\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} \gist{% 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 \[ |\cB(X)| \le \omega_1 \cdot \fc = \fc. \] }{Use strong induction. $|\Sigma^0_1(X)| \le \fc$, since $X$ is second countable. \[|\Sigma^0_\xi(X)| \le (\overbrace{\aleph_0}^{\mathclap{\{\xi': \xi' < \xi\} \text{ is countable~ ~ ~ ~}}} \cdot \underbrace{\fc}_{\mathclap{\text{inductive assumption}}})^{\overbrace{\aleph_0}^{\mathclap{\text{countable unions}}}}\] } \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. \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. \item $\Delta^0_\xi(X)$ is closed under finite unions and finite intersections. \end{itemize} \end{enumerate} \end{proposition} \gist{% \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% \gist{ (countable unions yield all open sets, but there are open sets that are not clopen)}{}. \end{example} \subsection{Turning Borel Sets into Clopens} \begin{theorem}% \gist{% \footnote{Whilst strikingly concise the verb ``\vocab[Clopenization™]{to clopenize}'' unfortunately seems to be non-standard vocabulary. Our tutor repeatedly advised against using it in the final exam. Contrary to popular belief the very same tutor was \textit{not} the one first to introduce it, as it would certainly be spelled ``to clopenise'' if that were the case. }% }{}% \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} \gist{% 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. }{% Clopenize $B$. We can embed $2^{ \omega}$ into Polish spaces. Clopenization makes the topology finer, so this is still continuous wrt.~the original topology. $2^{\omega}$ is compact, so this is an embedding. } \end{proof} \begin{refproof}{thm:clopenize} \gist{% 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. }{% Let $A$ be the set of clopenizable sets. We show that $A = \cB(X)$. } \gist{The proof rests on two lemmata:}{} \begin{lemma} \label{thm:clopenize:l1} \gist{% 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 $\cT_F$. }{% Closed sets can be clopenized. } \end{lemma} \begin{proof} \gist{% Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$. Both are Polish spaces. Take the coproduct% \footnote{In the lecture, this was called the \vocab{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. }{Consider $(F, \cT\defon{F}) \oplus (X \setminus F, \cT\defon{X \setminus F})$.} \end{proof} \gist{% So all closed sets are in $A$. Furthermore $A$ is closed under complements, since complements of clopen sets are clopen. }{So $\Sigma^0_1(X), \Pi^0_1(X) \subseteq A$.} \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 Polish and $\cB(\cT_\infty) = \cB(T)$. \end{lemma} \begin{refproof}{thm:clopenize:l2} \gist{% 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 $\cT_n$). }{} % Proof was finished in lecture 8 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} \gist{% \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 \[ V = U_{n_1} \cap U_{n_2} \cap \ldots \cap U_{n_u} \] 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} }{} \begin{claim} $D = \{(x,x,\ldots) \in Y : x \in X\} \overset{\text{closed}}{\subseteq} Y.$ \end{claim} \gist{% \begin{subproof} 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. \end{subproof} It follows that $D$ is Polish. }{} \end{refproof} \gist{% 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}