Metric Space is Hausdorff

Theorem
Let $$M = \left\{{A, d}\right\}$$ be a metric space.

Then $$M$$ is a Hausdorff space.

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

Then $$d \left({x, y}\right) > 0$$.

Put $$\epsilon = \frac {d \left({x, y}\right)} 2$$.

Then the $\epsilon$-neighborhoods $$N_\epsilon \left({x}\right)$$ and $$N_\epsilon \left({y}\right)$$ are disjoint metric spaces containing $$x$$ and $$y$$ respectively.

Hence the result by the definition of Hausdorff space