Say that a second-order sentence $\varphi$ is averageable iff there exists some infinite cardinal $\kappa$ and some nonprincipal ultrafilter $\mathcal{U}$ on $\kappa$ such that for every $\kappa$-tuple of structures $(\mathfrak{A}_\eta)_{\eta<\kappa}$ we have $$\prod_{\eta<\kappa}\mathfrak{A}_\eta/\mathcal{U}\models\varphi\quad\iff\quad\{\eta<\kappa:\mathfrak{A}_\eta\models\varphi\}\in\mathcal{U}.$$

Expanding on an observation of Will Boney, I'm curious whether it is consistent with large cardinals that only the "(somewhat) obviously averageable" sentences are averageable:

Is $\mathsf{ZFC+VP}$ consistent with the following statement: "For every signature $S$ and every averageable $\varphi\in\mathsf{SOL}[S]$, there is some $\mathcal{L}_{\infty,\infty}[S\sqcup\{R_1,...,R_n\}]$-sentence $\hat{\varphi}$ such that the models of $\varphi$ are exactly the reducts of models of $\hat{\varphi}$?

(Basically, is it consistent with large cardinals that all averageable second-order sentences are "infinitary $\Delta^1_1$?") The point is that if $\mathcal{U}$ is a $\kappa$-complete nonprincipal ultrafilter on $\kappa$ then $\mathcal{U}$ witnesses the averageability of all "$\Delta^1_1(\mathcal{L}_{\kappa,\kappa})$"-sentences, and in the (current) absence of a candidate averageable sentence not in such a class it's natural to ask whether there need be any at all.

Here "$\mathsf{VP}$" is (schematic) Vopenka's principle. My current read on $\mathsf{VP}$ is that it's the natural large cardinal axiom for making abstract logics behave as nicely as possible; I'd also be interested in answers relative to other systems, as long as they are at least as strong as $\mathsf{ZFC}$ + a proper class of measurables (to ensure that there are enough interesting ultrafilters in the first place). I would also be interested in the version of this question with "ultrafilter" replaced with "extender," but since I know less about extenders I'm focusing on ultrafilters for now.

  • $\begingroup$ I think that ZFC + VP + "there is a measurable cardinal $\kappa$ and an OD normal measure on $\kappa$" proves that there is a second order statement $\varphi$ which is averageable but not (somewhat) obviously averageable. Also, it's easy to force over a model of ZFC + VP to arrange the OD measure assumption. So we get the consistency in the other direction...I don't know if that's already clear, but if you're interested I can write something on that. $\endgroup$
    – Farmer S
    Dec 23, 2021 at 22:45
  • $\begingroup$ I am definitely interested in that direction, please elaborate! (Honestly I go back and forth about which direction would be most exciting.) $\endgroup$ Dec 23, 2021 at 22:52

1 Answer 1


This doesn't answer your question, but gives some information that might be useful.

Main Claim: ZFC + "There is a proper class of inaccessible limits of measurable cardinals" + "There is some measurable cardinal $\kappa$ and an ordinal definable normal measure on $\kappa$" disproves the statement in question.

Corollary: If ZFC + VP is consistent, then so is ZFC + VP + the negation of the statement in question.

Proof of corollary: Let $\kappa$ be a measurable cardinal. Let $\mu$ be a normal measure on $\kappa$. In some interval of form $(\kappa^+,\eta)$, where $\eta$ is ${<}$ the next measurable $>\kappa$, we can force an encoding of $(\mathcal{P}(\kappa),\mu)$ into the GCH pattern, without adding $\kappa$-sequences. This preserves the fact that $\mu$ is a normal measure on $\kappa$, makes $\mu$ OD, and preserves all large cardinal structure for critical points $>\kappa$, hence VP.

Proof of Main Claim: We deal here with linear iterations of $V$ of length ${<\omega}$ which use only normal measures. That is, a sequence $\left<M_n,\mu_n\right>_{n\leq N}$ such that $N<\omega$, $M_0=V$,and for $n<N$ we have $\mu_n\in M_n$ and $M_n\models$"$\mu_n$ is a normal measure on some measurable cardinal", and $M_{n+1}=\mathrm{Ult}_0(M_n,\mu_n)$, where the $\mathrm{Ult}_0$ denotes the ultrapower constructed using only functions in $M_n$. The critical points of the iteration are the ordinals $\kappa_n$ such that $\mu_n$ is a measure on $\kappa_n$. (So $\kappa_n=\mathrm{crit}(i_{n,n+1})$ where $i_{n,n+1}:M_n\to M_{n+1}$ is the ultrapower map, and $i_{mn}:M_m\to M_n$ is defined by composing these, for $m< n$.)

