# Weak forms of the Axiom of Choice

Let $n\geq 2$ be a natural number and consider the following:

$AC(n)$: For each family $\{X_i\}_{i \in I}$ of $n$-element sets the product $\prod_{i\in I}X_i$ is non-empty.

Is it known that for which values of $m$ and $n$, $AC(m)$ and $AC(n)$ are equivalent !?

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The classic reference is A. Mostowski,Axiom of choice for finite sets, Fund. Math. 33 (1945), 137-168. The matter is also discussed on pp. 101-108 of W. Sierpiński, Cardinal and Ordinal Numbers, second edition revised, Warszawa 1965. –  bof Feb 9 at 19:45

There is about half a chapter devoted to this in Jech The Axiom of Choice, the key point is Theorem 7.15 which gives a condition for $\mathsf{AC}(n)\implies\mathsf{AC}(m)$. You may want to look around there (p.111 for the theorem).

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A reference of course is great, but would it be possible for you to mention what is the result? –  Joel David Hamkins Feb 9 at 12:53
@JoelDavidHamkins See the my answer (my comment in fact). –  Rita J Feb 9 at 13:56
Joel, I have an excellent excuse why I didn't post it in my answer. I was going to when I'd get home but it's moot now... –  Asaf Karagila Feb 9 at 14:30

Thanks to @AsafKaragila, I found the Jech's book very useful. Here is the result in page 111 concerning to my question: (Note that in what follows $C_n$ is what I have called $AC(n)$.) I don't know how to put this as a comment. So I am writing this as a new answer.

(S) $\qquad$ There is no decomposition of $n$ in a sum of primes,
$$n = p_1 + \ldots + p_s,$$ $\qquad\quad\;\:$ such that $p_i > m$ for all $i = 1, \dots, s$.

(In particular, if $n > m$, then $n$ is not a prime.)

Theorem 7.15. If $m, n$ satisfy condition (S) and if $C_k$ holds for every $k \leq m$, then $C_n$ holds.

Theorem 7.16. If $m, n$ do not satisfy condition (S), then there is a model of set theory in which $C_k$ holds for every $k \leq m$ but $C_n$ fails.

(Images: Theorem 7.15, Theorem 7.16)

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Since the question asked in particular about equivalence between $AC(m)$ and $AC(n)$ with $m,n\geq2$, it seems worth stating explicitly that the only non-trivial equivalence of this simple form is between $AC(2)$ and $AC(4)$. It is easy to see (though I overlooked it in the first version of this answer) that $AC(m)$ implies $AC(n)$ whenever $n$ divides $m$. The non-trivial direction, from $AC(2)$ to $AC(4)$, is a beautiful theorem of Tarski. (Thanks to bof for pointing out the error in the original version of this answer.)

It should perhaps also be mentioned that, although Mostowski proved a lot of results about implications of the form "$[AC(m_1)\land\cdots\land AC(m_k)]\implies AC(n)$, the complete characterization of which of these implications are provable in ZF was done by Gauntt. I don't think Gauntt ever published more than abstracts of this work, but there's a detailed exposition in a paper of John Truss, "Finite axioms of choice" [Annals of mathematical logic 6 (1973/4) 147-176].

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@bof You're right. I'll edit the answer to correct this point. thanks. –  Andreas Blass Feb 10 at 14:22

Isn't the point that this is really all about finite groups? Given a group $G$, the $G$-axiom of choice says that for any collection of sets with a simply transitive action of $G$, we may choose from each set a coset of a proper subgroup.

Then $AC(n)$ is equivalent to the axiom of choice for all groups $G$ with a fixed-point-free action on $n$ objects.

The axiom of choice for a subgroup or a quotient group of a group implies the axiom of choice for that group.

I believe that all implications between different finite axioms of choice follow from these facts.

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