Regular Space is T2 Space

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

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

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

From the definition of $T_3$ space:


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

Let $x, y \in X$.

As $T$ is $T_0$, it follows that:


 * $\exists V \in \vartheta: x \in V, y \notin V$
 * $\exists V \in \vartheta: y \in V, x \notin V$

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

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:


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

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

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

Alternative Proof
We have that a $T_3$ Space is an Urysohn space.

We also have that an an Urysohn space is a Hausdorff ($T_2$) space.

Hence the result.