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, $d(x,y)\neq 0$ . 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, $d(z,x)<\frac{d(x,y)}{2}$ and $d(z,y)<\frac{d(x,y)}{2}$ . But then $d(z,x)+d(z,y)<d(x,y)$ , contradicting the triangle inequality.

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

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