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An ordinal $\alpha$ is admissible if $L_\alpha\vDash KP$ (Kripke–Platek set theory). $\omega_1^{CK}$ is the least non-recursive ordinal; the set of all recursive ordinals. It is known that $\omega_1^{CK}$ is admissible.

Is there any proof that $\omega_1^{CK}$ is admissible which uses only the fact that $\omega_1^{CK}$ is the supremum of the definable (In $\mathbb{N}$) well-orderings of the natural numbers?

Motivation: I have been thinking about this problem and am hoping to understand why $\mathsf{Def}(L_\omega)=\mathsf{Ad}(L_\omega)=\omega_1^{CK}$: Can two versions of $\omega_1^{CK}(\mathsf{Ord})$ ever coincide?

Master
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    I'm not sure I quite understand this question. Are you asking for a proof that the least arithmetically-undefinable ordinal is admissible which does not involve showing that every arithmetically-definable ordinal is computable? (Note that computable $\implies$ arithmetically definable.) – Noah Schweber Jan 03 '21 at 08:14
  • Yes, that is the question. I was hoping not to use computability theory at all, ideally. – Master Jan 03 '21 at 17:20
  • Not really going to get you anything different. You just substituted arithmetic definability for computable definability which is just a slightly different part of computability theory (one tends to prove things about both notions using the same kind fo techniques). In particular, in this case you could probably just take apart the usual proof and use arithmetic instead of computable and everything would shake out the same. – Peter Gerdes Jan 07 '21 at 20:31
  • In other words I'm suggesting that you could technically avoid using the result that every arithmetic ordinal notation has the same height as some computable ordinal notation but all you'd be really doing is running the exact same argument in terms of arithmetic ordinal notations (in effect using the equivalence to transform the proof rather than in the proof). – Peter Gerdes Jan 07 '21 at 20:34

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