\tutorial{11}{2024-01-09}{}

An equivalent definition of subflows can be given as   follows:
\begin{definition}
    Let $(X,T)$ be a flow with action $\alpha_x$.
    Let $Y \subseteq X$ be a compact subspace of $X$.
    If $Y$ is invariant under $\alpha_x$, we say that
    $(Y,T)$ (with action  $\alpha_x\defon{T \times Y}$
    is a subflow of $(X,T)$.
\end{definition}

\begin{example}[Flows with a non-closed orbit]
    \begin{enumerate}[1.]
        \item Consider $(S^1, \Z)$
            with action given by $1 \cdot x = x + c$ for
            a fixed $c \in \R\setminus\Q$.\footnote{We identify $S^1$ and $\faktor{\R}{\Z}$.}
            Then the orbit of $0$, $\{nc : n \in \Z\}$ is dense but consists only of irrationals
            (except $0$),
            so it is not closed.
        \item Consider $(S^1, \Q)$ with action $qx \coloneqq x + q$.
            The orbit of $0$, $\faktor{\Q}{\Z} \subseteq S^1$,
            is dense but not closed.

            $(S^1,\Q)$ is minimal.
    \end{enumerate}
\end{example}
\begin{example}[\vocab{Left Bernoulli shift}]
    Consider $(\{0,1\}^{\Z},  T)$,
    where $T = \Z$ and the action is given by
    \begin{IEEEeqnarray*}{rCl}
       \Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
       (m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
    \end{IEEEeqnarray*}

    The orbit of $z \coloneqq (0)_{n \in \Z}$ consist of only on point.
    In particular it is closed.

    Let $x \coloneqq  ( [n = 0])_{n \in \Z}$.
    Then $Tx = \{([n = m])_{n \in \Z} | m \in \Z\}$.
    Clearly $z \not\in Tx$.
    \begin{claim}
        $z \in \overline{Tx}$
    \end{claim}
    \begin{proof}
        Consider a basic open $z \in U_I = \{y : y_i = 0, i \in I\}$
        where $I \subseteq \Z$ is finite.
        Then $U_I \cap Tx \neq \emptyset$
        as we can shift the $1$ out of $I$,
        i.e.~$(\max I + 1) x \in U_I$.
    \end{proof}
\end{example}

Flows are always on non-empty spaces $X$.
\begin{fact}
    Consider a flow $(X,T)$.
    The following are equivalent:
    \begin{enumerate}[(i)]
        \item Every $T$-orbit is dense.
        \item There is no proper subflow,
    \end{enumerate}
    If these conditions hold, the flow is called \vocab{minimal}.
\end{fact}
\begin{proof}
    (i) $\implies$ (ii):
    Let $(Y,T)$ be a subflow of $(X,T)$.
    take  $y \in Y$. Then $Ty$ is dense in $X$.
    But $Ty \subseteq Y$, so $Y$ is dense in $X$.
    Since $Y$ is closed, we get $Y = X$.

    (ii) $\implies$ (i):
    Take $x \in X$. Consider $Tx$.
    It suffices to show that $\overline{Tx}$ is a subflow.
    Clearly $\overline{Tx}$ is closed,
    so it suffices to show that it is  $T$-invariant.
    Let $y \in \overline{Tx}$ and $t \in T$.
    Take $ty \in U \overset{\text{open}}{\subseteq} X$.
    Since $t^{-1}$ acts as a homeomorphism
    we have $y \in t^{-1} U \overset{\text{open}}{\subseteq} X$.
    We find some $t'x \in t^{-1}U$ since $y \in \overline{Tx}$.
    So $tt'x \in Tx \cap U$.
\end{proof}

\begin{fact}
    Every flow $(X,T)$ contains a minimal subflow.
\end{fact}
\begin{proof}
    We use Zorn's lemma:
    Let $S$ be the set of all subflows of $(X,T)$
    ordered by  $Y \le Y' :\iff Y \supseteq Y'$.
    We need to show that for a chain $\langle Y_i : i \in I \rangle$,
    there exists a lower bound.
    Consider $\bigcap_{i \in I} Y_i$. This a subflow:
    \begin{itemize}
        \item It is closed as it is an intersection of closed sets.
        \item It is $T$-invariant, since each of the $Y_i$ is.
        \item It is non-empty by \yaref{tut10fact}.
    \end{itemize}
\end{proof}
\begin{fact}
    \label{tut10fact}
    Let $X$ be a topological space.
    Then $X$ is compact iff every family of closed sets with
    FIP\footnote{finite intersection property, i.e.~the intersection of every finite sub-family is non-empty}
    has non-empty intersection.
\end{fact}
\begin{proof}
    Note that families of
    closed sets correspond to families of open sets by taking complements.
    A family of open sets is a cover iff the corresponding family
    has empty intersection,
    and is admits a finite subcover iff the corresponding family
    has the FIP.
\end{proof}