If sets $A, B$ in Euclidean space are closed sets, they have the same boundary and their interior's intersection is non-empty, can we say $A=B$? If sets $A, B$ in Euclidean space are closed sets, they have the same boundary and their interiors intersection is non-empty, can we say $A=B$? Any suggestions and comments are welcome!
 A: No.  For example, $A=\overline{\bigcup_{n\in\mathbb{Z}} [\arctan(2n-1),\arctan(2n)]}\cup [10,11]$, $B=\overline{([-\pi/2,\pi/2]-\bigcup [\arctan(2n-1),\arctan(2n)]}\cup[10,11]$.
To come up with such an example, note we can WLOG forget the interior intersection is nonempty by adding a requirement $A,B$ bounded (since we can take union with a closed ball far far away) and $\mathbb{R}$ is homeomorphic to a bounded open interval.  So we want to come up with two different nonempty closed sets in $\mathbb{R}$ with the same boundary, which is easy enough: $\bigcup[2n-1,2n]$ and $\bigcup [2n,2n+1]$ both have boundary $\mathbb{Z}$.
A: Counterexample in $\mathbb R^2$:
                                               
\begin{align*}
A&=\text{red disk}\bigcup\text{black vertical line}\\
B&=\text{red disk}\bigcup\text{black vertical line}\bigcup\text{blue area}
\end{align*}
The red disk includes the circle on its boundary, and the blue area extends to infinity towards north, south, and east.
\begin{align*}
\partial A&=\partial B=\text{red circle}\bigcup\text{black vertical line}\\
A^{\circ}&=\text{interior of red disk}\\
B^{\circ}&=\text{interior of red disk}\bigcup\text{blue area}\\
A^{\circ}\cap B^{\circ}&=\text{interior of red disk}\neq\varnothing\\
A&\neq B
\end{align*}
