Regular Space is T2 Space
Theorem
Let $\struct {S, \tau}$ be a regular space.
Then $\struct {S, \tau}$ is also a $T_2$ (Hausdorff) space.
Proof
Let $T = \struct {S, \tau}$ be a regular space.
From the definition of regular space:
- $\struct {S, \tau}$ is a $T_3$ space
- $\struct {S, \tau}$ 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:
- $x \in \relcomp S V$
Let $F := \relcomp S V$.
As $V$ is open, by definition of closed set we have that $F = \relcomp S V$ is closed.
That is:
- $\relcomp S V \in \tau$
As $y \in V$ it follows that $y \notin F$, that is:
- $y \in \relcomp S F$
Now $\struct {S, \tau}$ is a $T_3$ space, and so:
- $\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$
So we have that:
- $x \in F \subseteq U \implies x \in U$
- $y \notin F, y \in V$
such that $U \cap V = \O$.
So:
- $\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
which is precisely the definition of a $T_2$ (Hausdorff) space.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Regular and Normal Spaces