Regular Space is Preserved under Homeomorphism

Theorem

Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.

Let $\phi: T_A \to T_B$ be a homeomorphism.


If $T_A$ is a regular space, then so is $T_B$.


Proof

We have that $\struct {S, \tau}$ is a regular space if and only if:

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


From $T_3$ Space is Preserved under Homeomorphism:

If $T_A$ is a $T_3$ space, then so is $T_B$.

From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism:

If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.


Hence the result.

$\blacksquare$


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: Functions, Products, and Subspaces
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