# 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 $\u03f5>0$. For each $k\in K$ choose ${\delta}_{k}$ such that $d(k,x)\le {\delta}_{k}$ implies $d(f(k),f(x))\le \frac{\u03f5}{2}$. 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

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

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 $ | $\frac{\u03f5}{2}}+{\displaystyle \frac{\u03f5}{2}$ | (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 |