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

Definition. A class $\mathcal{C}$ of pairwise elementary equivalent $\mathcal{L}$-structures is unifiable by ultrapower if there is an index set $I$ and an ultrafilter $F$ on it such that $\forall M,N\in \mathcal{C}~~~\prod_{F}M\cong \prod_{F}N$.

A theorem by Shelah and Keisler says that each set of elementary equivalent structures with size $2$ is unifiable by ultrapower.

Question. For which cardinal $\kappa$ the following assertion is true?

Each set $\mathcal{C}$ of size $\kappa$ of pairwise elementary equivalent $\mathcal{L}$-structures is unifiable by ultrapower.

What about when $\mathcal{C}$ is a proper class of pairwise elementary equivalent $\mathcal{L}$-structures?

The question is closely related to the problem of iterated ultrapowers.

share|improve this question
    
No proper class can ever be unifiable, since the class must have elements of cardinality greater than (least cardinality of an elt. of the class)$^{\text{cardinality of the index set}}$. I'm pretty sure that every set is unifiable, though. –  Noah S Feb 27 at 6:06
    
As far as I recall, the ultrafilter in Shelah's proof only depends on the cardinality of the structures and the language. So, yes, every set is unifiable. –  Emil Jeřábek Feb 27 at 10:12
2  
@NoahS, isn't your comment over-stated? We could have a proper class consisting of copies of a single structure, and this is surely unifiable by an ultrapower. I guess you are thinking of a proper class of pairwise non-isomorphic structures. –  Joel David Hamkins Feb 27 at 13:19
    
@Joel, you're right of course. I need to be more careful when I talk about "structures" vs. "isomorphism -types of structures." –  Noah S Mar 13 at 3:10

1 Answer 1

A class $C$ of elementarily equivalent structures is unifiable by ultrapower if and only if the structures in $C$ have bounded cardinality (in other words, if they have a set of representatives up to isomorphism).

The left-to-right implication was given in Noah’s comment above: if we fix $M\in C$, then every $N\in C$ satisfies $$|N|\le\Bigl|\prod_FN\Bigr|=\Bigl|\prod_FM\Bigr|=:\kappa.$$

As for the converse implication, Shelah proved in Every two elementarily equivalent models have isomorphic ultrapowers something considerably stronger than just the Shelah–Keisler theorem:

Theorem (Shelah). Let $\lambda$ be an infinite cardinal, and $\mu=\min\{\mu:\lambda^\mu>\lambda\}$. There exists an ultrafilter $F$ on $\lambda$ with the following property.

Whenever $\{M_\alpha:\alpha<\lambda\}$ and $\{N_\alpha:\alpha<\lambda\}$ are sequences of models of cardinality $\le\kappa<\mu$ in the same language such that the ultraproducts $\prod_{\alpha<\lambda}M_\alpha/F$ and $\prod_{\alpha<\lambda}N_\alpha/F$ are elementarily equivalent, then they are isomorphic.

In particular, if $M,N$ are elementarily equivalent models of cardinality $<\mu$, then the ultrapowers $M^\lambda/F$ and $N^\lambda/F$ are isomorphic.

Note that for a given $\kappa$, we can take e.g. $\lambda=2^\kappa$ to make $\mu>\kappa$.

share|improve this answer

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.