Maximal chains and antichains of statements weaker than AC

Question

First, I would like to say that I asked this question (a more general one actually) on math.stackexchange.com.

Consider the set $\varPhi$ of statements in the language of ZF that are weaker than AC (assuming ZF). If we define $\varphi\preceq\psi$ by $\sf ZF\vdash\varphi\rightarrow\psi$ then $\preceq$ is a preorder on $\varPhi$. This induces a partial order $\leq$ on $\varPhi/\sim$, where $\sim$ is the equivalence relation defined by $\sf ZF\vdash\varphi\leftrightarrow\psi$. The theorems of ZF form a single class in this equivalence relation, and it is the greatest element $\gamma$ under this partial order. The class of statements equivalent to AC under ZF (let's call it $\lambda$) is the least element.

I would like to know if it's possible to prove that there are no finite maximal chains or finite maximal antichains (other then the trivial $\{\lambda\}$ and $\{\gamma\}$). More generally, what are the possible (finite) cardinalities of maximal chains and antichains?

If there are other non-trivial things that can be said about this order, I would appreciate those too.

Answer 1

It is not possible to prove there are no finite maximal antichains, because there are.

Paul Howard proved that Łoś's theorem and the Boolean Prime Ideal theorem imply in conjunction the Axiom of Choice. However there is a model in which every ultrafilter is principal, so Łoś's theorem holds trivially and therefore does not imply the Boolean Prime Ideal theorem; and there are models of the Boolean Prime Ideal theorem where Łoś's theorem fails (since the axiom of choice fails there).

Bell and Fremlin proved all sort of similar conjunctions related to Hahn-Banach, Krein-Milman, and other geometric theorems about Banach spaces, where the conjunction of two imply the axiom of choice. These constitute of finite maximal antichains as well.

And there are probably others, which are "less interesting", like "Every set can be injected into $\Bbb R\times\alpha$ for some ordinal $\alpha$" and "The real numbers can be well-ordered", which in conjunction imply the axiom of choice, but can hold separately without it.

Note that all the suggestions here are of size $2$. This is because any finite antichain can be reduced to one of size $2$; but we can always break principles up. We can replace "$\Bbb R$" by some other set which is the product of several distinct sets and then argue that only one of them can be well-ordered each time.

Edit: Note that the above treats "antichain" in the Boolean algebraic sense. It is certainly not the case that any choice principle either implies/follows from Łoś's theorem or from the Boolean Prime Ideal theorem. If we are interested in comparability rather than compatibility, then the answer is that every maximal antichain is either trivial or [countably] infinite, as remarked by Emil below Joel's answer.

As for finite maximal chains, the situation here is a bit less interesting and a bit more strained. For a finite antichain to be maximal it means that given any cardinality $X$, $\sf AC_X$, must be compatible with one of the principles in the chain, or imply the axiom of choice in conjunction. And that means that somehow we are in a situation where you already have quite a lot of choice, and any amount of choice above a specific set will readily imply the axiom. Similarly for comparing cardinalities of sets and $\aleph$ numbers.

Edit: As the algebra is in fact atomless, it follows that it is dense, so every finite chain can be extended by at least one more element which lies between two successive elements of the chain. Therefore every maximal chain must be infinite.

Answer 2

The partial order of statements weaker than the axiom of choice is simply the upper-cone above AC in the Lindenbaum algebra of ZF statements. That is, it is a quotient of the Lindenbaum algebra by the principal ideal generated by the statement AC. In particular, it is a Boolean algebra, where in this case, the equivalence class of AC is $0$. With this perspective, one quickly realizes that every nontrivial statement $\varphi$ strictly weaker than AC, and therefore strictly above AC in the Lindenbaum algebra, forms a maximal antichain with its relative complement $\text{AC}\vee\neg\varphi$. So of course, every nontrivial statement is part of a finite maximal antichain. (Asaf had noted a few instances of this in his answer.)

Meanwhile, on the other side of the question, let me mention that since it is known, assuming the consistency of ZF, that there are infinitely many inequivalent statements weaker than AC (such as choice for families of various-sized finite sets, or DC for increasingly long transfinite sequences), it follows that the Boolean algebra is infinite. From this, it follows that there must be infinite antichains, simply because every infinite Boolean algebra has an infinite antichain.