpicture} \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/tikz_stuff.tex b/research/02 Trees vc-density/tikz_stuff.tex index 90781539..eb0c78d6 100644 --- a/research/02 Trees vc-density/tikz_stuff.tex +++ b/research/02 Trees vc-density/tikz_stuff.tex @@ -1,59 +1,59 @@ -\documentclass{amsart} - -\usepackage{tikz} - -\begin{document} - \begin{tikzpicture}[scale=.5] - \path [fill=blue] (-10,-15) -- (0,0) -- (10,-15); - \path [fill=orange] (3.3,-4) -- (6,0) -- (8.6,-4); - \path [fill=brown] (9.3,-4) -- (12,0) -- (14.6,-4); - - \path [fill=green] (-3.3,-15) -- (0,-5) -- (3.3, -15); - \path [fill=red] (3.3,-15) -- (0,-5) -- (6.6,-15); - \path [fill=cyan] (-3.3,-15) -- (0,-5) -- (-6.6,-15); - \path [fill=yellow] (0,-9) -- (-2,-15) -- (2, -15); - - %\draw [gray] (0,0) -- (-10,-15); - %\draw [gray] (0,0) -- (10, -15); - - %\draw [gray] (0,-5) -- (-6.6,-15); - %\draw [gray] (0,-5) -- (6.6, -15); - - %\draw [gray] (0,-5) -- (-3.3,-15); - %\draw [gray] (0,-5) -- (3.3, -15); - - %\draw [gray] (0,-9) -- (-2,-15); - %\draw [gray] (0,-9) -- (2, -15); - - - \draw (0,0) -- (0,-3) node{}; - \draw (0,-3) --(1,-4) node{}; - \draw (0,-3) --(-2,-5) node{}; - - \draw (0,0) --(0,-5) node{}; - - \draw (0,-5) --(0,-9) node{}; - - \draw (0,-9) --(0,-11) node{}; - \draw (0,-11) --(-1,-14) node{}; - \draw (0,-11) --(1,-14) node{}; - - \draw (0,-5) --(2,-9) node{}; - \draw (2,-9) --(2,-10) node{}; - \draw (2,-9) --(3,-11) node{}; - - - \draw (6,0) -- (6,-2) node{}; - \draw (6,-2) --(7,-3.5) node{}; - \draw (6,-2) --(5,-3) node{}; - \end{tikzpicture} - - \begin{tikzpicture} - \node {root} - child {node {left}} - child {node {right} - child {node {child}} - child {node {child}} - }; - \end{tikzpicture} +\documentclass{amsart} + +\usepackage{tikz} + +\begin{document} + \begin{tikzpicture}[scale=.5] + \path [fill=blue] (-10,-15) -- (0,0) -- (10,-15); + \path [fill=orange] (3.3,-4) -- (6,0) -- (8.6,-4); + \path [fill=brown] (9.3,-4) -- (12,0) -- (14.6,-4); + + \path [fill=green] (-3.3,-15) -- (0,-5) -- (3.3, -15); + \path [fill=red] (3.3,-15) -- (0,-5) -- (6.6,-15); + \path [fill=cyan] (-3.3,-15) -- (0,-5) -- (-6.6,-15); + \path [fill=yellow] (0,-9) -- (-2,-15) -- (2, -15); + + %\draw [gray] (0,0) -- (-10,-15); + %\draw [gray] (0,0) -- (10, -15); + + %\draw [gray] (0,-5) -- (-6.6,-15); + %\draw [gray] (0,-5) -- (6.6, -15); + + %\draw [gray] (0,-5) -- (-3.3,-15); + %\draw [gray] (0,-5) -- (3.3, -15); + + %\draw [gray] (0,-9) -- (-2,-15); + %\draw [gray] (0,-9) -- (2, -15); + + + \draw (0,0) -- (0,-3) node{}; + \draw (0,-3) --(1,-4) node{}; + \draw (0,-3) --(-2,-5) node{}; + + \draw (0,0) --(0,-5) node{}; + + \draw (0,-5) --(0,-9) node{}; + + \draw (0,-9) --(0,-11) node{}; + \draw (0,-11) --(-1,-14) node{}; + \draw (0,-11) --(1,-14) node{}; + + \draw (0,-5) --(2,-9) node{}; + \draw (2,-9) --(2,-10) node{}; + \draw (2,-9) --(3,-11) node{}; + + + \draw (6,0) -- (6,-2) node{}; + \draw (6,-2) --(7,-3.5) node{}; + \draw (6,-2) --(5,-3) node{}; + \end{tikzpicture} + + \begin{tikzpicture} + \node {root} + child {node {left}} + child {node {right} + child {node {child}} + child {node {child}} + }; + \end{tikzpicture} \end{document} \ No newline at end of file diff --git a/research/02 Trees vc-density/vc-trees-all_figures.tex b/research/02 Trees vc-density/vc-trees-all_figures.tex index 44877a36..7f80340f 100644 --- a/research/02 Trees vc-density/vc-trees-all_figures.tex +++ b/research/02 Trees vc-density/vc-trees-all_figures.tex @@ -1,58 +1,58 @@ - - \tikzstyle{node}=[circle, draw] - - \tikzstyle{up}=[node, fill = white] - \tikzstyle{c1}=[node, fill = white] - \tikzstyle{md}=[node, fill = lightgray] - \tikzstyle{c2}=[node, fill = white] - \tikzstyle{dn}=[node, fill = white] - \tikzstyle{ds}=[node, fill = white] - \tikzstyle{ex}=[node, fill = white] - \tikzstyle{nd}=[rectangle, draw] - - \begin{figure}[p] - \input {vc-trees-fig_1} - \caption{Proper subdivision for $(A, B) = (A^{c_1}_{c_2}, B^{c_1}_{c_2})$} - \end{figure} - - \tikzstyle{up}=[node, fill = lightgray] - \tikzstyle{c1}=[node, fill = white] - \tikzstyle{md}=[node, fill = white] - \tikzstyle{c2}=[node, fill = white] - \tikzstyle{dn}=[node, fill = white] - \tikzstyle{ds}=[node, fill = white] - \tikzstyle{ex}=[node, fill = white] - \tikzstyle{nd}=[rectangle, draw] - - \begin{figure}[p] - \input {vc-trees-fig_1} - \caption{Proper subdivision for $(A, B) = (A_{c_1}, B_{c_1})$} - \end{figure} - - \tikzstyle{up}=[node, fill = white] - \tikzstyle{c1}=[node, fill = white] - \tikzstyle{md}=[node, fill = white] - \tikzstyle{c2}=[node, fill = white] - \tikzstyle{dn}=[node, fill = white] - \tikzstyle{ds}=[node, fill = lightgray] - \tikzstyle{ex}=[node, fill = white] - \tikzstyle{nd}=[rectangle, draw] - - \begin{figure}[p] - \input {vc-trees-fig_1} - \caption{Proper subdivision for $(A, B) = (A^{c_1}_G, B^{c_1}_G)$ for $G = \{c_2\}$} - \end{figure} - - \tikzstyle{up}=[node, fill = white] - \tikzstyle{c1}=[node, fill = white] - \tikzstyle{md}=[node, fill = white] - \tikzstyle{c2}=[node, fill = white] - \tikzstyle{dn}=[node, fill = white] - \tikzstyle{ds}=[node, fill = white] - \tikzstyle{ex}=[node, fill = lightgray] - \tikzstyle{nd}=[rectangle, draw] - - \begin{figure}[p] - \input {vc-trees-fig_1} - \caption{Proper subdivision for $(A, B) = (A_G, B_G)$ for $G = \{c_1, c_2\}$} - \end{figure} + + \tikzstyle{node}=[circle, draw] + + \tikzstyle{up}=[node, fill = white] + \tikzstyle{c1}=[node, fill = white] + \tikzstyle{md}=[node, fill = lightgray] + \tikzstyle{c2}=[node, fill = white] + \tikzstyle{dn}=[node, fill = white] + \tikzstyle{ds}=[node, fill = white] + \tikzstyle{ex}=[node, fill = white] + \tikzstyle{nd}=[rectangle, draw] + + \begin{figure}[p] + \input {vc-trees-fig_1} + \caption{Proper subdivision for $(A, B) = (A^{c_1}_{c_2}, B^{c_1}_{c_2})$} + \end{figure} + + \tikzstyle{up}=[node, fill = lightgray] + \tikzstyle{c1}=[node, fill = white] + \tikzstyle{md}=[node, fill = white] + \tikzstyle{c2}=[node, fill = white] + \tikzstyle{dn}=[node, fill = white] + \tikzstyle{ds}=[node, fill = white] + \tikzstyle{ex}=[node, fill = white] + \tikzstyle{nd}=[rectangle, draw] + + \begin{figure}[p] + \input {vc-trees-fig_1} + \caption{Proper subdivision for $(A, B) = (A_{c_1}, B_{c_1})$} + \end{figure} + + \tikzstyle{up}=[node, fill = white] + \tikzstyle{c1}=[node, fill = white] + \tikzstyle{md}=[node, fill = white] + \tikzstyle{c2}=[node, fill = white] + \tikzstyle{dn}=[node, fill = white] + \tikzstyle{ds}=[node, fill = lightgray] + \tikzstyle{ex}=[node, fill = white] + \tikzstyle{nd}=[rectangle, draw] + + \begin{figure}[p] + \input {vc-trees-fig_1} + \caption{Proper subdivision for $(A, B) = (A^{c_1}_G, B^{c_1}_G)$ for $G = \{c_2\}$} + \end{figure} + + \tikzstyle{up}=[node, fill = white] + \tikzstyle{c1}=[node, fill = white] + \tikzstyle{md}=[node, fill = white] + \tikzstyle{c2}=[node, fill = white] + \tikzstyle{dn}=[node, fill = white] + \tikzstyle{ds}=[node, fill = white] + \tikzstyle{ex}=[node, fill = lightgray] + \tikzstyle{nd}=[rectangle, draw] + + \begin{figure}[p] + \input {vc-trees-fig_1} + \caption{Proper subdivision for $(A, B) = (A_G, B_G)$ for $G = \{c_1, c_2\}$} + \end{figure} diff --git a/research/02 Trees vc-density/vc-trees-fig_1.log b/research/02 Trees vc-density/vc-trees-fig_1.log deleted file mode 100644 index f185e6db..00000000 --- a/research/02 Trees vc-density/vc-trees-fig_1.log +++ /dev/null @@ -1,1192 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.9.3) 25 JUL 2014 12:29 -entering extended mode -**vc-trees-fig_1.tex -("C:\Users\Anton\SparkleShare\Research\research\02 Trees vc-density\vc-trees-fig_1.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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-Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no ] in font nullfont! -LaTeX Font Info: External font `cmex10' loaded for size -(Font) <7> on input line 9. -LaTeX Font Info: External font `cmex10' loaded for size -(Font) <5> on input line 9. -Missing character: There is no [ in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no - in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no 2 in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 6 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 3 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 6 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no w in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no r in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no , in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no v in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 1 in font nullfont! -Missing character: There is no 0 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no o in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no u in font nullfont! -Missing character: There is no p in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no [ in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no b in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no g in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no s in font nullfont! -Missing character: There is no t in font nullfont! -Missing character: There is no a in font nullfont! -Missing character: There is no n in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no e in font nullfont! -Missing character: There is no = in font nullfont! -Missing character: There is no 5 in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no m in font nullfont! -Missing character: There is no ] in font nullfont! -Missing character: There is no c in font nullfont! -Missing character: There is no h in font nullfont! -Missing character: There is no i in font nullfont! -Missing character: There is no l in font nullfont! -Missing character: There is no d in font nullfont! -Missing character: There is no n in font nullfont! 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LaTeX Error: \begin{document} ended by \end{tikzpicture}. - -See the LaTeX manual or LaTeX Companion for explanation. -Type H for immediate help. - ... - -l.66 \end{tikzpicture} - -Your command was ignored. -Type I to replace it with another command, -or to continue without it. - -) -! Emergency stop. -<*> vc-trees-fig_1.tex - -*** (job aborted, no legal \end found) - - -Here is how much of TeX's memory you used: - 11 strings out of 493921 - 373 string characters out of 3144877 - 49031 words of memory out of 3000000 - 3404 multiletter control sequences out of 15000+200000 - 3640 words of font info for 14 fonts, out of 3000000 for 9000 - 841 hyphenation exceptions out of 8191 - 18i,1n,12p,93b,64s stack positions out of 5000i,500n,10000p,200000b,50000s -! ==> Fatal error occurred, no output PDF file produced! diff --git a/research/02 Trees vc-density/vc-trees-fig_1.tex b/research/02 Trees vc-density/vc-trees-fig_1.tex index 4f0e9595..78fc65c1 100644 --- a/research/02 Trees vc-density/vc-trees-fig_1.tex +++ b/research/02 Trees vc-density/vc-trees-fig_1.tex @@ -1,66 +1,66 @@ -\begin{tikzpicture} - \node[up] {} - child[grow = north, level distance=10mm] {node[up] {} - child[grow = north west, level distance=10mm]{node[up]{} %left up - [sibling distance=5mm] - child{node[up]{}} - child{node[up]{}} - } - child[level distance=15mm]{node[c1]{$c_1$} %main up - [sibling distance=5mm] - child [grow = 120, level distance=12mm] {node[md]{} %1 - child[grow = 120]{node[md]{} %2 - child[grow = 150]{node[md]{} - child{node[md]{}} - child{node[md]{}} - } - child[grow = 120]{node[c2]{$c_2$} %3 - [grow = north] - child{node[dn]{}} - child{node[dn]{} - child{node[dn]{}} - child{node[dn]{}} - } - } - } - child[grow = north]{node[md]{} - child[grow = 60]{node[md]{}} - } - } - child[grow = 30, level distance=10mm]{node[ds]{} %aux1 middle - [sibling distance=5mm] - child{node[ds]{}} - child{node[ds]{}} - } - child [grow = 60, level distance=10mm] {node[ds]{} %aux2 middle - [sibling distance=5mm] - child{node[ds]{}} - child{node[ds]{}} - } - } - child [grow = north east, level distance=10mm] {node[up]{} %right up - [sibling distance=5mm] - child{node[up]{}} - child{node[up]{}} - } - } - child[grow = east, level distance=40mm, white]{node[black, ex]{} - [grow = north] - [level distance=10mm] - child[black]{node[ex]{} - [sibling distance=5mm] - child{node[ex]{}} - child{node[ex]{}} - } - child[black] {node[ex]{} - [sibling distance=5mm] - child{node[ex]{}} - child{node[ex]{}} - } - } - ; - \draw [black, fill=white] (70mm, 10mm) circle [radius=2mm]; - \node [right] at (72mm, 10mm) {$A$}; - \draw [black, fill=lightgray] (70mm, 0) circle [radius=2mm]; - \node [right] at (72mm, 0) {$B$}; -\end{tikzpicture} +\begin{tikzpicture} + \node[up] {} + child[grow = north, level distance=10mm] {node[up] {} + child[grow = north west, level distance=10mm]{node[up]{} %left up + [sibling distance=5mm] + child{node[up]{}} + child{node[up]{}} + } + child[level distance=15mm]{node[c1]{$c_1$} %main up + [sibling distance=5mm] + child [grow = 120, level distance=12mm] {node[md]{} %1 + child[grow = 120]{node[md]{} %2 + child[grow = 150]{node[md]{} + child{node[md]{}} + child{node[md]{}} + } + child[grow = 120]{node[c2]{$c_2$} %3 + [grow = north] + child{node[dn]{}} + child{node[dn]{} + child{node[dn]{}} + child{node[dn]{}} + } + } + } + child[grow = north]{node[md]{} + child[grow = 60]{node[md]{}} + } + } + child[grow = 30, level distance=10mm]{node[ds]{} %aux1 middle + [sibling distance=5mm] + child{node[ds]{}} + child{node[ds]{}} + } + child [grow = 60, level distance=10mm] {node[ds]{} %aux2 middle + [sibling distance=5mm] + child{node[ds]{}} + child{node[ds]{}} + } + } + child [grow = north east, level distance=10mm] {node[up]{} %right up + [sibling distance=5mm] + child{node[up]{}} + child{node[up]{}} + } + } + child[grow = east, level distance=40mm, white]{node[black, ex]{} + [grow = north] + [level distance=10mm] + child[black]{node[ex]{} + [sibling distance=5mm] + child{node[ex]{}} + child{node[ex]{}} + } + child[black] {node[ex]{} + [sibling distance=5mm] + child{node[ex]{}} + child{node[ex]{}} + } + } + ; + \draw [black, fill=white] (70mm, 10mm) circle [radius=2mm]; + \node [right] at (72mm, 10mm) {$A$}; + \draw [black, fill=lightgray] (70mm, 0) circle [radius=2mm]; + \node [right] at (72mm, 0) {$B$}; +\end{tikzpicture} diff --git a/research/03 NFCP trees/NFCP trees.aux b/research/03 NFCP trees/NFCP trees.aux deleted file mode 100644 index 49492c14..00000000 --- a/research/03 NFCP trees/NFCP trees.aux +++ /dev/null @@ -1,13 +0,0 @@ -\relax -\citation{parigot_trees} -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{NFCP for stable trees}}{1}} -\newlabel{lm_tree_code}{{1.4}{1}} -\newlabel{lm_categoricity}{{1.5}{1}} -\bibcite{parigot_trees}{1} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\newlabel{th_tree_nfcp}{{1.7}{2}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{2}} diff --git a/research/03 NFCP trees/NFCP trees.bbl b/research/03 NFCP trees/NFCP trees.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/03 NFCP trees/NFCP trees.blg b/research/03 NFCP trees/NFCP trees.blg deleted file mode 100644 index cdcb8773..00000000 --- a/research/03 NFCP trees/NFCP trees.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: NFCP trees.aux -I found no \bibdata command---while reading file NFCP trees.aux -I found no \bibstyle command---while reading file NFCP trees.aux -(There were 2 error messages) diff --git a/research/03 NFCP trees/NFCP trees.log b/research/03 NFCP trees/NFCP trees.log deleted file mode 100644 index bbb00fdd..00000000 --- a/research/03 NFCP trees/NFCP trees.log +++ /dev/null @@ -1,424 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.10.19) 30 JAN 2014 18:37 -entering extended mode -**NFCP*trees.tex -("C:\Users\Anton\SparkleShare\Research\research\03 NFCP trees\NFCP trees.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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Rerun to get cross-references right. - - ) -Here is how much of TeX's memory you used: - 10230 strings out of 493922 - 189894 string characters out of 3144898 - 233322 words of memory out of 3000000 - 13181 multiletter control sequences out of 15000+200000 - 10215 words of font info for 40 fonts, out of 3000000 for 9000 - 1002 hyphenation exceptions out of 8191 - 56i,13n,55p,673b,232s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on "NFCP trees.dvi" (2 pages, 15924 bytes). diff --git a/research/03 NFCP trees/NFCP trees.tex b/research/03 NFCP trees/NFCP trees.tex index 2d786d18..8c5190d3 100644 --- a/research/03 NFCP trees/NFCP trees.tex +++ b/research/03 NFCP trees/NFCP trees.tex @@ -1,128 +1,128 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{tikz} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} - -\begin{document} - -\title{NFCP trees} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} -%more info - -\begin{abstract} - We investigate stable NFCP trees. -\end{abstract} - -\maketitle - -\section{NFCP for stable trees} -Parigot in \cite{parigot_trees} proves that stable trees are exactly trees of finite height. Here we study them further, stating equivalent conditions for having NFCP property. Final result of this section is theorem \ref{th_tree_nfcp} where we show that having NFCP is the same as being $\aleph_0$-categorical and also provide a simple graph theoretic characterization of NFCP. - -Recall that we say that a formula $\phi(x, y)$ has FCP if for arbitrarily large $n$ there are $a_1, \ldots, a_n$ such that for any $I \subsetneq [n]$ we have $\{\phi(x, a_i)\}_{i \in I}$ is consistent but $\{\phi(x, a_i)\}_{i \in [n]}$ is inconsistent. Theory is NFCP if no formula has FCP. - -Another notion that we consider is uniform boundedness. A formula is $\phi(x,y)$ uniformly bounded if there is $N$ such that for all $b$, $\phi(x, b)$ has either infinitely many realizations or less that $N$ many. Theory has uniform boundedness property if every formula does. - -In this section all the considered trees $T$ are stable. - -\begin{Definition} -Consider a tree $T$ and $n \leq m$ with $n \in \N, m \in \N \cup \{\infty\}$. Define $T \midr [n,m]$ to be the induced subtree of $T$ obtained restricting nodes at depth $[n,m]$. Call $T \midr [n, \infty]$ the \emph{$n$th-slice} of $T$. -\end{Definition} - -\begin{Definition} - Consider a tree $T$ of height $N$. Let $T_n$ denote the collection of connected components in the $n$th-slice of $T$. $T$ is called \emph{almost finite} if for every $n \leq N$, $T_n$ contains finitely many elements up to a (poset) isomorphism. -\end{Definition} - -\begin{Definition} - Given an element $t \in T$ in a tree, let $A_t = \{a \in T \mid a \geq t\}$ denote all the elements below $t$. -\end{Definition} - -\begin{Definition} \label{df_coding} - Given $T$ suppose $T_{n+1}$ is finite. We associate a coding sequence to every node at depth $n$. - Denote all elments of $T_{n+1}$ as $C_1, \ldots, C_N$. - Fix a node $t$ at depth $n$ and let $S$ be all the nodes at depth $n+1$ that are below $t$. - Let $c_i \in \N \cup \{\infty\}$ be the number of elements $s \in S$ such that $A_s$ has the isomorphism class $C_i$. - We code $t$ by sequence $\langle s_1, s_2, \ldots, s_N \rangle$ -\end{Definition} - - -\begin{Lemma} \label{lm_tree_code} - Suppose we have a countable tree $T$ with $T_{n+1}$ finite. Two nodes $t,s$ at depth $n$ have the same code if and only if $A_t$ is isomorphic to $A_s$. -\end{Lemma} - -\begin{proof} - Clear. -\end{proof} - -\begin{Lemma} \label{lm_categoricity} - Suppose $T$ is almost finite and countable. Then all the non-isomorphic connected components have different theories. -\end{Lemma} - -\begin{Note} - For any sentence $\phi$ we can easily construct a formula $\psi(x)$ such that - \begin{align*} - T \models \psi(t) \iff A_t \models \phi - \end{align*} -\end{Note} - -\begin{proof} - Suppose the tree has depth $N$. We prove the statement by induction for $T_n$ with $n = 0..N$. Induction starts at $T_N$ and finishes at $T_0 = T$. For $T_N$ the statement is trivial as in $T_N$ all components are single nodes, so there is a unique isomorphism class. Now, suppose the statement holds for $T_{n+1}$. Label isomorphism classes in $T_{n+1}$ as $C_1, \ldots, C_N$ and pick distinguishing sentences $\phi_1, \phi_2, \ldots, \phi_N$. That is, for a node $t$ at depth $n+1$ we have - \begin{align*} - A_t \models \phi_i \iff A_t \text{ has isomorphism class } C_i - \end{align*} - %We can easily construct formulas $\psi_1(x), \psi_2(x), \ldots, \psi_N(x)$ such that - %\begin{align*} - % A_t \models \phi_i \iff T \models \psi_i(t) - %\end{align*} - Now pick two nodes $a,b$ at depth $n$ in different isomorphism classes. By Lemma \ref{lm_tree_code} they have different codes. That is there is $i \in [1..N]$ such that $a,b$ have different numbers cones with isomorphism class $C_i$, say $n_a \neq n_b$. Without loss of generality $n_a < n_b$. Property - \begin{align*} - %P(t) \iff \{s \text{ node under $t$} \midr T \models \psi_i(s)\} \text { has $n_a$ elements} - P(t) \iff \{s \text{ node under $t$} \midr A_s \models \phi_i\} \text { has $n_a$ elements} - \end{align*} - is first-order definable (see Note above). This shows that $A_a, A_b$ have different theories. -\end{proof} - -\begin{Theorem} \label{th_tree_nfcp} - Let $T$ be a stable tree of height $N$. The following are equivalent: - \begin{enumerate} - \item $T$ is $\aleph_0$-categorical - \item $T$ has NFCP - \item $T$ is almost finite - \item Every element $t \in T$ has finitely many child cones up to a (poset) isomorphism and $T_0$ is finite. - \end{enumerate} -\end{Theorem} - -Note that conditions (3) and (4) are purely combinatorial. - -\begin{proof} - $(1) \Rightarrow (2)$. This is true in arbitrary theories. For a given arity $\aleph_0$-categorical theories have finitely many types, all of them isolated. This means that for a given formula $\phi(x, y)$ we can list all the types of $y$ for which there are infinitely many $x$ such that $\phi(x, y)$ holds. There are finitely many such types and all of them are isolated, so we can code it by single formula - disjunction of isolating formulas. - - $(2) \Rightarrow (3)$. We prove this by contrapositive. Suppose $T$ is not almost finite. Pick largest $n$ such that $T_n$ has infinitely many isomorphism classes. Thus $T_{n+1}$ is almost finite, thus we can code elements at depth $n$ (see Definition \ref{df_coding}). There are infinitely isomorphism classes in $T_n$, so there are infinitely many possible codes. As every code is a finite sequence, there has to be a position in the sequence that takes infinitely many possible values. More precisely, there is an isomorphism class $C$ such that $|N(t)|$ can take arbitrarily large values where - \begin{align*} - N(t) = \{s \text{ node under $t$} \midr A_s \text { has isomorphism class $C$} \} - \end{align*} - Lemma \ref{lm_categoricity} applied to $T_{n+1}$ shows that $N(t)$ is first order definable, i.e. there is formula $\phi(x, y)$ such that $\phi(T, t) = N(t)$. This formula witnesses failure of NFTP. - - $(3) \Rightarrow (1)$ This is exactly Lemma \ref{lm_tree_code} % not really - - $(3) \Leftrightarrow (4)$ Forward direction is trivial. Converse we prove this by contrapositive. Suppose $T$ is not almost finite. Pick the largest $n$ such that $T_n$ is infinite. Then $T_{n+1}$ is almost finite. If $n=0$ this contradicts $T_0$ being finite. Otherwise there has to be a node at depth $n-1$ that has infinitely many non-isomorphic nodes under it. -\end{proof} - -\begin{thebibliography}{9} - -\bibitem{parigot_trees} - Michel Parigot. - Th\'eories d'arbres. - \textit{Journal of Symbolic Logic}, 47, 1982. - -\end{thebibliography} - - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{tikz} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} + +\begin{document} + +\title{NFCP trees} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} +%more info + +\begin{abstract} + We investigate stable NFCP trees. +\end{abstract} + +\maketitle + +\section{NFCP for stable trees} +Parigot in \cite{parigot_trees} proves that stable trees are exactly trees of finite height. Here we study them further, stating equivalent conditions for having NFCP property. Final result of this section is theorem \ref{th_tree_nfcp} where we show that having NFCP is the same as being $\aleph_0$-categorical and also provide a simple graph theoretic characterization of NFCP. + +Recall that we say that a formula $\phi(x, y)$ has FCP if for arbitrarily large $n$ there are $a_1, \ldots, a_n$ such that for any $I \subsetneq [n]$ we have $\{\phi(x, a_i)\}_{i \in I}$ is consistent but $\{\phi(x, a_i)\}_{i \in [n]}$ is inconsistent. Theory is NFCP if no formula has FCP. + +Another notion that we consider is uniform boundedness. A formula is $\phi(x,y)$ uniformly bounded if there is $N$ such that for all $b$, $\phi(x, b)$ has either infinitely many realizations or less that $N$ many. Theory has uniform boundedness property if every formula does. + +In this section all the considered trees $T$ are stable. + +\begin{Definition} +Consider a tree $T$ and $n \leq m$ with $n \in \N, m \in \N \cup \{\infty\}$. Define $T \midr [n,m]$ to be the induced subtree of $T$ obtained restricting nodes at depth $[n,m]$. Call $T \midr [n, \infty]$ the \emph{$n$th-slice} of $T$. +\end{Definition} + +\begin{Definition} + Consider a tree $T$ of height $N$. Let $T_n$ denote the collection of connected components in the $n$th-slice of $T$. $T$ is called \emph{almost finite} if for every $n \leq N$, $T_n$ contains finitely many elements up to a (poset) isomorphism. +\end{Definition} + +\begin{Definition} + Given an element $t \in T$ in a tree, let $A_t = \{a \in T \mid a \geq t\}$ denote all the elements below $t$. +\end{Definition} + +\begin{Definition} \label{df_coding} + Given $T$ suppose $T_{n+1}$ is finite. We associate a coding sequence to every node at depth $n$. + Denote all elments of $T_{n+1}$ as $C_1, \ldots, C_N$. + Fix a node $t$ at depth $n$ and let $S$ be all the nodes at depth $n+1$ that are below $t$. + Let $c_i \in \N \cup \{\infty\}$ be the number of elements $s \in S$ such that $A_s$ has the isomorphism class $C_i$. + We code $t$ by sequence $\langle s_1, s_2, \ldots, s_N \rangle$ +\end{Definition} + + +\begin{Lemma} \label{lm_tree_code} + Suppose we have a countable tree $T$ with $T_{n+1}$ finite. Two nodes $t,s$ at depth $n$ have the same code if and only if $A_t$ is isomorphic to $A_s$. +\end{Lemma} + +\begin{proof} + Clear. +\end{proof} + +\begin{Lemma} \label{lm_categoricity} + Suppose $T$ is almost finite and countable. Then all the non-isomorphic connected components have different theories. +\end{Lemma} + +\begin{Note} + For any sentence $\phi$ we can easily construct a formula $\psi(x)$ such that + \begin{align*} + T \models \psi(t) \iff A_t \models \phi + \end{align*} +\end{Note} + +\begin{proof} + Suppose the tree has depth $N$. We prove the statement by induction for $T_n$ with $n = 0..N$. Induction starts at $T_N$ and finishes at $T_0 = T$. For $T_N$ the statement is trivial as in $T_N$ all components are single nodes, so there is a unique isomorphism class. Now, suppose the statement holds for $T_{n+1}$. Label isomorphism classes in $T_{n+1}$ as $C_1, \ldots, C_N$ and pick distinguishing sentences $\phi_1, \phi_2, \ldots, \phi_N$. That is, for a node $t$ at depth $n+1$ we have + \begin{align*} + A_t \models \phi_i \iff A_t \text{ has isomorphism class } C_i + \end{align*} + %We can easily construct formulas $\psi_1(x), \psi_2(x), \ldots, \psi_N(x)$ such that + %\begin{align*} + % A_t \models \phi_i \iff T \models \psi_i(t) + %\end{align*} + Now pick two nodes $a,b$ at depth $n$ in different isomorphism classes. By Lemma \ref{lm_tree_code} they have different codes. That is there is $i \in [1..N]$ such that $a,b$ have different numbers cones with isomorphism class $C_i$, say $n_a \neq n_b$. Without loss of generality $n_a < n_b$. Property + \begin{align*} + %P(t) \iff \{s \text{ node under $t$} \midr T \models \psi_i(s)\} \text { has $n_a$ elements} + P(t) \iff \{s \text{ node under $t$} \midr A_s \models \phi_i\} \text { has $n_a$ elements} + \end{align*} + is first-order definable (see Note above). This shows that $A_a, A_b$ have different theories. +\end{proof} + +\begin{Theorem} \label{th_tree_nfcp} + Let $T$ be a stable tree of height $N$. The following are equivalent: + \begin{enumerate} + \item $T$ is $\aleph_0$-categorical + \item $T$ has NFCP + \item $T$ is almost finite + \item Every element $t \in T$ has finitely many child cones up to a (poset) isomorphism and $T_0$ is finite. + \end{enumerate} +\end{Theorem} + +Note that conditions (3) and (4) are purely combinatorial. + +\begin{proof} + $(1) \Rightarrow (2)$. This is true in arbitrary theories. For a given arity $\aleph_0$-categorical theories have finitely many types, all of them isolated. This means that for a given formula $\phi(x, y)$ we can list all the types of $y$ for which there are infinitely many $x$ such that $\phi(x, y)$ holds. There are finitely many such types and all of them are isolated, so we can code it by single formula - disjunction of isolating formulas. + + $(2) \Rightarrow (3)$. We prove this by contrapositive. Suppose $T$ is not almost finite. Pick largest $n$ such that $T_n$ has infinitely many isomorphism classes. Thus $T_{n+1}$ is almost finite, thus we can code elements at depth $n$ (see Definition \ref{df_coding}). There are infinitely isomorphism classes in $T_n$, so there are infinitely many possible codes. As every code is a finite sequence, there has to be a position in the sequence that takes infinitely many possible values. More precisely, there is an isomorphism class $C$ such that $|N(t)|$ can take arbitrarily large values where + \begin{align*} + N(t) = \{s \text{ node under $t$} \midr A_s \text { has isomorphism class $C$} \} + \end{align*} + Lemma \ref{lm_categoricity} applied to $T_{n+1}$ shows that $N(t)$ is first order definable, i.e. there is formula $\phi(x, y)$ such that $\phi(T, t) = N(t)$. This formula witnesses failure of NFTP. + + $(3) \Rightarrow (1)$ This is exactly Lemma \ref{lm_tree_code} % not really + + $(3) \Leftrightarrow (4)$ Forward direction is trivial. Converse we prove this by contrapositive. Suppose $T$ is not almost finite. Pick the largest $n$ such that $T_n$ is infinite. Then $T_{n+1}$ is almost finite. If $n=0$ this contradicts $T_0$ being finite. Otherwise there has to be a node at depth $n-1$ that has infinitely many non-isomorphic nodes under it. +\end{proof} + +\begin{thebibliography}{9} + +\bibitem{parigot_trees} + Michel Parigot. + Th\'eories d'arbres. + \textit{Journal of Symbolic Logic}, 47, 1982. + +\end{thebibliography} + + \end{document} \ No newline at end of file diff --git a/research/04 Parigots exposition/Parigots expo.aux b/research/04 Parigots exposition/Parigots expo.aux deleted file mode 100644 index 41c1bbb0..00000000 --- a/research/04 Parigots exposition/Parigots expo.aux +++ /dev/null @@ -1,7 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Definitions}}{1}} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/04 Parigots exposition/Parigots expo.bbl b/research/04 Parigots exposition/Parigots expo.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/04 Parigots exposition/Parigots expo.blg b/research/04 Parigots exposition/Parigots expo.blg deleted file mode 100644 index 5c5e8ac3..00000000 --- a/research/04 Parigots exposition/Parigots expo.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Parigots expo.aux -I found no \citation commands---while reading file Parigots expo.aux -I found no \bibdata command---while reading file Parigots expo.aux -I found no \bibstyle command---while reading file Parigots expo.aux -(There were 3 error messages) diff --git a/research/04 Parigots exposition/Parigots expo.log b/research/04 Parigots exposition/Parigots expo.log deleted file mode 100644 index 52685a3b..00000000 --- a/research/04 Parigots exposition/Parigots expo.log +++ /dev/null @@ -1,415 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.8.30) 2 JAN 2014 11:20 -entering extended mode -**Parigots*expo.tex -("C:\Users\Anton\SparkleShare\Research\research\04 Parigots exposition\Parigots expo.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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-\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{tikz} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} - -\begin{document} - -\title{NIP and stable trees} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} -%more info - -\maketitle - -\section{Definitions} - -\begin{Definition} Work inside a model of a tree $M$ - \begin{enumerate} - \item Given a set $A$ denote its \emph{initial closure} - \begin{align*} - I(A) = \{b \mid \exists a \in A \ b \leq a\} - \end{align*} - \item \emph{Initial segements} are sets that are initially closed and linearly ordered (chain) - \item Given set $A$ define its \emph{initial part} (which is an initial segment) - \begin{align*} - S(A) = \{b \mid b < A\} = \{b \mid \forall a \in A \ b < a\} - \end{align*} - \item Given initial segment $S$ define its \emph{final part} (also known as the closed cone of $S$) - \begin{align*} - F(S) = \{b \mid b > A\} = \{b \mid \forall a \in A \ b > a\} - \end{align*} - \item Given initial segment $S$ define its \emph{trunk part} - \begin{align*} - T(S) = M - F(S) - \end{align*} - \item Connected component $B$ of some final part $F(S)$ is called \emph{branch} of $S$ (also known as an open cone of $S$) - \end{enumerate} - All those definitions are understood to be inside of $M$. If the underlying model is unclear from the context we will specify it with an underscript such as, for example, $F_M(S)$ or $S_M(A)$ -\end{Definition} - -\begin{Definition} - We would like to study how constructions in the previous definition behave under elementary extensions. Let $M \preccurlyeq M'$, $S \subset M$, an initial segment, $F = F_M(S)$, its final part and $B$ some branch in $F$. - \begin{enumerate} - \item Define $S^*$ the \emph{weak extension} of $S$ to be initial closure of $S$ in $M'$, i.e. $S^* = I_{M'}(S)$ - \item Define $S'$ the \emph{strong extension} of $S$ to be initial part of $F$ in $M'$, i.e. $S' = S_{M'}(F)$ - \item Define $B^*$ the \emph{weak extension} of $B$ to be \emph{final closure} of $B$ in $M'$ - \begin{align*} - B^* = \{m \in M' \mid \exists a \in B \ b \leq m\} - \end{align*} - \item Collection of all such $B^*$ is called \emph{weak branch collection} of $S$ and is denoted $F_M(S)^*$ - \begin{align*} - F_M(S)^* = \bigcup \{B^* \mid B \text{ is a branch of } S\} - \end{align*} - \item Define $B'$ the \emph{strong extension} of $B$ to be the connected component of $F_{M'}(S')$ containing $B$ - \item Collection of all such $B'$ is called \emph{strong branch collection} of $S$ and is denoted $F_M(S)'$ - \begin{align*} - F_M(S)^* = \bigcup \{B' \mid B \text{ is a branch of } S\} - \end{align*} - \item \emph{Complete branch collection} of $S$ is defined to be the final part of $S'$, i.e. $F_{M'}(S')$ - \end{enumerate} -\end{Definition} - -\begin{Note} - \begin{enumerate} Observe following containments. All of them can be strict - \item $S^* \subseteq S'$ - \item $B^* \subseteq B'$ - \item $F_M(S)^* \subseteq F_M(S)' \subseteq F_{M'}(S')$ - \end{enumerate} -\end{Note} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{tikz} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} + +\begin{document} + +\title{NIP and stable trees} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} +%more info + +\maketitle + +\section{Definitions} + +\begin{Definition} Work inside a model of a tree $M$ + \begin{enumerate} + \item Given a set $A$ denote its \emph{initial closure} + \begin{align*} + I(A) = \{b \mid \exists a \in A \ b \leq a\} + \end{align*} + \item \emph{Initial segements} are sets that are initially closed and linearly ordered (chain) + \item Given set $A$ define its \emph{initial part} (which is an initial segment) + \begin{align*} + S(A) = \{b \mid b < A\} = \{b \mid \forall a \in A \ b < a\} + \end{align*} + \item Given initial segment $S$ define its \emph{final part} (also known as the closed cone of $S$) + \begin{align*} + F(S) = \{b \mid b > A\} = \{b \mid \forall a \in A \ b > a\} + \end{align*} + \item Given initial segment $S$ define its \emph{trunk part} + \begin{align*} + T(S) = M - F(S) + \end{align*} + \item Connected component $B$ of some final part $F(S)$ is called \emph{branch} of $S$ (also known as an open cone of $S$) + \end{enumerate} + All those definitions are understood to be inside of $M$. If the underlying model is unclear from the context we will specify it with an underscript such as, for example, $F_M(S)$ or $S_M(A)$ +\end{Definition} + +\begin{Definition} + We would like to study how constructions in the previous definition behave under elementary extensions. Let $M \preccurlyeq M'$, $S \subset M$, an initial segment, $F = F_M(S)$, its final part and $B$ some branch in $F$. + \begin{enumerate} + \item Define $S^*$ the \emph{weak extension} of $S$ to be initial closure of $S$ in $M'$, i.e. $S^* = I_{M'}(S)$ + \item Define $S'$ the \emph{strong extension} of $S$ to be initial part of $F$ in $M'$, i.e. $S' = S_{M'}(F)$ + \item Define $B^*$ the \emph{weak extension} of $B$ to be \emph{final closure} of $B$ in $M'$ + \begin{align*} + B^* = \{m \in M' \mid \exists a \in B \ b \leq m\} + \end{align*} + \item Collection of all such $B^*$ is called \emph{weak branch collection} of $S$ and is denoted $F_M(S)^*$ + \begin{align*} + F_M(S)^* = \bigcup \{B^* \mid B \text{ is a branch of } S\} + \end{align*} + \item Define $B'$ the \emph{strong extension} of $B$ to be the connected component of $F_{M'}(S')$ containing $B$ + \item Collection of all such $B'$ is called \emph{strong branch collection} of $S$ and is denoted $F_M(S)'$ + \begin{align*} + F_M(S)^* = \bigcup \{B' \mid B \text{ is a branch of } S\} + \end{align*} + \item \emph{Complete branch collection} of $S$ is defined to be the final part of $S'$, i.e. $F_{M'}(S')$ + \end{enumerate} +\end{Definition} + +\begin{Note} + \begin{enumerate} Observe following containments. All of them can be strict + \item $S^* \subseteq S'$ + \item $B^* \subseteq B'$ + \item $F_M(S)^* \subseteq F_M(S)' \subseteq F_{M'}(S')$ + \end{enumerate} +\end{Note} + \end{document} \ No newline at end of file diff --git a/research/04 Parigots exposition/problem 2.aux b/research/04 Parigots exposition/problem 2.aux deleted file mode 100644 index 1dd6dfcf..00000000 --- a/research/04 Parigots exposition/problem 2.aux +++ /dev/null @@ -1,6 +0,0 @@ -\relax -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{0pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/04 Parigots exposition/problem 2.bbl b/research/04 Parigots exposition/problem 2.