Tychonoff Space is Preserved under Homeomorphism

Theorem
Let $T_A = \left({S_A, \tau_A}\right), T_B = \left({S_B, \tau_B}\right)$ 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 $\left({S, \tau}\right)$ is a Tychonoff space :
 * $\left({S, \tau}\right)$ is a $T_{3 \frac 1 2}$ space
 * $\left({S, \tau}\right)$ is a $T_0$ (Kolmogorov) space.

From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism:
 * If $T_A$ is a $T_{3 \frac 1 2}$ 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.