Josia Pietsch a0cc70cadf
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\lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman}
% TODO read notes
% TODO def. almost distal
% From Lecture 23, you need to know the proposition on page 7 (with the proof), but I won't ask you for other proofs from that lecture
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}
A_\cP \coloneqq \{x \in X \colon \cP(x)\}
is comeager
and $x_0 \in A_\cP$.
For example let $X = \mathbb{K}_I$
and $\cP$ the property of being a distal minimal flow.
We will usually omit $\cP$.
Let $I$ be a linear order
\begin{theorem}[Beleznay and Foreman]
The set of distal minimal flows is $\Pi_1^1$-complete.
Consider $\WO(\N) \subset \LO(\N)$.
We know that this is $\Pi_1^1$-complete. % TODO ref
S & \coloneqq & \{ x \in \LO(\N) :& x \text{ has a least element},\\
&&& \text{for any $t$, there is $t \oplus 1$, the successor of $t$.}\}
$S$ is Borel.\footnote{cf.~\yaref{s12e1}}
We will
construct a reduction
M \colon S &\longrightarrow & C(\mathbb{K}^\N,\mathbb{K})^\N. %\\
% \alpha &\longmapsto & M(\alpha)
We want that $\alpha \in \WO(\N) \iff M(\alpha)$
codes a distal minimal flow of rank $\alpha$.
\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 $\alpha$.
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 $(f_i)_{i \in I} \in \bigcap_{n} T_{n}^\alpha$.
Let $\{(X_\xi, T) : \xi \le \eta\}$ be a normal
quasi-isometric system
and $\{(Y_i, T) : i \in I\}$
such that
\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
{(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]
\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 $(X_\eta, T)$.
Then $I$ is well-ordered with $\otp(Y) = \eta + 1$.
\begin{theorem}[Beleznay Foreman]
The order %TODO (Furstenberg rank)
is a $\Pi^1_1$-rank.
For the proof one shows that $\le^\ast$ and $<^\ast$
are $\Pi^1_1$, where
\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 $\mathop{order}(p_1) < \mathop{order}(p_2)$.
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 $\pi_i(z_k) = \pi_i(x_1)$,
$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
The order of a minimal distal flow on a separable,
metric space is countable.
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 $\pi_\alpha^{-1}\left( \pi_\alpha(x_0) \right)
= F_\alpha \overset{\text{closed}}{\subseteq} X$.
\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 $\alpha$ it is $F_{\alpha+1} \subsetneq F_\alpha$,
because $\pi^{\alpha+1}_\alpha \colon (X_{\alpha+1},\Z) \to (X_\alpha,\Z)$
is not a bijection
and all the fibers are isomorphic.
So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
decreasing 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.