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/04 Parigots exposition/problem 2.blg b/research/04 Parigots exposition/problem 2.blg deleted file mode 100644 index f81c8a58..00000000 --- a/research/04 Parigots exposition/problem 2.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: problem 2.aux -I found no \citation commands---while reading file problem 2.aux -I found no \bibdata command---while reading file problem 2.aux -I found no \bibstyle command---while reading file problem 2.aux -(There were 3 error messages) diff --git a/research/04 Parigots exposition/problem 2.log b/research/04 Parigots exposition/problem 2.log deleted file mode 100644 index afffe344..00000000 --- a/research/04 Parigots exposition/problem 2.log +++ /dev/null @@ -1,161 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 19 JAN 2016 10:52 -entering extended mode -**problem*2.tex -("C:\Users\Anton\SparkleShare\Research\research\04 Parigots exposition\problem 2.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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1 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/research/04 Parigots exposition/problem 2.tex b/research/04 Parigots exposition/problem 2.tex index 4562b385..45c45bd9 100644 --- a/research/04 Parigots exposition/problem 2.tex +++ b/research/04 Parigots exposition/problem 2.tex @@ -1,73 +1,73 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} - -\begin{document} - - \begin{enumerate} - \item Show that if $A \sim B$ then $A^n \sim B^n$ - \item Show that - \begin{align*} - \begin{bmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 1 \\ - 0 & 0 & 0 - \end{bmatrix} - \nsim - \begin{bmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 0 \\ - 0 & 0 & 0 - \end{bmatrix} - \end{align*} - \item Show that - \begin{align*} - \begin{bmatrix} - 0 & 0 & 0 \\ - 0 & 0 & 1 \\ - 0 & 0 & 0 - \end{bmatrix} - \sim - \begin{bmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 0 \\ - 0 & 0 & 0 - \end{bmatrix} - \end{align*} - \item Show that - \begin{align*} - \begin{bmatrix} - 0 & 1 & 0 \\ - 0 & 0 & 1 \\ - 0 & 0 & 0 - \end{bmatrix} - \sim - \begin{bmatrix} - 0 & 1 & 1 \\ - 0 & 0 & 1 \\ - 0 & 0 & 0 - \end{bmatrix} - \end{align*} - \item Find $n$ non-similar $n\times n$ nilpotent matrices for arbitrary $n$. - \item Show that - \begin{align*} - \begin{bmatrix} - 1 & 1 & 0 \\ - 0 & 1 & 1 \\ - 0 & 0 & 1 - \end{bmatrix} - \nsim - \begin{bmatrix} - 1 & 1 & 0 \\ - 0 & 1 & 0 \\ - 0 & 0 & 1 - \end{bmatrix} - \end{align*} - \end{enumerate} +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} + +\begin{document} + + \begin{enumerate} + \item Show that if $A \sim B$ then $A^n \sim B^n$ + \item Show that + \begin{align*} + \begin{bmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 1 \\ + 0 & 0 & 0 + \end{bmatrix} + \nsim + \begin{bmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} + \end{align*} + \item Show that + \begin{align*} + \begin{bmatrix} + 0 & 0 & 0 \\ + 0 & 0 & 1 \\ + 0 & 0 & 0 + \end{bmatrix} + \sim + \begin{bmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 0 \\ + 0 & 0 & 0 + \end{bmatrix} + \end{align*} + \item Show that + \begin{align*} + \begin{bmatrix} + 0 & 1 & 0 \\ + 0 & 0 & 1 \\ + 0 & 0 & 0 + \end{bmatrix} + \sim + \begin{bmatrix} + 0 & 1 & 1 \\ + 0 & 0 & 1 \\ + 0 & 0 & 0 + \end{bmatrix} + \end{align*} + \item Find $n$ non-similar $n\times n$ nilpotent matrices for arbitrary $n$. + \item Show that + \begin{align*} + \begin{bmatrix} + 1 & 1 & 0 \\ + 0 & 1 & 1 \\ + 0 & 0 & 1 + \end{bmatrix} + \nsim + \begin{bmatrix} + 1 & 1 & 0 \\ + 0 & 1 & 0 \\ + 0 & 0 & 1 + \end{bmatrix} + \end{align*} + \end{enumerate} \end{document} \ No newline at end of file diff --git a/research/04 Parigots exposition/problem 3.aux b/research/04 Parigots exposition/problem 3.aux deleted file mode 100644 index 1dd6dfcf..00000000 --- a/research/04 Parigots exposition/problem 3.aux +++ /dev/null @@ -1,6 +0,0 @@ -\relax -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{0pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/04 Parigots exposition/problem 3.bbl b/research/04 Parigots exposition/problem 3.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/04 Parigots exposition/problem 3.blg b/research/04 Parigots exposition/problem 3.blg deleted file mode 100644 index ce30d565..00000000 --- a/research/04 Parigots exposition/problem 3.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: problem 3.aux -I found no \citation commands---while reading file problem 3.aux -I found no \bibdata command---while reading file problem 3.aux -I found no \bibstyle command---while reading file problem 3.aux -(There were 3 error messages) diff --git a/research/04 Parigots exposition/problem 3.log b/research/04 Parigots exposition/problem 3.log deleted file mode 100644 index 86f60201..00000000 --- a/research/04 Parigots exposition/problem 3.log +++ /dev/null @@ -1,161 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 26 JAN 2016 10:27 -entering extended mode -**problem*3.tex -("C:\Users\Anton\SparkleShare\Research\research\04 Parigots exposition\problem 3.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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1 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/research/04 Parigots exposition/problem 3.tex b/research/04 Parigots exposition/problem 3.tex index 8e2d9e7a..86568626 100644 --- a/research/04 Parigots exposition/problem 3.tex +++ b/research/04 Parigots exposition/problem 3.tex @@ -1,25 +1,25 @@ -\documentclass[12pt]{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} - -\begin{document} - - Let $T$ be a linear transformation $T: V \arr V$, that is $W$-invariant for some subspace $W$ - \begin{enumerate} - \item Suppose $T$ is invertible. Show that $(\bar T)^{-1} = \overline {T^{-1}} $. - \item Show that $T^t$ is $W^0$-invariant (don't assume invertible anymore). - \end{enumerate} - - Let $S$ be a linear transformation $S: V \arr W$ - - \begin{enumerate} - \item Show that if $S$ is injective then $S^t$ is surjective. - \item Show that $\ker S^t = (\im S)^0$. - \end{enumerate} +\documentclass[12pt]{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} + +\begin{document} + + Let $T$ be a linear transformation $T: V \arr V$, that is $W$-invariant for some subspace $W$ + \begin{enumerate} + \item Suppose $T$ is invertible. Show that $(\bar T)^{-1} = \overline {T^{-1}} $. + \item Show that $T^t$ is $W^0$-invariant (don't assume invertible anymore). + \end{enumerate} + + Let $S$ be a linear transformation $S: V \arr W$ + + \begin{enumerate} + \item Show that if $S$ is injective then $S^t$ is surjective. + \item Show that $\ker S^t = (\im S)^0$. + \end{enumerate} \end{document} \ No newline at end of file diff --git a/research/04 Parigots exposition/problem 4.aux b/research/04 Parigots exposition/problem 4.aux deleted file mode 100644 index 1dd6dfcf..00000000 --- a/research/04 Parigots exposition/problem 4.aux +++ /dev/null @@ -1,6 +0,0 @@ -\relax -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{0pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/04 Parigots exposition/problem 4.bbl b/research/04 Parigots exposition/problem 4.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/04 Parigots exposition/problem 4.blg b/research/04 Parigots exposition/problem 4.blg deleted file mode 100644 index c9ff234d..00000000 --- a/research/04 Parigots exposition/problem 4.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: problem 4.aux -I found no \citation commands---while reading file problem 4.aux -I found no \bibdata command---while reading file problem 4.aux -I found no \bibstyle command---while reading file problem 4.aux -(There were 3 error messages) diff --git a/research/04 Parigots exposition/problem 4.log b/research/04 Parigots exposition/problem 4.log deleted file mode 100644 index 021d5d1e..00000000 --- a/research/04 Parigots exposition/problem 4.log +++ /dev/null @@ -1,172 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 2 FEB 2016 10:49 -entering extended mode -**problem*4.tex -("C:\Users\Anton\SparkleShare\Research\research\04 Parigots exposition\problem 4.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - kannada, kurmanji, latin, latvian, lithuanian, malayalam, marathi, mongolian, mongolianlmc, monogreek, ngerman, ngerman --x-2012-05-30, nynorsk, oriya, panjabi, pinyin, polish, portuguese, romanian, russian, sanskrit, serbian, slovak, sloven -ian, spanish, swedish, swissgerman, tamil, telugu, turkish, turkmen, ukenglish, ukrainian, uppersorbian, usenglishmax, w -elsh, loaded. -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amscls\amsart.cls" -Document Class: amsart 2009/07/02 v2.20.1 -\linespacing=\dimen102 -\normalparindent=\dimen103 -\normaltopskip=\skip41 -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsmath\amsmath.sty" -Package: amsmath 2013/01/14 v2.14 AMS math features -\@mathmargin=\skip42 - -For additional information on amsmath, use the `?' option. -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsmath\amstext.sty" -Package: amstext 2000/06/29 v2.01 - -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsmath\amsgen.sty" -File: amsgen.sty 1999/11/30 v2.0 -\@emptytoks=\toks14 -\ex@=\dimen104 -)) -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsmath\amsbsy.sty" -Package: amsbsy 1999/11/29 v1.2d -\pmbraise@=\dimen105 -) -("C:\Program Files (x86)\MiKTeX 2.9\tex\latex\amsmath\amsopn.sty" -Package: amsopn 1999/12/14 v2.01 operator names -) -\inf@bad=\count79 -LaTeX Info: Redefining \frac on input line 210. -\uproot@=\count80 -\leftroot@=\count81 -LaTeX Info: Redefining \overline on input line 306. -\classnum@=\count82 -\DOTSCASE@=\count83 -LaTeX Info: Redefining \ldots on input line 378. -LaTeX Info: Redefining \dots on input line 381. -LaTeX Info: Redefining \cdots on input line 466. -\Mathstrutbox@=\box26 -\strutbox@=\box27 -\big@size=\dimen106 -LaTeX Font Info: Redeclaring font encoding OML on input line 566. -LaTeX Font Info: Redeclaring font encoding OMS on input line 567. -\macc@depth=\count84 -\c@MaxMatrixCols=\count85 -\dotsspace@=\muskip10 -\c@parentequation=\count86 -\dspbrk@lvl=\count87 -\tag@help=\toks15 -\row@=\count88 -\column@=\count89 -\maxfields@=\count90 -\andhelp@=\toks16 -\eqnshift@=\dimen107 -\alignsep@=\dimen108 -\tagshift@=\dimen109 -\tagwidth@=\dimen110 -\totwidth@=\dimen111 -\lineht@=\dimen112 -\@envbody=\toks17 -\multlinegap=\skip43 -\multlinetaggap=\skip44 -\mathdisplay@stack=\toks18 -LaTeX Info: Redefining \[ on input line 2665. -LaTeX Info: Redefining \] on input line 2666. -) -LaTeX Font Info: Try loading font information for U+msa on input line 388. - 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-Output written on "problem 4.pdf" (1 page, 62829 bytes). -PDF statistics: - 34 PDF objects out of 1000 (max. 8388607) - 0 named destinations out of 1000 (max. 500000) - 1 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/research/04 Parigots exposition/problem 4.tex b/research/04 Parigots exposition/problem 4.tex index f8b41397..5d819226 100644 --- a/research/04 Parigots exposition/problem 4.tex +++ b/research/04 Parigots exposition/problem 4.tex @@ -1,25 +1,25 @@ -\documentclass[12pt]{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\spn}{span} - -\begin{document} - - - \begin{enumerate} - \item Let $A \in M_n(F)$. Define $$V_A = \spn\{I, A, A^2, A^3, \ldots \} \subset M_n(F)$$ - Show that $\dim(V_A) = \deg(q_A)$ \\ - \item Now suppose that $A,B \in M_n(F)$ commute, that is $AB = BA$. Define - $$V_{A,B} = \spn\{A^i B^j \mid i,j \in \N\} \subset M_n(F)$$ - Show that $$\dim(V_{A,B}) \leq \dim(V_A) \dim(V_B)$$\\ - \item Using previous results show that - $$ \deg(q_{A+B}) \leq \deg(q_A) \deg(q_B) $$ - $$ \deg(q_{AB}) \leq \deg(q_A) \deg(q_B) $$ - \end{enumerate} - +\documentclass[12pt]{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\spn}{span} + +\begin{document} + + + \begin{enumerate} + \item Let $A \in M_n(F)$. Define $$V_A = \spn\{I, A, A^2, A^3, \ldots \} \subset M_n(F)$$ + Show that $\dim(V_A) = \deg(q_A)$ \\ + \item Now suppose that $A,B \in M_n(F)$ commute, that is $AB = BA$. Define + $$V_{A,B} = \spn\{A^i B^j \mid i,j \in \N\} \subset M_n(F)$$ + Show that $$\dim(V_{A,B}) \leq \dim(V_A) \dim(V_B)$$\\ + \item Using previous results show that + $$ \deg(q_{A+B}) \leq \deg(q_A) \deg(q_B) $$ + $$ \deg(q_{AB}) \leq \deg(q_A) \deg(q_B) $$ + \end{enumerate} + \end{document} \ No newline at end of file diff --git a/research/04 Parigots exposition/problem.aux b/research/04 Parigots exposition/problem.aux deleted file mode 100644 index 1dd6dfcf..00000000 --- a/research/04 Parigots exposition/problem.aux +++ /dev/null @@ -1,6 +0,0 @@ -\relax -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{0pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/04 Parigots exposition/problem.bbl b/research/04 Parigots exposition/problem.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/04 Parigots exposition/problem.blg b/research/04 Parigots exposition/problem.blg deleted file mode 100644 index 2a098f11..00000000 --- a/research/04 Parigots exposition/problem.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: problem.aux -I found no \citation commands---while reading file problem.aux -I found no \bibdata command---while reading file problem.aux -I found no \bibstyle command---while reading file problem.aux -(There were 3 error messages) diff --git a/research/04 Parigots exposition/problem.log b/research/04 Parigots exposition/problem.log deleted file mode 100644 index 9d6b2a34..00000000 --- a/research/04 Parigots exposition/problem.log +++ /dev/null @@ -1,160 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 12 JAN 2016 11:32 -entering extended mode -**problem.tex -("C:\Users\Anton\SparkleShare\Research\research\04 Parigots exposition\problem.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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1 words of extra memory for PDF output out of 10000 (max. 10000000) - diff --git a/research/04 Parigots exposition/problem.tex b/research/04 Parigots exposition/problem.tex index 9ffeaf81..4ee7b304 100644 --- a/research/04 Parigots exposition/problem.tex +++ b/research/04 Parigots exposition/problem.tex @@ -1,24 +1,24 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\DeclareMathOperator{\TT}{\boldface T} -\DeclareMathOperator{\A}{\boldface A} -\DeclareMathOperator{\B}{\boldface B} -\DeclareMathOperator{\PR}{P} - -\begin{document} - Let $S = (s_1, s_2, \ldots, s_n)$ be a collection of vectors in a vector space $V$. - Suppose we also have $W \subseteq V$ a subspace. - Denote $\bar S = (\bar s_1, \bar s_2, \ldots, \bar s_n)$ where $\bar s_i = s_i + W$. - So $\bar S$ is a collection of vectors in $V/W$. - \begin{enumerate} - \item Show that if $S$ spans $V$ then $\bar S$ spans $V/W$. - \item Give an example of $S, V, W$ where $\bar S$ spans $V/W$ but $S$ doesn't span $V$. - \item Is it true that if $S$ is linearly independent in $V$ then $\bar S$ is linearly independent in $V/W$? - Prove it or give a counterexample. - \item Is it true that if $\bar S$ is linearly independent in $V/W$ then $S$ is linearly independent in $V$? - Prove it or give a counterexample. - \end{enumerate} +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\DeclareMathOperator{\TT}{\boldface T} +\DeclareMathOperator{\A}{\boldface A} +\DeclareMathOperator{\B}{\boldface B} +\DeclareMathOperator{\PR}{P} + +\begin{document} + Let $S = (s_1, s_2, \ldots, s_n)$ be a collection of vectors in a vector space $V$. + Suppose we also have $W \subseteq V$ a subspace. + Denote $\bar S = (\bar s_1, \bar s_2, \ldots, \bar s_n)$ where $\bar s_i = s_i + W$. + So $\bar S$ is a collection of vectors in $V/W$. + \begin{enumerate} + \item Show that if $S$ spans $V$ then $\bar S$ spans $V/W$. + \item Give an example of $S, V, W$ where $\bar S$ spans $V/W$ but $S$ doesn't span $V$. + \item Is it true that if $S$ is linearly independent in $V$ then $\bar S$ is linearly independent in $V/W$? + Prove it or give a counterexample. + \item Is it true that if $\bar S$ is linearly independent in $V/W$ then $S$ is linearly independent in $V$? + Prove it or give a counterexample. + \end{enumerate} \end{document} \ No newline at end of file diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.aux b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.aux deleted file mode 100644 index df13b9be..00000000 --- a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.aux +++ /dev/null @@ -1,23 +0,0 @@ -\relax -\citation{stable_graphs} -\citation{infinite_megner} -\citation{stable_graphs} -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Preliminaries}}{1}} -\newlabel{cr_disjoint_paths}{{1.3}{1}} -\newlabel{cr_hull_finite}{{1.4}{1}} -\newlabel{th_superflat_equivalence}{{1.6}{2}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Indiscernible sequences}}{2}} -\newlabel{lm_bump}{{2.2}{2}} -\newlabel{lm_uniform}{{2.4}{2}} -\citation{simon_dp_minimal} -\newlabel{cr_bump}{{2.7}{3}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Superflat graphs are dp-minimal}}{3}} -\bibcite{stable_graphs}{1} -\bibcite{infinite_megner}{2} -\bibcite{simon_dp_minimal}{3} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{4}} diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.bbl b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.blg b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.blg deleted file mode 100644 index cd5acc8a..00000000 --- a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Flat graphs and dp-rank.aux -I found no \bibdata command---while reading file Flat graphs and dp-rank.aux -I found no \bibstyle command---while reading file Flat graphs and dp-rank.aux -(There were 2 error messages) diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.log b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.log deleted file mode 100644 index 86f01c6a..00000000 --- a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.log +++ /dev/null @@ -1,172 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=latex 2013.8.30) 27 JAN 2014 19:02 -entering extended mode -**Flat*graphs*and*dp-rank.tex -("C:\Users\Anton\SparkleShare\Research\research\06 Flat graphs and dp-rank\Flat graphs and dp-rank.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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Otherwise just continue, -and I'll forget about whatever was undefined. - -[2] [3] -Missing character: There is no â in font cmr8! -Missing character: There is no € in font cmr8! -Missing character: There is no “ in font cmr8! -Missing character: There is no â in font cmr8! -Missing character: There is no € in font cmr8! -Missing character: There is no “ in font cmr8! - [4] ("C:\Users\Anton\SparkleShare\Research\research\06 Flat graphs and dp-rank\Flat graphs and dp-rank.aux") ) -Here is how much of TeX's memory you used: - 1682 strings out of 493922 - 19007 string characters out of 3144899 - 73869 words of memory out of 3000000 - 4845 multiletter control sequences out of 15000+200000 - 10215 words of font info for 40 fonts, out of 3000000 for 9000 - 1002 hyphenation exceptions out of 8191 - 34i,10n,32p,541b,280s stack positions out of 5000i,500n,10000p,200000b,50000s - -Output written on "Flat graphs and dp-rank.dvi" (4 pages, 20380 bytes). diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.old_no_megner.tex b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.old_no_megner.tex index 18d6b151..0716c8db 100644 --- a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.old_no_megner.tex +++ b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.old_no_megner.tex @@ -1,187 +1,187 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - - -\begin{document} - -\title{Superflat graphs are dp-minimal} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We show that the theory of superflat graphs is dp-minimal. -\end{abstract} - -\maketitle - -\section{Preliminaries} - -Superflat graphs in model theoretic context were first introduced in \cite{stable_graphs} as a natural class of stable graphs. - -We work with an infinite graph $G$ and a subset of vertices $V \subset V(G)$. Say that $V$ is $n$-connected if there aren't a set of $n-1$ vertices removing which disconnects every pair of vertices in $V$. Connectivity of $V$ is the smallest $n$ such that $V$ are $n$-connected. - -\begin{Definition} - Suppose $V \subset V(G)$ has finite connectivity $n+1$. Let connectivity hull of $V$ to be union of all $n$-point sets that disconnect it. -\end{Definition} - -\begin{Note} - If $V$ has finite connectivity $n+1$, then having finite connectivity hull is the same as having finitely many $n$-point sets that disconnect it. -\end{Note} - -\begin{Definition} - Let $V \subset V(G)$. We say that a set is minimal with respect to $V$ when that set disconnects $V$ and is minimal among such sets (ordered by inclusion). -\end{Definition} - -\section{Connectivity hull is finite} - -Here we show our main technical lemma. This section is purely combinatorial, with no mention of model theory. - -\begin{Lemma} - Suppose $V = \{a,b\}$ is subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. -\end{Lemma} - -\begin{Corollary} - Suppose $V$ is a finite subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. -\end{Corollary} - -\begin{proof} - Fix set $P = {p_1, \ldots, p_m}$ of all unordered pairs from $V$. Every pair $p_i$ has connectivity $n_i$ and by previous lemma has finitely many sets of $n_i$ points that disconnect it, denoted by $S_i$. Consider a finite minimal set $D$ that disconnects $V$. Define $D_i$ to be one of the sets in $S_i$ such that $D_i \subset D$ (one always exists). Then $D = \cup_{i \in I} D_i$ (if it was larger that would contradict minimality). Thus minimal sets are determined by unions of sets in $\{S_1 \ldots S_m\}$ and there are finitely many of such possible sets as each $S_i$ is finite. Connectivity hull of $V$ is a union of some finite minimal sets thus it has to be finite. -\end{proof} - -\begin{Corollary} - Suppose $V$ is a countable subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. -\end{Corollary} - -% no compactness! -\begin{proof} - Let $n+1$ be connectivity of $V$. Order $V = \{v_1, v_2, \ldots\}$ and consider increasing finite parts $V_i = \{v_1, \dots, v_i\}$. Connectivity of $V_i$ as a function of $i$ is non-decreasing and is bounded by connectivity of $V$ which is $n+1$. Suppose connectivity of all $V_i$ is less than $n+1$ for all $i$ then by compactness connectivity of $V$ would be less than $n+1$ as well, which is a contradiction. Thus for large enough $i$, let's say $\forall i \geq N$, connectivity of $V_i$ is $n+1$. Let $H_N$ denote connectivity hull of $V_N$ and $H$ denote connectivity hull of $V$. As $V_N \subset V$ we have that $H \subset V_N$ as any $n$-point set disconnecting $V$ would also disconnect smaller set $V_N$. $V_N$ is finite by the previous corollary, so $V$ has to be finite as well. - Let $n$ be connectivity of $V$. - To every $V_i$ we associate two values. $n_i \in \N$ is connectivity of $V_i$ and $S_i$ collection of $n_i-1$-element sets that disconnect it, which is finite by previous lemma. For $i \leq j$ we have $n_i \leq n_j \leq n$. If $n_i = n_j$, then $S_i \supseteq S_j$ as any $(n_i-1)$-point set disconnecting $V_j$ would also disconnect smaller set $V_i$. Thus pair $(n_i, S_i)$ stabilizes at some $(m, S)$ as $i$ goes to infinity. $\bigcup S$ is the connectivity hull of $V$, and it is finite as needed. -\end{proof} - -\section{Application to indiscernible sequences} - -In this section we work in a flat graph. It is stable so all the indiscernible sequences are totally indiscernible. Also note that by indiscernibility all pairwise distances between points are the same. - -Superflat graphs are the graphs that for every $m \in \N$ there is $n \in \N$ such that avoid $K^m_n$ is avoided, where $K^m_n$ is $m$-subdvision of complete graph on $n$ vertices. Equivalent characterization given in \cite{stable_graphs}, Theorem 2 - -\begin{Theorem} - In a flat Graphs for every $n \in \N$ and an infinite set $A$ of vertices, there exists a finite $B$ and infinite $A' \subseteq A$ such that for all $x,y \in A'$ we have $d_{B}(x, y) > n$. -\end{Theorem} - -Roughly, the intuition is that from every infinite set we can extract a sparse infinite subset (after throwing away finitely many nodes). - -We will need a refined notion of connectivity for the following argument to work. Suppose we have two points $a,b$ distance $n$ apart. Denote $P(a,b)$ union of all paths of length $n$ going from $a$ to $b$. If we have a collection of vertices $V$ such that every two have distance $n$ between them, denote - -Let $V \subset V(G)$. Denote $P_n(V)$ a union of all paths of length $\leq n$ between points of $V$. It is a subgraph of $G$. - -\begin{align*} - P(V) = \bigcup_{a \neq b \in V} P(a,b) -\end{align*} - -\begin{Lemma} - Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Fix $n \in \N$. There exists a finite set $B$ such that - \begin{align*} - \forall i \neq j \ d_B(a_i, a_j) > n - \end{align*} -\end{Lemma} - -\begin{proof} - By a flatness result we can find an infinite $J \subset I$ and a finite set $B'$ such that each pair from $(a_j)_{j \in J}$ have distance $>n$ over $B'$. Using total indiscernibility we have an automorphism sending $(a_j)_{j \in J}$ to $(a_i)_{i \in I}$. Image of $B'$ under this automorphism is the required set $B$. -\end{proof} - -In other words, $B$ disconnects $P_n(\{a_i\})$. This shows that $\{a_i\}$ has finite connectivity in $P(\{a_i\})$. Applying lemma from last section we obtain that connectivity hull of $\{a_i\}$ in $P_n(\{a_i\})$ is finite. - -\begin{Lemma} - Connectivity hull of $\{a_i\}$ in $P_n(\{a_i\})$ is $\{a_i\}$-definable as a subset of $G$. -\end{Lemma} - -\begin{Definition} - Given a graph $G$ and $V \subset V(G)$ define $H(G, V)$ to be connectivity hull of $V$ in $G$. -\end{Definition} - -\begin{Note} - Given a finite $V$ we have $H(P_n(V), V)$ is definable. -\end{Note} - -\begin{proof} - Consider finite parts of the sequence $I_i = \{a_1, a_2, \ldots, a_i\}$. We study $H_i = H(P_n(I_i), I_i)$ as a function of $i$ as approximations of the hull in question. We have the following properties - \begin{align*} - \forall i \ H(P_n(I_i), I_i) &\subseteq H(P_n(I), I_i) \\ - \forall i \leq j \ H(P_n(I), I_i) &\supseteq H(P_n(I), I_j) - \end{align*} - By proof of previous lemma $H(P_n(I), I_i)$ stabilizes to a finite $H(P_n(I), I)$. Eventually $H_i(P_n(I), I_i)$ stabilizes to that value as well. This shows that for large enough $i>N$ we have $H_i = H_{i+m}$. By symmetry of indiscernible sequence we have that any subset of size $N$ defines the connectivity hull. -\end{proof} - -\begin{Lemma} - $I$ is indiscernible over the $A \cup H(P_n(I), I)$. -\end{Lemma} - -\begin{proof} - Denote the hull by $H$. Fix an $A$-formula $\phi(x,y)$. Consider a collection of traces $\phi(\vec a, H^\{|y|\})$ for $\vec a \in I^{|x|}$. As $H$ is $I$ definable those are either all distinct or all the same. Finiteness of $H$ forces latter. This shows indiscernability. -\end{proof} - -\begin{Corollary} - Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Then there is a countable $B$ such that $(a_i)$ is indiscernible over $A \cup B$ and - \begin{align*} - \forall i \neq j \ d_B(a_i, a_j) = \infty - \end{align*} -\end{Corollary} - -\begin{proof} - Let $B_i = H(P_i(I), I)$. Successive applications of previous lemma yield the appropriate set $B = \bigcup B_i$. -\end{proof} - -That is every indiscernible sequence can be upgraded to have infinite distance over its parameter set. - -\section{Superflat graphs are dp-minimal} - -\begin{Lemma} - Suppose $x \equiv_A y$ and $d_A(x, c) = d_A(y, c) = \infty$. Then $x \equiv_{Ac} y$ -\end{Lemma} - -\begin{proof} - Define an equivalence relation $G - A$. Two points $p, q$ are equivalent if $d_A(p,q)$ is finite. There is an automorphism $f$ of $G$ fixing $A$ sending $x$ to $y$. Denote by $X$ and $Y$ equivalence classes of $x$ and $y$ respectively. It's easy to see that $f(X) = Y$. Define the following function - \begin{align*} - &g = f \text { on } X \\ - &g = f^{-1} \text { on } Y \\ - &\text{identity otherwise} - \end{align*} - It is easy to see that $g$ is an automorphism fixing $Ac$ that sends $x$ to $y$. -\end{proof} - -\begin{Theorem} - Let $G$ be a flat graph with $(a_i)_{i\in\Q}$ indiscernible over $A$ and $b \in G$. There exists $c \in \Q$ such that all $(a_i)_{i\in\{\Q - c\}}$ have the same type over $Ab$. -\end{Theorem} - -\begin{proof} - Find $B \supseteq A$ such that $(a_i)$ is indiscernible over $B$ and has infinite distance over $B$. All the elements of the indiscernible sequence fall into distinct equivalence classes. $b$ can be in at most one of them. Exclude that element from the sequence. Remaining sequence elements are all infinitely far away from $b$. By previous lemma we have that elements of indiscernible sequence all have the same type over $Bb$. -\end{proof} - -But this is exactly what it means to be dp-minimal, as given, say, in \cite{simon_dp_minimal} Lemma 1.4.4 - -\begin{Corollary} - Flat graphs are dp-minimal. -\end{Corollary} - - -\begin{thebibliography}{9} - -\bibitem{diestel} - Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. - -\bibitem{infinite_megner} - Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 - -\bibitem{stable_graphs} - Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. - -\bibitem{simon_dp_minimal} - P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. - - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + + +\begin{document} + +\title{Superflat graphs are dp-minimal} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We show that the theory of superflat graphs is dp-minimal. +\end{abstract} + +\maketitle + +\section{Preliminaries} + +Superflat graphs in model theoretic context were first introduced in \cite{stable_graphs} as a natural class of stable graphs. + +We work with an infinite graph $G$ and a subset of vertices $V \subset V(G)$. Say that $V$ is $n$-connected if there aren't a set of $n-1$ vertices removing which disconnects every pair of vertices in $V$. Connectivity of $V$ is the smallest $n$ such that $V$ are $n$-connected. + +\begin{Definition} + Suppose $V \subset V(G)$ has finite connectivity $n+1$. Let connectivity hull of $V$ to be union of all $n$-point sets that disconnect it. +\end{Definition} + +\begin{Note} + If $V$ has finite connectivity $n+1$, then having finite connectivity hull is the same as having finitely many $n$-point sets that disconnect it. +\end{Note} + +\begin{Definition} + Let $V \subset V(G)$. We say that a set is minimal with respect to $V$ when that set disconnects $V$ and is minimal among such sets (ordered by inclusion). +\end{Definition} + +\section{Connectivity hull is finite} + +Here we show our main technical lemma. This section is purely combinatorial, with no mention of model theory. + +\begin{Lemma} + Suppose $V = \{a,b\}$ is subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. +\end{Lemma} + +\begin{Corollary} + Suppose $V$ is a finite subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. +\end{Corollary} + +\begin{proof} + Fix set $P = {p_1, \ldots, p_m}$ of all unordered pairs from $V$. Every pair $p_i$ has connectivity $n_i$ and by previous lemma has finitely many sets of $n_i$ points that disconnect it, denoted by $S_i$. Consider a finite minimal set $D$ that disconnects $V$. Define $D_i$ to be one of the sets in $S_i$ such that $D_i \subset D$ (one always exists). Then $D = \cup_{i \in I} D_i$ (if it was larger that would contradict minimality). Thus minimal sets are determined by unions of sets in $\{S_1 \ldots S_m\}$ and there are finitely many of such possible sets as each $S_i$ is finite. Connectivity hull of $V$ is a union of some finite minimal sets thus it has to be finite. +\end{proof} + +\begin{Corollary} + Suppose $V$ is a countable subset of $V(G)$ with finite connectivity. Then its connectivity hull is finite. +\end{Corollary} + +% no compactness! +\begin{proof} + Let $n+1$ be connectivity of $V$. Order $V = \{v_1, v_2, \ldots\}$ and consider increasing finite parts $V_i = \{v_1, \dots, v_i\}$. Connectivity of $V_i$ as a function of $i$ is non-decreasing and is bounded by connectivity of $V$ which is $n+1$. Suppose connectivity of all $V_i$ is less than $n+1$ for all $i$ then by compactness connectivity of $V$ would be less than $n+1$ as well, which is a contradiction. Thus for large enough $i$, let's say $\forall i \geq N$, connectivity of $V_i$ is $n+1$. Let $H_N$ denote connectivity hull of $V_N$ and $H$ denote connectivity hull of $V$. As $V_N \subset V$ we have that $H \subset V_N$ as any $n$-point set disconnecting $V$ would also disconnect smaller set $V_N$. $V_N$ is finite by the previous corollary, so $V$ has to be finite as well. + Let $n$ be connectivity of $V$. + To every $V_i$ we associate two values. $n_i \in \N$ is connectivity of $V_i$ and $S_i$ collection of $n_i-1$-element sets that disconnect it, which is finite by previous lemma. For $i \leq j$ we have $n_i \leq n_j \leq n$. If $n_i = n_j$, then $S_i \supseteq S_j$ as any $(n_i-1)$-point set disconnecting $V_j$ would also disconnect smaller set $V_i$. Thus pair $(n_i, S_i)$ stabilizes at some $(m, S)$ as $i$ goes to infinity. $\bigcup S$ is the connectivity hull of $V$, and it is finite as needed. +\end{proof} + +\section{Application to indiscernible sequences} + +In this section we work in a flat graph. It is stable so all the indiscernible sequences are totally indiscernible. Also note that by indiscernibility all pairwise distances between points are the same. + +Superflat graphs are the graphs that for every $m \in \N$ there is $n \in \N$ such that avoid $K^m_n$ is avoided, where $K^m_n$ is $m$-subdvision of complete graph on $n$ vertices. Equivalent characterization given in \cite{stable_graphs}, Theorem 2 + +\begin{Theorem} + In a flat Graphs for every $n \in \N$ and an infinite set $A$ of vertices, there exists a finite $B$ and infinite $A' \subseteq A$ such that for all $x,y \in A'$ we have $d_{B}(x, y) > n$. +\end{Theorem} + +Roughly, the intuition is that from every infinite set we can extract a sparse infinite subset (after throwing away finitely