Theorem: Union of Boundaries

Let $T = \struct {S, \tau}$ be a topological space.
Let $A, B$ be subsets of $S$.
Then:
:$\partial A \cup \partial B = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
where $\partial A$ denotes the boundary of $A$.

Proof:
First we will prove that
:$\partial A \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
Let $x \in \partial A$.
{{AimForCont}} that
:$x \notin \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
Then by definition of union:
:$x \notin \map \partial {A \cup B} \land x \notin \map \partial {A \cap B} \land x \notin \paren {\partial A \cap \partial B}$
By Characterization of Boundary by Open Sets:
:$\exists Q \in \tau: x \in Q \land \paren {\paren {A \cup B} \cap Q = \O \lor \relcomp S {A \cup B} \cap Q = \O}$
By Intersection Distributes over Union. Complement of Union:
:$\paren {A \cap Q} \cup \paren {B \cap Q} = \O \lor \relcomp S A \cap \relcomp S B \cap Q = \O$
Because $x \notin \paren {\partial A \cap \partial B}$ therefore by definition of intersection:
:$x \notin \partial B$
By Characterization of Boundary by Open Sets:
:$\exists U \in \tau: x \in U \land \paren {B \cap U = \O \lor \relcomp S B \cap U = \O}$
As $x \in \partial A$ by Characterization of Boundary by Open Sets:
:$A \cap Q \ne \O$
Then by Union is Empty iff Sets are Empty:
:$\paren {A \cap Q} \cup \paren {B \cap Q} \ne \O$
Hence:
:$\relcomp S A \cap \relcomp S B \cap Q = \O$
We will show that:
:$B \cap U = \O \implies \relcomp S A \cap Q \cap U = \O$
Assume:
:$B \cap U = \O$
Then:
:$U \subseteq \relcomp S B$
By Intersection with Empty Set:
:$\relcomp S A \cap Q \cap \relcomp S B \cap U = \O \cap U = \O$
Thus by Intersection with Subset is Subset:
:$\relcomp S A \cap Q \cap U = \O$
By definition of intersection:
:$x \in Q \cap U$
By definition of topological space:
:$Q \cap U$ is open.
Then by Characterization of Boundary by Open Sets:
:$\relcomp S A \cap Q \cap U \ne \O$
Hence:
:$B \cap U \ne \O$
Then:
:$\relcomp S B \cap U = \O$
Therefore:
:$U \subseteq B$
Because $x \notin \map \partial {A \cap B}$ by Characterization of Boundary by Open Sets:
:$\exists V \in \tau: x \in V \land \paren {A \cap B \cap V = \O \lor \relcomp S {A \cap B} \cap V = \O}$
By Complement of Intersection:
:$A \cap B \cap V = \O \lor \paren {\relcomp S A \cup \relcomp S B} \cap V = \O$
By Intersection Distributes over Union:
:$A \cap V \cap B = \O \lor \paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} = \O$
Because $x \in \partial A$ therefore by Characterization of Boundary by Open Sets:
:$\relcomp S A \cap V \ne \O$
Then by Union is Empty iff Sets are Empty:
:$\paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} \ne \O$
Hence:
:$A \cap V \cap B = \O$
By Intersection with Empty Set:
:$A \cap V \cap B \cap U = \O \cap U = \O$
By Intersection with Subset is Subset:
:$A \cap V \cap U = \O$
By definition of intersection:
:$x \in V \cap U$
By definition of topological space:
:$V \cap U$ is open.
Then by Characterization of Boundary by Open Sets:
:$A \cap Q \cap V \ne \O$
This contradicts with $A \cap V \cap U = \O$
Thus the inclusion is proved.
{{qed|lemma}}
Analogically:
:$\partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
By Union of Subsets is Subset:
:$\partial A \cup \partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
By Boundary of Union is Subset of Union of Boundaries:
:$\map \partial {A \cup B} \subseteq \partial A \cup \partial B$
By Boundary of Intersection is Subset of Union of Boundaries:
:$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$
By Intersection is Subset of Union:
:$\partial A \cap \partial B \subseteq \partial A \cup \partial B$
Hence by Union of Subsets is Subset:
:$\map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B} \subseteq \partial A \cup \partial B$
Thus by definition of set equality the result follows:
:$\partial A \cup \partial B = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
{{qed}}