This is a "forcing-absolute" followup to this question, whose answer was largely unsatisfying. The question is:

Suppose $V=L$. Is there an $\mathcal{L}_{\infty,\omega}$-sentence $\varphi$ such that in no forcing extension is $\varphi$ equivalent to a second-order sentence?

Throughout, by "forcing" I mean "set forcing," although the (tame) class forcing version also seems potentially interesting.

EDIT: I forgot to add that I'm restricting attention to infinite structures here (which is key to my comment below that every projectively-definable infinitary sentence is second-order expressible). As Fedor Pakhomov commented below, without this restriction the problem is trivial since second-order theories of finite structures can't be changed by forcing. I do not, however, want to restrict attention to countable structures.

I've decided to focus on models of $\mathsf{V=L}$ since that hypothesis seems to add interesting flavor to the question in a few ways:

  • It implies that such a $\varphi$ must not be in $\mathcal{L}_{\omega_1,\omega}$. This is because $(1)$ every constructible real is projective in some forcing extension and $(2)$ every projectively definable $\mathcal{L}_{\omega_1,\omega}$-sentence is equivalent to a second-order sentence. So a positive answer would have to crucially rely on uncountable Boolean combinations - which seems a bit odd, because each specific $\mathcal{L}_{\infty,\omega}$-sentence is equivalent to some $(\mathcal{L}_{\omega_1,\omega})^{V[G]}$-sentence in an appropriate forcing extension $V[G]$ (just collapse the size of the sentence), but isn't an obvious contradiction since the "potential projectivity" fact about $L$ doesn't seem to lift to arbitrary forcing extensions of $L$.

  • It rules out the "silly" solution provided by large cardinals. If large cardinals exist - specifically, enough to guarantee projective absoluteness - then any infinitary sentence which is not equivalent to a second-order sentence in $V$ remains so in all forcing extensions. (Note that this would give us an example in $\mathcal{L}_{\omega_1,\omega}$ for that matter.) But $\mathsf{V=L}$ breaks this "hammer," so that we seem to be forced to do some actual work.

  • If the answer is yes, there is in fact a definable example, namely the least such sentence with respect to the $L$-ordering. Of course this is silly, but it suggests that there might be canonical examples in a more interesting sense. By contrast I could imagine models of $\mathsf{ZFC+V\not=L}$ where the existence of such a sentence is guaranteed nonconstructively (e.g. by a more intricate counting argument), and so no canonical example need exist.

That said, since it seems plausible that the $\mathsf{V=L}$-situation is more difficult to attack than I'd hoped, I'm also interested in results for other extensions of $\mathsf{ZFC}$.

  • $\begingroup$ The answer is no for fairly boring reason. Reasoning in $\mathsf{ZFC}+V=L$ consider some set $C$ of finite models in the signature of pure equality such that for no second-order sentence $\varphi$ we have $M\models \varphi\iff M \in C$ ($C$ exists by cardinality argument). Since validity of second-order sentences in finite models isn't changed by forcing any $\mathcal{L}_{\omega_1,\omega}$-formula $\psi$ axiomatizing $C$ gives a counterexample. I guess the right question should be about infinite models. $\endgroup$ Commented Dec 3, 2021 at 11:45
  • $\begingroup$ @FedorPakhomov Yes I forgot to include that - note that the restriction to infinite models shows up implicitly in my claim that "projective infinitary = second-order" (without an infinite domain to work with this doesn't hold). Fixed, thanks! $\endgroup$ Commented Dec 3, 2021 at 16:23

1 Answer 1


For any given finite signature $\Omega$ there is a second-order sentence $\varphi$ of the signature $\Omega$ such that $\mathsf{ZFC}+V=L$ proves that for any $\mathcal{L}_{\infty,\omega}$-formula $\psi$ of the signature $\Omega$ there is a poset $P$ for which it is $\Vdash_P$-forced that for any infinite model $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$ iff $\mathfrak{M}\models \psi$.

$\varphi$ should be a second-order sentence expressing in all infinite models $\mathfrak{M}$ that the following holds (see below for a more explicit construction of $\varphi$):

  1. there exists the greatest ordinal $\alpha$ such that the value $\aleph_\alpha^{L_\beta}$ is the same for all large enough countable limit ordinals $\beta$;
  2. $\alpha$ is the position in $<_L$ of some $\mathcal{L}_{\omega_1,\omega}$-sentence $\psi$
  3. $\psi$ is true in $\mathfrak{M}$.

