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README.md
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# Logic 3: Abstract Topological Dynamics and Descriptive Set Theory
These are my notes on the lecture Logic 3: Abstract Topological Dynamics and Descriptive Set Theory,
taught by [TODO]
taught by Aleksandra Kwiatkowska
in the winter 23/24 at the University Münster.
**This is not an official script.**

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These are my notes on the lecture Probability Theory,
taught by \textsc{Aleksandra Kwiatkowska}
in the summer term 2023 at the University Münster.
\begin{warning}
This is not an official script.
The official lecture notes can be found on
\href{https://sites.google.com/site/akwiatkmath/teaching/logic-3-abstract-topological-dynamics-and-descriptive-set-theory}{here}.
\end{warning}
If you find errors or want to improve something,
please send me a message:\\
\texttt{lecturenotes@jrpie.de}.
This notes follow the way the material was presented in the lecture rather
closely. Additions (e.g.~from exercise sheets)
and slight modifications have been marked with $\dagger$.

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\lecture{01}{2023-10-10}{Introduction}
\section{Introduction}
\begin{definition}
Let $X$ be a nonempty topological space.
We say that $X$ is a \vocab{Polish space}
if $X$ is
\begin{itemize}
\item \vocab{separable},
i.e.~there exists a countable dense subset, and
\item \vocab{completely metrisable},
i.e.~there exists a complete metric on $X$
which induces the topology.
\end{itemize}
\end{definition}
Note that Polishness is preserved under homeomorphisms,
i.e.~it is really a topological property.
\begin{example}
\begin{itemize}
\item $\R$ is a Polish space,
\item $\R^n$ for finite $n$ is Polish,
\item $[0,1]$,
\item any countable discrete topological space,
\item the completion of any separable metric space
considered as a topological space.
\end{itemize}
\end{example}
Polish spaces behave very nicely.
We will see that uncountable polish spaces have size $2^{\aleph_0}$.
There are good notions of big (comeager)
and small (meager).
\subsection{Topology background}
Recall the following notions:
\begin{definition}[\vocab{product topology}]
Let $(X_i)_{i \in I}$ be a family of topological spaces.
Consider the set $\prod_{i \in I} X_i$
and the topology induced by basic open sets
$\prod_{i \in I} U_i$ with $U_i \subseteq X_i$ open
and $U_i \subsetneq X_i$ for only finitely many $i$.
\end{definition}
\begin{fact}
Countable products of separable spaces are separable,
\end{fact}
\begin{definition}
A topological space $X$ is \vocab{second countable},
if it has a countable base.
\end{definition}
If $X$ is a topological space.
Then if $X$ is second countable, it is also separable.
However the converse of this does not hold.
\begin{example}
Let $X$ be an uncountable set.
Take $x_0 \in X$ and consider the topology given by
\[
\tau = \{U \subseteq X | U \ni x_0\} \cup \{\emptyset\}.
\]
Then $\{x_0\}$ is dense in $X$, but $X$ is not second countable.
\end{example}
\begin{example}[Sorgenfrey line]
\todo{Counterexamples in Topology}
\end{example}
\begin{fact}
For metric spaces, the following are equivalent:
\begin{itemize}
\item separable,
\item second-countable,
\item \vocab{Lindelöf} (every open cover has a countable subcover).
\end{itemize}
\end{fact}
\begin{fact}
Compact\footnote{It is not clear whether compact means compact and Hausdorff in this lecture.} Hausdorff spaces are \vocab{normal} (T4)
i.e.~two disjoint closed subsets can be separated
by open sets.
\end{fact}
\begin{fact}
For a metric space, the following are equivalent:
\begin{itemize}
\item compact,
\item \vocab{sequentially compact} (every sequence has a convergent subsequence),
\item complete and \vocab{totally bounded}
(for all $\epsilon > 0$ we can cover the space with finitely many $\epsilon$-balls).
\end{itemize}
\end{fact}
\begin{theorem}[Urysohn's metrisation theorem]
Let $X$ be a topological space.
If $X$ is
\begin{itemize}
\item second countable,
\item Hausdorff and
\item regular (T3)
\end{itemize}
then $X$ is metrisable.
