\tutorial{10}{2023-12-19}{Sheet 9}
\subsection{Sheet 9}

\nr 1
$(X, \tau') \xrightarrow{x \mapsto x} (X, \tau)$ is Borel
(by one of the equivalent definitions of being Borel).
Thus $\cB(X, \tau) \subseteq \cB(X, \tau')$
(by the other equivalent definition of being Borel).
Let $U \subseteq (X, \tau')$ be Borel.
$\id\defon{U}$ is injective,
hence $U$ is Borel in $(X, \tau)$ by Lusin-Suslin.



\paragraph{Related stuff}

\begin{fact}
    Let $X,Y$ be Polish.
    $f\colon  X \to Y$
    is Borel iff its graph $\Gamma_f$ is Borel.
\end{fact}
\begin{proof}
    Take a countable open base
    $V_0, V_1, \ldots$ of $Y$.
    Then $\Gamma_f = \{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}$
    (because the space is Hausdorff).
    If $f$ is Borel,
    then clearly the RHS is Borel
    since
    \begin{IEEEeqnarray*}{rCl}
    &&\{(x,y) : \forall n < \omega.~f(x) \in V_n \implies y \in V_n\}\\
    &=& \bigcap_{n < \omega} (f^{-1}(V_n)^{c}Y \cup f^{-1}(V_n) \times V_n\}\\
    \end{IEEEeqnarray*}

    On the other hand suppose that $\Gamma_f$ is Borel.
    Then
    \[
    f^{-1}(B) = \pi_X(X \times B \cap \Gamma_f)
    \]
    is analytic.\footnote{Note that the projection of a Borel set is not necessarily Borel.
    Moreover note that we only used that $\Gamma_f$ is analytic.}
    On the other hand
    \[
    f^{-1}(B)^c = f^{-1}(B^c)
    \]
    is analytic
    and we know that $\Sigma_1^1 \cap \Pi_1^1 = \cB$
    by the \yaref{cor:lusinseparation}.
\end{proof}
In fact we have shown
\begin{fact}
    The following are equivalent
    \begin{itemize}
        \item $f$ is Borel,
        \item $\Gamma_f$ is Borel,
        \item $\Gamma_f$ is analytic.
    \end{itemize}
\end{fact}



\nr 2

\begin{definition}
    Let $X$ be a topological space.
    Let $K(X)$ be the set of all compact subspaces of $X$.
    The \vocab{Vietoris Topology}, $\tau_V$, on $K(X)$
    is the topology with basic open sets
    \[
        [U_0; U_1, \ldots, U_n] = \{K \in K(X) : K \subseteq U_0 \land \forall 1 \le i \le n .~K \cap U_i \neq \emptyset\}
    \]
    for $U_i \overset{\text{open}}{\subseteq} X$.
\end{definition}
\begin{definition}
    Let $(X,d)$ be a matric space with $d \le 1$.
    We define a metric $d_H$ on $K(X)$ as follows:
    $d_H(\emptyset, \emptyset) \coloneqq  0$, $d_H(K, \emptyset) \coloneqq 1$ for $K \neq \emptyset$
    and
    \[
    d_H(K_0, K_1) \coloneqq  \max \{\max_{x \in K_0} d(x,K_1), \max_{x \in K_1} d(x,K_0)\}
    \]
    for $K_0, K_1 \neq \emptyset$.
\end{definition}
\begin{fact}
    $d_H$ is indeed a metric.
\end{fact}
\begin{proof}
    Let $\delta(K, L) \coloneqq \max_{x \in K} d(x,L)$.
    It suffices to show
    $\delta(X,Z) \le \delta(X,Y) + \delta(Y,Z)$,
    since then
    \begin{IEEEeqnarray*}{rCl}
        d_H(X,Z) &\le & \max \{\delta(X,Y) + \delta(Y,Z), \delta(Z,Y) + \delta(Y,X)\}\\
                 &\le & d_H(X,Y) + d_H(Y,Z).
    \end{IEEEeqnarray*}

    Using the fact that $d(\cdot , Z)$ is uniformly continuous,
    specifically
    \[
        |d(x,Z) - d(y,Z)| \le  d(x,y), % TODO REF SHEET 1
    \]
    we get
    \begin{IEEEeqnarray*}{lrCl}
        &d(x,Z) &\le & d(x,y) + d(y, Z)\\
        &       &\le & d(x,y) + \delta(Y,Z)\\
        \implies& d(x,Z) - \delta(Y,Z) &\le & d(x,Y)\\
        \implies& d(x,Z) &\le & \delta(X,Y) + \delta(Y,Z)\\
        \implies & \delta(X,Z) &\le & \delta(X,Y) + \delta(Y,Z).
    \end{IEEEeqnarray*}
\end{proof}

