Josia Pietsch e887f46a5d
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An equivalent definition of subflows can be given as follows:
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 $(X,T)$.
\begin{example}[Flows with a non-closed orbit]
\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.
\begin{example}[\vocab{Left Bernoulli shift}]
Consider $(\{0,1\}^{\Z}, T)$,
where $T = \Z$ and the action is given by
\Z \times \{0,1\}^{\Z}&\longrightarrow & \Z \\
(m, (x_n)_{n \in \Z})&\longmapsto & (x_{n+m})_{n \in \Z}.
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 $z \not\in Tx$.
$z \in \overline{Tx}$
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.~$(\max I + 1) x \in U_I$.
Flows are always on non-empty spaces $X$.
Consider a flow $(X,T)$.
The following are equivalent:
\item Every $T$-orbit is dense.
\item There is no proper subflow,
If these conditions hold, the flow is called \vocab{minimal}.
(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 $tt'x \in Tx \cap U$.
Every flow $(X,T)$ contains a minimal subflow.
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:
\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}.
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.
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.