# Question on a limit of admissible ordinals

Let there be an omega sequence of ordinals such that the first is the least $\Sigma_1$-admissible ordinal and the $n+1$st is the least $\Sigma_{n+1}$-admissible ordinal. What is the name, if any, of the union of all of these? Incidentally, $L$ at the level of this ordinal would be the minimal model of ZFC minus the power set axiom.

-
I don't see why the last sentence is true. – François G. Dorais Jul 21 '13 at 21:49

To see this, let $\gamma_n$ be the least $\Sigma_n$-admissible ordinal (which I understand you to mean that $L_{\gamma_n}$ satisfies KP plus $\Sigma_n$-replacement, or equivalently, $\Sigma_n$-collection); and then the ordinal in question is $\gamma=\sup_n\gamma_n$ and the corresponding structure $L_\gamma$. But this structure does not satisfy KP, since the map $n\mapsto\gamma_n$ on domain $n\in \omega$ is $\Sigma_1$-definable in $L_\gamma$. The point is that whether some $L_\beta$ satisfies a certain sentence is a $\Delta_0$ espressible assertion about $L_\beta$, since all quantifiers are bounded by $L_\beta$, and whether $\beta=\gamma_n$ is a $\Sigma_1$-property of $\beta$ inside $L_\gamma$. Namely, $\beta=\gamma_n$ if and only if there is a transitive set $X$ that thinks it is $L_\beta$ and there is a satisfaction predicate on this set fulfilling Tarski's recursive definition of truth and according to this satisfaction predicate, the set $X$ satisfies KP and every instance of $\Sigma_n$-collection, and finally no smaller $\beta'\lt\beta$ satisfies all those.
It seems that one can get it down to $\Delta_1$, since the structure $L_\beta$ is unique and so is its satisfaction predicate. So the relation $\beta=\gamma_n$ is a $\Delta_1$ property of $\beta$ and $n$ inside $L_\gamma$. – Joel David Hamkins Jul 21 '13 at 22:06
@FrodeBjørdal Unless I'm very confused, the next admissible ordinal strictly above any given $\alpha$ is never $\Sigma_2$-admissible. – Andreas Blass Jul 21 '13 at 22:55
I would expect that the answer is negative. Note that your $f$ need not grow very rapidly. If $A$ is any set of natural numbers that isn't definable in ZFC minus power set, then you could take $S$ to be $\{2n:n\in A\}\cup\{2n+1:n\notin A\}$. – Andreas Blass Jul 22 '13 at 23:00