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I am given that $A \subset B \subset \mathbb{R}$, $A$ is open, $B$ is closed, and that $\partial A = \partial B$.

Can I prove from this that $B$ is either equal to the closure of $A$ or it is unbounded or some stronger result relating $A$ and $B$?

Okay so I know now that $\bar{A} \neq B$ in general. Is there any statement I can say about them though?

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Here is an example that illustrates that what is suggested in the question does not hold, even with reasonable additional hypotheses, as long as one allows the equal boundaries to be infinite.

Let $U=\bigcup_{n\in\Bbb N}U_n$ be a union of bounded open intervals $U_n$ with disjoint closures, which are chosen such that $(-1)^n+(-\frac12)^n\in U_n$ for all$~n$. Then $\overline U$ of course also contains the boundary points of all those intervals, but also the limit points $-1$ and $1$ which are not of that form. Now $C=\overline U\cup[-1,1]$ is another closed set with the same boundary as $U$ (and as $\overline U$), in particular $\partial C$ contains $\{-1,1\}$ since these are not interior points of $C$.

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  • $\begingroup$ Is there any significant result you can derive from the fact that the boundaries are equal? $\endgroup$ – ruler501 Mar 28 '14 at 3:07

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