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
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