lecture 6

inputs/lecture_04.tex

@@ -95,7 +95,7 @@
     It is immediate that $r$ is a retraction.
 \end{refproof}
 
-\subsection{Meager and Comeager Sets}
+\section{Meager and Comeager Sets}
 
 \begin{definition}
     Let $X$ be a topological space, $A \subseteq  X$.

inputs/lecture_05.tex

@@ -121,7 +121,8 @@
 % TODO Fubini
 \begin{notation}
     Let $X ,Y$ be topological spaces,
-    $A \subseteq  X \times Y$,
+    $A \subseteq  X \times Y$
+    and
     $x \in X, y \in Y$.
 
     Let
@@ -151,12 +152,12 @@ but for meager sets:
             and similarly for $y$.
         \item $A$ is meager
             \begin{IEEEeqnarray*}{rll}
-                \iff &\{x \in X : A_x \text{ is meager }\}&\text{ is comeager}\\
+                \iff &\{x \in X : A_x \text{ is meager}\}&\text{ is comeager}\\
-                \iff &\{y \in Y : A^y \text{ is meager }\}& \text{ is comeager}.
+                \iff &\{y \in Y : A^y \text{ is meager}\}& \text{ is comeager}.
             \end{IEEEeqnarray*}
         \item $A$ is comeager
             \begin{IEEEeqnarray*}{rll}
-                \iff & \{x \in X: A_x \text{ is comeager }\} &\text{ is comeager}\\
+                \iff & \{x \in X: A_x \text{ is comeager}\} &\text{ is comeager}\\
                 \iff & \{y \in Y: A^y \text{ is comeager}\} & \text{ is comeager}.
             \end{IEEEeqnarray*}
     \end{enumerate}
@@ -211,7 +212,6 @@ but for meager sets:
         U \cap U_n = \proj_x(W \cap (U \times V_n))
         \]
         nonempty since $W$ is dense.
-        
     \end{refproof}
 
     \begin{claim} % [1a']
@@ -257,9 +257,9 @@ but for meager sets:
         as a countable intersection of comeager sets.
     \end{refproof}
 
-    \todo{Finish proof}
+% \phantom\qedhere
-    \phantom\qedhere
+% \end{refproof}
-\end{refproof}
+% TODO fix claim numbers
 
 \begin{remark}
     Suppose that $A$ has the BP.
@@ -267,5 +267,3 @@ but for meager sets:
     $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
     Then $A = U \symdif M$.
 \end{remark}
-
-

