From bfcb894c89be5bcae6729aa14fe03bd4968664d6 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Mon, 13 Nov 2023 00:23:51 +0100 Subject: [PATCH] lecture 08 --- inputs/lecture_07.tex | 192 ++++++++++++++++++++++++++++++++++++++++++ inputs/lecture_08.tex | 143 +++++++++++++++++++++++++++++++ logic.sty | 1 + logic3.tex | 2 + 4 files changed, 338 insertions(+) create mode 100644 inputs/lecture_07.tex create mode 100644 inputs/lecture_08.tex diff --git a/inputs/lecture_07.tex b/inputs/lecture_07.tex new file mode 100644 index 0000000..09fc8be --- /dev/null +++ b/inputs/lecture_07.tex @@ -0,0 +1,192 @@ +\lecture{07}{2023-11-07}{} + +\begin{proposition} + Let $X$ be second countable. + Then $|\cB(X)| \le \fc$. + % $\fc := 2^{\aleph_0}$ +\end{proposition} +\begin{proof} + We use strong induction on $\xi < \omega_1$. + We have $\Sigma^0_1(X) \le \fc$ + (for every element of the basis, we can decide + whether to use it in the union or not). + + Suppose that $\forall \xi' < \xi.~|\Sigma^0_{\xi'}(X)| \le \fc$. + Then $|\Pi^0_{\xi'}(X)| \le \fc$. + We have that + \[ + \Sigma^0_\xi(X) = \{\bigcup_{n} A_n : n \in \N, A_n \in \Pi^0_{\xi_n}(X), \xi_n < \xi\}. + \] + Hence $|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}$. + + We have + \[ + \cB(X) = \bigcup_{\xi < \omega_1} \Sigma^0_\xi(X). + \] + Hence + \[ + |\cB(X)| \le \omega_1 \cdot \fc = \fc. + \] +\end{proof} + +\begin{proposition}[Closure properties] + Suppose that $X$ is metrizable. + Let $1 \le \xi < \omega_1$. + Then + \begin{enumerate}[(a)] + \item \begin{itemize} + \item $\Sigma^0_\xi(X)$ is closed under countable unions. + \item $\Pi^0_\xi(X)$ is closed under countable intersections. + \item $\Delta^0_\xi(X)$ is closed under complements, + countable unions and + countable intersections. + \end{itemize} + \item \begin{itemize} + \item $\Sigma^0_\xi(X)$ is closed under \emph{finite} intersections. + \item $\Pi^0_\xi(X)$ is closed under \emph{finite} unions. + \end{itemize} + + \end{enumerate} +\end{proposition} +\begin{proof} + \begin{enumerate}[(a)] + \item This follows directly from the definition. + Note that a countable intersection can be written + as a complement of the countable union of complements: + \[ + \bigcap_{n} B_n = \left( \bigcup_{n} B_n^{c} \right)^{c}. + \] + \item If suffices to check this for $\Sigma^0_{\xi}(X)$. + Let $A = \bigcup_{n} A_n$ for $A_n \in \Pi^0_{\xi_n}(X)$ + and $B = \bigcup_{m} B_m$ for $B_m \in \Pi^0_{\xi'_m}(X)$. + Then + \[ + A \cap B = \bigcup_{n,m} \left( A_n \cap B_m \right) + \] + and $A_n \cap B_m \in \Pi^{0}_{\max(\xi_n, \xi'_m)}(X)$. + + \end{enumerate} +\end{proof} +\begin{example} + Consider the cantor space $2^{\omega}$. + We have that $\Delta^0_1(2^{\omega})$ + is not closed under countable unions + (countable unions yield all open sets, but there are open + sets that are not clopen). +\end{example} + +\subsection{Turning Borels Sets into Clopens} + +\begin{theorem} + \label{thm:clopenize} + Let $(X, \cT)$ be a Polish space. + For any Borel set $A \subseteq X$, + there is a finer Polish topology, + \footnote{i.e.~$\cT_A \supseteq \cT$ and $(X, \cT_A)$ is Polish} + such that + \begin{itemize} + \item $A$ is clopen in $\cT_A$, + \item the Borel sets do not change, + i.e.~$\cB(X, \cT) = \cB(X, \cT_A)$. + \end{itemize} +\end{theorem} +\begin{corollary}[Perfect set property] + Let $(X, \cT)$ be Polish, + and let $B \subseteq X$ be Borel and uncountable. + Then there is an embedding + of the cantor space $2^{\omega}$ + into $B$. +\end{corollary} +\begin{proof} + Pick $\cT_B \supset \cT$ + such that $(X, \cT_B)$ is Polish, + $B$ is clopen in $\cT_B$ and + $\cB(X,\cT) = \cB(X, \cT_B)$. + + Therefore $(\cB, \cT_B\defon{B})$ is Polish. + We know that there is an embedding + $f\colon 2^{\omega} \to (B, \cT_{B}\defon{B})$. + + Consider $f\colon 2^{\omega} \to B \subseteq (X, \cT)$. + This is still continuous as $\cT \subseteq \cT_B$. + Since $2^{\omega}$ is compact, $f$ is an embedding. + %\todo{Think about this} +\end{proof} + +\begin{refproof}{thm:clopenize} + We show that + \begin{IEEEeqnarray*}{rCl} + A \coloneqq \{B \subseteq \cB(X, \cT)&:\exists & \cT_B \supseteq \cT .