Regular Space is Semiregular Space
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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