Can we treat proper classes in a well-defined way that has them behave as ordinal numbers?

Question

(1) Can we treat proper classes in a well-defined way that has them behave as ordinal numbers, where we can have arithmetic and some new notion of cardinality unique to proper classes, going higher and higher? (2) Can we do such within surreal numbers as a proper class? (3) If so, can we, like we do with ordinals, imagine infinitely more proper classes, that might aid in realizing the scope of how far we can make extendable the surreals, and proper classes themselves for that matter, indefinitely, beyond a notion of proper class?

Answer 1

A set theorist may have more to say, but here's my understanding:

(1) The only proper class that's like an ordinal would be the class of all ordinals. And then you can't take the successor. If you wanted to define cardinality as "the least ordinal such that..." then you only get one more "cardinality" for proper classes from this.

If you have something like the axiom of limitation of size then that one new cardinality is enough for all the proper classes. But if you don't have such an axiom, then you could end up with proper classes that the class of all ordinals doesn't "count", even if you had the axiom of choice at the level of sets.

(2) I'm not sure what you mean, but you could put the ordinals (as surreals) in the left position to form a "gap" greater than all surreals.

(3) As in the answer to (1), it doesn't really make sense to go any farther.

Answer 2

To complete Mark S.' answer, there are (say in NBG set theory) proper classes equipped with a well-ordering, which contain a proper initial segment isomorphic to the class $\mathbf{On}$ of ordinals.

For instance take the class $\mathbf{On}\times \mathbf{On}$ of ordered pairs of ordinals, with the lexicographic order. So $\mathbf{On} \times \{0\}$ is an initial segment of $\mathbf{On}\times \mathbf{On}$.

But since $\mathbf{On}$ cannot lie in a class (because classes only contain sets), there is no "generalized" ordinal which is isomorphic the class $\mathbf{On}\times \mathbf{On}$.

I don't know if there are ways to generalize the notion of ordinals so that for instance there exist a unique ordinal isomorphic to $\mathbf{On} \times \mathbf{On}$ as previously defined.

Answer 3

A 'solution' to the problem nombre pointed out is to just define an ordinal as a set or class with a well-order, and then we can define equivalence, and the ordinal operations. The downside of this approach is that every ordinal has multiple representations, and the standard solution "let an ordinal be an equivalence class" doesn't work, because classes can't contain other classes. However, this approach allows one to define more ordinals, such as $\mathrm{On}^2$.

Another approach, which isn't obvious and I haven't seen before, is to use not a class theory, but Tarski-Grothendieck set theory. If we let $V$ be a Grothendieck universe, then $P(V)$ (I think) is a model of NBG, and thus $\{x\in V:x\text{ is an ordinal}\}$ behaves like $\mathrm{On}$. This approach allows defining generalized ordinals even further. For example, there isn't even a bijection from the ordinal $\{x\in P(V):x\text{ is an ordinal}\}$ to $\{x\in V:x\text{ is an ordinal}\}$. And then there's always a Grothendieck universe containing $V$...