Regular Space is T2 Space

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

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

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

From the definition of regular space:


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

Let $x, y \in S$.

As $T$ is $T_0$, it follows that either:
 * $\exists V \in \tau: x \in V, y \notin V$

or:
 * $\exists V \in \tau: y \in V, x \notin V$

that is, there exists $V$, an open set, containing one but not the other.

, suppose that $\exists V \in \tau: y \in V, x \notin V$.

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

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:


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

So we have that:
 * $x \in F \subseteq U \implies x \in U$
 * $y \notin F, y \in V$

such that $U \cap V = \varnothing$.

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

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