35 lines
1.8 KiB
TeX
35 lines
1.8 KiB
TeX
\section{Analytic Sets}
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\label{section:analytic-sets}
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\begin{definition}[Analytic Set]
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\label{definition:analytic-set}
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Let $X$ be a Polish space and $A \subset X$, then $A$ is \textbf{analytic} if there exists a Polish space $Z$ and $f \in C(Z; X)$ such that $f(Z) = A$.
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\end{definition}
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\begin{proposition}
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\label{proposition:analytic-sets}
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Let $X$ be a Polish space, then:
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\begin{enumerate}
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\item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcup_{n \in \natp}A_n$ is analytic.
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\item For each family $\seq{A_n} \subset X$ of analytic sets, $\bigcap_{n \in \natp}A_n$ is analytic.
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\item For any $A \in \cb_X$, $A$ is analytic.
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\end{enumerate}
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\end{proposition}
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\begin{proof}[Proof, {{\cite[Proposition 8.2.1-8.2.3]{CohnMeasure}}}. ]
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For each $n \in \natp$, let $Z_n$ be a Polish space and $f_n \in C(Z_n; X)$ such that $A_n = f_n(Z_n)$.
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(1): By \autoref{proposition:polish-space-extension}, $Z = \bigsqcup_{n \in \natp}Z_n$ is a Polish space. Let $f \in C(Z; X)$ be the gluing of $\seq{f_n}$, then $\bigcup_{n \in \natp}A_n = f(Z)$.
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(2): By \autoref{proposition:polish-space-extension}, $Z = \prod_{n \in \natp}Z_n$ is a Polish space. For each $m, n \in \natp$, $\bracs{f_m \circ \pi_m = f_n \circ \pi_n}$ is closed. Thus
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\[
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\Delta := \bigcap_{m \in \natp}\bigcap_{n \in \natp}\bracs{f_m \circ \pi_m = f_n \circ \pi_n}
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\]
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is closed, and Polish by \autoref{proposition:polish-space-extension}. Hence $\bigcap_{n \in \natp}A_n = f_1 \circ \pi_1(\Delta)$ is also analytic.
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(3): By \autoref{proposition:polish-space-extension}, every open and closed subset of $X$ is analytic. Thus (1), (2), and \autoref{lemma:monotone-borel-characterisation} imply that every element of $\cb_X$ is analytic.
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\end{proof}
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