# mixing action

Let $X$ be a topological space^{} and let $G$ be a semigroup.
An action $\varphi ={\{{\varphi}_{g}\}}_{g\in G}$ of $G$ on $X$ is (topologically) mixing
if, given any two *open* subsets $U$, $V$ of $X$,
the intersection^{} $U\cap {\varphi}_{g}(V)$ 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 ${\varphi}_{n}(x)={F}^{n}(x)$
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

$${\varphi}_{g}(c)(z)=c(g\cdot z)\forall c:G\to X$$ |

is mixing.

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

$U$ | $=$ | $\mathrm{\{}c\in {X}^{G}\mid c{|}_{E}=u{|}_{E}\}$ | ||

$V$ | $=$ | $\mathrm{\{}c\in {X}^{G}\mid c{|}_{F}=v{|}_{F}\}$ |

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