\\ + && (X, \cT_B) \text{ is Polish},\\ + && \cB(X, \cT) = \cB(X, \cT_B)\\ + && B \text{ is clopen in $\cT_B$}\\ + \} + \end{IEEEeqnarray*} + is equal to the set of Borel sets. + + The proof rests on two lemmata: + \begin{lemma} + \label{thm:clopenize:l1} + Let $(X,\cT)$ be a Polish space. + Then for any $F \overset{\text{closed}}{\subseteq} X$ (wrt. $\cT$) + there is $\cT_F \supseteq \cT$ + such that $\cT_F$ is Polish, + $\cB(\cT) = \cB(\cT_F)$ + and $F$ is clopen in $\cT_F$. + \end{lemma} + \begin{proof} + Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$. + Both are Polish spaces. + Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$ + of these spaces. + This space is Polish, + and the topology is generated by $\cT \cup \{F\}$, + hence we do not get any new Borel sets. + \end{proof} + So all closed sets are in $A$. + Furthermore $A$ is closed under complements, + since complements of clopen sets are clopen. + + \begin{lemma} + \label{thm:clopenize:l2} + Let $(X, \cT)$ be Polish. + Let $\{\cT_n\}_{n < \omega}$ + be Polish topologies + such that $\cT_n \supseteq \cT$ + and $\cB(\cT_n) = \cB(\cT)$. + Then the topology $\cT_\infty$ generated by $\bigcup_{n} \cT_n$ + is still Polish + and $\cB(\cT_\infty) = \cB(T)$. + \end{lemma} + \begin{proof} + We have that $\cT_\infty$ is the smallest + topology containing all $\cT_n$. + To get $\cT_\infty$ + consider + \[ + \cF \coloneqq \{A_1 \cap A_2 \cap \ldots \cap A_n : A_i \in \cT_i\}. + \] + Then + \[ + \cT_\infty = \{\bigcup_{i<\omega} B_i : B_i \in \cF\}. + \] + (It suffices to take countable unions, + since we may assume that the $A_1, \ldots, A_n$ in the + definition of $\cF$ belong to + a countable basis of the respective $\cT_n$). + + \todo{This proof will be finished in the next lecture} + \end{proof} + + We need to show that $A$ is closed under countable unions. + By \yaref{thm:clopenize:l2} there exists a topology + $\cT_\infty$ such that $A = \bigcup_{n < \omega} A_n$ is open in $\cT_\infty$ + and $\cB(\cT_\infty) = \cB(\cT)$. + Applying \yaref{thm:clopenize:l1} + yields a topology $\cT_\infty'$ such that + $(X, \cT_\infty')$ is Polish, + $\cB(\cT_\infty') = \cB(\cT)$ + and $A $ is clopen in $\cT_{\infty}'$. +\end{refproof} + + diff --git a/inputs/lecture_08.tex b/inputs/lecture_08.tex new file mode 100644 index 0000000..162b50c --- /dev/null +++ b/inputs/lecture_08.tex @@ -0,0 +1,143 @@ +\lecture{08}{2023-11-10}{} + +\todo{put this lemma in the right place} +\begin{lemma}[Lemma 2] + Let $(X, \cT)$ be a Polish space. + Let $\cT_n \supseteq \cT$ be Polish + with $\cB(X, \cT_n) = \cB(X, \cT)$. + Let $\cT_\infty$ be the topology generated + by $\bigcup_n \cT_n$. + Then $(X, \cT_\infty)$ is Polish + and $\cB(X, \cT_\infty) = \cB(X, \cT)$. +\end{lemma} +\begin{proof} + Let $Y = \prod_{n \in \N} (X, \cT_n)$. + Then $Y$ is Polish. + Let $\delta\colon (X, \cT_\infty) \to Y$ + defined by $\delta(x) = (x,x,x,\ldots)$. + \begin{claim} + $\delta$ is a homeomorphism. + \end{claim} + \begin{subproof} + Clearly $\delta$ is a bijection. + We need to show that it is continuous and open. + + Let $U \in \cT_i$. + Then + \[ + \delta^{-1}(D \cap \left( X \times X \times \ldots\times U \times \ldots) \right)) = U \in \cT_i \subseteq \cT_\infty, + \] + hence $\delta$ is continuous. + Let $U \in \cT_\infty$. + Then $U$ is the union of sets of the form + \[ + V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{nu} + \] + for some $n_1 < n_2 < \ldots < n_u$ + and $U_{n_i} \in \cT_i$. + + Thus is suffices to consider sets of this form. + We have that + \[ + \delta(V) = D \cap (X \times X \times \ldots \times U_{n_1} \times \ldots \times U_{n_2} \times \ldots \times U_{n_u} \times X \times \ldots) \overset{\text{open}}{\subseteq} D. + \] + \end{subproof} + + This will finish the proof since + \[ + D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y + \] + Why? Let $(x_n) \in Y \setminus D$. + Then there are $i < j$ such that $x_i \neq x_j$. + Take disjoint open $x_i \in U$, $x_j \in V$. + Then + \[(x_n) \in X \times X \times \ldots \times U \times \ldots \times X \times \ldots \times V \times X \times \ldots\] + is open in $Y\setminus D$. + Hence $Y \setminus D$ is open, thus $D$ is closed. + It follows that $D$ is Polish. +\end{proof} + +\subsection{Parametrizations} +\todo{choose better title} + + +Let $\Gamma$ denote a collection of sets in some space. +For us $\Gamma$ will be one of $\Sigma^0_\xi(X), \Pi^0_\xi(X), \Delta^0_\xi(X), \cB(X)$, +where $X$ is a metrizable, usually second countable space. + +\begin{definition} + We say that $\cU \subseteq Y \times X$ + is \vocab{$Y$-universal} for $\Gamma(X)$ / + $\cU$ \vocab{parametrizes} $\Gamma(X)$ + iff: + \begin{itemize} + \item $\cU \in \Gamma$, + \item $\{U_y : y \in Y\} = \Gamma(X)$. + \end{itemize} +\end{definition} + +\begin{example} + Let $X = \omega^\omega$, $Y = 2^{\omega}$ + and consider $\Gamma = \Sigma^0_{\omega+5}(\omega^\omega)$. + We will show that there is a $2^{\omega}$-universal + set for $\Gamma$. +\end{example} + +\begin{theorem} + Let $X$ be a separable, metrizable space. + Then for every $\xi \ge 1$, + there is a $2^{\omega}$-universal + set for $\Sigma^0_\xi(X)$ and + similarly for $\Pi^0_\xi(X)$. +\end{theorem} +\begin{proof} + Note that if $\cU$ is $2^{\omega}$ universal for + $\Sigma^0_\xi(X)$, then $(2^{\omega} \times X) \setminus \cU$ + is $2^{\omega}$-universal for $\Pi^0_\xi(X)$. + Thus it suffices to consider $\Sigma^0_\xi(X)$. + + First let $\xi = 1$. + We construct $\cU \overset{\text{open}}{\subseteq} 2^{\omega} \times X$ + such that + \[ + \{U_y : y \in 2^\omega\} = \Sigma^0_1(X). + \] + + Let $(V_n)$ be a basis of open sets of $X$. + For all $y \in 2^\omega$ and $x \in X$ + put $(y,x) \in \cU$ iff + $x \in \bigcup \{V_n : y_n = 1\}$. + $\cU$ is open. + Let $V = \bigcup \{V_n : V_n \subseteq V\}$. + Pick $y \in 2^\omega$ + and let $y_n = 1$ iff $V_n \subseteq V$. + Then $\cU_y = V$. + + + Now suppose that there exists a + $2^{\omega}$-universal set for $\Sigma^0_{\eta}(X)$ + for all $\eta < \xi$. + Fix $\xi_0 \le \xi_1 \le \ldots < \xi$ + such that $\xi_n \to \xi$ if $\xi$ is a limit, + or $\xi_n = \xi'$ if $\xi = \xi' +1$ is a successor. + + Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$. + + Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$ + some $\eta_n < \xi$, + we also have + $A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$. + + We construct a $(2^\omega)^\omega \cong 2^\omega$-universal set + for $\Sigma^0_\xi(X)$. + For $(y_n) \in (2^\omega)^\omega$ + and $x \in X$ + we set $((y_n), x) \in \cU$ + iff $\exists n.~(y_n, x) \in U_{\xi_n}$, + i.e.~iff $\exists n.~x \in (U_{\xi_n})_{y_n}$. + + Let $A \in \Sigma^0_\xi(X)$. + Then $A = \bigcup_{n} B_n$ for some $B_n \in \Pi^0_{\xi_n}(X)$. + % TODO + Furthermore $\cU \in \Sigma^0_{\xi}((2^\omega)^\omega \times X)$. +\end{proof} diff --git a/logic.sty b/logic.sty index 69a4c37..5fbc0b7 100644 --- a/logic.sty +++ b/logic.sty @@ -134,5 +134,6 @@ \DeclareMathOperator{\hght}{height} \DeclareMathOperator{\symdif}{\triangle} \DeclareSimpleMathOperator{proj} +\newcommand{\fc}{\mathfrak{c}} \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}} diff --git a/logic3.tex b/logic3.tex index ef7bffc..3c1360b 100644 --- a/logic3.tex +++ b/logic3.tex @@ -30,6 +30,8 @@ \input{inputs/lecture_04} \input{inputs/lecture_05} \input{inputs/lecture_06} +\input{inputs/lecture_07} +\input{inputs/lecture_08}