Let $\{M_i\}_{i\in I}$ be a family of $L$-structures such that for each $i\in I$, $T_i=Th(M_i)$ is X where $X\in\{\text{stable, simple, NIP, NSOP}, \dots\}$. Let $U$ be a non-principal ultrafilter on $I$, and $M=\prod_U M_i$. Also, let $T=Th(M)$.

  1. Is $T$ an X theory as well?

  2. What can we say about $T$ in general?

  • 4
    $\begingroup$ This is equivalent to asking whether or not $X$ is an elementary class in the language $L$. For any sufficiently non-trivial language this is going to fail. A simple example would be to consider Boolean algebras. Theories of finite Boolean algebras are stable (all theories of finite structures are), but no theory of an infinite Boolean algebra is going to be anything nice, so a pseudo-finite Boolean algebra would give a counterexample to your question 1. $\endgroup$ – James Hanson Jul 23 at 14:05

I think, in general, it would be hard to say somethings about $T$. For example, finite linear orders satisfy stability (and hence simplicity). However a non principal ultraproduct of finite linear orders (except the trivial case) has the strict order property (SOP) and hence is not simple (and in particular, is not stable).


As already noted by others, for most “Shelah-style” dividing lines the answers to the questions are “no” and “probably not much”, respectively.

But I think this question is still nice for fueling interesting examples. For instance, here is an example of the “opposite kind” where the ultra product has better behavior.

Let $M_n$ be the Fraisse limit of finite $K_n$-free graphs, for $n\geq3$. Then $Th(M_n)$ is SOP3 for any $n$. On the other hand, the theory of a nonprincipal ultraproduct of $\{M_n\}_{n\geq 3}$ is that of the random graph. So this theory is simple.

In this example, of course the point is that SOP3 in the individual structures is not witnessed uniformly by the same formula.

EDIT: Here is a proof of the claim above. Let $M=\prod_{U}M_n$ where $U_n$ is nonprincipal. We show that $M$ satisfies the axiomatization of the theory of the random graph. Clearly, $M$ is a graph so we just need to verify the extension axioms. In particular, fix $k\geq 1$ and let $\phi$ be the axiom saying that for any two disjoint sets of $k$ vertices, there is some element connected to everything in the first set and nothing in the second set. Then $M_n\models \phi$ for any $n>k+1$, and so $M\models\phi$.

  • $\begingroup$ Is it clear that the ultraproduct of $\{M_n\}_{n\geq3}$ is the random graph? $\endgroup$ – Lajos Jul 23 at 15:16
  • $\begingroup$ @Lajos yes, via the extension axioms for the random graph. I will edit to add this. $\endgroup$ – Gabe Conant Jul 23 at 15:21
  • $\begingroup$ So, it would be like the ultraproduct of Paley graphs. right? $\endgroup$ – Lajos Jul 23 at 15:22
  • $\begingroup$ @GabeConant Hi Gabe, do you have any other example, just off the top of your head, for the case that the ultra product has a better behavior than the original structures? $\endgroup$ – Mostafa Mirabi Jul 24 at 14:29
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    $\begingroup$ @MostafaMirabi There are other examples that feel more like “cheating” (but perhaps illustrate the point better). Let $L$ be a language with binary relation symbols $R_n$ for $n\geq0$. Let $M_n$ be an $L$-structure such that the reduct to $R_n$ is a model of ZFC, and $R_m$ is equality for $m\neq n$. Then a nonprincipal ultraproduct $\prod_U M_n$ is just an infinite set with no structure. $\endgroup$ – Gabe Conant Jul 24 at 14:56

I'm going to assume we're talking about $\omega$-incomplete ultrafilters.

Others have pointed out that none of the major dividing lines are elementary; that is, these properties are not equivalent to a countable conjunction of sentences.

Some dividing lines, like stability and NIP, have the form "for each formula $\phi$, there is a numeric n such that a sentence $\sigma_{n,\phi}$ holds". For instance, NIP holds if each $\phi$ has bounded VC dimension. Such a property holds in an ultraproduct exactly when it holds uniformly in the ground models.

That is, $\prod_{\mathcal{U}}M_i$ is NIP if and only if, for each formula $\phi(x;y)$, there is a bound $d$ such that, for most $i$, $\phi$ has VC dimension $d$ in $M_i$. That means an ultraproduct of NIP models can be IP (if the VC dimension is unbounded) and an ultraproduct of IP models can be NIP (if each model has an IP formula, but each individual formula is IP in most models)

I'm not sure if every dividing line can be expressed in that form. But (at the risk of self-promotion) there's a more general framework for describing properties of ultraproducts in terms of "higher order uniformity", so it will, in general, be true that a dividing line holds in the ultraproduct if it holds uniformly in the ground models, for a suitable notion of uniformity.


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