\lecture{15}{2023-12-05}{}

\begin{theorem}[Boundedness Theorem]
    \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness}
    \gist{%
        Let $X$ be Polish, $C \subseteq X$ coanalytic,
        $\phi\colon  C \to \omega_1$ a coanalytic rank on $C$,
        $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$.
    }{%
        Let $X$ Polish, $C \in \Pi^1_1(X)$, $A \in \Sigma^1_1(X)$,
        $A \subseteq C$, $\phi\colon C \to \omega_1$ a coanalytic rank.
    }
    Then $\sup \{\phi(x) : x \in A\} < \omega_1$.

    Moreover for all $\xi < \omega_1$,
    \[
    D_\xi \coloneqq  \{x \in C : \phi(x) < \xi\}
    \]
    and
    \[
    E_\xi \coloneqq  \{x \in C : \phi(x) \le \xi\}
    \]
    are Borel subsets of $X$.
\end{theorem}
\begin{proof}
    Let
    \begin{IEEEeqnarray*}{rCl}
        x \prec y&:\iff& x,y \in A \land \phi(x) < \phi(y)\\
                 &\iff& x,y \in A \land y \not\le_\phi^\ast x.
    \end{IEEEeqnarray*}
    Since $A$ is analytic,
    this relation is analytic and wellfounded on $X$.
    By \yaref{thm:kunenmartin}
    we get $\rho(\prec) < \omega_1$.
    Thus $\sup \{\phi(x) : x \in A\} < \omega_1$.

    Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$,
    it suffices to check $E_\xi \in \Sigma_1^1(X)$.
    Let $\alpha \coloneqq \sup \{\phi(x) : x \in C\}$.
    Then $E_\xi = E_\alpha$ for all $\alpha \le \xi < \omega_1$.

    Consider $\xi \le \alpha$.
    \begin{itemize}
        \item If there exists $x_0 \in C$ with $\phi(x_0) \ge \xi$,
            pick such $x_0$ of minimal rank.
            Then for all $y \in X$ we have
            \begin{IEEEeqnarray*}{rClr}
                y \in E_\xi &\iff& y \in C \land \phi(y) \le \xi\\
                            &\iff& y \le^\ast_\phi x_0 & ~ \text{ coanalytic}\\
                            &\iff& x_0 \not<^\ast_\phi y & ~ \text{ analytic}\\
            \end{IEEEeqnarray*}
            So $E_\xi$ is Borel.
            % TODO If $\alpha < \omega_1$, this also shows that $E_\alpha$ is Borel?
        \item If there exists no such $x_0$
            then $\xi = \alpha$
            and
            \[
            E_\xi = E_\alpha = \bigcup_{\eta < \alpha}
            \]
            is a countable union of Borel sets by the previous case.
    \end{itemize}
\end{proof}

\pagebreak
\section{Abstract Topological Dynamics}

% \subsection*{Basic Definitions}
% TODO: move to appendix?

Recall:
\begin{definition}+
    Let $X$ be a set.
    A \vocab{group action} of a group $G$ on $X$
    is a function
    $\alpha\colon G \times X \to X$
    such that
    \begin{itemize}
        \item $\forall x \in X.~\alpha(1_G,x) = x$,
        \item $\forall g,h \in G, x \in X.~\alpha(gh,x) = \alpha(g,\alpha(h,x))$.
    \end{itemize}

    Often we will abbreviate $\alpha(g,x)$ as $g\cdot x$.


    For $x \in X$,
    the \vocab{orbit} of $x$ is defined as
    \[
    G\cdot x \coloneqq \{g\cdot x : g \in G\}.
    \]

    A group action is called \vocab{transitive}
    iff $g \mapsto g \cdot x$ is surjective for all $x \in X$,
    i.e.~iff the action has exactly one orbit.


