email about exam (2024-01-22)
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\lecture{23}{2024-01-19}{More sketches of ideas of Beleznay and Foreman}
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% TODO read notes
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% TODO def. almost distal
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% 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
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\begin{notation}
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Let $X$ be a Polish space and $\cP$ a property of elements of $X$,
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then we say that $x_0 \in X$ is \vocab{generic}
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@ -104,9 +108,11 @@ Let $I$ be a linear order
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The order %TODO (Furstenberg rank)
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is a $\Pi^1_1$-rank.
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\end{theorem}
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For the proof one shows that $\le^\ast$ and $<^\ast$
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are $\Pi^1_1$, where
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\begin{enumerate}[(1)]
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\begin{proof}[sketch]
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\notexaminable{
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For the proof one shows that $\le^\ast$ and $<^\ast$
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are $\Pi^1_1$, where
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\begin{enumerate}[(1)]
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\item $p_1 \le^\ast p_2$ iff $p_1$ codes
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a distal minimal flow and if
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$p_2$ also codes a distal minimal flow,
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@ -115,17 +121,18 @@ are $\Pi^1_1$, where
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a distal minimal flow and if
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$p_2$ also codes a distal minimal flow,
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then $\mathop{order}(p_1) < \mathop{order}(p_2)$.
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\end{enumerate}
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One uses that $(Y_{i+1}, T)$ is a maximal
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isometric extension of $(Y_i,T)$
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ind $(X,T)$
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iff for all $x_1,x_2$ from a fixed countable dense set
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in $X$,
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for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$,
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there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$,
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$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
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\end{enumerate}
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One uses that $(Y_{i+1}, T)$ is a maximal
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isometric extension of $(Y_i,T)$
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ind $(X,T)$
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iff for all $x_1,x_2$ from a fixed countable dense set
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in $X$,
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for all $i$ with $\pi_{i\oplus 1}(x_1) = \pi_{i \oplus 1}(x_2)$,
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there is a sequence $(z_k)$ such that $\pi_i(z_k) = \pi_i(x_1)$,
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$F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
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}
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\end{proof}
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\begin{proposition}
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The order of a minimal distal flow on a separable,
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metric space is countable.
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@ -171,4 +178,3 @@ $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
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Then $\alpha \mapsto U_\alpha$ is an injection.
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\end{proof}
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