$(A\_{\alpha})\_{\alpha\in B}$ a family of sets indexed by a set $B$. Is $\Pi \_ {\alpha\in B} \ A\_{\alpha}$ a set? I can see that it is a set if $A\_{\alpha}=A \ \ \forall\alpha$ because in that case the product is a subset of the power set ${\cal P}\ (A\times B)$. As far as I understand, the axiom of choice only says the product is not the empty set, and it doesn't say if it is a set at all. Or am I wrong?



Is an arbitrary product of sets a set?
The standard definition of $(A\_\alpha)\_{\alpha \in B}$ a family of sets indexed by a set $B$ is that one is given a function $A$ with domain $B$ so that for $\alpha \in B$ we understand $A\_\alpha$ as $A(\alpha)$. Then, yes, the product is a set. 


The function $A$ itself is an element of the powerset of $B \times \text{ran}(A)$ (where $\text{ran}(A)$ is the range of $A$) and the product $\prod\_{\alpha \in B} A\_\alpha$ is by definition the set of functions $f$ with domain $B$ having the property that $(\forall \alpha \in B) f(\alpha) \in A(\alpha)$. As such, one proves that the product exists by applying the comprehension axiom to the above defining formula observing that any such $f$ will be an element of the powerset of $B \times \bigcup \text{ran} A$.


[And yes, the Axiom of Choice merely says that the product is nonempty if all of the $A\_\alpha$ are nonempty.]


