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Consider Lebesgue's covering lemma in the following form:

Let $(X,d)$ be a compact metric space and let $\{U_i\}_{i\in I}$ be an open cover of $X$. Then there exists $\delta>0 $ such that each subset $Y$ of $X$ of diameter less than or equal to $\delta$ lies within some $U_i$.

What are possibly the important and striking uses of this lemma named after a famous mathematician? I have seen only one use and that was in the derivation of the fundamental group of the circle, using $\mathbb R$ as the universal cover. However I can't imagine that this is the only one, especially because in the more general setting of covering spaces, it is possible to do without this lemma. Is this lemma more fundamentally important befitting its name, and if so, what are some uses to convince myself?

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I recall $X$ being a compact metric space in the lemma. – Asaf Karagila Sep 19 '11 at 6:25
Oh yes, sorry, fixed. Thanks. – Lit Sep 19 '11 at 6:29

If I remember correctly, the basic application of this in my topology class was to prove that continuous maps from compact metric spaces were uniformly continuous.

However, the lemma is really important in algebraic topology. It is used almost everywhere where you need to cut up your domain (the interval for a path or the square for a homotopy) into sufficiently small parts. From the top of my head, we used it in the proof of Seifert-van Kampen's theorem, various results on covering spaces, probably the excision theorem for singular homology etc.

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This lemma is the key ingredient in the proof that, for metrizable spaces, topological compactness is the same as sequential compactness. Cfr. Munkres Topology, chapter 3, theorem 28.2.

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It’s not needed at all for that proof. A sequentially compact space is easily shown to be countably compact. If it’s metric, it’s totally bounded, hence separable, hence Lindelöf, and a countably compact Lindelöf space is compact. Total boundedness is easy. If for some $\epsilon>0$ $X$ has no finite $\epsilon$-net, recursively construct an infinite sequence whose points are pairwise at least $\epsilon$ apart; clearly it has no convergent subsequence. – Brian M. Scott Sep 19 '11 at 22:45
@BrianM.Scott: In the proof I'm thinking of, Lebesgue lemma is used to show that (sequentially compact) => (compact): indeed, fist one shows that a seq. compact space is totally bounded, then uses this to find a finite number of small enough balls that cover the whole space. By Lebesgue lemma each ball is contained in one set of the cover, enabling us to choose a finite subcover. – Giuseppe Negro Sep 20 '11 at 13:21
Yours is a different approach that I didn't know. I will think about it, thank you! – Giuseppe Negro Sep 20 '11 at 13:23

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