# What's the thorny issue on: “If all $S\in \ell$ are nonempty, does it follow that $\prod_{S\in \ell} S$ is nonempty? when $\ell$ is infinite?”

I'm reading Paolo Aluffi's ALGEBRA, Chapter 0.

Here he proposes that there's a thorny issue:

What is this thorny issue?

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The thorny issue is whether or not the product is nonempty in general. In order to say so, we need to accept the axiom of choice. –  anon Dec 7 '12 at 13:06
An element in $\prod_\mathcal{S} S$ is a tuple with each entry corresponding to one element in each $S$. How can you pick one element of each $S$ without the axiom of choice? –  Neal Dec 7 '12 at 13:07
If the product is nonempty when $S\in \ell$ is finite, why it wouldn't be nonempty when $S\in \ell$ is infinite? –  Vÿska Dec 7 '12 at 13:08
Do you mean, "if the product is nonempty when $\mathcal{I}$ is finite, why wouldn't it be nonempty when $\mathcal{I}$ is infinite"? The questions is, why would it? The ability to choose a finite number of selections does not automatically give one the ability to fast forward through an infinite number of selections - we must accept this independently. –  anon Dec 7 '12 at 13:11
@GustavoBandeira Let $\phi(X)$ be the proposition "$X$ is a finite set". Obviously every finite set $X$ satisfies $\phi(X)$. So are you saying that all sets $X$ satisfy $\phi(X)$ as well? There's absolutely no reason why that kind of reasoning should be accepted. –  Zhen Lin Dec 7 '12 at 13:14

The assumption that the product of non-empty sets is non-empty is known as the axiom of choice (note that every function in the product is a choice function, so the assertion that the product is non-empty is exactly the assertion that there is a choice function).

The axiom of choice has a myriad of counterintuitive consequences, like the Banach-Tarski paradox for example. There has been quite some troubles with the axiom of choice when it was first proposed, but with time it was shown that new no contradictions are added if we assume it, or if we assume that it fails. Since then a lot of mathematicians rather just assume it in order to tame the way infinite objects behave.

If the collection $\ell$ is finite, then we can prove that the product is non-empty without the axiom of choice. However if the collection itself is infinite, then we need the axiom of choice for the proof. Of course there are some collections which we can prove to have non-empty products without the axiom of choice, for example a collection of singletons, or finite sets of real numbers, but generally speaking one needs the axiom of choice.

Also related:

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The non-emptyness of this set cannot be deduced from the usual nine axioms of Zermelo-Fraenkel set theory. Instead, it needs to be postulated as a separate axiom, the axiom of choice. Some (but these days only very few) people do not like to assume the axiom of choice in their mathematics, because it has some counter-intuitive consequences (but omitting it does, too!).

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Look this question. –  Vÿska Dec 7 '12 at 13:10

As Aluffi hints at, the thorny issue is that the nonemptiness of the Cartesian product of an arbitrary family of nonempty sets is equivalent to the Axiom of Choice. There are situations where the Axiom of Choice is not required to show that the Cartesian product is nonempty:

• for finite such families (meaning that $\mathscr{S}$ is finite);
• if the union $\bigcup_{S \in \mathscr{S}} S$ is well-ordered;
• more generally, if a unique element of each $S \in \mathscr{S}$ can be described in a uniform manner (e.g., if each $S \in \mathscr{S}$ is a finite set of reals, so "the minimum element of $S$" is a definition of a unique element of each $S$);

However, it can happen (if Choice fails) that $S_n$ is nonempty and finite (even of cardinality $2$) for each $n \in \mathbb{N}$, yet $\prod_{n} S_n$ is empty.

I don't think I say too much by declaring that the vast majority of mathematicians today fully accept the Axiom of Choice, as it has many helpful consequences/equivalents (e.g., every vector space has a basis, cardinalities of sets are always comparable, the product of compact spaces is compact) and by itself it cannot add contradictions to set theory (if ZFC is inconsistent, then so is ZF).

For more information, consider looking at any number of answers by Asaf from this list of questions.

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I LOL'd at the last line there. :-) –  Asaf Karagila Dec 7 '12 at 13:30
@Asaf: I was certain that you would provide a helpful list of associated answers; I thought I'd provide an ever growing list of associated answers. For instance, this answer now shows up in the list. –  Arthur Fischer Dec 7 '12 at 13:36
Well, yeah. And I did provide one related link (so far), but I think that it is a relatively soft question about the historical issue. That question was asked once, and I only recall dealing with it in-depth (i.e. more than just mentioning it) in one or two other answers. –  Asaf Karagila Dec 7 '12 at 13:40