T3 Space is Semiregular

Theorem
Let $T = \struct {S, \tau}$ be a $T_3$ space.

Then $T$ is a semiregular space.

Proof
Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$.

In other words, $\BB$ is the collection of regular open sets contained in $S$.

It follows immediately from the definition of a basis that our theorem is proved if we can show that:
 * $\BB$ is a cover for $S$
 * $\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$

First we show that $\BB$ is a cover for $S$.

Since $S$ is open:
 * $S^\circ = S$

Since $S$ is closed:
 * $S^- = S$

Therefore:
 * $S^{- \circ} = S$

and so:
 * $S \in \BB$

From Set is Subset of Union:
 * $S \subseteq \bigcup \BB$

Thus $\BB$ covers $S$.

Next we demonstrate the second condition for $\BB$ to be a basis.

Let $U, V \in \BB$, $x \in U \cap V$.

We will show that for some $W \in \BB$:


 * $x \in W \subseteq U \cap V$

By General Intersection Property of Topological Space, $U \cap V \in \tau$.

Since $T$ is $T_3$, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$:


 * $\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$

Let $W := {N_x}^{- \circ}$.

Then by Interior of Closure is Regular Open, $W^{- \circ} = W$ and $W \in \BB$.

On the other hand, since $N_x$ is closed:


 * ${N_x}^- = N_x$, and thus $W = {N_x}^\circ$

By definition of interior, we have $Q \subseteq W \subseteq N_x$.

Therefore we have $x \in W \subseteq N_x \subseteq U \cap V$, as desired.

It follows that $\BB$ is a basis for $\tau$.

Hence the result by definition of semiregular space.