Given $\mathcal{L}_{\infty,\omega}$-formula $\psi$ we consider $\alpha_0$ that is the position of $\psi$ in $<_L$. Let $P$ be the Levy collapse of $\aleph_{\alpha_0}$ onto $\omega$. Let us check that for this $\varphi,\psi$, and $P$ we have the desired equivalence.

Indeed let us reason inside $\Vdash_P$. Clearly, $\psi$ became an $\mathcal{L}_{\omega_1,\omega}$-formula.

Observe that $\alpha$ from 1. will always be equal to $\alpha_0$. Indeed, let us choose countable limit $\beta_0>\alpha_0$ such that all ordinals $\gamma<\alpha_0$ that aren't $L$-cardinals aren't cardinals in $L_{\beta_0}$. Clearly, for any limit $\beta>\beta_0$, we have $\aleph_{\alpha}^{L_\beta}=\aleph_\alpha^L$. For any $\alpha'>\alpha_0$ since $\aleph_{\alpha'}^L$ isn't countable and any countable $\gamma$ we could find limit countable $\beta'$ such that $\aleph_{\alpha'}^{L_{\beta'}}$ is either undefined or has the value $>\gamma$. Hence $\alpha_0$ is the ordinal $\alpha$ from 1.

Using this the proof of semantical equivalence of $\varphi$ and $\psi$ in infinite models is trivial.

Now let me give more detailed description of $\varphi$. Let $\mathsf{KPUL}_2(\in,\in_u)$ be a second-order formula depending on element variable $\alpha$ and binary predicate symbols $\in,\in_u$ asserting that we have $(\mathsf{KPU}-\mathsf{Foundation})+\textsf{Second-Order-Foundation}+L[\mathfrak{M}]=V$ for the structure where each element of the underlying domain simultaniously represent itself (as an urelement) and some set, $\in$ gives membership of sets in sets, $\in_u$ gives membership of urelements in sets. Formula $\mathsf{Emb}(R,\in,\in_u,\in',\in_u')$ expresses that $\mathsf{KPUL}_2(\in,\in_u)\land \mathsf{KPUL}_2(\in',\in_u')$ and the unary function $f$ gives an end embedding of the admissible set given by $(\in,\in_u)$ into the admissible set $(\in',\in_u')$. Formula $\mathsf{St}(\alpha,\beta,\kappa,\in,\in_u)$ expresses that $\mathsf{KPUL}_2(\in,\in_u)$ and in the corresponding admissible set $\alpha$ is a countable ordinal, $\beta$ is a countable limit ordinal, $\kappa$ is the ordinal that is $\alpha$-th cardinal according to $L_\beta$, and for any $f,\in',\in_u',\beta'$ if $\mathsf{Emb}(f,\in,\in_u,\in',\in_u')$ and $\beta'>f(\beta)$ is a countable limit ordinal according to the admissible set given by $(\in',\in_u')$, then there is $f(\alpha)$-th cardinal according to $L_{\beta'}$ and it is equal to $f(\kappa)$. Formula $\mathsf{MSt}(\alpha,\in,\in_u)$ expresses that $\mathsf{St}(\alpha,\in,\in_u)$ and we don't have $\mathsf{St}(\alpha+1,\in,\in_u)$. Formula $\mathsf{LC}(\alpha,\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\alpha$ is an ordinal, $\psi$ is a constructible $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$, and its place in $<_L$-order is $\alpha$. Formula $\mathsf{Tr}(\psi,\in,\in_u)$ expresses that in the admissible set given by $(\in,\in_u)$, $\psi$ is an $\mathcal{L}_{\omega_1,\omega}$-sentence of the signature $\Omega$ that is true in the underlying model $\mathfrak{M}$. We put $\varphi$ to be: $$\exists \in,\in_u,\alpha,\psi(\mathsf{KPUL}_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$

  • $\begingroup$ I don't quite understand your $\varphi$. The first clause seems to be "referring to" the universe, not (the powerset of) the structure $\mathfrak{M}$, or am I missing something? $\endgroup$ Commented Dec 3, 2021 at 16:26
  • $\begingroup$ @NoahSchweber The first and second clauses indeed don't use structure of the model and simply are formulated in terms of quantification over predicates (specifically we need unary and binary predicates) on the domain of the model. Since the domain is infinite, we easily could express quantification over countable structures. I am busy right now, but a bit latter I'll expand on the definition of $\varphi$. $\endgroup$ Commented Dec 3, 2021 at 16:42
  • $\begingroup$ @NoahSchweber Added a more detailed definition of the sentence $\varphi$. $\endgroup$ Commented Dec 3, 2021 at 19:48

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