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lecture 6
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2
inputs/lecture_04.tex
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@ -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, $A \subseteq X$.

46
inputs/lecture_05.tex
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@ -3,7 +3,7 @@
\begin{fact}
A set $A$ is nwd iff $\overline{A}$ is nwd.
If $F$ is closed then
If $F$ is closed then
$F$ is nwd iff $X \setminus F$ is open and dense.
Any meager set $B$ is contained in
@ -45,7 +45,7 @@
Then
\[
\left( \bigcup_{n < \omega} A_n \right) \symdif \left( \bigcup_{n < \omega} U_n \right)
\]
\]
is meager,\todo{small exercise}
hence $\bigcup_{n < \omega} A_n \in \cA$.
@ -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 &\{y \in Y : A^y \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}.
\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}
@ -171,8 +172,8 @@ but for meager sets:
is nwd,
then
\[
\{x \in X : F_x \text{is nwd}\}
\]
\{x \in X : F_x \text{is nwd}\}
\]
is comeager.
\end{claim}
\begin{refproof}{thm:kuratowskiulam:c1a}
@ -187,12 +188,12 @@ but for meager sets:
We want to show that
\[
\{x \in X: \forall n.~ (W_x \cap V_n) \neq \emptyset\}
\]
\]
is a comeager set.
This is equivalent to
\[
\{x \in X : (W_x \cap V_n) \neq \emptyset\}
\]
\]
being comeager for all $n$,
because the intersection
of countably many comeager sets is comeager.
@ -209,9 +210,8 @@ but for meager sets:
It is
\[
U \cap U_n = \proj_x(W \cap (U \times V_n))
\]
\]
nonempty since $W$ is dense.
\end{refproof}
\begin{claim} % [1a']
@ -220,10 +220,10 @@ but for meager sets:
is nwd,
then
\[
\{x \in X : F_x \text{is nwd}\}
\]
\{x \in X : F_x \text{is nwd}\}
\]
is comeager.
\end{claim}
\begin{refproof}{thm:kuratowskiulam:c1ap}
We have that $\overline{F}$ is nwd.
@ -231,8 +231,8 @@ but for meager sets:
the set
\[
\{x \in X: \overline{F_x} \text{ is nwd}\} \subseteq
\{x \in X: F_x \text{ is nwd}\}
\]
\{x \in X: F_x \text{ is nwd}\}
\]
is comeager.
\end{refproof}
@ -242,8 +242,8 @@ but for meager sets:
If $M \subseteq X \times Y$ is meager,
then
\[
\{x \in X : M_x \text{ is meager}\}
\]
\{x \in X : M_x \text{ is meager}\}
\]
is comeager.
\end{claim}
\begin{refproof}{thm:kuratowskiulam:c1b}
@ -257,9 +257,9 @@ but for meager sets:
as a countable intersection of comeager sets.
\end{refproof}
\todo{Finish proof}
\phantom\qedhere
\end{refproof}
% \phantom\qedhere
% \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}

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inputs/lecture_06.tex Normal file
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@ -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}

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

1
logic3.tex
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@ -29,6 +29,7 @@
\input{inputs/lecture_03}
\input{inputs/lecture_04}
\input{inputs/lecture_05}
\input{inputs/lecture_06}

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quiver.sty Normal file
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@ -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