Regular Space is Completely Hausdorff Space

Theorem
Let $\left({X, \vartheta}\right)$ be a $T_3$ space.

Then $\left({X, \vartheta}\right)$ is also an Urysohn ($T_{2 \frac 1 2}$) space.

Proof
Let $T = \left({X, \vartheta}\right)$ be an $T_3$ space.

From the definition:


 * $\left({X, \vartheta}\right)$ is a regular space
 * $\left({X, \vartheta}\right)$ is a Kolmogorov ($T_0$) space.

Let $x, y \in X$.

From T3 Space is T2 Space and T2 Space is T1 Space we have that a $T_3$ space is a $T_1$ space.

So from Equivalent Definitions for T1 Space both $\left\{{x}\right\}$ and $\left\{{y}\right\}$ are closed sets.

Suppose WLOG that $\exists V \in \vartheta: y \in V, x \notin V$.

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

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

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

That is, $F \in \complement \left({\vartheta}\right)$.

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

Now $\left({X, \vartheta}\right)$ is a regular space, and so:


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

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

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