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I've been looking for an example of an empty Cartesian product whose factors are non-empty. From what I've gathered so far, this statement is equivalent to the negation of AC, ie. AC fails. So constructing an example means finding a collection of sets for which no choice function exists. But I haven't a clue how to go about such a proof.

Can you actually construct an empty cartesian product where each factor is nonempty, or is it not possible by virtue of the definition? If the latter, is there a proof of such a statement? Obviously if it is, it's probably over my head, but I'd be interested nonetheless.

Thanks in advance!

Metta World Peace
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    Well as you know this is equivalent to $\neg$AC, so if you want to construct an example, you had better start with some set $X$ for which you have no choice function and move from there. – James Jun 22 '13 at 23:16

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No. You cannot really construct something like that.

The negation of the axiom of choice is as non-constructive as the axiom of choice itself. All it tells us is that somewhere in the universe of sets there exists a family of non-empty sets whose product is empty.

If you want more, you will have to assume more. What does that mean? For example we know that we can assume that there is a family of sets of real numbers whose product is empty, and therefore the product of all non-empty sets of real numbers is empty. But it is also consistent that the real numbers can be well-ordered, and so the collection of all non-empty sets of real numbers does admit a choice function. So you have to assume one way or another.

But generally, you cannot point out at a non-well orderable set, or a family of non-empty sets whose product is empty.

Asaf Karagila
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  • I vaguely remembered you mentioned Russell set in some other related questions. – Metta World Peace Jun 22 '13 at 23:25
  • @Metta: Yes, there are many examples, one of which is the statement that there is a countable family of disjoint pairs which does not admit a choice function on any infinite sub-family. – Asaf Karagila Jun 22 '13 at 23:29
  • Alright, thanks, I was afraid of that. I did a quick search for 'choice function' and the first answer [here][1] gives the result that $P(\mathbb{R})-\emptyset$ does not have a choice function. Of course it doesn't hint to any intuition of why the cartesian product of those sets would be empty, but as you said that isn't possible. Thanks for your help! [1]: http://math.stackexchange.com/questions/190747/finding-a-choice-function-without-the-choice-axiom ((edit:I guess linking doesn't work in the comments?)). –  Jun 22 '13 at 23:54
  • @Bleys: The point is that any set which can be shown to exist (without using choice) and is not provably well-orderable is also not provable non-well orderable. Similarly any family of provably (without choice) non-empty sets which does not provably admit a choice function, also cannot provably have an empty product. (Linking in comments is in the [title](url) format.) – Asaf Karagila Jun 23 '13 at 00:05