\begin{itemize}
    \item We have
        \begin{IEEEeqnarray*}{rCl}
            d_H(K_0, K_1) < \epsilon & \iff & \max \{\max_{x \in K_0}d(x, K_1), \max_{x \in K_1} d(x,K_0)\} < \epsilon\\
                                     &\iff& \max_{x \in K_0} d(x, K_1) < \epsilon \land \max_{x \in K_1} d(x, K_0) < \epsilon\\
                                     &\iff& K_0 \subseteq B_{\epsilon}(K_1) \land K_1 \subseteq B_\epsilon(K_0).
        \end{IEEEeqnarray*}
    \item Note that a subbase of $\tau_V$
        is given by  $[U]$ and $\langle U \rangle \coloneqq [X;U]$ for $U \overset{\text{open}}{\subseteq} X$.

        Let $K \in [U]$.
        Then $d(\cdot , U^c)\colon  U \to \R_{\ge 0}$
        is always non-zero and continuous.
        So $d(K,U^c)$ attains a minimum  $\epsilon > 0$.
        Then $B_{\epsilon}^H_\epsilon(K) \subseteq U$,
        so $[U]$ is open in $\tau_V$.

        Let $K \in \langle U \rangle$.
        Take some $k \in K \cap U$.
        Then there is some $\epsilon > 0$
        such that $B_\epsilon(k) \subseteq U$.
        Then $K \in B_{\epsilon}^H(K) \subseteq \langle U \rangle$.

        \todo{Other direction}
        % $\tau_H \subseteq \tau_V$

    \item Consider a countable dense subset of $X$.
        Let  $\cK$ be the set of finite subsets of that countable dense subset.
        Then $\cK \subseteq K(X)$ is dense:
        Take $K \in K(X)$ an let $\epsilon > 0$.
        $K$ can be covered with finitely many $\epsilon$-balls
        with centers from the countable dense subsets.
        Let $K' \in \cK$ be the set of the centers.
        Then $d_H(K, K') \le \epsilon$.
\end{itemize}

\nr 3
\begin{itemize}
    \item By transfinite induction we get that $\alpha$ is an ordinal,
        since $\prec$ is well-founded and the supremum of a sets
        of ordinals is an ordinal.
        Since $\rho_{\prec}\colon  X \to \alpha$
        is a surjection, it follows that $\alpha \le |X|$,
        i.e.~$\alpha < |X|^+$.
    \item By induction on $\rho_{\prec_X}(x)$ we show that
        $\rho_{\prec_{X}}(x) \le \rho_{\prec_Y}(f(x))$.
        For $0$ this is trivial.
        Suppose that $\rho_{\prec_{X}}(x) = \alpha$
        and the statement was shown for all $\beta < \alpha$.
        Then
        \begin{IEEEeqnarray*}{rCl}
            \rho_{\prec_Y}(f(x)) &=& \sup \{\rho_{\prec_Y}(y') + 1 | y' \prec f(x)\}\\
                                     &\ge& \sup \{\rho_{\prec_Y}(f(x')) + 1 | f(x') \prec f(x)\}\\
                                     &\ge & \sup \{\rho_{\prec_Y}(f(x')) + 1 | x' \prec x\}\\
                                     &\ge & \sup \{\rho_{\prec_X}(x') + 1 | x' \prec x\}\\
                                     &=& \rho_{\prec_X}(x).
            \end{IEEEeqnarray*}
    \item Infinite branches of $T_\prec$ correspond
        to infinite descending chain of $\prec$,
        hence $T_{\prec}$ is well-founded iff $\prec$ is well-founded.

        % Unwarp$^{\text{tm}}$ definitions.
        Suppose that $\prec$ is well-founded.
        Note that $\rho_T(s)$ depends only on the last element of  $s$,
        as for $s, s' \in T$ with the same last element,
        we have $s \concat x \in T \iff s' \concat x \in T$.

        Let $s = (s_0, \ldots, s_n)$.
        Let us show that $\rho_T(s) = \rho_{\prec}(s_n)$.
        We use induction on $\rho_T(s)$.
        For leaves this is immediate.
        From the last exercise sheet we know that
        \[
            \rho_T(s) = \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}.
        \]
        Hence
        \begin{IEEEeqnarray*}{rCl}
            \rho_T(s) &=& \sup \{\rho_T(s \concat a) + 1 | s \concat a \in T\}\\
                      &=& \sup \{\rho_{\prec}(a) + 1 | s \concat a \in T\}\\
                      &=& \sup \{\rho_{\prec}(a) + 1 | a \prec s_n\}\\
                      &=& \rho_\prec(s_n).
        \end{IEEEeqnarray*}
\end{itemize}



\nr 4

\todo{Next time}

