Size of infinite products without the axiom of choice Obviously, by definition, without $AC$, there may be non empty sets $(X_i)_{i\in I}$ such that $\prod_{i\in I}X_i = \emptyset$. 
However, there may also be some sets that have "the same size" as with $AC$, for instance if we take the $X_i$'s to be $\{0,1\}$, then with or without $AC$, one has $\prod_{i\in I}X_i \sim \mathcal{P}(I)$. 
I was wondering if there is anything less "extreme", that is not "$0$ or the same size as with $AC$".
Obviously this is not well defined and so the question of how to define it is part of my question. 
But for instance, to make you understand what I mean and maybe help you make it precise, may there be (finite ? But not singletons)sets $(X_i)_{i\in I}$, with $I$ infinite and with $\prod_{i\in I}X_i \sim 3$ or $4$, etc., or even $\aleph_0$ ?
So I guess there are two questions :
1- is there a sensible way to define size more generally than just giving examples as I did ?
2- If so, can infinite products have any size without $AC$ ?
 A: $\sf ZF$ proves that the following things are impossible together for any infinite family of sets $\{X_i\mid i\in I\}$:


*

*$\prod X_i=\aleph_\alpha$; and 

*$\operatorname{cf}(\aleph_\alpha)=\aleph_0$.


The reason is simple. First note that $\sf ZFC$ itself disproves the whole thing, then note that if $\prod X_i$ is not empty, then for all $i$, $|X_i|\leq|\prod X_i|$. So by (1) it follows that all the $X_i$'s can be uniformly well-ordered, and $I$ as well as a consequence.
You can now either repeat the usual proof that shows that an infinite product has an uncountable cofinality; or just code the whole thing into a set of ordinals $A$ and note that in $L[A]$ we have the same family of sets (or an isomorphic family) with a product whose size is a cardinal of uncountable cofinality $\kappa$. Now we might argue that $\kappa$ is not a cardinal in $V$, or that it has changed its cofinality, but then we can add one more set of ordinals which codes that information to show that in fact in $V$ we have more than $\kappa$ elements in the product.

So we have that an infinite product is either empty or infinite, and if it is infinite it is not both well-ordered and has countable cofinality.
The closest you could probably get is to take a countable family of pairwise disjoint sets $S_i$ such that the product of any infinitely many of them is empty. For example $S_i$ is a partition of a Russell set into pairs.
Now take $X_i=S_i\cup\{\varnothing\}$. Thus we add a canonical choice function ($f(i)=\varnothing$), and therefore we get $|\bigcup X_i|^{<\omega}+1$. This will invariably be strictly more than $\aleph_0$, but as far as things go, this is a somewhat minimal example, since it is just a finite change from being the canonical choice function.
