Can we only have a Cartesian product of a countable number of sets? Can we only have a Cartesian product of a countable number of sets? 
I suspect the answer to this is yes. This is because the resulting tuple of the Cartesian product will be ordered, and each position on that tuple will have a number- as a mark of position. This number will undoubtedly be a natural numbers. This number (position) can also be given to the set from which the element at that position is taken, hence making the number of sets countable. 
I'm not sure of the above explanation. Verification would be extremely helpful. Thanks
 A: You can take the cartesian product of any collection of sets. If $\{S_i\}_{i\in I}$ is a collection of sets indexed by a set $I$, then the cartesian product is defined as the set of all functions $f:I\to \bigcup \{S_i\}$ such that $f(i)\in S_i$ for all $i\in I$. 
The finite and countable cases are easily seen to agree with this definition. 
Remark: The axiom of choice is equivalent to the claim that the cartesian product of a non-empty collection of non-empty sets is non-empty. 
A: As I wrote in my answer to your question about the axiom of choice. If you define the Cartesian product as an inductive process where at each step you have ordered pairs (and not ordered triplets, and so on) then you can only take products a finite number of times; any finite number; but finite nonetheless.
Otherwise you have something of the form $(a_1,(a_2,(a_3,\ldots$ and it is not hard to see how this would contradict the axiom of regularity. (Let $P_n$ the pair whose right coordinate is $a_n$ then $\{P_{n+1}\}\in P_n$.)
On the other hand, if you define a Cartesian product as a set of choice functions, then it is possible to write an infinite product with any index set (finite, countable or otherwise), and assuming the axiom of choice we can also prove that that this product is never empty (if none of the sets is empty, anyway).
