proof of Lebesgue number lemma
By way of contradiction, suppose that no Lebesgue number existed. Then there exists an open cover of such that for all there exists an such that no contains (the open ball of radius around ). Specifically, for each , since we can choose an such that no contains . Now, is compact so there exists a subsequence of the sequence of points that converges to some . Also, being an open cover of implies that there exists and such that . Since the sequence converges to , for large enough it is true that ( is the metric on ) and . Thus after an application of the triangle inequality, it follows that
contradicting the assumption that no contains . Hence a Lebesgue number for does exist.
|Title||proof of Lebesgue number lemma|
|Date of creation||2013-03-22 13:09:20|
|Last modified on||2013-03-22 13:09:20|
|Last modified by||scanez (1021)|