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Are Sigma(n)-admissible ordinals for n>0, i.e. ordinals such that Gödelś constructible hierarchy at that level is a model of Sigma(n)-KP, recursively inaccessible? According to Wikipedia on large countable ordinals it seems to follow that already Sigma(1)-admissible ordinals are nonprojectible and highly recursively inaccessible beyond recursively Mahlo. More information and clarification would be greatly appreciated.

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For $n\gt 1$, the answer is yes. An ordinal is recursively inaccessible if it is admissible and a limit of admissible ordinals. And every $\Sigma_{n+1}$-admissible ordinal is a limit of $\Sigma_n$-admissible ordinals, by a reflection type argument, and hence recursively inaccessible.

But for $n=1$, the answer is no, since the $\Sigma_0$-admissible and $\Sigma_1$-admissible ordinals are the same. The reason is that in KP, defined with $\Sigma_0$-collection, we can already prove $\Sigma_1$-collection, since if if every $a\in A$ has a $b$ with $\exists x \varphi(a,b,x,z)$, where $\varphi$ is $\Sigma_0$, then we can apply collection to the assertion that every $a\in A$ has a pair $(b,x)$ such that $\varphi(a,b,x,z)$, and then take the projection of the collecting set on these pairs to collect the desired $b$'s. This argument shows that $\Pi_n$ collection implies $\Sigma_{n+1}$-collection. But we don't climb further, since $\Sigma_n$-collection does not generally imply $\Pi_n$-collection.

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  • $\begingroup$ But perhaps you mean something different by $\Sigma_n$-admissible than I thought. $\endgroup$ Jul 19, 2013 at 22:08
  • $\begingroup$ Thanks. According to Wikipedia KP + Sigma (1)-separation requires a recursively inaccessible ordinal - a nonprojectible one. Is that right? $\endgroup$ Jul 20, 2013 at 9:17

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