Regular Space is T2 Space

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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.

Without loss of generality, 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.

$\blacksquare$


Sources