\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. \end{proof}