w23-logic-3/inputs/lecture_23.tex

\lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman}

\begin{notation}
    Let $X$ be a Polish space and $\cP$ a property of elements of $X$,
    then we say that $x_0 \in X$ is \vocab{generic} 
    if
    \[
    A_\cP \coloneqq \{x  \in X \colon  \cP(x)\}
    \]
    is comeager
    and $x_0 \in A_\cP$.
\end{notation}
For example let $X = \mathbb{K}_I$
and $\cP$ the property of being a distal minimal flow.
\begin{abuse}
    We will usually omit $\cP$.
\end{abuse}


Let $I$ be a linear order

\begin{theorem}[Beleznay and Foreman]
    The set of distal minimal flows is $\Pi_1^1$-complete.
\end{theorem}

\begin{proof}[sketch]
    Consider $\WO(\N) \subset \LO(\N)$.
    We know that this is $\Pi_1^1$-complete. % TODO ref

    Let
    \begin{IEEEeqnarray*}{rCll}
        S & \coloneqq  & \{ x \in \LO(\N) :& x \text{ has a least element},\\
          &&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
    \end{IEEEeqnarray*}
    \todo{Exercise sheet 12}
    $S$ is Borel.

    We will % TODO ? 
    construct a reduction
    \begin{IEEEeqnarray*}{rCl}
        M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
        % \alpha &\longmapsto &  M(\alpha)
    \end{IEEEeqnarray*}
    We want that $\alpha \in \WO(\N) \iff M(\alpha)$
    codes a distal minimal flow of rank $\alpha$.

    \begin{enumerate}[1.]
        \item For any $\alpha \in S$, $M(\alpha)$ is a code for
            a flow which is coded by a generic $(f_i)_{i \in I}$.
            Specifically we will take a flow
            corresponding to some $(f_i)_{i \in I}$
            which is in the intersection of all
            $U_n$, $V_{j,m,n,\frac{p}{q}}$
            (cf.~proof of \yaref{thm:distalminimalofallranks}).

        \item If $\alpha \in \WO(\N)$,
            then additionally $(f_i)_{i \in I}$ will code
            a distal minimal flow of ordertype $\alpha$.
    \end{enumerate}
    
    One can get a Borel map $S \ni \alpha \mapsto \{T_n^{\alpha} : n \in \N\}$,
    such that $T^{\alpha}_n$ is closed,
    $T^{\alpha}_n \neq \emptyset$, $\diam(T^\alpha_n) \xrightarrow{n \to  \infty} 0$,
    $T^\alpha_{n+1} \subseteq  T^\alpha_n$,
    $T^{\alpha}_n \subseteq  W^{\alpha}_n$,
    where $W^{\alpha}_n$ is an enumeration of $U_m^\alpha$,$V^\alpha_{j,m,n,\frac{p}{q}}$.
    Then $(f_i)_{i \in I} \in  \bigcap_{n} T_{n}^\alpha$.
\end{proof}
\begin{lemma}
    Let $\{(X_\xi, T) : \xi  \le \eta\}$ be a normal
    quasi-isometric system
    and $\{(Y_i, T) : i \in I\}$
    such that
    \begin{enumerate}[(i)]
        \item $I \in S$ and additionally $I$ has a largest element.
        \item $Y_0$ is the trivial flow and  $Y_\infty = X_\eta$,
            where $0$ and $\infty$ denote the minimal
            resp.~maximal element of $I$.
        \item $\forall i < j$
            % https://q.uiver.app/#q=WzAsMyxbMCwwLCIoWF9cXGV0YSwgVCkiXSxbMSwwLCJZX2oiXSxbMSwxLCJZX2kiXSxbMCwxLCJcXHBpX2oiXSxbMCwyLCJcXHBpX2kiLDJdLFsxLDIsIlxccGleal9pIl1d
\[\begin{tikzcd}
	{(X_\eta, T)} & {Y_j} \\
	& {Y_i}
	\arrow["{\pi_j}", from=1-1, to=1-2]
	\arrow["{\pi_i}"', from=1-1, to=2-2]
	\arrow["{\pi^j_i}", from=1-2, to=2-2]
\end{tikzcd}\]
        \item If $i \in I$ is a limit (i.e.~there does not exist
            an immediate predecessor),
            then $(Y_i,T)$ is the inverse limit
            of  $\{(Y_j,T) : j < i\}$
            with respect to the factor maps.
        \item $(Y_{i\oplus 1}, T)$ is a maximal isometric
            extension of $(Y_i, T)$
            in $(X_\eta, T)$.
    \end{enumerate}

