In ZFC set theory (or better in NBG set theory, where the language is more flexible with proper classes), we have that every unbounded class of ordinal numbers is a proper subclass of the class On of all ordinals and that every such proper class has a unique bijection (by the enumeration function) with the proper class On. More generally, every proper class that is (class) wellorderable (that is, that is wellorderable, with every descending section being a set) is also bijective with On. So that, every proper class is bijective with On iff V (the class of all sets) is (class) wellorderable (this is known as the global choice axiom). So, if we do not have the global choice axiom there is no bijection between On and V, and we are left with at least two bijectiveequivalent collections of proper classes. Question 1: Does ZFC (or NBG) prove more that that concerning the bijectiveequivalent collections of proper classes ? Question 2: If we pass to a set theory whose language allows for quantification over proper classes (like KM, the KelleyMorse set theory), do we know more about these bijectiveequivalent collections of proper classes ?. Gérard Lang

$\begingroup$ There surely won't be a uique bijection between POrd and a proper subclass, but maybe you meant that there is a preferred one in which they are listed in order. $\endgroup$– Adam EpsteinMar 14, 2013 at 10:40

$\begingroup$ What sort of axiomatization are you thinking of for NBG/MK? I think the most common axiomatizations include (or imply) Limitation of Size, which says that a class is proper iff it admits an injection from V, and from which you then get Global Choice. $\endgroup$– user642796Mar 14, 2013 at 10:43

$\begingroup$ Very nice question. I interpret you as asking: how many proper class cardinalities can there be? Under global choice, there is only one. But without GC, there are at least two. Are there more? By the way, you indicate that NBG does not allow quantification over classes, but this is not really accurate, since we can form such assertions in the language of NBG, and prove them or refute them and so on. What NBG does not allow, which is what I think you meant to say, is for us to use such formulas in the replacement scheme as an axiom. $\endgroup$– Joel David HamkinsMar 14, 2013 at 10:57

$\begingroup$ 1/ Maybe there will not be a unique bijection, then I prefer to take the one given by the enumeration function. But how could I build another one ? 2/ Concerning axiomatization: for NBG, I would take what is given by Gödel in the Princeton monography on the independance of the continuum hypothesis; for KM, I would take the appendix of Leley's book on General Topology. Gérard Lang $\endgroup$– Gérard LangMar 14, 2013 at 10:59

$\begingroup$ Dear Joel David Hamkins, thank you very much for your comment. Naturally, I agree. Notably, the real question is about how many proper class "cardinalities" can there be. Gérard Lang $\endgroup$– Gérard LangMar 14, 2013 at 11:02
1 Answer
I will give a partial answer for models of $NBG$ in which local choice ($AC$) holds, namely:
Theorem. There is a model $M$ of $ZFC$ whose natural expansion $(M,\cal{A})$ to a model of $NBG$ (where $\cal{A}$ is the collection of all parametrically definable classes of $M$) has at least 3 bijectioninequivalent classes, namely: $Ord$ (the class of ordinals), $\cal{P}$$(Ord)$ (the class of all subsets of ordinals), and $V$ (of class of sets).
Moreover, it is a theorem of $NBG$ plus $AC$ that there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$.
Proof. Let $M$ be a model of $ZFC$ which has a definable class of pairs with no no choice function (see, e.g., the answers given by Hamkins and myself to this MO question).
It is clear that there is no injection from $V$ into $Ord$ that is canonically coded in $\cal{A}$.
We next observe that there is no injection from $V$ into $\cal{P}$$(Ord)$ that is coded in $\cal{A}$. This is because there is an obvious definable linear ordering of $\cal{P}$$ (Ord)$ (by viewing each subset of ordinals as a binary sequence), and the existence of such an injection would contradict the fact that there is no definable choice function on a class of pairs.
Finally, to show that there is no injection from $\cal{P}$$(Ord)$ into $Ord$ in our model of $NBG$, we take advantage of the axiom of choice within $M$. More specifically, recall that in the presence of the axiom of choice, every set is definable from a subset of ordinals, the idea being: given any set $x$, let TC($x$) be the transitive closure of {$x$}. Then by AC there is bijection $f$ between TC($x$) and some ordinal $\alpha$. This allows us to copy the $\in$ relation on TC($x$) to a binary relation $r$ on $\alpha$, and then we can use a canonical pairing function on ordinals to code $r$ as some subset $s$ of ordinals. Then $x$ is definable from $s$ via "$x$ is the top element of the transitive collapse of $r$".
Note that the above procedure gives rise to a definable surjection $F$ from $\cal{P}$$(Ord)$ onto $V$: given a subset $s$ of ordinals, first decode it as a binary relation $r$ on ordinals, and then ask whether $r$ is an extensional, wellfounded relation with a top element. If no, let $F(s)$ be $0$, and if yes, then let $F(s)$ be the top element of the transitive collapse of $r$.
Now if there exists a definable injection $F$ from $\cal{P}$$(Ord)$ into $Ord$, then in light of the fact that $\cal{P}$$(Ord)$ would then be definably wellordered, the above map $F$ could be "inverted" by an injection $G$ from $V$ into $\cal{P}$$(Ord)$, where $G(x)$ is the first $s$ such that $F(s)=x$. This, in turn would enable us to definably wellorder $V$, contradiction.
Finally, to verify the last claim of the theorem: the coding explained above allows us to define an injection $H$ from $V$ into $\cal{P}(\cal{P}$$(Ord))$, namely $H(x)$ is the collection of all subsets $s$ of $\kappa$, where $\kappa$ is the cardinality of the transitive closure of {$x$}, such that when $s$ is canonically decoded as a relation $r$ on $\kappa$, then $r$ is an extensional wellfounded relation with a top element such that $x$ is equal to the top element of the transitive closure of $r$.
Therefore, since the class version of the SchröderBernstein theorem holds in models of $NBG$, there is a bijection between $V$ and $\cal{P}(\cal{P}$$(Ord))$ in models of $NBG$ in which the (local) axiom of choice holds.

$\begingroup$ Dear Ali Enayat, let me thank you very heartily for your answer. This proves that 1 and 2 are not the only possible cardinalities concerning models of NBG, and that 3 is another one. This is a partial answer, because it does not rule out the possibility for others cardinalities. But I consider that this is a plain answer to question 1 as I formulated it, and will credit you for this. I would be most interested to know if question 2 (as interpreted by Joel Hamkins) gives more. Gérard Lang $\endgroup$ Mar 15, 2013 at 10:21