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As in title - I'm looking for a proof of the existence of a countable recursively inaccessible or recursively Mahlo ordinals, especially the first one. When looking for it in all the papers I stumbled across their existence wasn't questioned, and no proofs were referenced. On Wikipedia (where I found the first mention of provability of their existence in ZFC) there is also no direct reference.

Thank you in advance for any help!

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up vote 4 down vote accepted

An ordinal $\xi$ is recursively inaccessible when it is an admissible limit of admissible ordinals. To be admissible means that $L_\alpha$ is a model of the Kripke-Platek axioms of set theory. It is not difficult to see that $\omega_1$ itself is admissible, since $L_{\omega_1}\models\text{KP}$, and by taking elementary substructures, it follows that there is a closed unbounded set of ordinals $\alpha$ with $L_\alpha\prec L_{\omega_1}$. Thus, $\omega_1$ is recursively inaccessible, and indeed a limit of recursively inaccessible ordinals, because this is reflected to $\alpha$ whenever $L_\alpha\prec L_{\omega_1}$. So we have an abundance of these ordinals.

I'm not exactly sure what you mean by recursively Mahlo, but I suppose you want that the set of smaller admissible ordinals $\alpha$ is stationary with respect to definable clubs of some complexity. But this property is also true of true $\omega_1$, since as we said, there is a full closed unbounded set of countable admissible ordinals $\alpha$. And so it follows again by taking elementary substructures $L_\alpha\prec L_{\omega_1}$ that there will be many countable ordinals with this property.

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Recursively Mahlo ordinals come from ordinal analysis; the easiest definition is an ordinal satisfying KP together with $\Pi_2$ reflection on admissibles (i.e., if $L_\alpha\vDash\forall x\exists y\phi$ then there is an admissible $\beta<\alpha$ such that $L_\beta\vDash\forall x\exists y\phi$). – Henry Towsner Apr 19 '14 at 22:33
Thanks very much, Henry. The Wikipedia page… defines it in terms of every (sufficiently) definable club containing an admissible ordinal, which aligns with classical Mahloness, and I guess the point is that this is equivalent to your property by thinking about the club of ordinals closed under witnesses to the $\Pi_2$ property. – Joel David Hamkins Apr 19 '14 at 23:02
@JoelDavidHamkins I can see how we get an unbounded set of such $\alpha$ (making $\omega_1$ recursively inaccessible) but I can't really see why this set is closed. – Wojowu Apr 20 '14 at 8:08
@Wojowu The set of $\alpha<\omega_1$ with $L_\alpha\prec L_{\omega_1}$ is closed, because it's limit points are unions of elementary chains, and therefore also elementary. – Joel David Hamkins Apr 20 '14 at 12:13

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