proof of Heine-Cantor theorem

We seek to show that $f:K\to X$ is continuous with $K$ a compact metric space, then $f$ is uniformly continuous. Recall that for $f:K\to X$, uniform continuity is the condition that for any $\epsilon >0$, there exists $\delta$ such that

for all $x,y\in K$

Suppose $K$ is a compact metric space, $f$ continuous on $K$. Let $ϵ>0$. For each $k\in K$ choose ${\delta }_k$ such that $d(k,x)\le {\delta }_k$ implies $d(f(k),f(x))\le \frac{ϵ}{2}$. Note that the collection of balls $B(k,\frac{{\delta }_k}{2})$ covers $K$, so by compactness there is a finite subcover, say involving $k_1,\mathrm{\dots },k_n$. Take

$$\delta =\underset{i=1,\mathrm{\dots },n}{\min }\frac{{\delta }_{k_i}}{2}$$

Then, suppose $d(x,y)\le \delta$. By the choice of $k_1,\mathrm{\dots },k_n$ and the triangle inequality, there exists an $i$ such that $d(x,k_i),d(y,k_i)\le {\delta }_{k_i}$. Hence,

$d(f(x),f(y))$ $\le$ $d(f(x),f(k_i))+d(f(y),f(k_i))$ (1)

$\le$ ${\displaystyle \frac{ϵ}{2}}+{\displaystyle \frac{ϵ}{2}}$ (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