# proof of Heine-Cantor theorem

We prove this theorem in the case when $X$ and $Y$ are metric spaces.

Suppose $f$ is not uniformly continuous. Then

$$ |

In particular by letting $\delta =1/k$ we can construct two sequences ${x}_{k}$ and ${y}_{k}$ such that

$$ |

Since $X$ is compact^{} the two sequence have convergent subsequences i.e.

$${x}_{{k}_{j}}\to \overline{x}\in X,{y}_{{k}_{j}}\to \overline{y}\in X.$$ |

Since $d({x}_{k},{y}_{k})\to 0$ we have $\overline{x}=\overline{y}$. Being $f$ continuous^{} we hence conclude $d(f({x}_{{k}_{j}}),f({y}_{{k}_{j}}))\to 0$ which is a contradiction^{} being $d(f({x}_{k}),f({y}_{k}))\ge \u03f5$.

Title | proof of Heine-Cantor theorem |
---|---|

Canonical name | ProofOfHeineCantorTheorem |

Date of creation | 2013-03-22 13:31:26 |

Last modified on | 2013-03-22 13:31:26 |

Owner | paolini (1187) |

Last modified by | paolini (1187) |

Numerical id | 5 |

Author | paolini (1187) |

Entry type | Proof |

Classification | msc 46A99 |