T4 and T3 Space is T 3 1/2

Theorem
Let $T = \left({X, \vartheta}\right)$ be:
 * a $T_4$ space,
 * a $T_3$ space.

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

Proof
Let $T = \left({X, \vartheta}\right)$ be a $T_4$ space which is also a $T_3$ space.

From it being $T_3$:
 * $\forall F \in \complement \left({\vartheta}\right), y \in \complement_X \left({F}\right): \exists U, V \in \vartheta: F \subseteq U, y \in V: U \cap V = \varnothing$

Consider this $U \in \vartheta$, which is disjoint from $\left\{{y}\right\}$.

Then $\complement_X \left({U}\right)$ is a closed set which is disjoint from $F$ but such that $\left\{{y}\right\} \subseteq \complement_X \left({U}\right)$.

As $T$ is a $T_4$ space, we have that from Urysohn's Lemma there exists an Urysohn function $f$ for $F$ and $\complement_X \left({U}\right)$.

As $\left\{{y}\right\} \subseteq \complement_X \left({U}\right)$, this function $f$ is a Urysohn function for $F$ and $\left\{{y}\right\}$ as well.

So:
 * For any closed set $F \subseteq X$ and any point $y \in X$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\left\{{y}\right\}$.

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