# mixing action

Let $X$ be a topological space and let $G$ be a semigroup. An action $\phi=\{\phi_{g}\}_{g\in G}$ of $G$ on $X$ is (topologically) mixing if, given any two open subsets $U$, $V$ of $X$, the intersection $U\cap\phi_{g}(V)$ is nonempty for all $g\in G$ except at most finitely many.

Example 1. Let $F:X\to X$ be a continuous function. Then $F$ is topologically mixing if and only if the action of the monoid $\mathbb{N}$ on $X$ defined by $\phi_{n}(x)=F^{n}(x)$ is mixing according to the definition given above.

Example 2. Suppose $X$ is a discrete nonempty set and $G$ is a group; endow $X^{G}$ with the product topology. The action of $G$ on $X^{G}$ defined by

 $\phi_{g}(c)(z)=c(g\cdot z)\;\;\forall c:G\to X$

is mixing.

To prove this fact, we may suppose without loss of generality that $U$ and $V$ are two cylindric sets of the form:

 $\displaystyle U$ $\displaystyle=$ $\displaystyle\{c\in X^{G}\mid\left.{c}\right|_{E}=\left.{u}\right|_{E}\}$ $\displaystyle V$ $\displaystyle=$ $\displaystyle\{c\in X^{G}\mid\left.{c}\right|_{F}=\left.{v}\right|_{F}\}$

for suitable finite subsets $E,F\subseteq G$ and functions $u,v:G\to X$. Then the only chance for $U\cap\phi_{g}(V)$ to be empty, is that $e=gf$ for some $e\in E$, $f\in F$ such that $u(e)\neq v(f)$: but then, $g\in EF^{-1}$, which is finite.

Title mixing action MixingAction 2013-03-22 19:19:32 2013-03-22 19:19:32 Ziosilvio (18733) Ziosilvio (18733) 5 Ziosilvio (18733) Definition msc 37B05