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.
$\blacksquare$