T3 1/2 Space is T3 Space

Theorem
Let $T$ be a $T_{3 \frac 1 2}$ space.

Then $T$ is also a $T_3$ space.

Proof
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space.

From the definition of $T_{3 \frac 1 2}$ space:


 * 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 $\set y$.

Let $F \subseteq S$ be a closed set in $T$ and let $y \in \relcomp S F$.

An Urysohn function for $F$ and $\set y$ is a continuous mapping $f: S \to \closedint 0 1$ where:
 * $\forall a \in F: \map f a = 0$
 * $\map f y = 1$

Let:
 * $U = \map {f^{-1} } 0$


 * $V = \map {f^{-1} } 1$

As $f$ is continuous, both $U$ and $V$ are open in $T$.

Suppose $x \in U \cap V$.

Then we would have:
 * $x \in U \implies \map f x = 0$


 * $x \in V \implies \map f x = 1$

which contradicts the many-to-one nature of a mapping.

So $U \cap V = \O$.

Thus we have:


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

That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.

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