\tutorial{14}{2024-01-30}{}

\subsection{Sheet 12}
\nr 1
% Examinable

Let $\LO(\N) \overset{\text{closed}}{\subseteq} 2^{\N\times \N}$ denote the set of linear orders on $\N$.

Let $S \subseteq \LO(\N)$ be the set of orders having a least
element and such that every element has an immediate successor.
\begin{itemize}
    \item $S$ is Borel in $\LO(\N)$:

        Let $M_n \subseteq \LO(\N)$ be the set of orders with minimal element $n$.
        Let $I_{n,m} \subseteq \LO(\N)$ be the set of orders such
        that $m$ is the immediate successor of $n$.

        Clearly $S = \left(\bigcap_n \bigcup_{m\neq n} I_{n,m}\right) \cap \bigcup_n M_n$,
        so it suffices to show that $M_n$ and $I_{n,m}$ are Borel.
        It is $M_n = \bigcap_{m\neq n} \{\prec : m \not\prec n\}$
        and $I_{n,m} = \{\prec: n \prec m\} \cap \bigcap_{i} \{\prec : n \preceq i \preceq m \implies n = i \lor n = m \}$.
    \item Give an example of an element of  $S$ which is not well-ordered:

        Consider $\{1 - \frac{1}{n} : n \in \N^+\} \cup \{1 +   \frac{1}{n} : n \in \N^{+}\} \subseteq \R$
        with the order $<_\R$.
        This is an element of $S$,
        but $\{x \in S: x \ge 1\}$ has no minimal element,
        hence it is not well-ordered.

\end{itemize}

\nr 2
% Examinable
Recall the definition of the circle shift flow $(\R / \Z, \Z)$
with parameter $\alpha \in \R$, $1 \cdot x \coloneqq  x + \alpha$.

\begin{itemize}
    \item If $\alpha \not\in  \Q$, then $(\R / \Z, \Z)$ is minimal:

        This is known as \href{https://en.wikipedia.org/wiki/Dirichlet's_approximation_theorem}{Dirichlet's Approximation Theorem}.

    \item Consider $\R/\Z$ as a topological group.
        Any subgroup $H$ of $\R / \Z$ is dense in $\R / \Z$
        or of the form $H = \{ x \in \R / \Z | mx = 0\}$
        for some $m \in \Z$.


        If $H$ contains an irrational element $\alpha$, then
        it is dense by the previous point.

        Suppose that $H \subseteq \Q / \Z$.
        Let $D$ be the set of denominators of elements of $H$
        written as irreducible fractions.
        If $D$ is finite,
        then  $H = \{x \in \R / \Z : \mathop{lcm}(D)x = 0\}$.
        Otherwise $H$ is dense, as it contains
        elements of arbitrarily large denominator.
\end{itemize}


\nr 3
% somewhat examinable (for 1.0)
% TODO

\begin{enumerate}[(a)]
    \item $(X,T)$ is distal iff it does not have a proximal pair,
        i.e.~$a\neq b$, $c$ such that $t_n \in T$,
        $t_na, t_nb \to c$.

        Equivalently,
        for all $a,b$ there exists an  $\epsilon$,
        such that for all $t \in T$, $d(ta,tb) > \epsilon$.


    \item \todo{TODO}% TODO (not too hard)
        % (b)
        % Let $(X,T)$ be distal with a dense orbit,
        % then it is distal minimal.
        % Sheet 8: has dense orbit is Borel
        % Distal flow decomposes into distal minimal flows.
\end{enumerate}


\nr 4

% Examinable!
% TODO THINK!

\gist{%
% RECAP
Let $X$ be a metrizable topological space
and let $K(X) \coloneqq  \{ K \subseteq  X : K \text{ compact}\}$.

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}
    \label{fact:s12e4}
Let $(X,d)$ be a complete metric space.
Then so is  $(K(X), d_H)$.
\end{fact}
\begin{refproof}{fact:s12e4}
    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{ $U \cap K_n \neq \emptyset$ 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}
        \label{fact:s12e4:c1}
        $K_n \to K$.
    \end{claim}
    \begin{refproof}{fact:s12e4:c1}
    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{refproof}
\end{refproof}

\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