    Then $I$ is well-ordered with $\otp(Y) = \eta + 1$.
\end{lemma}

\begin{theorem}[Beleznay Foreman]
    The order %TODO (Furstenberg rank)
    is a $\Pi^1_1$-rank.
\end{theorem}
For the proof one shows that $\le^\ast$ and $<^\ast$
are  $\Pi^1_1$, where
\begin{enumerate}[(1)]
    \item $p_1 \le^\ast p_2$ iff $p_1$ codes
        a distal minimal flow and if
        $p_2$ also codes a distal minimal flow,
        then $\mathop{order}(p_1) \le \mathop{order}(p_2)$.
    \item $p_1 <^\ast p_2$ iff $p_1$ codes
        a distal minimal flow and if
        $p_2$ also codes a distal minimal flow,
        then $\mathop{order}(p_1) < \mathop{order}(p_2)$.
\end{enumerate}

One uses that $(Y_{i+1}, T)$ is a maximal
isometric extension of $(Y_i,T)$
ind $(X,T)$
iff for all $x_1,x_2$ from a fixed countable dense set
in $X$,
for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$,
there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$,
$F(z_k, x_1) \to  0$, $F(z_k, x_2) \to 0$.

\begin{proposition}
    The order of a minimal distal flow on a separable,
    metric space is countable.
\end{proposition}
\begin{proof}
    Let $(X,\Z)$ be such a flow,
    i.e.~ $X$ is separable, metric and compact.

    Produce a normal quasi-isometric system
    \[
    \{(X_\alpha, \Z) : \alpha \le \beta\}
    \]
    with $(X_\beta, \Z) = (X,\Z)$.
    We need to show that $\beta < \omega_1$.

    Let $\pi_\alpha\colon (X,\Z) \to (X_\alpha, \Z)$.
    Fix $x_0 \in X$.
    For every $\alpha$
    consider $\pi_\alpha^{-1}\left( \pi_\alpha(x_0) \right)
    = F_\alpha \overset{\text{closed}}{\subseteq} X$.

    \begin{itemize}
        \item For $\alpha_1 < \alpha_2 \le \beta$
            we have that $F_{\alpha_1} \supseteq F_{\alpha_2}$.
        \item For limits $\gamma \le \beta$,
            we have that $F_\gamma = \bigcap_{\alpha < \gamma} F_\alpha$,
            since $(X_\gamma,\Z)$ is the inverse limit of
            $\{(X_{\alpha}, \Z):\alpha < \gamma\}$.
        \item For all $\alpha < \beta$, $F_{\alpha+1} \subsetneq F_\alpha$,
            because $\pi^{\alpha+1}_\alpha \colon (X_{\alpha+1},\Z) \to (X_\alpha,\Z)$
            is not a bijection
            and all the fibers are isomorphic.
    \end{itemize}

    So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
    increasing chain of closed subsets.
    But $X$ is second countable,
    so $\beta$ is countable:
    Let $\{U_n\} = \cB$ be a countable basis
    and for $\alpha$ let $U_\alpha \in \cB$
    be such that $U_\alpha \cap F_\alpha = \emptyset$
    and $U_\alpha \cap F_{\alpha+1} \neq \emptyset$.
    Then $\alpha \mapsto U_\alpha$ is an injection.
\end{proof}