Tychonoff Space is Preserved under Homeomorphism

Theorem
Let $T_A = \left({X_A, \vartheta_A}\right), T_B = \left({X_B, \vartheta_B}\right)$ be topological spaces.

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

If $T_A$ is a $T_{3 \frac 1 2}$ (Tychonoff) space, then so is $T_B$.

Proof
We have that $\left({X, \vartheta}\right)$ is a $T_{3 \frac 1 2}$ (Tychonoff) space iff:
 * $\left({X, \vartheta}\right)$ is a completely regular space
 * $\left({X, \vartheta}\right)$ is a Kolmogorov ($T_0$) space.

From Completely Regular Space is Preserved under Homeomorphism:
 * If $T_A$ is a completely regular 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.