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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.
This is not an official script.
The official lecture notes can be found on
If you find errors or want to improve something,
please send me a message:\\
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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Let $X$ be a nonempty topological space.
We say that $X$ is a \vocab{Polish space}
if $X$ is
\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.
Note that Polishness is preserved under homeomorphisms,
i.e.~it is really a topological property.
\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.
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 $i$.
Countable products of separable spaces are separable,
A topological space $X$ is \vocab{second countable},
if it has a countable base.
If $X$ is a topological space.
Then if $X$ is second countable, it is also separable.
However the converse of this does not hold.
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.
\begin{example}[Sorgenfrey line]
\todo{Counterexamples in Topology}
For metric spaces, the following are equivalent:
\item separable,
\item second-countable,
\item \vocab{Lindelöf} (every open cover has a countable subcover).
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.
For a metric space, the following are equivalent:
\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).
\begin{theorem}[Urysohn's metrisation theorem]
Let $X$ be a topological space.
If $X$ is
\item second countable,
\item Hausdorff and
\item regular (T3)
then $X$ is metrisable.
If $X$ is a compact Hausdorff space,
the following are equivalent:
\item $X$ is Polish,
\item $X$ is metrisable,
\item $X$ is second countable.
\subsection{Some facts about polish spaces}
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
\item every open $d$-ball is in $\tau$ ($\implies \tau_d \subseteq d$ )
\item every open set in $\tau$ is a union of open $d$-balls.
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.
We sometimes denote $\min(a,b)$ by $a \wedge b$.
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 $\tau$.
To check the triangle inequality:
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).
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.
Let $A$ be a Polish space.
Then $A^{\omega}$ Polish.
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 $(x_n)^{(k)} \xrightarrow{k \to \infty} (a_n)$.
\begin{definition}[Our favourite Polish spaces]
\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)
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.
$X$ 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
f\colon X &\longrightarrow & [0,1]^{\omega} \\
x&\longmapsto & (d(x,x_n))_{n < \omega}
$f$ is injective.
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 $d(x,y) = 0$.
$f$ is continuous.
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.
$f^{-1}$ is continuous.
Countable disjoint unions of Polish spaces are Polish.
Define a metric in the obvious way.
Closed subspaces of Polish spaces are Polish.
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.
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.
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}
A subspace of a Polish space is Polish
iff it is $G_{\delta}$.
Closed subsets of a metric space $(X, d )$
are $G_{\delta}$.
Let $C \subseteq X$ be closed.
Let $U_{\frac{1}{n}} \coloneqq \{x | d(x, C) < \frac{1}{n}\}$.
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 $G_{\delta}$.
Let $ X$ be Polish.
Let $d$ be a complete metric on $X$.
\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 $((0,1), d)$.
d_1(x,y) \coloneqq | x -y|
+ \left|\frac{1}{\min(x, 1- x)}
- \frac{1}{\min(y, 1-y)}
also is a complete metric on $(0,1)$
which is compatible with $d$.
We want to generalize this idea.
If $Y \subseteq (X,d)$ is $G_{\delta}$,
then there exists a complete metric on $Y$.
Let $Y = U$be open in $X$.
Consider the map
f_U\colon U &\longrightarrow &
\underbrace{X}_{d} \times \underbrace{\R}_{|\cdot |} \\
x &\longmapsto & \left( x, \frac{1}{d(x, U^c)} \right).
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 $X \times \R$:
\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.
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 $(X \times \R, d_1)$.
Let $Y = \bigcap_{n \in \N} U_n$ be $G_{\delta}$.
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)
As for an open $U$, $f_Y$ is an embedding.
Since $X \times \R^{\N}$
is completely metrizable,
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
If $Y \subseteq (X,d)$ is completely metrizable,
then $Y$ is a $G_{\delta}$ subspace.
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
\item $U \cap Y \neq \emptyset$,
\item $\diam_d(U) \le \frac{1}{n}$,
\item $\diam_{d_Y}(U \cap Y) \le \frac{1}{n}$.
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 $x$.

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\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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\course{Logic 3: Abstract Topological Dynamics and Descriptive Set Theory}
\lecturer{Aleksandra Kwiatkowska}
\author{Josia Pietsch}
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