proof of Lebesgue number lemma

By way of contradiction, suppose that no Lebesgue number existed. Then there exists an open cover $\mathcal{U}$ of $X$ such that for all $\delta>0$ there exists an $x\in X$ such that no $U\in\mathcal{U}$ contains $B_{\delta}(x)$ (the open ball of radius $\delta$ around $x$ ). Specifically, for each $n\in\mathbb{N}$ , since $1/n>0$ we can choose an $x_{n}\in X$ such that no $U\in\mathcal{U}$ contains $B_{1/n}(x_{n})$ . Now, $X$ is compact so there exists a subsequence $(x_{n_{k}})$ of the sequence of points $(x_{n})$ that converges to some $y\in X$ . Also, $\mathcal{U}$ being an open cover of $X$ implies that there exists $\lambda>0$ and $U\in\mathcal{U}$ such that $B_{\lambda}(y)\subseteq U$ . Since the sequence $(x_{n_{k}})$ converges to $y$ , for $k$ large enough it is true that $d(x_{n_{k}},y)<\lambda/2$ ( $d$ is the metric on $X$ ) and $1/n_{k}<\lambda/2$ . Thus after an application of the triangle inequality, it follows that

B_{1/n_{k}}(x_{n_{k}})\subseteq B_{\lambda}(y)\subseteq U,

contradicting the assumption that no $U\in\mathcal{U}$ contains $B_{1/n}(x_{n})$ . Hence a Lebesgue number for $\mathcal{U}$ does exist.

Title	proof of Lebesgue number lemma
Canonical name	ProofOfLebesgueNumberLemma
Date of creation	2013-03-22 13:09:20
Last modified on	2013-03-22 13:09:20
Owner	scanez (1021)
Last modified by	scanez (1021)
Numerical id	7
Author	scanez (1021)
Entry type	Proof
Classification	msc 54E45