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For each $n$, consider the following property $\text{ACF}_{n}$ (axiom of choice on finite sets of size $n$):

For any family of sets $\mathcal{F}$ such that for all $S \in \mathcal{F}$, $|S| = n$, there is a choice function on $\mathcal{F}$ (i.e. a function $f$ such that $f(S) \in S$ for all $S \in \mathcal{F}$).

I'm currently interested in the situation where $n \in \mathbb{N}$ (and $n \geq 2$, since $\text{ACF}_0$ is just false and $\text{ACF}_1$ is a theorem of ZF), although $n$ being an infinite cardinal is interesting as well. For example, $\text{ACF}_2$ is the axiom of choice used in the famous example of picking a sock from an infinite family of pairs of socks.

Despite this being such a natural weakening of the axiom of choice (in my opinion), I couldn't find a name for this family of axioms. I believe that it is no stronger than the Boolean Prime ideal theorem; $\text{ACF}_n$ is true if there is a total order on $\bigcup_{i \in I} S_i$ (since in this case, the choice function can simply take the minimum of the set), and BPIT implies that every set can be totally ordered.

My main question is: are the $\text{ACF}_n$ independent of each other? At first, I thought $\text{ACF}_n$ would be weaker -- and probably strictly weaker -- than $\text{ACF}_{n+1}$, but actually I couldn't even figure out how to prove $\text{ACF}_n$ from $\text{ACF}_{n+1}$.

There are some implications, however. For example, $\text{ACF}_2$ and $\text{ACF}_3$ together imply $\text{ACF}_4$. Given a set $S$ of size $4$, we can pick a canonical member by first collecting all $6$ size-2 subsets, and applying the $\text{ACF}_2$ choice function on each one, picking the most frequently picked element. There can only be two-way or three-way ties, not four-way ties (since $6$ is not a multiple of $4$), so ties can be resolved by applying $\text{ACF}_2$ or $\text{ACF}_3$ again.

So what exactly is the relationship between all the $\text{ACF}_n$?

Edit: I realized that $\text{ACF}_m$ implies $\text{ACF}_n$ if $m$ is a multiple of $n$ (e.g. $\text{ACF}_4$ implies $\text{ACF}_2$) just by duplicating the elements. My new conjecture is that this is precise in terms of pairwise implications i.e. $\text{ACF}_m$ proves $\text{ACF}_n$ in ZF if and only if $n \mid m$. Is this actually true?

soktinpk
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    Tarski proved that ACF${}_2$ implies ACF${}_4$. The proof is essentially what you wrote in the question, where you also assumed ACF${}_3$. The point is that, in case of a $3$-way tie, instead of invoking ACF${}_3$, you can just choose the one element not involved in the tie. – Andreas Blass Aug 05 '23 at 21:21
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    Two references with a lot of information about implications between (conjunctions of) ACF${}_n$ axioms are: Andrzej Mostowski, "Axiom of choice for finite sets,'' Fund. Math. 33 (1945) 137--168 and John Truss, "Finite axioms of choice,'' Ann. Math. Logic 6 (1973) 147--176. – Andreas Blass Aug 05 '23 at 21:26
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    More generally your idea for $ACF_4$ shows that if $n$ is not prime then $ACF_n$ follows from $ACF_m$ for all $m\leq n/2$ (pick a prime factor $p$ of $n$ and choose an element of each subset of size $p$; since $\binom n p$ is not divisible by $n$ the elements will not be chosen equally often and you can use this to choose a subset of size at most $n/2$). – Eric Wofsey Aug 05 '23 at 22:25
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    Related: Finite choice without AC AND Axiom of choice for sets of finite sets AND the two paragraphs beginning with "Chapter VI (pp. 95-96 & 99)" in (and my comment to) this MSE answer AND this 30 April 2007 sci.math post. – Dave L. Renfro Aug 05 '23 at 22:32
  • There is a discussion of these choice axioms for $n$-element sets in Sierpinski's book Cardinal and Ordinal Numbers. – bof Aug 06 '23 at 03:02

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