T3 Space is Semiregular

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Theorem

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


Then $T$ is a semiregular space.


Proof

Let $\mathcal B = \left\{{B \subseteq S: B^{- \circ} = B }\right\}$.

In other words, $\mathcal B$ 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:

$\mathcal B$ is a cover for $S$
$\forall U, V \in \mathcal B: \forall x \in U \cap V: \exists W \in \mathcal B: x \in W \subseteq U \cap V$


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

Since $S$ is open, $S^\circ = S$.

Since $S$ is closed, $S^- = S$.

Therefore $S^{- \circ} = S$ and $S \in \mathcal B$.

From Set is Subset of Union:

$S \subseteq \bigcup \mathcal B$

Thus $\mathcal B$ covers $S$.

$\Box$


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

Let $U, V \in \mathcal B$, $x \in U \cap V$.

We will show that for some $W \in \mathcal B$:

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


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 $U \cap V$:

$\exists N_x: \complement_S \left({N_x}\right) \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \left({U \cap V}\right)$

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


It follows that $\mathcal B$ is a basis for $\tau$.

Hence the result by definition of semiregular space.

$\blacksquare$


Sources