We would like to find a second order formula $\varphi$ which is averageable, but not (somewhat) obviously averageable. We will look for $\varphi$ in the language of set theory. Note that such a formula is averageable as witnessed by ultrafilter $U$ iff $M=\mathrm{Ult}(V,U)$ is $\varphi$-correct; that is, iff we have $M\models\varphi(X)\iff V\models\varphi(X)$ for all structures $X\in M$ for the language of set theory. Of course $\varphi$ can be taken to demand that $X$ is transitive, and so if desired, one can restrict attention to such things, and also further to sets of the form $V_\alpha^M$, or even ordinals (we will do the latter below). To arrange $\varphi$ we will look at a kind of closure point of forming ultrapowers -- produced by considering finite iterations via normal measures, and roughly, $\varphi(\lambda)$ will say that $\lambda$ is inaccessible in some such finite iterate. However, in order for the correctness proof to work, we will need to slightly modify this plan, by restricting the possible nature of the iterations further, and this is where the OD measure comes in.

Let $\lambda_0$ be the least inaccessible cardinal such that there is a normal measure $U\in V_{\lambda_0}$ such that $U$ is (first order) definable from ordinal parameters over $V_{\lambda_0}$. Then by minimizing on ordinal parameters, actually there is one which is outright definable from no parameters over $V_{\lambda_0}$. Let $U_0$ be such, and we fix a definition of $U_0$ over $V_{\lambda_0}$. Note that $\lambda_0$ and $U_0$ are uniformly second-order describable over any ordinal $\gamma\geq\lambda_0$ (one just uses the 2nd order quantifiers to request a subset of $\gamma$ encoding the right sort of information, so as to recover $V_{\lambda_0},U_0$). Let $\xi_0$ be the corresponding measurable cardinal, i.e. $U_0$ is a normal measure on $\xi_0$.

Let $\varphi(\lambda)$ be the second order statement in the language of set theory asserting the following clauses:

  • $\lambda$ is (isomorphic to) an ordinal,

  • $\lambda\geq\lambda_0$,

  • there is a finite linear iteration $\left<M_n,\mu_n\right>_{n\leq N}$ of $V$ (so $N<\omega$) which uses only normal measures, with strictly increasing critical points $\left<\kappa_n\right>_{n< N}$ (i.e. $\kappa_n<\kappa_{n+1}$),

  • if $N>0$ then $\xi_0\leq\kappa_0$,

  • there is $N_0\leq N$ such that: (i) $\left<M_n,\mu_n\right>_{n\leq N_0}$ is just given by iterating $U_0$ and its resulting images, i.e. if $0<N_0$ then $\mu_0=U_0$, and if $0<m<N_0$ then $\mu_m=i_{0m}(U_0)=(U_0)^{M_m}$ (note possibly $N_0=0$); (ii) if $N_0<N$ then $i_{0N_0}(\xi_0)<\kappa_{N_0}$ and $\kappa_n$ is not a limit of measurable cardinals in $M_n$, for $N_0\leq n<N$.

  • $M_N\models$"$\lambda$ is an inaccessible limit of measurables", and $i_{0N}(\xi_0)<\lambda$ and $\kappa_n<\lambda$ for all $n<N$.

(The requirement that $\kappa_n<\lambda$ for all $n<N$ actually makes no difference to the truth of $\varphi(\lambda)$, since otherwise we can always cut an iteration off at the first point a critical point goes $\geq\lambda$.)

The fact that $\varphi$ is second order is because we can code all the information needed to verify that $\varphi(\lambda)$ holds in a subset of $\lambda$, and given an appropriate such code, it's simply definable to say that it works. For suppose $\left<M_n,\mu_n\right>_{n\leq N}$ is an iteration as above, and $\lambda$ is an inaccessible limit of measurables of $M_N$ as above. If $\lambda\in\mathrm{rg}(i_{0N})$, say $i_{0N}(\bar{\lambda})=\lambda$, then note $\xi_0<\bar{\lambda}$ and $\kappa_n< i_{0n}(\bar{\lambda})$ for each $n<N$ (we can't have $\kappa_n=i_{0n}(\bar{\lambda})$ because $\kappa_n$ is only a successor measurable of $M_n$). Note then that in fact $\bar{\lambda}=i_{0n}(\bar{\lambda})=\lambda$ (using the inaccessibility of $\bar{\lambda}$ in $V$). But also the iteration can be considered as an iteration starting with first model $M_0=V_{\bar{\lambda}}$ (instead of first model $M_0=V$). By inaccessibility, $|V_\lambda|=\lambda$, so there is $A\subseteq\lambda$ coding $V_\lambda$ and the iteration, which is enough. Now suppose instead that $\lambda\notin\mathrm{rg}(i_{0N})$, and let $\theta$ be the least ordinal such that $i_{0N}(\theta)>\lambda$. Then note $\theta$ is a limit of inaccessible limits of measurable cardinals of $V$, so $\theta=|V_\theta|$. The iteration is equivalent to one on $V_{\theta+1}$; we have $\kappa_n<i_{0n}(\theta)$ since otherwise $\theta$ would be a successor measurable in $V$, which it isn't. (Therefore $\theta$ is in fact singular of measurable cofinality, and $i_{0N}``\theta$ is bounded in $i_{0N}(\theta)$.) So it suffices to see that $|V_{\theta+1}|\leq\lambda$, as then we can code $V_{\theta+1}$ through a subset of $\lambda$. But $i_{0N}``V_\theta\subseteq V_{\theta'}^{M_N}$ where $\theta'=\sup i_{0N}``\theta<\lambda$, and we get an injection $V_{\theta+1}\to V_{\theta'+1}^{M_N}$ by sending $X\mapsto i_{0N}(X)\cap V_{\theta'}^{M_N}$. (Note that if $X\neq Y$ then $i_{0N}(X),i_{0N}(Y)$ disagree somewhere within $i_{0N}``V_{\theta}$). But $M_N$ has a bijection between $V_{\theta'+1}^{M_N}$ and some ordinal ${<\lambda}$, since $M_N\models$"$\lambda$ is inaccessible", which suffices.)

