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Fixed typo.

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Bokuan Li
2026-09-16 13:27:08 -04:00
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src/dyn/top/def.tex
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@@ -149,7 +149,7 @@
\begin{example}[The Shift] \begin{example}[The Shift]
\label{example:bernoulli-shift-z} \label{example:bernoulli-shift-z}
Let $\integer \curvearrowright 2^\integer$ be the Bernoulli shift of $G$, then: Let $\integer \curvearrowright 2^\integer$ be the Bernoulli shift of $\integer$, then:
\begin{enumerate} \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 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 $\integer \curvearrowright 2^\integer$ is point transitive, and hence topologically transitive\footnote{It is easier to show topological transitivity: move the trees until they are disjoint}.