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We work in the set theory NBG (with local choice, but not global choice), because if there is global choice, every proper class is well-ordered, so that every proper class is bijective with the class On of ordinals and the class V universe, and we only have one level of bijective equivalent proper classes.

As it is known that the Schroeder-Bernstein is valid for proper classes, we have a proper class bijection between two proper classes A and B if we have an injection of A into B and also an injection of B into A, so that we have bijective-equivalent levels of proper classes and that we have a preorder on proper classes corresponding to injections.

From the answer given by Ali Enayat on 15/03/2013 to a question of mine, it is possible to have at least three distinct and ordered levels, namely the level of On, the level of P(On) and the level of P(P(On)), that is also the level of V.

From the answer of JD Hamkins today to another question of me, it is not true that the class On can be injected in every proper class (I hoped that the level of On could be the minimal level in the order on proper classes). This does not completely prove that we have a fourth class under On, because I do not know if it is true that given two proper classes A and B either A injects in to B or B injects into A.

Question 1: Is it true that given two proper classes there is an injection between them?

Question 2: If the answer to question 1 is true, is there a minimal level for the proper classes levels under injection?

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In a previous question Joel Hamkins linked to a model obtained by adding Cohen subsets to every regular cardinal. This model had the property that the universe could not be linearly ordered.

In particular, there was a class of pairs that witnessed that. Namely there was a class of pairs without a choice function. Consider now the class of partial choice functions. Of course there is no injection from the ordinals to that class, since from such injection we could have engineered a choice function from the class of pairs.

But on the other hand, of course there is no injection this class into the ordinals, since in that case we could have easily defined a choice function using the well-ordering obtained from such injection.

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With regard to your first question, I claim that global choice is equivalent to the assertion that any two classes are comparable under injectivity.

If global choice holds, then this is clearly the case. Conversely, consider the classes Ord and the class $W$ that I used in my answer to your previous question. If Ord injects into $W$, then global choice holds, since that is what I had argued there. And if $W$ injects into Ord, then we can well-order $W$ and from this we may construct a well-ordering of $V$.

Thus, if global choice fails, there must be incomparable classes.

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  • $\begingroup$ For once, I got here before you! :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 2, 2014 at 21:23
  • $\begingroup$ Yes, indeed. I was napping, and I had to get the bread out of the oven... $\endgroup$
    – Joel David Hamkins
    Commented Dec 2, 2014 at 21:26
  • $\begingroup$ Well, we switch places. It's late, and I have an early morning followed by a long day full of math! :-P $\endgroup$
    – Asaf Karagila ♦
    Commented Dec 2, 2014 at 21:27
  • $\begingroup$ Thank you very much to both of you. If I correctly understand, the situation is that if global choice fails, we have the class of V and at least two incomparable classes (and we can explicitly name On and W as such) that both inject in V. Moreover in the example of Ali Enayat we have a fourth class P(On) that is between V and On. It would be most interesting to know more about the kind of (not total) orders with a maximal element that are possible. Would it necessarily be a tree ? $\endgroup$
    – Gérard Lang
    Commented Dec 2, 2014 at 22:45
  • $\begingroup$ Yes, that is interesting. If one includes the sets, then it is not a tree, since every set injects into Ord and W, and these both inject into V. $\endgroup$
    – Joel David Hamkins
    Commented Dec 2, 2014 at 23:02
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The answer to question 2 is Negative. This is because, answering to my question "Injection of the proper class of ordinals into every proper class", J.D. Hamkins proved on 02/12/2014 the existence in NBG of a proper class W that does not inject into On.He also proved that On does not inject into W. So that if A were to be the minimal proper class for injection into proper classes, A would inject into On, so would then be bijective with On and well-orderable. But A would also be injective into W; but by chaining an injection from On into A and an injection from A into W, we would get an injection from On into W, that is impossible. Gérard Lang

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