T4 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 $T_4$ space, then so is $T_B$.

Proof
Suppose that $T_A$ is a $T_4$ space.

Let $B_1$ and $B_2$ be closed in $T_B$ that are disjoint.

Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the preimages of $B_1$ and $B_2$ under $\phi$, respectively.

From Preimage of Intersection under Mapping it follows that $A_1$ and $A_2$ are disjoint.

Also, as $\phi$ is a homeomorphism, it is a fortiori continuous.

Thus Continuity Defined from Closed Sets applies to yield that both $A_1$ and $A_2$ are closed.

Now as $T_A$ is a $T_4$ space, we find disjoint open sets $U_1$ containing $A_1$ and $U_2$ containing $A_2$.

From Image of Subset is Subset of Image, we have $A_1 = \phi \sqbrk {\phi^{-1} \sqbrk {A_1} } \subseteq \phi \sqbrk {U_1}$.

Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism.

Mutatis mutandis, we deduce also $A_2 \subseteq \phi \sqbrk {U_2}$.

From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint.

Since $\phi$ is a homeomorphism, they are also both open in $T_B$.

Therewith, we have construed two disjoint opens in $T_B$, one containing $A_1$, and the other containing $A_2$.

Hence $T_B$ is shown to be a $T_4$ space as well.