Metric Space is Hausdorff/Proof 2

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Theorem

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

Then $M$ is a Hausdorff space.


Proof

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

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

Thus, $M$ has to be Hausdorff.

$\blacksquare$