# 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 $\u03f5>0$ and $\delta >0$ there is $N>0$ such that for each pair $x,y$ of points of $X$ such that $d(x,y)>\u03f5$ 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 $K=\{(x,y)\in X\times X:d(x,y)\ge \u03f5/2\}$. 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 $$ for each $(u,v)\in {U}_{(x,y)}$. Since $K$ is compact and $\{{U}_{(x,y)}:(x,y)\in K\}$ is an open cover of $K$, there is a finite subcover $\{{U}_{({x}_{i},{y}_{i})}:1\le i\le m\}$. Let $N=\mathrm{max}\{|{n}_{({x}_{i},{y}_{i})}|:1\le i\le m\}$. If $d(x,y)>\u03f5$, then $(x,y)\in K$, so that $(x,y)\in {U}_{({x}_{i},{y}_{i})}$ for some $i\in \{1,\mathrm{\dots},m\}$. Thus for $n={n}_{({x}_{i},{y}_{i})}$ we have $$ and $|n|\le N$ as requred.

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 |