lecture 02 fixed proof of 'polish subspace of polish space is G_delta'
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@ -130,15 +130,19 @@
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Now let $x \in \bigcap_{n \in \N} V_n$.
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For each $n$ pick $x \in U_n \subseteq X$ open
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satisfying (i), (ii), (iii).
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W.l.o.g. the $U_n$ are decreasing.
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From (i) and (ii) it follows that $x \in \overline{Y}$,
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since we can consider a sequence of points $y_n \in U_n \cap Y$
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and get $y_n \xrightarrow{d} x$.
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On the other hand $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$,
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so the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$,
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hence $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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$y_n$ converges to the unique point in $\bigcap_{n} \overline{U_n \cap Y}$.
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For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
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is an open set containing $x$,
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hence $U_n' \cap Y \neq \emptyset$.
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Thus we may assume that the $U_i$ form a decreasing sequence.
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We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
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If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
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since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
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and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
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The sequence $y_n$ converges to the unique point in
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$\bigcap_{n} \overline{U_n \cap Y}$.
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Since the topologies agree, this point is $x$.
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\end{refproof}
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\end{refproof}
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