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The usual proof that "Dedekind finite = finite" from $\mathsf{ZF+AC_\omega}$ goes by, given a Dedekind-infinite set $X$, applying countable choice to the sequence $(X^n)_{n\in\omega}$. It just occurred to me that I don't know whether using such "increasingly large" sets is actually needed here.

Specifically, given an infinite set $X$, say that an $X$-small sequence is an $\omega$-indexed sequence of nonempty sets $(A_i)_{i\in\omega}$ such that, for some finite $n$, no $A_i$ surjects onto $X^n$. My question is the following:

Is the theory $\mathsf{ZF}$ + "There is an infinite Dedekind-finite $X$ such that every $X$-small sequence has a choice function" consistent?

A related question is whether $\mathsf{ZFA}$ + "The class $\mathscr{A}$ of atoms is an infinite Dedekind-finite set" + "Every $\mathscr{A}$-small sequence has a choice function" is consistent. At a glance I don't think Pincus' embedding theorem (let alone Jech/Sochor) would apply here, so a positive answer wouldn't address the actual question, but it would still be interesting.

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  • $\begingroup$ Do the embedding theorems not apply because the class of atoms is not necessarily a set? $\endgroup$
    – James E Hanson
    Commented Sep 18, 2022 at 18:10
  • $\begingroup$ @JamesHanson No (and note that I've stated that the class of atoms is a set in the case I'm interested in), the issue is that the embedding theorems only work if the statement in question is sufficiently syntactically "nice." And, at a glance, I don't think this one is (although I could be wrong, I'm notoriously bad at this sort of thing). $\endgroup$
    – Noah Schweber
    Commented Sep 18, 2022 at 18:17
  • $\begingroup$ Ah, it's because surjection-based bounds on cardinality are wonky. In particular, '$X$-small sets' could have arbitrarily high rank, right? $\endgroup$
    – James E Hanson
    Commented Sep 18, 2022 at 18:31
  • $\begingroup$ Add $\omega\times\omega$ Cohen reals with the permutation group $S_\omega\wr S_\omega$ and with the filter generated by fixing finitely many 'blocks', then, I strongly suspect for no particularly good reason, you may be able to prove that any set surjects onto the amorphous set or blocks (up to a finite error) or can be well-ordered. Since the power set of an amorphous set is Dedekind-finite, this means there is no infinite sequence of distinct subsets; and so one may be able to deduce that any such sequence must be actually finitely many sets rotating around, or have a well-orderable union. $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 19, 2022 at 2:31
  • $\begingroup$ (Again, just a strategy for someone who wants to have a go at this.) $\endgroup$
    – Asaf Karagila ♦
    Commented Sep 19, 2022 at 2:32

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