### Forcing classes to have the same size

#### by kamerynwilliams

Work in second-order set theory. If Global Choice holds, then all proper classes have the same size. There is a bijection between and so by restricting that bijection to we get a bijection between and a proper-class sized subclass of , which is of course in bijection with . But if Global Choice fails then there can be classes of different size. For example, consider an inaccessible cardinal so that does not have a definable global well-order. Then equipped with its definable subsets as classes gives a model of Gödel–Bernays set theory (without Global Choice) whose (= ) is not in bijective correspondence with its (= ).

Suppose we start out with Global Choice failing, and hence there being classes of different size. What happens if we try to force classes to have the same size?

Let’s start with a warm-up.

Proposition:Work in , i.e. Gödel–Bernays set theory with neither Global Choice nor Powerset. If is generic for the forcing to add a Cohen-generic subclass of , then from can be defined (in a first-order way) a global well-order of ordertype . In particular, all classes have the same size in the Cohen-extension.

Some remarks before the proof. First, to be clear, the Cohen poclass consists of set-sized partial functions from to , ordered by reverse inclusion. Second, in the absence of Powerset the various formulations of Global Choice are not equivalent. The version obtained from a Cohen-generic subclass of is the strongest. Having a global well-order of ordertype (equivalently, having a bijection from to ) is stronger than having a global well-order (of any ordertype), stronger than cardinal comparability for classes, and stronger than the existence of a global choice function. Whereas with Powerset they are all equivalent.

*Proof:* Every set is coded as a set of ordinals. This is because we can code a set by taking an isomorphic copy of as a set of pairs of ordinals. Since we have a definable pairing function for ordinals, we can code a set of pairs of ordinals as a set of ordinals. Now observe that by genericity every set of ordinals is coded into a Cohen-generic subclass of : if is a condition in the forcing then extend by adding a copy of (the characteristic function for) the set of ordinals starting above . So we can define a global well-order of order-type by saying that if the first place in the generic where is coded comes before the first place in the generic where is coded. ∎

The upshot of this proposition is that once you add a Cohen-generic subclass of , you’ve forced every class to have the same size. So if you want classes of different sizes you cannot add Cohen-generic subclasses of .

Let’s consider a more direct way to force two classes to have the same size. Let and be proper classes. Define the partially-ordered class to consist of all set-sized partial injections from to , ordered by reverse inclusion. By genericity, if is generic for then every member of is in its domain and every member of is in its range. So will be a bijection between and , forcing them to have the same size.

Before saying something about this forcing, I need a definition.

Definition:A proper class isamorphousif there is no way to decompose into a disjoint union of two proper classes.

This is modeled upon the definition of an amorphous set, i.e. an infinite set which cannot be decomposed into a disjoint union of two infinite sets. (Of course, a fragment of choice rules out the existence of amorphous sets.) Indeed, we can get models with amorphous classes by considering amorphous sets. Work in a universe with amorphous subsets of . Then, is a model of which has amorphous classes.

Proposition:Suppose that the proper class is not amorphous. Then forcing with adds a Cohen-generic subclass of . Put as a slogan: forcing a non-amorphous class to be the same size as forces all classes to have the same size.

*Proof: *Let and be disjoint proper classes whose union is . We will use them to extract a Cohen-generic subclass of from a generic for . Specifically, define a function as iff . More generally, any condition defines a condition in the Cohen poclass; namely for define iff . Because and are both proper classes this map is onto the Cohen poclass.

Let us see that this is Cohen-generic. Take any dense subclass of the Cohen poclass. Let . This is a dense subclass of . Otherwise, there would be with no extension in . But then would have no extension in , contradicting the density of . Thus, since meets , we get that meets . As was arbitrary, is generic for the Cohen poclass. ∎

We can generalize this result. Instead of considering the class we can consider any class . We then work similarly, but define by first restricting to and then checking whether is in or . And it clearly doesn’t matter which coordinate is first or second. So forcing with for either or (and both non-amorphous) forces all classes to have the same size.

In summary, for a large collection of classes we would like to force to have the same size, doing so via the obvious poclass has the effect of making all classes have the same size. Does this always happen, even if neither class contains ?

Question:Over , must always force all classes to have the same size?

I’m not 100% clear on the role of amorphous classes here. I needed that my classes were not amorphous, but I suspect that this comes for free in my context. I strongly suspect the following question has a positive answer, but don’t have the choiceless set theory know-how to easily see a proof.

Question:Does prove that there are no amorphous classes? That is, does having the axiom of choice forsetsimply there are no amorphous classes?

A positive answer would be analogous to the ZF fact that countable choice implies there are no amorphous sets.

I disagree with the last line. The analog of set-choice in a second order context should be finitary choice in ZF, not countable choice (even though there is a strong difference between the two).

The question is very interesting, though. I don’t think I know enough about forcing over models without a von Neumann hierarchy, though. As any counterexample must come from one.