uniform expansivity

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

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

Proof. Let $K=\{(x,y)\in X\times X:d(x,y)\ge ϵ/2\}$. Then K is closed, and hence compact. For each pair $(x,y)\in K$, there is $n_{(x,y)}\in \mathbb{Z}$ such that $d(f^{n_{(x,y)}}(x),f^{n_{(x,y)}}(y))\ge c$. Since the mapping $F:X\times X\to X\times X$ 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 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 $\{U_{(x_i,y_i)}:1\le i\le m\}$. Let $N=\max \{|n_{(x_i,y_i)}|:1\le i\le m\}$. If $d(x,y)>ϵ$, then $(x,y)\in K$, so that $(x,y)\in U_{(x_i,y_i)}$ for some $i\in \{1,\mathrm{\dots },m\}$. Thus for $n=n_{(x_i,y_i)}$ we have and $|n|\le N$ as requred.