Dilation situation.

src/dyn/top/def.tex

@@ -34,14 +34,65 @@
 \begin{definition}[Transitive]
 \label{definition:tds-transitive}
     Let $G \curvearrowright X$ be a $G$-flow and $x \in X$, then $x$ is \textbf{topologically transitive} with respect to $G$ if $Gx$ is dense in $X$. 
+    
+    The flow $G \curvearrowright X$ is \textbf{point transitive} if it admits a topologically transitive point. 
+
+    Finally, $G \curvearrowright X$ is \textbf{topologically transitive} if for any $U, V \subset X$ non-empty and open, there exists $g \in G$ such that $gU \cap V \ne \emptyset$\footnote{Approximations of points may be translated into approximations of other points. }. 
 \end{definition}
 
+\begin{proposition}
+\label{proposition:tds-point-transitive}
+    Let $G \curvearrowright X$ be a point transitive $G$-flow, then $G \curvearrowright X$ is also topologically transitive. 
+\end{proposition}
+\begin{proof}
+    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}[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. 
+    \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 $Y = X$. 
+\end{proof}
 
 
 
+\begin{definition}[Bernoulli Shift]
+\label{definition:bernoulli-shift}
+    Let $G$ be a group, $X = 2^G$ equipped with the product topology, and
+    \[
+    \alpha: X \to X \quad \alpha_g(x)(h) = x(g^{-1}h)
+    \]
+
+    Viewing $2^G$ as the power set of $G$, $\alpha_g(A) = gA$ for any $A \subset G$. The $G$-flow $G \curvearrowright X$ is the \textbf{Bernoulli shift} of $G$, and $G$-subflows of $X$ are \textbf{subshifts} of $G$. 
+\end{definition}
+
+\begin{example}[The Shift]
+    \label{example:bernoulli-shift-z}
+
+    Let $\integer \curvearrowright 2^\integer$ be the Bernoulli shift of $G$, then:
+    \begin{enumerate}
+        \item For any $x, y \in 2^\integer$, $y \in \ol{\integer x}$ if and only if for each $A \subset \integer$, there exists $n \in \integer$ such that $y|_A = nx|_A$. In other words, $x$ contains a copy of any segment of $y$\footnote{Topological conditions in Bernoulli shifts translate into combinatorial information. }. 
+        \item $\integer \curvearrowright 2^\integer$ is point transitive, and hence topologically transitive\footnote{It is easier to show topological transitivity: move the trees until they are disjoint}. 
+        \item 
+    \end{enumerate}
+    
+\end{example}
+
+