Regular Space is Semiregular Space

Theorem

Let $\struct {S, \tau}$ be a regular space.


Then $\struct {S, \tau}$ is also a semiregular space.


Proof

Let $T = \struct {S, \tau}$ be a regular space.

From the definition:

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

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

Hence the result, by definition of semiregular space.

$\blacksquare$


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
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