tutorial 12
Some checks failed
Build latex and deploy / checkout (push) Failing after 15m22s

This commit is contained in:
Josia Pietsch 2024-01-16 23:14:56 +01:00
parent b0a943abb1
commit f074e8841b
Signed by: josia
GPG key ID: E70B571D66986A2D
6 changed files with 211 additions and 6 deletions

View file

@ -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}

View file

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

View file

@ -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.

155
inputs/tutorial_12.tex Normal file
View file

@ -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}

34
inputs/tutorial_12b.tex Normal file
View file

@ -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.

View file

@ -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}