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

inputs/lecture_02.tex

@@ -130,15 +130,19 @@
          Now let $x \in \bigcap_{n \in \N} V_n$.
          For each $n$ pick $x \in U_n \subseteq X$ open
          satisfying (i), (ii), (iii).
-         W.l.o.g. the $U_n$ are decreasing.
          From (i) and (ii) it follows that $x \in \overline{Y}$,
          since we can consider a sequence of points $y_n \in U_n \cap Y$
          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 $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$.
+         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}