# 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 ProofOfHeineCantorTheorem 2013-03-22 15:09:43 2013-03-22 15:09:43 drini (3) drini (3) 10 drini (3) Proof msc 46A99