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Let $\mathsf{ZFC_2}$ denote $\mathsf{ZFC}$ with a second-order replacement axiom. It has been discussed in some other answers that every model of $\mathsf{ZFC_2}$ is isomorphic to $V_\kappa$ for some inaccessible cardinal $\kappa$. But it seems to me that the intended "model" of $\mathsf{ZFC}$ is supposed to be the proper class $V$, so it seems natural to ask whether $V$ is unique as a "class model" of $\mathsf{ZFC_2}$ in some sense.

Let's define "class model" to be a model whose underlying "set" is a proper class. And let's define "class categorical" to mean that all class models are isomorphic. Then is $\mathsf{ZFC_2}$ class categorical?

The intuition behind this idea is an analogy with $\mathsf{PA}$. First-order $\mathsf{PA}$ has many distinct non-standard models. But $\mathsf{PA_2}$ is categorical and singles out the informally "intended" model uniquely (up to isomorphism). So the thought is, maybe $\mathsf{ZFC_2}$ does the same for the "intended model" of $\mathsf{ZFC}$. That said, I have no idea what sorts of thorny issues might arise when we try extending models and categoricity to encompass proper classes.

WillG
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    Are you also requiring your class-sized models to be "locally set-like," in the sense that for each such $\mathcal{M}$ and each $m\in\mathcal{M}$ the class ${n\in\mathcal{M}: n\in^\mathcal{M}m}$ is a set? If so the answer is yes, but if not things are more complicated. – Noah Schweber Mar 07 '24 at 15:19
  • @NoahSchweber In other words, you're asking if I'm only considering models whose elements are sets? Yes, that's my main interest. I didn't even realize non-set elements was a possibility. – WillG Mar 07 '24 at 16:47
  • Second-order replacement, but not separation? I don’t think first-order separation and second-order replacement imply second-order separation… – user3840170 Mar 07 '24 at 17:41
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    @user3840170: 2nd-order replacement implies 2nd-order separation. If $U$ is a unary predicate, let $R$ be the binary predicate $R(x,y):=(Ux\land y={x})\lor(\neg Ux\land y=\emptyset)$. Then $\forall x\exists! y R(x,y)$ and for a set $T$, ${x\in T\colon Ux}=\bigcup {y\colon \exists x\in T, R(x,y)}$. This along with 1st order powerset implies 2nd order powerset. Also, 2nd order regularity as follows - if $U$ is a unary predicate and $Ux_0$ for some $x_0$ then an $\in$-minimal element of $\text{trcl}(x_0)\cap U$ is $\in$-minimal for $U$. – Chad K Mar 07 '24 at 18:31

1 Answers1

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There are some subtleties around class-sized models, but any reasonable approach to handling them will give an affirmative answer to your question if you restrict attention to "locally set-like" models, i.e. models $\mathcal{M}$ for whichevery class of the form $$\{n\in\mathcal{M}: n\in^\mathcal{M}m\}$$ for $m\in\mathcal{M}$ is actually a set.

To see this, we just follow the usual "quasi-categoricity" argument for the set-sized case. By second-order Foundation, $\mathcal{M}$ must be well-founded; by second-order Powerset, after taking a Mostowski collapse (permitted by the well-foundedness we've just established) we must have $V_\alpha^\mathcal{M}=V_\alpha$ for each ordinal $\alpha$. And since $\mathcal{M}$ is class-sized, that leaves only one candidate: $\mathcal{M}=V$.

But this analysis breaks down if we allow $\mathcal{M}$ to not be locally set-like. In particular, the set-like part of $\mathcal{M}\models^{class}\mathsf{ZFC_2}$ will always be (isomorphic to) $V$ itself, but it may be a "proper top extension" of $V$. On the other hand, no such $\mathcal{M}$ will be first-order definable in $V$, so if you restrict attention to definable class models we recover categoricity.

(Note that there are $V$-definable well-founded non-locally-set-like structures; e.g. take $\mathit{Ord}$ with the order that puts $0$ above everything else and otherwise is ordered as usual. So the definability barrier here is nontrivial.)

Noah Schweber
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  • I didn't realize that second-order versions of power set and foundation were important for any of this. What exactly are the second-order versions of these? And why do they "need" to be second order (to make this answer go through)? (I thought the main motivation for making replacement second order is that the first-order version is restricted by only allowing first-order definable functions.) – WillG Mar 07 '24 at 18:49
  • @WillG They're corollaries of second-order replacement, so not every presentation of second-order ZFC lists them separately. The key point is just that models of second-order ZFC are well-founded and compute powersets correctly; this is already needed for the original set-sized quasicategoricity proof, so there's nothing new here. – Noah Schweber Mar 07 '24 at 19:00
  • Ok, but I'm curious about what they actually say. For example, I would guess the second-order power set sentence is something like "For every class there exists a 'power class'." But wouldn't such a power-class then be a third-order object, requiring third-order logic to state? – WillG Mar 07 '24 at 19:24
  • @WillG Second-order powerset intuitively says "Every set's powerclass is a set." In more detail, using lowercase variables for sets and uppercase ones for classes, second-order powerset says: "For every $x$ there is a $y$ such that for every $Z\subseteq x$ there is a $u\in y$ with $u=Z$" (or if you prefer, "$\forall q(q\in u\iff q\in Z)$" for that last clause). – Noah Schweber Mar 07 '24 at 19:28
  • (Actually this $y$ might "overshoot" the powerset, but we can always cut down via separation later.) Similarly, second-order foundation says "For every nonempty $X$ there is a $y\in X$ such that no $z\in y$ has $z\in X$." – Noah Schweber Mar 07 '24 at 19:30
  • @WillG Actually, my comment four previously isn't quite right: these are best thought of as corollaries of second-order separation, not replacement. Second-order separation is the sentence "Every subclass of a set is (coextensional with) a set," which immediately gives you that powersets are computed correctly and that the universe is truly well-founded. – Noah Schweber Mar 07 '24 at 20:23
  • I see. Although as Chad K. shows in a comment on the post, second-order separation follows from second-order replacement. – WillG Mar 08 '24 at 04:16