# What can I do with proper classes?

There are standard tricks, constructions and techniques in ZFC when working with proper classes; for instance one can form the cartesian product of a pair of classes without difficulty, or more generally a fibred product. What I am after is a reference that is a summary of these constructions/techniques.

From a category-theoretic point of view, one should be able to say what the category of classes in ZFC and class functions is 'like' (working purely with the first-order theory of categories). I am aware of Joyal and Moerdijk's Algebraic Set Theory, and that they give as an example the category of classes in NBG. However, I would like to stick within ZFC.

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Bonus marks (figuratively speaking) if said reference deals with classes in ZF, too (I'm looking at you, Asaf!) –  David Roberts Nov 6 '12 at 0:30
I'm not sure what you're looking to do. The idea is simply to re-express the action you want to do; this also has nothing to do with choice. If you want union, you take the disjunction of the formulas; for intersection conjuction; for "$\bigcup A$" simply apply the formula of union over the elements of $A$... –  Asaf Karagila Nov 6 '12 at 8:35
I think I'd just like to know how much I can get away with pretending my classes are sets. Clearly I cannot do everything that ZF(C) allows for sets, but I feel there should be some fragment of the axioms that work. –  David Roberts Nov 6 '12 at 8:47

I'm not sure if there is a written reference for that.

Let me give a quick summary of what I can tell. We know that classes are actually formulas with one free variable. We can therefore manipulate classes the same way we manipulate formulas, with the added fact that we can encode all sort of structure into the sets.

For example, if $A=\{x\mid A(x)\}$ and $B=\{x\mid B(x)\}$ then the union would be disjunction; the intersection would be conjuction. So far this is really just the same as formulas being manipulated.

If we wish to take a Cartesian product we can use the fact that we can define ordered pairs, and so $$A\times B=\{z\mid z=\langle x,y\rangle\land A(x)\land B(y)\}$$

This can become intensely complex using all sort of crazy formulas. What you can't do is talk about classes as actual objects. You cannot say that "there exists a subclass", or "the collection of all subclasses" - although you can talk about collection of all subsets: $$\mathcal P(A)=\{x\mid\forall y(y\in x\rightarrow A(y))\}=\{x\mid x\subseteq A\}$$

You can talk about the existence of subclasses externally, and if you're lucky enough you can prove they are definable internally as well. For example, you can talk about the class of ordinals or sets of ordinals externally, and since these are simple enough definitions this is also internally definable.

I suppose that if you wish to give some recent example which bothered you I could try and help to figure it out.

In the comments David asks about equivalence relations and quotient of equivalence relations. The problem here is that often the quotient space is a collection of equivalence classes, and if the equivalence classes are not sets then this object is not even definable in ZFC.

However we can use Scott's trick to overcome this issue. Scott's trick makes heavy use of the axiom of regularity, and its two common uses are defining cardinals in ZF and proving that one can talk about ultrapowers internally.

Suppose that $\varphi(x,y)$ is a formula defining an equivalence relation. We define $$A_\varphi=\{x\mid \varphi(x,A)\land\mathrm{rank}(x)\text{ is minimal for this property}\}$$ Namely if $\varphi(x,A)$ is true and $x\in A_\varphi$ then there is no $y$ whose rank is smaller than that of $x$ and $\varphi(y,A)$.

By the fact that all the sets in $A_\varphi$ have the same rank we can now show that this i a set, since it is bounded by some $V_\alpha$. It is possible that there is a canonical choice of representative, or it is possible that there is none. But we don't care.

Now we can talk about $C/\varphi = \{A_\varphi\mid A\in C\}$, as a collection of equivalence classes, and $A_\varphi=B_\varphi$ if and only if $\varphi(A,B)$ is true as wanted.

If $C/\varphi$ is a set (for example if you can prove there can only be set-many of equivalence classes) then using the axiom of choice - if it exists - it is possible to just pick a system of representatives, but since the equivalence classes themselves need not be sets, you would still have to use Scott's trick to define the collection from which you are choosing.

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Well, if I consider the category of classes (really certain formulas etc) and class functions, it has finite limits, as you mention. But I would like to construct a quotient if a class by an equivalence relation, or say that a given set is a quotient of a proper class... –  David Roberts Nov 6 '12 at 11:48
David, so to be clear you want to say that you have a class equivalence relation $R$ and you want to define a quotient? Well, generally quotient space is a collection of equivalence classes. If the equiv. classes are sets then it's easy to define, but if they are sets you need to do some sort of Scott's trick kind. I will add this to the answer as well. –  Asaf Karagila Nov 6 '12 at 12:03
I think that's about right. I have a class and an equivalence relation such that each equivalence class is a proper class, and there are only set-many of them. If you like, I have a set with a surjective map from a class, but I can't exactly check universal properties of quotients, so I need a canonical construction to make sure my set really is the quotient. –  David Roberts Nov 6 '12 at 23:55
I think that Scott's trick should really do the trick in this case. –  Asaf Karagila Nov 7 '12 at 0:34
Cool, thanks. I'll see if anyone else jumps in, but this is probably what I'm after. –  David Roberts Nov 7 '12 at 1:42

OK, here is a category-theoretic approach, just come out on the arXiv. Namely, we can define the syntactic category for ZF (which is equivalent to the category of definable classes) and this is a Boolean pretopos with a subobject classifier which is not cartesian closed.

So essentially you can do anything for classes that you can do for sets except form 'function classes' and (e.g. power classes).

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You know I actually had this question in mind when I posted this "quote"? :-) –  Asaf Karagila Dec 14 '12 at 8:20
@AsafKaragila - nice to know I provided some inspiration! –  David Roberts Dec 18 '12 at 2:11

I think that Devlin's The Joy of Sets does that. He deals mostly with ZFC in this book, but in some chapter in the middle (I believe it is in the chapter on ordinals), he uses the convenience of classes to prove a general recursion theorem. He even explains why the use of classes is justified, although he emphasizes that classes are not valid objects of ZFC.

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