From f074e8841bfb3d00d2cc5bd84e3f99aaa2b9b393 Mon Sep 17 00:00:00 2001 From: Josia Pietsch Date: Tue, 16 Jan 2024 23:14:56 +0100 Subject: [PATCH] tutorial 12 --- inputs/lecture_19.tex | 4 +- inputs/lecture_22.tex | 3 - inputs/tutorial_09.tex | 19 ++++- inputs/tutorial_12.tex | 155 ++++++++++++++++++++++++++++++++++++++++ inputs/tutorial_12b.tex | 34 +++++++++ logic3.tex | 2 + 6 files changed, 211 insertions(+), 6 deletions(-) create mode 100644 inputs/tutorial_12.tex create mode 100644 inputs/tutorial_12b.tex diff --git a/inputs/lecture_19.tex b/inputs/lecture_19.tex index 8cecc60..3c1d7fd 100644 --- a/inputs/lecture_19.tex +++ b/inputs/lecture_19.tex @@ -23,13 +23,15 @@ Note that since $X$ compact the notions of equicontinuity and uniform equicontinuity coincide. \begin{fact}+[{\cite[Lecture 6, Exercise 1]{tao}}] + \label{fact:isometriciffequicontinuous} A flow $(X,T)$ is isometric iff it is equicontinuous. \end{fact} \begin{proof} Clearly an isometric flow is equicontinuous. On the other hand suppose that $T$ is uniformly equicontinuous. Define a metric $\tilde{d}$ on $X$ by setting - $\tilde{d}(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$. + $\tilde{d}(x,y) \coloneqq \sup_{t \in T} d(tx,ty) \le 1$ + (wlog.~$d \le 1$). By equicontinuity of $T$ we get that $\tilde{d}$ and $d$ induce the same topology on $X$. \end{proof} diff --git a/inputs/lecture_22.tex b/inputs/lecture_22.tex index 33cabc4..2cffaf6 100644 --- a/inputs/lecture_22.tex +++ b/inputs/lecture_22.tex @@ -189,6 +189,3 @@ For this we define % TODO similarities to the lemma used today \end{itemize} \end{proof} - - - diff --git a/inputs/tutorial_09.tex b/inputs/tutorial_09.tex index 2acc229..8b5b5ee 100644 --- a/inputs/tutorial_09.tex +++ b/inputs/tutorial_09.tex @@ -21,7 +21,6 @@ Material on topological dynamics: \end{itemize} - \nr 1 \begin{remark} @@ -126,6 +125,8 @@ amounts to a finite number of conditions on the preimage. In the lecture we only look at metrizable flows, so the definitions from the exercise sheet and from the lecture don't agree. + + Everywhere but here we will use the definition from the lecture. \end{remark} \begin{itemize} @@ -148,7 +149,21 @@ amounts to a finite number of conditions on the preimage. \iff&\exists x \in X.~\forall U \overset{\text{open}}{\subseteq} X.~ \exists z \in \Z.~f^z(x) \in U. \end{IEEEeqnarray*} - \item \todo{TODO} + \item Let $X$ be a compact Polish space. + What is the Borel complexity of $\Homeo(X)$ inside $\cC(X,X)$? + + Recall that $\cC(X,X)$ is a Polish space with the uniform topology. + We have + \begin{IEEEeqnarray*}{rCl} + \Homeo(X) &=& \{f \in \cC(X,X) : f \text{ is bijective and } f^{-1} \text{ is continuous}\}\\ + &=& \{f \in \cC(X,X) : f \text{ is bijective}\} + \end{IEEEeqnarray*} + by the general fact + \begin{fact} + Let $X$ be comapct and $Y$ Hausdorff, + $f\colon X \to Y$ a continuous bijection. + Then $f$ is a homeomorphism. + \end{fact} \item It suffices to check the condition from part (b) for open sets $U$ of a countable basis and points $x \in X$ belonging to a countable dense subset. diff --git a/inputs/tutorial_12.tex b/inputs/tutorial_12.tex new file mode 100644 index 0000000..abb2752 --- /dev/null +++ b/inputs/tutorial_12.tex @@ -0,0 +1,155 @@ +\tutorial{12}{2024-01-16T12:00}{} + + +\begin{question} + What is an example of a flow with a dense orbit + that isn't minimal. +\end{question} + +\begin{example} + Consider the Bernoulli shift. + $T = \Z \acts \{0,1\}^{\Z}$. + $(0)$ is a subflow. + Let $\phi\colon \Z \to \{0,1\}^{< \omega}$ + be an enumeration of all finite binary sequences. + Consider the concatenation + \[ + \ldots \concat \phi(-2) \concat \phi(-1) \concat \phi(0) \concat \phi(1) \concat \phi(2) \concat \ldots + \] +\end{example} + +\subsection{Sheet 11} + +\begin{fact} + If $A$, $B$ are topological spaces, + then $f\colon A \to B$, + is continuous iff + $f(\overline{S}) \subseteq \overline{f(S)}$ + for all $S \subseteq A$. +\end{fact} +\begin{proof} + Suppose that $f$ is continuous. + Take $a \in \overline{S}$. + Take any $f(a) \in U \overset{\text{open}}{\subseteq} B$. + $f^{-1}(U)$ is open and $f^{-1}(U) \ni a$. + So there exists $s \in S$ such that $s \in f^{-1}(U)$ + and $f(s) \in U$. + + + On the other hand suppose $f(\overline{S}) \subseteq \overline{f(S)}$ + for all $S \subseteq A$. + It suffices to show that preimages of closed sets are closed. + Let $V \overset{\text{closed}}{\subseteq} B$. + Then $f(\overline{f^{-1}(V)}) \subseteq \overline{ff^{-1}(V)} \subseteq V$, + hence + \[ + \overline{f^{-1}(V)} \subseteq f^{-1}(f(\overline{f^{-1}(V)})) \subseteq f^{-1}(V). + \] +\end{proof} +\begin{fact} + \label{fact:t12:2} + Let $A$ be compact and $B$ Hausdorff. + Let $f\colon A \to B$ be continuous + and $S \subseteq A$. + Then $f(\overline{S}) = \overline{f(S)}$. +\end{fact} +\begin{subproof} + We have already shown $f(\overline{S}) \subseteq \overline{f(S)}$. + Since $A$ is compact, $f(\overline{S})$ is compact + and since $B$ is Hausdorff, compact subsets of $B$ are closed. +\end{subproof} + +Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup. +Let $d$ be a compatible metric on $X$. +\begin{enumerate}[(a)] + \item Let $f_0 \in X^X$ be a continuous function. + Then $L_{f_0}\colon X^X \to X^X, f \mapsto f_0 \circ f$ is continuous. + + Consider $\{f : f_0 \circ f \in U_{\epsilon}(x,y)\}$. + We have + \begin{IEEEeqnarray*}{rCl} + &&f_0 \circ f \in U_{\epsilon}(x,y)\\ + &\iff& d(x, f_0(f(y))) < \epsilon\\ + &\iff& f(y) \in f_0^{-1}(B_{\epsilon}(x))\\ + &\iff& f \in \bigcup_{\tilde{x} \in f_0^{-1}(B_{\epsilon}(x))}U_{\epsilon_{\tilde{x}}}(\tilde{x}, y) + \end{IEEEeqnarray*} + where $\epsilon_{\tilde{x}}$ is such that $B_{\epsilon_{\tilde{x}}}(\tilde{x}) \subseteq f_0^{-1}(B_{\epsilon}(x))$. + (This is possible since $f_0$ is continuous, hence $f_0^{-1}(B_{\epsilon}(x))$ + is open.) + Clearly the RHS is open. + \item If $f_0$ is not continuous, then $L_{f_0}$ is in general not continuous: + Let $X = [0,1]$, and $f_0 \coloneqq \One_{\Q}$. + Consider $U \coloneqq U_{\frac{1}{2}}(1,1)$. + Then $\{f : f_0 \circ f \in U\} = \{f : f(1) \in \Q\}$ + is not open. + + \item Let $x_0 \in X$. The evaluation map + \begin{IEEEeqnarray*}{rCl} + \ev_{x_0}\colon X^X &\longrightarrow & X \\ + f &\longmapsto & f(x_0) + \end{IEEEeqnarray*} + is continuous: + + Let $y \in X$ and consider $B_\epsilon(y) \subseteq X$. + By definition $\ev_{x_0}(B_{\epsilon}(y)) = U_{\epsilon}(y,x_0)$ + is open. + + \item For any $x \in X$, we have $Gx = \overline{Tx}$: + + By definition $G = \overline{\{x \mapsto tx : t \in T\}}$. + Consider $\ev_x \colon X^X \to X$. + $X^X$ is compact and $X$ is Hausdorff. + Hence we can apply + \label{fact:t12:2}. + + \item Let $x_0 \neq x_1 \in X$. + Then $(x_0,x_1)$ is a proximal pair iff $d(gx_0,gx_1) = 0$ + for some $g \in G$: + + Let $(x_0,x_1)$ be proximal. + Consider $(\ev_{x_0}, \ev_{x_1})\colon X^X \to X \times X$ + and $d\colon X \times X \to \R$. + Both maps are continuous. + Consider $D \coloneqq \{d(gx_0, gx_1) : g \in G\}$. + $G$ is compact, hence $D$ is compact. + $D$ contains elements arbitrarily close to $0$ + and $D$ is closed, so $0 \in D$. + + On the other hand let $g \in G$ be such that $d(gx_0,gx_1) = 0$. + We want to show that $(x_0,x_1)$ is proximal. + + As $gx_0 = gx_1$, we have that there eixsts $\epsilon > 0$ + such that $g \in U_{\epsilon}(gx_0, x_1) \cap U_\epsilon(gx_1, x_0)$ + As $g \in \overline{T}$ for all $\epsilon > 0$, + there exists $t \in T$ with $t \in U_\epsilon(gx_0,x_1) \cap U_\epsilon(gx_1,x_0)$. + Hence $d(tx_1, gx_0) < \epsilon$ and $d(tx_0, gx_1) < \epsilon$. + + + \item $(X,T)$ contains a minimal subflow: + + + We apply Zorn's lemma. + It suffices to show that a chain of subflows $X \supseteq X_1 \supseteq X_2 \supseteq \ldots$ + has a limit. + We claim that $(\bigcap_n X_n, T)$ is a subflow, i.e.~ $\bigcap_n X_n$ + is $T$-invariant. + Indeed, since all the $X_n$ are $T$-invariant, + we have $T(\bigcap_n X_n) \subseteq \bigcap_n TX_n \subseteq \bigcap_n X_n$. + + It is clear that $\bigcap_{n} X_n$ is closed. + Since $X$ is compact the intersection is also non-empty. + \item Show that if $T$ is a compact metrisable topological flow, + then $(X,T)$ is equicontinuous. + + Suppose that $(X,T)$ is not equicontinuous. + Then there exists $\epsilon> 0$ + such that + \[\forall \delta > 0.~ \exists x,y \in X, t \in Y.~d(x,y) < \delta \land d(tx,ty) \ge \epsilon.\] + Take $\delta_n = \frac{1}{n}$. + Choose bad $x_n$, $y_n$, $t_n$. + Since $X$ and $T$ are compact, + wlog.~$x_n \to x'$, $y_n \to y'$, $t_n \to t'$. + So $d(t'x', t'y') > \epsilon$, + but $x' = y'$ $\lightning$ +\end{enumerate} + diff --git a/inputs/tutorial_12b.tex b/inputs/tutorial_12b.tex new file mode 100644 index 0000000..acf7793 --- /dev/null +++ b/inputs/tutorial_12b.tex @@ -0,0 +1,34 @@ +\tutorial{12b}{2024-01-16T13:09:02}{} + +\subsection{Sheet 10} + +\nr 2 + +\todo{Def skew shift flow (on $(\R / \Z)^2$!)} +The Bernoulli shift, $\Z \acts \{0,1\}^{\Z}$, is not distal. +Let $x = (0)$ and $y = (\delta_{0,i})_{i \in \Z}$. +Let $t_n \to \infty$. +Then $t_n y \to (0) = t_n x$. + +The skew shift flow is distal: +This is tedious but probably not too hard. + +The skew shift flow is not equicontinuous: + + + + + + +\subsection{Sheet 11} + +We did \yaref{fact:isometriciffequicontinuous}. + +$d$ and $d'(x,y) \coloneqq \sup_{t \in T} d(tx,ty)$ +induce the same topology. +Let $\tau, \tau'$ be the corresponding topologies. + +$\tau \subseteq \tau'$ easy, +$\tau' \subseteq \tau'$ : use equicontinuity. + + diff --git a/logic3.tex b/logic3.tex index a3ea118..0268c53 100644 --- a/logic3.tex +++ b/logic3.tex @@ -64,6 +64,8 @@ \input{inputs/tutorial_09} \input{inputs/tutorial_10} \input{inputs/tutorial_11} +\input{inputs/tutorial_12b} +\input{inputs/tutorial_12} \section{Facts} \input{inputs/facts}