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