email about exam (2024-01-22)
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@ -1,5 +1,9 @@
\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
\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}
@ -104,9 +108,11 @@ Let $I$ be a linear order
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)]
\begin{proof}[sketch]
\notexaminable{
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,
@ -115,17 +121,18 @@ are $\Pi^1_1$, where
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 $\pi_i(z_k) = \pi_i(x_1)$,
$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
\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 $\pi_i(z_k) = \pi_i(x_1)$,
$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
}
\end{proof}
\begin{proposition}
The order of a minimal distal flow on a separable,
metric space is countable.
@ -171,4 +178,3 @@ $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
Then $\alpha \mapsto U_\alpha$ is an injection.
\end{proof}