tutorial 12

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}

inputs/lecture_22.tex

@@ -189,6 +189,3 @@ For this we define
             % TODO similarities to the lemma used today
     \end{itemize}
 \end{proof}
-
-
-

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.

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}
+

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.
+
+

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}