\lecture{06}{2023-11-03}{} \gist{% % \begin{refproof}{thm:kuratowskiulam} \begin{enumerate}[(i)] \item Let $A$ be a set with the Baire property. Write $A = U \symdif M$ for $U$ open and $M$ meager. Then for all $x$, we have that $A_x = U_x \symdif M_x$, where $U_x$ is open, and $\{x : M_x \text{ is meager}\}$ is comeager. Therefore $\{x : U_x \text{ open } \land M_x \text{ meager }\}$ is comeager, and for those $x$, $A_x$ has the Baire property. \end{enumerate} % TODO: fix counter \begin{claim} % Claim 2 \label{thm:kuratowskiulam:c2} For $P \subseteq X$, $Q \subseteq Y$ with the Baire property, let $R \coloneqq P \times Q$. Then $R$ is meager iff at least one of $P$ or $Q$ is meager. \end{claim} \begin{refproof}{thm:kuratowskiulam:c2} Suppose that $R$ is meager. Then by \yaref{thm:kuratowskiulam:c1b}, we have that $C = \{x : R_x \text{ is meager }\}$ is comeager. \begin{itemize} \item If $P$ is meager, the statement holds trivially. \item If $P$ is not meager, then $P \cap C \neq \emptyset$. For $x \in P \cap C$ we have that $R_x$ is meager and $R_x = Q$, hence $Q$ is meager. \end{itemize} On the other hand suppose that $P$ is meager. Then $P = \bigcup_{n} F_n$ for nwd sets $F_n$. Note that $F_n \times Y$ is nwd. So $F_n \times Q$ is also nwd. Hence $P \times Q$ is a countable union of nwd sets, so it is meager. \end{refproof} \begin{enumerate}[(i)] \item[(ii)] ``$\impliedby$'' Let $A$ be a set with the Baire property such that $\{x : A_x \text{ is meager}\}$ is comeager. Let $A = U \symdif M$ for $U$ open and $M$ meager. Towards a contradiction suppose that $A$ is not meager. Then $U$ is not meager. Since $X \times Y$ is second countable, we have that $U$ is a countable union of open rectangles. At least one of them, say $G \times H \subseteq U$, is not meager. By \yaref{thm:kuratowskiulam:c2}, both $G$ and $H$ are not meager. Since $\{x\colon A_x \text{ is meager} \land M_x \text{ is meager}\}$ is comeager (using \yaref{thm:kuratowskiulam:c1b}), there is $x_0 \in G$ such that $A_{x_0}$ is meager and $M_{x_0}$ is meager. But then $H$ is meager as \[ H \setminus M_{x_0} \subseteq U_{x_0} \setminus M_{x_0} \subseteq U_{x_0} \symdif M_{x_0} = A_{x_0} \] and $M_{x_0}$ is meager $\lightning$. ``$\implies$'' This is \yaref{thm:kuratowskiulam:c1b}. \end{enumerate} }{% \begin{itemize} \item (ii) $\iff$ (iii): pass to complement. \item $F \overset{\text{closed}}{\subseteq} X \times Y$ nwd. $\implies \{x \in X : F_x \text{ nwd}\} $ comeager: \begin{itemize} \item $W = F^c$ is open and dense, show that $\{x : W_x \text{ dense}\}$ is comeager. \item $(V_n)$ enumeration of basis. Show that $U_n \coloneqq \{x : V_n \cap W_x \neq \emptyset\}$ is comeager for all $n$. \item $U_n$ is open (projection of open) and dense ($W$ is dense, hence $W \cap ( U \times V_n) \neq \emptyset$ for $U$ open). \end{itemize} \item $F \subseteq X \times Y$ is nwd $\implies \{x \in X: F_x \text{ nwd}\}$ comeager. (consider $\overline{F}$). \item (ii) $\implies$: $M \subseteq X \times Y$ meager $\implies \{x \in X: M_x \text{ meager}\}$ comeager (write $M$ as ctbl. union of nwd.) \item (i): If $A$ has the Baire Property, then $A = U \symdif M$, $A_x = U_x \symdif M_x$, $U_x$ open and $\{x : M_x \text{ meager}\}$ comeager $\implies$ (i). \item $P \subseteq X$, $Q \subseteq Y$ BP, then $P \times Q$ meager $\iff$ $P$ or $Q$ meager. \begin{itemize} \item $\impliedby$ easy \item $\implies$ Suppose $P \times Q$ meager, $P$ not meager. $\emptyset\neq P \cap \underbrace{\{x : (P \times Q)_x \text{ meager} \}}_{\text{comeager}} \ni x$. $(P \times Q)_x = Q$ is meager. \end{itemize} \item (ii) $\impliedby$: \begin{itemize} \item $A$ BP, $\{x : A_x \text{ meager}\}$ comeager. \item $A = U \symdif M$. \item Suppose $A$ not meager $\leadsto$ $U$ not meager $\leadsto \exists G \times H \subseteq U$ not meager. \item $G$ and $H$ are not meager. \item $\exists x_0 \in G \cap \underbrace{\{x: A_x \text{ meager } \land M_x \text{ meager}\}}_\text{comeager}$. \item $H$ meager, as \[ H \subseteq U_{x_0} \subseteq A_{x_0} \cup M_{x_0}. \] \end{itemize} \end{itemize} } \end{refproof} \gist{% \begin{remark} Suppose that $A$ has the BP. Then there is an open $U$ such that $A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager. Then $A = U \symdif M$. \end{remark} }{} \section{Borel sets} % TODO: fix chapters \begin{definition} Let $X$ be a topological space. Let $\cB(X)$ denote the smallest $\sigma$-algebra, that contains all open sets. Elements of $\cB(X)$ are called \vocab{Borel sets}. \end{definition} \begin{remark} Note that all Borel sets have the Baire property. \end{remark} \subsection{The hierarchy of Borel sets} \gist{Let $\omega_1$ be the first uncountable ordinal.}{} For every $d < \omega_1$, we define by transfinite recursion classes $\Sigma^0_\alpha$ and $\Pi^0_\alpha$ (or $\Sigma^0_\alpha(X)$ and $\Pi^0_\alpha(X)$ for a topological space $X$). Let $X$ be a topological space. Then define \[\Sigma^0_1(X) \coloneqq \{U \overset{\text{open}}{\subseteq} X\},\] \[ \Pi^0_\alpha(X) \coloneqq \lnot \Sigma^0_\alpha(X) \coloneqq \{X \setminus A | A \in \Sigma^0_\alpha(X)\}, \] and for $\alpha > 1$ \[ \Sigma^0_\alpha \coloneqq \{\bigcup_{n < \omega} A_n : A_n \in \Pi^0_{\alpha_n}(X) \text{ for some $\alpha_n < \alpha$}\}. \] Note that $\Pi_1^0$ is the set of closed sets, $\Sigma^0_2 = F_\sigma$, and $\Pi^0_2 = G_\delta$. Furthermore define \[ \Delta^0_\alpha(X) \coloneqq \Sigma^0_\alpha(X) \cap \Pi^0_\alpha(X), \] i.e.~$\Delta^0_1$ is the set of clopen sets. \adjustbox{width=\textwidth, center}{% % https://q.uiver.app/#q=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 \begin{tikzcd} & {\substack{\Sigma_1^0\\\text{open}}} && {\substack{\Sigma^0_2\\F_\sigma}} &&&& {\Sigma^0_\xi} \\ {\substack{\Delta_1^0\\\text{clopen}}} && {\Delta^0_2} && {\Delta^0_3} & \ldots & {\Delta^0_\xi} && {\Delta^0_{\xi + 1}} & \ldots \\ & {\substack{\Pi^0_1\\\text{closed}}} && {\substack{\Pi_2^0\\G_\delta}} &&&& {\Pi^0_\xi} \arrow["\subseteq", hook, from=2-1, to=1-2] \arrow["\subseteq"', hook, from=2-1, to=3-2] \arrow["\subseteq"', hook, from=3-2, to=2-3] \arrow["\subseteq", hook, from=1-2, to=2-3] \arrow["\subseteq", hook, from=2-3, to=1-4] \arrow["\subseteq"', hook, from=2-3, to=3-4] \arrow["\subseteq", hook, from=2-7, to=1-8] \arrow["\subseteq"', hook, from=2-7, to=3-8] \arrow["\subseteq", hook, from=1-4, to=2-5] \arrow["\subseteq"', hook, from=3-4, to=2-5] \arrow["\subseteq", hook, from=1-8, to=2-9] \arrow["\subseteq"', hook, from=3-8, to=2-9] \end{tikzcd}% } \begin{proposition} Let $X$ be a metrizable space. Then \begin{enumerate}[(a)] \item $\Sigma^0_\eta(X) \cup \Pi^0_\eta(X) \subseteq \Delta^0_\xi(X)$ for all $1 \le \eta < \xi < \omega_1$. \item $\cB(X) = \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$. \end{enumerate} \end{proposition} \begin{proof} \gist{% \begin{enumerate}[(a)] \item \begin{observe} \label{ob:sigmasuffices} For all $1 \le \alpha < \beta < \omega_1$, we have $\Pi^0_\alpha(X) \subseteq \Sigma^0_\beta(X)$ by taking ``unions'' of singleton sets. Furthermore $\Sigma^0_\alpha(X) \subseteq \Pi^0_\beta(X)$ by passing to complements. \end{observe} It suffices to show $\Sigma^0_\eta(X) \subseteq \Delta^0_\xi(X)$, since $\Delta^0_\eta(X)$ is closed under complements. Furthermore, it suffices to show $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$, by the observation (since $\Sigma^0_\eta(X) \subseteq \Pi^0_\xi(X)$ and $\Delta^0_\xi(X) = \Sigma^0_\xi(X) \cap \Pi^0_\xi(X)$). So to prove (a) it suffices to show that for all $1 \le \eta < \xi < \omega_1$, we have $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$. For $\eta = 1, \xi = 2$ this holds, since every open set is $F_\sigma$.% \footnote{Here we use that $X$ is metrizable!} % \todo{REF} For $\eta > 1, \xi > \eta$, we have \begin{IEEEeqnarray*}{rCl} \Sigma^0_\eta(X) &=& \{ \bigcup_{n} A_n : A_n \in \Pi^0_{\alpha_n}(X), \alpha_n < \eta\}\\ &\subseteq& \{\bigcup_{n}B_n : B_n \in \Pi^0_{\beta_n}(X), \beta_n < \xi\} = \Sigma^0_\xi(X). \end{IEEEeqnarray*} \item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Sigma^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Pi^0_\alpha(X) = \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$. We need to show that $\cB_0 = \cB(X)$. Clearly $\cB_0 \subseteq \cB(X)$. It suffices to notice that $\cB_0$ is a $\sigma$-algebra containing all open sets. Consider $\bigcup_{n < \omega} A_n$ for some $A_n \in B_0$. Then $A_n \in \Pi^0_{\alpha_n}(X)$ for some $\alpha_n < \omega_1$. Let $\alpha = \sup \alpha_n < \omega_1$. Then $\bigcup_{n < \omega} A_n \in \Sigma^0_\alpha(X)$. It is clear that $\cB_0$ is closed under complements. \end{enumerate} }{ \begin{enumerate}[(a)] \item It suffices to show that $\Sigma^0_\eta(X) \subseteq \Sigma^0_\xi(X)$ for all $1 \le \eta < \xi < \omega_1$. For $\eta = 1, \xi = 2$ this holds, since open sets of a metrizable space are $F_\sigma$. Induction. \item Let $\cB_0 \coloneqq \bigcup_{\alpha < \omega_1} \Delta^0_\alpha(X)$. This is a $\sigma$-algebra containing all open sets. \end{enumerate} } \end{proof} \begin{example} % TODO move to counter examples. Consider the cofinite topology on $\omega_1$. Then the non-empty open sets of this are not $F_\sigma$. \end{example}