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\begin{proof}
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Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
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Both are Polish spaces.
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Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$
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of these spaces.
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Take the coproduct%
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\footnote{In the lecture, this was called the \vocab{topological sum}.}
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$F \oplus (X \setminus F)$ of these spaces.
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This space is Polish,
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and the topology is generated by $\cT \cup \{F\}$,
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hence we do not get any new Borel sets.
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Let $X$ be a Polish space.
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A set $A \subseteq X$
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is called \vocab{analytic}
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if
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iff
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\[
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\exists Y \text{ Polish}.~\exists B \in \cB(Y).~
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\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~
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