T3 1/2 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 $T_{3 \frac 1 2}$ 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 $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$.

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

Define $g: S_B \to \closedint 0 1$ by:


 * $\map g x = \map f {\map {\phi^{-1} } x}$

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

Also, for all $x \in F$:


 * $\map g x = 0$

as $\map {\phi^{-1} } x \in G$.

Similarly, because $\map {\phi^{-1} } y = z$:


 * $\map g y = 1$

Hence $g$ is an Urysohn function for $F$ and $\set y$.

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