Metric Space is Hausdorff/Proof 1

Theorem

Let $M = \struct {A, d}$ 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 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.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $5$: Metric Spaces