From daec11591d8309a8fd1d537258d129bb43509c3d Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Fri, 8 Dec 2023 01:39:20 +0100 Subject: [PATCH] lecture 15 --- inputs/lecture_12.tex | 2 +- inputs/lecture_13.tex | 4 +- inputs/lecture_14.tex | 1 + inputs/lecture_15.tex | 160 +++++++++++++++++++++++++++++++++++++++++- logic.sty | 1 + 5 files changed, 164 insertions(+), 4 deletions(-) diff --git a/inputs/lecture_12.tex b/inputs/lecture_12.tex index fd37caf..ec91c05 100644 --- a/inputs/lecture_12.tex +++ b/inputs/lecture_12.tex @@ -119,7 +119,7 @@ For the proof we need some prerequisites: \end{refproof} \begin{corollary} -\label{cor:ifsi11c} +\label{cor:ifs11c} $\IF$ is $\Sigma^1_1$-complete. \end{corollary} \begin{proof} diff --git a/inputs/lecture_13.tex b/inputs/lecture_13.tex index 27ac96d..ab38e1e 100644 --- a/inputs/lecture_13.tex +++ b/inputs/lecture_13.tex @@ -51,7 +51,7 @@ with $(f^{-1}(\{1\}), <)$. \begin{proposition} Suppose that $(A, <)$ is a countable well ordering. Then for a tree $T \subseteq A^{<\N}$ on $A$, - Then $T$ is well-fonuded iff + Then $T$ is well-founded iff $(T, <_{KB}\defon{T})$ is well ordered. \end{proposition} \begin{proof} @@ -87,7 +87,7 @@ with $(f^{-1}(\{1\}), <)$. \] (equivalently $x \in \IF \iff f(x) \in \LO \setminus \WO$). This suffices, since $\IF \subseteq \Tr$ is $\Sigma^1_1$-complete - (see \yaref{cor:ifiss11c}). + (see \yaref{cor:ifs11c}). Fix a bijection $b\colon \N \to \N^{<\N}$. diff --git a/inputs/lecture_14.tex b/inputs/lecture_14.tex index baef8a4..3976165 100644 --- a/inputs/lecture_14.tex +++ b/inputs/lecture_14.tex @@ -134,6 +134,7 @@ Let $\rho(\prec) \coloneqq \sup \{\rho_{\prec}(x) + 1 : x \in X\}$. \todo{TODO} \end{exercise} \begin{theorem}[Kunen-Martin] + \yalabel{Kunen-Martin}{Kunen-Martin}{thm:kunenmartin} If $(X, \prec)$ is wellfounded and $\prec \subseteq X^2$ is $\Sigma^1_1$ then $\rho(\prec) < \omega_1$. diff --git a/inputs/lecture_15.tex b/inputs/lecture_15.tex index 13f2138..d408219 100644 --- a/inputs/lecture_15.tex +++ b/inputs/lecture_15.tex @@ -1,3 +1,161 @@ \lecture{15}{2023-12-05}{} -\todo{Lecture 15 is missing} +\begin{theorem}[The Boundedness Theorem] + \yalabel{Boundedness Theorem}{Boundedness}{thm:boundedness} + + Let $X$ be Polish, $C \subseteq X$ coanalytic, + $\phi\colon C \to \omega_1$ a coanalytic rank on $C$, + $A \subseteq C$ analytic, i.e.~$A \in \Sigma^1_1(X)$. + Then $\sup \{\phi(x) : x \in A\} < \omega_1$. + + Moreover for all $\xi < \omega_1$, + \[ + D_\xi \coloneqq \{x \in C : \phi(x) < \xi\} + \] + and + \[ + E_\xi \coloneqq \{x \in C : \phi(x) \le \xi\} + \] + are Borel subsets of $X$. +\end{theorem} +\begin{proof} +% Let +% \begin{IEEEeqnarray*}{rCl} +% x \prec y&:\iff& x,y \in A \land \phi(x) < \phi(y)\\ +% &\iff& x,y \in A \land y \not\le_\phi^\ast x. +% \end{IEEEeqnarray*} +% Since $A$ is analytic, +% this relation is analytic and wellfounded on $X$. +% By \yaref{thm:kunenmartin} +% we get $\rho(\prec) < \omega_1$. +% Thus $\sup \{\phi(x) : x \in A\} < \omega_1$. +% +% Since $D_\xi = \bigcup_{\eta < \xi} E_\xi$, +% it suffices to check $E_\xi \in \Sigma_1^1(X)$. +% If $\alpha = \sup \{\phi(x) : x \in C\}$, +% we have $E_\xi = E_\alpha$ for all $\alpha < \xi < \omega_1$. + + \todo{TODO: Copy from official notes} +\end{proof} + +\section{Abstract Topological Dynamics} + +\begin{definition} + Let $T$ be a topological group\footnote{usually $T = \Z$ with the discrete topology} + and let $X$ be a compact metrizable space. + + A \vocab{flow} $(X, T)$, sometimes denoted $T \acts X$ + is a continuous action + \begin{IEEEeqnarray*}{rCl} + T \times X&\longrightarrow & X \\ + (t,x) &\longmapsto & tx. + \end{IEEEeqnarray*} + + A flow is \vocab{minimal} iff every orbit is dense. + + $(Y,T)$ is a \vocab{subflow} of $(X,T)$ if $Y \subseteq X$ + and $Y$ is invariant under $T$, + i.e.~$\forall t \in T,y \in Y.