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 HausdorffPlanetmathPlanetmath.

Say we’ve got distinct x,yX. Since d is a metric, d(x,y)0. Then the open balls Bx=B(x,d(x,y)2) and By=B(y,d(x,y)2) are open sets in the metric topology which contain x and y respectively. If we could show Bx and By 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 Bx and By. Suppose there is a z in both, and we’ll derive a contradictionMathworldPlanetmathPlanetmath. Since z is in these open balls, d(z,x)<d(x,y)2 and d(z,y)<d(x,y)2. But then d(z,x)+d(z,y)<d(x,y), contradicting the triangle inequalityMathworldMathworldPlanetmath.

So Bx and By are disjoint, and X is Hausdorff.

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