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Is every $\Delta_2$ set a 'finite boolean combination' of $\Sigma_1$ sets? (I.e. is it a member of the smallest collection of sets closed under finite union, intersection and complement that contains all $\Sigma_1$ sets?)

Probably not. But is there an easy counterexample?

Same question for $\Delta_3$ and $\Sigma_2$.

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    But $\Sigma_0$ is closed under all such operations.2012-01-04
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    Huh? Also under complement?2012-01-04
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    The definition I know is $\Sigma_0=\Pi_0=\Delta_0$ are the sets describable by formulas in which all quantifiers are bounded.2012-01-04
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    Oops; question edited.2012-01-04
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    The topological version (boldface version)... Given a set, which is both $F_\sigma$ and $G_\delta$, is it necessarily a finite Boolean combination of open sets? Or (next level): Given a set, which is both $F_{\sigma\delta}$ and $G_{\delta\sigma}$, is it necessarily a finite Boolean combination of $F_\sigma$ sets?2012-01-04

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