Josia Pietsch
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\tutorial{11}{20240109}{}




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 nonclosed 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 nonempty 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 nonempty 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 subfamily is nonempty}


has nonempty 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}
