# 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)) 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)) and $|n|\leq N$ as requred.

Title uniform expansivity UniformExpansivity 2013-03-22 13:55:15 2013-03-22 13:55:15 Koro (127) Koro (127) 7 Koro (127) Theorem msc 37B99