Normal Space is T3 Space

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

Then $\struct {S, \tau}$ is also a $T_3$ space.

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

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

Let $F$ be any closed set in $T$.

Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$.

As $T$ is a Fréchet ($T_1$) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed.

As $T = \struct {S, \tau}$ is a normal space, we have that:


 * $\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$

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

But $F$ and $\set y$ are disjoint closed sets.

So:
 * $\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$

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