Consider the following curious statement:
$(S)$ $\;$ Let $X$ be a non-empty set and let $f:X \to X$ be fixpoint-free (that is $f(x) \neq x$ for all $x\in X$). Then there are subsets $X_1, X_2, X_3 \subseteq X$ with $X_1\cup X_2\cup X_3 = X$ and $$X_i \cap f(X_i) = \emptyset$$ for $i \in \{1,2,3\}$.
There are easy examples showing that one cannot get by using $2$ subsets only. Statement $(S)$ can be proved using the axiom of choice.
Question. Does $(S)$ imply (AC)?
Best Answer
The three-set lemma is listed as form 285 in Howard and Rubin's "Consequences of the axiom of choice". According to their book, the earliest appearance seems to be a problem in a 1963 issue of the American Mathematical Monthly (problem 5077).
As mentioned already in the comments by Emil Jerabek, this form of choice is not equivalent to full AC, but already follows from the Boolean prime ideal theorem (BPI). However, it is not equivalent to BPI either, since it already follows from the ordering principle (every set can be linearly ordered), which readily implies the axiom of choice for families of finite sets, and this latter implies the three-set lemma in question as shown by Wisniewski ("On functions without fixed points", Comment. Math. Prace Mat vol 17, pp. 227-228, 1973); as proved by Mathias turning a Fraenkel-Mostowski model into a model of ZF, the ordering principle does not imply BPI.