Proper class if and only if there is a bijection with the class of ordinals.

Question

I am looking for a way to formalize the sentence "$A$ is a proper class", and so I was wondering if I could write it like "a proper class is something that is in bijection with the class of all ordinals.", or if there is another way to characterize class that are not sets.

Otherwise, would this be correct? If $A = \{ x \mid \varphi(x)\}$ for some formula $x$, then would this be a condition for $A$ to be a proper class?

$$\forall \, X \, \lnot \left( x \in X \leftrightarrow \varphi(x)\right)$$

Answer 1

First of all, to state that $\varphi$ defines a proper class, you only need to state that it does not define a set. Namely, there is no set which is exactly all those elements that have property $\varphi$. This can be done in several ways, direct or indirect:

1. For every set $x$ there is $y$ such that $y\notin x$ and $\varphi(y)$.
2. For every ordinal $\alpha$, there is $y$ such that $y\notin V_\alpha$ and $\varphi(y)$.
3. There does not exist a set $x$ such that for all $y$, if $\varphi(y)$, then $y\in x$.

(Yes, 1 and 3 are the same thing.)

Secondly, you can't quite quantify over proper classes. So to say that "there exists a bijection between two classes" is problematic. If you have a formula which defines a bijection, that's great. Use that one. But you can't state "Oh, there is one", because that would be a second-order statement.

And finally, it is not true that a proper class is in bijection with the set of ordinals. That implies Global Choice, and we know that there are models of $\sf ZFC$ where global choice does not hold. Therefore, it is not true that a proper class will necessarily be in bijection with the class of ordinals. In fact, if you are willing to drop the axiom of choice for sets, it isn't even true that there is an injection from $\omega$ into a proper class!

However, there is one grain of truth to the whole thing. Because of the von Neumann hierarchy, we can prove that a class is proper if and only if it surjects onto the ordinals. And that is a $\sf ZF$ proof right there. But that's more of less the best you can do (even in $\sf ZFC$).