Answer by Noah Schweber for What is $\omega_1^{CK}(\mathsf{Ord})$?
Bumping an old question, I believe that Andreas' bound of the next admissible ordinal is in general (and I suspect always) not sharp. The crucial point is not the theory of the model in question, but...
View ArticleAnswer by Andreas Blass for What is $\omega_1^{CK}(\mathsf{Ord})$?
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 $\text{Ord}^M$...
View ArticleWhat is $\omega_1^{CK}(\mathsf{Ord})$?
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...
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