\end{theorem}
\begin{fact}
If $X$ is a compact Hausdorff space,
the following are equivalent:
\begin{itemize}
\item $X$ is Polish,
\item $X$ is metrisable,
\item $X$ is second countable.
\end{itemize}
\end{fact}
\subsection{Some facts about polish spaces}
\begin{fact}
Let $(X, \tau)$ be a topological space.
Let $d$ be a metric on $X$.
We will denote the topology induces by this metric as $\tau_d$.
To show that $\tau = \tau_d$,
it is equivalent to show that
\begin{itemize}
\item every open $d$-ball is in $\tau$ ($\implies \tau_d \subseteq d$ )
and
\item every open set in $\tau$ is a union of open $d$-balls.
\end{itemize}
To show that $\tau_d = \tau_{d'}$
for two metrics $d, d'$,
suffices to show that open balls in one metric are unions of open balls in the other.
\end{fact}
\begin{notation}
We sometimes denote $\min(a,b)$ by $a \wedge b$.
\end{notation}
\begin{proposition}
\label{prop:boundedmetric}
Let $(X, \tau)$ be a topological space,
$d$ a metric on $ X$ compatible with $\tau$ (i.e.~it induces $\tau$).
Then $d' \coloneqq \min(d,1)$ is also a metric compatible with $\tau$.
\end{proposition}
\begin{proof}
To check the triangle inequality:
\begin{IEEEeqnarray*}{rCl}
d(x,y) \wedge 1 &\le & \left( d(x,z) + d(y,z) \right) \wedge 1\\
&\le & \left( d(x,z) \wedge 1 \right) + \left( d(y,z) \wedge 1 \right).
\end{IEEEeqnarray*}
For $\epsilon \le 1$ we have $B_\epsilon'(x) = B_\epsilon(x)$
and for $\epsilon > 1$, $B'_\epsilon(x) = X$.
Since $d$ is complete, we have that $d'$ is complete.
\end{proof}
\begin{proposition}
Let $A$ be a Polish space.
Then $A^{\omega}$ Polish.
\end{proposition}
\begin{proof}
Let $A$ be separable.
Then $A^{\omega}$ is separable.
(Consider the basic open sets of the product topology).
Let $d \le 1$ be a complete metric on $A$.
Define $D$ on $A^\omega$ by
\[
D\left( (x_n), (y_n) \right) \coloneqq
\sum_{n< \omega} 2^{-(n+1)} d(x_n, y_n).
\]
Clearly $D \le 1$.
It is also clear, that $D$ is a metric.
We need to check that $D$ is complete:
Let $(x_n)^{(k)}$ be a Cauchy sequence in $A^{\omega}$.
Consider the pointwise limit $(a_n)$.
This exists since $x_n^{(k)}$ is Cauchy for every fixed $n$.
Then $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
\end{proof}
\begin{definition}[Our favourite Polish spaces]
\begin{itemize}
\item $2^{\omega}$ is called the \vocab{Cantor set}.
(Consider $2$ with the discrete topology)
\item $\omega^{\omega}$ is called the \vocab{Baire space}.
($\omega$ with descrete topology)
\item $[0,1]^{\omega}$ is called the \vocab{Hilbert cube}.
($[0,1] \subseteq \R$ with the usual topology)
\end{itemize}
\end{definition}
\begin{proposition}
Let $X$ be a separable, metrisable topological space.
Then $X$ topologically embeds into the
\vocab{Hilbert cube},
i.e. there is an injective $f: X \hookrightarrow [0,1]^{\omega}$
such that $f: X \to f(X)$ is a homeomorphism.
\end{proposition}
\begin{proof}
$X$ is separable, so it has some countable dense subset,
which we order as a sequence $(x_n)_{n \in \omega}$.
Let $d$ be a metric on $X$ which is compatible with the topology.
W.l.o.g.~$d \le 1$ (by \yaref{prop:boundedmetric}).
Let $d$ be the metric of $X$ and define
\begin{IEEEeqnarray*}{rCl}
f\colon X &\longrightarrow & [0,1]^{\omega} \\
x&\longmapsto & (d(x,x_n))_{n < \omega}
\end{IEEEeqnarray*}
\begin{claim}
$f$ is injective.
