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 :
 * $\struct {S, \tau}$ 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.