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
Canonical name	MixingAction
Date of creation	2013-03-22 19:19:32
Last modified on	2013-03-22 19:19:32
Owner	Ziosilvio (18733)
Last modified by	Ziosilvio (18733)
Numerical id	5
Author	Ziosilvio (18733)
Entry type	Definition
Classification	msc 37B05