\end{claim}
\begin{subproof}
Suppose that $f(x) = f(y)$.
Then $d(x,x_n) = d(y,y_n)$ for all $n$.
Hence $d(x,y) \le d(x,x_n) + d(y,x_n) = 2 d(x,x_n)$.
Since $(x_n)$ is dense, we get $d(x,y) = 0$.
\end{subproof}
\begin{claim}
$f$ is continuous.
\end{claim}
\begin{subproof}
Consider a basic open set in $[0,1]^{\omega}$,
i.e. specify open sets $U_1, \ldots, U_n$ on finitely many coordinates.
$f^{-1}(U_1 \times \ldots \times U_n \times \ldots)$
is a finite intersection of open sets,
hence it is open.
\end{subproof}
\begin{claim}
$f^{-1}$ is continuous.
\end{claim}
\begin{subproof}
\todo{Exercise!}
\end{subproof}
\end{proof}
\begin{proposition}
Countable disjoint unions of Polish spaces are Polish.
\end{proposition}
\begin{proof}
Define a metric in the obvious way.
\end{proof}
\begin{proposition}
Closed subspaces of Polish spaces are Polish.
\end{proposition}
\begin{proof}
Let $X$ be Polish and $V \subseteq X$ closed.
Let $d$ be a complete metric on $X$.
Then $d\defon{V}$ is complete.
Subspaces of second countable spaces
are second countable.
\end{proof}
\begin{definition}
Let $X$ be a topological space.
A subspace $A \subseteq X$ is called
$G_\delta$\footnote{Gebietdurchschnitt}, if it is a countable intersection of open sets.
\end{definition}
Next time: Closed sets are $G_\delta$.
A subspace of a Polish space is Polish iff it is $G_{\delta}$

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\lecture{02}{2023-10-13}{Subsets of Polish spaces}
\begin{theorem}
\label{subspacegdelta}
A subspace of a Polish space is Polish
iff it is $G_{\delta}$.
\end{theorem}
\begin{remark}
Closed subsets of a metric space $(X, d )$
are $G_{\delta}$.
\end{remark}
\begin{proof}
Let $C \subseteq X$ be closed.
Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
\todo{Exercise}
Clearly $C \subseteq \bigcap U_{\frac{1}{n}}$.
Let $x \in \bigcap U_{\frac{1}{n}}$.
Then $\forall n .~ \exists x_n\in C.~d(x,x_n) < \frac{1}{n}$.
The $x_n$ converge to $x$ and since $C$ is closed,
we get $x \in C$.
Hence $C = \bigcap U_{\frac{1}{n}}$
is $G_{\delta}$.
\end{proof}
\begin{example}
Let $ X$ be Polish.
Let $d$ be a complete metric on $X$.
\begin{enumerate}[a)]
\item If $Y \subseteq X$ is closed,
then $(Y,d\defon{Y})$ is complete.
\item $Y = (0,1) \subseteq \R$
with the usual metric $d(x,y) = |x-y|$.
Then $x_n \to 0$ is Cauchy in $((0,1), d)$.
But
\[
d_1(x,y) \coloneqq | x -y|
+ \left|\frac{1}{\min(x, 1- x)}
- \frac{1}{\min(y, 1-y)}
\right|
\]
also is a complete metric on $(0,1)$
which is compatible with $d$.
We want to generalize this idea.
\end{enumerate}
\end{example}
\begin{refproof}{subspacegdelta}
\begin{claim}
\label{psubspacegdelta:c1}
If $Y \subseteq (X,d)$ is $G_{\delta}$,
then there exists a complete metric on $Y$.
\end{claim}
\begin{refproof}{psubspacegdelta:c1}
Let $Y = U$be open in $X$.
Consider the map
\begin{IEEEeqnarray*}{rCl}
f_U\colon U &\longrightarrow &
\underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
\end{IEEEeqnarray*}
Note that $X \times \R$ with the
\[d_1((x_1,y_1), (x_2, y_2)) \coloneqq d(x_1,x_2) + |y_1 - y_2|\]
metric is complete.
