Normal Space is T3 Space

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

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

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

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

Let $F$ be any closed set in $T$, and let $y \in \complement_S \left({F}\right)$, that is, $y \in S$ 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 $T = \left({S, \tau}\right)$ is a normal space, we have that:


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

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

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

So:
 * $\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$

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