Boundary of the intersection of two open sets in $\mathbb{R}^n$ Let $A,B$ be open subsets of $\mathbb{R}^n$. 
Does the following equality hold?
$$\partial(A\cap B)= (\bar A \cap \partial B) \cup (\partial A \cap \bar B)$$
Edit: Thanks for showing me in the answers that above formula fails if $A$ and $B$ are disjoint but their boundaries still intersect. I was able to come up with a similar formula which avoids this case
$$[\partial(A\cap B)]\setminus(\partial A \cap \partial B)= (A \cap \partial B) \cup (\partial A \cap B),$$
which I was able to prove and suffices for what I need to do.
However, when showing that $ (A \cap \partial B) \cup (\partial A \cap B)\subseteq \partial(A\cap B)$, I needed to assume that the topology is induced by a metric. I wonder if the formula still holds in an arbitrary topological space.
 A: This is not true generally unless $\overline{A\cap B}=\overline{A}\cap \overline{B}$.
\begin{align}
\partial (A\cap B)&= \overline{A\cap B}-(A\cap B)^{o}
\\
&=(\overline{A}\cap \overline{B})-(A^{o}\cap B^{o})
\\
&=(\overline{A}\cap \overline{B})\cap(A^{o}\cap B^{o})^c
\\
&=(\overline{A}\cap \overline{B})\cap(A^{o^c}\cup B^{o^c})
\\
&=(\overline{A}\cap \overline{B}\cap A^{o^c})\cup(\overline{A}\cap \overline{B}\cap B^{o^c})
\\
&=(\overline{A}\cap \partial B)\cup (\overline{B}\cap \partial A)
\end{align}
By this post, $\overline{A\cap B}=\overline{A}\cap \overline{B}$ implies discrete space. So this is impossible in $\Bbb{R}^n$. However, we can prove generally 
$$
\partial(A\cap B)\subset (\bar A \cap \partial B) \cup (\partial A \cap \bar B)
$$
for it is always true that $\overline{A\cap B}\subset \overline{A}\cap \overline{B}$. This can be done easily by replacing "$=$" with "$\subset $" at 2nd line of above proof and rest follows.
A: It does not hold. Consider for example $$A=\{x\in \mathbb{R}^n:|x|<1\}$$ and 
$$B=\{x\in \mathbb{R}^n:|x-(2,0,\dots,0)|<1\}.$$
Since $A\cap B=\varnothing$, $\partial(A\cap B)=\varnothing$, but the RHS in your formula is the set $\{(1,0,\dots,0)\}$.
A: If $A$ is dense and co-dense in the non-empty space $X$ (that is, $X$ \ $A$ is also  dense in $X$), suppose $B=X$ \ $A.$ Then $\emptyset=A\cap B=\partial (A\cap B)$ but $\bar A=\bar B=\partial A=\partial B=X\ne \emptyset.$
For example, with $X= \mathbb R^n$ let $A$ be the set of points with rational co-ordinates.
