uniform expansivity

Let $(X,d)$ be a compact metric space and let $f\colon X\to X$ be an expansive homeomorphism.

Theorem (uniform expansivity). For every $\epsilon>0$ and $\delta>0$ there is $N>0$ such that for each pair $x, y$ of points of $X$ such that $d(x,y)>\epsilon$ there is $n\in\mathbb{Z}$ with $|n|\leq N$ such that $d(f^{n}(x),f^{n}(y))>c-\delta$ , where $c$ is the expansivity constant of $f$ .

Proof. Let $K=\{(x,y)\in X\times X:d(x,y)\geq\epsilon/2\}$ . Then K is closed, and hence compact. For each pair $(x,y)\in K$ , there is $n_{(x,y)}\in\mathbb{Z}$ such that $d(f^{n_{(x,y)}}(x),f^{n_{(x,y)}}(y))\geq c$ . Since the mapping $F\colon X\times X\to X\times X$ defined by $F(x,y)=(f(x),f(y))$ is continuous, $F^{n_{x}}$ is also continuous and there is a neighborhood $U_{(x,y)}$ of each $(x,y)\in K$ such that $d(f^{n_{(x,y)}}(u),f^{n_{(x,y)}}(v))<c-\delta$ for each $(u,v)\in U_{(x,y)}$ . Since $K$ is compact and $\{U_{(x,y)}:(x,y)\in K\}$ is an open cover of $K$ , there is a finite subcover $\{U_{(x_{i},y_{i})}:1\leq i\leq m\}$ . Let $N=\max\{|n_{(x_{i},y_{i})}|:1\leq i\leq m\}$ . If $d(x,y)>\epsilon$ , then $(x,y)\in K$ , so that $(x,y)\in U_{(x_{i},y_{i})}$ for some $i\in\{1,\dots,m\}$ . Thus for $n=n_{(x_{i},y_{i})}$ we have $d(f^{n}(x),f^{n}(y))<c-\delta$ and $|n|\leq N$ as requred.

Title	uniform expansivity
Canonical name	UniformExpansivity
Date of creation	2013-03-22 13:55:15
Last modified on	2013-03-22 13:55:15
Owner	Koro (127)
Last modified by	Koro (127)
Numerical id	7
Author	Koro (127)
Entry type	Theorem
Classification	msc 37B99