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Let I be a set, and for each $\alpha \in I$, let $X_\alpha$ be a non-empty set. Then $\prod_{\alpha \in I}X_\alpha$ is also non-empty.

There is (a) the possibly-infinite set $I$ and also (b) all the possibly-infinite sets $X_\alpha$. If one of either (a) or (b) is restricted to finite sets only, does the statement follow from other common axioms? Is one of these requirements (a or b) more important than the other?

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The rule of thumb for when AC is needed is when you need to make an infinite number of arbitrary choices. To get an element of $\prod_{i\in I}X_i$ you need to choose an element of $X_i$ for each $i\in I.$

If $I$ is finite, that is only a finite number of choices, so AC is not required.

If $I$ is infinite then there is a potential need for AC. Note we didn't say "an infinite number of arbitrary choices from infinitely many outcomes," so $X_i$ being finite doesn't necessarily help us here.

However, if there is some well-defined way of choosing the elements (i.e. the choice is not "arbitrary"), then choice is not required. For instance, if the $X_i$ are all sets of natural numbers, we can just say "choose the least element in $X_i$". More generally, if the $X_i$ are well-ordered, we can do the same thing.

But do not confuse "subset of $\mathbb N$" with "countable" and "well-ordered" with "well-orderable". If all we know is that there is some well-ordering, we may still need to make an arbitrary choice of which well-ordering to use. So we can't just say "the $X_i$ are countable, thus we can list them, so just take the first one on the list"... the question "which list?" requires an infinite number of arbitrary choices. In fact, it is consistent in the absence of AC that there is a countably infinite $I$ and the $X_i$ each have two elements, but $\prod_{i\in I}X_i$ is empty.

But on the other hand , if $\bigcup_iX_i$ is well-orderable then we can well-order that (that's just making one choice) and then choose the element of $X_i$ that is least in that well-ordering.

(Ok, I'll stop.)

So it's a bit tricky, but you get used to it.

Your cardinality restrictions lead to natural weakenings of the axiom of choice. For instance, the statement that $\prod_{i\in I} X_i$ is nonempty for all countable $I$ is called "countable choice." This is enough to develop most of analysis, although a somewhat stronger variant called dependent choice is more convenient. (Some choice is needed to prove some "obvious" things like that a countable union of countable sets is countable and that every infinite set has a countable subset. Countable choice suffices for this.) If we say it holds for $X_i$ finite, that is usually called AC(Fin) or something like that (it's not quite as well-established as countable choice).

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