TITLE: The boundary of union is the union of boundaries when the sets have disjoint closures
QUESTION [12 upvotes]: Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$? 
I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result.

REPLY [2 votes]: Here is a proof using only set operations, to complement the other answers.

Proposition. Suppose $\overline{A} \cap B = A \cap \overline{B} = \varnothing$. Then $\mathrm{Int}(A \cup B) = \mathrm{Int}(A) \cup \mathrm{Int}(B)$.
Proof. Since $\mathrm{Int}(A) \cup \mathrm{Int}(B)$ is an open set contained in $A \cup B$, we have
$$ \mathrm{Int}(A) \cup \mathrm{Int}(B) \subseteq \mathrm{Int}(A \cup B). $$
For the converse, note that $A \subseteq \overline{B}^c = \mathrm{Int}(B^c)$ and $B \subseteq \overline{A}^c = \mathrm{Int}(A^c)$, hence $\mathrm{Int}(A \cup B) \subseteq A \cup B \subseteq \mathrm{Int}(B^c) \cup \mathrm{Int}(A^c)$. Therefore
$$ \mathrm{Int}(A \cup B) = \big(\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c)\big) \cup \big(\mathrm{Int}(A \cup B) \cap \mathrm{Int}(A^c)\big). \tag*{$(1)$} $$
Note that $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c) \subseteq (A \cup B) \cap B^c = A \setminus B = A$. Since the LHS is an open set contained in $A$, it follows that $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(B^c) \subseteq \mathrm{Int}(A)$. Analogously, $\mathrm{Int}(A \cup B) \cap \mathrm{Int}(A^c) \subseteq \mathrm{Int}(B)$, so it follows from $(1)$ that $\mathrm{Int}(A \cup B) \subseteq \mathrm{Int}(A) \cup \mathrm{Int}(B)$.$\quad\Box$

To prove the same for the boundary, recall that $\partial S = \overline{S} \setminus \mathrm{Int}(S)$. We get the following:

Corollary. Suppose $\overline{A} \cap B = A \cap \overline{B} = \varnothing$. Then $\partial(A \cup B) = \partial A \cup \partial B$.
Proof. Recall that $\overline{A \cup B} = \overline{A} \cup \overline{B}$, even for arbitrary sets. It follows that
\begin{align*}
       \partial(A \cup B) &= \overline{A \cup B} \setminus \mathrm{Int}(A \cup B) \\[1ex]
       &= (\overline{A} \cup \overline{B}) \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B)) \\[1ex]
       &= \big(\overline{A} \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B))\big) \cup \big(\overline{B} \setminus (\mathrm{Int}(A) \cup \mathrm{Int}(B))\big) \\[1ex]
       &= (\overline{A} \setminus \mathrm{Int}(A)) \cup (\overline{B} \setminus \mathrm{Int}(B)) \tag*{$(2)$}\\[1ex]
       &= \partial A \cup \partial B,
\end{align*}
where to deduce $(2)$ we use that $\overline{A} \cap \mathrm{Int}(B) \subseteq \overline{A} \cap B = \varnothing$ and $\overline{B} \cap \mathrm{Int}(A) \subseteq \overline{B} \cap A = \varnothing$.$\quad\Box$

Under OP's stronger assumption that $\overline{A} \cap \overline{B} = \varnothing$, we have the following stronger result.

Theorem. Suppose $\overline{A} \cap \overline{B} = \varnothing$. Then $\partial(A \cup B)$ is the topological disjoint union of $\partial A$ and $\partial B$.
Proof. Since $\partial A \cap \partial B \subseteq \overline{A} \cap \overline{B} = \varnothing$, it follows from the preceding Corollary that $\partial(A \cup B)$ is the disjoint set union of $\partial A$ and $\partial B$. To see that it is also a topological disjoint union, note that $\partial A$ and $\partial B$ are closed sets, so both are clopen in $\partial(A \cup B)$.$\quad\Box$