Say we have a topological space $X$.
I'm trying to show that one can express the union of the boundaries of two sets $A$ and $B$ as
$$\partial{A} \cup \partial{B}=\partial(A \cup B) \cup \partial(A \cap B) \cup (\partial{A} \cap \partial{B})$$
It wasn't too difficult to prove the inclusion
$$\partial(A \cup B) \cup \partial(A \cap B) \cup (\partial{A} \cap \partial{B}) \subseteq \partial{A} \cup \partial{B}$$
But I'm not sure how to prove it in the other direction. Here's what I have so far:
Let $x \in \partial{A}$ and for the sake of contradiction assume that $x \notin \partial{(A \cup B)} \cup \partial{(A \cap B)} \cup (\partial{A} \cap \partial{B})$. Then
$$x \notin \partial{(A \cup B)} \land x \notin \partial{(A \cap B)} \land x \notin (\partial{A} \cap \partial{B})$$
From the first proposition, it follows that there exists an open neighborhood $U_x$ of $x$ such that
$$U_x \cap (A \cup B)=\varnothing \lor U_x \cap \complement_X(A \cup B)=\varnothing$$
$$ \Rightarrow (U_x \cap A) \cup (U_x \cap B)=\varnothing \lor U_x \cap \complement_X(A) \cap \complement_X(B)=\varnothing$$
From $x \notin (\partial A \cap \partial B)$, it follows that $x \notin \partial B$. Thus there's some open neighborhood $V_x$ of $x$ such that
$$V_x \cap B=\varnothing \lor V_x \cap \complement_X(B)=\varnothing$$
But as $x \in \partial A$, we must have $U_x \cap A \neq \varnothing$ and $V_x \cap A \neq \varnothing$, hence $(U_x \cap A) \cup (U_x \cap B)$ cannot be empty. Therefore $U_x \cap \complement_X(A) \cap \complement_X(B)=\varnothing$. We also know that there exists an open neighborhood $W_x$ of $x$ such that
$$W_x \cap A \cap B=\varnothing \lor W_x \cap \complement_X(A \cap B)=\varnothing$$
$$\Rightarrow W_x \cap A \cap B=\varnothing \lor (W_x \cap \complement_X(A)) \cup (W_x \cap \complement_X(B))=\varnothing$$
But since $x \in \partial A$, the set $W_x \cap \complement_X(A)$ must be non-empty. Therefore the above union must also be non-empty, and we conclude that $W_x \cap A \cap B=\varnothing$.
This is where I get stuck, and I'm not sure how to proceed from there. Any hints would be appreciated.
Best Answer
Notation: $S^c=X$ \ $S$ for any $S.$
Let $f(A,B)=\partial A\cup \partial B.$
Let $g(A,B)=\partial (A\cup B)\cup \partial (A\cap B)\cup (\partial A\cap \partial B).$
Observations: $f(A,B)=f(B,A)=f(A^c,B^c)$ and $g(A,B)=g(B,A)=g(A^c,B^c).$
To show $g(A,B)\supset f(A,B)$ it suffices to show $g(A,B)\supset \partial A,$ because then $g(A,B)=g(B,A)\supset \partial B.$
For any $p\in \partial A:$
(i). If $p\in \partial B$ then $p\in \partial A\cap \partial B\subset g(A,B).$
(ii). If $p\not \in \partial B$ then $p\in int(B)$ or $p\in int (B^c).$
(ii-a). If $p\in int(B):$ If $U$ is any nbhd of $p$ then $U\cap int(B)$ is a nbhd of $p, $ and $p\in \partial A.$ So there exist $p'\in A\cap U\cap int (B)\subset A\cap B$ and $p''\in A^c\cap U\cap int(B)\subset A^c\cap B\subset (A\cap B)^c.$
So $p\in \partial (A\cap B)\subset g(A,B).$
(ii-b). If $p \in int (B^c):$ Then $p\in \partial A^c$ and $p\in int(B^c).$
By (ii-a) applied to $A^c$ and $B^c,$ this implies $p\in \partial (A^c\cap B^c)$.... And we have $\partial (A^c\cap B^c)=\partial ((A^c\cap B^c)^c)=\partial (A\cup B)\subset g(A,B).$