\lecture{05}{2023-10-31}{} \begin{fact} \begin{itemize} \item A set $A$ is nwd iff $\overline{A}$ is nwd. \item If $F$ is closed then $F$ is nwd iff $X \setminus F$ is open and dense. \item Any meager set $B$ is contained in a meager $F_{\sigma}$-set. \end{itemize} \end{fact} \gist{% \begin{proof} \begin{itemize} \item This follows from the definition as $\overline{\overline{A}} = \overline{A}$. \item Trivial. \item Let $B = \bigcup_{n < \omega} B_n$ be a union of nwd sets. Then $B \subseteq \bigcup_{n < \omega} \overline{B_n}$. \end{itemize} \end{proof} }{} \gist{% \begin{definition} A \vocab{$\sigma$-algebra} on a set $X$ is a collection of subsets of $X$ such that: \begin{itemize} \item $\emptyset, X \in \cA$, \item $ A \in \cA \implies X \setminus A \in \cA$, \item $(A_i)_{i < \omega}, A_i \in \cA \implies \bigcup_{i < \omega} A_i \in \cA$. \end{itemize} \end{definition} \begin{fact} Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$ we have that $\sigma$-algebras are closed under countable intersections. \end{fact} }{} \begin{theorem} \label{thm:bairesigma} Let $X$ be a topological space. Then the collection of sets with the Baire property is \gist{a $\sigma$-algebra on $X$. It is}{} the smallest $\sigma$-algebra containing all meager and open sets. \end{theorem} \begin{refproof}{thm:bairesigma} Let $\cA$ be the collection of sets with the Baire property. Since open sets have the Baire property, we have $\emptyset, X \in \cA$. Let $A_n \in \cA$ for all $n < \omega$. Take $U_n$ such that $A_n \symdif U_n$ is meager. Then \[ \left( \bigcup_{n < \omega} A_n \right) \symdif \left( \bigcup_{n < \omega} U_n \right) \] is meager, hence $\bigcup_{n < \omega} A_n \in \cA$. Let $A \in \cA$. Take some open $U$ such that $U \symdif A$ is meager. We have $(X \setminus U) \symdif (X \setminus A) = U \symdif A$. \begin{claim} \label{thm:bairesigma:c1} If $F$ is closed, then $F \setminus \inter(F)$ is nwd. In particular, $F \symdif \inter(F)$ is nwd. \end{claim} \begin{refproof}{thm:bairesigma:c1} $F \setminus \inter(F)$ is closed, hence $\overline{F \setminus \inter(F)} = F \setminus \inter(F)$. Clearly $\inter(F\setminus\inter(F)) = \emptyset$. \end{refproof} From the claim we get that $X \setminus A =^\ast X \setminus U =^\ast \inter(X \setminus U)$. Hence $X \setminus A \in \cA$. It is clear that all meager sets have the Baire property. Let $A \in \cA$. Then $A = (A \setminus U) \cup (A \cap U)$ for some open $U$ such that $A \setminus U$ is meager. We have $A \cap U = U \setminus (U \setminus A)$. Thus we get that $\cA$ is the minimal $\sigma$-algebra containing all meager and all open sets. \end{refproof} %\begin{example} % Nwd set of positive measure. % TODO % remove open intervals such that their length does not add to 0 % %\end{example} \begin{theorem}[Baire Category theorem] \yalabel{Baire Category Theorem}{Baire Category Theorem}{thm:bct} Let $X$ be a completely metrizable space. Then every comeager set of $X$ is dense in $X$. \end{theorem} \gist{% \todo{Proof (copy from some other lecture)} }{Not proved in the lecture.} \begin{theoremdef} Let $X$ be a topological space. The following are equivalent: \begin{enumerate}[(i)] \item Every nonempty open set is non-meager in $X$. \item Every comeager set is dense. \item The intersection of countable many open dense sets is dense. \end{enumerate} In this case $X$ is called a \vocab{Baire space}.% \footnote{cf.~\yaref{s5e1}} \end{theoremdef} \begin{proof} (i) $\implies$ (ii) \gist{% Consider a comeager set $A$. Let $U\neq \emptyset$ be any open set. Since $U$ is non-meager, we have $A \cap U \neq \emptyset$. }{The intersection of a comeager and a non-meager set is nonempty.