It is well-known (forgive the pun) that the axiom of choice (which states that the product of every non-empty family of non-empty sets is non-empty) implies that the proper class of all cardinal numbers is well-ordered, at least in the presence of the ZF axioms.

Does the converse hold? If not, is the statement that the proper class of all cardinal numbers is well-ordered consistent with axioms like determinacy, which, while interesting, contradict full-blown choice?


This is known as the trichotomy principle. If every two cardinals are comparable then the axiom of choice holds.

So in fact the axiom of choice is equivalent to the slightly weaker claim that the cardinals are linearly ordered.

To see this is true, let $A$ be a set, and let $\kappa$ be some ordinal such that $\kappa\nleq|A|$. Such ordinal exists and in fact we can ensure that $\mathcal{P(P(P(}A)))$ is strictly larger than such $\kappa$.

Since $\kappa\nleq|A|$ we have that $A$ can be injected into $\kappa$ and therefore be well-ordered. So the ordering principle holds, and therefore the axiom of choice holds.

This is known as Hartogs' theorem, and the least $\kappa$ cannot be injected into $A$ is known as the Hartogs number of $A$, often denoted as $\aleph(A)$.

What happens when the axiom of choice fails, then? What is the ordering of the cardinals? We do not know much. We do know how to produce models in which many partial orders can be embedded into the cardinals, but we don't know a whole lot more in $\sf ZF$.


To address the part of your question about determinacy, I will mention that there is a simple counterexample to trichotomy under $\mathsf{AD}$. Namely, $\mathbb{R}$ does not inject into $\omega_1$ and $\omega_1$ does not inject into $\mathbb{R}$. These statements both follow from the fact that under $\mathsf{AD}$ any well-ordered set of reals is countable.

For more information about the structure of cardinals under $\mathsf{AD}$ you may want to see the paper A trichotomy theorem in natural models of $\mathsf{AD^+}$. (Note that the trichotomy here is unrelated to that which was disproved in the above paragraph.)

  • $\begingroup$ Under reasonable assumptions, $\sf AD$ implies that every countable poset embeds to the cardinals below $\mathcal P(\Bbb R)$. That's quite the opposite of linear ordering. :-) $\endgroup$ – Asaf Karagila May 8 '13 at 21:27
  • $\begingroup$ @Asaf That sounds familiar. But can you point me to a reference? $\endgroup$ – Trevor Wilson May 8 '13 at 22:16
  • $\begingroup$ I actually don't. I think it may appear in Woodin's big book. I think that ${\sf AD}+V=L(\Bbb R)$ is enough. $\endgroup$ – Asaf Karagila May 8 '13 at 22:43
  • $\begingroup$ @Trevor: Woodin's paper The cardinals below $|[\omega_1]^{<\omega_1}|$, Annals of Pure and Applied Logic 140 (1-3), (2006), 161-232, has much more under $\mathsf{AD}_{\mathbb R}+\mathsf{DC}$. (Cont.) $\endgroup$ – Andrés E. Caicedo May 8 '13 at 22:44
  • $\begingroup$ Under $\mathsf{AD}^+$ alone, we should be able to reconstruct the full complexity of results of Alekos and Greg on Rigidity theorems for actions of product groups and countable Borel equivalence relations, using Ben's soft approach, and showing Asaf's claim by just working on cardinals that are quotients of $\mathbb R$ by "nice" equivalence relations (coming from actions of nice countable groups). I do not know if there is an explicit reference to this in print. $\endgroup$ – Andrés E. Caicedo May 8 '13 at 22:45

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