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@ -91,7 +91,6 @@
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because $\tilde{f_U}$ is continuous.
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It is closed}{} in $X \times \R$ \gist{because
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$\tilde{f_U} \to \infty$ for $d(x, U^c) \to 0$}{}.
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\todo{Make this precise}
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Therefore we identified $U$ with a closed subspace of
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the Polish space $(X \times \R, d_1)$.
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@ -96,7 +96,7 @@
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Polish space and $\cC = 2^\omega$ the Cantor space,
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then they are Borel isomorphic.
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There is $2^\omega \hookrightarrow X$ Borel
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(continuous wrt.~to the topology of $X$)
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(continuous wrt.~the topology of $X$)
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On the other hand
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\[
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X \hookrightarrow\cN \overset{\text{continuous embedding\footnotemark}}{\hookrightarrow}\cC
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@ -119,5 +119,3 @@
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Then $f(B) = U_{(x_0,\ldots,x_{n-1})}$ is open,
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hence $f$ is open.
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\end{enumerate}
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@ -95,7 +95,7 @@ for some $B_i \in \cB(Y_i)$.
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Then $\bigcap A_i$ is the image of $D$
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under $Z \xrightarrow{(y_n) \mapsto f_0(y_0)} X$.
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\paragraph{Other solution}
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\emph{Other solution:}
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Let $F_n \subseteq X \times \cN$ be closed,
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and $C \subseteq X \times \cN^{\N}$ defined by
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