Boundary of Union of Separated Sets equals Union of Boundaries

Theorem
Let $T$ be a topological space.

Let $A, B$ be subsets of $T$.

Let $A$ and $B$ are separated.

Then:
 * $\map \partial {A \cup B} = \partial A \cup \partial B$

where:
 * $\partial A$ denotes the boundary of $A$
 * $A \cup B$ denotes the union of $A$ and $B$.

Proof
By definition of separated sets:
 * $(1): \quad A^- \cap B = A \cap B^- = \O$

By Separated Sets are Disjoint:
 * $A \cap B = \O$

We will prove that:
 * $\partial A \cap \partial B \subseteq \map \partial {A \cup B}$

Let $x \in \partial A \cap \partial B$.

Then by definition of intersection:
 * $x \in \partial A \land x \in \partial B$

Hence by Boundary is Intersection of Closure with Closure of Complement:
 * $x \in A^- \cap \paren {\relcomp T A}^- \land x \in B^- \cap \paren {\relcomp T B}^-$

where:
 * $A^-$ denotes the closure of $A$
 * $\relcomp T A = T \setminus A$ denotes the relative complement of $A$ in $T$.

Then by definition of intersection:
 * $x \in A^- \land x \in B^-$

Therefore by definition of empty set, intersection, and $(1)$:
 * $x \notin B \land x \notin A$

Then by definition of union:
 * $x \notin A \cup B$

By definition of relative complement:
 * $x \in \relcomp T {A \cup B}$

By definition of closure:
 * $\relcomp T {A \cup B} \subseteq \paren {\relcomp T {A \cup B} }^-$

Then by definition of subset:
 * $x \in \paren {\relcomp T {A \cup B} }^-$

By Closure of Finite Union equals Union of Closures:
 * $A^- \cup B^- = \paren {A \cup B}^-$

By definition of union:
 * $x \in \paren {A \cup B}^-$

Then by definition of intersection:
 * $x \in \paren {A \cup B}^- \cap \paren {\relcomp T {A \cup B} }^-$

Thus by Boundary is Intersection of Closure with Closure of Complement:
 * $x \in \map \partial {A \cup B}$

This ends the proof of inclusion.

Thus by Union with Superset is Superset the result:
 * $\partial A \cup \partial B = \map \partial {A \cup B}$