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 from Distinct Points in Metric Space have Disjoint Neighborhoods, there exist $\epsilon$-neighborhoods $N_\epsilon \left({x}\right)$ and $N_\epsilon \left({y}\right)$ which are disjoint metric spaces containing $x$ and $y$ respectively.

Hence the result by the definition of Hausdorff space.

$\blacksquare$

Sources

• Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{I}: \ \S 5$