# Tychonoff 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 Tychonoff (completely regular) space, then so is $T_B$.

## Proof

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

$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space
$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space.
If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$.
If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.

Hence the result.

$\blacksquare$