T3 1/2 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 $T_{3 \frac 1 2}$ space, then so is $T_B$.

Proof
Let $F \subseteq S_B$ be closed, and let $y \in S_B$ such that $y \notin F$.

Let $G = \phi^{-1} \left[{F}\right]$ and let $z = \phi^{-1} \left({y}\right)$.

Since $T_A$ is a $T_{3 \frac 1 2}$ space, there exists an Urysohn function $f: X_A \to \left[{0 \,.\,.\, 1}\right]$ for $G$ and $\left\{{z}\right\}$.

Define $g: S_B \to \left[{0 \,.\,.\, 1}\right]$ by:


 * $g \left({x}\right) = f \left({\phi^{-1} \left({x}\right)}\right)$

By Composite of Continuous Mappings is Continuous, $g$ is continuous.

Also, for all $x \in F$:


 * $g \left({x}\right) = 0$

since $\phi^{-1} \left({x}\right) \in G$.

Similarly, since $\phi^{-1} \left({y}\right) = z$:


 * $g \left({y}\right) = 1$

Hence $g$ is an Urysohn function for $F$ and $\left\{{y}\right\}$.

Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a $T_{3 \frac 1 2}$ space.