Dynamics cleanup.

src/dyn/top/def.tex

@@ -48,41 +48,6 @@
     Let $x \in G$ be a topologically transitive point and $U, V \subset X$ be non-empty and open. Since $Gx$ is dense in $G$, $Gx \cap U \ne \emptyset$ and $Gx \cap V \ne \emptyset$. After translating $x$, assume without loss of generality that $x \in U$, then $GU \supset Gx \cap V \ne \emptyset$. Therefore there exists $g \in G$ such that $gU \cap V \ne \emptyset$. 
 \end{proof}
 
-\begin{definition}[Periodic]
-\label{definition:tds-periodic}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{periodic} with respect to $G$ if $Gx$ is finite. 
-\end{definition}
-
-\begin{definition}[Return Set]
-\label{definition:tds-return-set}
-    Let $G \curvearrowright X$ be a $G$-flow, $x \in X$, and $\emptyset \ne U \subset X$, then the \textbf{return set} of $x$ to $U$ is
-    \[
-    N(x, U) = \bracsn{g \in G|gx \in U}
-    \]
-\end{definition}
-
-\begin{definition}[Recurrence]
-\label{definition:tds-recurrence}
-    Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
-\end{definition}
-
-\begin{definition}[Uniformly Recurrent]
-\label{definition:tds-uniformly-recurrent}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
-\end{definition}
-
-
-\begin{definition}[Syndetic]
-\label{definition:group-syndetic}
-    Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
-\end{definition}
-
-\begin{lemma}
-\label{lemma:subgroup-syndetic}
-    Let $G$ be a group and $H \subset G$ be a subgroup, then $H$ is syndetic if and only if the index of $H$ in $G$ is finite. 
-\end{lemma}
-% Proof omitted. 
-
 \begin{example}
     \label{example:z-syndetic}
     Let $S \subset \integer$, then $S$ has \textbf{bounded gaps} if
@@ -94,69 +59,6 @@
 \end{example}
 % Proof omitted. 
 
-\begin{definition}[Minimality]
-\label{definition:tds-minimal}
-    Let $G \curvearrowright X$ be a $G$-flow, then the following are equivalent:
-    \begin{enumerate}
-        \item $G \curvearrowright X$ admits no proper subflows.
-        \item Every point in $X$ is topologically transitive. 
-        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
-    \end{enumerate}
-    
-
-    If the above holds, then $G \curvearrowright X$ is \textbf{minimal}.
-\end{definition}
-\begin{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 $g \in hN(x, U) \subset FN(x, U)$.  
-\end{proof}
-
-\begin{lemma}
-\label{lemma:tds-minimal-recurrent}
-    Let $G$ be an infinite group and $G \curvearrowright X$ be minimal $G$-flow, then every point in $X$ is recurrent. 
-\end{lemma}
-\begin{proof}
-    Let $x \in X$ and $U \in \cn_X(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$. 
-\end{proof}
-
-
-\begin{definition}[Minimal Point]
-\label{definition:tds-minimal-point}
-    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
-    \begin{enumerate}
-        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
-        \item $x$ is uniformly recurrent. 
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
-
-    (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. 
-\end{proof}
-
-\begin{theorem}[Existence of Minimal Subflows]
-\label{theorem:tds-minimal-existence}
-    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
-\end{theorem}
-% Proof omitted. 
-
-\begin{corollary}
-\label{corollary:tds-recurrent-existence}
-    Let $G$ be an infinite group and $G \curvearrowright X$ be a $G$-flow, then $X$ admits a recurrent point. 
-\end{corollary}
-\begin{proof}
-    By \autoref{theorem:tds-minimal-existence}, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, \autoref{lemma:tds-minimal-recurrent} implies that every point in $Y$ is recurrent. 
-\end{proof}
-
-
-
 \begin{definition}[Bernoulli Shift]
 \label{definition:bernoulli-shift}
     Let $G$ be a group, $X = 2^G$ equipped with the product topology, and

