# Can a proper class be a member of a set?

There exists a definition in NBG stating that a class can be considered a set if it belongs to another class, and it is considered a proper class if it does not belong to any class. I see that this definition put forth by NBG is a restriction placed on the proper class to make it unmanageable, whether using union, intersection, power relations, or otherwise. It is merely an attempt to patch together non-contradictory formal expressions that could have been avoided by using ZFC alone.

Is there anything forcing us to adopt this definition, given that, starting from sets whose existence doesn't cause contradictions, we might eventually obtain proper classes whose existence does lead to contradiction, with previous sets being members of them? As demonstrated in Cantor's paradox, which proved that the set of all sets is not a set. What prevents us once again from obtaining new sets whose existence is logically acceptable, based on these controversial entities known as proper classes, and these proper classes become members of them?

In other words, can there be a set $$A$$ where $$(\ldots \in x \in \ldots \in A)$$ , where $$x$$ is a proper class, and the points represent a series of memberships that may be finite or infinite?

• No; see Class: "the notion of "proper class", e.g., as entities that are not members of another entity." Commented May 21 at 6:48
• If a class belongs to another class we have the Russell's paradox again applied to classes instead of sets... Commented May 21 at 7:06
• Cantor insisted on the freedom in mathematics (though he did not apply such a principle uniformly), so you can certainly develop any logical theory you like, including the possibility you mentioned of proper classes being elements of sets. But such an approach would be incompatible with ZFC since in ZFC by definition an element of a set is itself a set. Commented May 22 at 14:03
• @BezinaTaki, that's an interesting idea, but this also cannot occur in ZFC. Indeed, the axiom of well-foundedness rules out an infinite descending chain of the kind you mentioned. If the chain is finite, then all the intermediate steps are sets by transitivity. Though in axiomatic nonstandard analysis you do have a kind of infinite descending chain of inclusions, but that's another story :-) Commented May 23 at 9:02
• @Lukrau, I agree, I am not sure what we are arguing about exactly :-) Let me give you an example of a situation where such a chain is possible. In axiomatic nonstandard analysis, the integers are $\mathbb N$ as usual, but there are two types of integers: standard and nonstandard. All standard ones precede all nonstandard ones. The collection $\omega$ of all standard integers is not a set but can be viewed as a class. Then you have the chain of command $1\in2\in3\cdots\in\omega$ and $\omega$ is included in $\mathbb N$. Commented May 27 at 9:14