# 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:

$(1): \quad A^- \cap B = A \cap B^- = \O$
$A \cap B = \O$
 $\ds \partial A \cup \partial B$ $=$ $\ds \map \partial {A \cup B} \cup \map \partial {A \cap B} \cup \paren {\partial A \cap \partial B}$ Union of Boundaries $\ds$ $=$ $\ds \map \partial {A \cup B}\cup \O \cup \paren {\partial A \cap \partial B}$ Boundary of Empty Set is Empty $\ds$ $=$ $\ds \map \partial {A \cup B} \cup \paren {\partial A \cap \partial B}$ Union with Empty Set

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:

$x \in \partial A \land x \in \partial B$
$x \in A^- \cap \paren {\relcomp T A}^- \land x \in B^- \cap \paren {\relcomp T B}^-$

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:

$x \in \paren {\relcomp T {A \cup B} }^-$
$A^- \cup B^- = \paren {A \cup B}^-$

By definition of union:

$x \in \paren {A \cup B}^-$

Then by definition of intersection:

$x \in \paren {A \cup B}^- \cap \paren {\relcomp T {A \cup B} }^-$
$x \in \map \partial {A \cup B}$

This ends the proof of inclusion.

Thus by Union with Superset is Superset the result:

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

$\blacksquare$