} (ii) $\implies$ (iii) The complement of an open dense set is nwd. \gist{% Hence the intersection of countable many open dense sets is comeager. }{} (iii) $\implies$ (i) Let us first show that $X$ is non-meager. Suppose that $X$ is meager. Then $X = \bigcup_{n} A_n = \bigcup_{n} \overline{A_n}$ is the countable union of nwd sets. We have that \[ \emptyset = \bigcap_{n} (X \setminus \overline{A_n}) \] is dense by (iii). This proof can be adapted to other open sets $X$. \end{proof} \begin{notation} Let $X ,Y$ be topological spaces, $A \subseteq X \times Y$ and $x \in X, y \in Y$. Let \[ A_x \coloneqq \{y \in Y : (x,y) \in A\} \] and \[ A^y \coloneqq \{x \in X : (x,y) \in A\} . \] \end{notation} \gist{% The following similar to Fubini, but for meager sets: }{} \begin{theorem}[Kuratowski-Ulam] \yalabel{Kuratowski-Ulam}{Kuratowski-Ulam}{thm:kuratowskiulam} Let $X,Y$ be second countable topological spaces. Let $A \subseteq X \times Y$ be a set with the Baire property.% \footnote{It is important that $A$ has the Baire property (cf. \yaref{s5e4}).} Then \begin{enumerate}[(i)] \item $\{x \in X : A_x \text{ has the BP }\}$ is comeager\footnote{Note that not necessarily all sections have the BP. For example $\{x\} \times Y$ is meager in $X \times Y$} and similarly for $y$. \item $A$ is meager \begin{IEEEeqnarray*}{rll} \iff &\{x \in X : A_x \text{ is meager}\}&\text{ is comeager}\\ \iff &\{y \in Y : A^y \text{ is meager}\}& \text{ is comeager}. \end{IEEEeqnarray*} \item $A$ is comeager \begin{IEEEeqnarray*}{rll} \iff & \{x \in X: A_x \text{ is comeager}\} &\text{ is comeager}\\ \iff & \{y \in Y: A^y \text{ is comeager}\} & \text{ is comeager}. \end{IEEEeqnarray*} \end{enumerate} \end{theorem} \begin{refproof}{thm:kuratowskiulam} \gist{ (ii) and (iii) are equivalent by passing to the complement. \begin{claim}%[1a] \label{thm:kuratowskiulam:c1a} If $F \overset{\text{closed}}{\subseteq} X \times Y$ is nwd, then \[ \{x \in X : F_x \text{is nwd}\} \] is comeager. \end{claim} \begin{refproof}{thm:kuratowskiulam:c1a} Put $W = F^c$. This is open and dense in $X \times Y$. It suffices to show that $\{x \in X : W_x \text{ is dense}\}$ is comeager. Note that $W_x$ is open for all $x$. Fix a countable basis $(V_n)$ of $Y$ with $V_n$ non-empty. We want to show that \[ \{x \in X: \forall n.~ (W_x \cap V_n) \neq \emptyset\} \] is a comeager set. This is equivalent to \[ \{x \in X : (W_x \cap V_n) \neq \emptyset\} \] being comeager for all $n$, because the intersection of countably many comeager sets is comeager. Fix $n$ and let $U_n \coloneqq \{x \in X: (W_x \cap V_n) = \emptyset\}$. We will show that $U_n$ is open and dense, hence it is comeager. $U_n = \proj_x(W \cap (X \times V_n))$ is open since projections of open sets are open. Let $U \subseteq X$ be nonempty and open. We need to show that $U \cap U_n \neq \emptyset$. It is \[ U \cap U_n = \proj_x(W \cap (U \times V_n)) \] nonempty since $W$ is dense. \end{refproof} \begin{claim} % [1a'] \label{thm:kuratowskiulam:c1ap} If $F \subseteq X \times Y$ is nwd, then \[ \{x \in X : F_x \text{is nwd}\} \] is comeager. \end{claim} \begin{refproof}{thm:kuratowskiulam:c1ap} We have that $\overline{F}$ is nwd. Hence by \yaref{thm:kuratowskiulam:c1a} the set \[ \{x \in X: \overline{F_x} \text{ is nwd}\} \subseteq \{x \in X: F_x \text{ is nwd}\} \] is comeager. \end{refproof} \begin{claim}% [1b] \label{thm:kuratowskiulam:c1b} If $M \subseteq X \times Y$ is meager, then \[ \{x \in X : M_x \text{ is meager}\} \] is comeager. \end{claim} \begin{refproof}{thm:kuratowskiulam:c1b} This follows from \yaref{thm:kuratowskiulam:c1ap}: Let $M = \bigcup_{n < \omega} F_n$ where the $F_n$ are nwd. Apply \yaref{thm:kuratowskiulam:c1ap} to each $F_n$. We get that $M_x$ is comeager as a countable intersection of comeager sets. \end{refproof} }{} % \phantom\qedhere % \end{refproof} % TODO fix claim numbers