# Metric Space is Hausdorff

## Theorem

Let $M = \struct {A, d}$ be a metric space.

Then $M$ is a Hausdorff space.

## Proof 1

Let $x, y \in A: x \ne y$.

Then from Distinct Points in Metric Space have Disjoint Open Balls, there exist open $\epsilon$-balls $\map {B_\epsilon} x$ and $\map {B_\epsilon} y$ which are disjoint open sets containing $x$ and $y$ respectively.

Hence the result by the definition of Hausdorff space.

$\blacksquare$

## Proof 2

Aiming for a contradiction, suppose $M$ is not Hausdorff.

That is, there are $x, y \in A: x \ne y$ such that:

$\forall \epsilon \in \R_{>0}: \exists z \in \map {B_\epsilon} x \cap \map {B_\epsilon} y$

where $\map {B_\epsilon} x$ denote the open $\epsilon$-ball of $x$ in $M$.

Let $r = \dfrac {\map d {x, y} } 2$.

Let $z \in \map {B_r} x \cap \map {B_r} y$.

Then:

 $\ds z$ $\in$ $\ds \map {B_r} x \cap \map {B_r} y$ $\ds \leadsto \ \$ $\ds z$ $\in$ $\ds \map {B_r} x$ Definition of Set Intersection $\, \ds \land \,$ $\ds z$ $\in$ $\ds \map {B_r} y$ $\ds \leadsto \ \$ $\ds \map d {x, z}$ $<$ $\ds r$ Definition of Open Ball $\, \ds \land \,$ $\ds \map d {y, z}$ $<$ $\ds r$ $\ds \leadsto \ \$ $\ds \map d {x, z} + \map d {y, z}$ $<$ $\ds 2 r$ $\ds \leadsto \ \$ $\ds \map d {x, z} + \map d {y, z}$ $<$ $\ds \map d {x, y}$

Thus, $M$ has to be Hausdorff.

$\blacksquare$