lecture 05
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@ -119,7 +119,6 @@
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The complement of a meager set is called
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\vocab{comeager}.
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\end{definition}
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\begin{example}
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$\Q \subseteq \R$ is meager.
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Let $A, B \subseteq X$.
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We write $A =^\ast B$
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iff the \vocab{symmetric difference},
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$A \mathop{\triangle} B \coloneqq (A\setminus B) \cup (B \setminus A)$,
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$A \symdif B \coloneqq (A\setminus B) \cup (B \setminus A)$,
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is meager.
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\end{notation}
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\begin{remark}
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@ -144,13 +143,14 @@ Note that open sets and meager sets have the Baire property.
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% \begin{example}
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% $\Q \subseteq \R$ is $F_\sigma$.
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%
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% $\R \setminus \Q \subseteq \R$ is $G_\delta$.
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%
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% $\Q \subseteq \R$ is not $G_{\delta}$.
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% (It is dense and meager,
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% hence it can not be $G_\delta$,
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% by the Baire category theorem).
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% \end{example}
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\begin{example}
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\begin{itemize}
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\item $\Q \subseteq \R$ is $F_\sigma$.
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\item $\R \setminus \Q \subseteq \R$ is $G_\delta$.
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\item $\Q \subseteq \R$ is not $G_{\delta}$.
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(It is dense and meager,
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hence it can not be $G_\delta$
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by the Baire category theorem).
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\end{itemize}
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\end{example}
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271
inputs/lecture_05.tex
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271
inputs/lecture_05.tex
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\lecture{05}{2023-10-31}{}
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\begin{fact}
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A set $A$ is nwd iff $\overline{A}$ is nwd.
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If $F$ is closed then
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$F$ is nwd iff $X \setminus F$ is open and dense.
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Any meager set $B$ is contained in
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a meager $F_{\sigma}$-set.
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\end{fact}
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\begin{definition}
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A \vocab{$\sigma$-algebra} on a set $X$
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is a collection of subsets of $X$
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such that:
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\begin{itemize}
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\item $\emptyset, X \in \cA$,
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\item $ A \in \cA \implies X \setminus A \in \cA$,
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\item $(A_i)_{i < \omega}, A_i \in \cA \implies \bigcup_{i < \omega} A_i \in \cA$.
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\end{itemize}
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\end{definition}
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\begin{fact}
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Since $\bigcap_{i < \omega} A_i = \left( \bigcup_{i < \omega} A_i^c \right)^c$
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we have that $\sigma$-algebras are closed under countable intersections.
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\end{fact}
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\begin{theorem}
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\label{thm:bairesigma}
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Let $X$ be a topological space.
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Then the collection of sets with the Baire property
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is a $\sigma$-algebra on $X$.
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It is the smallest $\sigma$-algebra
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containing all meager and open sets.
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\end{theorem}
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\begin{refproof}{thm:bairesigma}
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Let $\cA$ be the collection of sets with the Baire property.
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Since open sets have the Baire property,
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we have $\emptyset, X \in \cA$.
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Let $A_n \in \cA$ for all $n < \omega$.
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Take $U_n$ such that $A_n \symdif U_n$ is meager.
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Then
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\[
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\left( \bigcup_{n < \omega} A_n \right) \symdif \left( \bigcup_{n < \omega} U_n \right)
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\]
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is meager,\todo{small exercise}
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hence $\bigcup_{n < \omega} A_n \in \cA$.
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Let $A \in \cA$.
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Take some open $U$ such that $U \symdif A$ is meager.
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We have $(X \setminus U) \symdif (X \setminus A) = U \symdif A$.
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\begin{claim}
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\label{thm:bairesigma:c1}
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If $F$ is closed,
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then $F \setminus \inter(F)$
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is nwd.
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In particular, $F \symdif \inter(F)$ is nwd.
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\end{claim}
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\begin{refproof}{thm:bairesigma:c1}
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\todo{TODO}
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\end{refproof}
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From the claim we get that
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$X \setminus A =^\ast X \setminus U =^\ast \inter(X \setminus U)$.
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Hence $X \setminus A \in \cA$.
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It is clear that all meager sets have the Baire property.
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Let $A \in \cA$. Then $A = (A \setminus U) \cup (A \cap U)$
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for some open $U$
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such that $A \setminus U$ is meager.
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We have $A \cap U = U \setminus (U \setminus A)$.
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Thus we get that $\cA$ is the minimal $\sigma$-algebra
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containing all meager and all open sets.
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\end{refproof}
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%\begin{example}
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% Nwd set of positive measure.
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% TODO
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% remove open intervals such that their length does not add to 0
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%
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%\end{example}
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\begin{theorem}[Baire Category theorem]
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Let $X$ be a completely metrizable space.
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Then every comeager set of $X$ is dense in $X$.
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\end{theorem}
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\todo{Proof (copy from some other lecture)}
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\begin{proposition}
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Let $X$ be a topological space.
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The following are equivalent:
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\begin{enumerate}[(i)]
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\item Every nonempty open set
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is non-meager in $X$.
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\item Every comeager set is dense.
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\item The intersection of countable many
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open dense sets is dense.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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\todo{Proof (short)}
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(iii) $\implies$ (i)
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Let us first show that $X$ is non-meager.
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Suppose that $X$ is meager. Then $X = \bigcup_{n} A_n = \bigcup_{n} \overline{A_n}$
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is the countable union of nwd sets.
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We have that
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\[
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\emptyset = \bigcap_{n} (X \setminus \overline{A_n})
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\]
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is dense by (iii).