inputs/lecture_06.tex

@@ -0,0 +1,208 @@
+\lecture{06}{2023-11-03}{}
+
+% \begin{refproof}{thm:kuratowskiulam}
+    \begin{enumerate}[(i)]
+        \item Let $A$ be a set with the Baire Property.
+            Write $A = U \symdif M$
+            for $U$ open and $M$ meager.
+            Then for all $x$,
+            we have that $A_x = U_x \symdif M_x$,
+            where $U_x$ is open,
+            and $\{x : M_x \text{ is meager}\}$ is comeager.
+            Therefore $\{x : U_x \text{ open } \land M_x \text{ meager }\}$
+            is comeager,
+            and for those $x$, $A_x$ has the Baire property.
+    \end{enumerate}
+    % TODO: fix counter
+    \begin{claim} % Claim 2
+        \label{thm:kuratowskiulam:c2}
+        For $P \subseteq X$, $Q \subseteq Y$ with the Baire
+        property, let $R \coloneqq P \times Q$.
+        Then $R$ is meager iff at least one of $P$ or $Q$ is meager.
+    \end{claim}
+    \begin{refproof}{thm:kuratowskiulam:c2}
+        Suppose that $R$ is meager.
+        Then by \yaref{thm:kuratowskiulam:c1b},
+        we have that $C = \{x : R_x \text{ is meager }\}$ is comeager.
+        \begin{itemize}
+            \item If $P$ is meager, the statement holds trivially.
+            \item If $P$ is not meager,
+                then  $P \cap C \neq  \emptyset$.
+                For $x \in  P \cap C$
+                we have that $R_x$ is meager
+                and $R_x = Q$,
+                hence $Q$ is meager.
+        \end{itemize}
+
+        On the other hand suppose that $P$ is meager.
+        Then $P = \bigcup_{n} F_n$ for nwd sets $F_n$.
+        Note that $F_n \times Y$ is nwd.
+        So $F_n \times Q$ is also nwd.
+        Hence $P \times Q$ is a countable union
+        of nwd sets,
+        so it is meager.
+    \end{refproof}
+    \begin{enumerate}[(i)]
+        \item[(ii)]
+            ``$\impliedby$''
+            Let $A$ be a set with the Baire property
+            such that $\{x : A_x \text{ is meager}\}$ is comeager.
+            Let $A = U \symdif M$ for $U$ open and $M$ meager.
+            Towards a contradiction suppose that  $A$ is not meager.
+            Then $U$ is not meager.
+            Since $X \times  Y$ is second countable,
+            we have that $A$ is a countable union of open rectangles.
+            At least one of them, say $G \times H \subseteq A$,
+            is not meager.
+            By \yaref{thm:kuratowskiulam:c2},
+            both $G$ and $H$ are not meager.
+            Since
+            $\{x\colon A_x \text{ is meager} \land M_x \text{ is meager}\}$
+            is comeager (using \yaref{thm:kuratowskiulam:c1b}),
+            there is $x_0 \in G$ such that $A_{x_0}$ is meager and $M_{x_0}$
+            is meager.
+            But then $H$ is meager as
+            \[
+                H \setminus M_{x_0} \subseteq  U_{x_0} \setminus M_{x_0}
+                \subseteq  U_{x_0} \symdif M_{x_0} = A_{x_0}
+            \] 
+            and $M_{x_0}$ is meager $\lightning$.
+
+            ``$\implies$''
+            This is \yaref{thm:kuratowskiulam:c1b}.
+    \end{enumerate}
+\end{refproof}
+
+\section{Borel sets} % TODO: fix chapters
+
+\begin{definition}
+    Let $X$ be a topological space.
+    Let $\cB(X)$ denote the smallest $\sigma$-algebra,
+    that contains all open sets.
+    Elements of $\cB(X)$ are called \vocab{Borel sets}.
+\end{definition}
+\begin{remark}
+    Note that all Borel sets have the Baire property.
+\end{remark}
+
+\subsection{The hierarchy of Borel sets}
+
+Let $\omega_1$ be the first uncountable ordinal.
+For every $d < \omega_1$,
+we define by transfinite recursion
+classes $\Sigma^0_\alpha$
+and $\Pi^0_\alpha$
+(or $\Sigma^0_\alpha(X)$ and $\Pi^0_\alpha(X)$ for a topological space $X$).
+
+Let $X$ be a topological space.
+Then define
+\[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\]
+\[
+\Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq 
+\{X \setminus A | A \in \Sigma^0_\alpha(X)\},
+\] 
+% \todo{Define $\lnot$ (element-wise complement)}
+and for $\alpha > 1$
+\[
+\Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}.
+\] 
+
+Note that $\Pi_1^0$ is the set of closed sets,
+$\Sigma^0_2 = F_\sigma$,
+and $\Pi^0_2 = G_\delta$.
+
+Furthermore define
+\[
+\Delta^0_\alpha(X(X)) \coloneqq \Sigma^0_\alpha(X) \cap \Pi^0_\alpha(X),
+\]
+i.e.~$\Delta^0_1$ is the set of clopen sets.
+
+\iffalse % TODO Fix this!
+\resizebox{\textwidth}{!}{
+% https://q.uiver.app/#q=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
+\[\begin{tikzcd}
+	& {\Sigma_1^0} && {\Sigma^0_2} &&&& {\Sigma^0_\xi} \\
+	{\Delta_1^0} && {\Delta^0_2} && {\Delta^0_3} & \ldots & {\Delta^0_\xi} && {\Delta^0_{\xi + 1}} & \ldots \\
+	& {\Pi^0_1} && {\Pi_2^0} &&&& {\Pi^0_\xi}
+	\arrow["\subseteq", hook, from=2-1, to=1-2]
+	\arrow["\subseteq"', hook', from=2-1, to=3-2]
+	\arrow[hook, from=3-2, to=2-3]
+	\arrow[hook, from=1-2, to=2-3]
+	\arrow[hook, from=2-3, to=1-4]
+	\arrow[hook', from=2-3, to=3-4]
+	\arrow[hook, from=2-7, to=1-8]
+	\arrow[hook', from=2-7, to=3-8]
+	\arrow[hook, from=1-4, to=2-5]
+	\arrow[hook', from=3-4, to=2-5]
+	\arrow[hook, from=1-8, to=2-9]
+	\arrow[hook', from=3-8, to=2-9]
+\end{tikzcd}\]%
+}
+\fi
+
+\begin{proposition}
+    Let $X$ be a metrizable space.
+    Then
+    \begin{enumerate}[(a)]
+        \item $\Sigma^0_\eta(X) \cup \Pi^0_\eta(X) \subseteq \Delta^0_\xi(X)$
+            for all $1 \le \eta < \xi < \omega_1$.
+        \item $\cB(X) = \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    \begin{enumerate}[(a)]
+        \item \begin{observe}
+                \label{ob:sigmasuffices}
+                For all $1 \le  \alpha < \beta < \omega_1$,
+                we have $\Pi^0_\alpha(X) \subseteq \Sigma^0_\beta(X)$
+                by taking ``unions'' of singleton sets.
+
+                Furthermore $\Sigma^0_\alpha(X) \subseteq \Pi^0_\beta(X)$
+                by passing to complements.
+            \end{observe}
+            It suffices to show $\Sigma^0_\eta(X) \subseteq \Delta^0_\xi(X)$,
+            since $\Delta^0_\eta(X)$ is closed under complements.
+
+            Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$,
+            by \yaref{ob:sigmasuffices}
+            (since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$
+            and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$).
+
+            So to prove (a) it suffices to show that for all $1 \le  \eta < \xi < \omega_1$,
+            we have $\Sigma^0_\eta(X) \subseteq  \Sigma^0_\xi(X)$.
+            For $\eta = 1, \xi = 2$
+            this holds, since every open set is $F_\sigma$.%
+            \footnote{Here we use that $X$ is metrizable!}
+            % \todo{REF}
+
+            For $\eta > 1, \xi > \eta$,
+            we have
+            \begin{IEEEeqnarray*}{rCl}
+                \Sigma^0_\eta(X) &=&
+                    \{ \bigcup_{n}  A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \eta\}\\
+                    &\subseteq&
+                    \{\bigcup_{n}B_n : B_n \in \Pi^0_{\beta_n}(X), \beta_n < \xi\}
+                    = \Sigma^0_\xi(X).
+            \end{IEEEeqnarray*}
+        \item Let $\cB_0 \coloneqq  \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1}  \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$.
+            We need to show that $\cB_0 = \cB(X)$.
+            Clearly $\cB_0 \subseteq \cB(X)$.
+            It suffices to notice that $\cB_0$ is a $\sigma$-algebra
+            containing all open sets.
+            Consider $\bigcup_{n < \omega} A_n$ for some $A_n \in B_0$.
+            Then $A_n \in \Pi^0_{\alpha_n}(X)$ for some $\alpha_n < \omega_1$.
+            Let $\alpha = \sup \alpha_n < \omega_1$.
+            Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$.
+            It is clear that $\cB_0$ is closed under complements.
+    \end{enumerate}
+
+\end{proof}
+
+\begin{example}
+    % TODO move to counter examples.
+    Consider the cofinite topology on $\omega_1$.
+    Then the non-empty open sets of this are not $F_\sigma$.
+\end{example}
+
+
+