$f_U$ is an embedding of $U$ into $X \times \R$:
\begin{itemize}
\item It is injective because of the first coordinate.
\item It is continuous since $d(x, U^c)$ is continuous
and only takes strictly positive values. % TODO
\item The inverse is continuous because projections
are continuous.
\end{itemize}
So we have shown that $U$ is homeomorphic to % TODO with ?
the graph of $\tilde{f_U}\colon x \mapsto \frac{1}{d(x, U^c)}$.
The graph is closed in $U \times \R$,
because $\tilde{f_U}$ is continuous.
It is closed in $X \times \R$ because
$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$.
\todo{Make this precise}
Therefore we identified $U$ with a closed subspace of
the Polish space $(X \times \R, d_1)$.
\end{refproof}
Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
Take
\begin{IEEEeqnarray*}{rCl}
f_Y\colon Y &\longrightarrow & X \times \R^{\N} \\
x &\longmapsto &
\left(x, \left( \frac{1}{\delta(x,U_n^c)} \right)_{n \in \N}\right)
\end{IEEEeqnarray*}
As for an open $U$, $f_Y$ is an embedding.
Since $X \times \R^{\N}$
is completely metrizable,
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
\begin{claim}
\label{psubspacegdelta:c2}
If $Y \subseteq (X,d)$ is completely metrizable,
then $Y$ is a $G_{\delta}$ subspace.
\end{claim}
\begin{refproof}{psubspacegdelta:c2}
There exists a complete metric $d_Y$ on $Y$.
For every $n$,
let $V_n \subseteq X$ be the union
of all open sets $U \subseteq X$ such that
\begin{enumerate}[(i)]
\item $U \cap Y \neq \emptyset$,
\item $\diam_d(U) \le \frac{1}{n}$,
\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
\end{enumerate}
We want to show that $Y = \bigcap_{n \in \N} V_n$.
For $x \in Y$, $n \in \N$ we have $x \in V_n$,
as we can choose two neighbourhoods
$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
such that $\diam_{d_Y}(U) < \frac{1}{n}$
and $U_2 \cap Y = U_1$.
Additionally choose $x \in U_3$ open in $X$
with $\diam_{d}(U_3) < \frac{1}{n}$.
Then consider $U_2 \cap U_3 \subseteq V_n$.
Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
Now let $x \in \bigcap_{n \in \N} V_n$.
For each $n$ pick $x \in U_n \subseteq X$ open
satisfying (i), (ii), (iii).
W.l.o.g. the $U_n$ are decreasing.
From (i) and (ii) it follows that $x \in \overline{Y}$,
since we can consider a sequence of points $y_n \in U_n \cap Y$
and get $y_n \xrightarrow{d} x$.
On the other hand $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$,
so the $y_n$ form a Cauchy sequence with respect to $d_Y$,
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$,
hence $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
$y_n$ converges to the unique point in $\bigcap_{n} \overline{U_n \cap Y}$.
Since the topologies agree, this point is $x$.
\end{refproof}
\end{refproof}

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logic.sty
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\usepackage[index]{mkessler-vocab}
\usepackage{mkessler-code}
\usepackage{jrpie-math}
\usepackage{jrpie-yaref}
\usepackage[normalem]{ulem}
\usepackage{pdflscape}
\usepackage{longtable}
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\newcommand{\concat}{{}^\frown}
\DeclareMathOperator{\hght}{height}
\DeclareSimpleMathOperator{Prod} % TODO Remove this. Did I mean \prod ?
\newcommand{\lecture}[2]{Lecture #1 #2}
\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}

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logic3.tex
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\documentclass[10pt,ngerman,a4paper,fancyfoot,git]{mkessler-script}
\course{Logic 3: Abstract Topological Dynamics and Descriptive Set Theory}
\lecturer{}
\assistant{}
\author{}
\lecturer{Aleksandra Kwiatkowska}
%\assistant{}
\author{Josia Pietsch}
\usepackage{logic}
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\newpage
\input{inputs/lecture_01}
\input{inputs/lecture_02}
\cleardoublepage