src/dyn/top/index.tex

@@ -2,3 +2,5 @@
 \label{chap:topological-dynamics}
 
 \input{./def.tex}
+\input{./minimal.tex}
+\input{./recurrence.tex}

src/dyn/top/minimal.tex

@@ -0,0 +1,64 @@
+\section{Minimal Flows}
+\label{section:minimal-flow}
+
+\begin{definition}[Minimality]
+\label{definition:tds-minimal}
+    Let $G \curvearrowright X$ be a $G$-flow, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright X$ admits no proper subflows.
+        \item Every point in $X$ is topologically transitive. 
+        \item For every $x \in X$ and $\emptyset \ne U \subset X$ open, $N(x, U)$ is syndetic in $G$. In particular, $x$ is uniformly recurrent. 
+    \end{enumerate}
+    
+
+    If the above holds, then $G \curvearrowright X$ is \textbf{minimal}.
+\end{definition}
+\begin{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 $g \in hN(x, U) \subset FN(x, U)$.  
+\end{proof}
+
+\begin{lemma}
+\label{lemma:tds-minimal-recurrent}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be minimal $G$-flow, then every point in $X$ is recurrent. 
+\end{lemma}
+\begin{proof}
+    Let $x \in X$ and $U \in \cn_X(x)$, then $N(x, U)$ is syndetic. As $G$ is infinite, so is $N(x, U)$. 
+\end{proof}
+
+
+\begin{definition}[Minimal Point]
+\label{definition:tds-minimal-point}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then the following are equivalent:
+    \begin{enumerate}
+        \item $G \curvearrowright \ol{Gx}$ is a minimal $G$-subflow.
+        \item $x$ is uniformly recurrent. 
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): See \autoref{definition:tds-minimal}. 
+
+    (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. 
+\end{proof}
+
+\begin{theorem}[Existence of Minimal Subflows]
+\label{theorem:tds-minimal-existence}
+    Let $G \curvearrowright X$ be a $G$-flow, then there exists $Y \subset X$ closed such that $G \curvearrowright Y$ is minimal. In which case, every point in $Y$ is a minimal. 
+\end{theorem}
+% Proof omitted. 
+
+\begin{corollary}
+\label{corollary:tds-recurrent-existence}
+    Let $G$ be an infinite group and $G \curvearrowright X$ be a $G$-flow, then $X$ admits a recurrent point. 
+\end{corollary}
+\begin{proof}
+    By \autoref{theorem:tds-minimal-existence}, there exists $Y \subset X$ closed such that $G \curvearrowright X$ is minimal. In which case, \autoref{lemma:tds-minimal-recurrent} implies that every point in $Y$ is recurrent. 
+\end{proof}
+

src/dyn/top/recurrence.tex

@@ -0,0 +1,36 @@
+\section{Topological Recurrence}
+\label{section:tds-recurrence}
+
+\begin{definition}[Periodic]
+\label{definition:tds-periodic}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{periodic} with respect to $G$ if $Gx$ is finite. 
+\end{definition}
+
+\begin{definition}[Return Set]
+\label{definition:tds-return-set}
+    Let $G \curvearrowright X$ be a $G$-flow, $x \in X$, and $\emptyset \ne U \subset X$, then the \textbf{return set} of $x$ to $U$ is
+    \[
+    N(x, U) = \bracsn{g \in G|gx \in U}
+    \]
+\end{definition}
+
+\begin{definition}[Recurrence]
+\label{definition:tds-recurrence}
+    Let $G$ be an infinite group, $G \curvearrowright X$ be a $G$-flow, and $x \in X$, then $x$ is \textbf{recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is infinite.
+\end{definition}
+
+\begin{definition}[Syndetic]
+\label{definition:group-syndetic}
+    Let $G$ be a group and $S \subset G$, then $S$ is \textbf{(left) syndetic} if there exists $F \subset G$ finite $FS = G$. 
+\end{definition}
+
+\begin{lemma}
+\label{lemma:subgroup-syndetic}
+    Let $G$ be a group and $H \subset G$ be a subgroup, then $H$ is syndetic if and only if the index of $H$ in $G$ is finite. 
+\end{lemma}
+% Proof omitted. 
+
+\begin{definition}[Uniformly Recurrent]
+\label{definition:tds-uniformly-recurrent}
+    Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{uniformly recurrent} if for every $U \in \cn_X(x)$, $N(x, U)$ is syndetic. 
+\end{definition}