logic.sty

@@ -22,6 +22,7 @@
 \usepackage{listings}
 \usepackage{multirow}
 \usepackage{float}
+\usepackage{quiver}
 %\usepackage{algorithmicx}
 
 \newcounter{subsubsubsection}[subsubsection]

logic3.tex

@@ -29,6 +29,7 @@
 \input{inputs/lecture_03}
 \input{inputs/lecture_04}
 \input{inputs/lecture_05}
+\input{inputs/lecture_06}
 
 
 

quiver.sty

@@ -0,0 +1,40 @@
+% *** quiver ***
+% A package for drawing commutative diagrams exported from https://q.uiver.app.
+%
+% This package is currently a wrapper around the `tikz-cd` package, importing necessary TikZ
+% libraries, and defining a new TikZ style for curves of a fixed height.
+%
+% Version: 1.4.0
+% Authors:
+% - varkor (https://github.com/varkor)
+% - AndréC (https://tex.stackexchange.com/users/138900/andr%C3%A9c)
+
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesPackage{quiver}[2021/01/11 quiver]
+
+% `tikz-cd` is necessary to draw commutative diagrams.
+\RequirePackage{tikz-cd}
+% `amssymb` is necessary for `\lrcorner` and `\ulcorner`.
+\RequirePackage{amssymb}
+% `calc` is necessary to draw curved arrows.
+\usetikzlibrary{calc}
+% `pathmorphing` is necessary to draw squiggly arrows.
+\usetikzlibrary{decorations.pathmorphing}
+
+% A TikZ style for curved arrows of a fixed height, due to AndréC.
+\tikzset{curve/.style={settings={#1},to path={(\tikztostart)
+    .. controls ($(\tikztostart)!\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
+    and ($(\tikztostart)!1-\pv{pos}!(\tikztotarget)!\pv{height}!270:(\tikztotarget)$)
+    .. (\tikztotarget)\tikztonodes}},
+    settings/.code={\tikzset{quiver/.cd,#1}
+        \def\pv##1{\pgfkeysvalueof{/tikz/quiver/##1}}},
+    quiver/.cd,pos/.initial=0.35,height/.initial=0}
+
+% TikZ arrowhead/tail styles.
+\tikzset{tail reversed/.code={\pgfsetarrowsstart{tikzcd to}}}
+\tikzset{2tail/.code={\pgfsetarrowsstart{Implies[reversed]}}}
+\tikzset{2tail reversed/.code={\pgfsetarrowsstart{Implies}}}
+% TikZ arrow styles.
+\tikzset{no body/.style={/tikz/dash pattern=on 0 off 1mm}}
+
+\endinput