From Wikipedia: The axiom of countable choice or axiom of denumerable choice, denoted $AC\omega$, is an axiom of set theory that states that every countable collection of non-empty sets must have a choice function. I.e., given a function $A$ with domain $\mathbb{N}$ (where $\mathbb{N}$ denotes the set of natural numbers) such that $A(n)$ is a non-empty set for every $n\in\mathbb{N}$, then there exists a function $f$ with domain $\mathbb{N}$ such that $f(n)\in A(n)$ for every $n\in\mathbb{N}$.
Question Does function $A$ necessarily need to be defined on $\mathbb{N}$ or can it also be defined on a set $M$ where $|M|=\aleph_0$ or $|M|<+\infty$?
Thanks!
No, the restriction that the index set (= domain of $A$) be $\mathbb{N}$ is inessential: from the axiom of countable choice as stated we can deduce the apparently-more-general (arbitrary countably-infinite-or-finite index set) form as a corollary.
Below, I'll use the inclusive definition of countability: $X$ is countable if there is an injection from $X$ into $\mathbb{N}$ (this is equivalent to demanding that $X$ admit a surjection from $\mathbb{N}$ or be empty). Suppose $A$ is a function with domain a countable set $I$ such that $A(i)\not=\emptyset$ for each $i\in I$; I'll show that there is a choice function for $A$, that is, a function $f$ with domain $I$ such that $f(i)\in A(i)$ for each $i\in I$.
Since $I$ is countable, fix an injection $j:I\rightarrow\mathbb{N}$. Let $B(n)=A(j^{-1}(n))$ if $n\in ran(j)$ and let $B(n)=\{0\}$ (say) otherwise. By AC$\omega$ we get a choice function $g$ for $B$. Now define $f$ by setting $$f(i)=g(j(i)).$$