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 $T = \left({X, \vartheta}\right)$ be a $T_{3 \frac 1 2}$ space.

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


 * 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 $\left\{{y}\right\}$.

Let $F \subseteq X$ be a closed set in $T$ and let $y \in \complement_X \left({F}\right)$.

An Urysohn function for $F$ and $\left\{{y}\right\}$ is a continuous mapping $f: X \to \left[{0. . 1}\right]$ where:
 * $\forall a \in F: f \left({a}\right) = 0$
 * $f \left({y}\right) = 1$

Let:
 * $U = f^{-1} \left({0}\right)$


 * $V = f^{-1} \left({1}\right)$

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

Suppose $x \in U \cap V$.

Then we would have:
 * $x \in U \implies f \left({x}\right) = 0$


 * $x \in V \implies f \left({x}\right) = 1$

which contradicts the functional nature of a mapping

So $U \cap V = \varnothing$

Thus we have:


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

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

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