proof of Heine-Cantor theorem

We seek to show that f:KX is continuousMathworldPlanetmath with K a compactPlanetmathPlanetmath metric space, then f is uniformly continuousPlanetmathPlanetmath. Recall that for f:KX, uniform continuity is the condition that for any ε>0, there exists δ such that


for all x,yK

Suppose K is a compact metric space, f continuous on K. Let ϵ>0. For each kK choose δk such that d(k,x)δk implies d(f(k),f(x))ϵ2. Note that the collection of balls B(k,δk2) covers K, so by compactness there is a finite subcover, say involving k1,,kn. Take


Then, suppose d(x,y)δ. By the choice of k1,,kn and the triangle inequalityMathworldMathworldPlanetmath, there exists an i such that d(x,ki),d(y,ki)δki. Hence,

d(f(x),f(y)) d(f(x),f(ki))+d(f(y),f(ki)) (1)
ϵ2+ϵ2 (2)

As x,y were arbitrary, we have that f is uniformly continuous.
This proof is similar to one found in Mathematical Principles of Analysis, Rudin.

Title proof of Heine-Cantor theorem
Canonical name ProofOfHeineCantorTheorem
Date of creation 2013-03-22 15:09:43
Last modified on 2013-03-22 15:09:43
Owner drini (3)
Last modified by drini (3)
Numerical id 10
Author drini (3)
Entry type Proof
Classification msc 46A99