Completely Hausdorff Space is Hausdorff Space

Theorem
Let $\left({X, \vartheta}\right)$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Then $\left({X, \vartheta}\right)$ is also a $T_2$ (Hausdorff) space.

Proof
Let $T = \left({X, \vartheta}\right)$ be a completely Hausdorff space.

From the definition:


 * $\forall x, y \in X: x \ne y: \exists U, V \in \vartheta: x \in U, y \in V: U^- \cap V^- = \varnothing$

We have that Set is Subset of Closure and so $U \subseteq U^-$ and $V \subseteq V^-$.

This leads to:


 * $\forall x, y \in X: x \ne y: \exists U, V \in \vartheta: x \in U, y \in V: U \cap V = \varnothing$

which is precisely the definition of a Hausdorff ($T_2$) space.