\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}

