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 $S$.
$\Box$
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.
$\blacksquare$
Sources
- 1964: Steven A. Gaal: Point Set Topology: $\text{II}: \ \S 2$: $(T_3)$ Spaces, Regular and Semiregular 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: Additional Separation Properties