w23-logic-3/inputs/lecture_18.tex

\subsection{Sketch of proof of \yaref{thm:l16:3}}
\lecture{18}{2023-12-15}{Sketch of proof of \yaref{thm:l16:3}}

The goal for this lecture is to give a very rough
sketch of \yaref{thm:l16:3} in the case of $|Z| = 1$.
% \begin{theorem}[Furstenberg]
%     Let $(X, T)$ be a minimal distal flow
%     and let $(Z,T)$ be a proper factor of $X$%
%     \footnote{i.e.~$(X,T)$ and $(Z,T)$ are not isomorphic.}
%     Then three is another factor $(Y,T)$ of $(X,T)$
%     which is a proper isometric extension of  $Z$.
% \end{theorem}


Let $(X,T)$ be a distal flow.
Then $G \coloneqq  E(X,T)$ is a group.
\begin{definition}
    \label{def:F}
For $x, x' \in X$ define
\[
F(x,x') \coloneqq \inf \{d(gx, gx') : g \in G\}.
\]
\end{definition}
\begin{fact}
\begin{enumerate}[(a)]
    \item $F(x,x') = F(x', x)$,
    \item $F(x,x') \ge 0$ and $F(x,x') = 0$ iff $x = x'$.
    \item $F(gx, gx') = F(x,x')$ since  $G$ is a group.
    \item $F$ is an upper semi-continuous function on $X^2$,
        i.e.~$\forall a \in R.~\{(x,x') \in X^2 : F(x,x') < a\} \overset{\text{open}}{\subseteq} X^2$.

        This holds because $F$ is the infimum of continuous functions
        \begin{IEEEeqnarray*}{rCl}
            f_g\colon X^2 &\longrightarrow & \R \\
            (x,x') &\longmapsto & d(gx, gx')
        \end{IEEEeqnarray*}
        for $g \in G$.
\end{enumerate}
\end{fact}
\begin{theoremdef}
    \label{def:ftop}
    The sets
    \[
    U_a(x) \coloneqq  \{x' : F(x,x') < a\}
    \]
    form the basis of a topology in $X$.
    This topology is called the  \vocab{F-topology} on $X$.
    In this setting, the original topology
    is also called the \vocab{E-topology}.
\end{theoremdef}
This will follow from the following lemma:
\begin{lemma}
    \label{lem:ftophelper}
    Let $F(x,x') < a$.
    Then there exists $\epsilon > 0$ such that
    whenever $F(x',x'') < \epsilon$, then $F(x,x'') < a$.
\end{lemma}
\begin{refproof}{def:ftop}
    We have to show that if $U_a(x_1) \cap U_b(x_2) \neq \emptyset$,
    then this intersection is the union
    of sets of this kind.
    Let $x' \in U_a(x_1)$.
    Then by \yaref{lem:ftophelper},
    there exists $\epsilon_1 > 0$ with $U_{\epsilon_1}(x') \subseteq U_a(x_1)$.
    Similarly there exists  $\epsilon_2 > 0$
    such that $U_{\epsilon_2}(x') \subseteq  U_b(x_2)$.
    So for $\epsilon \le  \epsilon_1, \epsilon_2$,
    we get $U_{\epsilon}(x') \subseteq U_a(x_1) \cap U_b(x_2)$.
\end{refproof}
\begin{refproof}{lem:ftophelper}%
    \footnote{This was not covered in class.}

    Let $T = \bigcup_n  T_n$,% TODO Why does this exist?
    $T_n$ compact, wlog.~$T_n \subseteq T_{n+1}$, and
    let $G(x,x') \coloneqq  \{(gx,gx') : g \in G\} \subseteq  X \times X$.
    Take $b$ such that $F(x,x') < b < a$.
    Then $U = \{(u,u') \in G(x,x') : d(u,u') < b\}$
    is open in $G(x,x')$
    and since $F(x,x') < b$ we have $U \neq \emptyset$.
    \begin{claim}
        There exists $n$ such that
        \[
        \forall  (u,u') \in G(x,x').~T_n(u,u')\cap U \neq \emptyset.
        \]
    \end{claim}
    \begin{subproof}
        Suppose not.
        Then for all $n$, there is $(u_n, u_n') \in G(x,x')$
        with
        \[T_n(u_n, u_n') \subseteq G(x,x') \setminus U.\]
        Note that the RHS is closed.
        For $m > n$ we have
        $T_n(u_m, u'_m) \subseteq G(x,x') \setminus U$
        since $T_n \subseteq T_m$.
        By compactness of $X$,
        there exists $v,v'$ and some subsequence
        such that $(u_{n_k}, u'_{n_k}) \to  (v,v')$.

        So for all $n$ we have $T_n(v,v') \subseteq  G(x,x') \setminus U$,
        hence $T(v,v') \cap U = \emptyset$,
        so $G(v,v') \cap U = \emptyset$.
        But this is a contradiction as $\emptyset\neq U \subseteq G(v,v')$.
    \end{subproof}
    The map
    \begin{IEEEeqnarray*}{rCl}
        T\times X&\longrightarrow & X \\
        (t,x) &\longmapsto & tx
    \end{IEEEeqnarray*}
    is continuous.
    Since $T_n$ is compact,
    we have that $\{(x,t) \mapsto tx : t \in T_n\}$
    is equicontinuous.\todo{Sheet 11}
    So there is $\epsilon > 0$ such that
    $d(x_1,x_2) < \epsilon \implies d(tx_1, tx_2) < a -b$
    for all $t \in T_n$.

