T3 Space is Semiregular

Theorem
Let $T = \left({S, \tau}\right)$ be a $T_3$ space.

Then every open set of $T$ can be expressed as the union of regular open sets.

Proof
Let $\mathcal B = \left\{{B \subseteq S: B^{- \circ} = B }\right\}$. In other words, $\mathcal B$ is the regular open sets contained in $S$.

It follows immediately from the definition of a basis that our theorem is proved if:
 * $\mathcal B$ is a cover for $S$
 * $\forall U, V \in \mathcal B: \forall x \in \left({U \cap V}\right): \exists W \in \mathcal B: x \in W \subseteq \left({ U \cap V }\right)$

First we show that $\mathcal B$ is a cover for $S$. Let $x \in S$.

Since $S$ is open $S^ \circ = S$, and since $S$ is closed $S^- = S$. Therefore $S^{- \circ} = S$ and $S \in \mathcal{B}$.

We've now shown that $x$ is in an element of $\mathcal B$, so $x \in \bigcup \mathcal B$ and therefore $S \subseteq \bigcup \mathcal B$.

Thus $\mathcal B$ covers $S$.

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

Let $U, V \in \mathcal B$, $x \in \left({U \cap V}\right)$. We'll show that $\exists W \in \mathcal B: x \in W \subseteq \left({ U \cap V}\right)$.

Note that $U \cap V$ is a finite intersection of open sets, so $\left({ U \cap V }\right) \in \tau$.

Since $T$ is $T_3$, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$:
 * $\exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \left({U \cap V}\right)$

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

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 $x \in W \subseteq N_x \subseteq \left( U \cap V \right)$, as desired.