# 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:

$\partial \left({ A \cup B }\right) = \partial A \cup \partial B$

where

$\partial A$ denotes the boundary of $A$,
$A \cup B$ denotes the union of $A$ and $B$.

## Proof

By definition of separated sets:

$(1): \quad A^- \cap B = A \cap B^- = \varnothing$
$A \cap B = \varnothing$
 $\displaystyle \partial A \cup \partial B$ $=$ $\displaystyle \partial \left({A \cup B}\right) \cup \partial \left({A \cap B}\right) \cup \left({\partial A \cap \partial B}\right)$ Union of Boundaries $\displaystyle$ $=$ $\displaystyle \partial \left({A \cup B}\right) \cup \varnothing \cup \left({\partial A \cap \partial B}\right)$ Boundary of Empty Set is Empty $\displaystyle$ $=$ $\displaystyle \partial \left({A \cup B}\right) \cup \left({\partial A \cap \partial B}\right)$ Union with Empty Set

We will proof that

$\partial A \cap \partial B \subseteq \partial \left({A \cup B}\right)$

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

Then by definition of intersection:

$x \in \partial A \land x \in \partial B$
$x \in A^- \cap \left({\complement_T \left({A}\right)}\right)^- \land x \in B^- \cap \left({\complement_T \left({B}\right)}\right)^-$

where

$A^-$ denotes the closure of $A$,
$\complement_T \left({A}\right) = T \setminus A$ 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:

$x \in \complement_T \left({A \cup B}\right)$

By definition of closure:

$\complement_T \left({A \cup B}\right) \subseteq \left({\complement_T \left({A \cup B}\right)}\right)^-$

Then by definition of subset:

$x \in \left({\complement_T \left({A \cup B}\right)}\right)^-$
$A^- \cup B^- = \left({A \cup B}\right)^-$

By definition of union:

$x \in \left({A \cup B}\right)^-$

Then by definition of intersection:

$x \in \left({A \cup B}\right)^- \cap \left({\complement_T \left({A \cup B}\right)}\right)^-$
$x \in \partial \left({A \cup B}\right)$

This ends the proof of inclusion.

Thus by Union with Superset is Superset the result:

$\partial A \cup \partial B = \partial \left({A \cup B}\right)$

$\blacksquare$