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单词 ProofOfHeineCantorTheorem1
释义

proof of Heine-Cantor theorem


We prove this theorem in the case when X and Y are metric spaces.

Suppose f is not uniformly continuous. Then

ϵ>0δ>0x,yXd(x,y)<δbutd(f(x),f(y))ϵ.

In particular by letting δ=1/k we can construct two sequences xk and yk such that

d(xk,yk)<1/kandd(f(xk),f(yk)ϵ.

Since X is compactPlanetmathPlanetmath the two sequence have convergent subsequences i.e.

xkjx¯X,ykjy¯X.

Since d(xk,yk)0 we have x¯=y¯. Being f continuousPlanetmathPlanetmath we hence conclude d(f(xkj),f(ykj))0 which is a contradictionMathworldPlanetmathPlanetmath being d(f(xk),f(yk))ϵ.

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更新时间:2025/5/4 5:55:50