Regular Space is Semiregular Space

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Theorem

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


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


Proof

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

From the definition:

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

We also have that a $T_3$ Space is Semiregular.

Hence the result, by definition of semiregular space.

$\blacksquare$


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