diff --git a/src/dyn/top/def.tex b/src/dyn/top/def.tex index a33c960..6952c71 100644 --- a/src/dyn/top/def.tex +++ b/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 diff --git a/src/dyn/top/index.tex b/src/dyn/top/index.tex index 38b357e..76a6106 100644 --- a/src/dyn/top/index.tex +++ b/src/dyn/top/index.tex @@ -2,3 +2,5 @@ \label{chap:topological-dynamics} \input{./def.tex} +\input{./minimal.tex} +\input{./recurrence.tex} diff --git a/src/dyn/top/minimal.tex b/src/dyn/top/minimal.tex new file mode 100644 index 0000000..55f2e60 --- /dev/null +++ b/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} + diff --git a/src/dyn/top/recurrence.tex b/src/dyn/top/recurrence.tex new file mode 100644 index 0000000..172c3c7 --- /dev/null +++ b/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}