Completely Normal Space is Normal Space

Theorem

Let $\struct {S, \tau}$ be a completely normal space.

Then $\struct {S, \tau}$ is also a normal space.

Proof

Let $\struct {S, \tau}$ be a completely normal space.

From the definition, $\struct {S, \tau}$ is a completely normal space if and only if:

$\struct {S, \tau}$ is a $T_5$ space

$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.

We have that a $T_5$ space is a $T_4$ space.

So:

$\struct {S, \tau}$ is a $T_4$ space

$\struct {S, \tau}$ is a $T_1$ (Fréchet) space.

which is precisely the definition of a normal 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 $2$: Separation Axioms: Regular and Normal Spaces