T4 and T3 Space is T 3 1/2

Theorem

Let $T = \struct {S, \tau}$ be:

a $T_4$ space

and also:

a $T_3$ space.

Then $T$ is also a $T_{3 \frac 1 2}$ space.

Proof

Let $T = \struct {S, \tau}$ be a $T_4$ space which is also a $T_3$ space.

From it being $T_3$:

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

Consider this $U \in \tau$, which is disjoint from $\set y$.

Then $\relcomp S U$ is a closed set which is disjoint from $F$ but such that $\set y \subseteq \relcomp S U$.

As $T$ is a $T_4$ space, we have that from Urysohn's Lemma there exists an Urysohn function $f$ for $F$ and $\relcomp S U$.

As $\set y \subseteq \relcomp S U$, this function $f$ is a Urysohn function for $F$ and $\set y$ as well.

So:

For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$.

which is precisely the definition of a $T_{3 \frac 1 2}$ space.

$\blacksquare$