Urysohn's Lemma Converse

Lemma
Let $T = \struct {S, \tau}$ be a topological space.

Let there exist an Urysohn function for any two $A, B \subseteq S$ which are closed sets in $T$ such that $A \cap B = \O$.

Then $T = \struct {S, \tau}$ is a $T_4$ space.

Proof
Let $A$ and $B$ be arbitrary closed sets of $T$ and let $f$ be an Urysohn function for $A$ and $B$.

Let $C = \hointr 0 {\dfrac 1 4}$ and $D = \hointl {\dfrac 3 4} 1$.

Then $C$ and $D$ are open in $\closedint 0 1$.

Hence, $\map {f^{-1} } C$ and $\map {f^{-1} } D$ are open in $T$.

Furthermore, by definition of Urysohn function, $A \subset \map {f^{-1} } C$ and $B \subset \map {f^{-1} } D$.

Also, from Preimage of Intersection under Mapping:
 * $C \cap D = \O \implies \map {f^{-1} } C \cap \map {f^{-1} } D = \O$

Therefore, $T$ is a $T_4$ space.

Also see

 * Urysohn's Lemma