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@ -58,7 +58,7 @@
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\end{proof}
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\end{proof}
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\subsection{Parametrizations}
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\subsection{Parametrizations}
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\todo{choose better title}
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%\todo{choose better title}
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Let $\Gamma$ denote a collection of sets in some space.
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Let $\Gamma$ denote a collection of sets in some space.
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@ -109,9 +109,9 @@ where $X$ is a metrizable, usually second countable space.
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put $(y,x) \in \cU$ iff
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put $(y,x) \in \cU$ iff
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$x \in \bigcup \{V_n : y_n = 1\}$.
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$x \in \bigcup \{V_n : y_n = 1\}$.
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$\cU$ is open.
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$\cU$ is open.
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Let $V = \bigcup \{V_n : V_n \subseteq V\}$.
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For any $V \overset{\text{open}}{\subseteq} X$,
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Pick $y \in 2^\omega$
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define $y \in 2^\omega$
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and let $y_n = 1$ iff $V_n \subseteq V$.
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by $y_n = 1$ iff $V_n \subseteq V$.
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Then $\cU_y = V$.
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Then $\cU_y = V$.
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@ -125,7 +125,7 @@ where $X$ is a metrizable, usually second countable space.
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Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
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Recall that $\eta_1 \le \eta_2 \implies \Pi^0_{\eta_1}(X) \subseteq \Pi^0_{\eta_2}(X)$.
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Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$
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Note that if $A = \bigcup_n A_n$, with $A_n \in \Pi^0_{\eta_n}(X)$
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some $\eta_n < \xi$,
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for some $\eta_n < \xi$,
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we also have
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we also have
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$A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$.
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$A = \bigcup_n A_n'$ with $A'_n \in \Pi^0_{\xi_n}(X)$.
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@ -146,7 +146,7 @@ where $X$ is a metrizable, usually second countable space.
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Since $2^{\omega}$ embeds
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Since $2^{\omega}$ embeds
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into any uncountable polish space $Y$
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into any uncountable polish space $Y$
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such that the image is closed,
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such that the image is closed,
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we can $2^{\omega}$ by $Y$
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we can replace $2^{\omega}$ by $Y$
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in the statement of the theorem.%
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in the statement of the theorem.%
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\footnote{By definition of the subspace topology
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\footnote{By definition of the subspace topology
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and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
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and transfinite induction, $\Sigma^0_\xi(Y)\defon{2^\omega} = \Sigma^0_\xi(2^\omega)$.}
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@ -10,19 +10,18 @@ Let $x = (0)$ and $y = (\delta_{0,i})_{i \in \Z}$.
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Let $t_n \to \infty$.
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Let $t_n \to \infty$.
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Then $t_n y \to (0) = t_n x$.
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Then $t_n y \to (0) = t_n x$.
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The skew shift flow is distal:
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% The skew shift flow is distal:
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This is tedious but probably not too hard.
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% This is tedious but probably not too hard.
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%
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The skew shift flow is not equicontinuous:
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% The skew shift flow is not equicontinuous:
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\subsection{Sheet 11}
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We did \yaref{fact:isometriciffequicontinuous}.
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\begin{refproof}{fact:isometriciffequicontinuous}.
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$d$ and $d'(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$
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$d$ and $d'(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$
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induce the same topology.
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induce the same topology.
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@ -32,3 +31,4 @@ $\tau \subseteq \tau'$ easy,
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$\tau' \subseteq \tau'$ : use equicontinuity.
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$\tau' \subseteq \tau'$ : use equicontinuity.
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\end{refproof}
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