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Forgot to assume ultrafilter.

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Bokuan Li
2026-09-23 15:48:33 -04:00
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src/dyn/top/def.tex
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@@ -138,7 +138,7 @@
(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 $G = F^{-1}N(x, V)$. (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 a filter $\fF \subset 2^G$ 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 $Gy \cap U \supset Gy \cao f^{-1}\ol{V} \ne \emptyset$. 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. 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 $Gy \cap U \supset Gy \cao f^{-1}\ol{V} \ne \emptyset$. 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.
\end{proof} \end{proof}
\begin{theorem}[Existence of Minimal Subflows] \begin{theorem}[Existence of Minimal Subflows]