Regular Space is Completely Hausdorff Space

Theorem
Let $\struct {S, \tau}$ be a regular space.

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

Proof
Let $T = \struct {S, \tau}$ be a regular space.

Let $x, y \in S$ with $x \ne y$.

From the definition of a completely Hausdorff Space:
 * $T$ is completely Hausdorff if $x$ and $y$ are separated by closed neighborhoods.

From the definition:


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

From Regular Space is $T_2$ Space, there exist disjoint open sets containing $x$ and $y$ respectively:
 * $\exists U_x, U_y \in \tau: x \in U_x, y \in U_y, U_x \cap U_y = \O$

From definition of $T_3$ space, each open set contains a closed neighborhood around each of its points, specifically for $x$ and $y$:
 * $\exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U_x$
 * $\exists N_y: \relcomp S {N_y} \in \tau: \exists V_y \in \tau: y \in V_y \subseteq N_y \subseteq U_y$

From Subsets of Disjoint Sets are Disjoint:
 * $N_x \cap N_y = \O$

Hence $x$ and $y$ are separated by closed neighborhoods.

The result follows.