# 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), contradicting the triangle inequality.

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

Title metric spaces are Hausdorff MetricSpacesAreHausdorff 2013-03-22 14:21:29 2013-03-22 14:21:29 waj (4416) waj (4416) 4 waj (4416) Proof msc 54D10 msc 54E35 MetricSpace SeparationAxioms