~ty \in Y$. + + A flow $(X,T)$ is \vocab{isometric} + iff there is a metric $d$ on $X$ such that + for all $t \in T$ the map + \begin{IEEEeqnarray*}{rCl} + a_t\colon X &\longrightarrow & X \\ + x &\longmapsto & tx + \end{IEEEeqnarray*} + is an \vocab{isometry}, + i.e.~$\forall t \in T.~\forall x,y \in X.~d(a_t(x),a_t(y)) = d(x,y)$. + + If $(X,T)$ is a flow, then a pair $(x,y)$, $x \neq y$ + is \vocab{proximal} iff + \[ + \exists z \in X.~\exists (t_n)_{n < } \in T^{\omega}.~t_n x \xrightarrow{n \to \infty} z \land t_n y \xrightarrow{n \to \infty} z. + \] + + A flow is \vocab{distal} iff + it has no proximal pair. +\end{definition} +\begin{remark} + Note that a flow is minimal iff it has no proper subflows. +\end{remark} +% \begin{example} +% Recall that $S_1 = \{z \in \C : |z| = 1\}$. +% Let $X = S_1$, $T = S_1$ +% $(\alpha,\beta) \mapsto \alpha + \beta$ is isometric. +% % TODO: In the official notes it says \alpha + \beta, but this is no group action. +% % Maybe \alpha * \beta ? +% \end{example} + +\begin{definition} + Let $X,Y$ be compact metric spaces + and $\pi\colon (X,T) \to (Y,T)$ a factor map. + Then $(X,T)$ is an \vocab{isometric extension} + of $(Y,T)$ if there is a real valued $\rho:$ + defined on the pullback, $\{(x_1,x_2) \in X^2 : \pi(x_1) = \pi(x_2)\}$, % TODO nice notation? + such that + \begin{enumerate}[(a)] + \item $\rho$ is continuous. + \item For each $y \in Y$, $\rho$ is a metric on the fibre + $X_y \coloneqq \{x \in X: \pi(x) = y\}$. + \item $\forall t \in T.~\rho(tx_1,tx_2) = \rho(x_1,x_2)$. + \item $\forall y,y' \in Y.~$ + the metric spaces $(X_y, \rho)$ and $(X_{y'}, \rho)$ + are isometric. + \end{enumerate} +\end{definition} +\begin{remark} + A flow is isometric iff it is an isometric extension + of the trivial flow, + i.e.~the flow acting on a singleton. + % TODO THINK ABOUT THIS! +\end{remark} +\begin{proposition} + An isometric extension of a distal flow is distal. +\end{proposition} +\begin{proof} + Let $\pi\colon X\to Y$ be an isometric extension. + Towards a contradiction, + suppose that $x_1,x_2 \in X$ are proximal. + Take $z \in X$ and $(g_n) \in T^{\omega}$ + such that $g_n x_1 \to z$ and $g_n x_2 \to z$. + + Then $g_n \pi(x_1) \to \pi(z)$ + and $g_n \pi(x_2) \to \pi(z)$, + so by distality of $Y$ + we have $\pi(x_1) = \pi(x_{2})$. + Then $\rho(g_n x_1, g_n x_2)$ + is defined and equal to $\rho(x_1,x_2)$. + By the continuity of $\rho$, + we get $\rho(g_n x_1, g_n x_2) \to \rho(z,z) = 0$. + Therefore $\rho(x_1,x_2) = 0$. + Hence $x_1 = x_2$ $\lightning$. +\end{proof} + +\begin{definition} + Let $\Sigma = \{(X_i, T) : i \in I\} $ + be a collection of factors of $(X,T)$. % TODO State precise definition of a factor + Let $\pi_i\colon (X,T) \to (X_i, T)$ denote the factor map. + Then $(X, T)$ is the \vocab{limit} of $\Sigma$ + iff + \[ + \forall x_1,x_2 \in X.~\exists i \in I.~\pi_i(x_1) \neq \pi_i(x_2). + \] + % TODO think about abstract nonsense +\end{definition} + +\begin{proposition} + A limit of distal flows is distal. +\end{proposition} +\begin{proof} + Let $(X,T)$ be a limit of $\Sigma = \{(X_i, T) : i \in I\}$. + Suppose that each $(X_i, T)$ is distal. + If $(X,T)$ was not distal, + then there were $x_1, x_2, z \in X$ + and a sequence $(g_n)$ in $T$ + with $g_n x_1 \to z$ and $g_n x_2 \to z$. + Take $i \in I$ such that $\pi_i(x_1) \neq \pi_i(x_2)$. + But then $g_n \pi_i(x_1) \to \pi_i(z)$ + and $g_n \pi_i(x_2) \to \pi_i(z)$, + which is a contradiction since $(X_i, T)$ is distal. +\end{proof} diff --git a/logic.sty b/logic.sty index 81f2429..eb78b62 100644 --- a/logic.sty +++ b/logic.sty @@ -145,6 +145,7 @@ \DeclareMathOperator{\symdif}{\triangle} \DeclareSimpleMathOperator{proj} \newcommand{\fc}{\mathfrak{c}} +\DeclareMathOperator{\acts}{\curvearrowright} \newcommand\lecture[3]{\hrule{\color{darkgray}\hfill{\tiny[Lecture #1, #2]}}} \newcommand\tutorial[3]{\hrule{\color{darkgray}\hfill{\tiny[Tutorial #1, #2]}}}