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 $\varepsilon>0$ , there exists $\delta$ such that

d_{K}(x,y)<\delta\implies d_{X}(f(x),f(y))<\epsilon

for all $x,y\in K$

Suppose $K$ is a compact metric space, $f$ continuous on $K$ . Let $\epsilon>0$ . For each $k\in K$ choose $\delta_{k}$ such that $d(k,x)\leq\delta_{k}$ implies $d(f(k),f(x))\leq\frac{\epsilon}{2}$ . Note that the collection of balls $B(k,\frac{\delta_{k}}{2})$ covers $K$ , so by compactness there is a finite subcover, say involving $k_{1},\ldots,k_{n}$ . Take

\delta=\min_{i=1,\ldots,n}\frac{\delta_{k_{i}}}{2}

Then, suppose $d(x,y)\leq\delta$ . By the choice of $k_{1},\ldots,k_{n}$ and the triangle inequality, there exists an $i$ such that $d(x,k_{i}),d(y,k_{i})\leq\delta_{k_{i}}$ . Hence,

	$\displaystyle d(f(x),f(y))$	$\displaystyle\leq$	$\displaystyle d(f(x),f(k_{i}))+d(f(y),f(k_{i}))$		(1)
		$\displaystyle\leq$	$\displaystyle\frac{\epsilon}{2}+\frac{\epsilon}{2}$		(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