Metric Space is Hausdorff

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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.


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


\(\ds z\) \(\in\) \(\ds \map {B_r} x \cap \map {B_r} y\)
\(\ds \leadsto \ \ \) \(\ds z\) \(\in\) \(\ds \map {B_r} x\) 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
\(\, \ds \land \, \) \(\ds \map d {y, z}\) \(<\) \(\ds r\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, z} + \map d {y, z}\) \(<\) \(\ds 2 r\)
\(\ds \leadsto \ \ \) \(\ds \map d {x, z} + \map d {y, z}\) \(<\) \(\ds \map d {x, y}\)

This contradicts Metric Space Axiom $(\text M 2)$: Triangle Inequality.

Thus, $M$ has to be Hausdorff.