many nodes). + +We will need a refined notion of connectivity for the following argument to work. Suppose we have two points $a,b$ distance $n$ apart. Denote $P(a,b)$ union of all paths of length $n$ going from $a$ to $b$. If we have a collection of vertices $V$ such that every two have distance $n$ between them, denote + +Let $V \subset V(G)$. Denote $P_n(V)$ a union of all paths of length $\leq n$ between points of $V$. It is a subgraph of $G$. + +\begin{align*} + P(V) = \bigcup_{a \neq b \in V} P(a,b) +\end{align*} + +\begin{Lemma} + Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Fix $n \in \N$. There exists a finite set $B$ such that + \begin{align*} + \forall i \neq j \ d_B(a_i, a_j) > n + \end{align*} +\end{Lemma} + +\begin{proof} + By a flatness result we can find an infinite $J \subset I$ and a finite set $B'$ such that each pair from $(a_j)_{j \in J}$ have distance $>n$ over $B'$. Using total indiscernibility we have an automorphism sending $(a_j)_{j \in J}$ to $(a_i)_{i \in I}$. Image of $B'$ under this automorphism is the required set $B$. +\end{proof} + +In other words, $B$ disconnects $P_n(\{a_i\})$. This shows that $\{a_i\}$ has finite connectivity in $P(\{a_i\})$. Applying lemma from last section we obtain that connectivity hull of $\{a_i\}$ in $P_n(\{a_i\})$ is finite. + +\begin{Lemma} + Connectivity hull of $\{a_i\}$ in $P_n(\{a_i\})$ is $\{a_i\}$-definable as a subset of $G$. +\end{Lemma} + +\begin{Definition} + Given a graph $G$ and $V \subset V(G)$ define $H(G, V)$ to be connectivity hull of $V$ in $G$. +\end{Definition} + +\begin{Note} + Given a finite $V$ we have $H(P_n(V), V)$ is definable. +\end{Note} + +\begin{proof} + Consider finite parts of the sequence $I_i = \{a_1, a_2, \ldots, a_i\}$. We study $H_i = H(P_n(I_i), I_i)$ as a function of $i$ as approximations of the hull in question. We have the following properties + \begin{align*} + \forall i \ H(P_n(I_i), I_i) &\subseteq H(P_n(I), I_i) \\ + \forall i \leq j \ H(P_n(I), I_i) &\supseteq H(P_n(I), I_j) + \end{align*} + By proof of previous lemma $H(P_n(I), I_i)$ stabilizes to a finite $H(P_n(I), I)$. Eventually $H_i(P_n(I), I_i)$ stabilizes to that value as well. This shows that for large enough $i>N$ we have $H_i = H_{i+m}$. By symmetry of indiscernible sequence we have that any subset of size $N$ defines the connectivity hull. +\end{proof} + +\begin{Lemma} + $I$ is indiscernible over the $A \cup H(P_n(I), I)$. +\end{Lemma} + +\begin{proof} + Denote the hull by $H$. Fix an $A$-formula $\phi(x,y)$. Consider a collection of traces $\phi(\vec a, H^\{|y|\})$ for $\vec a \in I^{|x|}$. As $H$ is $I$ definable those are either all distinct or all the same. Finiteness of $H$ forces latter. This shows indiscernability. +\end{proof} + +\begin{Corollary} + Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Then there is a countable $B$ such that $(a_i)$ is indiscernible over $A \cup B$ and + \begin{align*} + \forall i \neq j \ d_B(a_i, a_j) = \infty + \end{align*} +\end{Corollary} + +\begin{proof} + Let $B_i = H(P_i(I), I)$. Successive applications of previous lemma yield the appropriate set $B = \bigcup B_i$. +\end{proof} + +That is every indiscernible sequence can be upgraded to have infinite distance over its parameter set. + +\section{Superflat graphs are dp-minimal} + +\begin{Lemma} + Suppose $x \equiv_A y$ and $d_A(x, c) = d_A(y, c) = \infty$. Then $x \equiv_{Ac} y$ +\end{Lemma} + +\begin{proof} + Define an equivalence relation $G - A$. Two points $p, q$ are equivalent if $d_A(p,q)$ is finite. There is an automorphism $f$ of $G$ fixing $A$ sending $x$ to $y$. Denote by $X$ and $Y$ equivalence classes of $x$ and $y$ respectively. It's easy to see that $f(X) = Y$. Define the following function + \begin{align*} + &g = f \text { on } X \\ + &g = f^{-1} \text { on } Y \\ + &\text{identity otherwise} + \end{align*} + It is easy to see that $g$ is an automorphism fixing $Ac$ that sends $x$ to $y$. +\end{proof} + +\begin{Theorem} + Let $G$ be a flat graph with $(a_i)_{i\in\Q}$ indiscernible over $A$ and $b \in G$. There exists $c \in \Q$ such that all $(a_i)_{i\in\{\Q - c\}}$ have the same type over $Ab$. +\end{Theorem} + +\begin{proof} + Find $B \supseteq A$ such that $(a_i)$ is indiscernible over $B$ and has infinite distance over $B$. All the elements of the indiscernible sequence fall into distinct equivalence classes. $b$ can be in at most one of them. Exclude that element from the sequence. Remaining sequence elements are all infinitely far away from $b$. By previous lemma we have that elements of indiscernible sequence all have the same type over $Bb$. +\end{proof} + +But this is exactly what it means to be dp-minimal, as given, say, in \cite{simon_dp_minimal} Lemma 1.4.4 + +\begin{Corollary} + Flat graphs are dp-minimal. +\end{Corollary} + + +\begin{thebibliography}{9} + +\bibitem{diestel} + Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. + +\bibitem{infinite_megner} + Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 + +\bibitem{stable_graphs} + Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. + +\bibitem{simon_dp_minimal} + P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. + + +\end{thebibliography} + \end{document} diff --git a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.tex b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.tex index ba7e9397..886dffdf 100644 --- a/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.tex +++ b/research/06 Flat graphs and dp-rank/Flat graphs and dp-rank.tex @@ -1,221 +1,221 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\DeclareMathOperator{\I}{\mathcal I} -\DeclareMathOperator{\J}{\mathcal J} -\DeclareMathOperator{\acl}{acl} - -\begin{document} - -\title{Superflat graphs are dp-minimal} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We show that the theory of superflat graphs is dp-minimal. -\end{abstract} - -\maketitle - -\section{Preliminaries} - -Superflat graphs were introduced in \cite{stable_graphs} as a natural class of stable graphs. Here we present a direct proof showing dp-minimality. - -First, we introduce some basic graph-theoretic definitions. -\begin{Definition} - Work in an infinite graph $G$. Let $A, V \subset V(G)$ (where $V(G)$ denotes vertices of $G$) - \begin{enumerate} - \item $G' = G[V]$ is called \emph{induced} subgraph of $G$ \emph{spanned} by $V$ if it is obtained as a subgraph of $G$ by taking all edges between vertices in $V$. - \item For $a,b \in V(G)$ define the \emph{distance} $d(a,b)$ to be the length of the shortest path between $a$ and $b$ in $G$. - \item For $a,b \in V(G)$ define $d_A(a,b)$ to be the distance between $a$ and $b$ in $G[V(G) - A]$. Equivalently it is the shortest path between $a$ and $b$ that avoids vertices in $A$. - \item We say that $A$ \emph{separates} $V$ if for all $a,b \in V$, $d_A(a,b) = \infty$. - \item We say that $V$ has \emph{connectivity} $n$ if there are no sets of size $n-1$ in $V(G)$ that separates $V$. - \item Suppose $V$ has finite connectivity $n$. \emph{Connectivity hull} of $V$ is defined to be the union of all sets separating $V$ of size $n-1$. - \end{enumerate} -\end{Definition} - -In \cite{infinite_megner} we find a generalization of Megner's Theorem for infinite graphs - -\begin{Theorem} [Megner, Erd\H os, Aharoni, Berger] - Let $A$ and $B$ be two sets of vertices in a possibly infinite digraph. Then there exist a set $P$ of disjoint $A–B$ paths, and a set $S$ of vertices separating $A$ from $B$, such that $S$ consists of a choice of precisely one vertex from each path in $P$. -\end{Theorem} - -We use the following easy consequence - -\begin{Corollary} \label{cr_disjoint_paths} - Let $V$ be a subset of a graph $G$ with connectivity $n$. Then there exists a set of $n$ disjoint paths from $V$ into itself. -\end{Corollary} - -\begin{Corollary} \label{cr_hull_finite} - With assumptions as above, connectivity hull of $V$ is finite. -\end{Corollary} - -\begin{proof} - All the separating sets have to have exactly one vertex in each of those paths. -\end{proof} - -\begin{Definition} - Denote by $K^m_n$ an $m$-subdivision of the complete graph on $n$ vertices. - Graph is called superflat if for every $m \in \N$ there is $n \in \N$ such that the graph avoids $K^m_n$ as a subgraph. -\end{Definition} - -Theorem 2 in \cite{stable_graphs} gives a useful characterization of superflat graphs. - -\begin{Theorem} \label{th_superflat_equivalence} - The following are equivalent - \begin{enumerate} - \item $G$ is superflat - \item For every $n \in \N$ and an infinite set $A \subset V(G)$, there exists a finite $B \subset V(G)$ and infinite $A' \subseteq A$ such that for all $x,y \in A'$ we have $d_{B}(x, y) > n$. - \end{enumerate} -\end{Theorem} - -Roughly, in superflat graphs every infinite set contains a sparse infinite subset (possibly after throwing away finitely many nodes). - -\section{Indiscernible sequences} - -In this section we work in a superflat graph $G$. Stability implies that all the indiscernible sequences are totally indiscernible. - -\begin{Definition} - Let $V \subset V(G)$. $P_n(V)$, a subgraph of $G$ denotes a union of all paths of length $\leq n$ between points of $V$. -\end{Definition} - -\begin{Lemma} \label{lm_bump} - Let $I = (a_i : i \in \I)$ be a countable indiscernible sequence over $A$. Fix $n \in \N$. - There exists a finite set $B$ such that - \begin{align*} - \forall i \neq j \ d_B(a_i, a_j) > n - \end{align*} -\end{Lemma} - -\begin{proof} - By a \ref{th_superflat_equivalence} we can find an infinite $\J \subset \I$ and a finite set $B'$ such that each pair from $J = (a_j : j \in \J)$ have distance $>n$ over $B'$. - Using total indiscernibility we have an automorphism sending $J$ to $I$ fixing $A$. - Image of $B'$ under this automorphism is the required set $B$. -\end{proof} - -In other words, $B$ separates $I$ when viewed inside subgraph $P_n(I)$. -This shows that $I$ has finite connectivity in $P_n(I)$. -Applying Corollary \ref{cr_hull_finite} we obtain that connectivity hull of $I$ in $P_n(I)$ is finite. - -\begin{Definition} - We call a set $H \subseteq V(G)$ uniformly definable from an indiscernible sequence $I$ if there is a formula $\phi(x, y)$ such that for every $J \subset I$ of size $|y|$ we have - \begin{align*} - H = \{g \in G \mid \phi(g, J)\} - \end{align*} - where $J$ is considered a tuple. -\end{Definition} - -\begin{Lemma} \label{lm_uniform} - Fix a countable indiscernible sequence $I = (a_i : i \in \I)$. - Let $H$ be its connectivity hull inside of graph $P_n(I)$. - Then $H$ is uniformly definable from $I$ in $G$. -\end{Lemma} - -\begin{Definition} - Given a graph $G$ and $V \subset V(G)$ define $H(G, V)$ to be connectivity hull of $V$ in $G$. -\end{Definition} - -\begin{Note} - Given a finite $V$ we have $H(P_n(V), V)$ is $V$-definable. -\end{Note} - -\begin{proof} - Corollary \ref{cr_disjoint_paths} tells us that there finitely many paths between elements of $V$ such that $H(P_n(I), I)$ is inside those paths. - Take $J \subset I$ be the endpoints of those paths. - $H(P_n(J), J)$ is $J$-definable as noted above. - Moreover we have $H(P_n(I), I) \subseteq H(P_n(J), J)$, both finite sets. - In particular we have $H \subset \acl(J)$. - $I-J$ is indiscernible over $A \cup J$. - -\end{proof} - -\begin{proof}(of \ref{lm_uniform}) - Consider finite parts of the sequence $I_i = \{a_1, a_2, \ldots, a_i\}$. - Define $H_i = H(P_n(I_i), I_i)$. - It is $I_i$-definable. %and we have $H(P_n(I_i), I_i) \subseteq H(P_n(I), I)$. - Corollary \ref{cr_disjoint_paths} tells us that there finitely many paths between elements of $V$ such that $H(P_n(I), I)$ is inside those paths. - But for large enough $i$, say $i \geq N$, $P_n(I_i)$ will contain all of those paths. - Thus for $i \geq N$ we have $H(P_n(I), I) \subseteq P_n(I_i)$. - If a set separates $I$ then would be inside $P_n(I_i)$ and would separate $I_i$ as well. - Thus for $i \geq N$ we have $H(P_n(I), I) \subseteq H(P_n(I_i), I_i)$. - If the two sets are not equal, it is due to some elements in $H(P_n(I_i), I_i)$ failing to separate entire $I$. There are finitely many of those, so for large enough $i$, say $i \geq M$ we have $H(P_n(I_i), I_i) = H(P_n(I), I)$ stabilizing. - This shows that for $i \geq M$ we have $H_i = H_{i+m} = H(P_n(I), I)$. By symmetry of indiscernible sequence we have that any subset of size $N$ defines the connectivity hull. -\end{proof} - -\begin{Lemma} \label{cr_bump} - $I$ is indiscernible over the $A \cup H(P_n(I), I)$. -\end{Lemma} - -\begin{proof} - Denote the hull by $H$. Fix an $A$-formula $\phi(x,y)$. Consider a collection of traces $\phi(a, H^\{|y|\})$ for $a \in I^{|x|}$. If two of them are distinct, then by indiscernibility all of them are. But that is impossible as $H$ has finitely many subsets. Thus all the traces are identical. This shows that for any $a,b \in I^{|x|}$ and $h \in H^\{|y|\}$ we have $\phi(a, h) \iff \phi(b, h)$. As choice of $\phi$ was arbitrary, this shows that $I$ is indiscernible over $A \cup H(P_n(I), I)$. -\end{proof} - -\begin{Corollary} - Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Then there is a countable $B$ such that $(a_i)$ is indiscernible over $A \cup B$ and - \begin{align*} - \forall i \neq j \ d_B(a_i, a_j) = \infty - \end{align*} -\end{Corollary} - -\begin{proof} - Let $B_n = H(P_n(I), I)$. This is well defined by Lemma \ref{lm_bump} and has property - \begin{align*} - \forall i \neq j \ d_{B_n}(a_i, a_j) > n - \end{align*} - % does limiting work? - and $I$ is indiscernible over $A \cup B_i$ by Corollary \ref{cr_bump}. Let $B = \bigcup_{n \in \N} B_n$. -\end{proof} - -That is every indiscernible sequence can be upgraded to have infinite distance over its parameter set. - -\section{Superflat graphs are dp-minimal} - -% monster model? -\begin{Lemma} - Suppose $x \equiv_A y$ and $d_A(x, c) = d_A(y, c) = \infty$. Then $x \equiv_{Ac} y$ -\end{Lemma} - -\begin{proof} - Define an equivalence relation $G - A$. Two points $p, q$ are equivalent if $d_A(p,q)$ is finite. There is an automorphism $f$ of $G$ fixing $A$ sending $x$ to $y$. Denote by $X$ and $Y$ equivalence classes of $x$ and $y$ respectively. It's easy to see that $f(X) = Y$. Define the following function - \begin{align*} - &g = f \text { on } X \\ - &g = f^{-1} \text { on } Y \\ - &\text{identity otherwise} - \end{align*} - It is easy to see that $g$ is an automorphism fixing $Ac$ that sends $x$ to $y$. -\end{proof} - -\begin{Theorem} - Let $G$ be a flat graph with $(a_i)_{i\in\Q}$ indiscernible over $A$ and $b \in G$. There exists $c \in \Q$ such that all $(a_i)_{i\in\{\Q - c\}}$ have the same type over $Ab$. -\end{Theorem} - -\begin{proof} - Find $B \supseteq A$ such that $(a_i)$ is indiscernible over $B$ and has infinite distance over $B$. All the elements of the indiscernible sequence fall into distinct equivalence classes. $b$ can be in at most one of them. Exclude that element from the sequence. Remaining sequence elements are all infinitely far away from $b$. By previous lemma we have that elements of indiscernible sequence all have the same type over $Bb$. -\end{proof} - -But this is exactly what it means to be dp-minimal, as given, say, in \cite{simon_dp_minimal} Lemma 1.4.4 - -\begin{Corollary} - Flat graphs are dp-minimal. -\end{Corollary} - - -\begin{thebibliography}{9} - -\bibitem{stable_graphs} - Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. - -\bibitem{infinite_megner} - Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 - -\bibitem{simon_dp_minimal} - P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. - -%\bibitem{diestel} -% Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\DeclareMathOperator{\I}{\mathcal I} +\DeclareMathOperator{\J}{\mathcal J} +\DeclareMathOperator{\acl}{acl} + +\begin{document} + +\title{Superflat graphs are dp-minimal} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We show that the theory of superflat graphs is dp-minimal. +\end{abstract} + +\maketitle + +\section{Preliminaries} + +Superflat graphs were introduced in \cite{stable_graphs} as a natural class of stable graphs. Here we present a direct proof showing dp-minimality. + +First, we introduce some basic graph-theoretic definitions. +\begin{Definition} + Work in an infinite graph $G$. Let $A, V \subset V(G)$ (where $V(G)$ denotes vertices of $G$) + \begin{enumerate} + \item $G' = G[V]$ is called \emph{induced} subgraph of $G$ \emph{spanned} by $V$ if it is obtained as a subgraph of $G$ by taking all edges between vertices in $V$. + \item For $a,b \in V(G)$ define the \emph{distance} $d(a,b)$ to be the length of the shortest path between $a$ and $b$ in $G$. + \item For $a,b \in V(G)$ define $d_A(a,b)$ to be the distance between $a$ and $b$ in $G[V(G) - A]$. Equivalently it is the shortest path between $a$ and $b$ that avoids vertices in $A$. + \item We say that $A$ \emph{separates} $V$ if for all $a,b \in V$, $d_A(a,b) = \infty$. + \item We say that $V$ has \emph{connectivity} $n$ if there are no sets of size $n-1$ in $V(G)$ that separates $V$. + \item Suppose $V$ has finite connectivity $n$. \emph{Connectivity hull} of $V$ is defined to be the union of all sets separating $V$ of size $n-1$. + \end{enumerate} +\end{Definition} + +In \cite{infinite_megner} we find a generalization of Megner's Theorem for infinite graphs + +\begin{Theorem} [Megner, Erd\H os, Aharoni, Berger] + Let $A$ and $B$ be two sets of vertices in a possibly infinite digraph. Then there exist a set $P$ of disjoint $A–B$ paths, and a set $S$ of vertices separating $A$ from $B$, such that $S$ consists of a choice of precisely one vertex from each path in $P$. +\end{Theorem} + +We use the following easy consequence + +\begin{Corollary} \label{cr_disjoint_paths} + Let $V$ be a subset of a graph $G$ with connectivity $n$. Then there exists a set of $n$ disjoint paths from $V$ into itself. +\end{Corollary} + +\begin{Corollary} \label{cr_hull_finite} + With assumptions as above, connectivity hull of $V$ is finite. +\end{Corollary} + +\begin{proof} + All the separating sets have to have exactly one vertex in each of those paths. +\end{proof} + +\begin{Definition} + Denote by $K^m_n$ an $m$-subdivision of the complete graph on $n$ vertices. + Graph is called superflat if for every $m \in \N$ there is $n \in \N$ such that the graph avoids $K^m_n$ as a subgraph. +\end{Definition} + +Theorem 2 in \cite{stable_graphs} gives a useful characterization of superflat graphs. + +\begin{Theorem} \label{th_superflat_equivalence} + The following are equivalent + \begin{enumerate} + \item $G$ is superflat + \item For every $n \in \N$ and an infinite set $A \subset V(G)$, there exists a finite $B \subset V(G)$ and infinite $A' \subseteq A$ such that for all $x,y \in A'$ we have $d_{B}(x, y) > n$. + \end{enumerate} +\end{Theorem} + +Roughly, in superflat graphs every infinite set contains a sparse infinite subset (possibly after throwing away finitely many nodes). + +\section{Indiscernible sequences} + +In this section we work in a superflat graph $G$. Stability implies that all the indiscernible sequences are totally indiscernible. + +\begin{Definition} + Let $V \subset V(G)$. $P_n(V)$, a subgraph of $G$ denotes a union of all paths of length $\leq n$ between points of $V$. +\end{Definition} + +\begin{Lemma} \label{lm_bump} + Let $I = (a_i : i \in \I)$ be a countable indiscernible sequence over $A$. Fix $n \in \N$. + There exists a finite set $B$ such that + \begin{align*} + \forall i \neq j \ d_B(a_i, a_j) > n + \end{align*} +\end{Lemma} + +\begin{proof} + By a \ref{th_superflat_equivalence} we can find an infinite $\J \subset \I$ and a finite set $B'$ such that each pair from $J = (a_j : j \in \J)$ have distance $>n$ over $B'$. + Using total indiscernibility we have an automorphism sending $J$ to $I$ fixing $A$. + Image of $B'$ under this automorphism is the required set $B$. +\end{proof} + +In other words, $B$ separates $I$ when viewed inside subgraph $P_n(I)$. +This shows that $I$ has finite connectivity in $P_n(I)$. +Applying Corollary \ref{cr_hull_finite} we obtain that connectivity hull of $I$ in $P_n(I)$ is finite. + +\begin{Definition} + We call a set $H \subseteq V(G)$ uniformly definable from an indiscernible sequence $I$ if there is a formula $\phi(x, y)$ such that for every $J \subset I$ of size $|y|$ we have + \begin{align*} + H = \{g \in G \mid \phi(g, J)\} + \end{align*} + where $J$ is considered a tuple. +\end{Definition} + +\begin{Lemma} \label{lm_uniform} + Fix a countable indiscernible sequence $I = (a_i : i \in \I)$. + Let $H$ be its connectivity hull inside of graph $P_n(I)$. + Then $H$ is uniformly definable from $I$ in $G$. +\end{Lemma} + +\begin{Definition} + Given a graph $G$ and $V \subset V(G)$ define $H(G, V)$ to be connectivity hull of $V$ in $G$. +\end{Definition} + +\begin{Note} + Given a finite $V$ we have $H(P_n(V), V)$ is $V$-definable. +\end{Note} + +\begin{proof} + Corollary \ref{cr_disjoint_paths} tells us that there finitely many paths between elements of $V$ such that $H(P_n(I), I)$ is inside those paths. + Take $J \subset I$ be the endpoints of those paths. + $H(P_n(J), J)$ is $J$-definable as noted above. + Moreover we have $H(P_n(I), I) \subseteq H(P_n(J), J)$, both finite sets. + In particular we have $H \subset \acl(J)$. + $I-J$ is indiscernible over $A \cup J$. + +\end{proof} + +\begin{proof}(of \ref{lm_uniform}) + Consider finite parts of the sequence $I_i = \{a_1, a_2, \ldots, a_i\}$. + Define $H_i = H(P_n(I_i), I_i)$. + It is $I_i$-definable. %and we have $H(P_n(I_i), I_i) \subseteq H(P_n(I), I)$. + Corollary \ref{cr_disjoint_paths} tells us that there finitely many paths between elements of $V$ such that $H(P_n(I), I)$ is inside those paths. + But for large enough $i$, say $i \geq N$, $P_n(I_i)$ will contain all of those paths. + Thus for $i \geq N$ we have $H(P_n(I), I) \subseteq P_n(I_i)$. + If a set separates $I$ then would be inside $P_n(I_i)$ and would separate $I_i$ as well. + Thus for $i \geq N$ we have $H(P_n(I), I) \subseteq H(P_n(I_i), I_i)$. + If the two sets are not equal, it is due to some elements in $H(P_n(I_i), I_i)$ failing to separate entire $I$. There are finitely many of those, so for large enough $i$, say $i \geq M$ we have $H(P_n(I_i), I_i) = H(P_n(I), I)$ stabilizing. + This shows that for $i \geq M$ we have $H_i = H_{i+m} = H(P_n(I), I)$. By symmetry of indiscernible sequence we have that any subset of size $N$ defines the connectivity hull. +\end{proof} + +\begin{Lemma} \label{cr_bump} + $I$ is indiscernible over the $A \cup H(P_n(I), I)$. +\end{Lemma} + +\begin{proof} + Denote the hull by $H$. Fix an $A$-formula $\phi(x,y)$. Consider a collection of traces $\phi(a, H^\{|y|\})$ for $a \in I^{|x|}$. If two of them are distinct, then by indiscernibility all of them are. But that is impossible as $H$ has finitely many subsets. Thus all the traces are identical. This shows that for any $a,b \in I^{|x|}$ and $h \in H^\{|y|\}$ we have $\phi(a, h) \iff \phi(b, h)$. As choice of $\phi$ was arbitrary, this shows that $I$ is indiscernible over $A \cup H(P_n(I), I)$. +\end{proof} + +\begin{Corollary} + Let $(a_i)_{i \in I}$ be a countable indiscernible sequence over $A$. Then there is a countable $B$ such that $(a_i)$ is indiscernible over $A \cup B$ and + \begin{align*} + \forall i \neq j \ d_B(a_i, a_j) = \infty + \end{align*} +\end{Corollary} + +\begin{proof} + Let $B_n = H(P_n(I), I)$. This is well defined by Lemma \ref{lm_bump} and has property + \begin{align*} + \forall i \neq j \ d_{B_n}(a_i, a_j) > n + \end{align*} + % does limiting work? + and $I$ is indiscernible over $A \cup B_i$ by Corollary \ref{cr_bump}. Let $B = \bigcup_{n \in \N} B_n$. +\end{proof} + +That is every indiscernible sequence can be upgraded to have infinite distance over its parameter set. + +\section{Superflat graphs are dp-minimal} + +% monster model? +\begin{Lemma} + Suppose $x \equiv_A y$ and $d_A(x, c) = d_A(y, c) = \infty$. Then $x \equiv_{Ac} y$ +\end{Lemma} + +\begin{proof} + Define an equivalence relation $G - A$. Two points $p, q$ are equivalent if $d_A(p,q)$ is finite. There is an automorphism $f$ of $G$ fixing $A$ sending $x$ to $y$. Denote by $X$ and $Y$ equivalence classes of $x$ and $y$ respectively. It's easy to see that $f(X) = Y$. Define the following function + \begin{align*} + &g = f \text { on } X \\ + &g = f^{-1} \text { on } Y \\ + &\text{identity otherwise} + \end{align*} + It is easy to see that $g$ is an automorphism fixing $Ac$ that sends $x$ to $y$. +\end{proof} + +\begin{Theorem} + Let $G$ be a flat graph with $(a_i)_{i\in\Q}$ indiscernible over $A$ and $b \in G$. There exists $c \in \Q$ such that all $(a_i)_{i\in\{\Q - c\}}$ have the same type over $Ab$. +\end{Theorem} + +\begin{proof} + Find $B \supseteq A$ such that $(a_i)$ is indiscernible over $B$ and has infinite distance over $B$. All the elements of the indiscernible sequence fall into distinct equivalence classes. $b$ can be in at most one of them. Exclude that element from the sequence. Remaining sequence elements are all infinitely far away from $b$. By previous lemma we have that elements of indiscernible sequence all have the same type over $Bb$. +\end{proof} + +But this is exactly what it means to be dp-minimal, as given, say, in \cite{simon_dp_minimal} Lemma 1.4.4 + +\begin{Corollary} + Flat graphs are dp-minimal. +\end{Corollary} + + +\begin{thebibliography}{9} + +\bibitem{stable_graphs} + Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. + +\bibitem{infinite_megner} + Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 + +\bibitem{simon_dp_minimal} + P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. + +%\bibitem{diestel} +% Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. + +\end{thebibliography} + \end{document} diff --git a/research/07 shelah-spencer QE/Shelah-Spencer QE.aux b/research/07 shelah-spencer QE/Shelah-Spencer QE.aux deleted file mode 100644 index 956f01ef..00000000 --- a/research/07 shelah-spencer QE/Shelah-Spencer QE.aux +++ /dev/null @@ -1,23 +0,0 @@ -\relax -\citation{Laskowski} -\citation{Laskowski} -\citation{Laskowski} -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Introduction}}{1}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Omitting lemma}}{1}} -\newlabel{AA}{{2.2}{1}} -\newlabel{A}{{2.4}{1}} -\citation{Laskowski} -\citation{Laskowski} -\citation{Laskowski} -\citation{Laskowski} -\newlabel{B}{{2.5}{2}} -\newlabel{C}{{2.7}{2}} -\citation{Laskowski} -\bibcite{Laskowski}{1} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{3}{Quantifier Elimination}}{3}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{3}} diff --git a/research/07 shelah-spencer QE/Shelah-Spencer QE.bbl b/research/07 shelah-spencer QE/Shelah-Spencer QE.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/07 shelah-spencer QE/Shelah-Spencer QE.blg b/research/07 shelah-spencer QE/Shelah-Spencer QE.blg deleted file mode 100644 index 2544deda..00000000 --- a/research/07 shelah-spencer QE/Shelah-Spencer QE.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Shelah-Spencer QE.aux -I found no \bibdata command---while reading file Shelah-Spencer QE.aux -I found no \bibstyle command---while reading file Shelah-Spencer QE.aux -(There were 2 error messages) diff --git a/research/07 shelah-spencer QE/Shelah-Spencer QE.log b/research/07 shelah-spencer QE/Shelah-Spencer QE.log deleted file mode 100644 index 764c6dd6..00000000 --- a/research/07 shelah-spencer QE/Shelah-Spencer QE.log +++ /dev/null @@ -1,244 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 21 MAY 2014 11:23 -entering extended mode -**Shelah-Spencer*QE.tex -("C:\Users\Anton\SparkleShare\Research\research\07 shelah-spencer QE\Shelah-Spencer QE.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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-\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{diagrams} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\begin{document} - -\title{A note on quantifier elimination in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We simplify \cite{Laskowski}'s proof of quantifier elimination in Shelah-Spencer graphs. -\end{abstract} - -\maketitle - -\section{Introduction} - -Laskowski's paper \cite{Laskowski} provides a combinatorial proof of quantifier elimination in Shelah-Spencer graphs. Here we provide a simplification of the proof using only maximal chains and avoiding the use of proposition 3.1 and technical lemmas of section 4. - -We will use notation of \cite{Laskowski}, in particular things like $\K$, $\delta(\A/\B)$, $X_m(\A)$, $\S$, $\B^* \sqsubseteq \B'$,maximal embedding, $\Delta_\A(x)$, $\Psi_{\A,\B}(x)$ etc. However we will give a different definition of $Y(\ldots)$. When we write formulas $\theta(x,y)$ we may have $x,y$ to be tuples. - -\section{Omitting lemma} - -\begin{Definition} - Let $\M \models \S$, $\B \in \K$, embedding $f \colon \B \to \M$, $\Phi$ finite subset of $\K$ - \begin{enumerate} - \item Say that $f$ \textsl{omits} $\Phi$ if there are no $\C \in \Phi$ and $g \colon \C \to \M$ extending $f$. - \item Say that $f$ \textsl{admits} $\Phi$ if for every $\C \in \Phi$ there is $g \colon \C \to \M$ extending $f$. - \end{enumerate} -\end{Definition} - -\begin{Note} \label{AA} - Take notation as above and a structure $\C \in \K$ extending $\B$. Then $f$ doesn't omit $\{\C\}$ iff $f$ admits $\{\C\}$. -\end{Note} - -\begin{Definition} - Fix $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Define $Z(\B, \C, m)$ to be all $\B^* \in X_m(\B)$ such that there are no $\H$ with $|H \backslash B^*| < m$ satisfying - \begin{diagram} - & &\H \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\C \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} -\end{Definition} - -\begin{Lemma} \label{A} - Let $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Also let $\M \models \S$ and $f \colon \B \to \M$ an embedding. The following are equivalent: - \begin{enumerate} - \item $f$ omits $\{\C\}$. - \item There exists $\B^* \in Z(\B, \C, m)$ maximally embeddable into $\M$ over $f$. - \end{enumerate} -\end{Lemma} - -\begin{proof} - For the proof we identify $\B$ with $f(\B)$, i.e. for ease of notation assume that $\B \subset \M$. - - $(1) \Rightarrow (2)$ By remark 5.3 of \cite{Laskowski} there is some $B^* \in X_m(\B)$ maximally embeddable in $\M$ over $f$. Such embedding is unique by Lemma 3.8 of \cite{Laskowski}. Again, we identify $B^*$ with its maximal embedding into $\M$. To show $(2)$ we need to verify that $\B^* \in Z(\B, \C, m)$. Suppose not. Then there is $\H$ with $|H \backslash B^*| < m$ satisfying - \begin{diagram} - & &\H \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\C \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} - As $\B^* \leq \H$ and $\B \subset \M$ we can embed $\H$ into $\M$ (as $\M \models \S$). But this would witness $\C$ extending $\B$ in $\M$ which is impossible as we assumed that $f$ omits $\{\Phi\}$. - - $(2) \Rightarrow (1)$ Suppose $f$ doesn't omit $\{C\}$. Then by the note \ref{AA} $f$ admits $\{C\}$, i.e. there is an embedding of $\C$ into $M$ over $f$. We identify $\C$ with the image of that embedding. Similarly we identify $\B^*$ with the image of its maximal embedding over $f$. That is we may assume $\C, \B^* \subset \M$. Let $H$ be the substructure of $M$ induced by vertices $C \cup B^*$. As $|C \backslash B| < m$ we have $|H \backslash B^*| < m$. $\B^*$ is $m$-strong by remark 5.3 of \cite{Laskowski}. This forces $\B^* \leq H$. But this contradicts the fact that $\B^* \in Z(\B, \C, m)$. -\end{proof} - -\begin{Corollary} \label{B} - With the setup of the previous lemma, the following are equivalent: - \begin{enumerate} - \item $f$ admits $\{\C\}$. - \item There exists $\B^* \in X_m(\B) \backslash Z(\B, \C, m)$ maximally embeddable into $\M$ over $f$. - \end{enumerate} -\end{Corollary} - -For quantifier elimination we need to track multiple structures being admitted and omitted, hence the following definition. - -\begin{Definition} - Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. Define - \begin{align*} - Y(\B, \Phi, \Gamma, m) = \{B^* \in X_m(\B) \mid \ &\forall \C \in \Phi \ B^* \in Z(\B, \C, m) \text{ and}\\ - &\forall \D \in \Gamma \ B^* \notin Z(\B, \D, m)\} - \end{align*} -\end{Definition} - -\begin{Lemma} \label{C} - Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: - \begin{enumerate} - \item $f$ omits $\Phi$ and admits $\Gamma$. - \item There exists $\B^* \in Y(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. - \end{enumerate} -\end{Lemma} -\begin{proof} - Easy corollary of \ref{A} and \ref{B}. -\end{proof} - -\section{Quantifier Elimination} - -Following proof of 5.6 in \cite{Laskowski}, we have a formula $\theta(x, y)$, some $\A \subseteq \B \in \K$ with $\theta(x, y) \vdash \Delta_A(x) \wedge \Delta_\B(x, y)$. We may also assume that $\theta(x, y)$ is a conjunction of formulas of the type $\Psi_{\B, \C}(x,y)$ and their negations. More precisely -\begin{align*} - \theta(x, y) \Leftrightarrow &\bigwedge_{\C \in \Phi} \Psi_{\B, \C}(x,y) \ \wedge \\ - &\bigwedge_{\D \in \Gamma} \neg \Psi_{\B, \D}(x,y) -\end{align*} -for some finite subsets $\Phi, \Gamma$ of $\K$. Let $m = \max\{ |C \backslash B| \colon \C \in \Phi \text{ or } \C \in \Gamma\}$. We claim that in $\M \models \S$ -\begin{align*} - \exists y \theta(x, y) &\Leftrightarrow \bigvee_{\B^* \in Y(\B, \Phi, \Gamma, m)} (\text{$\B^*$ maximally embeds into $\M$ over $\A$}) \\ - &\Leftrightarrow \bigvee_{\B^* \in Y(\B, \Phi, \Gamma, m)} \left(\Psi_{\A, \B^*}(x) \wedge \bigwedge_{\B* \sqsubseteq B', B' \in X_m(\B)} \neg \Psi_{\A, \B'}(x)\right) -\end{align*} - -\begin{proof} - $(\Rightarrow)$ Fix $\B \subset \M$ witnessing existential statement. By remark 5.3 and lemma 3.8 in \cite{Laskowski} there is a unique $\B^* \in X_m$ maximally embeddable (with unique image) into $\M$ over $\B$ . By lemma \ref{C} $\B^* \in Y(\B, \Phi, \Gamma, m)$. - - $(\Leftarrow)$ Take the embedding $g\colon B^* \to \M$ and restrict it to $\B \subseteq \B^*$ i.e. $f = g \mid \B$. As $\B^* \in Y(\B, \Phi, \Gamma, m)$ by lemma \ref{C} $f$ omits $\Phi$ and admits $\Gamma$. Thus is is a witness to $\exists y \theta(x, y)$. -\end{proof} - -\begin{thebibliography}{9} - -\bibitem{Laskowski} - Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} - Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{diagrams} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\begin{document} + +\title{A note on quantifier elimination in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We simplify \cite{Laskowski}'s proof of quantifier elimination in Shelah-Spencer graphs. +\end{abstract} + +\maketitle + +\section{Introduction} + +Laskowski's paper \cite{Laskowski} provides a combinatorial proof of quantifier elimination in Shelah-Spencer graphs. Here we provide a simplification of the proof using only maximal chains and avoiding the use of proposition 3.1 and technical lemmas of section 4. + +We will use notation of \cite{Laskowski}, in particular things like $\K$, $\delta(\A/\B)$, $X_m(\A)$, $\S$, $\B^* \sqsubseteq \B'$,maximal embedding, $\Delta_\A(x)$, $\Psi_{\A,\B}(x)$ etc. However we will give a different definition of $Y(\ldots)$. When we write formulas $\theta(x,y)$ we may have $x,y$ to be tuples. + +\section{Omitting lemma} + +\begin{Definition} + Let $\M \models \S$, $\B \in \K$, embedding $f \colon \B \to \M$, $\Phi$ finite subset of $\K$ + \begin{enumerate} + \item Say that $f$ \textsl{omits} $\Phi$ if there are no $\C \in \Phi$ and $g \colon \C \to \M$ extending $f$. + \item Say that $f$ \textsl{admits} $\Phi$ if for every $\C \in \Phi$ there is $g \colon \C \to \M$ extending $f$. + \end{enumerate} +\end{Definition} + +\begin{Note} \label{AA} + Take notation as above and a structure $\C \in \K$ extending $\B$. Then $f$ doesn't omit $\{\C\}$ iff $f$ admits $\{\C\}$. +\end{Note} + +\begin{Definition} + Fix $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Define $Z(\B, \C, m)$ to be all $\B^* \in X_m(\B)$ such that there are no $\H$ with $|H \backslash B^*| < m$ satisfying + \begin{diagram} + & &\H \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\C \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} +\end{Definition} + +\begin{Lemma} \label{A} + Let $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Also let $\M \models \S$ and $f \colon \B \to \M$ an embedding. The following are equivalent: + \begin{enumerate} + \item $f$ omits $\{\C\}$. + \item There exists $\B^* \in Z(\B, \C, m)$ maximally embeddable into $\M$ over $f$. + \end{enumerate} +\end{Lemma} + +\begin{proof} + For the proof we identify $\B$ with $f(\B)$, i.e. for ease of notation assume that $\B \subset \M$. + + $(1) \Rightarrow (2)$ By remark 5.3 of \cite{Laskowski} there is some $B^* \in X_m(\B)$ maximally embeddable in $\M$ over $f$. Such embedding is unique by Lemma 3.8 of \cite{Laskowski}. Again, we identify $B^*$ with its maximal embedding into $\M$. To show $(2)$ we need to verify that $\B^* \in Z(\B, \C, m)$. Suppose not. Then there is $\H$ with $|H \backslash B^*| < m$ satisfying + \begin{diagram} + & &\H \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\C \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} + As $\B^* \leq \H$ and $\B \subset \M$ we can embed $\H$ into $\M$ (as $\M \models \S$). But this would witness $\C$ extending $\B$ in $\M$ which is impossible as we assumed that $f$ omits $\{\Phi\}$. + + $(2) \Rightarrow (1)$ Suppose $f$ doesn't omit $\{C\}$. Then by the note \ref{AA} $f$ admits $\{C\}$, i.e. there is an embedding of $\C$ into $M$ over $f$. We identify $\C$ with the image of that embedding. Similarly we identify $\B^*$ with the image of its maximal embedding over $f$. That is we may assume $\C, \B^* \subset \M$. Let $H$ be the substructure of $M$ induced by vertices $C \cup B^*$. As $|C \backslash B| < m$ we have $|H \backslash B^*| < m$. $\B^*$ is $m$-strong by remark 5.3 of \cite{Laskowski}. This forces $\B^* \leq H$. But this contradicts the fact that $\B^* \in Z(\B, \C, m)$. +\end{proof} + +\begin{Corollary} \label{B} + With the setup of the previous lemma, the following are equivalent: + \begin{enumerate} + \item $f$ admits $\{\C\}$. + \item There exists $\B^* \in X_m(\B) \backslash Z(\B, \C, m)$ maximally embeddable into $\M$ over $f$. + \end{enumerate} +\end{Corollary} + +For quantifier elimination we need to track multiple structures being admitted and omitted, hence the following definition. + +\begin{Definition} + Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. Define + \begin{align*} + Y(\B, \Phi, \Gamma, m) = \{B^* \in X_m(\B) \mid \ &\forall \C \in \Phi \ B^* \in Z(\B, \C, m) \text{ and}\\ + &\forall \D \in \Gamma \ B^* \notin Z(\B, \D, m)\} + \end{align*} +\end{Definition} + +\begin{Lemma} \label{C} + Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: + \begin{enumerate} + \item $f$ omits $\Phi$ and admits $\Gamma$. + \item There exists $\B^* \in Y(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. + \end{enumerate} +\end{Lemma} +\begin{proof} + Easy corollary of \ref{A} and \ref{B}. +\end{proof} + +\section{Quantifier Elimination} + +Following proof of 5.6 in \cite{Laskowski}, we have a formula $\theta(x, y)$, some $\A \subseteq \B \in \K$ with $\theta(x, y) \vdash \Delta_A(x) \wedge \Delta_\B(x, y)$. We may also assume that $\theta(x, y)$ is a conjunction of formulas of the type $\Psi_{\B, \C}(x,y)$ and their negations. More precisely +\begin{align*} + \theta(x, y) \Leftrightarrow &\bigwedge_{\C \in \Phi} \Psi_{\B, \C}(x,y) \ \wedge \\ + &\bigwedge_{\D \in \Gamma} \neg \Psi_{\B, \D}(x,y) +\end{align*} +for some finite subsets $\Phi, \Gamma$ of $\K$. Let $m = \max\{ |C \backslash B| \colon \C \in \Phi \text{ or } \C \in \Gamma\}$. We claim that in $\M \models \S$ +\begin{align*} + \exists y \theta(x, y) &\Leftrightarrow \bigvee_{\B^* \in Y(\B, \Phi, \Gamma, m)} (\text{$\B^*$ maximally embeds into $\M$ over $\A$}) \\ + &\Leftrightarrow \bigvee_{\B^* \in Y(\B, \Phi, \Gamma, m)} \left(\Psi_{\A, \B^*}(x) \wedge \bigwedge_{\B* \sqsubseteq B', B' \in X_m(\B)} \neg \Psi_{\A, \B'}(x)\right) +\end{align*} + +\begin{proof} + $(\Rightarrow)$ Fix $\B \subset \M$ witnessing existential statement. By remark 5.3 and lemma 3.8 in \cite{Laskowski} there is a unique $\B^* \in X_m$ maximally embeddable (with unique image) into $\M$ over $\B$ . By lemma \ref{C} $\B^* \in Y(\B, \Phi, \Gamma, m)$. + + $(\Leftarrow)$ Take the embedding $g\colon B^* \to \M$ and restrict it to $\B \subseteq \B^*$ i.e. $f = g \mid \B$. As $\B^* \in Y(\B, \Phi, \Gamma, m)$ by lemma \ref{C} $f$ omits $\Phi$ and admits $\Gamma$. Thus is is a witness to $\exists y \theta(x, y)$. +\end{proof} + +\begin{thebibliography}{9} + +\bibitem{Laskowski} + Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} + Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 + +\end{thebibliography} + \end{document} diff --git a/research/07 shelah-spencer QE/Shelah-Spencer QE_old.tex b/research/07 shelah-spencer QE/Shelah-Spencer QE_old.tex index 5f850bdc..3ac27c7f 100644 --- a/research/07 shelah-spencer QE/Shelah-Spencer QE_old.tex +++ b/research/07 shelah-spencer QE/Shelah-Spencer QE_old.tex @@ -1,131 +1,131 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} - -\usepackage{diagrams} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\begin{document} - -\title{Quantifier elimination in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{abstract} - We simplify \cite{Laskowski} proof of quantifier elimination in Shelah-Spencer graphs. -\end{abstract} - -\maketitle - -\section{Introduction} - -Laskowski's paper \cite{Laskowski} provides a combinatorial proof of quantifier elimination in Shelah-Spencer graphs. Here we provide a simplification of the proof using only maximal chains and avoiding technical lemmas of sections 3 and 4. - -We will use notation of \cite{Laskowski}, in particular things like $\K$, $\delta(\A/\B)$, $X_m(\A)$, $\S$, maximal embedding, etc. - -\section{Omitting lemma} - -\begin{Definition} - Let $\M \models \S$, $\B \in \K$, embedding $f \colon \B \to \M$, $\Phi$ finite subset of $\K$ - \begin{enumerate} - \item Say that $f$ \textsl{omits} $\Phi$ if there are no $\C \in \Phi$ and $g \colon \C \to \M$ extending $f$. - \item Say that $f$ \textsl{admits} $\Phi$ if for every $\C \in \Phi$ there is $g \colon \C \to \M$ extending $f$. - \end{enumerate} -\end{Definition} - -\begin{Definition} - Fix $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Define $Z(\B, \C, m)$ to be all $\B^* \in X_m(\B)$ such that there are no $\H$ with $|H \backslash B^*| < m$ satisfying - \begin{diagram} - & &\H \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\C \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} -\end{Definition} - -\begin{Definition} - Fix $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. Define $Z(\B, \Phi, \Gamma, m)$ to be all $\B^* \in X_m(\B)$ such that - \begin{enumerate} - \item For every $\C \in \Phi$ there \textsl{are no} $\H$ with $|H \backslash B^*| < m$ satisfying - \begin{diagram} - & &\H \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\C \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} - \item For every $\D \in \Gamma$ there \textsl{is some} $\G$ with $|G \backslash B^*| < m$ satisfying - \begin{diagram} - & &\G \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\D \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} - \end{enumerate} -\end{Definition} - -\begin{Lemma} - Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: - \begin{enumerate} - \item $f$ omits $\Phi$ and admits $\Gamma$. - \item There exists $\B^* \in Z(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. - \end{enumerate} -\end{Lemma} - -\begin{Lemma} - Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: - \begin{enumerate} - \item $f$ omits $\Phi$ and admits $\Gamma$. - \item There exists $\B^* \in Z(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. - \end{enumerate} -\end{Lemma} -\begin{proof} - $(1) \Rightarrow (2)$ Identify $\B$ with $f(\B)$, i.e. for ease of notation assume that $\B \subset \M$. By remark 5.3 of \cite{Laskowski} there is some $B^* \in X_m(\B)$ maximally embeddable in $\M$ over $f$. Such embedding is unique by Lemma 3.8 of \cite{Laskowski}. Again, we identify $B^*$ with its maximal embedding into $\M$. To show $(2)$ we need to verify that $\B^* \in Z(\B, \Phi, \Gamma, m)$. Suppose not. Two things can go wrong. First, there can be $\H$ with $|H \backslash B^*| < m$ and $\C \in \Phi$ satisfying - \begin{diagram} - & &\H \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\C \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} - As $\B^* \leq \H$ and $\B \subset \M$ we can embed $\H$ into $\M$ (as $\M \models \S$). But this would witness $\C$ extending $\B$ in $\M$ which is impossible as we assumed that $f$ omits $\Phi$. Another thing that could go wrong is that there could be $\D \in \Gamma$ and no $\G$ with $|G \backslash B^*| < m$ satisfying - \begin{diagram} - & &\G \\ - &\ruLine^\leq & &\luLine \\ - \B^* & & & &\D \\ - &\luLine & &\ruLine \\ - & &\B - \end{diagram} - As $f$ admits -\end{proof} - -\begin{thebibliography}{9} - -\bibitem{stable_graphs} - Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. - -\bibitem{infinite_megner} - Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 - -\bibitem{simon_dp_minimal} - P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. - -%\bibitem{diestel} -% Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{diagrams} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\begin{document} + +\title{Quantifier elimination in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We simplify \cite{Laskowski} proof of quantifier elimination in Shelah-Spencer graphs. +\end{abstract} + +\maketitle + +\section{Introduction} + +Laskowski's paper \cite{Laskowski} provides a combinatorial proof of quantifier elimination in Shelah-Spencer graphs. Here we provide a simplification of the proof using only maximal chains and avoiding technical lemmas of sections 3 and 4. + +We will use notation of \cite{Laskowski}, in particular things like $\K$, $\delta(\A/\B)$, $X_m(\A)$, $\S$, maximal embedding, etc. + +\section{Omitting lemma} + +\begin{Definition} + Let $\M \models \S$, $\B \in \K$, embedding $f \colon \B \to \M$, $\Phi$ finite subset of $\K$ + \begin{enumerate} + \item Say that $f$ \textsl{omits} $\Phi$ if there are no $\C \in \Phi$ and $g \colon \C \to \M$ extending $f$. + \item Say that $f$ \textsl{admits} $\Phi$ if for every $\C \in \Phi$ there is $g \colon \C \to \M$ extending $f$. + \end{enumerate} +\end{Definition} + +\begin{Definition} + Fix $\B, \C \in \K$, and $m \in \omega$ such that $|C \backslash B| < m$. Define $Z(\B, \C, m)$ to be all $\B^* \in X_m(\B)$ such that there are no $\H$ with $|H \backslash B^*| < m$ satisfying + \begin{diagram} + & &\H \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\C \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} +\end{Definition} + +\begin{Definition} + Fix $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. Define $Z(\B, \Phi, \Gamma, m)$ to be all $\B^* \in X_m(\B)$ such that + \begin{enumerate} + \item For every $\C \in \Phi$ there \textsl{are no} $\H$ with $|H \backslash B^*| < m$ satisfying + \begin{diagram} + & &\H \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\C \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} + \item For every $\D \in \Gamma$ there \textsl{is some} $\G$ with $|G \backslash B^*| < m$ satisfying + \begin{diagram} + & &\G \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\D \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} + \end{enumerate} +\end{Definition} + +\begin{Lemma} + Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: + \begin{enumerate} + \item $f$ omits $\Phi$ and admits $\Gamma$. + \item There exists $\B^* \in Z(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. + \end{enumerate} +\end{Lemma} + +\begin{Lemma} + Let $\B \in \K$, $\Phi, \Gamma$ finite subsets of $\K$, and $m \in \omega$ such that for each $\C \in \Phi$ or $\C \in \Gamma$ we have $\B \subseteq \C$ and $|C \backslash B| < m$. The following are equivalent: + \begin{enumerate} + \item $f$ omits $\Phi$ and admits $\Gamma$. + \item There exists $\B^* \in Z(\B, \Phi, \Gamma, m)$ maximally embeddable into $\M$ over $f$. + \end{enumerate} +\end{Lemma} +\begin{proof} + $(1) \Rightarrow (2)$ Identify $\B$ with $f(\B)$, i.e. for ease of notation assume that $\B \subset \M$. By remark 5.3 of \cite{Laskowski} there is some $B^* \in X_m(\B)$ maximally embeddable in $\M$ over $f$. Such embedding is unique by Lemma 3.8 of \cite{Laskowski}. Again, we identify $B^*$ with its maximal embedding into $\M$. To show $(2)$ we need to verify that $\B^* \in Z(\B, \Phi, \Gamma, m)$. Suppose not. Two things can go wrong. First, there can be $\H$ with $|H \backslash B^*| < m$ and $\C \in \Phi$ satisfying + \begin{diagram} + & &\H \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\C \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} + As $\B^* \leq \H$ and $\B \subset \M$ we can embed $\H$ into $\M$ (as $\M \models \S$). But this would witness $\C$ extending $\B$ in $\M$ which is impossible as we assumed that $f$ omits $\Phi$. Another thing that could go wrong is that there could be $\D \in \Gamma$ and no $\G$ with $|G \backslash B^*| < m$ satisfying + \begin{diagram} + & &\G \\ + &\ruLine^\leq & &\luLine \\ + \B^* & & & &\D \\ + &\luLine & &\ruLine \\ + & &\B + \end{diagram} + As $f$ admits +\end{proof} + +\begin{thebibliography}{9} + +\bibitem{stable_graphs} + Klaus-Peter Podewski and Martin Ziegler. Stable graphs. \textit{Fund. Math.}, 100:101-107, 1978. + +\bibitem{infinite_megner} + Aharoni, Ron and Berger, Eli (2009). "Menger's Theorem for infinite graphs". \textit{Inventiones Mathematicae} 176: 1–62 + +\bibitem{simon_dp_minimal} + P. Simon, \textit{On dp-minimal ordered structures}, J. Symbolic Logic 76 (2011), no. 2, 448–460. + +%\bibitem{diestel} +% Reinhard Diestel. \textit{Graph Theory}, volume 173 of \textit{Grad. Texts in Math.} Springer, 2005. + +\end{thebibliography} + \end{document} diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2 - backup.tex b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2 - backup.tex index eba39cc3..0f4c8613 100644 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2 - backup.tex +++ b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2 - backup.tex @@ -1,244 +1,244 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} -\usepackage{../Thm} - -\usepackage{mathrsfs} - - - -\renewcommand{\AA}{\mathscr A} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - - \newcommand{\U}{\mathcal U} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\newcommand{\curly}[1]{\left\{#1\right\}} -\newcommand{\paren}[1]{\left(#1\right)} -\newcommand{\abs}[1]{\left|#1\right|} - -\providecommand{\floor}[1]{\left \lfloor #1 \right \rfloor } - -%\DeclareMathOperator{\dim}{dim} - -\title{Some vc-density computations in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{document} - -\maketitle - -Fix a formula $\phi(x, y)$ that is a minimal chain extension $\curly{M_i}_{i \in [0..k]}$ with $M_0 = \{x, y\}$ with -\begin{itemize} - \item $\phi(x, y)$ determines edges and non-edges on $\curly{x, y}$. - \item there are no edges between $x$ and $y$. - \item there are no edges between $x$. - \item Let $\dim \paren{M_i/M_{i-1}} = -\epsilon_i$. - \item Let $\epsilon_L = \sum_{[1..k]} \epsilon_i$. - \item Let $\epsilon_U = \min_{[1..k]} \epsilon_i$. - \item Let $Y = \dim (y)$ considering $y$ as a graph. If $\{y\}$ are disconnected then $Y = |y|$. -\end{itemize} - -We work in special family of parameter sets -\begin{align*} - \AA = \curly{A \subset \U^{y} \mid \text{finite, disconnected, strongly embedded}} -\end{align*} - -We estimate $\vc_\AA(\phi)$, VC-density of $\phi$ restricted to parameter sets from $\AA$. - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section*{Lower bound on $\AA$} - -Let $n$ be the integer such that $n \epsilon_L < Y$ and $(n+1) \epsilon_L > Y$. - -Pick a finite $B \subset A^{|x|}$. - -Consider the graph $y$. -If $y$ is not positive, then $\phi$ has no realizations over $B$. -Otherwise, take an abstract realization of $y$, and label it by $b$. - -Fix $n$ arbitrary elements of $B$, label them $a_i$, with each $|a_i| = |x|$. -Abstractly adjoin $M_i/\curly{a_i, b} = M/\curly{x,y}$ for each $i$. -Let $\bar M = \bigcup M_i$ (disjointly). - -\begin{Claim} - $(A \cap \bar M) \leq \bar M$. -\end{Claim} -\begin{proof} - It's total dimension is $Y - n\epsilon_L > 0$ and all subextensions are positive as well. -\end{proof} - -Thus a copy of $\bar M$ can be embedded over $A$ into our ambient model. -Choice of elements of $B$ was arbitrary, thus showing that any $n$ elements can be traced out. -Thus we have $O(|B|^n)$ many traces showing $\vc$-density of at least $n$. - -\begin{align*} - \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_L}} -\end{align*} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section*{Lower bound} - -\begin{Claim} - In any random graph we can find arbitrarily large parameter set that belongs to $A$. -\end{Claim} - -This shows that - -\begin{align*} - \vc(\phi) \geq \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_L}} -\end{align*} - -\begin{Claim} - We can find a minimal extension $M / \{x, y\}$ with arbitrarily small dimension. -\end{Claim} - -This shows that vc function is infinite in Shelah-Spencer random graphs. - -\begin{align*} - \vc(n) = \infty -\end{align*} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section*{Upper bound on $\AA$} - -Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. - -Pick a trace of $\phi(x,y)$ on $A^{|x|}$ by a parameter $b$. - -\begin{align*} - B = \curly{a \in A^{|x|} \mid \phi(a, b)} -\end{align*} - -Pick $B' \subset B$, ordered $B' = \{a_i\}_{i \in I}$ such that -\begin{align*} - %a_i \cap \bigcup_{j \neq i} a_j \neq \emptyset - a_i \cap \bigcup_{j < i} a_j \neq \emptyset -\end{align*} -This is always possible by starting with $B$ and taking away elements one by one. -Call such a set a \emph{generating set} of $B$. - -Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \in I$. -Let $\bar M = \bigcup M_i$. -Consider $\bar M / A$. - -Pick $\bar M$ such that $\dim(\bar M / A)$ is maximized. - -$\bar M \cap A \leq \bar M$ as $A$ is strong. (Make sure $M$ is not too big!) -Let $\bar A = A - \curly{a_i}_{i \in I}$. -Suppose $\bar A \cap \bar M \neq \emptyset$. -Then we can abstractly reembed $\M$ over $A$ such that $\bar A \cap \bar M = \emptyset$. -This would increase the dimension, contradicting maximality. -Thus we can assume $A \cap \bar M = \{a_i\}_{i \in I}$ - -Let $\bar M_j = \bigcup_{i < j} M_i$. - -\begin{Lemma} - $\dim(\bar M_j / A) \leq j \cdot \epsilon_U$ -\end{Lemma} -\begin{proof} - Proceed by induction. - Base case is clear. - - For induction case apply lemma to $\bar M_j \cup \{a_j\}$ and $M_j / \{a_j, b\}$. - There are two cases - \begin{enumerate} - \item $M_j \subset \bar M' \cup \{a_j\}$. - In this case there are edges between $\{a_j\}$ and $M_j$ that contribute to dimension less than $-\epsilon_U$. - \item Otherwise $M_j$ adds extra dimension less than $-\epsilon_U$ - \end{enumerate} -\end{proof} - -Thus we have $\dim(\bar M / A) = \dim(\bar M_n / A) \leq -\epsilon_U n$. - -Thus as $A$ is strong we need $|B'| \epsilon_U < Y$. -This gives us $|B'| \leq n$. -Finally we need to relate $|B'|$ to $|B|$. - -Suppose we have $C \subset A^{|x|}$, finite with $|C| = N$. -A generating set for a trace has to have size $\leq n$. -Thus there are ${N \choose n} \leq N^n$ choices for a generating set. -A set generated from set of size $n$ can have at most $(x|n|)^{|x|}$ elements. -Thus a given set of size $n$ can generate at most -\begin{align*} - 2^{(x|n|)^{|x|}} -\end{align*} -sets. -Thus the number of possible traces on $C$ is bounded above by -\begin{align*} - 2^{(x|n|)^{|x|}} \cdot N^n = O(N^n) -\end{align*} -This bounds the vc-density by $n$. - -\begin{align*} - \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_U}} -\end{align*} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section*{Technical Lemmas} - -\begin{Lemma} - Suppose we have a set $B$ and a minimal pair $(M, A)$ with $A \subset B$ and $\dim(M/A) = -\epsilon$. -Then either $M \subseteq B$ or $\dim((M \cup B)/B) < -\epsilon$. -\end{Lemma} - -\begin{proof} - By diamond construction - - \begin{align*} - \dim((M \cup B)/B) \leq \dim(M / (M \cap B)) - \end{align*} - - and - - \begin{align*} - \dim(M / (M \cap B)) &= \dim (M/A) - \dim(M / (M \cap B)) \\ - \dim (M/A) &= -\epsilon \\ - \dim(M / (M \cap B)) &> 0 - \end{align*} -\end{proof} - - - -\begin{Lemma} - Suppose we have a set $B$ and a minimal chain $M_n$ with $M_0 \subset B$ and dimensions $-\epsilon_i$. -Let $\epsilon$ be the minimal of $\epsilon_i$. -Then either $M_n \subseteq B$ or $\dim((M_n \cup B)/B) < -\epsilon$. -\end{Lemma} - - -\begin{proof} - Let $\bar M_i = M_i \cup B$ - - \begin{align*} - \dim(\bar M_n/B) = \dim(\bar M_n/\bar M_{n-1}) + \ldots + \dim(\bar M_2/\bar M_1) + \dim(\bar M_1/B) - \end{align*} - - Either $M_n \subseteq B$ or one of the summands above is nonzero. - Apply previous lemma. -\end{proof} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section*{Counterexamples} - -% Add complete graph counterexample -% where we have a bunch of minimal extensions intersecting in a tiny way - -% \AA is indiscernible -% example of indiscernible sequence that is not strong -% example with non-strong embedding on every n-tuple of vertices? - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} +\usepackage{../Thm} + +\usepackage{mathrsfs} + + + +\renewcommand{\AA}{\mathscr A} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\U}{\mathcal U} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\newcommand{\curly}[1]{\left\{#1\right\}} +\newcommand{\paren}[1]{\left(#1\right)} +\newcommand{\abs}[1]{\left|#1\right|} + +\providecommand{\floor}[1]{\left \lfloor #1 \right \rfloor } + +%\DeclareMathOperator{\dim}{dim} + +\title{Some vc-density computations in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{document} + +\maketitle + +Fix a formula $\phi(x, y)$ that is a minimal chain extension $\curly{M_i}_{i \in [0..k]}$ with $M_0 = \{x, y\}$ with +\begin{itemize} + \item $\phi(x, y)$ determines edges and non-edges on $\curly{x, y}$. + \item there are no edges between $x$ and $y$. + \item there are no edges between $x$. + \item Let $\dim \paren{M_i/M_{i-1}} = -\epsilon_i$. + \item Let $\epsilon_L = \sum_{[1..k]} \epsilon_i$. + \item Let $\epsilon_U = \min_{[1..k]} \epsilon_i$. + \item Let $Y = \dim (y)$ considering $y$ as a graph. If $\{y\}$ are disconnected then $Y = |y|$. +\end{itemize} + +We work in special family of parameter sets +\begin{align*} + \AA = \curly{A \subset \U^{y} \mid \text{finite, disconnected, strongly embedded}} +\end{align*} + +We estimate $\vc_\AA(\phi)$, VC-density of $\phi$ restricted to parameter sets from $\AA$. + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section*{Lower bound on $\AA$} + +Let $n$ be the integer such that $n \epsilon_L < Y$ and $(n+1) \epsilon_L > Y$. + +Pick a finite $B \subset A^{|x|}$. + +Consider the graph $y$. +If $y$ is not positive, then $\phi$ has no realizations over $B$. +Otherwise, take an abstract realization of $y$, and label it by $b$. + +Fix $n$ arbitrary elements of $B$, label them $a_i$, with each $|a_i| = |x|$. +Abstractly adjoin $M_i/\curly{a_i, b} = M/\curly{x,y}$ for each $i$. +Let $\bar M = \bigcup M_i$ (disjointly). + +\begin{Claim} + $(A \cap \bar M) \leq \bar M$. +\end{Claim} +\begin{proof} + It's total dimension is $Y - n\epsilon_L > 0$ and all subextensions are positive as well. +\end{proof} + +Thus a copy of $\bar M$ can be embedded over $A$ into our ambient model. +Choice of elements of $B$ was arbitrary, thus showing that any $n$ elements can be traced out. +Thus we have $O(|B|^n)$ many traces showing $\vc$-density of at least $n$. + +\begin{align*} + \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_L}} +\end{align*} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section*{Lower bound} + +\begin{Claim} + In any random graph we can find arbitrarily large parameter set that belongs to $A$. +\end{Claim} + +This shows that + +\begin{align*} + \vc(\phi) \geq \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_L}} +\end{align*} + +\begin{Claim} + We can find a minimal extension $M / \{x, y\}$ with arbitrarily small dimension. +\end{Claim} + +This shows that vc function is infinite in Shelah-Spencer random graphs. + +\begin{align*} + \vc(n) = \infty +\end{align*} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section*{Upper bound on $\AA$} + +Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. + +Pick a trace of $\phi(x,y)$ on $A^{|x|}$ by a parameter $b$. + +\begin{align*} + B = \curly{a \in A^{|x|} \mid \phi(a, b)} +\end{align*} + +Pick $B' \subset B$, ordered $B' = \{a_i\}_{i \in I}$ such that +\begin{align*} + %a_i \cap \bigcup_{j \neq i} a_j \neq \emptyset + a_i \cap \bigcup_{j < i} a_j \neq \emptyset +\end{align*} +This is always possible by starting with $B$ and taking away elements one by one. +Call such a set a \emph{generating set} of $B$. + +Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \in I$. +Let $\bar M = \bigcup M_i$. +Consider $\bar M / A$. + +Pick $\bar M$ such that $\dim(\bar M / A)$ is maximized. + +$\bar M \cap A \leq \bar M$ as $A$ is strong. (Make sure $M$ is not too big!) +Let $\bar A = A - \curly{a_i}_{i \in I}$. +Suppose $\bar A \cap \bar M \neq \emptyset$. +Then we can abstractly reembed $\M$ over $A$ such that $\bar A \cap \bar M = \emptyset$. +This would increase the dimension, contradicting maximality. +Thus we can assume $A \cap \bar M = \{a_i\}_{i \in I}$ + +Let $\bar M_j = \bigcup_{i < j} M_i$. + +\begin{Lemma} + $\dim(\bar M_j / A) \leq j \cdot \epsilon_U$ +\end{Lemma} +\begin{proof} + Proceed by induction. + Base case is clear. + + For induction case apply lemma to $\bar M_j \cup \{a_j\}$ and $M_j / \{a_j, b\}$. + There are two cases + \begin{enumerate} + \item $M_j \subset \bar M' \cup \{a_j\}$. + In this case there are edges between $\{a_j\}$ and $M_j$ that contribute to dimension less than $-\epsilon_U$. + \item Otherwise $M_j$ adds extra dimension less than $-\epsilon_U$ + \end{enumerate} +\end{proof} + +Thus we have $\dim(\bar M / A) = \dim(\bar M_n / A) \leq -\epsilon_U n$. + +Thus as $A$ is strong we need $|B'| \epsilon_U < Y$. +This gives us $|B'| \leq n$. +Finally we need to relate $|B'|$ to $|B|$. + +Suppose we have $C \subset A^{|x|}$, finite with $|C| = N$. +A generating set for a trace has to have size $\leq n$. +Thus there are ${N \choose n} \leq N^n$ choices for a generating set. +A set generated from set of size $n$ can have at most $(x|n|)^{|x|}$ elements. +Thus a given set of size $n$ can generate at most +\begin{align*} + 2^{(x|n|)^{|x|}} +\end{align*} +sets. +Thus the number of possible traces on $C$ is bounded above by +\begin{align*} + 2^{(x|n|)^{|x|}} \cdot N^n = O(N^n) +\end{align*} +This bounds the vc-density by $n$. + +\begin{align*} + \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_U}} +\end{align*} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section*{Technical Lemmas} + +\begin{Lemma} + Suppose we have a set $B$ and a minimal pair $(M, A)$ with $A \subset B$ and $\dim(M/A) = -\epsilon$. +Then either $M \subseteq B$ or $\dim((M \cup B)/B) < -\epsilon$. +\end{Lemma} + +\begin{proof} + By diamond construction + + \begin{align*} + \dim((M \cup B)/B) \leq \dim(M / (M \cap B)) + \end{align*} + + and + + \begin{align*} + \dim(M / (M \cap B)) &= \dim (M/A) - \dim(M / (M \cap B)) \\ + \dim (M/A) &= -\epsilon \\ + \dim(M / (M \cap B)) &> 0 + \end{align*} +\end{proof} + + + +\begin{Lemma} + Suppose we have a set $B$ and a minimal chain $M_n$ with $M_0 \subset B$ and dimensions $-\epsilon_i$. +Let $\epsilon$ be the minimal of $\epsilon_i$. +Then either $M_n \subseteq B$ or $\dim((M_n \cup B)/B) < -\epsilon$. +\end{Lemma} + + +\begin{proof} + Let $\bar M_i = M_i \cup B$ + + \begin{align*} + \dim(\bar M_n/B) = \dim(\bar M_n/\bar M_{n-1}) + \ldots + \dim(\bar M_2/\bar M_1) + \dim(\bar M_1/B) + \end{align*} + + Either $M_n \subseteq B$ or one of the summands above is nonzero. + Apply previous lemma. +\end{proof} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section*{Counterexamples} + +% Add complete graph counterexample +% where we have a bunch of minimal extensions intersecting in a tiny way + +% \AA is indiscernible +% example of indiscernible sequence that is not strong +% example with non-strong embedding on every n-tuple of vertices? + \end{document} - + % Include edges between y as a chain minimal extension \ No newline at end of file diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.bbl b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.blg b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.blg deleted file mode 100644 index e4010112..00000000 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Shelah-Spencer VC - 2.aux -I found no \bibdata command---while reading file Shelah-Spencer VC - 2.aux -I found no \bibstyle command---while reading file Shelah-Spencer VC - 2.aux -(There were 2 error messages) diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tcp b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tcp index f7044c01..eb26f36d 100644 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tcp +++ b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tcp @@ -1,12 +1,12 @@ -[FormatInfo] -Type=TeXnicCenterProjectInformation -Version=4 - -[ProjectInfo] -MainFile=Shelah-Spencer VC - 2.tex -UseBibTeX=0 -UseMakeIndex=0 -ActiveProfile=LaTeX ⇨ DVI -ProjectLanguage=en -ProjectDialect=US - +[FormatInfo] +Type=TeXnicCenterProjectInformation +Version=4 + +[ProjectInfo] +MainFile=Shelah-Spencer VC - 2.tex +UseBibTeX=0 +UseMakeIndex=0 +ActiveProfile=LaTeX ⇨ DVI +ProjectLanguage=en +ProjectDialect=US + diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tex b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tex index 1493b56b..8ff3356e 100644 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tex +++ b/research/08 shelah-spencer VC/Shelah-Spencer VC - 2.tex @@ -1,758 +1,758 @@ -\documentclass{amsart} - -\usepackage{../AMC_style} -\usepackage{../Research} -\usepackage{../Thm} - -\usepackage{mathrsfs} - - - -\renewcommand{\AA}{\mathscr A} - \newcommand{\II}{\mathscr I} - \newcommand{\MM}{\mathscr M} - - \newcommand{\A}{\mathcal A} - \newcommand{\B}{\mathcal B} -\renewcommand{\C}{\mathcal C} - \newcommand{\D}{\mathcal D} -\renewcommand{\H}{\mathcal H} - \newcommand{\G}{\mathcal G} - \newcommand{\M}{\mathcal M} - \newcommand{\U}{\mathcal U} - \newcommand{\X}{\mathcal X} - \newcommand{\Y}{\mathcal Y} - - \newcommand{\K}{\boldface K_\alpha} -\renewcommand{\S}{S_\alpha} - -\newcommand{\curly}[1]{\left\{#1\right\}} -\newcommand{\paren}[1]{\left(#1\right)} -\newcommand{\abs}[1]{\left|#1\right|} -\newcommand{\agl}[1]{\left\langle #1 \right\rangle} - -\providecommand{\floor}[1]{\left \lfloor #1 \right \rfloor } - -%\DeclareMathOperator{\dim}{dim} - -\title{Some vc-density computations in Shelah-Spencer graphs} -\author{Anton Bobkov} -\email{bobkov@math.ucla.edu} - -\begin{document} - -\maketitle - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Preliminaries} - -VC density was introduced in \cite{vc_density} by Aschenbrenner, Dolich, Haskell, MacPherson, and Starchenko as a natural notion of dimension for NIP theories. In a NIP theory we can define a VC function - -\begin{align*} - \vc : \N \arr \N -\end{align*} - -Where $vc(n)$ measures complexity of definable sets in an $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. - -In this paper, we investigate vc-density of definable sets in Shelah-Spencer structures. -We follow notations in \cite{laskowski}. -In this paper we work with limit of random structure $G(n, n^{-\alpha})$ for $\alpha \in (0,1)$, irrational. -This structure is axiomatized by $S_\alpha$. -Our ambient model is $\MM$. -Notations we use are $\delta(\A), \delta(\A/\B), \A \leq \B$ as well as notions of $N$-strong substructure, minimal extension, chain minimal extension, minimal pair, and $N$-strong closure. - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Graph Combinatorics} - -We denote graph by $\A$, set of its vertices by $A$. -When we say $\A \subseteq \B$ we mean that $A \subseteq B$ and edges of $\A$ are also edges of $\B$. -However $\B$ may add new edges between vertices of $\A$. - -Fix $\alpha \in (0,1)$, irrational. -For a finite graph $\A$ let -\begin{align*} - \delta(\A) = |A| - \alpha e(\A) -\end{align*} - -where $e(\A)$ is the number of edges in $\A$. - -For finite $\A,\B$ with $\A \subseteq \B$ define $\delta(\B/\A) = \delta(\B) - \delta(\A)$. -We say that $\A \leq \B$ if $\A \subseteq \B$ and $\delta(\A'/\B) > 0$ for all $\A \subseteq \A' \subsetneq \B$. - -We say that finite $\A$ is positive if for all $\A' \subseteq \A$ we have $\delta(\A') \geq 0$. - -\begin{Definition} - We work in theory $S_\alpha$ axiomatized by - \begin{itemize} - \item Every finite substructure is positive - \item For a model $\MM$ given $\A \leq \B$ every embedding $f : \A \arr \MM$ extends to $g: \B \arr \MM$. - \end{itemize} -\end{Definition} - -For $\A, \B$ positive, $(\A, \B)$ is called a minimal pair if $\A \subseteq \B$, $\delta(\B/\A) < 0$ but $\delta(\A'/\A) \geq 0$ for all proper $\A \subseteq \A' \subsetneq \B$. - -$\agl{\A_i}_{i \leq m}$ is called a minimal chain if $(\A_i, \A_i+1)$ is a minimal pair (for all $i < m$). - -For a positive $\A$ let $\delta_\A(\bar x)$ be the atomic diagram of $\A$. For positive $\A \subset \B$ let - -\begin{align*} - \Psi_{\A,\B}(\bar x) = \delta_\A(\bar x) \wedge \exists \bar y \; \delta_\B(\bar x, \bar y) -\end{align*} - -Such formula is called chain-minimal extension formula if in addition we have that there is a minimal chain starting at $\A$ and ending in $\B$. Denote such formulas as $\Psi_{\agl{\M_i}}$ - -\begin{Theorem} [5.6 in \cite{laskowski}] - $S_\alpha$ admits quantifier elimination down to boolean combination of chain-minimal extension formulas. -\end{Theorem} - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Definitions} - -%\begin{Definition} - %Let $x = (x_1, \ldots x_n), y = (y_1, \ldots, y_m)$ be variable tuples. - %We call a formula $\phi(x, y)$ \emph{basic} when - %\begin{itemize} - %\item $\phi(x, y)$ is a minimal chain extension, denoted by $\curly{M_i}_{i \in [0..k]}$ with $M_0 = \{x, y\}$. - %\item $\phi(x, y)$ determines edges and non-edges on its variables $\{x_1, \ldots x_n\} \cup \{y_1, \ldots y_m\}$. - %\item there is no edge between $x_i$ and $y_j$ for all $i,j$. (see note \ref{note_edges}) - %\item Define $\mathbf x$ to be the graph on vertices $\{x_i\}$ with edges as defined by $\phi$. - %Similarly define $\mathbf y$. - %We require $\mathbf x$ and $\mathbf y$ to be positive. (see note \ref{note_positive})\ - %%\item all elements of $y$ that are connected to $M_k - \{x,y\}$. (see note \ref{note_special}) - %\end{itemize} -%\end{Definition} - -% issue: multiple possible chain decompositions? - -Fix tuples $x = (x_1, \ldots x_n), y = (y_1, \ldots, y_m)$. -We refer to chain-minimal extension formulas as basic formulas. -Let $\phi_{\agl{\M_i}}(x, y)$ be a basic formula. - -\begin{Definition} - Define $\X$ to be the graph on vertices $\{x_i\}$ with edges as defined by $\phi_{\agl{\M_i}}$. - Similarly define $\Y$. - We define those abstractly, i.e. on a new set of vertices disjoint from $\MM$. -\end{Definition} - -Note that $\X$, $\Y$ are positive as they are subgraphs of $\M_0$. -As usual $X, Y$ will refer to vertices of those graphs. - -We restrict our attention to formulas that define no edges between $X$ and $Y$. - -\begin{Note} \label{note_edges} - We can handle edges between $x$ and $y$ as separate elements of the minimal chain extension. -\end{Note} - -%\begin{Note} \label{note_positive} - %If either graph $\mathbf x$ or $\mathbf y$ is negative, then the formula would have no realizations as negative graphs cannot be embedded into our ambient model. -%\end{Note} - -%\begin{Note} \label{note_special} - %We add the final condition to simplify our analysis. Similar techniques can be used to acquire bounds on formulas not subject to that condition. -%\end{Note} - -\begin{Definition} \label{def_basic} - For a basic formula $\phi = \phi_{\agl{\M_i}_{i \leq k}}(x, y)$ let - \begin{itemize} - \item $\epsilon_i(\phi) = -\dim \paren{M_i/M_{i-1}}$. - \item $\epsilon_L(\phi) = \sum_{[1..k]} \epsilon_i(\phi)$. - \item $\epsilon_U(\phi) = \min_{[1..k]} \epsilon_i(\phi)$. - \item Let $\Y'$ be a subgraph of $\Y$ induced by vertices of $\Y$ that are connected to $M_k - (X \cup Y)$. - \item Let $Y(\phi) = \dim (\Y')$. - In particular if $\Y = \Y'$ and $\Y$ is disconnected then $Y(\phi)$ is just the arity of the tuple $y$. - \end{itemize} -\end{Definition} - -%\begin{Definition} -%\begin{align*} - %\epsilon_L(\neg \phi) &= \epsilon_L(\phi) \\ - %\epsilon_U(\neg \phi) &= \epsilon_U(\phi) \\ - %\epsilon_L(\phi \wedge \psi) &= \epsilon_L(\phi) + \epsilon_L(\psi) \\ - %\epsilon_U(\phi \wedge \psi) &= \min(\epsilon_U(\phi), \epsilon_U(\psi)) \\ - %\epsilon_L(\phi \vee \psi) &= \min(\epsilon_L(\phi), \epsilon_L(\psi)) \\ - %\epsilon_U(\phi \vee \psi) &= \min(\epsilon_U(\phi), \epsilon_U(\psi)) -%\end{align*} -%\end{Definition} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Lower bound} - -As a simplification for our lower bound computation we assume that all the basic formulas involved we have $\Y' = \Y$ (see Definition \ref{def_basic}). - -We work with formulas that are boolean combinations of basic formulas written in disjunctive-conjunctive form. -First, we extend our definition of $\epsilon$. - -\begin{Definition}[Negation] - If $\phi$ is a basic formula, then define - \begin{align*} - \epsilon_L(\neg \phi) &= \epsilon_L(\phi) - \end{align*} -\end{Definition} - -\begin{Definition}[Conjunction] - Take a collection of formulas $\phi_i(x, y)$ where each $\phi_i$ is positive or negative basic formula. - If both positive and negative formulas are present then $\epsilon_L(\phi) = \infty$. - We don't have a lower bound for that case. - If different formulas define $\X$ or $\Y$ differently then $\epsilon_L(\phi) = \infty$. - In that case of the conflicting definitions would make the formula have no realizations. - Otherwise - \begin{align*} - \epsilon_L(\bigwedge \phi_i) &= \sum \epsilon_L(\phi_i) - \end{align*} -\end{Definition} - -\begin{Definition} [Disjunction] - Take a collection of formulas $\psi_i$ where each instance is a conjunction of positive and negative instances of basic formulas that agree on $\X$ and $\Y$. % can generalize? - \begin{align*} - \epsilon_L(\bigvee \psi_i) &= \min \epsilon_L(\psi_i) - \end{align*} -\end{Definition} - -\begin{Theorem} - For a formula $\phi$ as above - \begin{align*} - \vc \phi \geq \floor{\frac{Y(\phi)}{\epsilon_L(\phi)}} - \end{align*} - where $Y(\phi)$ is $Y(\psi)$ for $\psi$ one the basic components of $\phi$ (all basic componenets agree on $\Y$). -\end{Theorem} - -\begin{proof} - First work with a formula that is a conjunction of positive basic formulas. - - \begin{align*} - \psi = \bigwedge_{j \leq J} \phi_j - \end{align*} - Then as we defined above - \begin{align*} - \epsilon_L(\psi) = \sum \epsilon_L(\phi_j) - \end{align*} - - Let $\phi$ be one of the basic formulas in $\psi$ with a chain $\agl{M_i}_{i \leq k}$. - Let $K_\phi = |M_k|$ i.e. the size of the extension. - Let $K$ be the largest such size among all $\phi_i$. - - Let $n$ be the integer such that $n \epsilon_L(\psi) < Y$ and $(n+1) \epsilon_L(\psi) > Y$. - - %Take an abstract realization of $y$ as dictated by $\psi$, and label it by $b$ ($b \in \MM$) - Label $\Y$ by an tuple $b$. - - Pick parameter set $A \subset \MM$ such that - - \begin{align*} - A = \bigcup_{i 0$ as needed. - \end{proof} - - $|\bar M| \leq N \cdot I \cdot K$ and $A$ is $\leq N \cdot I \cdot K$-strong. - Thus a copy of $\bar M$ can be embedded over $A$ into our ambient model $\MM$. - Our choice of $b_i$'s was arbitrary, so we get ${N \choose n}$ choices out of $N|x|$ many elements. - Thus we have $O(|A|^n)$ many traces. - - \begin{Lemma} - There are arbitrarily large sets with properties of $A$. - \end{Lemma} - - \begin{proof} - \textit{proof goes here. use lemma in laskoswki paper} - \end{proof} - - This shows - - \begin{align*} - \vc \psi \geq n = \floor{\frac{Y}{\epsilon_L}} - \end{align*} - - Now consider the formula which is a conjunction consists of negative basic formulas - \begin{align*} - \psi = \bigwedge \neg \phi_i - \end{align*} - Let - \begin{align*} - \bar \psi = \bigwedge \phi_i - \end{align*} - - Do the construction above for $\bar \psi$ and suppose its trace is $X \subset A$ for some $b$. - Then over $b$ the same construction gives trace $(A - X)$ for $\psi$. Thus we get as many traces. - - Finally consider a formula which is a disjunction of formulas considered above. - Choose the one with the smallest $\epsilon_L$, this yields the lower bound for the entire formula. - %explain! non-trivial. thing of disjunction of two formulas - % also! what if disjunction has formulas disagreeing on x and y - % aslo! what if disjuction where one of the formulas is mixed positive/negative formulas -\end{proof} - -\begin{Claim} - We can find a minimal extension $M / \{x, y\}$ with arbitrarily small dimension. -\end{Claim} - -\begin{proof} - Put proof in here. Follow construction in Laskowski paper. -\end{proof} - -This shows that vc function is infinite in Shelah-Spencer random graphs. - -\begin{align*} - \vc(n) = \infty -\end{align*} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Upper bound} - - -%\begin{Definition} - %Let $\phi$ be a basic formula with $M_i$ a minimal chain, - %$\epsilon_i$ its corresponding dimensions, - %and $M$ its total size. - %\begin{align*} - %U_\phi = \frac{M}{\min \epsilon_i} - %\end{align*} -%\end{Definition} - - - -%\begin{Definition} [Negation] - %Let $\phi$ be basic - %\begin{align*} - %U_{\neg \phi} = U_{\phi} - %\end{align*} -%\end{Definition} -% -%\begin{Definition} [Conjunction and Disjunction] - %Let $\phi_{ij}$ be basic or a negation of a basic formula. - %\begin{align*} - %\psi = \bigvee \bigwedge \phi_{ij} - %\end{align*} - %\begin{align*} - %U_\psi = \max U_{\phi_{ij}} - %\end{align*} -%\end{Definition} - -%Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. - -%Consider a formula which is a conjunction of positive basic formulas. - -Consider a case of a single basic formula $\phi(\vec x, \vec y)$. - -Suppose it defines a minimal chain extension over $\{x, y\}$. -Record the size of that extension as $K(\phi)$ and its total dimension $\epsilon(\phi) = \epsilon_U(\phi)$. - -In general we have parameter set $A \subset \MM^{|x|}$, however without loss of generality we may work with -a parameter set $A^{|x|}$, with $A \subset \MM$. - -Let $S = \floor{\frac{K(\phi)Y(\phi)}{\epsilon(\phi)}}$ (dependent only on $\phi$). - -For our proof to work we also need $A$ to be $S$-strong. -We can achieve this by taking (the unique) $S$-strong closure of $A$. -If size of $A$ is $N$ then the size of its closure is $O(N)$. %elaborate -So without loss of generality we can assume that $A$ is $S$-strong. - -\begin{Definition} - Define a $b$-trace of $\phi$ on $A$ - \begin{align*} - A_b = \phi(A, b) = \curly{a \in A^{|x|} \mid \phi(a, b)} - \end{align*} -\end{Definition} - -Let $\bar A = A \cup b$. % what about A \cap b??? - -\begin{Definition} - For a set $C$ define the boundary of $C$ over $\bar A$ - \begin{align*} - \partial(C, \bar A) = \curly{a \in \bar A \mid \text{there is an edge between $a$ and element of $C - \bar A$}} - \end{align*} -\end{Definition} - -\begin{Definition} - A \emph{witness} of $\phi(a, b)$ is a realization of the existential formula together with ${a_i, b}$. -\end{Definition} - -\begin{Definition} - For a trace $A_b = \{a_1, \ldots, a_I\}$ for each $\phi(a_i, b)$ pick a witness and then take a union of all those witnesses. Call this a witness of the trace $A_b$. %uniqueness -\end{Definition} - -\begin{Definition} - For each $b$ pick some $\bar M_b$ to be a witness of $A_b$. Define two quantities - \begin{itemize} - \item $\partial_b$ is the boundary $\partial(\bar M_b, \bar A)$ - \item Call $f \colon G_1 \arr G_2$ a $\partial$-isomorphism if it is a graph isomorphism and $f$ and $f^{-1}$ are constant on $\bar A$. - Define $\II_b$ as the $\partial$-isomorphism class of the graph induced on vertices $(\bar M_b - \bar A) \cup \partial_b$. %depends on witness choice - \end{itemize} -\end{Definition} - -\begin{Lemma} \label {bound_trace} - If $\II_{b_1} = \II_{b_2}$ then $A_{b_1} = A_{b_2}$. -\end{Lemma} - -\begin{proof} - Fix witnesses $\bar M_{b_1}, \bar M_{b_2}$. - Suppose we have $\phi(a, b_1)$ for some $a$. - Pick its witness $M_1 \subset \bar M_{b_1}$. - This gives us a witness $M_2$ via the $\partial$-isomorphism. -\end{proof} - -Thus to bound the number of traces it is sufficient to bound the number of possibilities for $\II_b$. - -\begin{Theorem} \label{main_bound} - \begin{align*} - |\partial_b| &\leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)}\\ - |\bar M_b - \bar A| &\leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)} - \end{align*} -\end{Theorem} - -\begin{Corollary} - \begin{align*} - \vc(\phi) \leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)} - \end{align*} -\end{Corollary} - -\begin{proof} - We count possible $\partial$-isomorphism classes $\II_b$. - Let $W = K(\phi) \frac{Y(\phi)}{\epsilon(\phi)}$. - If the parameter set $A$ is of size $N$ then there are $N \choose W$ choices for boundary $\partial_b$. - On top of the boundary there are at most $W$ extra vertices and $(2W)^2$ extra edges. - Thus there are at most - \begin{align*} - W \cdot 2^{(2W)^2} - \end{align*} - configurations up to a graph isomorphism. - In total this gives us - \begin{align*} - {N \choose W} \cdot W \cdot 2^{(2W)^2} = O(N^W) - \end{align*} - options for $\partial$-isomorphism classes. - By Lemma \ref{bound_trace} there are at most $O(N^W)$ many traces, giving the required bound. -\end{proof} - -\begin{proof} \textit{(of Theorem \ref{main_bound})} - Fix some $b$-trace $A_b$. Enumerate $A_b = \{a_1, \ldots, a_I\}$. - - Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \leq I$. - Let $\bar M_i = \bigcup_{j < i} M_j$. - Let $\bar M = \bigcup M_i$, a witness of $A_b$ - - \begin{Claim} - \begin{align*} - &\abs{\partial(M_i M, \bar A) - \partial(M, \bar A)} \leq |M_i| = K(\phi)\\ - &\dim(M_i M \bar A / M \bar A) > -\epsilon(\phi) - \end{align*} - \end{Claim} - - \begin{Definition} - $(j-1, j)$ is called a \emph{jump} if some of the following conditions happen - \begin{itemize} - \item New vertices are added outside of $\bar A$ i.e. - \begin{align*} - \bar M_j - \bar A \neq \bar M_{j-1} - \bar A - \end{align*} - \item New vertices are added to the boundary, i.e. - \begin{align*} - \partial(\bar M_j, \bar A) \neq \partial(\bar M_{j-1}, \bar A) - \end{align*} - \end{itemize} - \end{Definition} - - \begin{Definition} - We now let $m_i$ count all jumps below $i$ - %Let $d_i = \dim(\bar M_i/A)$. - \begin{align*} - m_i = \abs{\curly{j < i \mid (j-1, j) \text{ is a jump}}} - \end{align*} - \end{Definition} - - \begin{Lemma} \label{ub_lemma} - \begin{align*} - \dim(\bar M_i / \bar A) &\leq -m_i \cdot \epsilon(\phi) \\ - |\partial(\bar M_i, \bar A)| &\leq m_i \cdot K(\phi) \\ - |\bar M_j - \bar A| &\leq m_i \cdot K(\phi) - \end{align*} - \end{Lemma} - - \begin{proof} \textit{(of Lemma \ref{ub_lemma})} - Proceed by induction. - Second and third propositions are clear. - For the first proposition base case is clear. - - Induction step. - Suppose $\bar M_j \cap (A \cup b) = \bar M_{j+1}$ and $\partial(\bar M_j, A) = \partial(\bar M_{j+1}, A)$. - Then $m_i = m_{i+1}$ and the quantities don't change. - Thus assume at least one of these equalities fails. - - Apply Lemma \ref{chain_lemma} to $\bar M_j \cup (A \cup b)$ and $(M_{j+1}, a_{j+1}b)$. - There are two options - - \begin{itemize} - \item $\dim(\bar M_{j+1} \cup (A \cup b) / \bar M_i \cup (A \cup b)) \leq -\epsilon_U$. - This implies the proposition. - \item $M_{j+1} \subset \bar M_j \cup (A \cup b)$. - Then by our assumption it has to be $\partial(\bar M_j, A) \neq \partial(\bar M_{j+1}, A)$. - There are edges between $M_{j+1} \cap (\partial(\bar M_{j+1}, A) - \partial(\bar M_j, A))$ so they contribute some negative dimension $\leq \epsilon_U$. - \end{itemize} - This ends the proof for Lemma \ref{ub_lemma}. - \end{proof} - \textit{(Proof of Theorem \ref{main_bound} continued)} - First part of lemma \ref{ub_lemma} implies that we have $\dim(\bar M / \bar A) \leq -m_I \cdot \epsilon(\phi)$. - The requirement of $A$ to be $S$-strong forces - \begin{align*} - m_I \cdot \epsilon(\phi) &< Y(\phi) \\ - m_I &< \frac{Y(\phi)}{\epsilon(\phi)} \\ - \end{align*} - %Let $W = \frac{K(\phi)Y(\phi)}{\epsilon(\phi)}$ - Applying the rest of \ref{ub_lemma} gives us - \begin{align*} - |\partial(\bar M, A)| &\leq m_I \cdot K(\phi) \leq \frac{K(\phi)Y(\phi)}{\epsilon(\phi)} \\ - |\bar M \cap A| &\leq m_I \cdot K(\phi) \leq \frac{K(\phi)Y(\phi)}{\epsilon(\phi)} - \end{align*} - as needed. - This ends the proof for Theorem \ref{main_bound}. -\end{proof} - -So far we have computed an upper bound for a single basic formula $\phi$. - -To bound an arbitrary formula, write it as a boolean combination of basic formulas $\phi_i$ (via quantifier elimination) -It suffices to bound vc-density for collection of formulas $\{\phi_i\}$ to obtain a bound for the original formula. - -In general work with a collection of basic formulas $\{\phi_i\}_{i \in I}$. -The proof generalizes in a straightforward manner. -Instead of $A^{|x|}$ we now work with $A^{|x|} \times I$ separating traces of different formulas. -Formula with the largest quantity $Y(\phi)\frac{K(\phi)}{\epsilon(\phi)}$ contributes the most to the vc-density. -Thus we have -\begin{align*} - \Phi &= \{\phi_i\}_{i \in I} \\ - \vc(\Phi) &= \max_{i \in I} Y(\phi_i) \frac{K(\phi_i)}{\epsilon_{\phi_i}} -\end{align*} - - -%\begin{Definition} - %\begin{align*} - %d &= \dim \bar M / \bar A \\ - %s &= |\bar M - \bar A| \\ - %b &= |\partial(\bar M, \bar A)| - %\end{align*} -%\end{Definition} - - -%Thus as we consider $\bar M$ as an increasing union of witnesses to chain-minimal extensions, we see the extension with the largest ratio can contribute most to the boundary. -%Thus is our upper bound for the boundary. - - - - - - -%Now, classify every trace by the isomorphism class of $\bar M - A \cup \partial(\bar M, A)$ and by $\partial(\bar M, A)$. -%\begin{Lemma} - %Suppose we have traces $b_1, b_2$ with the same components as above. - %Then $A_{b_1} = A_{b_2}$. -%\end{Lemma} -% -%Consider $\bar M - A \cup \partial(\bar M, A)$. -%Number of vertices is $\leq (2W)^2$. -%Thus number of isomorphism classes $\leq 2^{(2W)^2}$. -% -%Consider $\partial(\bar M, A)$. -%Let $N = |A|$. -%Order matters, so the total number of choices for it is -%\begin{align*} - %N \cdot (N-1) \cdot \ldots \cdot (N - W + 1) = \frac{N!}{(N-W)!} -%\end{align*} -% -%Thus the number of possible different traces is bounded by -%\begin{align*} - %2^{(2W)^2} \cdot \frac{N!}{(N-W)!} = O(N^W) -%\end{align*} -% -%Since choice of $A$ was arbitrary, this gives -%\begin{align*} - %\vc{\phi} \leq W = \frac{|M|Y}{\epsilon_U} -%\end{align*} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Technical Lemmas} - -\begin{Lemma} - Suppose we have a set $B$ and a minimal pair $(M, A)$ with $A \subset B$ and $\dim(M/A) = -\epsilon$. -Then either $M \subseteq B$ or $\dim((M \cup B)/B) < -\epsilon$. -\end{Lemma} - -\begin{proof} - By diamond construction - - \begin{align*} - \dim((M \cup B)/B) \leq \dim(M / (M \cap B)) - \end{align*} - - and - - \begin{align*} - \dim(M / (M \cap B)) &= \dim (M/A) - \dim(M / (M \cap B)) \\ - \dim (M/A) &= -\epsilon \\ - \dim(M / (M \cap B)) &> 0 - \end{align*} -\end{proof} - - - -\begin{Lemma} \label{chain_lemma} - Suppose we have a set $B$ and a minimal chain $M_n$ with $M_0 \subset B$ and dimensions $-\epsilon_i$. -Let $\epsilon$ be the minimal of $\epsilon_i$. -Then either $M_n \subseteq B$ or $\dim((M_n \cup B)/B) < -\epsilon$. -\end{Lemma} - - -\begin{proof} - Let $\bar M_i = M_i \cup B$ - - \begin{align*} - \dim(\bar M_n/B) = \dim(\bar M_n/\bar M_{n-1}) + \ldots + \dim(\bar M_2/\bar M_1) + \dim(\bar M_1/B) - \end{align*} - - Either $M_n \subseteq B$ or one of the summands above is nonzero. - Apply previous lemma. -\end{proof} - -\begin{Lemma} \label{chain_intersect} - Suppose we have a minimal chain $M_n$ with dimensions $-\epsilon_i$. - Let $\epsilon$ be the sum of all $\epsilon_i$. - Suppose we have some $B$ with $B \subseteq M_n$. - Then $\dim B / (M_0 \cap B) \geq -\epsilon$. -\end{Lemma} - -\begin{proof} - Let $B_i = B \cap M_i$. - We have $\dim B_{i+1}/B_i \geq \dim M_{i+1}/M_i$ by minimality. - $\dim B / (M_0 \cap B) = \dim B_n / B_0 = \sum \dim B_{i+1}/B_i \geq -\epsilon$. -\end{proof} - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Counterexamples} - -% Add complete graph counterexample -% where we have a bunch of minimal extensions intersecting in a tiny way -% or something similar? - -% \AA is indiscernible -% example of indiscernible sequence that is not strong -% example with non-strong embedding on every n-tuple of vertices? - -% 2 chain minimal extension that is larger than the lower bound -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Upper bound on $\AA$} - -\begin{Definition} - \begin{align*} - \AA = \curly{A \subset \U^{y} \mid \text{finite, disconnected, strongly embedded}} - \end{align*} -\end{Definition} - -Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. - -Pick a trace of $\phi(x,y)$ on $A^{|x|}$ by a parameter $b$. - -\begin{align*} - B = \curly{a \in A^{|x|} \mid \phi(a, b)} -\end{align*} - -Pick $B' \subset B$, ordered $B' = \{a_i\}_{i \in I}$ such that -\begin{align*} - %a_i \cap \bigcup_{j \neq i} a_j \neq \emptyset - a_i \cap \bigcup_{j < i} a_j \neq \emptyset -\end{align*} -This is always possible by starting with $B$ and taking away elements one by one. -Call such a set a \emph{generating set} of $B$. - -Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \in I$. -Let $\bar M = \bigcup M_i$. -Consider $\bar M / A$. - -Pick $\bar M$ such that $\dim(\bar M / A)$ is maximized. - -$\bar M \cap A \leq \bar M$ as $A$ is strong. (Make sure $M$ is not too big!) -Let $\bar A = A - \curly{a_i}_{i \in I}$. -Suppose $\bar A \cap \bar M \neq \emptyset$. -Then we can abstractly reembed $\M$ over $A$ such that $\bar A \cap \bar M = \emptyset$. -This would increase the dimension, contradicting maximality. -Thus we can assume $A \cap \bar M = \{a_i\}_{i \in I}$ - -Let $\bar M_j = \bigcup_{i < j} M_i$. - -\begin{Lemma} - $\dim(\bar M_j / A) \leq j \cdot \epsilon_U$ -\end{Lemma} -\begin{proof} - Proceed by induction. - Base case is clear. - - For induction case apply lemma to $\bar M_j \cup \{a_j\}$ and $M_j / \{a_j, b\}$. - There are two cases - \begin{enumerate} - \item $M_j \subset \bar M' \cup \{a_j\}$. - In this case there are edges between $\{a_j\}$ and $M_j$ that contribute to dimension less than $-\epsilon_U$. - \item Otherwise $M_j$ adds extra dimension less than $-\epsilon_U$ - \end{enumerate} -\end{proof} - -Thus we have $\dim(\bar M / A) = \dim(\bar M_n / A) \leq -\epsilon_U n$. - -Thus as $A$ is strong we need $|B'| \epsilon_U < Y$. -This gives us $|B'| \leq n$. -Finally we need to relate $|B'|$ to $|B|$. - -Suppose we have $C \subset A^{|x|}$, finite with $|C| = N$. -A generating set for a trace has to have size $\leq n$. -Thus there are ${N \choose n} \leq N^n$ choices for a generating set. -A set generated from set of size $n$ can have at most $(x|n|)^{|x|}$ elements. -Thus a given set of size $n$ can generate at most -\begin{align*} - 2^{(x|n|)^{|x|}} -\end{align*} -sets. -Thus the number of possible traces on $C$ is bounded above by -\begin{align*} - 2^{(x|n|)^{|x|}} \cdot N^n = O(N^n) -\end{align*} -This bounds the vc-density by $n$. - -\begin{align*} - \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_U}} -\end{align*} - - -\begin{thebibliography}{9} - -\bibitem{vc_density} - M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, - \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) - -\bibitem{laskowski} - Michael C. Laskowski, \textit{A simpler axiomatization of the Shelah-Spencer almost sure theories}, - Israel J. Math. \textbf{161} (2007), 157–186. MR MR2}350161 - -\end{thebibliography} - -\end{document} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} +\usepackage{../Thm} + +\usepackage{mathrsfs} + + + +\renewcommand{\AA}{\mathscr A} + \newcommand{\II}{\mathscr I} + \newcommand{\MM}{\mathscr M} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + \newcommand{\U}{\mathcal U} + \newcommand{\X}{\mathcal X} + \newcommand{\Y}{\mathcal Y} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\newcommand{\curly}[1]{\left\{#1\right\}} +\newcommand{\paren}[1]{\left(#1\right)} +\newcommand{\abs}[1]{\left|#1\right|} +\newcommand{\agl}[1]{\left\langle #1 \right\rangle} + +\providecommand{\floor}[1]{\left \lfloor #1 \right \rfloor } + +%\DeclareMathOperator{\dim}{dim} + +\title{Some vc-density computations in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{document} + +\maketitle + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Preliminaries} + +VC density was introduced in \cite{vc_density} by Aschenbrenner, Dolich, Haskell, MacPherson, and Starchenko as a natural notion of dimension for NIP theories. In a NIP theory we can define a VC function + +\begin{align*} + \vc : \N \arr \N +\end{align*} + +Where $vc(n)$ measures complexity of definable sets in an $n$-dimensional space. Simplest possible behavior is $\vc(n) = n$ for all $n$. Theories with that property are known to be dp-minimal, i.e. having the smallest possible dp-rank. In general, it is not known whether there can be a dp-minimal theory which doesn't satisfy $\vc(n)=n$. + +In this paper, we investigate vc-density of definable sets in Shelah-Spencer structures. +We follow notations in \cite{laskowski}. +In this paper we work with limit of random structure $G(n, n^{-\alpha})$ for $\alpha \in (0,1)$, irrational. +This structure is axiomatized by $S_\alpha$. +Our ambient model is $\MM$. +Notations we use are $\delta(\A), \delta(\A/\B), \A \leq \B$ as well as notions of $N$-strong substructure, minimal extension, chain minimal extension, minimal pair, and $N$-strong closure. + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Graph Combinatorics} + +We denote graph by $\A$, set of its vertices by $A$. +When we say $\A \subseteq \B$ we mean that $A \subseteq B$ and edges of $\A$ are also edges of $\B$. +However $\B$ may add new edges between vertices of $\A$. + +Fix $\alpha \in (0,1)$, irrational. +For a finite graph $\A$ let +\begin{align*} + \delta(\A) = |A| - \alpha e(\A) +\end{align*} + +where $e(\A)$ is the number of edges in $\A$. + +For finite $\A,\B$ with $\A \subseteq \B$ define $\delta(\B/\A) = \delta(\B) - \delta(\A)$. +We say that $\A \leq \B$ if $\A \subseteq \B$ and $\delta(\A'/\B) > 0$ for all $\A \subseteq \A' \subsetneq \B$. + +We say that finite $\A$ is positive if for all $\A' \subseteq \A$ we have $\delta(\A') \geq 0$. + +\begin{Definition} + We work in theory $S_\alpha$ axiomatized by + \begin{itemize} + \item Every finite substructure is positive + \item For a model $\MM$ given $\A \leq \B$ every embedding $f : \A \arr \MM$ extends to $g: \B \arr \MM$. + \end{itemize} +\end{Definition} + +For $\A, \B$ positive, $(\A, \B)$ is called a minimal pair if $\A \subseteq \B$, $\delta(\B/\A) < 0$ but $\delta(\A'/\A) \geq 0$ for all proper $\A \subseteq \A' \subsetneq \B$. + +$\agl{\A_i}_{i \leq m}$ is called a minimal chain if $(\A_i, \A_i+1)$ is a minimal pair (for all $i < m$). + +For a positive $\A$ let $\delta_\A(\bar x)$ be the atomic diagram of $\A$. For positive $\A \subset \B$ let + +\begin{align*} + \Psi_{\A,\B}(\bar x) = \delta_\A(\bar x) \wedge \exists \bar y \; \delta_\B(\bar x, \bar y) +\end{align*} + +Such formula is called chain-minimal extension formula if in addition we have that there is a minimal chain starting at $\A$ and ending in $\B$. Denote such formulas as $\Psi_{\agl{\M_i}}$ + +\begin{Theorem} [5.6 in \cite{laskowski}] + $S_\alpha$ admits quantifier elimination down to boolean combination of chain-minimal extension formulas. +\end{Theorem} + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Definitions} + +%\begin{Definition} + %Let $x = (x_1, \ldots x_n), y = (y_1, \ldots, y_m)$ be variable tuples. + %We call a formula $\phi(x, y)$ \emph{basic} when + %\begin{itemize} + %\item $\phi(x, y)$ is a minimal chain extension, denoted by $\curly{M_i}_{i \in [0..k]}$ with $M_0 = \{x, y\}$. + %\item $\phi(x, y)$ determines edges and non-edges on its variables $\{x_1, \ldots x_n\} \cup \{y_1, \ldots y_m\}$. + %\item there is no edge between $x_i$ and $y_j$ for all $i,j$. (see note \ref{note_edges}) + %\item Define $\mathbf x$ to be the graph on vertices $\{x_i\}$ with edges as defined by $\phi$. + %Similarly define $\mathbf y$. + %We require $\mathbf x$ and $\mathbf y$ to be positive. (see note \ref{note_positive})\ + %%\item all elements of $y$ that are connected to $M_k - \{x,y\}$. (see note \ref{note_special}) + %\end{itemize} +%\end{Definition} + +% issue: multiple possible chain decompositions? + +Fix tuples $x = (x_1, \ldots x_n), y = (y_1, \ldots, y_m)$. +We refer to chain-minimal extension formulas as basic formulas. +Let $\phi_{\agl{\M_i}}(x, y)$ be a basic formula. + +\begin{Definition} + Define $\X$ to be the graph on vertices $\{x_i\}$ with edges as defined by $\phi_{\agl{\M_i}}$. + Similarly define $\Y$. + We define those abstractly, i.e. on a new set of vertices disjoint from $\MM$. +\end{Definition} + +Note that $\X$, $\Y$ are positive as they are subgraphs of $\M_0$. +As usual $X, Y$ will refer to vertices of those graphs. + +We restrict our attention to formulas that define no edges between $X$ and $Y$. + +\begin{Note} \label{note_edges} + We can handle edges between $x$ and $y$ as separate elements of the minimal chain extension. +\end{Note} + +%\begin{Note} \label{note_positive} + %If either graph $\mathbf x$ or $\mathbf y$ is negative, then the formula would have no realizations as negative graphs cannot be embedded into our ambient model. +%\end{Note} + +%\begin{Note} \label{note_special} + %We add the final condition to simplify our analysis. Similar techniques can be used to acquire bounds on formulas not subject to that condition. +%\end{Note} + +\begin{Definition} \label{def_basic} + For a basic formula $\phi = \phi_{\agl{\M_i}_{i \leq k}}(x, y)$ let + \begin{itemize} + \item $\epsilon_i(\phi) = -\dim \paren{M_i/M_{i-1}}$. + \item $\epsilon_L(\phi) = \sum_{[1..k]} \epsilon_i(\phi)$. + \item $\epsilon_U(\phi) = \min_{[1..k]} \epsilon_i(\phi)$. + \item Let $\Y'$ be a subgraph of $\Y$ induced by vertices of $\Y$ that are connected to $M_k - (X \cup Y)$. + \item Let $Y(\phi) = \dim (\Y')$. + In particular if $\Y = \Y'$ and $\Y$ is disconnected then $Y(\phi)$ is just the arity of the tuple $y$. + \end{itemize} +\end{Definition} + +%\begin{Definition} +%\begin{align*} + %\epsilon_L(\neg \phi) &= \epsilon_L(\phi) \\ + %\epsilon_U(\neg \phi) &= \epsilon_U(\phi) \\ + %\epsilon_L(\phi \wedge \psi) &= \epsilon_L(\phi) + \epsilon_L(\psi) \\ + %\epsilon_U(\phi \wedge \psi) &= \min(\epsilon_U(\phi), \epsilon_U(\psi)) \\ + %\epsilon_L(\phi \vee \psi) &= \min(\epsilon_L(\phi), \epsilon_L(\psi)) \\ + %\epsilon_U(\phi \vee \psi) &= \min(\epsilon_U(\phi), \epsilon_U(\psi)) +%\end{align*} +%\end{Definition} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Lower bound} + +As a simplification for our lower bound computation we assume that all the basic formulas involved we have $\Y' = \Y$ (see Definition \ref{def_basic}). + +We work with formulas that are boolean combinations of basic formulas written in disjunctive-conjunctive form. +First, we extend our definition of $\epsilon$. + +\begin{Definition}[Negation] + If $\phi$ is a basic formula, then define + \begin{align*} + \epsilon_L(\neg \phi) &= \epsilon_L(\phi) + \end{align*} +\end{Definition} + +\begin{Definition}[Conjunction] + Take a collection of formulas $\phi_i(x, y)$ where each $\phi_i$ is positive or negative basic formula. + If both positive and negative formulas are present then $\epsilon_L(\phi) = \infty$. + We don't have a lower bound for that case. + If different formulas define $\X$ or $\Y$ differently then $\epsilon_L(\phi) = \infty$. + In that case of the conflicting definitions would make the formula have no realizations. + Otherwise + \begin{align*} + \epsilon_L(\bigwedge \phi_i) &= \sum \epsilon_L(\phi_i) + \end{align*} +\end{Definition} + +\begin{Definition} [Disjunction] + Take a collection of formulas $\psi_i$ where each instance is a conjunction of positive and negative instances of basic formulas that agree on $\X$ and $\Y$. % can generalize? + \begin{align*} + \epsilon_L(\bigvee \psi_i) &= \min \epsilon_L(\psi_i) + \end{align*} +\end{Definition} + +\begin{Theorem} + For a formula $\phi$ as above + \begin{align*} + \vc \phi \geq \floor{\frac{Y(\phi)}{\epsilon_L(\phi)}} + \end{align*} + where $Y(\phi)$ is $Y(\psi)$ for $\psi$ one the basic components of $\phi$ (all basic componenets agree on $\Y$). +\end{Theorem} + +\begin{proof} + First work with a formula that is a conjunction of positive basic formulas. + + \begin{align*} + \psi = \bigwedge_{j \leq J} \phi_j + \end{align*} + Then as we defined above + \begin{align*} + \epsilon_L(\psi) = \sum \epsilon_L(\phi_j) + \end{align*} + + Let $\phi$ be one of the basic formulas in $\psi$ with a chain $\agl{M_i}_{i \leq k}$. + Let $K_\phi = |M_k|$ i.e. the size of the extension. + Let $K$ be the largest such size among all $\phi_i$. + + Let $n$ be the integer such that $n \epsilon_L(\psi) < Y$ and $(n+1) \epsilon_L(\psi) > Y$. + + %Take an abstract realization of $y$ as dictated by $\psi$, and label it by $b$ ($b \in \MM$) + Label $\Y$ by an tuple $b$. + + Pick parameter set $A \subset \MM$ such that + + \begin{align*} + A = \bigcup_{i 0$ as needed. + \end{proof} + + $|\bar M| \leq N \cdot I \cdot K$ and $A$ is $\leq N \cdot I \cdot K$-strong. + Thus a copy of $\bar M$ can be embedded over $A$ into our ambient model $\MM$. + Our choice of $b_i$'s was arbitrary, so we get ${N \choose n}$ choices out of $N|x|$ many elements. + Thus we have $O(|A|^n)$ many traces. + + \begin{Lemma} + There are arbitrarily large sets with properties of $A$. + \end{Lemma} + + \begin{proof} + \textit{proof goes here. use lemma in laskoswki paper} + \end{proof} + + This shows + + \begin{align*} + \vc \psi \geq n = \floor{\frac{Y}{\epsilon_L}} + \end{align*} + + Now consider the formula which is a conjunction consists of negative basic formulas + \begin{align*} + \psi = \bigwedge \neg \phi_i + \end{align*} + Let + \begin{align*} + \bar \psi = \bigwedge \phi_i + \end{align*} + + Do the construction above for $\bar \psi$ and suppose its trace is $X \subset A$ for some $b$. + Then over $b$ the same construction gives trace $(A - X)$ for $\psi$. Thus we get as many traces. + + Finally consider a formula which is a disjunction of formulas considered above. + Choose the one with the smallest $\epsilon_L$, this yields the lower bound for the entire formula. + %explain! non-trivial. thing of disjunction of two formulas + % also! what if disjunction has formulas disagreeing on x and y + % aslo! what if disjuction where one of the formulas is mixed positive/negative formulas +\end{proof} + +\begin{Claim} + We can find a minimal extension $M / \{x, y\}$ with arbitrarily small dimension. +\end{Claim} + +\begin{proof} + Put proof in here. Follow construction in Laskowski paper. +\end{proof} + +This shows that vc function is infinite in Shelah-Spencer random graphs. + +\begin{align*} + \vc(n) = \infty +\end{align*} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Upper bound} + + +%\begin{Definition} + %Let $\phi$ be a basic formula with $M_i$ a minimal chain, + %$\epsilon_i$ its corresponding dimensions, + %and $M$ its total size. + %\begin{align*} + %U_\phi = \frac{M}{\min \epsilon_i} + %\end{align*} +%\end{Definition} + + + +%\begin{Definition} [Negation] + %Let $\phi$ be basic + %\begin{align*} + %U_{\neg \phi} = U_{\phi} + %\end{align*} +%\end{Definition} +% +%\begin{Definition} [Conjunction and Disjunction] + %Let $\phi_{ij}$ be basic or a negation of a basic formula. + %\begin{align*} + %\psi = \bigvee \bigwedge \phi_{ij} + %\end{align*} + %\begin{align*} + %U_\psi = \max U_{\phi_{ij}} + %\end{align*} +%\end{Definition} + +%Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. + +%Consider a formula which is a conjunction of positive basic formulas. + +Consider a case of a single basic formula $\phi(\vec x, \vec y)$. + +Suppose it defines a minimal chain extension over $\{x, y\}$. +Record the size of that extension as $K(\phi)$ and its total dimension $\epsilon(\phi) = \epsilon_U(\phi)$. + +In general we have parameter set $A \subset \MM^{|x|}$, however without loss of generality we may work with +a parameter set $A^{|x|}$, with $A \subset \MM$. + +Let $S = \floor{\frac{K(\phi)Y(\phi)}{\epsilon(\phi)}}$ (dependent only on $\phi$). + +For our proof to work we also need $A$ to be $S$-strong. +We can achieve this by taking (the unique) $S$-strong closure of $A$. +If size of $A$ is $N$ then the size of its closure is $O(N)$. %elaborate +So without loss of generality we can assume that $A$ is $S$-strong. + +\begin{Definition} + Define a $b$-trace of $\phi$ on $A$ + \begin{align*} + A_b = \phi(A, b) = \curly{a \in A^{|x|} \mid \phi(a, b)} + \end{align*} +\end{Definition} + +Let $\bar A = A \cup b$. % what about A \cap b??? + +\begin{Definition} + For a set $C$ define the boundary of $C$ over $\bar A$ + \begin{align*} + \partial(C, \bar A) = \curly{a \in \bar A \mid \text{there is an edge between $a$ and element of $C - \bar A$}} + \end{align*} +\end{Definition} + +\begin{Definition} + A \emph{witness} of $\phi(a, b)$ is a realization of the existential formula together with ${a_i, b}$. +\end{Definition} + +\begin{Definition} + For a trace $A_b = \{a_1, \ldots, a_I\}$ for each $\phi(a_i, b)$ pick a witness and then take a union of all those witnesses. Call this a witness of the trace $A_b$. %uniqueness +\end{Definition} + +\begin{Definition} + For each $b$ pick some $\bar M_b$ to be a witness of $A_b$. Define two quantities + \begin{itemize} + \item $\partial_b$ is the boundary $\partial(\bar M_b, \bar A)$ + \item Call $f \colon G_1 \arr G_2$ a $\partial$-isomorphism if it is a graph isomorphism and $f$ and $f^{-1}$ are constant on $\bar A$. + Define $\II_b$ as the $\partial$-isomorphism class of the graph induced on vertices $(\bar M_b - \bar A) \cup \partial_b$. %depends on witness choice + \end{itemize} +\end{Definition} + +\begin{Lemma} \label {bound_trace} + If $\II_{b_1} = \II_{b_2}$ then $A_{b_1} = A_{b_2}$. +\end{Lemma} + +\begin{proof} + Fix witnesses $\bar M_{b_1}, \bar M_{b_2}$. + Suppose we have $\phi(a, b_1)$ for some $a$. + Pick its witness $M_1 \subset \bar M_{b_1}$. + This gives us a witness $M_2$ via the $\partial$-isomorphism. +\end{proof} + +Thus to bound the number of traces it is sufficient to bound the number of possibilities for $\II_b$. + +\begin{Theorem} \label{main_bound} + \begin{align*} + |\partial_b| &\leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)}\\ + |\bar M_b - \bar A| &\leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)} + \end{align*} +\end{Theorem} + +\begin{Corollary} + \begin{align*} + \vc(\phi) \leq K(\phi) \frac{Y(\phi)}{\epsilon(\phi)} + \end{align*} +\end{Corollary} + +\begin{proof} + We count possible $\partial$-isomorphism classes $\II_b$. + Let $W = K(\phi) \frac{Y(\phi)}{\epsilon(\phi)}$. + If the parameter set $A$ is of size $N$ then there are $N \choose W$ choices for boundary $\partial_b$. + On top of the boundary there are at most $W$ extra vertices and $(2W)^2$ extra edges. + Thus there are at most + \begin{align*} + W \cdot 2^{(2W)^2} + \end{align*} + configurations up to a graph isomorphism. + In total this gives us + \begin{align*} + {N \choose W} \cdot W \cdot 2^{(2W)^2} = O(N^W) + \end{align*} + options for $\partial$-isomorphism classes. + By Lemma \ref{bound_trace} there are at most $O(N^W)$ many traces, giving the required bound. +\end{proof} + +\begin{proof} \textit{(of Theorem \ref{main_bound})} + Fix some $b$-trace $A_b$. Enumerate $A_b = \{a_1, \ldots, a_I\}$. + + Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \leq I$. + Let $\bar M_i = \bigcup_{j < i} M_j$. + Let $\bar M = \bigcup M_i$, a witness of $A_b$ + + \begin{Claim} + \begin{align*} + &\abs{\partial(M_i M, \bar A) - \partial(M, \bar A)} \leq |M_i| = K(\phi)\\ + &\dim(M_i M \bar A / M \bar A) > -\epsilon(\phi) + \end{align*} + \end{Claim} + + \begin{Definition} + $(j-1, j)$ is called a \emph{jump} if some of the following conditions happen + \begin{itemize} + \item New vertices are added outside of $\bar A$ i.e. + \begin{align*} + \bar M_j - \bar A \neq \bar M_{j-1} - \bar A + \end{align*} + \item New vertices are added to the boundary, i.e. + \begin{align*} + \partial(\bar M_j, \bar A) \neq \partial(\bar M_{j-1}, \bar A) + \end{align*} + \end{itemize} + \end{Definition} + + \begin{Definition} + We now let $m_i$ count all jumps below $i$ + %Let $d_i = \dim(\bar M_i/A)$. + \begin{align*} + m_i = \abs{\curly{j < i \mid (j-1, j) \text{ is a jump}}} + \end{align*} + \end{Definition} + + \begin{Lemma} \label{ub_lemma} + \begin{align*} + \dim(\bar M_i / \bar A) &\leq -m_i \cdot \epsilon(\phi) \\ + |\partial(\bar M_i, \bar A)| &\leq m_i \cdot K(\phi) \\ + |\bar M_j - \bar A| &\leq m_i \cdot K(\phi) + \end{align*} + \end{Lemma} + + \begin{proof} \textit{(of Lemma \ref{ub_lemma})} + Proceed by induction. + Second and third propositions are clear. + For the first proposition base case is clear. + + Induction step. + Suppose $\bar M_j \cap (A \cup b) = \bar M_{j+1}$ and $\partial(\bar M_j, A) = \partial(\bar M_{j+1}, A)$. + Then $m_i = m_{i+1}$ and the quantities don't change. + Thus assume at least one of these equalities fails. + + Apply Lemma \ref{chain_lemma} to $\bar M_j \cup (A \cup b)$ and $(M_{j+1}, a_{j+1}b)$. + There are two options + + \begin{itemize} + \item $\dim(\bar M_{j+1} \cup (A \cup b) / \bar M_i \cup (A \cup b)) \leq -\epsilon_U$. + This implies the proposition. + \item $M_{j+1} \subset \bar M_j \cup (A \cup b)$. + Then by our assumption it has to be $\partial(\bar M_j, A) \neq \partial(\bar M_{j+1}, A)$. + There are edges between $M_{j+1} \cap (\partial(\bar M_{j+1}, A) - \partial(\bar M_j, A))$ so they contribute some negative dimension $\leq \epsilon_U$. + \end{itemize} + This ends the proof for Lemma \ref{ub_lemma}. + \end{proof} + \textit{(Proof of Theorem \ref{main_bound} continued)} + First part of lemma \ref{ub_lemma} implies that we have $\dim(\bar M / \bar A) \leq -m_I \cdot \epsilon(\phi)$. + The requirement of $A$ to be $S$-strong forces + \begin{align*} + m_I \cdot \epsilon(\phi) &< Y(\phi) \\ + m_I &< \frac{Y(\phi)}{\epsilon(\phi)} \\ + \end{align*} + %Let $W = \frac{K(\phi)Y(\phi)}{\epsilon(\phi)}$ + Applying the rest of \ref{ub_lemma} gives us + \begin{align*} + |\partial(\bar M, A)| &\leq m_I \cdot K(\phi) \leq \frac{K(\phi)Y(\phi)}{\epsilon(\phi)} \\ + |\bar M \cap A| &\leq m_I \cdot K(\phi) \leq \frac{K(\phi)Y(\phi)}{\epsilon(\phi)} + \end{align*} + as needed. + This ends the proof for Theorem \ref{main_bound}. +\end{proof} + +So far we have computed an upper bound for a single basic formula $\phi$. + +To bound an arbitrary formula, write it as a boolean combination of basic formulas $\phi_i$ (via quantifier elimination) +It suffices to bound vc-density for collection of formulas $\{\phi_i\}$ to obtain a bound for the original formula. + +In general work with a collection of basic formulas $\{\phi_i\}_{i \in I}$. +The proof generalizes in a straightforward manner. +Instead of $A^{|x|}$ we now work with $A^{|x|} \times I$ separating traces of different formulas. +Formula with the largest quantity $Y(\phi)\frac{K(\phi)}{\epsilon(\phi)}$ contributes the most to the vc-density. +Thus we have +\begin{align*} + \Phi &= \{\phi_i\}_{i \in I} \\ + \vc(\Phi) &= \max_{i \in I} Y(\phi_i) \frac{K(\phi_i)}{\epsilon_{\phi_i}} +\end{align*} + + +%\begin{Definition} + %\begin{align*} + %d &= \dim \bar M / \bar A \\ + %s &= |\bar M - \bar A| \\ + %b &= |\partial(\bar M, \bar A)| + %\end{align*} +%\end{Definition} + + +%Thus as we consider $\bar M$ as an increasing union of witnesses to chain-minimal extensions, we see the extension with the largest ratio can contribute most to the boundary. +%Thus is our upper bound for the boundary. + + + + + + +%Now, classify every trace by the isomorphism class of $\bar M - A \cup \partial(\bar M, A)$ and by $\partial(\bar M, A)$. +%\begin{Lemma} + %Suppose we have traces $b_1, b_2$ with the same components as above. + %Then $A_{b_1} = A_{b_2}$. +%\end{Lemma} +% +%Consider $\bar M - A \cup \partial(\bar M, A)$. +%Number of vertices is $\leq (2W)^2$. +%Thus number of isomorphism classes $\leq 2^{(2W)^2}$. +% +%Consider $\partial(\bar M, A)$. +%Let $N = |A|$. +%Order matters, so the total number of choices for it is +%\begin{align*} + %N \cdot (N-1) \cdot \ldots \cdot (N - W + 1) = \frac{N!