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