w23-logic-3/inputs/tutorial_14.tex

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\tutorial{14}{2024-01-30}{}
\subsection{Sheet 12}
\nr 1
% Examinable
% TODO (there is a more direct way to do it, not using analytic / coanalytic)
\nr 2
% Examinable
\nr 3
% somewhat examinable (for 1.0)
\nr 4
% Examinable!
% RECAP
Let $X$ be a metrizable topological space.
Let $K(X) \coloneqq \{ K \subseteq X : \text{ compact}\}$.
The Vietoris topology has a basis given by
$\{K \subseteq U\}$, $U$ open (type 1)
and $\{K : K \cap U \neq \emptyset\}$, $U$ open (type 2).
The Hausdorff metric on $K(X)$,
$d_H(K,L)$ is the smallest $\epsilon$
such that $K \subseteq B_{\epsilon}(L) \land L \subseteq B_\epsilon(K)$.
This is equal to the maximal point to set distance,
$\max_{a \in A} d(a,B)$.
On previous sheets, we checked that $d_H$ is a metric.
If $X$ is separable, then so is $K(X)$.
% END RECAP
\begin{fact}
Let $(X,d)$ be a complete metric space.
Then so is $(K(X), d_H)$.
\end{fact}
\begin{proof}
We need to show that $(K(X), d_H)$ is complete.
Let $(K_n)_{ n< \omega}$ be Cauchy in $(K(X), d_H)$.
Wlog.~$K_n \neq \emptyset$ for all $n$.
Let $K = \{ x \in X : \forall x \in U \overset{\text{open}}{\subseteq} X.~
\text{ $X$ intersects $K_n$ for infinitely many $n$}\}$.
Equivalently,
$K = \{x : x \text{ is a cluster point of some subsequence $(x_n)$ with $x_n \in K_n$ for all $K_n$}\}$.
(A cluster point is a limit of some subsequence).
\begin{claim}
$K_n \to K$.
\end{claim}
\begin{subproof}
Note that $K$ is closed (the complement is open).
\begin{claim}
$K \neq \emptyset$.
\end{claim}
\begin{subproof}
As $(K_n)$ is Cauchy,
there exists a sequence $(x_n)$ with $x_n \in K_n$
such that there exists a subsequence $(x_{n_i})$
with $d(x_{n_i}, x_{n_{i+1}}) < \frac{1}{2^{i+1}}$.
Let $n_0,n_1,\ldots$
be such that $d_H(K_a, K_b) < 2^{-i-1}$
for $a,b \ge n_i$.
Pick $x_{n_0} \in K_{n_0}$.
Then let $x_{n_{i+1}} \in K_{n_{i+1}}$ be such that
$d(x_{n_i}, x_{n_{i+1}})$ is minimal.
Then $x_{n_i} \xrightarrow{i \to \infty} x$
and we have $x \in K$.
\end{subproof}
\begin{claim}
$K$ is compact.
\end{claim}
\begin{subproof}
We show that $K$ is complete and totally bounded.
Since $K$ is a closed subset of a complete
space, it is complete.
So it suffices to show that $K$ is totally bounded.
Let $\epsilon > 0$
Take $N$ such that $d_H(K_i,K_j) < \epsilon$
for all $i,j \ge N$.
Cover $K_N$ with finitely many $\epsilon$-balls
with centers $z_i$.
Take $x \in K$.
Then the $\epsilon$-ball around $x$ intersects $K_j$
for some $j \ge N$, so
there exists $z_i$ such that $d(x,z_i) < 3\epsilon$.
Note that a subset of a bigger space is totally
bounded iff it is totally bounded in itself.
\end{subproof}
Now we show that $K_n \to K$
in $K(X)$.
Let $\epsilon > 0$.
Take $N$ such that for all $m,n \ge N$,
$d_H(K_m,K_n) < \frac{\epsilon}{2}$.
We'll first show that $\delta(K, K_n) < \epsilon$ for all $n > N$.
Let $x \in K$.
Take $(x_{n_i})$ with $x_{n_i} \in K_{n_i}, x_{n_i} \to x$.
Then for large $i$,
we have $n_i \ge N$ and $d(x_{n_i}, x) < \frac{\epsilon}{2}$.
Take $n \ge N$.
Then there exists $y_n \in K_n$
with $d(y_n, x_{n_i}) < \frac{\epsilon}{2}$.
So $d(x,y_n) < \epsilon$.
Now show that $\delta(K_n, K) < \epsilon$ for all $n \ge N$.
Take $y \in K_n$.
Show that $d(y,K) < \epsilon$.
To do this, construct a sequence of $y_{n_i} \in K_{n_i}$
starting with $y$ such that $d(y_{n_i}, y_{n_{i+1}}) < \frac{\epsilon}{2^{i+2}}$.
(same trick as before).
\end{subproof}
\end{proof}
\begin{fact}
If $X$ is compact metrisable,
then so is $K(X)$.
\end{fact}
\begin{proof}
We have just shown that $X$ is complete.
So it suffices to show that it is totally bounded.
Let $\epsilon > 0$.
Cover $X$ with finitely many $\epsilon$-balls.
Let $F$ be the set of the centers of these balls.
Consider $\cP(F) \setminus \{\emptyset\}$.
Clearly $\{B_x^{d_H} : x \in \cP(F) \setminus \{\emptyset\} \}$
is a finite cover of $K(X)$.
\end{proof}
% TODO complete and totally bounded Sutherland metric and topological spaces