From 8359683da62efdf6b4eeac4b564b54e42e8cd8e6 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 13 Oct 2023 23:54:02 +0200 Subject: [PATCH] initial commit --- README.md | 2 +- inputs/intro.tex | 17 +++ inputs/lecture_01.tex | 260 ++++++++++++++++++++++++++++++++++++++++++ inputs/lecture_02.tex | 145 +++++++++++++++++++++++ logic.sty | 5 +- logic3.tex | 9 +- 6 files changed, 431 insertions(+), 7 deletions(-) create mode 100644 inputs/intro.tex create mode 100644 inputs/lecture_01.tex create mode 100644 inputs/lecture_02.tex diff --git a/README.md b/README.md index 847a01a..245701e 100644 --- a/README.md +++ b/README.md @@ -1,7 +1,7 @@ # 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.** diff --git a/inputs/intro.tex b/inputs/intro.tex new file mode 100644 index 0000000..8f35813 --- /dev/null +++ b/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 $\dagger$. diff --git a/inputs/lecture_01.tex b/inputs/lecture_01.tex new file mode 100644 index 0000000..72d5da4 --- /dev/null +++ b/inputs/lecture_01.tex @@ -0,0 +1,260 @@ +\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 $G_{\delta}$ + + + diff --git a/inputs/lecture_02.tex b/inputs/lecture_02.tex new file mode 100644 index 0000000..76a3cc2 --- /dev/null +++ b/inputs/lecture_02.tex @@ -0,0 +1,145 @@ +\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} + diff --git a/logic.sty b/logic.sty index 8e76166..2a864c8 100644 --- a/logic.sty +++ b/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]}}} diff --git a/logic3.tex b/logic3.tex index a313599..b3f0a57 100644 --- a/logic3.tex +++ b/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