Is it true that if two open sets in a topological space intersect each other and share the same boundary, they are the same? Suppose we have a topological space $(X,\mathcal T)$. Open sets $A,B$ have the following properties:
$A\cap B\ne\emptyset$
$\partial A=\partial B$
Then $A=B$. Is it correct? If so, how to prove it? 
 A: As pointed by Josh Keneda, $A=\{(x,y):x^2+y^2<1\}$ and $B=\mathbb{R}^2\setminus\partial A$ give a counter-example.
You may have a more convoluted conter-example by considering the basins of attraction, in $\mathbb{C}$, for the Newton's map applied to $p(x)=x^3-1$. Assuming that for some $w\in\mathbb{C}$ the iterates $\varphi^{(n)}(w)=\varphi(\varphi^{(n-1)}(w))$ converge, with: 
$$\varphi(x) = \frac{1}{3x^2}+\frac{2x}{3},$$
they converge to some third root of unity, but $A_1,A_{\omega},A_{\omega^2}$ (with the obvious meaning) share the same boundary:
$\hspace2in$
so $A_1\cup A_\omega$ and $A_1\cup A_{\omega^2}$ give another counter-example.
A: Other answers have shown this need not be true in general.  However, it is true if you assume $A$ and $B$ are both connected.  Indeed, $\partial A\subseteq \partial B$ implies $A\cap B$ is closed as a subset of $B$, and hence must be all of $B$ since it is also open in $B$ and nonempty.  Similarly, $A\cap B$ must be all of $A$ as well.
A: Let $A=B \cup C$ where $C$ is open and closed, $B$ is open, $B$ and $C$ are disjoint and neither $B$ nor $C$ are empty. Then $A$ and $B$ have the same boundary.
A: Take the discrete topology on say the natural numbers, and let $A$ and $B$ be distinct sets with non-empty intersection. Their boundaries are empty, so they have the same boundary.
