# metric spaces are Hausdorff

Suppose we have a space $X$ and a metric $d$ on $X$. We’d like to show that the metric topology that $d$ gives $X$ is Hausdorff^{}.

Say we’ve got distinct $x,y\in X$. Since $d$ is a metric, $d(x,y)\ne 0$. Then the open balls ${B}_{x}=B(x,\frac{d(x,y)}{2})$ and ${B}_{y}=B(y,\frac{d(x,y)}{2})$ are open sets in the metric topology which contain $x$ and $y$ respectively. If we could show ${B}_{x}$ and ${B}_{y}$ are disjoint, we’d have shown that $X$ is Hausdorff.

We’d like to show that an arbitrary point $z$ can’t be in both ${B}_{x}$ and ${B}_{y}$. Suppose there is a $z$ in both, and we’ll derive a contradiction^{}. Since $z$ is in these open balls, $$ and $$. But then $$, contradicting the triangle inequality^{}.

So ${B}_{x}$ and ${B}_{y}$ are disjoint, and $X$ is Hausdorff.$\mathrm{\square}$

Title | metric spaces are Hausdorff |
---|---|

Canonical name | MetricSpacesAreHausdorff |

Date of creation | 2013-03-22 14:21:29 |

Last modified on | 2013-03-22 14:21:29 |

Owner | waj (4416) |

Last modified by | waj (4416) |

Numerical id | 4 |

Author | waj (4416) |

Entry type | Proof |

Classification | msc 54D10 |

Classification | msc 54E35 |

Related topic | MetricSpace |

Related topic | SeparationAxioms |