# Does it make sense to form a set whose elements are proper classes?

In chapter 4 of Handbook of Categorical Algebra, vol 1, the author defines a "subobject of $A$" as "an equivalence class of monomorphisms with codomain $A$" (for a suitable notion of equivalence). He then defines what it means for a category to be well-powered: "$\mathcal{A}$ is well-powered when the subobjects of every object constitute a set". Thus, for instance, the category of sets is well-powered.

I'm having trouble understanding exactly what it means to have such a set of subobjects. As far as I can tell, each element of a set should also be a set, but an equivalence class of monomorphisms could be a proper class: for instance, the class of singleton sets is not a set. On the other hand, it seems that one can cheat by defining a subobject to be a class containing one representative of each equivalence class of monomorphisms, even though this is not, strictly speaking, what's stated in the book.

How can one solve this problem? Is there a "normal" set theory where such a set of subobjects can contain proper classes? Or does one need to cheat like suggested above?

• Elements of a set must be a set so there is no set which has a proper class as element. You need a another word to denote `family of proper classes'... – Hanul Jeon Nov 5 '14 at 14:59
• That's what I suspected indeed, but how to make sense of the definition of "well-powered", then? – Arthur Azevedo De Amorim Nov 5 '14 at 15:20
• If classes are objects, then it seems to make just as much sense to have a set of them as it does to have a set of cats. We need a slightly different theory of sets, see en.wikipedia.org/wiki/Urelement, though. – GME Nov 5 '14 at 15:22
• This just makes sense with a global choice. Maybe he's assuming locally small categories... – user40276 Nov 13 '14 at 21:10
• @user40276 Yes, he is assuming locally small categories, but even with this assumption it seems a bit strange, because the monomorphisms in the definition need not have the same domain. Each hom set is indeed a set, but when you add up hom sets for lots of different objects, you can get something that is not one. – Arthur Azevedo De Amorim Nov 14 '14 at 14:57

## 1 Answer

No, proper classes are not elements of other sets (or other classes, for that matters).

But this is the great power of universes. Classes of one universes are just sets in a larger universe. So when you want to talk about collections of classes, you move to a larger universe, where you can treat them formally as sets.

Another, more complicated method of solving this issue, is to talk about schema of definitions when it comes to equivalence, and then the collections are represented by one of the classes (and you can sometimes prove that this representative doesn't matter). Then the whole thing becomes a much more technical and involved from a formal point of view, which is not a bad thing. But I think that if you're interested in category theory and want to talk about larger and larger categories, then universes are probably the way to go.

• If $M$ is a model of set theory, then $M$ thinks that $\varphi(y)$ defines a set if and only if $M\models\exists x\forall y(y\in x\leftrightarrow \varphi(y))$. Do you agree with that? – Asaf Karagila Nov 5 '14 at 15:53
• Good. So what does it mean when $\varphi$ defines a proper class? It means that $M$ does not satisfy the statement "There exists $x$ ...", so proper classes are collections of objects in $M$ which do not exist in $M$. But being elements, from the point of view of $M$, requires first to exist, because the $\in$ is only defined on the objects inside $M$. Ergo, proper classes are collections of objects of $M$ which do not exist in $M$. So they cannot be elements of other sets, since in this context we limit "set" to mean "an object inside $M$". – Asaf Karagila Nov 5 '14 at 15:58
• @GME: No, it means exactly that. If categories are not sets, they don't exist in the universe. I really don't know how to make this any clearer than that. As for your second remark, that might depend on your axioms, but in the usual axiomatization of $\sf ZFA+AC$ you can prove that every set is in bijection with a pure set (more specifically, an ordinal). So this remark is certainly false there. – Asaf Karagila Nov 5 '14 at 16:16
• @GME: You can't say that something is "a version of ZFC" if replacement fails. – Asaf Karagila Nov 5 '14 at 16:26
• @ArthurAzevedoDeAmorim "Power set" has a well-established meaning in ZF which does not run into these problems. In general, yes, a subobject in the sense of Borceux will be a proper class. I agree with Asaf in that (Grothendieck) universes are the most convenient way of avoiding trivial set-theoretic difficulties of this kind. – Zhen Lin Nov 5 '14 at 17:50