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This proof can be adapted to other open sets $X$.
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\end{proof}
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% TODO Fubini
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\begin{notation}
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Let $X ,Y$ be topological spaces,
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$A \subseteq X \times Y$,
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$x \in X, y \in Y$.
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Let
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\[
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A_x \coloneqq \{y \in Y : (x,y) \in A\}
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\]
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and
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\[
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A^y \coloneqq \{x \in X : (x,y) \in A\} .
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\]
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\end{notation}
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The following similar to Fubini,
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but for meager sets:
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\begin{theorem}[Kuratowski-Ulam]
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\label{thm:kuratowskiulam}
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Let $X,Y$ be second countable topological spaces.
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Let $A \subseteq X \times Y$
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be a set with the Baire property.
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Then
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\begin{enumerate}[(i)]
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\item $\{x \in X : A_x \text{ has the BP }\}$
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is comeager\footnote{Note that not necessarily all sections
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have the BP. For example $\{x\} \times Y$ is meager in $X \times Y$}
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and similarly for $y$.
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\item $A$ is meager
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\begin{IEEEeqnarray*}{rll}
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\iff &\{x \in X : A_x \text{ is meager }\}&\text{ is comeager}\\
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\iff &\{y \in Y : A^y \text{ is meager }\}& \text{ is comeager}.
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\end{IEEEeqnarray*}
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\item $A$ is comeager
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\begin{IEEEeqnarray*}{rll}
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\iff & \{x \in X: A_x \text{ is comeager }\} &\text{ is comeager}\\
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\iff & \{y \in Y: A^y \text{ is comeager}\} & \text{ is comeager}.
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\end{IEEEeqnarray*}
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\end{enumerate}
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\end{theorem}
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\begin{refproof}{thm:kuratowskiulam}
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(ii) and (iii) are equivalent by passing to the complement.
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\begin{claim}%[1a]
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\label{thm:kuratowskiulam:c1a}
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If $F \overset{\text{closed}}{\subseteq} X \times Y$
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is nwd,
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then
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\[
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\{x \in X : F_x \text{is nwd}\}
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\]
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is comeager.
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\end{claim}
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\begin{refproof}{thm:kuratowskiulam:c1a}
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Put $W = F^c$.
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This is open and dense in $X \times Y$.
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It suffices to show that $\{x \in X : W_x \text{ is dense}\}$
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is comeager.
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Note that $W_x$ is open for all $x$.
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Fix a countable basis $(V_n)$ of $Y$
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with $V_n$ non-empty.
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We want to show that
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\[
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\{x \in X: \forall n.~ (W_x \cap V_n) \neq \emptyset\}
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\]
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is a comeager set.
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This is equivalent to
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\[
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\{x \in X : (W_x \cap V_n) \neq \emptyset\}
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\]
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being comeager for all $n$,
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because the intersection
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of countably many comeager sets is comeager.
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Fix $n$ and let $U_n \coloneqq \{x \in X: (W_x \cap V_n) = \emptyset\}$.
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We will show that $U_n$ is open and dense,
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hence it is comeager.
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$U_n = \proj_x(W \cap (X \times V_n))$ is open
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since projections of open sets are open.
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Let $U \subseteq X$ be nonempty and open.
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We need to show that $U \cap U_n \neq \emptyset$.
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It is
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\[
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U \cap U_n = \proj_x(W \cap (U \times V_n))
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\]
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nonempty since $W$ is dense.
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\end{refproof}
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\begin{claim} % [1a']
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\label{thm:kuratowskiulam:c1ap}
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If $F \subseteq X \times Y$
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is nwd,
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then
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\[
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\{x \in X : F_x \text{is nwd}\}
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\]
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is comeager.
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\end{claim}
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\begin{refproof}{thm:kuratowskiulam:c1ap}
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We have that $\overline{F}$ is nwd.
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Hence by \yaref{thm:kuratowskiulam:c1a}
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the set
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\[
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\{x \in X: \overline{F_x} \text{ is nwd}\} \subseteq
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\{x \in X: F_x \text{ is nwd}\}
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\]
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is comeager.
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\end{refproof}
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\begin{claim}% [1b]
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\label{thm:kuratowskiulam:c1b}
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If $M \subseteq X \times Y$ is meager,
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then
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\[
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\{x \in X : M_x \text{ is meager}\}
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\]
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is comeager.
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\end{claim}
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\begin{refproof}{thm:kuratowskiulam:c1b}
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This follows from \yaref{thm:kuratowskiulam:c1ap}:
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Let $M = \bigcup_{n < \omega} F_n$
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where the $F_n$ are nwd.
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Apply \yaref{thm:kuratowskiulam:c1ap}
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to each $F_n$.
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We get that
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$M_x$ is comeager
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as a countable intersection of comeager sets.
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\end{refproof}
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\todo{Finish proof}
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\phantom\qedhere
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\end{refproof}
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\begin{remark}
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Suppose that $A$ has the BP.
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Then there is an open $U$ such that
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$A \symdif U \mathbin{\text{\reflectbox{$\coloneqq$}}} M$ is meager.
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Then $A = U \symdif M$.
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\end{remark}
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@ -131,5 +131,7 @@
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%\mathbin{\raisebox{1ex}{\scalebox{.7}{$\frown$}}}%
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\DeclareMathOperator{\hght}{height}
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\DeclareMathOperator{\symdif}{\triangle}
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\DeclareSimpleMathOperator{proj}
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\newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}}
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\input{inputs/lecture_02}
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\input{inputs/lecture_03}
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\input{inputs/lecture_04}
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\input{inputs/lecture_05}
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