Boundary of Union is Subset of Union of Boundaries

Theorem
Let $T = \left({S, \tau}\right)$ be a topological space.

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

Then:
 * $\partial \left({A \cup B}\right) \subseteq \partial A \cup \partial B$

where $\partial A$ denotes the boundary of $A$.

Proof
By Intersection is Subset:
 * $\complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({A}\right) \land \complement_S \left({A}\right) \cap \complement_S \left({B}\right) \subseteq \complement_S \left({B}\right)$

Then by Topological Closure of Subset is Subset of Topological Closure:
 * $\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({A}\right)}\right)^- \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \subseteq \left({\complement_S \left({B}\right)}\right)^-$

Hence by Boundary is Intersection of Closure with Closure of Complement:
 * $\left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap A^- \subseteq \partial A \land \left({\complement_S \left({A}\right) \cap \complement_S \left({B}\right)}\right)^- \cap B^- \subseteq \partial B$

Thus