initial commit

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.**

inputs/intro.tex

@@ -0,0 +1,17 @@
+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 $\dagger$.

inputs/lecture_01.tex

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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 $G_{\delta}$
+
+
+

inputs/lecture_02.tex

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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}
+

logic.sty

@@ -8,6 +8,7 @@
 \usepackage[index]{mkessler-vocab}
 \usepackage{mkessler-code}
 \usepackage{jrpie-math}
+\usepackage{jrpie-yaref}
 \usepackage[normalem]{ulem}
 \usepackage{pdflscape}
 \usepackage{longtable}
@@ -126,6 +127,4 @@
 \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]}}}

logic3.tex

@@ -1,9 +1,9 @@
 \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}
 
@@ -24,6 +24,9 @@
 
 \newpage
 
+\input{inputs/lecture_01}
+\input{inputs/lecture_02}
+
 
 
 \cleardoublepage