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Josia Pietsch 2024-02-10 05:20:35 +01:00
parent 6c5eda59e4
commit 3df55b6516
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4 changed files with 2 additions and 5 deletions

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@ -91,7 +91,6 @@
because $\tilde{f_U}$ is continuous.
It is closed}{} in $X \times \R$ \gist{because
$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
\todo{Make this precise}
Therefore we identified $U$ with a closed subspace of
the Polish space $(X \times \R, d_1)$.

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@ -96,7 +96,7 @@
Polish space and $\cC = 2^\omega$ the Cantor space,
then they are Borel isomorphic.
There is $2^\omega \hookrightarrow X$ Borel
(continuous wrt.~to the topology of $X$)
(continuous wrt.~the topology of $X$)
On the other hand
\[
X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC

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@ -119,5 +119,3 @@
Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
hence $f$ is open.
\end{enumerate}

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@ -95,7 +95,7 @@ for some $B_i \in \cB(Y_i)$.
Then $\bigcap A_i$ is the image of $D$
under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
\paragraph{Other solution}
\emph{Other solution:}
Let $F_n \subseteq X \times \cN$ be closed,
and $C \subseteq X \times \cN^{\N}$ defined by