}{(N-W)!} +%\end{align*} +% +%Thus the number of possible different traces is bounded by +%\begin{align*} + %2^{(2W)^2} \cdot \frac{N!}{(N-W)!} = O(N^W) +%\end{align*} +% +%Since choice of $A$ was arbitrary, this gives +%\begin{align*} + %\vc{\phi} \leq W = \frac{|M|Y}{\epsilon_U} +%\end{align*} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Technical Lemmas} + +\begin{Lemma} + Suppose we have a set $B$ and a minimal pair $(M, A)$ with $A \subset B$ and $\dim(M/A) = -\epsilon$. +Then either $M \subseteq B$ or $\dim((M \cup B)/B) < -\epsilon$. +\end{Lemma} + +\begin{proof} + By diamond construction + + \begin{align*} + \dim((M \cup B)/B) \leq \dim(M / (M \cap B)) + \end{align*} + + and + + \begin{align*} + \dim(M / (M \cap B)) &= \dim (M/A) - \dim(M / (M \cap B)) \\ + \dim (M/A) &= -\epsilon \\ + \dim(M / (M \cap B)) &> 0 + \end{align*} +\end{proof} + + + +\begin{Lemma} \label{chain_lemma} + Suppose we have a set $B$ and a minimal chain $M_n$ with $M_0 \subset B$ and dimensions $-\epsilon_i$. +Let $\epsilon$ be the minimal of $\epsilon_i$. +Then either $M_n \subseteq B$ or $\dim((M_n \cup B)/B) < -\epsilon$. +\end{Lemma} + + +\begin{proof} + Let $\bar M_i = M_i \cup B$ + + \begin{align*} + \dim(\bar M_n/B) = \dim(\bar M_n/\bar M_{n-1}) + \ldots + \dim(\bar M_2/\bar M_1) + \dim(\bar M_1/B) + \end{align*} + + Either $M_n \subseteq B$ or one of the summands above is nonzero. + Apply previous lemma. +\end{proof} + +\begin{Lemma} \label{chain_intersect} + Suppose we have a minimal chain $M_n$ with dimensions $-\epsilon_i$. + Let $\epsilon$ be the sum of all $\epsilon_i$. + Suppose we have some $B$ with $B \subseteq M_n$. + Then $\dim B / (M_0 \cap B) \geq -\epsilon$. +\end{Lemma} + +\begin{proof} + Let $B_i = B \cap M_i$. + We have $\dim B_{i+1}/B_i \geq \dim M_{i+1}/M_i$ by minimality. + $\dim B / (M_0 \cap B) = \dim B_n / B_0 = \sum \dim B_{i+1}/B_i \geq -\epsilon$. +\end{proof} + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Counterexamples} + +% Add complete graph counterexample +% where we have a bunch of minimal extensions intersecting in a tiny way +% or something similar? + +% \AA is indiscernible +% example of indiscernible sequence that is not strong +% example with non-strong embedding on every n-tuple of vertices? + +% 2 chain minimal extension that is larger than the lower bound +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Upper bound on $\AA$} + +\begin{Definition} + \begin{align*} + \AA = \curly{A \subset \U^{y} \mid \text{finite, disconnected, strongly embedded}} + \end{align*} +\end{Definition} + +Let $n$ be the integer such that $n \epsilon_U < Y$ and $(n+1) \epsilon_U > Y$. + +Pick a trace of $\phi(x,y)$ on $A^{|x|}$ by a parameter $b$. + +\begin{align*} + B = \curly{a \in A^{|x|} \mid \phi(a, b)} +\end{align*} + +Pick $B' \subset B$, ordered $B' = \{a_i\}_{i \in I}$ such that +\begin{align*} + %a_i \cap \bigcup_{j \neq i} a_j \neq \emptyset + a_i \cap \bigcup_{j < i} a_j \neq \emptyset +\end{align*} +This is always possible by starting with $B$ and taking away elements one by one. +Call such a set a \emph{generating set} of $B$. + +Let $M_i / \{a_i, b\}$ be a witness of $\phi(a_i, b)$ for each $i \in I$. +Let $\bar M = \bigcup M_i$. +Consider $\bar M / A$. + +Pick $\bar M$ such that $\dim(\bar M / A)$ is maximized. + +$\bar M \cap A \leq \bar M$ as $A$ is strong. (Make sure $M$ is not too big!) +Let $\bar A = A - \curly{a_i}_{i \in I}$. +Suppose $\bar A \cap \bar M \neq \emptyset$. +Then we can abstractly reembed $\M$ over $A$ such that $\bar A \cap \bar M = \emptyset$. +This would increase the dimension, contradicting maximality. +Thus we can assume $A \cap \bar M = \{a_i\}_{i \in I}$ + +Let $\bar M_j = \bigcup_{i < j} M_i$. + +\begin{Lemma} + $\dim(\bar M_j / A) \leq j \cdot \epsilon_U$ +\end{Lemma} +\begin{proof} + Proceed by induction. + Base case is clear. + + For induction case apply lemma to $\bar M_j \cup \{a_j\}$ and $M_j / \{a_j, b\}$. + There are two cases + \begin{enumerate} + \item $M_j \subset \bar M' \cup \{a_j\}$. + In this case there are edges between $\{a_j\}$ and $M_j$ that contribute to dimension less than $-\epsilon_U$. + \item Otherwise $M_j$ adds extra dimension less than $-\epsilon_U$ + \end{enumerate} +\end{proof} + +Thus we have $\dim(\bar M / A) = \dim(\bar M_n / A) \leq -\epsilon_U n$. + +Thus as $A$ is strong we need $|B'| \epsilon_U < Y$. +This gives us $|B'| \leq n$. +Finally we need to relate $|B'|$ to $|B|$. + +Suppose we have $C \subset A^{|x|}$, finite with $|C| = N$. +A generating set for a trace has to have size $\leq n$. +Thus there are ${N \choose n} \leq N^n$ choices for a generating set. +A set generated from set of size $n$ can have at most $(x|n|)^{|x|}$ elements. +Thus a given set of size $n$ can generate at most +\begin{align*} + 2^{(x|n|)^{|x|}} +\end{align*} +sets. +Thus the number of possible traces on $C$ is bounded above by +\begin{align*} + 2^{(x|n|)^{|x|}} \cdot N^n = O(N^n) +\end{align*} +This bounds the vc-density by $n$. + +\begin{align*} + \vc_\AA(\phi) \geq \floor{\frac{Y}{\epsilon_U}} +\end{align*} + + +\begin{thebibliography}{9} + +\bibitem{vc_density} + M. Aschenbrenner, A. Dolich, D. Haskell, D. Macpherson, S. Starchenko, + \textit{Vapnik-Chervonenkis density in some theories without the independence property}, I, preprint (2011) + +\bibitem{laskowski} + Michael C. Laskowski, \textit{A simpler axiomatization of the Shelah-Spencer almost sure theories}, + Israel J. Math. \textbf{161} (2007), 157–186. MR MR2}350161 + +\end{thebibliography} + +\end{document} + % Include edges between y as a chain minimal extension \ No newline at end of file diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC.aux b/research/08 shelah-spencer VC/Shelah-Spencer VC.aux deleted file mode 100644 index c21ea74f..00000000 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC.aux +++ /dev/null @@ -1,9 +0,0 @@ -\relax -\citation{Laskowski} -\bibcite{Laskowski}{1} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{12.7778pt} -\newlabel{tocindent1}{0pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} -\@writefile{toc}{\contentsline {section}{\tocsection {}{}{References}}{3}} diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC.bbl b/research/08 shelah-spencer VC/Shelah-Spencer VC.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC.blg b/research/08 shelah-spencer VC/Shelah-Spencer VC.blg deleted file mode 100644 index 0e137702..00000000 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC.blg +++ /dev/null @@ -1,4 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: Shelah-Spencer VC.aux -I found no \bibdata command---while reading file Shelah-Spencer VC.aux -I found no \bibstyle command---while reading file Shelah-Spencer VC.aux -(There were 2 error messages) diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC.log b/research/08 shelah-spencer VC/Shelah-Spencer VC.log deleted file mode 100644 index 29669fc9..00000000 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC.log +++ /dev/null @@ -1,255 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.5-1.40.14 (MiKTeX 2.9) (preloaded format=pdflatex 2014.9.19) 3 NOV 2014 13:21 -entering extended mode -**Shelah-Spencer*VC.tex -("C:\Users\Anton\SparkleShare\Research\research\08 shelah-spencer VC\Shelah-Spencer VC.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2013-05-26, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. -\end{proof} - -\begin{Lemma} - Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair with $\epsilon = \delta (\B_2/\A_2)$. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Assume that it is not that case that $\A_2 \subset \B_2$ and $\A_1$ is disjoint from $\A_2$ (No!). Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \epsilon$ -\end{Lemma} - -Fix an embedding $f$. It induces the following substructure diagram in $M$. Denote -\begin{align*} - \A &= \A_1 \cup \A_2 \\ - \B_f^* &= \B_1 \cup f(\B_2) \\ - \B_1^* &= \B_1 \cup \A \\ - \B_2^* &= f(\B_2) \cup \A \\ - \B^* &= \B_1^* \cap \B_2^* -\end{align*} - -\begin{diagram} - & &\B_f \\ - &\ruLine & &\luLine \\ - \B_1^* & & & &\B_2^* \\ - &\luLine & &\ruLine \\ - & &\B^* \\ - & &\uLine \\ - & &\A\\ -\end{diagram} - -From the diagram we see that -\begin{align*} - \delta(\B_f/\A) \leq \delta(\B_1^*/\A) + \delta(\B_2^*/\B^*) -\end{align*} -Thus all we need to do is to verify that -\begin{align*} - \delta(\B_2^*/\B^*) \leq \epsilon -\end{align*} -Let $\B'$ denote graph induced on all the vertices in $(f(B_2) / B_1) \cup A_2$. -Then $\B'$ is a substructure of $\B_2$ over $\A_2$. By minimality we get that $\delta(\B'/\A_2) \leq \epsilon$. -We need to show $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$. -Do the vertex computation -\begin{align*} - B_2^* - B^* &= \\ - f(B_2) - (B_1 \cap f(B_2)) - A &= \\ - f(B_2) - B_1 - A &= \\ - f(B_2) - B_1 - A_2 -\end{align*} -and -\begin{align*} - B' - A_2 &= - f(B_2) - B_1 - A_2 -\end{align*} - -So the sets of the extra vertices in the extension are the same. The base $\B_2^*/\B^*$ is larger so we can introduce some extra edges but no new vertices. This means that $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$ giving us the original statement. - - -Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists $\D$ over $\C = \{x,y\}$ such that $(\D, \C)$ is minimal with relative dimension $\epsilon$. Let $N$ be such that $N\epsilon < 1 < (N+1)\epsilon$. Then we argue that $vc(\phi) = N$. - -Fix a $m$-strong (for any $m > |D|$) set of non-connected vertices $A$. Fix some $a*$. We invesitgate the trace of $\phi(x, a*)$ on $A$. Suppose we have $a_1, \ldots, a_k$ satisfying $\phi(a_i, a^*)$ as witnessed by $D_i / \{a_i, a*\}$. Let $\D^* = \bigcap \D_i$ and $\C^*$ - -Call $\M$ $n$-composite embedding if there are distinct vertices $a_1, \ldots a_n$ and $a*$ in $M$ and there are an embeddings $\D \arr \M$ with $\C$ going to $\{a_i, a^*\}$. Image of $i$-th embedding is denoted $\D_i$. Note that images of embeddings can intersect each other or $a_j$'s. Consider $\D^* = \bigcap \D_i$ and $\C^* = \{a_1, \ldots a_n, a^*\}$. Dimension of $M$ is $\delta(\D^*/\C^*)$. - -Lemma: Dimension of $n$-composite embedding is at most $-n\epsilon$. - -Note: if $\D_i$ are disjoint over $\C^*$ then the dimension is exactly $-n\epsilon$. - -Take $n$-composite embedding with maximal dimension. Suppose it is larger than $-n\epsilon$. -Without loss of generality we may assume $\D_n$ intersects with $\D_1 \cup \ldots \cup \D_{n-1}$ over $\C^*$. -Consider two cases. -First, suppose that there is some element in $\D_n$ outside of $\D_1 \cup \ldots \cup \D_{n-1}$. -Let $\B_1 = \D_1 \cup \ldots \cup \D_{n-1}$. -Let $\A_1 = \{a_1, \ldots a_{n-1}\} \cup \{a*\}$. -Let $\B_2 = \D_n$. -Let $\A_2 = \{a_n, a*\}$. - -Lemma applies to the above. Above dimension is minimized when $\D_n$ is disjoint. Contradiction. - -Second, suppose that $\D_n \subseteq \B_1$. In particular $a_n \in \B_1$. Consider - -Consider sets $\B_1 \ldots \B_n$ with -\begin{enumerate} - \item $a_i \in \B_i$ - \item $a_i \in A$ - \item $a_i \neq a_j$ - \item $a* \in \bigcap \B_i$ -\end{enumerate} - and s.t. $\B_i / \{a*, a_i\}$ is isomorphic to $\B/\A$. We look at all the possible embeddings with those properties. We argue that a disjoint configuration minimizes total dimension of the whole construction. - -We argue by induction on $n$. Fix an embedding $\B_1, \ldots \B_n$ and consider possible choices for $\B_{n+1}, a_{n+1}$. We can pick $a_n$ to be an element of $A$ not used so far and embed $\B_{n+1}$ over $\{a*, a_i\}$ disjoint from the entire construction. On the other hand suppose it is embedded such that there is an intersection. We set up to apply the previous lemma. Let -\begin{align*} - \B_1 &= \bigcup_{1..n} \B_i \\ - \A_1 &= \{a_1, \ldots a_n\} \\ - \B_2 &= \B_{n+1} \\ - \A_2 &= \{a^*, a_{n+1}\} -\end{align*} -Applying the lemma say that the extra dimension cannot be larger then $\epsilon$. - -\begin{thebibliography}{9} - -\bibitem{Laskowski} - Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} - Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 - -\end{thebibliography} - +\documentclass{amsart} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\usepackage{diagrams} + + \newcommand{\A}{\mathcal A} + \newcommand{\B}{\mathcal B} +\renewcommand{\C}{\mathcal C} + \newcommand{\D}{\mathcal D} +\renewcommand{\H}{\mathcal H} + \newcommand{\G}{\mathcal G} + \newcommand{\M}{\mathcal M} + + \newcommand{\K}{\boldface K_\alpha} +\renewcommand{\S}{S_\alpha} + +\begin{document} + +\title{Some vc-density computations in Shelah-Spencer graphs} +\author{Anton Bobkov} +\email{bobkov@math.ucla.edu} + +\begin{abstract} + We compute vc-densities of minimal extension formulas in Shelah-Spencer random graphs. +\end{abstract} + +\maketitle + +We fix the density of the graph $\alpha$. + +\begin{Lemma} + For any $\A \in \K$ and $\epsilon > 0$ there exists an $\B$ such that $(\A, \B)$ is minimal and $\delta(\B/\A) < \epsilon$. +\end{Lemma} + +\begin{proof} + Let $m$ be an integer such that $m\alpha < 1 < (m+1)\alpha$. Suppose $\A$ has less than $m+1$ vertices. Make a construction $\A_0 = \A$ and $\A_{i+1}$ is $\A_i$ with one extra vertex connected to every single vertex of $A_i$. Stop when the total number of vertices is $m+1$. Proceed as in \cite{Laskowski} 4.1. Resulting construction is still minimal. +\end{proof} + +\begin{Lemma} + Let $\A_1 \subset \B_1$ and $\A_2 \subset \B_2$ be $\K$ structures with $(\A_2, \B_2)$ a minimal pair with $\epsilon = \delta (\B_2/\A_2)$. Let $M$ be some ambient structure. Fix embeddings of $\A_1, \B_1, \A_2$ into $M$. Assume that it is not that case that $\A_2 \subset \B_2$ and $\A_1$ is disjoint from $\A_2$ (No!). Now consider all possible embeddings $f \colon \B_2 \to M$ over $\A_1$. Let $\A = \A_1 \cup \A_2$ and $\B_f = \B_1 \cup f(\B_2)$ with $\delta_f = \delta(\B_f/\A)$. Then $\delta_f$ is at most $\delta(\B_1 \cup \A/\A) + \epsilon$ +\end{Lemma} + +Fix an embedding $f$. It induces the following substructure diagram in $M$. Denote +\begin{align*} + \A &= \A_1 \cup \A_2 \\ + \B_f^* &= \B_1 \cup f(\B_2) \\ + \B_1^* &= \B_1 \cup \A \\ + \B_2^* &= f(\B_2) \cup \A \\ + \B^* &= \B_1^* \cap \B_2^* +\end{align*} + +\begin{diagram} + & &\B_f \\ + &\ruLine & &\luLine \\ + \B_1^* & & & &\B_2^* \\ + &\luLine & &\ruLine \\ + & &\B^* \\ + & &\uLine \\ + & &\A\\ +\end{diagram} + +From the diagram we see that +\begin{align*} + \delta(\B_f/\A) \leq \delta(\B_1^*/\A) + \delta(\B_2^*/\B^*) +\end{align*} +Thus all we need to do is to verify that +\begin{align*} + \delta(\B_2^*/\B^*) \leq \epsilon +\end{align*} +Let $\B'$ denote graph induced on all the vertices in $(f(B_2) / B_1) \cup A_2$. +Then $\B'$ is a substructure of $\B_2$ over $\A_2$. By minimality we get that $\delta(\B'/\A_2) \leq \epsilon$. +We need to show $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$. +Do the vertex computation +\begin{align*} + B_2^* - B^* &= \\ + f(B_2) - (B_1 \cap f(B_2)) - A &= \\ + f(B_2) - B_1 - A &= \\ + f(B_2) - B_1 - A_2 +\end{align*} +and +\begin{align*} + B' - A_2 &= + f(B_2) - B_1 - A_2 +\end{align*} + +So the sets of the extra vertices in the extension are the same. The base $\B_2^*/\B^*$ is larger so we can introduce some extra edges but no new vertices. This means that $\delta(\B_2^*/\B^*) \leq \delta(\B'/\A_2)$ giving us the original statement. + + +Let $\phi(x,y)$ be a formula in a random graph with $|x|=|y|=1$ saying that there exists $\D$ over $\C = \{x,y\}$ such that $(\D, \C)$ is minimal with relative dimension $\epsilon$. Let $N$ be such that $N\epsilon < 1 < (N+1)\epsilon$. Then we argue that $vc(\phi) = N$. + +Fix a $m$-strong (for any $m > |D|$) set of non-connected vertices $A$. Fix some $a*$. We invesitgate the trace of $\phi(x, a*)$ on $A$. Suppose we have $a_1, \ldots, a_k$ satisfying $\phi(a_i, a^*)$ as witnessed by $D_i / \{a_i, a*\}$. Let $\D^* = \bigcap \D_i$ and $\C^*$ + +Call $\M$ $n$-composite embedding if there are distinct vertices $a_1, \ldots a_n$ and $a*$ in $M$ and there are an embeddings $\D \arr \M$ with $\C$ going to $\{a_i, a^*\}$. Image of $i$-th embedding is denoted $\D_i$. Note that images of embeddings can intersect each other or $a_j$'s. Consider $\D^* = \bigcap \D_i$ and $\C^* = \{a_1, \ldots a_n, a^*\}$. Dimension of $M$ is $\delta(\D^*/\C^*)$. + +Lemma: Dimension of $n$-composite embedding is at most $-n\epsilon$. + +Note: if $\D_i$ are disjoint over $\C^*$ then the dimension is exactly $-n\epsilon$. + +Take $n$-composite embedding with maximal dimension. Suppose it is larger than $-n\epsilon$. +Without loss of generality we may assume $\D_n$ intersects with $\D_1 \cup \ldots \cup \D_{n-1}$ over $\C^*$. +Consider two cases. +First, suppose that there is some element in $\D_n$ outside of $\D_1 \cup \ldots \cup \D_{n-1}$. +Let $\B_1 = \D_1 \cup \ldots \cup \D_{n-1}$. +Let $\A_1 = \{a_1, \ldots a_{n-1}\} \cup \{a*\}$. +Let $\B_2 = \D_n$. +Let $\A_2 = \{a_n, a*\}$. + +Lemma applies to the above. Above dimension is minimized when $\D_n$ is disjoint. Contradiction. + +Second, suppose that $\D_n \subseteq \B_1$. In particular $a_n \in \B_1$. Consider + +Consider sets $\B_1 \ldots \B_n$ with +\begin{enumerate} + \item $a_i \in \B_i$ + \item $a_i \in A$ + \item $a_i \neq a_j$ + \item $a* \in \bigcap \B_i$ +\end{enumerate} + and s.t. $\B_i / \{a*, a_i\}$ is isomorphic to $\B/\A$. We look at all the possible embeddings with those properties. We argue that a disjoint configuration minimizes total dimension of the whole construction. + +We argue by induction on $n$. Fix an embedding $\B_1, \ldots \B_n$ and consider possible choices for $\B_{n+1}, a_{n+1}$. We can pick $a_n$ to be an element of $A$ not used so far and embed $\B_{n+1}$ over $\{a*, a_i\}$ disjoint from the entire construction. On the other hand suppose it is embedded such that there is an intersection. We set up to apply the previous lemma. Let +\begin{align*} + \B_1 &= \bigcup_{1..n} \B_i \\ + \A_1 &= \{a_1, \ldots a_n\} \\ + \B_2 &= \B_{n+1} \\ + \A_2 &= \{a^*, a_{n+1}\} +\end{align*} +Applying the lemma say that the extra dimension cannot be larger then $\epsilon$. + +\begin{thebibliography}{9} + +\bibitem{Laskowski} + Michael C. Laskowski, \textsl{A simpler axiomatization of the Shelah-Spencer almost sure theories,} + Israel J. Math. \textbf{161} (2007), 157-186. MR MR2350161 + +\end{thebibliography} + \end{document} diff --git a/research/08 shelah-spencer VC/Shelah-Spencer VC.tps b/research/08 shelah-spencer VC/Shelah-Spencer VC.tps deleted file mode 100644 index f030baee..00000000 --- a/research/08 shelah-spencer VC/Shelah-Spencer VC.tps +++ /dev/null @@ -1,26 +0,0 @@ -[FormatInfo] -Type=TeXnicCenterProjectSessionInformation -Version=2 - -[Frame0] -Flags=0 -ShowCmd=1 -MinPos.x=-1 -MinPos.y=-1 -MaxPos.x=-1 -MaxPos.y=-1 -NormalPos.left=4 -NormalPos.top=26 -NormalPos.right=1289 -NormalPos.bottom=728 -Class=LaTeXView -Document=Shelah-Spencer VC.tex - -[Frame0_View0,0] -TopLine=0 -Cursor=0 - -[SessionInfo] -FrameCount=1 -ActiveFrame=0 - diff --git a/research/09 ATC stuff/ATC background.aux b/research/09 ATC stuff/ATC background.aux deleted file mode 100644 index 1269c74d..00000000 --- a/research/09 ATC stuff/ATC background.aux +++ /dev/null @@ -1,8 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{Combinatorics}}{1}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Model Theory}}{2}} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/09 ATC stuff/ATC background.bbl b/research/09 ATC stuff/ATC background.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/09 ATC stuff/ATC background.blg b/research/09 ATC stuff/ATC background.blg deleted file mode 100644 index a59478e7..00000000 --- a/research/09 ATC stuff/ATC background.blg +++ /dev/null @@ -1,5 +0,0 @@ -This is BibTeX, Version 0.99dThe top-level auxiliary file: ATC background.aux -I found no \citation commands---while reading file ATC background.aux -I found no \bibdata command---while reading file ATC background.aux -I found no \bibstyle command---while reading file ATC background.aux -(There were 3 error messages) diff --git a/research/09 ATC stuff/ATC background.log b/research/09 ATC stuff/ATC background.log deleted file mode 100644 index 2d98b432..00000000 --- a/research/09 ATC stuff/ATC background.log +++ /dev/null @@ -1,256 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 2 JUN 2015 21:35 -entering extended mode -**ATC*background.tex -("C:\Users\Anton\SparkleShare\Research\research\09 ATC stuff\ATC background.tex" -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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-\begin{Definition} - \begin{align*} - \pi_\F(n) = \max \{ \text {\# of atoms in boolean algebra generated by $S$} \mid \\ S \subset \F \text{ and } |S| = n\} - \end{align*} -\end{Definition} - -\includegraphics[scale=0.75]{circle.png} -\begin{Example} - Let $\F$ consist of all discs in the plane. - \begin{align*} - \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 - \end{align*} - \begin{align*} - \pi_\F(n) = n^2 - n + 2 - \end{align*} -\end{Example} - -\includegraphics[scale=0.75]{lines.png} -\begin{Example} - Let $\F$ consist of all half-planes in the plane. - \begin{align*} - \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 7 \ \ \ \pi_\F(4) = 11 - \end{align*} - \begin{align*} - \pi_\F(n) = n^2/2 + n/2 + 1 - \end{align*} -\end{Example} - -\begin{Example} \ \\ - \begin{enumerate} - \item Let $\F$ be a set of lines on a plane. Then - \begin{align*} - \pi_\F(n) &= n(n+1)/2 + 1 - \end{align*} - \item Let $\F$ be a set of disks on a plane. Then - \begin{align*} - \pi_\F(n) &= n^2 - n + 2 - \end{align*} - \item Let $\F$ be a set of balls in $\R^3$. Then - \begin{align*} - \pi_\F(n) &= n(n^2 - 3n + 8)/3 - \end{align*} - \item Let $\F$ be a set of intervals on a line. Then - \begin{align*} - \pi_\F(n) &= 2n - \end{align*} - \item Let $\F$ be a set of half-planes. Then - \begin{align*} - \pi_\F(n) &= n(n+1)/2 + 1 - \end{align*} - \item Let $\F$ be a collection of finite subsets of $\N$. Then - \begin{align*} - \pi_\F(n) &= 2^n - \end{align*} - \item Let $\F$ be a collection of polygons in a plane. Then - \begin{align*} - \pi_\F(n) &= 2^n - \end{align*} - \end{enumerate} -\end{Example} - -\begin{Theorem} [Sauer-Shelah] - Shatter function is either $2^n$ or bounded by a polynomial. -\end{Theorem} - -\begin{Definition} - Suppose growth of shatter function for $\F$ is polynomial. - Let $r$ be the smallest real such that - \begin{align*} - \pi_\F(n) = O(n^r) - \end{align*} - We define such $r$ to be the vc-density of $\F$. - If shatter function grows exponentially, we let vc-density to be infinite. -\end{Definition} - -\section{Model Theory} - -Consider a structure with a language -\begin{align*} - (\R, 0, 1, +, \cdot, \leq) -\end{align*} - -We work with subsets definable by first-order formulas. -Those are called definable sets. - -\begin{align*} - \phi(x) &:= \paren{(5 \leq x) \wedge (x \leq 7.7)} \vee (x \leq 0) \text { (either $x$ is between 5 and 7.7 or it is non-negative)}\\ - \psi(x) &:= (\exists y \ y \cdot y = x) \text { (there exists $y$ such that $y$ squared is $x$) }\\ - \gamma(x) &:= (x \cdot x \cdot x \cdot x = 2) \text { ($x$ to the fourth power is 2) }\\ -\end{align*} - - -\begin{itemize} - \item $\phi(\R)$ defines the set $[5, 7.7] \cup (-\infty, 0]$ - \item $\psi(\R)$ defines the set $[0, \infty)$ -\end{itemize} - -\begin{enumerate} - \item in rationals $(\Q, \cdot)$, $\gamma(\Q)$ defines an empty subset - \item in reals $(\R, \cdot)$, $\gamma(\R)$ defines a subset with two elements - \item in complex numbers $(\C, \cdot)$, $\gamma(\C)$ defines a subset with four elements - \item in quaternions $(\mathbb H, \cdot)$, $\gamma(\mathbb H)$ defines an infinite subset -\end{enumerate} - -\begin{align*} - \theta(x) = &\forall y \exists z \ x \leq z \leq y \\ - &\text{for all $y$ there exists $z$ such that $x \leq z \leq y$} -\end{align*} - -\begin{enumerate} - \item in $(\Q, \leq)$, $\theta(\Q)$ defines an empty subset - \item in $(\N, \leq)$, $\theta(\N)$ defines an empty subset - \item in $(\Q^{\geq 0}, \leq)$, $\theta(\Q^{\geq 0})$ defines the set with one element $\{0\}$ -\end{enumerate} - -\begin{Definition} - for a formula $\phi(x_1 \ldots x_n, y_1, \ldots y_m)$ we can plug in elements of our structure as parameters in places of $y$ variables. This gives us a collection of definable sets. -\end{Definition} - -\begin{Example} - \begin{align*} - \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 - \end{align*} -\end{Example} - -In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. \\ - -We say that all discs in $\R^2$ are defined uniformly by $\phi$. \\ - -What are the collection of sets we can consider when working with a model? \\ - -We can look at all definable subsets. That's not interesting, always has an infinite vc-density. -Uniformly definable families offer more interesting behavior. \\ - -A model is said to be NIP if all uniformly definable families have finite vc-density. \\ - -For a given structure $M$, let $\vc^M(n)$ be the largest $\vc$-density achieved by $n$-dimensional families of uniformly definable sets. - -\begin{align*} - \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} -\end{align*} - -It is easy to show that $\vc_M(n) \geq n$ for all models. - -Open questions about vc functions. - -Is $\vc_M(n) = n \vc_M(1)$? If not, is there a linear relationship? - +\documentclass{amsart} +\usepackage{graphicx} + +\usepackage{../AMC_style} +\usepackage{../Research} + +\newcommand{\F}{\mathcal F} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\paren}[1]{\left( #1 \right)} + +%\DeclareMathOperator{\vc}{vc} + +\begin{document} + +\section{Combinatorics} + +Suppose we have an infinite collection of sets $\F$. +Take $n$ many of those sets. +They generate a boolean algebra. +Count the number of atoms in it. +There can be at most $2^n$ atoms, though depending on the collection there may be much less. +For a given $n$, out of all choices of $n$ sets, record the highest possible number of atoms generated. +We define that to be a shatter function. + +\begin{Definition} + \begin{align*} + \pi_\F(n) = \max \{ \text {\# of atoms in boolean algebra generated by $S$} \mid \\ S \subset \F \text{ and } |S| = n\} + \end{align*} +\end{Definition} + +\includegraphics[scale=0.75]{circle.png} +\begin{Example} + Let $\F$ consist of all discs in the plane. + \begin{align*} + \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 + \end{align*} + \begin{align*} + \pi_\F(n) = n^2 - n + 2 + \end{align*} +\end{Example} + +\includegraphics[scale=0.75]{lines.png} +\begin{Example} + Let $\F$ consist of all half-planes in the plane. + \begin{align*} + \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 7 \ \ \ \pi_\F(4) = 11 + \end{align*} + \begin{align*} + \pi_\F(n) = n^2/2 + n/2 + 1 + \end{align*} +\end{Example} + +\begin{Example} \ \\ + \begin{enumerate} + \item Let $\F$ be a set of lines on a plane. Then + \begin{align*} + \pi_\F(n) &= n(n+1)/2 + 1 + \end{align*} + \item Let $\F$ be a set of disks on a plane. Then + \begin{align*} + \pi_\F(n) &= n^2 - n + 2 + \end{align*} + \item Let $\F$ be a set of balls in $\R^3$. Then + \begin{align*} + \pi_\F(n) &= n(n^2 - 3n + 8)/3 + \end{align*} + \item Let $\F$ be a set of intervals on a line. Then + \begin{align*} + \pi_\F(n) &= 2n + \end{align*} + \item Let $\F$ be a set of half-planes. Then + \begin{align*} + \pi_\F(n) &= n(n+1)/2 + 1 + \end{align*} + \item Let $\F$ be a collection of finite subsets of $\N$. Then + \begin{align*} + \pi_\F(n) &= 2^n + \end{align*} + \item Let $\F$ be a collection of polygons in a plane. Then + \begin{align*} + \pi_\F(n) &= 2^n + \end{align*} + \end{enumerate} +\end{Example} + +\begin{Theorem} [Sauer-Shelah] + Shatter function is either $2^n$ or bounded by a polynomial. +\end{Theorem} + +\begin{Definition} + Suppose growth of shatter function for $\F$ is polynomial. + Let $r$ be the smallest real such that + \begin{align*} + \pi_\F(n) = O(n^r) + \end{align*} + We define such $r$ to be the vc-density of $\F$. + If shatter function grows exponentially, we let vc-density to be infinite. +\end{Definition} + +\section{Model Theory} + +Consider a structure with a language +\begin{align*} + (\R, 0, 1, +, \cdot, \leq) +\end{align*} + +We work with subsets definable by first-order formulas. +Those are called definable sets. + +\begin{align*} + \phi(x) &:= \paren{(5 \leq x) \wedge (x \leq 7.7)} \vee (x \leq 0) \text { (either $x$ is between 5 and 7.7 or it is non-negative)}\\ + \psi(x) &:= (\exists y \ y \cdot y = x) \text { (there exists $y$ such that $y$ squared is $x$) }\\ + \gamma(x) &:= (x \cdot x \cdot x \cdot x = 2) \text { ($x$ to the fourth power is 2) }\\ +\end{align*} + + +\begin{itemize} + \item $\phi(\R)$ defines the set $[5, 7.7] \cup (-\infty, 0]$ + \item $\psi(\R)$ defines the set $[0, \infty)$ +\end{itemize} + +\begin{enumerate} + \item in rationals $(\Q, \cdot)$, $\gamma(\Q)$ defines an empty subset + \item in reals $(\R, \cdot)$, $\gamma(\R)$ defines a subset with two elements + \item in complex numbers $(\C, \cdot)$, $\gamma(\C)$ defines a subset with four elements + \item in quaternions $(\mathbb H, \cdot)$, $\gamma(\mathbb H)$ defines an infinite subset +\end{enumerate} + +\begin{align*} + \theta(x) = &\forall y \exists z \ x \leq z \leq y \\ + &\text{for all $y$ there exists $z$ such that $x \leq z \leq y$} +\end{align*} + +\begin{enumerate} + \item in $(\Q, \leq)$, $\theta(\Q)$ defines an empty subset + \item in $(\N, \leq)$, $\theta(\N)$ defines an empty subset + \item in $(\Q^{\geq 0}, \leq)$, $\theta(\Q^{\geq 0})$ defines the set with one element $\{0\}$ +\end{enumerate} + +\begin{Definition} + for a formula $\phi(x_1 \ldots x_n, y_1, \ldots y_m)$ we can plug in elements of our structure as parameters in places of $y$ variables. This gives us a collection of definable sets. +\end{Definition} + +\begin{Example} + \begin{align*} + \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 + \end{align*} +\end{Example} + +In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. \\ + +We say that all discs in $\R^2$ are defined uniformly by $\phi$. \\ + +What are the collection of sets we can consider when working with a model? \\ + +We can look at all definable subsets. That's not interesting, always has an infinite vc-density. +Uniformly definable families offer more interesting behavior. \\ + +A model is said to be NIP if all uniformly definable families have finite vc-density. \\ + +For a given structure $M$, let $\vc^M(n)$ be the largest $\vc$-density achieved by $n$-dimensional families of uniformly definable sets. + +\begin{align*} + \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} +\end{align*} + +It is easy to show that $\vc_M(n) \geq n$ for all models. + +Open questions about vc functions. + +Is $\vc_M(n) = n \vc_M(1)$? If not, is there a linear relationship? + \end{document} \ No newline at end of file diff --git a/research/09 ATC stuff/ATC basckground.aux b/research/09 ATC stuff/ATC basckground.aux deleted file mode 100644 index b36cc7b3..00000000 --- a/research/09 ATC stuff/ATC basckground.aux +++ /dev/null @@ -1,8 +0,0 @@ -\relax -\@writefile{toc}{\contentsline {section}{\tocsection {}{1}{VC-dimension}}{1}} -\@writefile{toc}{\contentsline {section}{\tocsection {}{2}{Model Theory}}{1}} -\newlabel{tocindent-1}{0pt} -\newlabel{tocindent0}{0pt} -\newlabel{tocindent1}{17.77782pt} -\newlabel{tocindent2}{0pt} -\newlabel{tocindent3}{0pt} diff --git a/research/09 ATC stuff/ATC basckground.bbl b/research/09 ATC stuff/ATC basckground.bbl deleted file mode 100644 index e69de29b..00000000 diff --git a/research/09 ATC stuff/ATC basckground.blg b/research/09 ATC stuff/ATC basckground.blg deleted file mode 100644 index a6a84d86..00000000 --- a/research/09 ATC stuff/ATC basckground.blg +++ /dev/null @@ -1,5 +0,0 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+\headcommand {\beamer@partpages {1}{21}} +\headcommand {\beamer@subsectionpages {1}{21}} +\headcommand {\beamer@sectionpages {1}{21}} +\headcommand {\beamer@documentpages {21}} +\headcommand {\def \inserttotalframenumber {21}} diff --git a/research/09 ATC stuff/ATC slides.out b/research/09 ATC stuff/ATC slides.out deleted file mode 100644 index e69de29b..00000000 diff --git a/research/09 ATC stuff/ATC slides.tex b/research/09 ATC stuff/ATC slides.tex index 54400638..54138312 100644 --- a/research/09 ATC stuff/ATC slides.tex +++ b/research/09 ATC stuff/ATC slides.tex @@ -1,269 +1,269 @@ -\documentclass{beamer} - -\usepackage{../Research} - -\newcommand{\F}{\mathcal F} -\newcommand{\curly}[1]{\left\{ #1 \right\}} -\newcommand{\paren}[1]{\left( #1 \right)} -\newcommand{\defn}{\underline} - -\title{VC-density in model theoretic structures} -\author{Anton Bobkov} -\date{June 3, 2015} - - -\begin{document} - -\maketitle - -\begin{frame} - Suppose we have an (infinite) collection of sets $\F$. \\ - We define the \defn{shatter function} $\pi_\F \colon \N \arr \N$ of $\F$ - - \begin{align*} - \pi_\F(n) = \max \{ &\text {\# of atoms in the boolean algebra generated by $\mathcal S$} \\ - &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} - \end{align*} -\end{frame} - -\begin{frame} - Example: Let $\F$ consist of all half-planes in the plane. - \begin{figure}[p] - \centering - \includegraphics[scale=0.75]{lines.png} - \end{figure} - \begin{align*} - \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 7 \ \ \ \pi_\F(4) = 11 - \end{align*} - \begin{align*} - \pi_\F(n) = n^2/2 + n/2 + 1 - \end{align*} -\end{frame} - -\begin{frame} -More examples: \\ - \begin{enumerate} - %\item Half-planes in the plane: $\pi_\F(n) = n^2/2 + n/2 + 1$ - \item Disks in the plane: $\pi_\F(n) = n^2 - n + 2$ - \item Balls in $\R^3$: $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ - \item Intervals in the line: $\pi_\F(n) = 2n$ - \item Finite subsets of $\N$: $\pi_\F(n) = 2^n$ - \item Convex polygons in the plane: $\pi_\F(n) = 2^n$ - \end{enumerate} -\end{frame} - -\begin{frame} - \begin{Theorem} [Sauer-Shelah '72] - The shatter function is either $2^n$ or bounded by a polynomial. - \end{Theorem} - \begin{Definition} - Suppose the growth of the shatter function of $\F$ is polynomial. - Let $\vc(\F)$ be the infimum of all positive reals $r$ such that - \begin{align*} - \pi_\F(n) = O(n^r) - \end{align*} - Call $\vc(\F)$ the \defn{vc-density} of $\F$. - If the shatter function grows exponentially, we let $\vc(\F) := \infty$. - \end{Definition} -\end{frame} - -\begin{frame} - \frametitle{Applications} - \begin{itemize} - \item Model Theory (NIP theories) - \item VC-Theorem in probability (Vapnik-Chervonenkis '71) - \item Computational learning theory (PAC learning, Warmuth conjecture) - \item Computational geometry - \item Functional analysis (Bourgain-Fremlin-Talagrand theory) - \item Abstract topological dynamics (tame dynamical systems) - \end{itemize} -\end{frame} - - -\begin{frame} - \frametitle{Model Theory} - Model Theory studies definable sets in first-order structures. - \begin{align*} - (\Q, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) := \paren{\exists y \ y \cdot y = x} - \end{align*} - $\phi(\Q)$ defines the set of rationals that are a square. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) := \paren{\exists y \ y \cdot y = x} - \end{align*} - $\phi(\R)$ defines the set $[0, \infty)$. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \psi(x_1, x_2) := (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1\cdot x_1 \leq x_2) - \end{align*} - $\psi(\R^2)$ defines the set in $\R^2$ that is an intersection of a disc with an inside of a parabola. -\end{frame} - -\begin{frame} - \begin{Definition} - Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$ and structure $M$. - Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^m$. - \begin{align*} - \F^M_\phi = \curly{\phi(M^m, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} - \end{align*} - $\F^M_\phi$ is a \defn{uniformly definable family}. \\ - Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$. - \end{Definition} -\end{frame} - -\begin{frame} - \begin{Example} - Consider the following formula in structure $(\R, 0, 1 +, \cdot, \leq)$ - \begin{align*} - \phi(x_1, x_2, y_1, y_2, y_3) := (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 - \end{align*} - For $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ and center $(a,b)$. \\ - Thus $\F^\R_\phi$ is a collection of all discs in $\R^2$. - \end{Example} -\end{frame} - -\begin{frame} - For a given structure $M$, possible numbers of isomorphism classes for structures elementarily equivalent to it were classified by Shelah ('78). - This classification involved defining important dividing lines describing complexity of structures. - One of those diving lines is whether the structure is NIP or not. - \begin{Definition} - Structure $M$ is said to be \defn{NIP} (no independence property) if all uniformly definable families in it have finite $\vc$-density. - \end{Definition} - Examples: - \begin{itemize} - \item $(\C, 0, 1, +, \cdot)$ is NIP - \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP - \item Random graph $(V, R)$ is not NIP - \item $(\Q, 0, 1, +, \cdot)$ is not NIP. - \end{itemize} - - \textbf{Open Question}: If $M$ is NIP, is it possible for $\vc^M(\phi)$ to be irrational? \\ -\end{frame} - -\begin{frame} - \frametitle{History} - \begin{itemize} - \item VC-dimension defined by Vapnik-Chervonenkis '71 - \item NIP theories studied by Shelah '71 - \item vc-density in model theoretic context introduced by Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13 - \end{itemize} -\end{frame} - -\begin{frame} - Given an NIP structure $M$ we define $\vc^M(n)$ to be the largest $\vc$-density achieved by uniformly definable families in $M^n$. - \begin{align*} - \vc^M(n) = \sup \curly{ \vc^M(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} - \end{align*} - Easy to show: - \begin{align*} - \vc^M(n) \geq n \cdot \vc^M(1) \geq n - \end{align*} - - \textbf{Open Question}: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? If $\vc(1) < \infty$ do we have $\vc(n) < \infty$? -\end{frame} - -\begin{frame} - Examples: - \begin{itemize} - \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ - \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ - \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{vc-density in trees} - Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. - \begin{itemize} - \item Trees are NIP (Parigot '82) - \item Trees are dp-minimal (Simon '11) - \item Trees have $vc(n) = n$ (B. '14) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{proof background} - $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\\ - Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have - \begin{align*} - \tp(a), \tp(b) \vdash \tp(ab) - \end{align*} - This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. -\end{frame} - -\begin{frame} - \frametitle{Shelah-Spencer graphs} - Let $\alpha$ irrational $\in (0,1)$. Consider a random graph on $n$ vertices where the probability of any given two vertices having an edge is $n^{-\alpha}$. Shelah-Spencer ('88) showed that 0-1 law holds for first-order sentences. An (infinite) structure satisfying those axioms is called a \defn{Shelah-Spencer graph}. - \begin{itemize} - \item VC-dimension computation for neighborhoods (Anthony-Brightwell-Cooper '95) - \item Shelah-Spencer graphs are stable thus NIP (Baldwin-Shi '96, Baldwin-Shelah '97) - \item Quantifier elimination (Laskowski '06) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Background} - \begin{itemize} - \item $B/A$ is called an \defn{extension} if $A$ is a subgraph of $B$. - \item For $B/A$ we define a \defn{dimension} - \begin{align*} - \delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A| - \end{align*} - %\item Extension $B/A$ is \defn{positive} if its dimension is positive. - \item $B/A$ is called \defn{minimal} if it has negative dimension, but its every subextension has a positive dimension. - \item $(A_0, \ldots A_n)$ is a \defn{minimal chain} if each $A_{i + 1}/A_i$ is minimal. - \item $B/A$ is a \defn{chain-minimal} extension if there exists a minimal chain $(A_0, \ldots A_n)$ such with $A_0 = A, A_n = B$. - \end{itemize} - For $B/A$ chain-minimal define - \begin{align*} - \phi_{A,B}(\vec x) = \exists \vec x^* \text { such that $\vec x^*/\vec x$ is isomorphic to $B/A$} - \end{align*} - \begin{Theorem} [quantifier elimination, Laskowski '06] - In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x)}$. - \end{Theorem} -\end{frame} - -\begin{frame} - \frametitle{vc-density in Shelah-Spencer graphs} - \begin{Theorem} [B. '15] - For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ explicitly computable from $\delta(B_i/A_i)$ such that - \begin{align*} - \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| - \end{align*} - \end{Theorem} - \begin{Corollary} - $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. - \end{Corollary} -\end{frame} - -\begin{frame} - \frametitle{Future work} - \begin{itemize} - \item Other partial orderings, lattices - \item Other graph structures, in particular flat graphs - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ - \end{itemize} -\end{frame} - -\end{document} - - -\begin{frame} - To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. - Any -\end{frame} - +\documentclass{beamer} + +\usepackage{../Research} + +\newcommand{\F}{\mathcal F} +\newcommand{\curly}[1]{\left\{ #1 \right\}} +\newcommand{\paren}[1]{\left( #1 \right)} +\newcommand{\defn}{\underline} + +\title{VC-density in model theoretic structures} +\author{Anton Bobkov} +\date{June 3, 2015} + + +\begin{document} + +\maketitle + +\begin{frame} + Suppose we have an (infinite) collection of sets $\F$. \\ + We define the \defn{shatter function} $\pi_\F \colon \N \arr \N$ of $\F$ + + \begin{align*} + \pi_\F(n) = \max \{ &\text {\# of atoms in the boolean algebra generated by $\mathcal S$} \\ + &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} + \end{align*} +\end{frame} + +\begin{frame} + Example: Let $\F$ consist of all half-planes in the plane. + \begin{figure}[p] + \centering + \includegraphics[scale=0.75]{lines.png} + \end{figure} + \begin{align*} + \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 7 \ \ \ \pi_\F(4) = 11 + \end{align*} + \begin{align*} + \pi_\F(n) = n^2/2 + n/2 + 1 + \end{align*} +\end{frame} + +\begin{frame} +More examples: \\ + \begin{enumerate} + %\item Half-planes in the plane: $\pi_\F(n) = n^2/2 + n/2 + 1$ + \item Disks in the plane: $\pi_\F(n) = n^2 - n + 2$ + \item Balls in $\R^3$: $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ + \item Intervals in the line: $\pi_\F(n) = 2n$ + \item Finite subsets of $\N$: $\pi_\F(n) = 2^n$ + \item Convex polygons in the plane: $\pi_\F(n) = 2^n$ + \end{enumerate} +\end{frame} + +\begin{frame} + \begin{Theorem} [Sauer-Shelah '72] + The shatter function is either $2^n$ or bounded by a polynomial. + \end{Theorem} + \begin{Definition} + Suppose the growth of the shatter function of $\F$ is polynomial. + Let $\vc(\F)$ be the infimum of all positive reals $r$ such that + \begin{align*} + \pi_\F(n) = O(n^r) + \end{align*} + Call $\vc(\F)$ the \defn{vc-density} of $\F$. + If the shatter function grows exponentially, we let $\vc(\F) := \infty$. + \end{Definition} +\end{frame} + +\begin{frame} + \frametitle{Applications} + \begin{itemize} + \item Model Theory (NIP theories) + \item VC-Theorem in probability (Vapnik-Chervonenkis '71) + \item Computational learning theory (PAC learning, Warmuth conjecture) + \item Computational geometry + \item Functional analysis (Bourgain-Fremlin-Talagrand theory) + \item Abstract topological dynamics (tame dynamical systems) + \end{itemize} +\end{frame} + + +\begin{frame} + \frametitle{Model Theory} + Model Theory studies definable sets in first-order structures. + \begin{align*} + (\Q, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) := \paren{\exists y \ y \cdot y = x} + \end{align*} + $\phi(\Q)$ defines the set of rationals that are a square. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) := \paren{\exists y \ y \cdot y = x} + \end{align*} + $\phi(\R)$ defines the set $[0, \infty)$. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \psi(x_1, x_2) := (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1\cdot x_1 \leq x_2) + \end{align*} + $\psi(\R^2)$ defines the set in $\R^2$ that is an intersection of a disc with an inside of a parabola. +\end{frame} + +\begin{frame} + \begin{Definition} + Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$ and structure $M$. + Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^m$. + \begin{align*} + \F^M_\phi = \curly{\phi(M^m, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} + \end{align*} + $\F^M_\phi$ is a \defn{uniformly definable family}. \\ + Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$. + \end{Definition} +\end{frame} + +\begin{frame} + \begin{Example} + Consider the following formula in structure $(\R, 0, 1 +, \cdot, \leq)$ + \begin{align*} + \phi(x_1, x_2, y_1, y_2, y_3) := (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 + \end{align*} + For $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ and center $(a,b)$. \\ + Thus $\F^\R_\phi$ is a collection of all discs in $\R^2$. + \end{Example} +\end{frame} + +\begin{frame} + For a given structure $M$, possible numbers of isomorphism classes for structures elementarily equivalent to it were classified by Shelah ('78). + This classification involved defining important dividing lines describing complexity of structures. + One of those diving lines is whether the structure is NIP or not. + \begin{Definition} + Structure $M$ is said to be \defn{NIP} (no independence property) if all uniformly definable families in it have finite $\vc$-density. + \end{Definition} + Examples: + \begin{itemize} + \item $(\C, 0, 1, +, \cdot)$ is NIP + \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP + \item Random graph $(V, R)$ is not NIP + \item $(\Q, 0, 1, +, \cdot)$ is not NIP. + \end{itemize} + + \textbf{Open Question}: If $M$ is NIP, is it possible for $\vc^M(\phi)$ to be irrational? \\ +\end{frame} + +\begin{frame} + \frametitle{History} + \begin{itemize} + \item VC-dimension defined by Vapnik-Chervonenkis '71 + \item NIP theories studied by Shelah '71 + \item vc-density in model theoretic context introduced by Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13 + \end{itemize} +\end{frame} + +\begin{frame} + Given an NIP structure $M$ we define $\vc^M(n)$ to be the largest $\vc$-density achieved by uniformly definable families in $M^n$. + \begin{align*} + \vc^M(n) = \sup \curly{ \vc^M(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} + \end{align*} + Easy to show: + \begin{align*} + \vc^M(n) \geq n \cdot \vc^M(1) \geq n + \end{align*} + + \textbf{Open Question}: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? If $\vc(1) < \infty$ do we have $\vc(n) < \infty$? +\end{frame} + +\begin{frame} + Examples: + \begin{itemize} + \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ + \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ + \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{vc-density in trees} + Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. + \begin{itemize} + \item Trees are NIP (Parigot '82) + \item Trees are dp-minimal (Simon '11) + \item Trees have $vc(n) = n$ (B. '14) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{proof background} + $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\\ + Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have + \begin{align*} + \tp(a), \tp(b) \vdash \tp(ab) + \end{align*} + This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. +\end{frame} + +\begin{frame} + \frametitle{Shelah-Spencer graphs} + Let $\alpha$ irrational $\in (0,1)$. Consider a random graph on $n$ vertices where the probability of any given two vertices having an edge is $n^{-\alpha}$. Shelah-Spencer ('88) showed that 0-1 law holds for first-order sentences. An (infinite) structure satisfying those axioms is called a \defn{Shelah-Spencer graph}. + \begin{itemize} + \item VC-dimension computation for neighborhoods (Anthony-Brightwell-Cooper '95) + \item Shelah-Spencer graphs are stable thus NIP (Baldwin-Shi '96, Baldwin-Shelah '97) + \item Quantifier elimination (Laskowski '06) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Background} + \begin{itemize} + \item $B/A$ is called an \defn{extension} if $A$ is a subgraph of $B$. + \item For $B/A$ we define a \defn{dimension} + \begin{align*} + \delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A| + \end{align*} + %\item Extension $B/A$ is \defn{positive} if its dimension is positive. + \item $B/A$ is called \defn{minimal} if it has negative dimension, but its every subextension has a positive dimension. + \item $(A_0, \ldots A_n)$ is a \defn{minimal chain} if each $A_{i + 1}/A_i$ is minimal. + \item $B/A$ is a \defn{chain-minimal} extension if there exists a minimal chain $(A_0, \ldots A_n)$ such with $A_0 = A, A_n = B$. + \end{itemize} + For $B/A$ chain-minimal define + \begin{align*} + \phi_{A,B}(\vec x) = \exists \vec x^* \text { such that $\vec x^*/\vec x$ is isomorphic to $B/A$} + \end{align*} + \begin{Theorem} [quantifier elimination, Laskowski '06] + In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x)}$. + \end{Theorem} +\end{frame} + +\begin{frame} + \frametitle{vc-density in Shelah-Spencer graphs} + \begin{Theorem} [B. '15] + For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ explicitly computable from $\delta(B_i/A_i)$ such that + \begin{align*} + \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| + \end{align*} + \end{Theorem} + \begin{Corollary} + $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. + \end{Corollary} +\end{frame} + +\begin{frame} + \frametitle{Future work} + \begin{itemize} + \item Other partial orderings, lattices + \item Other graph structures, in particular flat graphs + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ + \end{itemize} +\end{frame} + +\end{document} + + +\begin{frame} + To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. + Any +\end{frame} + diff --git a/research/09 ATC stuff/ATC slides.toc b/research/09 ATC stuff/ATC slides.toc deleted file mode 100644 index 4c18f8e5..00000000 --- a/research/09 ATC stuff/ATC slides.toc +++ /dev/null @@ -1 +0,0 @@ -\beamer@endinputifotherversion {3.33pt} diff --git a/research/09 ATC stuff/CRE1C9E.tmp b/research/09 ATC stuff/CRE1C9E.tmp index 8a5cd543..4e11a756 100644 --- a/research/09 ATC stuff/CRE1C9E.tmp +++ b/research/09 ATC stuff/CRE1C9E.tmp @@ -1,252 +1,252 @@ -\documentclass{beamer} - -\usepackage{../Research} - -\newcommand{\F}{\mathcal F} -\newcommand{\curly}[1]{\left\{ #1 \right\}} - -\title{VC-density in model theoretic structures} -\author{Anton Bobkov} -\date{June 3, 2015} - - -\begin{document} - -\maketitle - -\begin{frame} - Suppose we have an (infinite) collection of sets $\F$. \\ - We define the shatter function $\pi_\F \colon \N \arr \N$ of $\F$ - - \begin{align*} - \pi_\F(n) = \max \{ &\text {\# of atoms in boolean algebra generated by $\mathcal S$} \\ - &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} - \end{align*} -\end{frame} - -\begin{frame} - Example: Let $\F$ consist of all discs in the plane. - \begin{figure}[p] - \centering - \includegraphics[scale=0.75]{circle.png} - \end{figure} - \begin{align*} - \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 - \end{align*} - \begin{align*} - \pi_\F(n) = n^2 - n + 2 - \end{align*} -\end{frame} - -\begin{frame} -More examples: \\ - \begin{enumerate} - \item Lines in the plane $\pi_\F(n) = n^2/2 + n/2 + 1$ - \item Disks in the plane $\pi_\F(n) = n^2 - n + 2$ - \item Balls in $\R^3$ $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ - \item Intervals in the line $\pi_\F(n) = 2n$ - \item Half-planes in the plane $\pi_\F(n) = n(n+1)/2 + 1$ - \item Finite subsets of $\N$ $\pi_\F(n) = 2^n$ - \item Convex polygons in the plane $\pi_\F(n) = 2^n$ - \end{enumerate} -\end{frame} - -\begin{frame} - \begin{Theorem} [Sauer-Shelah '72] - The shatter function is either $2^n$ or bounded by a polynomial. - \end{Theorem} - \begin{Definition} - Suppose the growth of the shatter function of $\F$ is polynomial. - Let $\vc(\F)$ infimum of all positive reals $r$ such that - \begin{align*} - \pi_\F(n) = O(n^r) - \end{align*} - Call $\vc(\F)$ the vc-density of $\F$. - If the shatter function grows exponentially, we let $\vc(\F) := \infty$. - \end{Definition} -\end{frame} - -\begin{frame} - \frametitle{Applications} - \begin{itemize} - \item NIP theories - \item VC-Theorem in probability (Vapnik-Chervonenkis '71) - \item Computational learning theory (PAC learning) - \item Computational geometry - \item Functional analysis (Bourgain-Fremlin-Talagrand theory) - \item Abstract topological dynamics (tame dynamical systems) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{History} - \begin{itemize} - \item Vapnik-Chervonenkis '71 - introduced VC-dimension - \item NIP theories (Shelah '71, '90) - \item vc-density (Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Model Theory} - Model Theory studies definable sets in first-order structures. - \begin{align*} - \boldsymbol{\Q} = (\Q, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) = \exists y \ y \cdot y = x - \end{align*} - In the structure above $\phi(x)$ defines the set of rationals that are a square. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) = \exists y \ y \cdot y = x - \end{align*} - In the structure above $\phi(x)$ defines the set $[0, \infty)$. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \psi(x_1, x_2) = (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1^2 \leq x_2) - \end{align*} - This defines a set in $\R^2$. -\end{frame} - -\begin{frame} - We work with families of uniformly definable sets. - Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$. - Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^n$. - \begin{align*} - \F^M_\phi = \curly{\phi(M^n, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} - \end{align*} - Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$ \\ - Open Question: it is possible for $\vc^M(\phi)$ to be irrational? -\end{frame} - -\begin{frame} - \begin{align*} - \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 - \end{align*} - In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. - Thus $\F^\R_\phi$ is a collection of all disks in $\R^2$. -\end{frame} - -\begin{frame} - Shelah ('90) classified number of isomorphic classes for non-standard models. - Important groups of structures included: stable, NIP, simple. - A model $M$ is said to be NIP if all uniformly definable families in it have finite $\vc$-density. - \begin{itemize} - \item $(\C, 0, 1, +, \cdot)$ is stable (so both NIP and simple) - \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP and not stable - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP and not stable - \item Random graph $(V, R)$ is simple and not stable. - \item Pseudo-finite fields are simple and not stable. - \item $(\Q, 0, 1, +, \cdot)$ is in neither of those categories. - \end{itemize} -\end{frame} - -\begin{frame} - Given an NIP structure $M$ we define a vc-function of $n$ to be the largest $\vc$-density achieved by families of uniformly definable sets in $M^n$. - \begin{align*} - \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} - \end{align*} - Easy to show $\vc^M(n) \geq n \vc^M(1)$, $\vc^M(n) \geq n$ \\ - Open question: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? -\end{frame} - -\begin{frame} - Examples - \begin{itemize} - \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ (true for all quasi o-minimal structures) - \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ - \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ - \item ACVF has $\vc(n) \leq 2n$. - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{vc-density in trees} - Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. - \begin{itemize} - \item Trees are NIP (Parigot '82) - \item Trees are dp-minimal (Simon '11) - \item Trees have $vc(n) = n$ (B. '13) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{proof background} - $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\ - Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have - \begin{align*} - \tp(a), \tp(b) \vdash \tp(ab) - \end{align*} - This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. -\end{frame} - -\begin{frame} - \frametitle{Shelah-Spencer graphs} - Consider a random graph on $n$ vertices where the probability of the given two vertices having an edge is $n^{-\alpha}$. - Shelah-Spencer graph is a limit of such graphs for $\alpha$ irrational in $(0,1)$. - We view it in a language with a single binary relation. - \begin{itemize} - \item Shelah-Spencer graphs can be axiomatized (Shelah-Spencer '88) - \item Shelah-Spencer graphs are stable (Baldwin-Shi '96, Baldwin-Shelah '97 ) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Background} -\begin{Definition} - \begin{itemize} - \item To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. - \item $B/A$ is called a positive extension if quantity $\delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A|$ is positive. - \item $B/A$ is called minimal if its dimension is negative, but every subextension is positive. - \item $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. - \end{itemize} -\end{Definition} - For $B/A$ chain-minimal define - \begin{align*} - \phi_{A,B}(\vec x) = \exists \vec y \text { such that $\vec y/\vec x$ is isomorphic to $B/A$} - \end{align*} - \begin{Theorem} [quantifier elimination] - In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x, \vec y)}$. - \end{Theorem} -\end{frame} - -\begin{frame} - \frametitle{vc-density in Shelah-Spencer graphs} - - \begin{itemize} - \item $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. - \item For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ (with a simple dependence on $\delta(B_i/A_i)$) such that - \begin{align*} - \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| - \end{align*} - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Future work} - \begin{itemize} - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ - \item Other partial orderings, lattices - \item Other graph structures, in particular flat graphs - \end{itemize} -\end{frame} - -\end{document} - - -\begin{frame} - To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. - Any -\end{frame} - +\documentclass{beamer} + +\usepackage{../Research} + +\newcommand{\F}{\mathcal F} +\newcommand{\curly}[1]{\left\{ #1 \right\}} + +\title{VC-density in model theoretic structures} +\author{Anton Bobkov} +\date{June 3, 2015} + + +\begin{document} + +\maketitle + +\begin{frame} + Suppose we have an (infinite) collection of sets $\F$. \\ + We define the shatter function $\pi_\F \colon \N \arr \N$ of $\F$ + + \begin{align*} + \pi_\F(n) = \max \{ &\text {\# of atoms in boolean algebra generated by $\mathcal S$} \\ + &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} + \end{align*} +\end{frame} + +\begin{frame} + Example: Let $\F$ consist of all discs in the plane. + \begin{figure}[p] + \centering + \includegraphics[scale=0.75]{circle.png} + \end{figure} + \begin{align*} + \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 + \end{align*} + \begin{align*} + \pi_\F(n) = n^2 - n + 2 + \end{align*} +\end{frame} + +\begin{frame} +More examples: \\ + \begin{enumerate} + \item Lines in the plane $\pi_\F(n) = n^2/2 + n/2 + 1$ + \item Disks in the plane $\pi_\F(n) = n^2 - n + 2$ + \item Balls in $\R^3$ $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ + \item Intervals in the line $\pi_\F(n) = 2n$ + \item Half-planes in the plane $\pi_\F(n) = n(n+1)/2 + 1$ + \item Finite subsets of $\N$ $\pi_\F(n) = 2^n$ + \item Convex polygons in the plane $\pi_\F(n) = 2^n$ + \end{enumerate} +\end{frame} + +\begin{frame} + \begin{Theorem} [Sauer-Shelah '72] + The shatter function is either $2^n$ or bounded by a polynomial. + \end{Theorem} + \begin{Definition} + Suppose the growth of the shatter function of $\F$ is polynomial. + Let $\vc(\F)$ infimum of all positive reals $r$ such that + \begin{align*} + \pi_\F(n) = O(n^r) + \end{align*} + Call $\vc(\F)$ the vc-density of $\F$. + If the shatter function grows exponentially, we let $\vc(\F) := \infty$. + \end{Definition} +\end{frame} + +\begin{frame} + \frametitle{Applications} + \begin{itemize} + \item NIP theories + \item VC-Theorem in probability (Vapnik-Chervonenkis '71) + \item Computational learning theory (PAC learning) + \item Computational geometry + \item Functional analysis (Bourgain-Fremlin-Talagrand theory) + \item Abstract topological dynamics (tame dynamical systems) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{History} + \begin{itemize} + \item Vapnik-Chervonenkis '71 - introduced VC-dimension + \item NIP theories (Shelah '71, '90) + \item vc-density (Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Model Theory} + Model Theory studies definable sets in first-order structures. + \begin{align*} + \boldsymbol{\Q} = (\Q, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) = \exists y \ y \cdot y = x + \end{align*} + In the structure above $\phi(x)$ defines the set of rationals that are a square. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) = \exists y \ y \cdot y = x + \end{align*} + In the structure above $\phi(x)$ defines the set $[0, \infty)$. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \psi(x_1, x_2) = (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1^2 \leq x_2) + \end{align*} + This defines a set in $\R^2$. +\end{frame} + +\begin{frame} + We work with families of uniformly definable sets. + Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$. + Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^n$. + \begin{align*} + \F^M_\phi = \curly{\phi(M^n, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} + \end{align*} + Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$ \\ + Open Question: it is possible for $\vc^M(\phi)$ to be irrational? +\end{frame} + +\begin{frame} + \begin{align*} + \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 + \end{align*} + In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. + Thus $\F^\R_\phi$ is a collection of all disks in $\R^2$. +\end{frame} + +\begin{frame} + Shelah ('90) classified number of isomorphic classes for non-standard models. + Important groups of structures included: stable, NIP, simple. + A model $M$ is said to be NIP if all uniformly definable families in it have finite $\vc$-density. + \begin{itemize} + \item $(\C, 0, 1, +, \cdot)$ is stable (so both NIP and simple) + \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP and not stable + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP and not stable + \item Random graph $(V, R)$ is simple and not stable. + \item Pseudo-finite fields are simple and not stable. + \item $(\Q, 0, 1, +, \cdot)$ is in neither of those categories. + \end{itemize} +\end{frame} + +\begin{frame} + Given an NIP structure $M$ we define a vc-function of $n$ to be the largest $\vc$-density achieved by families of uniformly definable sets in $M^n$. + \begin{align*} + \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} + \end{align*} + Easy to show $\vc^M(n) \geq n \vc^M(1)$, $\vc^M(n) \geq n$ \\ + Open question: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? +\end{frame} + +\begin{frame} + Examples + \begin{itemize} + \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ (true for all quasi o-minimal structures) + \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ + \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ + \item ACVF has $\vc(n) \leq 2n$. + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{vc-density in trees} + Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. + \begin{itemize} + \item Trees are NIP (Parigot '82) + \item Trees are dp-minimal (Simon '11) + \item Trees have $vc(n) = n$ (B. '13) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{proof background} + $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\ + Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have + \begin{align*} + \tp(a), \tp(b) \vdash \tp(ab) + \end{align*} + This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. +\end{frame} + +\begin{frame} + \frametitle{Shelah-Spencer graphs} + Consider a random graph on $n$ vertices where the probability of the given two vertices having an edge is $n^{-\alpha}$. + Shelah-Spencer graph is a limit of such graphs for $\alpha$ irrational in $(0,1)$. + We view it in a language with a single binary relation. + \begin{itemize} + \item Shelah-Spencer graphs can be axiomatized (Shelah-Spencer '88) + \item Shelah-Spencer graphs are stable (Baldwin-Shi '96, Baldwin-Shelah '97 ) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Background} +\begin{Definition} + \begin{itemize} + \item To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. + \item $B/A$ is called a positive extension if quantity $\delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A|$ is positive. + \item $B/A$ is called minimal if its dimension is negative, but every subextension is positive. + \item $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. + \end{itemize} +\end{Definition} + For $B/A$ chain-minimal define + \begin{align*} + \phi_{A,B}(\vec x) = \exists \vec y \text { such that $\vec y/\vec x$ is isomorphic to $B/A$} + \end{align*} + \begin{Theorem} [quantifier elimination] + In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x, \vec y)}$. + \end{Theorem} +\end{frame} + +\begin{frame} + \frametitle{vc-density in Shelah-Spencer graphs} + + \begin{itemize} + \item $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. + \item For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ (with a simple dependence on $\delta(B_i/A_i)$) such that + \begin{align*} + \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| + \end{align*} + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Future work} + \begin{itemize} + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ + \item Other partial orderings, lattices + \item Other graph structures, in particular flat graphs + \end{itemize} +\end{frame} + +\end{document} + + +\begin{frame} + To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. + Any +\end{frame} + diff --git a/research/09 ATC stuff/CRE5D90.tmp b/research/09 ATC stuff/CRE5D90.tmp index 8a5cd543..4e11a756 100644 --- a/research/09 ATC stuff/CRE5D90.tmp +++ b/research/09 ATC stuff/CRE5D90.tmp @@ -1,252 +1,252 @@ -\documentclass{beamer} - -\usepackage{../Research} - -\newcommand{\F}{\mathcal F} -\newcommand{\curly}[1]{\left\{ #1 \right\}} - -\title{VC-density in model theoretic structures} -\author{Anton Bobkov} -\date{June 3, 2015} - - -\begin{document} - -\maketitle - -\begin{frame} - Suppose we have an (infinite) collection of sets $\F$. \\ - We define the shatter function $\pi_\F \colon \N \arr \N$ of $\F$ - - \begin{align*} - \pi_\F(n) = \max \{ &\text {\# of atoms in boolean algebra generated by $\mathcal S$} \\ - &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} - \end{align*} -\end{frame} - -\begin{frame} - Example: Let $\F$ consist of all discs in the plane. - \begin{figure}[p] - \centering - \includegraphics[scale=0.75]{circle.png} - \end{figure} - \begin{align*} - \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 - \end{align*} - \begin{align*} - \pi_\F(n) = n^2 - n + 2 - \end{align*} -\end{frame} - -\begin{frame} -More examples: \\ - \begin{enumerate} - \item Lines in the plane $\pi_\F(n) = n^2/2 + n/2 + 1$ - \item Disks in the plane $\pi_\F(n) = n^2 - n + 2$ - \item Balls in $\R^3$ $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ - \item Intervals in the line $\pi_\F(n) = 2n$ - \item Half-planes in the plane $\pi_\F(n) = n(n+1)/2 + 1$ - \item Finite subsets of $\N$ $\pi_\F(n) = 2^n$ - \item Convex polygons in the plane $\pi_\F(n) = 2^n$ - \end{enumerate} -\end{frame} - -\begin{frame} - \begin{Theorem} [Sauer-Shelah '72] - The shatter function is either $2^n$ or bounded by a polynomial. - \end{Theorem} - \begin{Definition} - Suppose the growth of the shatter function of $\F$ is polynomial. - Let $\vc(\F)$ infimum of all positive reals $r$ such that - \begin{align*} - \pi_\F(n) = O(n^r) - \end{align*} - Call $\vc(\F)$ the vc-density of $\F$. - If the shatter function grows exponentially, we let $\vc(\F) := \infty$. - \end{Definition} -\end{frame} - -\begin{frame} - \frametitle{Applications} - \begin{itemize} - \item NIP theories - \item VC-Theorem in probability (Vapnik-Chervonenkis '71) - \item Computational learning theory (PAC learning) - \item Computational geometry - \item Functional analysis (Bourgain-Fremlin-Talagrand theory) - \item Abstract topological dynamics (tame dynamical systems) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{History} - \begin{itemize} - \item Vapnik-Chervonenkis '71 - introduced VC-dimension - \item NIP theories (Shelah '71, '90) - \item vc-density (Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Model Theory} - Model Theory studies definable sets in first-order structures. - \begin{align*} - \boldsymbol{\Q} = (\Q, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) = \exists y \ y \cdot y = x - \end{align*} - In the structure above $\phi(x)$ defines the set of rationals that are a square. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \phi(x) = \exists y \ y \cdot y = x - \end{align*} - In the structure above $\phi(x)$ defines the set $[0, \infty)$. -\end{frame} - -\begin{frame} - \begin{align*} - (\R, 0, 1, +, \cdot, \leq) - \end{align*} - \begin{align*} - \psi(x_1, x_2) = (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1^2 \leq x_2) - \end{align*} - This defines a set in $\R^2$. -\end{frame} - -\begin{frame} - We work with families of uniformly definable sets. - Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$. - Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^n$. - \begin{align*} - \F^M_\phi = \curly{\phi(M^n, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} - \end{align*} - Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$ \\ - Open Question: it is possible for $\vc^M(\phi)$ to be irrational? -\end{frame} - -\begin{frame} - \begin{align*} - \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 - \end{align*} - In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. - Thus $\F^\R_\phi$ is a collection of all disks in $\R^2$. -\end{frame} - -\begin{frame} - Shelah ('90) classified number of isomorphic classes for non-standard models. - Important groups of structures included: stable, NIP, simple. - A model $M$ is said to be NIP if all uniformly definable families in it have finite $\vc$-density. - \begin{itemize} - \item $(\C, 0, 1, +, \cdot)$ is stable (so both NIP and simple) - \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP and not stable - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP and not stable - \item Random graph $(V, R)$ is simple and not stable. - \item Pseudo-finite fields are simple and not stable. - \item $(\Q, 0, 1, +, \cdot)$ is in neither of those categories. - \end{itemize} -\end{frame} - -\begin{frame} - Given an NIP structure $M$ we define a vc-function of $n$ to be the largest $\vc$-density achieved by families of uniformly definable sets in $M^n$. - \begin{align*} - \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} - \end{align*} - Easy to show $\vc^M(n) \geq n \vc^M(1)$, $\vc^M(n) \geq n$ \\ - Open question: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? -\end{frame} - -\begin{frame} - Examples - \begin{itemize} - \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ (true for all quasi o-minimal structures) - \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ - \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ - \item ACVF has $\vc(n) \leq 2n$. - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{vc-density in trees} - Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. - \begin{itemize} - \item Trees are NIP (Parigot '82) - \item Trees are dp-minimal (Simon '11) - \item Trees have $vc(n) = n$ (B. '13) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{proof background} - $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\ - Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have - \begin{align*} - \tp(a), \tp(b) \vdash \tp(ab) - \end{align*} - This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. -\end{frame} - -\begin{frame} - \frametitle{Shelah-Spencer graphs} - Consider a random graph on $n$ vertices where the probability of the given two vertices having an edge is $n^{-\alpha}$. - Shelah-Spencer graph is a limit of such graphs for $\alpha$ irrational in $(0,1)$. - We view it in a language with a single binary relation. - \begin{itemize} - \item Shelah-Spencer graphs can be axiomatized (Shelah-Spencer '88) - \item Shelah-Spencer graphs are stable (Baldwin-Shi '96, Baldwin-Shelah '97 ) - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Background} -\begin{Definition} - \begin{itemize} - \item To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. - \item $B/A$ is called a positive extension if quantity $\delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A|$ is positive. - \item $B/A$ is called minimal if its dimension is negative, but every subextension is positive. - \item $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. - \end{itemize} -\end{Definition} - For $B/A$ chain-minimal define - \begin{align*} - \phi_{A,B}(\vec x) = \exists \vec y \text { such that $\vec y/\vec x$ is isomorphic to $B/A$} - \end{align*} - \begin{Theorem} [quantifier elimination] - In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x, \vec y)}$. - \end{Theorem} -\end{frame} - -\begin{frame} - \frametitle{vc-density in Shelah-Spencer graphs} - - \begin{itemize} - \item $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. - \item For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ (with a simple dependence on $\delta(B_i/A_i)$) such that - \begin{align*} - \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| - \end{align*} - \end{itemize} -\end{frame} - -\begin{frame} - \frametitle{Future work} - \begin{itemize} - \item $(\Q_p, 0, 1, +, \cdot, \mid)$ - \item Other partial orderings, lattices - \item Other graph structures, in particular flat graphs - \end{itemize} -\end{frame} - -\end{document} - - -\begin{frame} - To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. - Any -\end{frame} - +\documentclass{beamer} + +\usepackage{../Research} + +\newcommand{\F}{\mathcal F} +\newcommand{\curly}[1]{\left\{ #1 \right\}} + +\title{VC-density in model theoretic structures} +\author{Anton Bobkov} +\date{June 3, 2015} + + +\begin{document} + +\maketitle + +\begin{frame} + Suppose we have an (infinite) collection of sets $\F$. \\ + We define the shatter function $\pi_\F \colon \N \arr \N$ of $\F$ + + \begin{align*} + \pi_\F(n) = \max \{ &\text {\# of atoms in boolean algebra generated by $\mathcal S$} \\ + &\mid \mathcal S \subset \F \text{ with } |\mathcal S| = n\} + \end{align*} +\end{frame} + +\begin{frame} + Example: Let $\F$ consist of all discs in the plane. + \begin{figure}[p] + \centering + \includegraphics[scale=0.75]{circle.png} + \end{figure} + \begin{align*} + \pi_\F(1) = 2 \ \ \ \pi_\F(2) = 4 \ \ \ \pi_\F(3) = 8 \ \ \ \pi_\F(4) = 14 + \end{align*} + \begin{align*} + \pi_\F(n) = n^2 - n + 2 + \end{align*} +\end{frame} + +\begin{frame} +More examples: \\ + \begin{enumerate} + \item Lines in the plane $\pi_\F(n) = n^2/2 + n/2 + 1$ + \item Disks in the plane $\pi_\F(n) = n^2 - n + 2$ + \item Balls in $\R^3$ $\pi_\F(n) = n^3/3 - n^2 + 8n/3$ + \item Intervals in the line $\pi_\F(n) = 2n$ + \item Half-planes in the plane $\pi_\F(n) = n(n+1)/2 + 1$ + \item Finite subsets of $\N$ $\pi_\F(n) = 2^n$ + \item Convex polygons in the plane $\pi_\F(n) = 2^n$ + \end{enumerate} +\end{frame} + +\begin{frame} + \begin{Theorem} [Sauer-Shelah '72] + The shatter function is either $2^n$ or bounded by a polynomial. + \end{Theorem} + \begin{Definition} + Suppose the growth of the shatter function of $\F$ is polynomial. + Let $\vc(\F)$ infimum of all positive reals $r$ such that + \begin{align*} + \pi_\F(n) = O(n^r) + \end{align*} + Call $\vc(\F)$ the vc-density of $\F$. + If the shatter function grows exponentially, we let $\vc(\F) := \infty$. + \end{Definition} +\end{frame} + +\begin{frame} + \frametitle{Applications} + \begin{itemize} + \item NIP theories + \item VC-Theorem in probability (Vapnik-Chervonenkis '71) + \item Computational learning theory (PAC learning) + \item Computational geometry + \item Functional analysis (Bourgain-Fremlin-Talagrand theory) + \item Abstract topological dynamics (tame dynamical systems) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{History} + \begin{itemize} + \item Vapnik-Chervonenkis '71 - introduced VC-dimension + \item NIP theories (Shelah '71, '90) + \item vc-density (Aschenbrenner, Dolich, Haskell, Macpherson, Starchenko '13) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Model Theory} + Model Theory studies definable sets in first-order structures. + \begin{align*} + \boldsymbol{\Q} = (\Q, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) = \exists y \ y \cdot y = x + \end{align*} + In the structure above $\phi(x)$ defines the set of rationals that are a square. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \phi(x) = \exists y \ y \cdot y = x + \end{align*} + In the structure above $\phi(x)$ defines the set $[0, \infty)$. +\end{frame} + +\begin{frame} + \begin{align*} + (\R, 0, 1, +, \cdot, \leq) + \end{align*} + \begin{align*} + \psi(x_1, x_2) = (x_1 \cdot x_1 + x_2 \cdot x_2 \leq 1.5) \wedge (x_1^2 \leq x_2) + \end{align*} + This defines a set in $\R^2$. +\end{frame} + +\begin{frame} + We work with families of uniformly definable sets. + Fix a formula $\phi(x_1 \ldots x_m, y_1, \ldots y_n) = \phi(\vec x, \vec y)$. + Plug in elements from $M$ for $y$ variables to get a family of definable sets in $M^n$. + \begin{align*} + \F^M_\phi = \curly{\phi(M^n, a_1, \ldots a_n) \mid a_1, \ldots a_n \in M} + \end{align*} + Define $\vc^M(\phi)$ to be the vc-density of the family $\F^M_\phi$ \\ + Open Question: it is possible for $\vc^M(\phi)$ to be irrational? +\end{frame} + +\begin{frame} + \begin{align*} + \phi(x_1, x_2, y_1, y_2, y_3) = (x_1 - y_1)^2 + (x_2 - y_2)^2 \leq y_3^2 + \end{align*} + In structure $(\R, +, \cdot, \leq)$ given $a,b,r \in \R$ the formula $\phi(x_1, x_2, a, b, r)$ defines a disk in $\R^2$ with radius $r$ with center $(a,b)$. + Thus $\F^\R_\phi$ is a collection of all disks in $\R^2$. +\end{frame} + +\begin{frame} + Shelah ('90) classified number of isomorphic classes for non-standard models. + Important groups of structures included: stable, NIP, simple. + A model $M$ is said to be NIP if all uniformly definable families in it have finite $\vc$-density. + \begin{itemize} + \item $(\C, 0, 1, +, \cdot)$ is stable (so both NIP and simple) + \item $(\R, 0, 1, +, \cdot, \leq)$ is NIP and not stable + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ is NIP and not stable + \item Random graph $(V, R)$ is simple and not stable. + \item Pseudo-finite fields are simple and not stable. + \item $(\Q, 0, 1, +, \cdot)$ is in neither of those categories. + \end{itemize} +\end{frame} + +\begin{frame} + Given an NIP structure $M$ we define a vc-function of $n$ to be the largest $\vc$-density achieved by families of uniformly definable sets in $M^n$. + \begin{align*} + \vc^M(n) = max \curly{ \vc(\phi) \mid \phi(\vec x, \vec y) \text{ with } |\vec x| = n} + \end{align*} + Easy to show $\vc^M(n) \geq n \vc^M(1)$, $\vc^M(n) \geq n$ \\ + Open question: Is $\vc^M(n) = n \vc^M(1)$? If not, is there a linear relationship? +\end{frame} + +\begin{frame} + Examples + \begin{itemize} + \item $(\R, 0, 1, +, \cdot, \leq)$ has $\vc(n) = n$ (true for all quasi o-minimal structures) + \item $(\C, 0, 1, +, \cdot)$ has $\vc(n) = n$ + \item $(\Q_p, 0, 1, +, \cdot)$ has $\vc(n) \leq 2n - 1$ + \item ACVF has $\vc(n) \leq 2n$. + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{vc-density in trees} + Consider structure $(T, \leq)$ where elements of $T$ are vertices of a rooted tree and we say that $a \leq b$ if $a$ is below $b$ in the tree. + \begin{itemize} + \item Trees are NIP (Parigot '82) + \item Trees are dp-minimal (Simon '11) + \item Trees have $vc(n) = n$ (B. '13) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{proof background} + $\tp(a)$, a type of an element $a$ is a set of all the formulas that that are true about $a$.\ + Parigot's observation: there is a natural way to split a tree into parts $A, B$ such that for $a \in A$ and $b \in B$ we have + \begin{align*} + \tp(a), \tp(b) \vdash \tp(ab) + \end{align*} + This allows us to decompose complex types into simple parts, which we can use to compute $\vc$-density. +\end{frame} + +\begin{frame} + \frametitle{Shelah-Spencer graphs} + Consider a random graph on $n$ vertices where the probability of the given two vertices having an edge is $n^{-\alpha}$. + Shelah-Spencer graph is a limit of such graphs for $\alpha$ irrational in $(0,1)$. + We view it in a language with a single binary relation. + \begin{itemize} + \item Shelah-Spencer graphs can be axiomatized (Shelah-Spencer '88) + \item Shelah-Spencer graphs are stable (Baldwin-Shi '96, Baldwin-Shelah '97 ) + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Background} +\begin{Definition} + \begin{itemize} + \item To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. + \item $B/A$ is called a positive extension if quantity $\delta(B/A) = |V_B/V_A| - \alpha |E_B/E_A|$ is positive. + \item $B/A$ is called minimal if its dimension is negative, but every subextension is positive. + \item $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. + \end{itemize} +\end{Definition} + For $B/A$ chain-minimal define + \begin{align*} + \phi_{A,B}(\vec x) = \exists \vec y \text { such that $\vec y/\vec x$ is isomorphic to $B/A$} + \end{align*} + \begin{Theorem} [quantifier elimination] + In Shelah-Spencer graph every definable set can be defined by a boolean combination of formulas $\phi_{A_i, B_i(\vec x, \vec y)}$. + \end{Theorem} +\end{frame} + +\begin{frame} + \frametitle{vc-density in Shelah-Spencer graphs} + + \begin{itemize} + \item $\vc(1) = \infty$, so vc-function is not well-behaved for this structure. + \item For a formula $\phi(\vec x, \vec y)$ we can define $\epsilon_L, \epsilon_U$ (with a simple dependence on $\delta(B_i/A_i)$) such that + \begin{align*} + \epsilon_L |\vec x| \leq \vc(\phi) \leq \epsilon_U |\vec x| + \end{align*} + \end{itemize} +\end{frame} + +\begin{frame} + \frametitle{Future work} + \begin{itemize} + \item $(\Q_p, 0, 1, +, \cdot, \mid)$ + \item Other partial orderings, lattices + \item Other graph structures, in particular flat graphs + \end{itemize} +\end{frame} + +\end{document} + + +\begin{frame} + To a finite graph $A$ assign a dimension $\delta(A) = |V| - \alpha |E|$. $B/A$ is an extension. $\delta(B/A)$ is $\delta(A) = |V_B/V_A| - \alpha |E_B/E_A|$. $B/A$ is called minimal if its dimension is negative, but every subextension is positive. $(A_0, \ldots A_n)$ is a minimal chain if each $A_{i + 1}/A_i$ is minimal. + Any +\end{frame} + diff --git a/research/AMC_style.sty b/research/AMC_style.sty index 86a48d3b..d5c83969 100644 --- a/research/AMC_style.sty +++ b/research/AMC_style.sty @@ -1,81 +1,81 @@ -%% -%% Some mathematical symbols are not included in the basic LaTeX -%% package. 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The +%%% numbering is still tied to the theorem counter. +%%% + + +\theoremstyle{remark} + +\newtheorem{Remark}[Theorem]{Remark} + + +%%% +%%% The following, if uncommented, numbers equations within sections. +%%% + \numberwithin{equation}{section} \ No newline at end of file diff --git a/research/Research.sty b/research/Research.sty index fc18aee7..ec33346e 100644 --- a/research/Research.sty +++ b/research/Research.sty @@ -1,96 +1,96 @@ -\newcommand{\Contradiction}{\ensuremath{\downlsquigarrow}} -%\newcommand{\Contradiction}{\ensuremath{\Rightarrow\Leftarrow}} -%\newcommand{\Contradiction}{\ensuremath{\bot}} -%\newcommand{\Contradiction}{\ensuremath{\nleftrightarrow}} -%\newcommand{\Contradiction}{\textreferencemark} -%\newcommand{\Contradiction}{\ensuremath{\rightleftarrows}} -%\newcommand{\Contradiction}{\blitza} - -\newcommand{\N}{\ensuremath{\mathbb{N}}} -\newcommand{\Z}{\ensuremath{\mathbb{Z}}} -\newcommand{\Q}{\ensuremath{\mathbb{Q}}} -\newcommand{\R}{\ensuremath{\mathbb{R}}} -\newcommand{\C}{\ensuremath{\mathbb{C}}} - -\newcommand{\imply}{\ensuremath{\ \Rightarrow \ }} -\newcommand{\eqdef}{\stackrel{\text{\tiny def}}{=}} -\newcommand{\invert}[1]{{#1}^{\text{\tiny -1}}} -\newcommand{\inv}[1]{{#1}^{\text{\tiny -1}}} -\newcommand{\ifff}{\ensuremath{\Leftrightarrow}} - -\newcommand{\la}{\langle} -\newcommand{\ra}{\rangle} - -\newcommand{\midr}{\upharpoonright} - -\newcommand{\boldface}{\boldsymbol} - - -\DeclareMathOperator{\arr}{\longrightarrow} -\DeclareMathOperator{\tr}{tr} -\DeclareMathOperator{\Span}{span} -\DeclareMathOperator{\im}{im} - - -\newcommand{\Mat}[1]{\mbox{Mat}_{#1 \times #1}} -\newcommand{\Gl}[2]{\mbox{GL}_{#1}(#2)} -\newcommand{\GL}{\Gl} - -\newcommand{\chs}[2] -{ - \left ( - \begin{array}{c} - #1 \\ - #2 - \end{array} - \right) -} - -\newcommand{\img}[1] -{ - \begin{center} - \includegraphics[scale=1.75]{#1} - \end{center} -} - -\newcommand{\twopartdef}[4] -{ - \left\{ - \begin{array}{ll} - #1 & \mbox{if } #2 \\ - #3 & \mbox{if } #4 - \end{array} - \right. -} - -\newcommand{\twopartdefotw}[3] -{ - \left\{ - \begin{array}{ll} - #1 & \mbox{if } #2 \\ - #3 & \mbox{otherwise} - \end{array} - \right. -} - -\newcommand{\threepartdef}[6] -{ - \left\{ - \begin{array}{lll} - #1 & \mbox{if } #2 \\ - #3 & \mbox{if } #4 \\ - #5 & \mbox{if } #6 - \end{array} - \right. -} - - -\newcommand{\LL}{\mathcal L} -\newcommand{\PP}{\mathcal P} -\DeclareMathOperator{\tp}{tp} -\DeclareMathOperator{\vc}{vc} -\DeclareMathOperator{\VC}{VC} -\DeclareMathOperator{\pp}{\boldface p} - - - +\newcommand{\Contradiction}{\ensuremath{\downlsquigarrow}} +%\newcommand{\Contradiction}{\ensuremath{\Rightarrow\Leftarrow}} +%\newcommand{\Contradiction}{\ensuremath{\bot}} +%\newcommand{\Contradiction}{\ensuremath{\nleftrightarrow}} +%\newcommand{\Contradiction}{\textreferencemark} +%\newcommand{\Contradiction}{\ensuremath{\rightleftarrows}} +%\newcommand{\Contradiction}{\blitza} + +\newcommand{\N}{\ensuremath{\mathbb{N}}} +\newcommand{\Z}{\ensuremath{\mathbb{Z}}} +\newcommand{\Q}{\ensuremath{\mathbb{Q}}} +\newcommand{\R}{\ensuremath{\mathbb{R}}} +\newcommand{\C}{\ensuremath{\mathbb{C}}} + +\newcommand{\imply}{\ensuremath{\ \Rightarrow \ }} +\newcommand{\eqdef}{\stackrel{\text{\tiny def}}{=}} +\newcommand{\invert}[1]{{#1}^{\text{\tiny -1}}} +\newcommand{\inv}[1]{{#1}^{\text{\tiny -1}}} +\newcommand{\ifff}{\ensuremath{\Leftrightarrow}} + +\newcommand{\la}{\langle} +\newcommand{\ra}{\rangle} + +\newcommand{\midr}{\upharpoonright} + +\newcommand{\boldface}{\boldsymbol} + + +\DeclareMathOperator{\arr}{\longrightarrow} +\DeclareMathOperator{\tr}{tr} +\DeclareMathOperator{\Span}{span} +\DeclareMathOperator{\im}{im} + + +\newcommand{\Mat}[1]{\mbox{Mat}_{#1 \times #1}} +\newcommand{\Gl}[2]{\mbox{GL}_{#1}(#2)} +\newcommand{\GL}{\Gl} + +\newcommand{\chs}[2] +{ + \left ( + \begin{array}{c} + #1 \\ + #2 + \end{array} + \right) +} + +\newcommand{\img}[1] +{ + \begin{center} + \includegraphics[scale=1.75]{#1} + \end{center} +} + +\newcommand{\twopartdef}[4] +{ + \left\{ + \begin{array}{ll} + #1 & \mbox{if } #2 \\ + #3 & \mbox{if } #4 + \end{array} + \right. +} + +\newcommand{\twopartdefotw}[3] +{ + \left\{ + \begin{array}{ll} + #1 & \mbox{if } #2 \\ + #3 & \mbox{otherwise} + \end{array} + \right. +} + +\newcommand{\threepartdef}[6] +{ + \left\{ + \begin{array}{lll} + #1 & \mbox{if } #2 \\ + #3 & \mbox{if } #4 \\ + #5 & \mbox{if } #6 + \end{array} + \right. +} + + +\newcommand{\LL}{\mathcal L} +\newcommand{\PP}{\mathcal P} +\DeclareMathOperator{\tp}{tp} +\DeclareMathOperator{\vc}{vc} +\DeclareMathOperator{\VC}{VC} +\DeclareMathOperator{\pp}{\boldface p} + + + diff --git a/research/Thm.sty b/research/Thm.sty index 1a049a98..d2fd6c5d 100644 --- a/research/Thm.sty +++ b/research/Thm.sty @@ -1,21 +1,21 @@ -\usepackage{amsthm} - -\theoremstyle{theorem} %bold title, italicized font - -%\newtheorem{Lemma}[theorem]{Lemma} -%\newtheorem{Corollary}[theorem]{Corollary} -%\newtheorem{Theorem}{Theorem}[section] -%\newtheorem{Proposition}[theorem]{Proposition} -\newtheorem{Claim}[Theorem]{Claim} - -%\theoremstyle{Definition} %bold title, regular font -%\newtheorem*{Problem}{Problem} -%\newtheorem{Definition}[theorem]{Definition} -%\newtheorem{Exercise}[theorem]{Exercise} -%\newtheorem{Question}[theorem]{Question} -%\newtheorem*{Solution}{Solution} -%\newtheorem{Note}{Solution} -%\newtheorem{Remark}{Remark} -%\newtheorem{Fact}{Remark} -%\newtheorem{Example}[theorem]{Example} -%\newtheorem{Warning}[theorem]{Warning} +\usepackage{amsthm} + +\theoremstyle{theorem} %bold title, italicized font + +%\newtheorem{Lemma}[theorem]{Lemma} +%\newtheorem{Corollary}[theorem]{Corollary} +%\newtheorem{Theorem}{Theorem}[section] +%\newtheorem{Proposition}[theorem]{Proposition} +\newtheorem{Claim}[Theorem]{Claim} + +%\theoremstyle{Definition} %bold title, regular font +%\newtheorem*{Problem}{Problem} +%\newtheorem{Definition}[theorem]{Definition} +%\newtheorem{Exercise}[theorem]{Exercise} +%\newtheorem{Question}[theorem]{Question} +%\newtheorem*{Solution}{Solution} +%\newtheorem{Note}{Solution} +%\newtheorem{Remark}{Remark} +%\newtheorem{Fact}{Remark} +%\newtheorem{Example}[theorem]{Example} +%\newtheorem{Warning}[theorem]{Warning} diff --git a/research/junk/sample.tex b/research/junk/sample.tex index d184ac6a..27c6f1fb 100644 --- a/research/junk/sample.tex +++ b/research/junk/sample.tex @@ -1,294 +1,294 @@ - -%% -%% This is LaTeX2e input. -%% - -%% The following tells LaTeX that we are using the -%% style file amsart.cls (That is the AMS article style -%% -\documentclass{amsart} - -%% This has a default type size 10pt. Other options are 11pt and 12pt -%% This are set by replacing the command above by -%% \documentclass[11pt]{amsart} -%% -%% or -%% -%% \documentclass[12pt]{amsart} -%% - -%% -%% Some mathematical symbols are not included in the basic LaTeX -%% package. Uncommenting the following makes more commands -%% available. -%% - -%\usepackage{amssymb} - -%% -%% The following is commands are used for importing various types of -%% grapics. -%% - -%\usepackage{epsfig} % For postscript -%\usepackage{epic,eepic} % For epic and eepic output from xfig - -%% -%% The following is very useful in keeping track of labels while -%% writing. The variant \usepackage[notcite]{showkeys} -%% does not show the labels on the \cite commands. -%% - -%\usepackageshowkeys} - - -%%%% -%%%% The next few commands set up the theorem type environments. -%%%% Here they are set up to be numbered section.number, but this can -%%%% be changed. -%%%% - -\newtheorem{thm}{Theorem}[section] -\newtheorem{prop}[thm]{Proposition} -\newtheorem{lem}[thm]{Lemma} -\newtheorem{cor}[thm]{Corollary} - - -%% -%% If some other type is need, say conjectures, then it is constructed -%% by editing and uncommenting the following. -%% - -%\newtheorem{conj}[thm]{Conjecture} - - -%%% -%%% The following gives definition type environments (which only differ -%%% from theorem type invironmants in the choices of fonts). The -%%% numbering is still tied to the theorem counter. -%%% - -\theoremstyle{definition} -\newtheorem{definition}[thm]{Definition} -\newtheorem{example}[thm]{Example} - -%% -%% Again more of these can be added by uncommenting and editing the -%% following. -%% - -%\newtheorem{note}[thm]{Note} - - -%%% -%%% The following gives remark type environments (which only differ -%%% from theorem type invironmants in the choices of fonts). The -%%% numbering is still tied to the theorem counter. -%%% - - -\theoremstyle{remark} - -\newtheorem{remark}[thm]{Remark} - - -%%% -%%% The following, if uncommented, numbers equations within sections. -%%% - -\numberwithin{equation}{section} - - -%%% -%%% The following show how to make definition (also called macros or -%%% abbreviations). For example to use get a bold face R for use to -%%% name the real numbers the command is \mathbf{R}. To save typing we -%%% can abbreviate as - -\newcommand{\R}{\mathbf{R}} % The real numbers. - -%% -%% The comment after the defintion is not required, but if you are -%% working with someone they will likely thank you for explaining your -%% definition. -%% -%% Now add you own definitions: -%% - -%%% -%%% Mathematical operators (things like sin and cos which are used as -%%% functions and have slightly different spacing when typeset than -%%% variables are defined as follows: -%%% - -\DeclareMathOperator{\dist}{dist} % The distance. - - - -%% -%% This is the end of the preamble. -%% - - -\begin{document} - -%% -%% The title of the paper goes here. Edit to your title. -%% - -\title{Mathematical Theorems} - -%% -%% Now edit the following to give your name and address: -%% - -\author{Ralph Howard} -\address{Department of Mathematics, University of South Carolina, -Columbia, SC 29208} -\email{howard@math.sc.edu} -\urladdr{www.math.sc.edu/$\sim$howard} % Delete if not wanted. - -%% -%% If there is another author uncomment and edit the following. -%% - -%\author{Second Author} -%\address{Department of Mathematics, University of South Carolina, -%Columbia, SC 29208} -%\email{second@math.sc.edu} -%\urladdr{www.math.sc.edu/$\sim$second} - -%% -%% If there are three of more authors they are added in the obvious -%% way. -%% - -%%% -%%% The following is for the abstract. The abstract is optional and -%%% if not used just delete, or comment out, the following. -%%% - -\begin{abstract} -Great stuff. -\end{abstract} - -%% -%% LaTeX will not make the title for the paper unless told to do so. -%% This is done by uncommenting the following. -%% - -\maketitle - -%% -%% LaTeX can automatically make a table of contents. This is done by -%% uncommenting the following: -%% - -%\tableofcontents - -%% -%% To enter text is easy. Just type it. A blank line starts a new -%% paragraph. -%% - - -Call me Ishmael. Some years ago --- never mind how long precisely --- -having little or no money in my purse, and nothing particular to -interest me on shore, I thought I would sail about a little and see -the watery part of the world. It is a way I have of driving off the -spleen, and regulating the circulation. Whenever I find myself -growing grim about the mouth; whenever it is a damp, drizzly November -in my soul; whenever I find myself involuntarily pausing before coffin -warehouses, and bringing up the rear of every funeral I meet; and -especially whenever my hypos get such an upper hand of me, that it -requires a strong moral principle to prevent me from deliberately -stepping into the street, and methodically knocking people's hats off ---- then, I account it high time to get to sea as soon as I can. This -is my substitute for pistol and ball. With a philosophical flourish -Cato throws himself upon his sword; I quietly take to the ship. There -is nothing surprising in this. If they but knew it, almost all men in -their degree, some time or other, cherish very nearly the same -feelings towards the ocean with me. - - -There now is your insular city of the Manhattoes, belted round by -wharves as Indian isles by coral reefs - commerce surrounds it with -her surf. Right and left, the streets take you waterward. Its extreme -down-town is the battery, where that noble mole is washed by waves, -and cooled by breezes, which a few hours previous were out of sight of -land. Look at the crowds of water-gazers there. - - -%% -%% To put mathematics in a line it is put between dollor signs. That -%% is $(x+y)^2=x^2+2xy+y^2$ -%% - -Anyone caught using formulas such as $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$ -or $\frac{1}{x+y}=\frac{1}{x}+\frac{1}{y}$ will fail. - -%% -%%% Displayed mathematics is put between double dollar signs. -%% - -The binomial theorem is -$$ -(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}. -$$ -A favorite sum of most mathematicians is -$$ -\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}. -$$ -Likewise a popular integral is -$$ -\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi} -$$ - - -%% -%% A Theorem is stated by -%% - -\begin{thm} The square of any real number is non-negative. -\end{thm} - -%% -%% Its proof is set off by -%% - -\begin{proof} -Any real number $x$ satisfies $x>0$, $x=0$, or $x<0$. -If $x=0$, then $x^2=0\ge 0$. If $x>0$ then as a positive time a -positive is positive we have $x^2=xx>0$. If $x<0$ then $-x>0$ and so -by what we have just done $x^2=(-x)^2>0$. So in all cases $x^2\ge0$. -\end{proof} - - -%% -%% A new section is started as follows: -%% - - -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% -\section{Introduction} -%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% - -This is a new section. - -%% -%% A subsection is started by -%% - -\subsection{Things that need to be done.} -Prove theorems. - - - - -\end{document} - - - - - - + +%% +%% This is LaTeX2e input. +%% + +%% The following tells LaTeX that we are using the +%% style file amsart.cls (That is the AMS article style +%% +\documentclass{amsart} + +%% This has a default type size 10pt. Other options are 11pt and 12pt +%% This are set by replacing the command above by +%% \documentclass[11pt]{amsart} +%% +%% or +%% +%% \documentclass[12pt]{amsart} +%% + +%% +%% Some mathematical symbols are not included in the basic LaTeX +%% package. Uncommenting the following makes more commands +%% available. +%% + +%\usepackage{amssymb} + +%% +%% The following is commands are used for importing various types of +%% grapics. +%% + +%\usepackage{epsfig} % For postscript +%\usepackage{epic,eepic} % For epic and eepic output from xfig + +%% +%% The following is very useful in keeping track of labels while +%% writing. The variant \usepackage[notcite]{showkeys} +%% does not show the labels on the \cite commands. +%% + +%\usepackageshowkeys} + + +%%%% +%%%% The next few commands set up the theorem type environments. +%%%% Here they are set up to be numbered section.number, but this can +%%%% be changed. +%%%% + +\newtheorem{thm}{Theorem}[section] +\newtheorem{prop}[thm]{Proposition} +\newtheorem{lem}[thm]{Lemma} +\newtheorem{cor}[thm]{Corollary} + + +%% +%% If some other type is need, say conjectures, then it is constructed +%% by editing and uncommenting the following. +%% + +%\newtheorem{conj}[thm]{Conjecture} + + +%%% +%%% The following gives definition type environments (which only differ +%%% from theorem type invironmants in the choices of fonts). The +%%% numbering is still tied to the theorem counter. +%%% + +\theoremstyle{definition} +\newtheorem{definition}[thm]{Definition} +\newtheorem{example}[thm]{Example} + +%% +%% Again more of these can be added by uncommenting and editing the +%% following. +%% + +%\newtheorem{note}[thm]{Note} + + +%%% +%%% The following gives remark type environments (which only differ +%%% from theorem type invironmants in the choices of fonts). The +%%% numbering is still tied to the theorem counter. +%%% + + +\theoremstyle{remark} + +\newtheorem{remark}[thm]{Remark} + + +%%% +%%% The following, if uncommented, numbers equations within sections. +%%% + +\numberwithin{equation}{section} + + +%%% +%%% The following show how to make definition (also called macros or +%%% abbreviations). For example to use get a bold face R for use to +%%% name the real numbers the command is \mathbf{R}. To save typing we +%%% can abbreviate as + +\newcommand{\R}{\mathbf{R}} % The real numbers. + +%% +%% The comment after the defintion is not required, but if you are +%% working with someone they will likely thank you for explaining your +%% definition. +%% +%% Now add you own definitions: +%% + +%%% +%%% Mathematical operators (things like sin and cos which are used as +%%% functions and have slightly different spacing when typeset than +%%% variables are defined as follows: +%%% + +\DeclareMathOperator{\dist}{dist} % The distance. + + + +%% +%% This is the end of the preamble. +%% + + +\begin{document} + +%% +%% The title of the paper goes here. Edit to your title. +%% + +\title{Mathematical Theorems} + +%% +%% Now edit the following to give your name and address: +%% + +\author{Ralph Howard} +\address{Department of Mathematics, University of South Carolina, +Columbia, SC 29208} +\email{howard@math.sc.edu} +\urladdr{www.math.sc.edu/$\sim$howard} % Delete if not wanted. + +%% +%% If there is another author uncomment and edit the following. +%% + +%\author{Second Author} +%\address{Department of Mathematics, University of South Carolina, +%Columbia, SC 29208} +%\email{second@math.sc.edu} +%\urladdr{www.math.sc.edu/$\sim$second} + +%% +%% If there are three of more authors they are added in the obvious +%% way. +%% + +%%% +%%% The following is for the abstract. The abstract is optional and +%%% if not used just delete, or comment out, the following. +%%% + +\begin{abstract} +Great stuff. +\end{abstract} + +%% +%% LaTeX will not make the title for the paper unless told to do so. +%% This is done by uncommenting the following. +%% + +\maketitle + +%% +%% LaTeX can automatically make a table of contents. This is done by +%% uncommenting the following: +%% + +%\tableofcontents + +%% +%% To enter text is easy. Just type it. A blank line starts a new +%% paragraph. +%% + + +Call me Ishmael. Some years ago --- never mind how long precisely --- +having little or no money in my purse, and nothing particular to +interest me on shore, I thought I would sail about a little and see +the watery part of the world. It is a way I have of driving off the +spleen, and regulating the circulation. Whenever I find myself +growing grim about the mouth; whenever it is a damp, drizzly November +in my soul; whenever I find myself involuntarily pausing before coffin +warehouses, and bringing up the rear of every funeral I meet; and +especially whenever my hypos get such an upper hand of me, that it +requires a strong moral principle to prevent me from deliberately +stepping into the street, and methodically knocking people's hats off +--- then, I account it high time to get to sea as soon as I can. This +is my substitute for pistol and ball. With a philosophical flourish +Cato throws himself upon his sword; I quietly take to the ship. There +is nothing surprising in this. If they but knew it, almost all men in +their degree, some time or other, cherish very nearly the same +feelings towards the ocean with me. + + +There now is your insular city of the Manhattoes, belted round by +wharves as Indian isles by coral reefs - commerce surrounds it with +her surf. Right and left, the streets take you waterward. Its extreme +down-town is the battery, where that noble mole is washed by waves, +and cooled by breezes, which a few hours previous were out of sight of +land. Look at the crowds of water-gazers there. + + +%% +%% To put mathematics in a line it is put between dollor signs. That +%% is $(x+y)^2=x^2+2xy+y^2$ +%% + +Anyone caught using formulas such as $\sqrt{x+y}=\sqrt{x}+\sqrt{y}$ +or $\frac{1}{x+y}=\frac{1}{x}+\frac{1}{y}$ will fail. + +%% +%%% Displayed mathematics is put between double dollar signs. +%% + +The binomial theorem is +$$ +(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}. +$$ +A favorite sum of most mathematicians is +$$ +\sum_{n=1}^\infty \frac{1}{n^2}=\frac{\pi^2}{6}. +$$ +Likewise a popular integral is +$$ +\int_{-\infty}^\infty e^{-x^2}\,dx=\sqrt{\pi} +$$ + + +%% +%% A Theorem is stated by +%% + +\begin{thm} The square of any real number is non-negative. +\end{thm} + +%% +%% Its proof is set off by +%% + +\begin{proof} +Any real number $x$ satisfies $x>0$, $x=0$, or $x<0$. +If $x=0$, then $x^2=0\ge 0$. If $x>0$ then as a positive time a +positive is positive we have $x^2=xx>0$. If $x<0$ then $-x>0$ and so +by what we have just done $x^2=(-x)^2>0$. So in all cases $x^2\ge0$. +\end{proof} + + +%% +%% A new section is started as follows: +%% + + +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% +\section{Introduction} +%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% + +This is a new section. + +%% +%% A subsection is started by +%% + +\subsection{Things that need to be done.} +Prove theorems. + + + + +\end{document} + + + + + + diff --git a/resume/cv.log b/resume/cv.log deleted file mode 100644 index b18fbeae..00000000 --- a/resume/cv.log +++ /dev/null @@ -1,318 +0,0 @@ -This is pdfTeX, Version 3.1415926-2.4-1.40.13 (MiKTeX 2.9) (preloaded format=pdflatex 2013.10.19) 14 MAR 2015 22:36 -entering extended mode -**cv.tex -(C:\Users\Anton\SparkleShare\Research\resume\cv.tex -LaTeX2e <2011/06/27> -Babel and hyphenation patterns for english, afrikaans, ancientgreek, arabic, armenian, assamese, basque, bengali -, bokmal, bulgarian, catalan, coptic, croatian, czech, danish, dutch, esperanto, estonian, farsi, finnish, french, galic -ian, german, german-x-2012-05-30, greek, gujarati, hindi, hungarian, icelandic, indonesian, interlingua, irish, italian, - 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