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Global choice is equivalent to saying $|V|=ON$, while ordinary choice is equivalent to saying that $V= H_{< ON}$. So the relative cardinalities of $V$ and $ON$ seems to affect the degree of choice available over classes. Now can this be affected by how much the cardinality of $ON$ is far from that of $V$?

I mean suppose we work in $\sf MK$ and axiomatize that $V=H_{<ON}$ [in other words all sets are injective to $ON$], and that $|X| \not \leq |ON| \leftrightarrow |X|=|V|$, then this says that the cardinality of the class of all set ordinals is immediately below that of the class of all sets, and that there are no other cardinalities other than those of sets, so the cardinalities of $V$ and $ON$ are very near.

On the other hand suppose instead of the above biconditional we stated that there are $|V|$ (or even strictly more than $|V|$) many proper class cardinalities$^\dagger$, then this would entail that the cardinality of the class of all set ordinals is quite distant from that of $V$ if not extremely lower.

Would those have different choice properties, that in some sense form a spectrum intermediate between set choice and global choice? For example, the former entailing more amount of choice than the latter?

$^\dagger$ this can be captured in $\sf MK$ by saying that there is a class $U$ that is the disjoint union of pairwise non-equinumerous proper classes such that the domain class of $U$ is equinumerous with $V$ [more clearly $U$ is a class of ordered pairs, where it's domain (the class of all first projections of elements of $U$) is equinumerous with $V$, and for each element $k$ in the domain of $U$, the class $K$ of all second projections of elements of $U$ whose first projection is $k$, is a proper class, we shall call such class $K$ as a proper class element of $U$, and if $K_1, K_2$ are distinct proper class elements of $U$ then they are not equinumerous]. If we further demand that any two proper class elements of $U$ have comparable cardinalities, then $ON$ would be hugely lower in cardinality than $V$.

We may even go further along that method by adding that no such union class $U$ can satisfy that for every proper class $X$ there is a proper class element of $U$ that is equinumerous to $X$; in other words there are more than $|V|$ many non-equinumerouse proper classes!

My question is not just about the consistency of those situations, which I think it's there, but more importantly about the use of those situations as choice properties over classes, do they have any known use in set\class theories as intermediate kinds of choice between set choice and global choice, since those would be so.

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  • $\begingroup$ As it stands, it is not quite clear what you are asking. I suggest rephrasing as follows. Is Morse-Kelley set theory without global choice (but with choice for sets) consistent with $|V|$ being the least cardinal above $|On|$? What about existence of $|V|$-many (or even $2^{|V|}$-many) proper classes, each of different cardinality? Similar question for NBG without global choice. $\endgroup$ Commented Oct 9, 2022 at 1:05
  • $\begingroup$ @DmytroTaranovsky, OK! Nice suggestion. The idea is the effect of those possibilities on choice over classes, I think we can have those versions, but by then can those be considered as various degrees of choice above set choice but below global choice, do they have any known consequences in theorization of set\class theory. $\endgroup$ Commented Oct 9, 2022 at 7:30
  • $\begingroup$ @DmytroTaranovsky The first is inconsistent: under NBG + local choice, if $|Ord| < |V|,$ then $Ord \cup \bigcup_{\alpha} \{\text{well-orderings of } V_{\alpha}\}$ is of strictly intermediate cardinality. $\endgroup$ Commented Oct 9, 2022 at 19:08
  • $\begingroup$ @ElliotGlazer, what's the lower bound on the cardinality of strictly intermediate cardinalities between $|Ord|$ and $|V|$ $\endgroup$ Commented Oct 9, 2022 at 19:23
  • $\begingroup$ In fact, local choice isn't even needed: just over NBG, global choice fails iff $|Ord| < |Ord \cup \bigcup_{\alpha} \{\text{well-orderings of } V_{\alpha} \}| < |V|.$ $\endgroup$ Commented Oct 9, 2022 at 19:23

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