# Can a weak second-order ZFC have countable models?

(This question is the "leftover" part of this old question of Mallik, which was substantially clarified in the comments. Throughout, "second-order logic" refers to the standard semantics.)

Let $$ZFC_2^{def}$$ be the theory in second-order logic consisting of:

• The usual (first-order) formulations of Infinity, Pairing, Union, Powerset, Extensionality, and Foundation.

• The Separation and Replacement schemes for second-order formulas.

(The "def" here stands for "definite," see the original question linked above.) My question is:

Is it consistent that $$ZFC_2^{def}$$ has a countable model?

(A bit more precisely: is the first-order statement "$$ZFC_2^{def}$$ has a countable model" consistent with first-order ZFC? It's perfectly kosher to reason about second-order logic inside a first-order system.)

• It's crucial that we're using first-order Powerset instead of second-order Powerset here, since of course Infinity + second-order Powerset ensures uncountability. On the other hand, it's not hard to show that we could replace first-order Foundation with second-order Foundation without changing the theory: that is, all models of $$ZFC_2^{def}$$ are well-founded.

• The Separation scheme for second-order formulas is not what's generally referred to as "second-order Separation:" the former is the scheme consisting of $$\forall \overline{a}\forall x\exists y\forall z(z\in y\leftrightarrow z\in x\wedge \varphi(\overline{a}, z))$$ for $$\varphi$$ a second-order formula, while the latter is the single axiom $$\forall x\forall A\exists y\forall z(z\in y\leftrightarrow z\in x\wedge z\in A).$$ Similarly, the Replacement scheme for second-order formulas is a priori weaker than the single axiom generally referred to as "second-order Replacement."

• It's not hard to show that $$ZFC_2^{def}$$ consistently doesn't have a countable model (as my answer to Mallik's original question does) but this uses an additional set-theoretic assumption: that there is a nice well-ordering of enough of the universe.

• In your last comment, $ZFC_2^{def}$ should replace $T$, right? Feb 17 '20 at 19:30
• @Z.A.K. Yup, fixed. Feb 17 '20 at 19:30
• Now when you say kosher, what exactly do you mean? Who was the rabbi overseeing the process? From which Jewish tradition he came? What is his approach to modernity? Feb 18 '20 at 9:51

Yes, it is possible that there are countable $$ZFC_2^{def}$$ models. We start with $$V=L$$ and an inaccessible cardinal $$\kappa$$. The model will be $$M=V_\kappa^L$$ in the forcing extension $$L[G]$$ where $$G$$ is $$\operatorname{Col}(\omega,\kappa)$$-generic over $$L$$. Of course the interesting axioms to check are the speraration and replacement schemes for 2nd order formulas. Lets do separation as it is notationally easier.
Assume $$a, p\in M$$ and that $$\varphi(x, y, z)$$ is a 2nd order $$\in$$-formula. Note that $$b=\{c\in a\mid (M, \mathcal P (M))\models \varphi(c, a, p)\}^{L[G]}$$ is ordinal definable in $$L[G]$$ since $$M=V_\kappa^L$$ (and thus $$\mathcal P(M)$$) is and $$a, p$$ are definable from there position in the canoncial wellorder of $$L$$. As $$a\subseteq HOD^{L[G]}$$ we have $$b\in HOD^{L[G]}$$. As is well known, $$\operatorname{Col}(\omega, \kappa)$$ is ordinal definable and cone homogeneous and thus $$HOD^{L[G]}\subseteq HOD^L=L$$ so that $$b\in M$$.
Replacement works similar, however there one needs to appeal to the regularity of $$\kappa$$ in $$L$$.
• Nice, +1! (FWIW it may be simpler to just say $M=L_\kappa$.) At a glance this looks right, I'll accept when I've had a few minutes to read it properly. Feb 19 '20 at 18:40