Now I claim that $\varphi$ is absolute between $V$ and $M=\mathrm{Ult}(V,U_0)$ (that is, for all $\lambda\in M$, we have $\varphi(\lambda)\iff M\models\varphi(\lambda)$), and note that it follows that $\varphi$ is averageable, as witnessed by $U_0$. (Letting $\lambda=\Pi_{\eta<\kappa}\kappa_\eta$, where the $\kappa_\eta$ are ordinals, we get $\varphi(\lambda)$ iff $M\models\varphi(\lambda)$ iff $\{\eta<\kappa\bigm|\varphi(\kappa_\eta)\}\in U_0$.)

Proof: It is immediate that $M\models\varphi(\lambda)\implies V\models\varphi(\lambda)$, since given the witnessing finite iteration in $M$, just insert $(V,U_0)$ at the front, and note this yields a witness which works for $\lambda$ in $V$. So suppose $V\models\varphi(\lambda)$ and let $\left<M_n,\mu_n\right>_{n\leq N}$ be a witnessing iteration. If $N>0$ and $\mu_0=U_0$ then the tail $\left<M_n,\mu_n\right>_{1\leq n\leq N}$ works in $M$. So suppose either $N=0$ or $\mu_0\neq U_0$, and hence $\xi_0<\kappa_0$ in the latter case. If $N=0$ it is easy: by $\varphi(\lambda)$, we have $\xi_0<\lambda$ and $\lambda$ inaccessible, so $i^V_{U_0}(\lambda)=\lambda$, so $M\models\varphi(\lambda)$. Suppose $N>0$. Then $i_{U_0}(\lambda)=\lambda$ still, so $M\models\varphi(\lambda)$. For $U_0\in M_N$, and $\lambda>\xi_0$ is inaccessible in $M_N$, so $i^{M_N}_{U_0}(\lambda)=\lambda$ where $i^{M_N}_{U_0}:M_N\to\mathrm{Ult}_0(M_N,U_0)$ is the ultrapower map associated to the internal ultrapower of $M_N$, i.e. formed using only functions in $M_N$. But since the iteration leading to $M_N$ has critical points $>\xi_0$, $M_N$ is $\xi_0$-closed, so this embedding is the same as $i^V_{U_0}\upharpoonright M_N$, so $i^V_{U_0}(\lambda)=i^{M_N}_{U_0}(\lambda)=\lambda$.

This proves the claim above. It remains to see that $\varphi$ is not (somewhat) obviously averageable. Suppose otherwise, and let $\widehat{\varphi}$ be the witness, and let $\widehat{\varphi}\in V_{\gamma_0}$ where $\gamma_0$ is a successor inaccessible with $\xi_0<\gamma_0$. Let $\lambda$ be the least inaccessible limit of measurables such that $\lambda>\gamma_0$. So $\varphi(\lambda)$ holds, so $\widehat{\varphi}((\lambda;R_1,\ldots,R_k))$ holds for some predicates $R_1,\ldots,R_k$. Now let $X\preccurlyeq V_{\lambda+1}$ with $V_{\gamma_0+1}\subseteq X$ and $X$ closed under $\gamma_0$-seqences and $|X|<\lambda$. Let $C$ be the transitive collapse of $X$. Note the fact that $\widehat{\varphi}((\lambda;R_1,\ldots,R_k))$ holds is first order over $V_{\lambda+1}$, and reflects into $C$ regarding the collapsed version $(\bar{\lambda},\bar{R}_1,\ldots,\bar{R}_n)$, which is correct about this assertion, because $C$ is closed under $\gamma_0$-sequences. So $\varphi(\bar{\lambda})$ holds. But $\gamma_0<\bar{\lambda}<\lambda$, and $\gamma_0$ is a successor inaccessible with $\xi_0<\gamma_0$. It follows that $\gamma_0$ is fixed by all the finite iteration maps under consideration, and hence so is $\lambda$. So no such iterate can witness $\varphi(\bar{\lambda})$, a contradiction.


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