What form of choice is "every Dedekind-finite set is finite" equivalent to?

Question

Halmos in his Naive Set Theory proves that every infinite set has a subset equivalent to $\omega$ using the axiom of choice with its full power. And this leads to the corollary that a set is infinite if and only if it is equivalent to some proper subset of it, which leads to each Dedekind-finite set being finite.

But I've also seen a proof (on Wikipedia) that this can also be proven with just countable choice. However Wikipedia also states that this result is strictly weaker than countable choice.

Question: It is clear that we do require some form of choice, not just ZF, to prove this result.$^1$ But it is even weaker than the countable choice. Can we explicitly state the form of this choice which is equivalent to this result?

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$^1$ I've come across the fact that there exists a model of ZF (whatever that means (sorry I've not done any model theory; this is just for your reference)) in which every infinite set is Dedekind-infinite, and yet the countable choice fails.

Answer 1

While answering this question: Strength of “Cofinite Choice”, I discovered that "every Dedekind-finite set is finite" is equivalent to the following "axiom of cofinite choice":

Let $A$ be a set of non-empty sets such that $(\bigcup A)\setminus X$ is finite for all $X\in A$. Then $A$ has a choice function.

See the linked answer for a proof. This seems to me to be a fairly natural choice principle.

Answer 2

Let us write D-finite for Dedekind-finite.

The following statements are equivalent in ZF:
(1) Every D-finite set is finite.
(2) If $\mathcal A$ is a family of nonempty D-finite sets and $|\mathcal A|\not\gt\aleph_0$, then $\mathcal A$ has a choice function.
(3) Every D-finite family of infinite D-finite sets has a choice function.
(4) Every countable family of infinite D-finite sets has a choice function.

(1)$\implies$(2): We may assume that the family is infinite and therefore countable. If $\{A_1,A_2,A_3,\dots\}$ is a countable family of nonempty D-finite sets without a choice function, then $\bigcup_{n=1}^\infty(A_1\times\cdots\times A_n)$ is an infinite D-finite set.

(2)$\implies$(3)$\land$(4): Obvious.

(3)$\implies$(1): Assume (3), and assume for a contradiction that $X$ is an infinite D-finite set. Let $S$ be the set of all finite sequences of distinct elements of $X$, and for $s\in S$ let $X_s=\{s\}\times(X\setminus\operatorname{range}(s))$. Then $S$ is a D-finite set, and $\{X_s:s\in S\}$ is a D-finite family of infinite D-finite sets. By (3) this family has a choice function; but with such a choice function we can recursively define an infinite sequence of distinct elements of $X$, contradicting our assumption that $X$ is D-finite.

(4)$\implies$(1): If $X$ is an infinite D-finite set, let $F_n=\{f\in X^n:f\text{ is injective}\}$; then $\{F_1,F_2,F_3,\dots\}$ is a countable family of infinite D-finite sets without a choice function.

P.S. The equivalence (1)$\iff$(3) is essentially Theorem 17 of Omar De la Cruz, Finiteness and choice, Fundamenta Mathematicae 173 (2002), 57–76.