    For $x \in X$,
    the \vocab{stabilizer subgroup}
    of $G$ with respect to $x$ is
    \[
        G_x \coloneqq  \{g \in G : g\cdot x = x\}.
    \]

\end{definition}
\begin{remark}+
    Group actions of a group $G$ on a set $X$
    correspond to group homomorphisms
    $G \to \Sym(X)$.
    Indeed for a group action $\alpha\colon G \times X \to X$
    consider
    \begin{IEEEeqnarray*}{rCl}
        G&\longrightarrow & \Sym(X) \\
        g&\longmapsto & (x \mapsto g \cdot x).
    \end{IEEEeqnarray*}
\end{remark}

\begin{definition}+
    A group $G$ with a topology
    is a \vocab{topological group}
    iff
    \begin{IEEEeqnarray*}{rCl}
        G \times G&\longrightarrow & G \\
        (x,y) &\longmapsto & x \cdot y
    \end{IEEEeqnarray*}
    and
    \begin{IEEEeqnarray*}{rCl}
        G&\longrightarrow & G \\
        x&\longmapsto & x^{-1}
    \end{IEEEeqnarray*}
    are continuous.
\end{definition}

\begin{definition}
    \label{def:flow}
    Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology}
    and let $X$ be a compact metrizable space.

    A \vocab{flow}  $(X, T)$, sometimes denoted $T \acts X$
    is a continuous action
    \begin{IEEEeqnarray*}{rCl}
        T \times X&\longrightarrow & X \\
        (t,x) &\longmapsto & tx.
    \end{IEEEeqnarray*}

    A flow is \vocab{minimal}  iff every orbit is dense.

    $(Y,T)$ is a \vocab{subflow}  of $(X,T)$ if $Y \subseteq X$
    and $Y$ is invariant under $T$,
    i.e.~$\forall t \in T,y \in Y.~ty \in Y$.

    A flow $(X,T)$ is \vocab{isometric}
    iff there is a metric $d$ on $X$ such that
    for all $t \in T$ the map
    \begin{IEEEeqnarray*}{rCl}
        a_t\colon X &\longrightarrow & X \\
        x &\longmapsto & tx
    \end{IEEEeqnarray*}
    is an \vocab{isometry},
    i.e.~$\forall t \in T.~\forall x,y \in X.~d(a_t(x),a_t(y)) = d(x,y)$.

    If $(X,T)$ is a flow, then a pair $(x,y)$, $x \neq y$
    is \vocab{proximal} iff
    \[
    \exists  z \in X.~\exists (t_n)_{n < } \in T^{\omega}.~t_n x \xrightarrow{n \to \infty}  z \land t_n y \xrightarrow{n \to \infty} z.
    \]

    A flow is \vocab{distal} iff
    it has no proximal pair.
\end{definition}
\begin{remark}
    Note that a flow is minimal iff it has no proper subflows.
\end{remark}


\begin{definition}+
    Let $(T,X)$ and $(T,Y)$ be flows.
    A \vocab{factor map} $\pi\colon (T,X) \to (T,Y)$
    is a continuous surjection $X \twoheadrightarrow Y$
    that is $T$-equivariant,
    i.e.~$\forall t \in T, x \in X.~\pi(t\cdot x) = t\cdot \pi(x)$.
    If such a factor map exists,
    we also say that $(T,Y)$ is a \vocab{factor}
    of $(T,X)$.

    An \vocab{isomorphism} from $(T,X)$ to $(T,Y)$ is
    a homeomorphism $X \leftrightarrow Y$
    commuting with the group action.
\end{definition}
\begin{warning}+
    What is called ``factor'' here is called ``subflow''
    by Furstenberg.
\end{warning}



\begin{example}
    Recall that $S_1 = \{z \in \C : |z| = 1\}$.
    Let $X = S_1$, $T = S_1$
    $(\alpha,\beta) \mapsto \alpha + \beta$ is isometric.%
    \footnote{Note that here we consider the abelian group structure of $S^1$
        and $\alpha + \beta$ denotes the addition of \emph{angles},
        i.e.~$\alpha \cdot \beta$ in complex numbers.}
\end{example}

