Theorem (uniform expansivity). For every and there is such that for each pair of points of such that there is with such that , where is the expansivity constant of .
Proof. Let . Then K is closed, and hence compact. For each pair , there is such that . Since the mapping defined by is continuous, is also continuous and there is a neighborhood of each such that for each . Since is compact and is an open cover of , there is a finite subcover . Let . If , then , so that for some . Thus for we have and as requred.
|Date of creation||2013-03-22 13:55:15|
|Last modified on||2013-03-22 13:55:15|
|Last modified by||Koro (127)|