Classes are often informally thought of as being "larger" than sets. Usually, the notion of "larger" is formalized via an injection: $B$ is "at least as large" as $A$ iff there is an injection from $A$ to $B$, and strictly "larger" if there is no injection going the other way.

Even though ZFC does not formalize the notion of proper classes, we can still speak sensibly of the notion of an injection from a set to a class, which is simply a function such that its image is within the class. So we can still say a set injects into some proper class if the relevant function exists.

It is known that without AC, we can have sets that are "incomparable" in size, because the universe is now only partially ordered rather than well-ordered. But it seems that we must now also have the situation where sets can be incomparable in size to classes!

Otherwise, if every set injected into every class, they would all inject into the ordinals, and hence be well-ordered. (Right?)

So without AC, we cannot say that classes are necessarily "larger" than sets -- just "different"!

My questions:

  1. Is this the right understanding?

  2. Is the axiom "every set injects into every class" equivalent to the axiom of choice?

  3. Is there some sensible way to interpret the notion of "class" other than as an entity too "large" to be a set, since this makes no sense without AC?

I would be happy to discuss within a theory like NBG as well that explicitly formalizes classes, although I tried to word this in such a way that it wasn't necessary.

  • $\begingroup$ It is interesting to know that if we extend the notion of cardinality to proper classes (see "Order in bijective equivalent collections of proper classes in set theory"), we find that even in NBG with choice, we do not obtain a well-order at all, because in fact we do not have a total order. $\endgroup$ – Gérard Lang May 1 '18 at 14:48
  • $\begingroup$ We can obtain a model with at least four non-equivalent proper classes On, V, A and B such that there is no injection of A into B and no injection of B into A (see"More on bijective-equivalent classes in NBG set theory (1) and (2) $\endgroup$ – Gérard Lang May 1 '18 at 14:51
  • $\begingroup$ You might find it interesting that the axiom in your second question requires Foundation in its prove (as well as AC, of course). I believe this axiom is called the injection principle. $\endgroup$ – Elliot Glazer May 1 '18 at 21:52
  1. Yes, your remarks about incomparability of sets and classes without the axiom of choice are correct.

  2. Yes, in ZF (or in GB), the axiom of choice is equivalent to the assertion that every set injects into every proper class (one must say proper class, since in the usual terminology, every set also counts as a class). The forward direction holds, since every proper class $A$ must have elements of unboundedly many ranks, and so it has arbitrarily large subclasses. Conversely, if a set $a$ injects into the ordinals, then it is well-orderable.

  3. In ZF or GB, that is, without the axiom of choice, one can interpret the notion of proper class as those classes that do not contain all their elements at some stage of the cumulative hierarchy. That is, they must contain elements of ranks unbounded in the ordinals. So the essence of what makes something a proper class is not its size, as such, but rather the fact that at no stage of the cumulative hierarchy did one finish completing the class in the sense of already having all of the elements of the class by that stage.

Meanwhile, one can recover the idea that proper classes are large, even in ZF or GB, by considering quotients of the class, for a class $A$ is a proper class if and only if there is an equivalence relation $\sim$ on $A$ (and one may freely assume that all equivalence classes are sets) such that the class $\text{Ord}$ injects into the quotient $A/\sim$. Indeed, the same equivalence relation $\sim$ works for all classes $A$: let $a\sim b$ if and only if they have the same rank. A class $A$ is a proper class if and only if it has elements from $\text{Ord}$ many distinct ranks.

  • $\begingroup$ Thank you Joel - I wonder, does any of this change if we use surjections rather than injections? I know that if two sets surject onto one another, this does not imply a bijection without AC. Still, suppose we simply declare that if A surjects onto B, this at least means A is "no smaller than" B. If A and B surject onto each other but there is no bijection, we could either just allow that to be the case, or we could even look at equivalence classes defined by having surjections both ways. Does anything like that lead to a neater picture, where the ordinals surject onto everything? $\endgroup$ – Mike Battaglia May 5 '18 at 0:30

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