# 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 ProofOfLebesgueNumberLemma 2013-03-22 13:09:20 2013-03-22 13:09:20 scanez (1021) scanez (1021) 7 scanez (1021) Proof msc 54E45