Normal Space is T3 Space

Theorem
Let $\left({X, \vartheta}\right)$ be a normal space.

Then $\left({X, \vartheta}\right)$ is also a $T_3$ space.

Corollary
Let $\left({X, \vartheta}\right)$ be a normal space.

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

Proof
Let $T = \left({X, \vartheta}\right)$ be a normal space.

From the definition of normal space:
 * $\left({X, \vartheta}\right)$ is a $T_4$ space
 * $\left({X, \vartheta}\right)$ is a Fréchet ($T_1$) space.

Let $F$ be any closed set in $T$, and let $y \in \complement_X \left({F}\right)$, that is, $y \in X$ such that $y \notin F$.

As $T$ is a Fréchet ($T_1$) space it follows from Equivalent Definitions for $T_1$ Space that $\left\{{y}\right\}$ is closed.

As $T = \left({X, \vartheta}\right)$ is a normal space, we have that:


 * $\forall A, B \in \complement \left({\vartheta}\right), A \cap B = \varnothing: \exists U, V \in \vartheta: A \subseteq U, B \subseteq V$

That is, for any two disjoint closed sets $A, B \subseteq X$ there exist open sets $U, V \in \vartheta$ containing $A$ and $B$ respectively.

But $F$ and $\left\{{y}\right\}$ are disjoint closed sets.

So:
 * $\forall F \in \complement \left({\vartheta}\right), y \in \complement_X \left({F}\right): \exists U, V \in \vartheta: F \subseteq U, y \in V: U \cap V = \varnothing$

which is precisely the definition of a $T_3$ space.

Proof of Corollary
Let $T = \left({X, \vartheta}\right)$ be a normal space.

From above, 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 $T_0$ Space

So $T$ is both a $T_3$ space and a $T_0$ space.

Hence $T$ is a regular space by definition.