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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/logic3abstracttopologicaldynamicsanddescriptivesettheory}{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}{20231010}{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 secondcountable,


\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}{20231013}{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 $d(x,y) = xy$.


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, 1y)}


\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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\usepackage[index]{mkesslervocab}


\usepackage{mkesslercode}


\usepackage{jrpiemath}


\usepackage{jrpieyaref}


\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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\documentclass[10pt,ngerman,a4paper,fancyfoot,git]{mkesslerscript}




\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



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