# Union of Boundaries

## Theorem

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 $x \in \partial A$.

Aiming for a contradiction, suppose 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:

$x \notin \map \partial {A \cup B} \land x \notin \map \partial {A \cap B} \land x \notin \paren {\partial A \cap \partial B}$
$\exists Q \in \tau: x \in Q \land \paren {\paren {A \cup B} \cap Q = \O \lor \relcomp S {A \cup B} \cap Q = \O}$
$\paren {A \cap Q} \cup \paren {B \cap Q} = \O \lor \relcomp S A \cap \relcomp S B \cap Q = \O$

Because $x \notin \paren {\partial A \cap \partial B}$ therefore by definition of intersection:

$x \notin \partial B$
$\exists U \in \tau: x \in U \land \paren {B \cap U = \O \lor \relcomp S B \cap U = \O}$

As $x \in \partial A$ by Characterization of Boundary by Open Sets:

$A \cap Q \ne \O$
$\paren {A \cap Q} \cup \paren {B \cap Q} \ne \O$

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:

$U \subseteq \relcomp S B$
$\relcomp S A \cap Q \cap \relcomp S B \cap U = \O \cap U = \O$
$\relcomp S A \cap Q \cap U = \O$

By definition of intersection:

$x \in Q \cap U$

By definition of topological space:

$Q \cap U$ is open.
$\relcomp S A \cap Q \cap U \ne \O$

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:

$\exists V \in \tau: x \in V \land \paren {A \cap B \cap V = \O \lor \relcomp S {A \cap B} \cap V = \O}$
$A \cap B \cap V = \O \lor \paren {\relcomp S A \cup \relcomp S B} \cap V = \O$
$A \cap V \cap B = \O \lor \paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} = \O$

Because $x \in \partial A$ therefore by Characterization of Boundary by Open Sets:

$\relcomp S A \cap V \ne \O$
$\paren {\relcomp S A \cap V} \cup \paren {\relcomp S B \cap V} \ne \O$

Hence:

$A \cap V \cap B = \O$
$A \cap V \cap B \cap U = \O \cap U = \O$
$A \cap V \cap U = \O$

By definition of intersection:

$x \in V \cap U$

By definition of topological space:

$V \cap U$ is open.
$A \cap Q \cap V \ne \O$

This contradicts with $A \cap V \cap U = \O$

Thus the inclusion is proved.

$\Box$

Analogically:

$\partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
$\partial A \cup \partial B \subseteq \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$
$\map \partial {A \cup B} \subseteq \partial A \cup \partial B$
$\map \partial {A \cap B} \subseteq \partial A \cup \partial B$
$\partial A \cap \partial B \subseteq \partial A \cup \partial B$

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:

$\partial A \cup \partial B = \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$

$\blacksquare$