We start with $ZF$. 


The axiom of countable choice, $AC\_\omega$, says that any set product of nonempty sets with a countable index set is nonempty. For any $ZF$-definable set $A$, we should be able to define $AC\_A$ in an analogous manner: any product of nonempty sets with index set $A$ is nonempty. 


What is the implication structure of these axioms? There is the obvious fact that if $A$ injects into $B$ (or rather, if that can be proven by $ZF$), then $AC\_B$ implies $AC\_A$ (Proof: choose $1$-element sets for $B\backslash A$, and restrict the choice function); is there anything else beyond the trivial?


Also, the axiom of choice implies that all of these axioms are true; is the converse true?



Relations of axioms of choice
Yes, the axiom of choice is equivalent to the assertion that $\text{AC}\_X$ holds for every definable $X$. 


One usually has to take a little care with foundational matters when definability is involved, since one cannot ordinarily express assertions of the form "every definable $X$ has a certain property", as the class of definable $X$ is not itself necessarily definable. 


Nevertheless, in your case, it turns out that we can get around this obstacle. Specifically, you may be surprised to hear that I claim that there is a single definable set $X$, such that $\text{ZF}$ proves that $\text{AC}\_X$ is equivalent to the axiom of choice. That is, we need only to consider one suitably defined set $X$, and $\text{AC}$ is provably equivalent to this special single instance of your axiom $\text{AC}\_X$ for this specific definable set $X$.


My set $X$ is: the smallest set $V\_\theta$, a rank initial segment of the cumulative hierarchy, such that there is an $X$-indexed family of nonempty sets with no choice function, if there is such a family, and $X=\emptyset$, otherwise. 


This is a perfectly good definition of a set $X$ in the theory $\text{ZF}$. If the axiom of choice holds, then $\text{AC}\_X$ holds for every set $X$, including the one I just defined. Suppose conversely that the axiom of choice fails. So there is some index set $I$ and indexed family $\langle A\_i\mid i\in I\rangle$ of nonempty sets $A\_i$ having no choice function. The set $I$ must be contained in some $V\_\theta$, since every set is a subset of some rank initial segment of the universe. Let $A\_i=\{0\}$ for all $i\in V\_\theta\setminus I$. So now we have a family $\langle A\_i\mid i\in V\_\theta\rangle$ of nonempty sets, and still there can be no choice function. So there is such a $\theta$ as in the definition of my set $X$. Thus, by design, it cannot be that $\text{AC}\_X$ holds for this set $X$. 


Of course, this is a kind of trick. It is like saying that every natural number has a certain property if a certain single number $n$ has the property, where $n$ is defined to be the smallest counterexample, if there is one, and $0$ otherwise. If that number $n$ has the property, then every number does. 


Many years ago I was at a party in Berkeley at the home of Ken Ribet and a mischevious Hendrik Lenstra mentioned to me and a small group of my fellow graduate students that although it was not known whether there are infinitely many prime pairs, nevertheless there is a number $N$ such that if there is any prime pair above $N$, then there are infinitely many prime pairs. "*If we could only find that $N$!*," he said, with a wink in his eye. The other students nodded at this profound-seeming statement, but I thought a bit and said: case 1 is that there are only finitely many pairs, and we can take any $N$ above them; case 2 is that there are infinitely many pairs, in which case we can take $N=17$ or anything. Lenstra shook my hand and said, yes, it is really trivial, isn't it? 


