### Forcing classes to have the same size

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

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?

Proposition: Work in $\mathsf{GB}^-$, i.e. Gödel–Bernays set theory with neither Global Choice nor Powerset. If $C \subseteq \mathrm{Ord}^M$ is generic for the forcing to add a Cohen-generic subclass of $\mathrm{Ord}$, then from $C$ can be defined (in a first-order way) a global well-order of ordertype $\mathrm{Ord}$. 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 $\mathrm{Ord}$ to $2$, 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 $\mathrm{Ord}$ is the strongest. Having a global well-order of ordertype $\mathrm{Ord}$ (equivalently, having a bijection from $\mathrm{Ord}$ to $V$) 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 $x$ by taking an isomorphic copy of $\mathord{\in} \upharpoonright \mathrm{TC}(\{x\})$ 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 $\mathrm{Ord}$: if $p$ is a condition in the forcing then extend $p$ by adding a copy of (the characteristic function for) the set of ordinals starting above $\sup \mathrm{dom}(p)$. So we can define a global well-order of order-type $\mathrm{Ord}$ by saying that $x < y$ if the first place in the generic $C$ where $x$ is coded comes before the first place in the generic where $y$ is coded. ∎

The upshot of this proposition is that once you add a Cohen-generic subclass of $\mathrm{Ord}$, 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 $\mathrm{Ord}$.

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

Before saying something about this $\mathbb B(X,Y)$ forcing, I need a definition.

Definition: A proper class $A$ is amorphous if there is no way to decompose $A$ 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 $H_{\omega_1}$. Then, $(H_{\omega_1}, \mathcal P(H_{\omega_1}))$ is a model of $\mathsf{GB}^- - \mathrm{Choice}$ which has amorphous classes.

Proposition: Suppose that the proper class $Y$ is not amorphous. Then forcing with $\mathbb B(\mathrm{Ord},Y)$ adds a Cohen-generic subclass of $\mathrm{Ord}$. Put as a slogan: forcing a non-amorphous class to be the same size as $\mathrm{Ord}$ forces all classes to have the same size.

Proof: Let $Y_0$ and $Y_1$ be disjoint proper classes whose union is $Y$. We will use them to extract a Cohen-generic subclass of $\mathrm{Ord}$ from a generic $B$ for $\mathbb B(\mathrm{Ord},Y)$. Specifically, define a function $C : \mathrm{Ord} \to 2$ as $C(\alpha) = i$ iff $B(\alpha) \in Y_i$. More generally, any condition $b \in \mathbb B(\mathrm{Ord},Y)$ defines a condition $p(b)$ in the Cohen poclass; namely for $\alpha \in \mathrm{dom}(b)$ define $p(b)(\alpha) = i$ iff $b(\alpha) \in Y_i$. Because $Y_0$ and $Y_1$ are both proper classes this map $b \mapsto p(b)$ is onto the Cohen poclass.

Let us see that this $C$ is Cohen-generic. Take any dense subclass $D$ of the Cohen poclass. Let $D' = \{ b \in \mathbb B(\mathrm{Ord},Y) : \exists p \in D\ p = p(b) \}$. This $D'$ is a dense subclass of $\mathbb B(\mathrm{Ord},Y)$. Otherwise, there would be $b \in \mathbb B(\mathrm{Ord},Y)$ with no extension in $D'$. But then $p(b)$ would have no extension in $D$, contradicting the density of $D$. Thus, since $B$ meets $D'$, we get that $C$ meets $D$. As $D$ was arbitrary, $C$ is generic for the Cohen poclass. ∎

We can generalize this result. Instead of considering the class $\mathrm{Ord}$ we can consider any class $X \supseteq \mathrm{Ord}$. We then work similarly, but define $p(b)$ by first restricting $b$ to $\mathrm{Ord}$ and then checking whether $b(\alpha)$ is in $Y_0$ or $Y_1$. And it clearly doesn’t matter which coordinate is first or second. So forcing with $\mathbb B(X,Y)$ for either $X \supseteq \mathrm{Ord}$ or $Y \supseteq \mathrm{Ord}$ (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 $\mathrm{Ord}$?

Question: Over $\mathsf{GB}^-$, must $\mathbb B(X,Y)$ 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 $\mathsf{GB}^-$ prove that there are no amorphous classes? That is, does having the axiom of choice for sets imply there are no amorphous classes?

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