Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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!

share|improve this question
add comment

1 Answer 1

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.

share|improve this answer
    
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 at 22:33
    
Thanks very much, Henry. The Wikipedia page en.wikipedia.org/wiki/… 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 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 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 at 12:13
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.