Josia Pietsch 6c2a76d838
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improved lecture 17
2024-02-05 20:30:13 +01:00

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What is an example of a flow with a dense orbit
that isn't minimal.
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
\subsection{Sheet 11}
If $A$, $B$ are topological spaces,
then $f\colon A \to B$,
is continuous iff
$f(\overline{S}) \subseteq \overline{f(S)}$
for all $S \subseteq A$.
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 $f(\overline{f^{-1}(V)}) \subseteq \overline{ff^{-1}(V)} \subseteq V$,
\overline{f^{-1}(V)} \subseteq f^{-1}(f(\overline{f^{-1}(V)})) \subseteq f^{-1}(V).
Let $A$ be compact and $B$ Hausdorff.
Let $f\colon A \to B$ be continuous
and $S \subseteq A$.
Then $f(\overline{S}) = \overline{f(S)}$.
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.
\nr 1
Let $(X,T)$ be a flow and $G = E(X,T)$ its Ellis semigroup.
Let $d$ be a compatible metric on $X$.
\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
&&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)
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
\ev_{x_0}\colon X^X &\longrightarrow & X \\
f &\longmapsto & f(x_0)
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
\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 $x' = y'$ $\lightning$