proof of Lebesgue number lemma

By way of contradictionMathworldPlanetmathPlanetmath, suppose that no Lebesgue number existed. Then there exists an open cover 𝒰 of X such that for all δ>0 there exists an xX such that no U𝒰 contains Bδ(x) (the open ballPlanetmathPlanetmath of radius δ around x). Specifically, for each n, since 1/n>0 we can choose an xnX such that no U𝒰 contains B1/n(xn). Now, X is compactPlanetmathPlanetmath so there exists a subsequence (xnk) of the sequence of points (xn) that convergesPlanetmathPlanetmath to some yX. Also, 𝒰 being an open cover of X implies that there exists λ>0 and U𝒰 such that Bλ(y)U. Since the sequence (xnk) converges to y, for k large enough it is true that d(xnk,y)<λ/2 (d is the metric on X) and 1/nk<λ/2. Thus after an application of the triangle inequalityMathworldMathworldPlanetmath, it follows that


contradicting the assumptionPlanetmathPlanetmath that no U𝒰 contains B1/n(xn). Hence a Lebesgue number for 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