Normal Space is T3 Space

Theorem
Let $\left({X, \vartheta}\right)$ be a $T_4$ space.

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

Corollary
Let $\left({X, \vartheta}\right)$ be a $T_4$ space.

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

Proof
Let $T = \left({X, \vartheta}\right)$ be a $T_4$ space.

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

Let $x \in F$.

Then as $T$ is a Fréchet ($T_1$) space it follows that:


 * $\exists U \in \vartheta: x \in U, y \notin U$
 * $\exists V \in \vartheta: y \in V, x \notin V$

that is, there exist open sets $U$ and $V$ both containing one but not the other.

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.

So we have our two [[Definition:Open Set (Topology)|open sets] $U, V$:
 * $F \subseteq U$
 * $y \in V$

such that $U \cap V = \varnothing$.

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 regular space.

Proof of Corollary
Let $T = \left({X, \vartheta}\right)$ be a $T_4$ space.

From above, we have that $T$ is a regular space.

We also have by definition of $T_4$ space that $T$ is a Fréchet ($T_1$) space.

From $T_1$ Space is $T_0$ Space we have that $T$ is a $T_0$ Space

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

Hence $T$ is a $T_3$ space by definition.