    Suppose now that $F(x', x'') < \epsilon$.
    Then there is $t_0 \in T$ such that $d(t_0x', t_0x'') < \epsilon$,
    hence $d(t t_0x', t t_0 x'') < a-b$ for all $t \in  T_n$.
    Since $(t_0x, t_0x') \in G(x,x')$,
    there is $t_1 \in T_n$
    with $(t_1t_0x, t_1t_0x') \in U$,
    i.e.~$d(t_1t_0x, t_1t_0x') < b$
    and therefore
    $F(x,x'') = d(t_1t_0x, t_1t_0x'') < a$.
\end{refproof}

Now assume $Z = \{\star\}$.
We want to sketch a proof of \yaref{thm:l16:3} in this case,
i.e.~show that if $(Z,T)$ is a proper factor of  a minimal distal flow
 $(X,T)$ then there is another factor  $(Y,T)$ of  $(X,T)$
 which is a proper isometric extension of $Z$.

\begin{proof}[sketch] % TODO: Think about this
\leavevmode
\begin{enumerate}[1.]
    \item For $x \in X$ define
        \begin{IEEEeqnarray*}{rCl}
            F_x\colon X &\longrightarrow & \R \\
            x' &\longmapsto & F(x,x').
        \end{IEEEeqnarray*}
    \item Define an equivalence relation on $X$,
        by $x_1 \sim x_2 :\iff \{x \in X : F_{x_1}(x) = F_{x_2}(x)\}$
        is comeager in $X$\footnote{with respect to the E-topology}.
        Then for all $g \in G$ we have
        $x_1 \sim x_2 \implies gx_1 \sim ~ gx_2$.

        Let $M \coloneqq  \{[x]_{\sim } : x \in X\} = \faktor{X}{\sim}$
        bet the quotient space.
        It is compact, second countable and Hausdorff.
        Let $\pi\colon X\to M$ denote the quotient map.
    \item $(Y,T) \mathbin{\text{\reflectbox{$\coloneqq$}}} (M,T)$
        is an isometric flow:
        \begin{enumerate}
            \item For $a > 0$, $x,x' \in X$ let
                \[
                    W(x,x') \coloneqq  \{g \in G : F(x, gx') < a\}.
                \]
                This turns out to be a subbasis of a topology
                which is coarser than the original topology on $G$.
                The new topology makes $G$ compact.
            \item Let $\theta(g)$ be the transformation of $M$
                defined by $\theta(g) \pi(x) = \pi(gx)$.
                This is well defined.
                Let $H = \theta(G)$.
                This is just a quotient of  $G$, $g \mapsto \theta(g)$
                may not be injective.
            \item One can show that $H$ is a topological group and $(M,H)$
                is a flow.\footnote{This is non-trivial.}
            \item Since $H$ is compact,
                $(M,H)$ is equicontinuous, %\todo{We didn't define this}
                i.e.~it is isometric.
                In particular, $(M,T)$ is isometric.
        \end{enumerate}
    \item $M \neq  \{\star\}$, i.e.~$(M,T)$ is non-trivial:

        Suppose towards a contradiction that $M = \{\star\}$,
        i.e.~$x_1 \sim x_2$ for all $x_1,x_2 \in X$.
        Fix $x_2$. For every $x_1 \in X$
        we have that
        \[
        \{x : F(x_1,x) = F(x_2,x)\}
        \]
        is comeager.
        Let $x_1$ be a point of continuity of $F_{x_2}$.
        Let $\langle a_n : n < \omega \rangle$ be a sequence
        of elements that set, i.e.~$F(x_1, a_n) = F(x_2, a_n)$,
        such that $a_n \to x_1$.
        So by the continuity of $F_{x_2}$ at $x_1$
        \begin{IEEEeqnarray*}{rCl}
            \lim_{n \to \infty} F(x_2, a_n) &=& F(x_2, x_1)
        \end{IEEEeqnarray*}
        and by the definition of $F$
        \begin{IEEEeqnarray*}{rCl}
            \lim_{n \to \infty} F(x_1,a_n) &=& F(x_1,x_1) = 0.
        \end{IEEEeqnarray*}
        So
        \[
        F(x_2,x_1) = \lim_{n \to \infty} F(x_2, a_n) = \lim_{n \to \infty}
        F(x_1,a_n) = 0
        \]
        and by distality we get $x_1 = x_2$.
        Since almost all points of $X$
        are points of continuity of $F_{x_2}$
        (\yaref{thm:usccomeagercont})
        this implies that $X \setminus \{x_2\}$ is meager.
        But then $X = \{\star\} \lightning$.
\end{enumerate}
\end{proof}

\begin{theorem}\footnote{Not covered in class}
    \label{thm:usccomeagercont}
    Let $X$ be a metric space
    and $\Gamma\colon X \to \R$ be upper semicontinuous.
    Then the set of continuity points of $\Gamma$ is comeager.
    \todo{Missing figure: upper semicontinuous function}
\end{theorem}
\begin{proof}
    Take $x$ such that $\Gamma$ is not continuous at $x$.
    Then there is an $\epsilon > 0$ 
    and $x_n \to x$ such that
    $\Gamma(x_n) + \epsilon \le \Gamma(x)$.
    Take $q \in \Q$ such that $\Gamma(x) - \epsilon < q < \Gamma(x)$.
    Then let
    \[
    B_q \coloneqq  \{a \in X : \Gamma(a) \ge  q\}.
    \]
    $X \setminus B_q = \{a \in X : \Gamma(a) <  q\}$
    is open, i.e.~$B_q$ is closed.
    Note that $x \in F_q \coloneqq B_q \setminus B_q^\circ$
    and $B_q \setminus B_q^\circ$ is nwd
    as it is closed and has empty interior,
    so $\bigcup_{q \in \Q} F_q$ is meager.
\end{proof}