\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