Is the Class Well Ordering principle "CWO" the maximal choice principle?

Question

In a prior posting, the Class Well-Ordering principle "$\sf CWO$" was presented, which simply states that there is a well-ordering over all classes of $\sf MK$. On the other hand, it is known that there is a plethora of Class Choice principles $\sf CC_\kappa (\mathcal F)$, that can extend $\sf MK$.

If we regard $\sf CWO$ as a choice principle, then is it the case that it is the maximal choice principle we can extend $\sf MK$ with? That is, all other choice principles extending $\sf MK$ are provable from it.

Of course, as mentioned in comments, the above question is not fully formal, since no formal definition of "choice principle" was presented. Here is a semiformal try:

A choice principle over theory $\sf T$ is an axiom (or schema) that when added to axioms of $\sf T$ it enables defining a uniform way of selections from some collections of objects that cannot be made in $\sf T$ alone. Here, "collections" can be captured as realized predicates (i.e.; fulfillable), and a "selection" is a function on realized predicates that sends them to objects realizing them. An additional criterion is that the addition of a choice principle must not result in increase in consistency strength over the original theory $\sf T$.

Accordingly $\sf CWO$ is a choice principle since it proves $\sf CC$ and doesn't increase consistency. But, is it the maximal choice principle in the sense given above?

Answer 1

According to the definition of choice principle that I've presented, $\sf CWO$ is not maximal. We can have a continuum principle, which would work here as a choice principle that proves $\sf CWO$ yet the converse is not true. This is:

To the language of $\sf MK$ add a primitive total unary function symbol $\mathcal F$ from classes to classes. Axiomatize:

• $\forall X \exists Y: Y \text { is a well-ordering } \land \mathcal F(X) = Y$
• $\forall X \forall Y: \mathcal F(X) \cong \mathcal F(Y) \to X=Y$

Where $\cong$ is order isomorphism.

The idea is that we can well order all classes after $\preceq$ defined as:

$X \preceq Y \iff \exists Z \subset \mathcal F(Y): \mathcal F(X) \cong Z$

The net result is that we'd have a continuum like situation over the class $V$ of all sets. That is, metatheoretically speaking, there is no predicate $P$ that has more classes fulfilling it than how many sets there are and at the same time has less classes fulfilling it than how many classes there are. The reason is because we can define a bijective predicate $\mathcal G$ from classes to superordinals, where the latter are $\preceq$-minimals of isomorphic well-orderings, where the usual $\sf Ord$ is the $\preceq$-minimal well-ordering isomorphic to the natural well-order on the class $\sf ORD$ of all ordinals in $V$. Now, the predicate "superordinal" is a smallest predicate strictly supernumerous to the set-hood predicate, so the predicate "class" which has as many objects fulfilling it as the predicate superordinal has (because of $\mathcal G$), would also be a smallest predicate larger than the set-hood predicate. So, we have one cardinality (metatheoretically speaking) that is strictly larger than the cardinality of $V$.