43.2 Minimal Flows

Proof. (1) $\Rightarrow$ (2): Let $x \in X$, then $\ol{Gx}$ with the restricted action is a $G$-subflow of $X$. As $\emptyset \ne \ol{Gx}$ and $G \curvearrowright X$ admits no proper subflows, $\ol{Gx}= x$.

(2) $\Rightarrow$ (1): Let $G \curvearrowright Y$ be a $G$-subflow of $G \curvearrowright X$. For any $x \in Y$, $Y \supset \ol{Gx}= X$, so $Y = X$.

(2) $\Rightarrow$ (3): Let $\emptyset \ne U \subset X$ be open. Since $G \curvearrowright X$ is minimal, $Gx \cap U \ne \emptyset$ for all $x \in X$. Thus $X \subset G^{-1}U = GU$. By compactness of $X$, there exists $F \subset G$ finite such that $FU = X$.

Let $x \in X$ be arbitrary, then for each $g \in G$, $gx \in X = FU$. Thus there exists $h \in F$ with $gx \in hU$ and $h^{-1}gx \in U$. Hence $h^{-1}g \in N(x, U)$, and $g \in hN(x, U) \subset FN(x, U)$.$\square$

Proof. Let $x \in X$ and $U \in \cn_{X}(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$.$\square$

Proof. (1) $\Rightarrow$ (2): See Definition 43.2.1.

(2) $\Rightarrow$ (1): Let $Y = \ol{Gx}$ and $y \in Y$. For any $U \in \cn_{X}(x)$, since $X$ is a compact Hausdorff space, there exists $V \in \cn_{X}(x)$ with $\ol{V}\subset U$. As $N(x, V)$ is syndetic in $G$, there exists $F \subset G$ finite such that $G = F^{-1}N(x, V)$.

Since $\ol{Gx}= Y$, there exists an ultrafilter $\fF \subset 2^{G}$ such that $\fF x \to y$. As $G = \bigcup_{f \in F}N(x, V)$, there exists $f \in F$ such that $f^{-1}N(x, V) \in \fF$. Thus $y = \lim \fF x \subset \ol{f^{-1}N(x, V)}\subset \ol{f^{-1}V}$. Equivalently, $fy \in \ol{V}$, and $Gy \cap U \supset Gy \cap f^{-1}\ol{V}\ne \emptyset$. As this holds for all $U \in \cn_{X}(x)$, $x \in \ol{Gy}$, and $\ol{Gy}= \ol{Gx}= Y$. Therefore every point in $Y$ is transitive, and $G \curvearrowright Y$ is a minimal subflow.$\square$

Proof. By Theorem 43.2.4, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, Lemma 43.2.2 implies that every point in $Y$ is recurrent.$\square$