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My answer is no because, $\mathbb{Q}^o = \emptyset$ and so $\overline{(\mathbb{Q}^o)} = \emptyset$ but $\overline{\mathbb{Q}} = \mathbb{R}$ and so $\big(\overline{\mathbb{Q}}\,\big)^o = \mathbb{R}$.

Is my example correct?

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    **Attention** The examples are corrects if you have usual topology2012-05-13

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Another simple example may be [0,1]. Its interior is (0,1), whose closure is [0,1] again. Now, closure of [0,1] is [0,1], whose interior is (0,1)! Your question may be more interesting, if you wanted all subsets of R,say, for which the terms closure and interior are commutative! Then your example of Q,the irrationals, the Cantor set and my example of any closed interval work.

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    I didnot notice the comment of Chris while I was typing my answer!2012-05-13