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@ -28,7 +28,7 @@ i.e.~it is really a topological property.
\end{example} \end{example}
Polish spaces behave very nicely. Polish spaces behave very nicely.
We will see that uncountable polish spaces have size $2^{\aleph_0}$. % TODO: mathfrak c for continuum We will see that uncountable polish spaces have size $2^{\aleph_0}$.
There are good notions of big (comeager) There are good notions of big (comeager)
and small (meager). and small (meager).

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@ -130,19 +130,15 @@
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 $y_n \xrightarrow{d} x$.
For all $n$ we have that $U_n' \coloneqq U_1 \cap \ldots \cap U_n$ On the other hand $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$,
is an open set containing $x$, so the $y_n$ form a Cauchy sequence with respect to $d_Y$,
hence $U_n' \cap Y \neq \emptyset$. since $\diam(U_n \cap Y) \xrightarrow{d_Y} 0$,
Thus we may assume that the $U_i$ form a decreasing sequence. hence $\diam(\overline{U_n \cap Y}) \xrightarrow{d_Y} 0$.
We have that $\diam_{d_Y}(U_n \cap Y) \le \frac{1}{n}$. $y_n$ converges to the unique point in $\bigcap_{n} \overline{U_n \cap Y}$.
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 $x$. Since the topologies agree, this point is $x$.
\end{refproof} \end{refproof}
\end{refproof} \end{refproof}