My apologies if this question has already been answered in the mathematical literature, but I am not a set theorist and this merely crossed my mind while I was studying measure theory.
The axiom of choice asserts that for any family of sets $F_i$ where $i\in\beta$ there exists at least one function $f:\beta\to\displaystyle\bigcup_{i\in\beta}F_i$ such that $f(x)\in F_x$ for all $x\in\beta$.
My question is this: in ZFC what do we know about the number of choice functions you can have?
For example, is ZFC consistent with the existence of an example where $\beta$ and $F_i$ are all infinite and where there are only countably many such choice functions?
No. There is a choice function $\tilde f$ that selects a two-element subset from each $F_i$. Let $f: \beta \to \cup F_i$ be a choice function that selects from each subset; i.e., $f(i) \in \tilde f(F_i)$. For each subset of $B \subseteq \beta$ there's a choice function $g$ with $g(i) \in \tilde f(F_i)$ that agrees with $f$ exactly on $B$, so there are at least $2^{\aleph_0}$ choice functions.