# How many sets can we get by taking interiors and closures?

I'm having following problem. I'm looking for the maximum number of different sets that we can generate by one set $B \subseteq \mathbb{R}$ by taking a finite amount of closures and interiors. For example $\{0\}$ generates the sets $\{0\}$ and $\emptyset$. At first I thought the answer was 3 (we can only generate B, the closure of B and the interior of B) because I thought $\overline{B^\circ}=\overline{B}$ and $\overline{B}^\circ=B^\circ$. But then I looked at the example $B=\mathbb{Q}$ and found that $\overline{\mathbb{Q}}^\circ=\mathbb{R}^\circ=\mathbb{R}\neq\emptyset=\mathbb{Q}^\circ$.
This is not what I would expect and I clearly need another method te search that maximum.

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Note that both these operations are idempotent, so the only way to achieve actual results is to intertwine them. –  Asaf Karagila May 29 '13 at 21:52
This isn't exactly what you're looking for, but it is related (and interesting): Kuratowski's Problem –  Jared May 29 '13 at 21:55
As suggested by @Jared's link, have a look at closure and interior repeatedly applied to $B=(0,1)\cup(1,2)\cup \{3\}\cup([4,5]\cap \mathbb Q)$. –  Hagen von Eitzen May 29 '13 at 21:59
@Jared: indeed related to my question and very interessenting. I'm suprised that the answer is 14 (I would absolutely not guess that answer). –  mvcouwen May 29 '13 at 22:03
Note that as the Wikipedia article on the 14 set problem says, that $S^\circ = \overline{S^c}^c$ (or $iS = ckcS$), so it seems that the problem reduces to the closure-complement question. –  kahen May 29 '13 at 22:32

Trying with the set $B=(0,1)\cup(1,2)\cup\{3\}\cup([4,5]\cap\mathbb{Q}$ (as suggested by Hagen von Eitzen) gives me that the answer has to be at least 8.