Let $\mathscr{L}$ be a recursive language. Let $\varphi$ be a $\mathscr{L}_{\omega_1 \omega}$-sentence and $\varphi \in L_{\omega_1^\emptyset}$. (Let $\varphi$ be a computably infinitary formula.) Let $\text{Mod}(\varphi)$ denote the set of countable models of $\varphi$.

Is there anything inherently wrong with the existence of the following:

(1) For every admissible ordinal $\alpha$, there is a $M \in \text{Mod}(\varphi)$ such that $\omega_1^M = \alpha$ and $\text{SR}(M)$ is $\alpha$ or $\alpha + 1$.

(2) For all $M \in \text{Mod}(\varphi)$, there is an admissible ordinal $\alpha$ such that $\text{SR}(M)$ is $\alpha$ or $\alpha + 1$.

(3) For each admissible ordinal $\alpha$, there is only one isomorphism class of Scott rank $\alpha$ or $\alpha + 1$.

Of course, a formula $\varphi$ with all these properties must be a counterexample to the Vaught's conjecture. So the real essence of this question is whether (regardless of the status of Vaught's conjecture), is it possible that the Scott rank of models of $\varphi$ take on all admissible values but no model of $\varphi$ has Scott rank in between two consecutive admissible ordinal.

The natural idea to try is to use Barwise compactness to produce a model of $\varphi$ of intermediate Scott rank. However, whenever one works in a countable admissible fragment $\mathscr{L}_A$, one can not really express non-isomorphism to the model whose Scott rank is the ordinal height of the admissible set $A$. So any model $M$ produced using a Barwise argument using $L_A$, could just be isomorphic to the possibly unique model with Scott rank the ordinal height of $A$.

The question is whether by some reason there must be a model of $\varphi$ whose Scott rank is between two consecutive admissible ordinals.

Thanks for any information on this question.


1 Answer 1


If $\phi$ is a counterexample to Vaught's Conjecture, then the Scott ranks of the models of $\phi$ include every limit ordinal below $\omega_{2}$ and above the quantifier depth of $\phi$. This follows from Theorem 10.6 in the following paper of mine: http://www.users.miamioh.edu/larsonpb/scott_proc_b1o.pdf

I attribute this theorem to Harrington there. Harrington showed in unpublished work that the Scott ranks of the models of a counterexample to Vaught's Conjecture are unbounded in $\omega_{2}$. I suspect that he knew the version in my paper, but I don't know for sure. At the June 2015 Berkeley meeting on Vaught's Conjecture, several people seemed to take it for a standard fact, and even to claim that it holds for the successor ordinals as well, although it may be that they were counting quantifiers or computing Scott ranks differently. I believe that I am following Hodges and Marker, and not Sacks, in my definition of Scott rank.

As for the general (non-counterexample) case, this paper characterizes the subsets of $\omega_{1}$ which can be the set of Scott ranks of countable models of a sentence in $L_{\omega_{1}, \omega}$, assuming Projective Determinacy:


  • 2
    $\begingroup$ Thanks. I actually found this result in your paper shortly after asking this question. $\endgroup$
    – William
    Jan 1, 2016 at 10:21

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