Metric Space is Fully Normal

Theorem

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

Then $M$ is a fully normal space.

Proof

We have that a metric space is fully $T_4$.

We also have that a metric space is a $T_1$ (Fréchet) space.

Hence the result, by definition of a fully normal space.

$\blacksquare$

Sources

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