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lecture 02 fixed proof of 'polish subspace of polish space is G_delta'

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Josia Pietsch 2023-10-17 18:25:37 +02:00
parent 66b7b06200
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inputs/lecture_02.tex
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@ -96,7 +96,7 @@
\end{IEEEeqnarray*} \end{IEEEeqnarray*}
As for an open $U$, $f_Y$ is an embedding. As for an open $U$, $f_Y$ is an embedding.
Since $X \times \R^{\N}$ Since $X \times \R^{\N}$
is completely metrizable, is completely metrizable,
so is the closed set $f_Y(Y) \subseteq X \times \R^\N$. so is the closed set $f_Y(Y) \subseteq X \times \R^\N$.
@ -122,7 +122,7 @@
$U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$, $U_1$ (open in $Y)$ and $U_2$ (open in $X$ ) of $x$,
such that $\diam_{d_Y}(U) < \frac{1}{n}$ such that $\diam_{d_Y}(U) < \frac{1}{n}$
and $U_2 \cap Y = U_1$. and $U_2 \cap Y = U_1$.
Additionally choose $x \in U_3$ open in $X$ Additionally choose $x \in U_3$ open in $X$
with $\diam_{d}(U_3) < \frac{1}{n}$. with $\diam_{d}(U_3) < \frac{1}{n}$.
Then consider $U_2 \cap U_3 \subseteq V_n$. Then consider $U_2 \cap U_3 \subseteq V_n$.
Hence $Y \subseteq \bigcap_{n \in \N} V_n$. Hence $Y \subseteq \bigcap_{n \in \N} V_n$.
@ -130,16 +130,20 @@
Now let $x \in \bigcap_{n \in \N} V_n$. Now let $x \in \bigcap_{n \in \N} V_n$.
For each $n$ pick $x \in U_n \subseteq X$ open For each $n$ pick $x \in U_n \subseteq X$ open
satisfying (i), (ii), (iii). satisfying (i), (ii), (iii).
W.l.o.g. the $U_n$ are decreasing.
From (i) and (ii) it follows that $x \in \overline{Y}$, From (i) and (ii) it follows that $x \in \overline{Y}$,
since we can consider a sequence of points $y_n \in U_n \cap Y$ since we can consider a sequence of points $y_n \in U_n \cap Y$
and get $y_n \xrightarrow{d} x$. and get $y_n \xrightarrow{d} x$.
On the other hand $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$, For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$
so the $y_n$ form a Cauchy sequence with respect to $d_Y$, is an open set containing $x$,
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$, hence $U_n' \cap Y \neq \emptyset$.
hence $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$. Thus we may assume that the $U_i$ form a decreasing sequence.
$y_n$ converges to the unique point in $\bigcap_{n} \overline{U_n \cap Y}$. We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
Since the topologies agree, this point is $x$. If follows that the $y_n$ form a Cauchy sequence with respect to $d_Y$,
since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$
and thus $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
The sequence $y_n$ converges to the unique point in
$\bigcap_{n} \overline{U_n \cap Y}$.
Since the topologies agree, this point is $x$.
\end{refproof} \end{refproof}
\end{refproof} \end{refproof}