5:36 PM
In the context of NBG, having a class model of some theory T only means that ZFC can interpret the theory T. For example, NBG has a class model of ZFC trivially. However, in the context of MK set theory, having a class model of some theory actually implies that the theory is consistent, since MK has stronger semantics for its classes. Classes in MK really just behave much more like sets.
Because of that, "class model" can have a very different meaning depending on the ambient theory. By contrast, having a set model always entails that the theory is consistent, in almost any reasonable ambient theory.
The reason for that is because, when you have a set model of a theory, you can define the truth of its formulae using a recursion on the quantifier complexity. This also works in MK when you have a class model, since MK is able to arbitrarily construct recursive sequences of classes. NBG cannot construct recursive sequences of proper classes, however, except in special cases.
Uhhmmmm it depends on what you mean, when you say that the theory is "first order" while also quantifying over classes
Well, actually it doesn't really depend. Under any reasonable definition of your question, the answer is yes, we can find theories which exclude such class models
You can always fine-tune the strength of set theory, by carefully defining the length of the cumulative hierarchy. For example, the theory of hereditarily finite sets (ZFC modified where the axiom of Infinity is swapped for its negation) cannot find any class model of ZFC. More specifically, that theory should prove that no class can be a standard model of ZFC.
6:12 PM
My preferred axiomatization of Sets appeals directly to the cumulative hierarchy. Include the axioms of Extensionality and Specification, and then define the following operation and classes.
UP(C) := {S : ∃(V∈C), S⊆V}
𝓗 := {C : ∀(V∈C), V=UP({u∈C : ¬[V⊆u]})}
𝓒 := {UP(C) : C∈𝓗}
𝓥 := UP(𝓒)
Here, UP(C) basically just denotes the union of the powersets of all the members in C. Using nothing but Specification and Extensionality, you can prove that 𝓒 is wellordered by the subset relation, where the powerset denotes the order theoretic successor. Moreover, every member of 𝓒 is equal to the u…
UP(C) := {S : ∃(V∈C), S⊆V}
𝓗 := {C : ∀(V∈C), V=UP({u∈C : ¬[V⊆u]})}
𝓒 := {UP(C) : C∈𝓗}
𝓥 := UP(𝓒)
Here, UP(C) basically just denotes the union of the powersets of all the members in C. Using nothing but Specification and Extensionality, you can prove that 𝓒 is wellordered by the subset relation, where the powerset denotes the order theoretic successor. Moreover, every member of 𝓒 is equal to the u…
You can get very very specific, with how long you want the cumulative hierarchy to be. For example, if you assert "every member of 𝓒 has a successor, and there's exactly one limit ordinal", then that's equivalent to saying that the length of the cumulative hierarchy is at least ω+ω
You could also assert "every rank in 𝓒 admits a limit above it", which would guarantee that the length of the cumulative hierarchy is at least ω^2.
Moreover, the length of the hierarchy often directly correlates to a theory that axiomatizes it. For example, if you only assert "there is at least one limit in 𝓒", then that theory is mutually interpretable with PA2 (second-order arithmetic).
If you also include Powerset and Replacement (e.g. using the order-theoretic formulation I stated previously), you get a theory mutually interpretable with ZFC.
To be clear, everything I just said about classes merely uses (something like) Henkin semantics, where "class" just means "definable collection". That is a conservative extension to any first-order theory. So, any statements I made about classes can be translated into a (schema of) first-order statement(s), which do not mention classes at all.
The theory of hereditarily finite sets is equivalent to the theory of the hierarchy of length exactly ω; that's also bi-interpretable with PA. Similarly, PA2 is bi-interpretable with the theory of the hierarchy of length ω+1. The second-order theory of Reals (where we can quantify over subsets of ℝ, with the full Specification schema) is bi-interpretable with the theory of hierarchy with length ω+2. Zermelo set theory is mutually interpretable with the hierarchy of length ω+ω.
6:54 PM
uhh correction to the correction; you actually do have that ZFC is "mutually interpretable", because ZF is mutually interpretable with ZFC, so my original correction was unnecessary. You do not have bi-interpretability with ZFC however, because Choice cannot be framed as an order-theoretic property of the cumulative hierarchy. However, you do have bi-interpretation between ZF and a certain extension of the theory of hierarchy, where the extended axioms are purely order-theoretic.
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