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In ZFC with class terms a class is merely a unary predicate $\varphi$ of the language which we then write as

$$\{X\mid\varphi(X)\}$$

which only suggests that we have built a collection out of these objects. We write $Y\in\{X\mid \varphi(X)\}$ for $\varphi(Y)$. Further we say that such a class (term) $\mathcal C$ is a set if

$$\exists X\forall Y[Y\in X\leftrightarrow Y\in\mathcal C].$$

In NBG and similar set theories (which build on classes instead of sets) every set is also a class. But is this true for this ZFC approach? For me it seems that a class is just a formula. And there are sets that I cannot fully describe using such a formula, e.g. a set given to me by the axiom of choice.

Are sets and classes in ZFC (with class terms) just different concept, each one not a "sub-concept" of the other?

After Asaf's answer:

I do not feel very well when writing $\{X\mid X\in A\}$ for arbitrary sets $A$ because for me $A$ is not a symbol of the language. However, when my set $A$ is definable by a predicate $\varphi$ with

$$\mathrm{ZFC}\vdash \exists A \varphi(A) \quad\text{ and }\quad \mathrm{ZFC}\vdash\varphi(A)\wedge \varphi(B)\to A=B,$$

then I could write $\{X\mid \forall A[\varphi(A)\to X\in A]\}$, and this would feel okay. But my problem are the sets for which there is no such $\varphi$. I do not know how I can include such an abbreviating notion (like $\{X\mid X\in A\}$) into a formal proof without feeling not quite sure what I am actually doing.

M. Winter
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  • As written, you are correct -- but usually in this formulation, the formula $\phi$ is allowed to have a parameter $A$; that is, you can actually, for each set $A$, form the class ${X \mid \phi(X,A)}$, and then $Y \in {X \mid \phi(X,A)}$ becomes a shorthand for $\phi(Y,A)$. In that case it is easy to show that every set is also a class: for a set $A$, you take the formula $\phi(X,A)$ to be $X \in A$. – Mees de Vries May 17 '17 at 09:33
  • Correct; in $\mathsf {ZFC}$ "domain" there is only one kind of objext: sets. The language of the theory uses formulas and we call a certain type of them "class-formula". – Mauro ALLEGRANZA May 17 '17 at 09:37
  • @Mees Doesn't this give me more of a "dependent class" $\mathcal C(A)$? I understand this approach in axioms like the axiom of specialization, where one quantifies universally over such $A$'s. But here, I have problems to see how $\varphi$ is still a valid formula of the language if I fix a set in there which is something more associated with the interpretation than the syntax. – M. Winter May 17 '17 at 09:40
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    Realted post: what-is-the-formal-way-to-define-class-in $\mathsf {ZFC}$. – Mauro ALLEGRANZA May 17 '17 at 09:43
  • @Mauron What certain type exactly? Unary predicates? – M. Winter May 17 '17 at 09:44
  • @M.Winter, I suppose you could call it a "dependent class" if you want to, but I feel like it does what you want it to. Perhaps I should ask a leading question: what do you want to do with these classes once you have internalized them to ZFC? What kind of properties do you want to express of them, or prove about them? – Mees de Vries May 17 '17 at 09:50
  • Yes; formulas $\varphi(x)$ with one free var (and zero or more parameters): we can say that the "basic task" of axiomatic set theory is to restrict attention to formulas $\varphi$ for which we can prove the existence of a set, say $a$, such that: $x ∈ a ↔ \varphi(x)$. – Mauro ALLEGRANZA May 17 '17 at 09:53
  • Thus, in my first comment above, is say "Correct" to your analysis. Regarding the title question: "[in $\mathsf {ZFC}$] is any set also a class?", the answer is: no; in $\mathsf {ZFC}$ there are no class. – Mauro ALLEGRANZA May 17 '17 at 09:55
  • @Mauro Ok I should have written: is there a class term for any set? – M. Winter May 17 '17 at 09:57

1 Answers1

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Yes. Sets are always classes in $\sf ZF$.

The reason is that while we often like to omit the parameters from our formulas, they can still be used when talking about classes. For example, "all the ordinals above $\kappa$" is a class definable with a parameter: $\kappa$.

Equally, a set $A$ is exactly the class $\{x\mid x\in A\}$.

Caveat lector: if we are talking about non-transitive models, then from the outside there is a difference between $x$ and $\{y\mid M\models y\in x\}$, but we can make this identification between sets and class so they are considered to be the same.

Asaf Karagila
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  • I do not feel very well when writing ${x\mid x\in A}$ because for me $A$ is not a symbol of the language. However, when my set $A$ is definable by a predicate $\varphi$ like $\mathrm{ZFC}\vdash \exists A \varphi(A)$ and $\mathrm{ZFC}\vdash\varphi(A)\wedge \varphi(B)\to A=B$, then I could write ${x\mid \forall A[\varphi(A)\to x\in A]}$, and this would feel ok. But my problem are the sets for which there is not such a $\varphi$. I do not know how I can include such an abbreviating notion into a formal proof without feeling not quite sure what I am doing. – M. Winter May 17 '17 at 10:30
  • But you do feel right writing ${\alpha<\kappa\mid\operatorname{cf}(\alpha)=\lambda}$ where $\kappa$ and $\lambda$ are two completely arbitrary ordinals? I get why this bothers you, because $x\in A$ is a very simple formula. But this is about semantics, especially if you think about proper classes as collections of a universe. If you want to think about this purely syntactic, then you don't have access to "most things", and then you can take $A$ as a free variable, and prove that $x\in A$ if and only if $x\in A$ to get the same thing, so $\forall A(x\in A\leftrightarrow x\in A)$ is provable – Asaf Karagila May 17 '17 at 10:53
  • Actually I do not feel well using this either (when being very formal of course). But, I could write $\forall \kappa,\lambda[\kappa,\lambda\in\mathrm{Ord}\to\exists X[X={\alpha<\kappa\mid\mathrm{cf}(\alpha)=\lambda}]]$ or something like this. Maybe I am just to unexperienced and maybe too careful because I do not want to lose the sight on how to translate my proof into an absolutely formal version if necessary. Every non-formal notation should be translatable one-to-one until the proof is completely formal. And I have no clue how to do this for something like ${x\mid x\in A}$. – M. Winter May 17 '17 at 11:30
  • But you are not defining a class. You are writing a closed sentence. – Asaf Karagila May 17 '17 at 12:00
  • Ah I see. Then for me this would be a class $\mathcal C(\kappa, \lambda)$ which depends on two ordinals. Not a single class, but more of a class function. – M. Winter May 17 '17 at 12:12
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    You can think about classes as "functions" from parameters to collections of objects in the universe using the "parameters assignment" function. Yes. – Asaf Karagila May 17 '17 at 12:30