# Normal Space is Regular Space

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## Theorem

Let $\left({S, \tau}\right)$ be a normal space.

Then $\left({S, \tau}\right)$ is also a regular space.

## Proof

Let $T = \left({S, \tau}\right)$ 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

- 1970: Lynn Arthur Steen and J. Arthur Seebach, Jr.:
*Counterexamples in Topology*... (previous) ... (next): $\text{I}: \ \S 2$: Regular and Normal Spaces