Normal Space is Regular Space

Theorem

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

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

Proof

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

From Normal Space is $T_3$ Space, we have that $T$ is a $T_3$ space.

We also have by definition of normal space that $T$ is a $T_1$ (Fréchet) space.

From $T_1$ Space is $T_0$ Space we have that $T$ is a $T_0$ (Kolmogorov) space

So $T$ is both a $T_3$ space and a $T_0$ (Kolmogorov) space.

Hence $T$ is a regular space by definition.

$\blacksquare$

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