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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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This is the first example that comes to mind. –  no identity May 13 '12 at 11:09
    
Yes, your example is correct. –  Matt N. May 13 '12 at 11:12
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It doesn't need to be so ill-behaved though. For example, take $(0,1)$ in $\mathbb{R}$. The closure of the interior is $[0,1]$, but the interior of the closure is $(0,1)$. –  Chris Eagle May 13 '12 at 11:19
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The title is somewhat misleading. Note that the closure of any set is closed, while the interior of any set is open. The only two sets (in $\mathbb R$) which are both closed and open are the empty set and $\mathbb R$. –  Asaf Karagila May 13 '12 at 11:32
    
Attention The examples are corrects if you have usual topology –  diofanto May 13 '12 at 11:35

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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! –  Somabha Mukherjee May 13 '12 at 11:30

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