Regular Space is Semiregular Space

Theorem
Let $\left({X, \tau}\right)$ be a regular space.

Then $\left({X, \tau}\right)$ is also a semiregular space.

Proof
Let $T = \left({X, \tau}\right)$ be a regular space.

From the definition:


 * $\left({X, \tau}\right)$ is a $T_3$ space
 * $\left({X, \tau}\right)$ is a $T_0$ (Kolmogorov) space.

We have that a regular space is a $T_2$ space.

We also have that an open set in a $T_3$ space is the union of regular open sets.

That is, from the definition of basis, the regular open sets of $T$ form a basis for $T$.

So:
 * $T = \left({X, \tau}\right)$ is a Hausdorff ($T_2$) space
 * The regular open sets of $T$ form a basis for $T$.

which is precisely the definition of a semiregular space.