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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?

  • 1
    This is the first example that comes to mind.2012-05-13
  • 0
    Yes, your example is correct.2012-05-13
  • 1
    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 $(0,1)$.2012-05-13
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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 $\mathbb R$.2012-05-13
  • 0
    **Attention** The examples are corrects if you have usual topology2012-05-13

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