Let $X$ be a topological space, and let $A$ and $B$ be any subsets of $X$. How to show that if $\overline{A} \cap \overline{B} = \emptyset$, then $b(A \cup B) = b(A) \cup b(B)$?
Here $b(A)$ denotes the boundary of $A$. Thus $p \in b(A)$ if and only if, for every open set $G$ containing $p$, we have $G \cap A \neq \emptyset$ and $G \cap (X \setminus A) \neq \emptyset$.
Moreover, we have
$$
\operatorname{Int} (A) \cap b(A) \cap \operatorname{Ext} (A) = \emptyset
$$
and
$$
\operatorname{Int} (A) \cup b(A) \cup \operatorname{Ext} (A) = X.
$$
And we also have
$$
\overline{A} =\operatorname{Int} (A) \cup b(A).
$$
What next?
Best Answer
For any $x\in b(A)\cup b(B), x\in b(A)$ or $x\in b(B)$. WLOG, let $x\in b(A)$. It follows that $x\in \overline A.$ So $x\notin \overline B$ (hence $x\in B^c$). Let $G_x$ be any open set containing $x$. Then, $G_x\cap (A\cup B)\ne \emptyset$ since $G_x\cap A$ is non empty; and $G_x\cap (A\cup B)^c=G_x\cap A^c\cap B^c\ne \emptyset$ (if not, then $G_x\subset A\cup B$. So if $V_x\subset G_x$ is any open set containing $x$ then $V_x\cap B$ can't be non empty else $V_x\subset A$ implying that $x$ is an interior point of $A$. So $x$ must be a limit point of $B$,i.e. $x\in \overline B$, which is a contradiction. ) It follows that $x\in b(A\cup B)$. So $$b(A)\cup b(B)\subset b(A\cup B)\tag 1$$
For any $x\in b(A\cup B), \text{ x is neither an interior point nor an exterior point of } A\cup B$ and $x\in \overline {A\cup B}=\overline A\cup \overline B.$ So wlog, let $x\in \overline A$. It follows that $x\notin \overline B$. Let $G_x$ be any open set containing $x$. $G_x\cap A\ne \emptyset$ holds because $x\in \overline A$.
Since $x$ is a boundary point of $A\cup B$, the following holds: $G_x\cap (A\cup B)^c=G_x\cap A^c\cap B^c\ne \emptyset$, whence it follows that $G_x\cap A^c\ne \emptyset.$
It follows that $$x\in b(A)\subset b(A)\cup b(B)\implies b(A\cup B)\subset b(A)\cup b(B)\tag 2$$
By $(1)$ and $(2)$, one gets the desired equality.