\begin{definition}
    \label{def:isometricextension}
    Let $X,Y$ be compact metric spaces
    and $\pi\colon (X,T) \to (Y,T)$ a factor map.
    Then $(X,T)$ is an \vocab{isometric extension}
    of $(Y,T)$ if there is
    $\rho\colon X\times_Y X \to \R$%
    \footnote{Recall that in the category of topological spaces
        the \vocab{fiber product} of
        $A \xrightarrow{f} C$, $B \xrightarrow{g} C$
        is $A \times_C B = \{(a,b) \in A \times B: f(a) = g(b)\}$,
        i.e. $X \times_Y X = \{(x_1,x_2) \in X^2 : \pi(x_1) = \pi(x_2)\}$.}
    such that
    \begin{enumerate}[(a)]
        \item $\rho$ is continuous.
        \item For each $y \in Y$, $\rho$ is a metric on the fiber
            $X_y \coloneqq \{x \in X: \pi(x) = y\}$.
        \item $\forall t \in T.~\rho(tx_1,tx_2) = \rho(x_1,x_2)$.
        \item $\forall y,y' \in Y.~$
            the metric spaces $(X_y, \rho)$ and  $(X_{y'}, \rho)$
            are isometric.
    \end{enumerate}
\end{definition}
\begin{remark}
    A flow is isometric iff it is an isometric extension
    of the trivial flow,
    i.e.~the flow acting on a singleton.
    Indeed maps $\rho\colon X\times_\star X = X^2 \to \R$
    as in \yaref{def:isometricextension}
    correspond to metrics witnessing that the flow is isometric.
\end{remark}
\begin{proposition}
    \label{prop:isomextdistal}
    An isometric extension of a distal flow is distal.
\end{proposition}
\begin{proof}
    Let $\pi\colon  X\to Y$ be an isometric extension.
    Towards a contradiction,
    suppose that $x_1,x_2 \in X$ are proximal.
    Take $z \in X$ and a sequence $(g_n)_{n < \omega}$ in $T$
    such that $g_n x_1 \to z$ and $g_n x_2 \to z$.

    Then $g_n \pi(x_1) \to  \pi(z)$
    and $g_n \pi(x_2) \to \pi(z)$,
    so by distality of $Y$
    we have $\pi(x_1) = \pi(x_{2})$.
    Then $\rho(g_n x_1, g_n x_2)$
    is defined and equal to $\rho(x_1,x_2)$.
    By the continuity of $\rho$,
    we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$.
    Therefore $\rho(x_1,x_2) = 0$.
    Hence $x_1 = x_2$ $\lightning$.
\end{proof}

\begin{definition}
    Let $\Sigma = \{(X_i, T) : i \in I\} $
    be a collection of factors of $(X,T)$.
    Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map.
    Then $(X, T)$ is a \vocab{limit}%
    \footnote{This is not a limit in the category theory sense and not uniquely determined.}
    % TODO THE inverse limit is A limit
    of $\Sigma$ iff
    \[
    \forall x_1 \neq x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2).
    \]
\end{definition}

\begin{proposition}
    \label{prop:limitdistal}
    A limit of distal flows is distal.
\end{proposition}
\begin{proof}
    Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$.
    Suppose that each $(X_i, T)$ is distal.
    If $(X,T)$ was not distal,
    then there were $x_1, x_2, z \in X$
    and a sequence $(g_n)$ in $T$
    with $g_n x_1 \to z$ and $g_n x_2 \to z$.
    Take $i \in I$ such that $\pi_i(x_1) \neq \pi_i(x_2)$.
    But then $g_n \pi_i(x_1) \to \pi_i(z)$
    and $g_n \pi_i(x_2) \to \pi_i(z)$,
    which is a contradiction since $(X_i, T)$ is distal.
\end{proof}
