Union of choice functions is a choice function

Question

I'm working on a problem and one part of my solution works only if a union of choice functions is a choice function. The best idea I have is to use that a choice function can be represented as $$ f = \{(\{a\}, a)\} $$ for any set (I used a Singleton for convenience) and that taking union of such choice functions will result in a well defined function that will be a choice function by definition but that really doesn't look like it works for me. Do I'm asking for any hint and/or help you can give me.

Answer 1

Your problem is that a union of functions is not necessarily a function. But indeed, if $\mathcal F=\{f_i\mid i\in I\}$ are all choice functions (from their respective domains) and $\bigcup\cal F$ is a function, then it is a choice function.

This is clear since $F(a)=f_i(a)$ for some $i$ such that $a\in\operatorname{dom} f_i$, and $f_i$ is a choice function, so $F(a)=f_i(a)\in a$.

---

Of course, in topics related to the axiom of choice, one has another caveat. Suppose that $A$ is a set of non-empty sets, you might be inclined to choose for each $a\in A$ a choice function, $f_a$ which chooses an element from $a$, and then take $\mathcal F=\{f_a\mid a\in A\}$, and appeal to the above.

But in order to prove that $\mathcal F$ even exists you already appealed to the axiom of choice in choosing $f_a$ for each $a\in A$.

Whether or not this is your proof strategy, this is something to keep an eye for.