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In a theory having proper classes it won't be that easy to phrase how many classes we have in comparison to the number of elements of some class. Here, I'll adopt the following method:

We'd say that: there are no more "classes satisfying a unary predicate ϕ" than "the elements of some class X", if and only if, there exists a class F of ordered pairs such that for each class C satisfying ϕ there is a unique element m of X such that: C={c∣(c,m)∈F}

We say that there are only countably many classes to mean there are no more classes than the elements of ω.

Concerning theory ZFC+Classes+Definability rule, if we add to it the axiom that each class has countably many elements, would it follow that there are only countably many classes?

Zuhair Al-Johar
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    Out of curiosity, do you have a guiding muse/heuristic you’re trying to chase down with all these different foundational approaches to proper classes? Some informal way you think proper classes ‘should’ behave that you’re trying to properly formalize? – Alec Rhea Dec 28 '22 at 02:01
  • @AlecRhea, I'm examining the line that all classes must be extensions of predicates, and all the latter ones must be definable in a clear cut manner, i.e. definable without parameters in a finitary first order language. That said then we can only have countably many classes. And since we are speaking about pure classes then all of them are countable. – Zuhair Al-Johar Dec 28 '22 at 05:32
  • Does this setting satisfy your intuition for how classes ‘should’ behave? I’m curious why you’re exploring all these different formalisms for the set/class dichotomy; if you could explain why a situation where we only have countably many countable classes is appealing to you I could better judge how interesting the question is. (I am not a downvoter, but could be an upvoter) – Alec Rhea Dec 28 '22 at 05:58
  • @AlecRhea, because of the intuition of all classes to be definable in a clear cut manner, if we maintain that ALL classes must be definable in a parameter free manner, then we only have countably many parameter free formulas, and so we can only have countably many classes definable after them; and since we are speaking of pure classes, then all would be countable. IF every model of ZFC can be extended to a pointwise (i.e. parameter free) definable model, then we can do everything in those countable models, then we can Occam razor the rest, that if there is a rest. – Zuhair Al-Johar Dec 28 '22 at 06:08
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    So, based in your usage of Occam, you view countable pointwise definable models as somehow the ‘simplest’ models of ZFC? Why? – Alec Rhea Dec 28 '22 at 06:23
  • @AlecRhea, pointwise models are always countable, they are the 'smallest' models of ZFC, and they are the most clear ones as far as perceiving the notion of 'set' as a subnotion of 'class', (i.e. sets as some kind of classes) is concerned. The clearest view of a class is as an extension of a predicate, and if we cannot describe a predicate by intention in a clear cut manner, then the class extending it would be dubious, since it would be an extension of an unclear intention. If we can do withtout them, then why use them? The matter is not technical, it is about the very concept of class itself – Zuhair Al-Johar Dec 28 '22 at 06:43

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As stated in the comments, I misread the definition of the theory and assumed it is completely in infinitary logic, while in reality the ZFC+classes fragment is still in finitary logic, so the answer bellow is wrong

I do not believe you need the additional axiom. The idea is to use the definability rule together with the fact we are using infinitary logic to get poor's man quantification over formuleas.

Let φ=X=X, and define F_0=\{(c,m) \mid φ_m(c)\} where the φ_i are the ones from the definability rule.

it is well defined because we are in L_{ω_1,ω}, if we want to be precise, let ψ_i(x)=x\text{ is the 2-tuple $(y,k)$ and $k=i$ and $φ_i(y)$} then F_0=\{x\mid \bigvee_{i\in\omega}ψ_i\}.

The definability rule exactly states that for each class C (and so each class C such that φ(C)) there exists m∈ω such that F_0^{-1}(\{m\})=C.

To get rid of duplicates entries simply note that "it wasn't previously defined" is definable in our language: define F=\{(c,m)∈F_0\mid n<m\implies \text{there exists $d$ such that $φ_m(d)⇔¬φ_n(d)$}\}.

Again, if we want to be precise, we can define the following sequence of formulaes, η_i(m)=\left(m=i⇒\bigwedge_{j<i}∃d\left(φ_i(d)⇔¬φ_j(d)\right)\right) then F=\{(c,m)\in F_0\mid \bigwedge_{i\in\omega}ν_i(m)\}.

Clearly each class is still the preimage of some natural number, and if m>n then F^{-1}(\{m\})≠F^{-1}(\{n\}).

Note that, assuming your theory is consistent, we get that F is single definable using a finitary formula.

Holo
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  • You are constructing classes using infinitary formulas, this cannot be assured to exist in \sf ZFC + Classes + \text{ Definability rule}. – Zuhair Al-Johar Dec 28 '22 at 05:39
  • @ZuhairAl-Johar you are right, I missed the part in your previous question that stated that ZFC+Classes is still in finitary logic, and assumed everything moved to infinitary logic. Sorry – Holo Dec 28 '22 at 06:48