(Throughout, all ultrafilters are nonprincipal.)

Given a property $P$ - really, a sentence in some appropriate logic - say that a ultrafilter $\mathcal{U}$ on a cardinal $\kappa$ averages $P$ iff for every $\kappa$-tuple of structures, the ultraproduct by $\mathcal{U}$ of that tuple has $P$ iff $\mathcal{U}$-many of the structures in the tuple have $P$. For example, Łoś's theorem says that all ultrafilters average all first-order-expressible properties. Meanwhile, non-first-order properties are "usually" not averaged by ultrafilters.

However, even looking at a single strong logic we may see genuinely different "levels of averageing" amongst ultrafilters: e.g. looking at $\mathsf{SOL}$, note that well-foundedness is averaged by exactly the countably complete ultrafilters. This motivates the separate notion of "averageability:" a property $P$ is averageable iff there is some ultrafilter averaging it, and a set $\Gamma$ of properties is averageable iff there is some single ultrafilter which averages each property in $\Gamma$.

Interestingly, I don't know offhand of any examples of divergent behavior in any natural logic. To keep things precise, I'll focus on $\mathsf{SOL}$:

Question 1: Is it consistent with $\mathsf{ZFC}$ + large cardinals that the union of two finite averageable sets of second-order sentences is again averageable?

Now question $1$ is of course extremely broad. We can try to sharpen it by restricting attention to a particular type of ultrafilter. There are, to my mind, two natural ways to do this:

Question 2: What happens if we restrict attention to nonprincipal ultrafilters on $\omega$?

Question 3: What happens if we restrict attention to nonprincipal countably closed ultrafilters (and assume that lots of measurable cardinals exist, to avoid the trivial answer)?

Personally I think that a positive answer to $(1)$ is highly implausible, but a positive answer to $(3)$ might follow from (very) large cardinals. $(2)$ is pretty mysterious to me; large cardinals don't really tame ultrafilters on $\omega$ as far as I know (nor does anything else to be fair), so I'm not really sure where to get started intuition-wise.


1 Answer 1


I'm not sure I have a definitive answer, but three nice observations that are too long for comments:

  1. If you relax from ultrafilters to extenders (seen as directed systems of ultrafilters), then there's a much stronger connection with large cardinals. In particular, the moral of Magidor's proof that extendibles are compactness cardinals for $\mathbb{L}^2$ is that the embeddings $V_\alpha \to V_\beta$ are $\mathbb{L}^2$-correct for structures in $V_\alpha$. So then all classes are averageable-by-extenders (although maybe this is slightly off from your definition because you need different extenders based on the rank of the models...)

  2. Say that $\phi \in \mathbb{L}^2$ is existential if all of the second-order quantifiers are existential (and there's no negation outside of them). Then it is the case that existential $\mathbb{L}^2$ properties are averageable upwards, that is, if $U$-many $M_i \vDash \phi$, then $\prod M_i/U \vDash \phi$. So if you say $\phi$ is $\Delta_1$ iff $\phi$ and $\neg \phi$ are existential, then the $\Delta_1$ properties should exactly be the `universally averageable' properties.

  3. Beyond the $\Delta_1$ statements, I don't think there's going to be too much that is averageable within the realm of second-order logic specifically. The example you mention of well-foundedness is really a phenomena where well-roundedness is $\mathbb{L}^2$-definable AND $\mathbb{L}_{\omega_1, \omega_1}$-definable, and this second one is what allows its preservation by countably complete ultrafilters.

To back up this intuition, one way to think about this is to imagine all of the $M_i$ as Henkin structures $(M_i, \mathcal{P}(M_i))$. Then a second-order statement is just first-order in this structure, and so is preserved in moving to the ultra product $\prod (M_i, \mathcal{P}(M_i))/U$. But there is a big gap between

  • $\prod \mathcal{P}(M_i)/U$, the subset sort of the Henkin structure and
  • $\mathcal{P}(\prod M_i/U)$, the actual power set of the ultraproduct that the second order statements will check

This is why universal second-order statements are not `averageable up', because they miss the extra subsets in $\mathcal{P}(\prod M_i/U) - \prod \mathcal{P}(M_i)/U$

I guess a question to help guide any of this intuition is the following: do you have any candidate properties in mind that are averageable but not a) uniformly averageable (by dint of being $\Delta_1$) or b) equivalent to some $\mathbb{L}_{\kappa,\kappa}$-property and are averageable only for $\kappa$-complete ultrafilters?

  • $\begingroup$ Nice observations, Bill. $\endgroup$
    – Asaf Karagila
    Commented Nov 17, 2021 at 21:53
  • $\begingroup$ Re: (2)/(3) and the end, I don't really have any such interesting candidates but I'll see if I can come up with one. (Maybe something from a fully-compact logic beyond FOL?) Re: (1), I'm happy to generalize to extenders but I really do care about not depending on the input structures, so I'm not fully satisfied with that. $\endgroup$ Commented Nov 18, 2021 at 4:33
  • $\begingroup$ An addendum to the $\Delta_1$ observation above: a better way of looking at it is that existential second-order sentences are preserved by ultraproducts is that the are PC-classes and reduct commutes with ultraproducts. $\endgroup$
    – Will Boney
    Commented Nov 18, 2021 at 22:33
  • $\begingroup$ One more addendum: I think averageability of some honest/unavoidable use of a universal quantifier for all structures is going to be very hard. As M gets big, the different between $P(\prod M)$ and $\prod P(M)$ is going to get bigger and bigger, so you'll keep missing out on more subsets that you need to universally quantify over. This is why I think an example would be very instructive, since you'll start to see where averageability fails. $\endgroup$
    – Will Boney
    Commented Nov 19, 2021 at 17:59
  • $\begingroup$ This is quickly getting out of my realm of competence, but this answer of Farmer S seems relevant. (Today's regret: why didn't I pay more attention in John's seminars?) $\endgroup$ Commented Dec 24, 2021 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.