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Josia Pietsch 2024-01-12 01:44:05 +01:00
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@ -144,8 +144,9 @@
\begin{proof}
Consider $(F, \cT\defon{F})$ and $(X \setminus F, \cT\defon{X \setminus F})$.
Both are Polish spaces.
Take the coproduct\footnote{topological sum} $F \oplus (X \setminus F)$
of these spaces.
Take the coproduct%
\footnote{In the lecture, this was called the \vocab{topological sum}.}
$F \oplus (X \setminus F)$ of these spaces.
This space is Polish,
and the topology is generated by $\cT \cup \{F\}$,
hence we do not get any new Borel sets.

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@ -26,7 +26,7 @@
Let $X$ be a Polish space.
A set $A \subseteq X$
is called \vocab{analytic}
if
iff
\[
\exists Y \text{ Polish}.~\exists B \in \cB(Y).~
\exists \underbrace{f\colon Y \to X}_{\text{continuous}}.~