Regular Space is Completely Hausdorff Space

Theorem

Let $\left({S, \tau}\right)$ be a regular space.

Then $\left({S, \tau}\right)$ is also a $T_{2 \frac 1 2}$ (completely Hausdorff) space.

Proof

Let $T = \left({S, \tau}\right)$ be a regular space.

From the definition:

$\left({S, \tau}\right)$ is a $T_3$ space
$\left({S, \tau}\right)$ is a $T_0$ (Kolmogorov) space.

Let $x, y \in S$.

From Regular Space is $T_2$ Space and $T_2$ Space is $T_1$ Space we have that a regular space is a $T_1$ space.

So from Equivalence of Definitions of $T_1$ Space both $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are closed sets.

Without loss of generality, suppose that $\exists V \in \tau: y \in V, x \notin V$.

Then $x \in \complement_S \left({V}\right)$ by definition of relative complement.

Let $F := \complement_S \left({V}\right)$.

As $V$ is open, by definition of closed set we have that $F = \complement_S \left({V}\right)$ is closed.

That is, $\complement_S \left({F}\right) \in \tau$.

As $y \in V$ it follows that $y \notin F$, that is, $y \in \complement_S \left({F}\right)$.

Now $\left({S, \tau}\right)$ is a $T_3$ space, and so:

$\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$

Thus:

$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \varnothing$

which is precisely the definition of an $T_{2 \frac 1 2}$ (completely Hausdorff) space.

$\blacksquare$