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

We know that if $\alpha<\omega_1^{CK}$ then there is some recursive $R$ such that $(\omega,R)$ has order type $\alpha$.

Let's consider now the ordinals, $\mathsf{Ord}$ with their natural order. This is a well-ordered class, and it behaves very much like $\omega$ (in fact $\mathsf{Ord}^{V_\omega}=\omega$). We can define new order types on $\mathsf{Ord}$ whose order type is strictly larger.

For example $\leq_0=\lbrace(\alpha,\beta)\mid 0<\alpha\land(\alpha<\beta\lor\beta=0)\rbrace$ would be a well-ordering of $\mathsf{Ord}$ which is isomorphic to $\mathsf{Ord}+1$.

We can do this with any set of ordinals, or even a class of ordinals: put all the limit ordinals strictly above all the successor ordinals (and zero), and in each part order by the natural order. We can even go further and define the following ordering:

  • Zero and successor ordinals first, order by the natural order.
  • Any limit ordinal is greater than zero and all the successors.
  • Given two limit ordinals with different cofinality then the order then by their cofinality.
  • Given two limit ordinals with the same cofinality then order them as usually.

And of course we can proceed more and more. And define even greater and more complicated order types (using parameters, of course).

My question is what sort of supremum these class-orders have? But of course this is too broad. Clearly this is going to depend on the universe itself, or rather the model. So let me put some constraints on this question.

Let $(M,\in)$ be a countable transitive model of ZFC, and suppose that $\mathsf{Ord}^M=\alpha$. Can we compute this $\omega_1^{CK}(\mathsf{Ord}^M)$ from $\alpha$ and $M$? Can we at least bound it from below and above (of course $\omega_1$ is an upper bound, but a more reasonable bound that is)?

I feel that my question is still quite broad, but is there anything intelligible we can say on those ordinals? Does the fact that $M$ is countable make any difference in the computation?

Two more points which crossed my mind as relevant, and perhaps helpful, to the above question.

  1. What happens if we add large cardinals assumptions into $M$. If for example there is a measurable in $M$, $\omega_1^{CK}(\mathsf{Ord}^M)>\omega_1^{CK}(\mathsf{Ord}^{L^M})$ (to make things interesting $M=L[U]$ maybe)?

  2. What happens between extensions which don't add ordinals? In particular what about forcing extensions? Can we change this ordinal by forcing? Or maybe give some intelligible conjecture as to when an inner model of $M$ has the same $\omega_1^{CK}(\mathsf{Ord})$?

share|improve this question
add comment

1 Answer

Although your notation $\omega_1^{CK}(\text{Ord}^M)$ suggests that you have some sort of generalized computability in mind, I'll take the question as being about all the well-orders of $\text{Ord}^M$ that are parametrically first-order definable over $M$. Such well-orders are elements of the next admissible set $M^+$, so the height of $M^+$ is an upper bound for their order-types. It might well be the least upper bound, but I'm not sure about that.

share|improve this answer
    
Thanks Andreas, is there anything large cardinals can do to affect the outcome? For example if $M\models\exists 0^\#$, or has a proper class of superhuge cardinals? I suppose that from that we can define more orderings... –  Asaf Karagila Dec 17 '12 at 18:38
    
Also you are spot on the first-order definable [with parameters] over $M$. I borrowed the notation because I found it somewhat relevant, $\omega_1^{CK}$ is the $\omega_1^{CK}(\mathsf{Ord}^{V_\omega})$ of the theory ZFC-Infinity+$\lnot$Infinity. –  Asaf Karagila Dec 17 '12 at 21:31
    
As far as I can see, the next admissible set remains an upper bound, no matter what large-cardinal axioms $M$ satisfies. What that next admissible set is depends, of course, on $M$, and I'd expect it may depend on other things than just large-cardinal properties. –  Andreas Blass Dec 18 '12 at 14:48
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.