Given a set $X$, we say that $X$ has choice sequences of length $|I|$, denoted $CS(|X|,|I|)$, if for any $f:I\to{\cal P}(X)\setminus\{\emptyset\}$ there is a function $g:I\to X$ such that $g(x)\in f(x)$ for all $x\in I$.
The notion of choice sequence is closely related to $AC_\kappa$, which asserts $\forall{\frak n},CS({\frak n},\kappa)$. It is easy to show
- $\forall n<\omega,CS({\frak m},n)$
- ${\frak m\le \frak n}\land{\frak p\le \frak r}\to CS({\frak n,r})\to CS({\frak m,p})$ (this also shows that $CS(|X|,|I|)$ does not depend on $X$ or $I$ individually but only their cardinals)
- $\forall{\frak m},CS({\frak n,m})\iff CS({\frak n},2^{\frak n})\iff\frak n$ is an aleph
My question concerns the status of $\forall{\frak n},CS({\frak n,n})$. My guess is that it is consistently false in $ZF$, but I haven't been able to find a counterexample. The obvious starting point is $CS(|X|,|X|)$ where $X=\bigsqcup_{n<\omega}2$ is a countable disjoint union of pairs with no choice function. Clearly for this set we have $\lnot\,CS(|X|,\omega)$, but since $\omega\not\le |X|$ I can't apply the first property above to immediately conclude, and most of the test functions on $X$ I can think of already break the pair symmetry in such a way that you can reconstruct a choice function. Does this counterexample work?
This statement is in fact equivalent to the axiom of choice.
Suppose that the axiom of choice fails, and let $\{A_i\mid i\in I\}$ be a family of non-empty sets not admitting a choice function. Take $X=\bigcup A_i\cup I$.
Now fix some $i_0\in I$ and consider the function $f(x)=\begin{cases}A_i& x=i\\ A_{i_0}& x\notin I\end{cases}$, if this admitted a choice function, then the restriction to $I$ will be a choice function from $\{A_i\mid i\in I\}$.