# Countable cartesian product and Axiom of Choice

In the A taste of Topology book, when talking about Cartesian product $\prod\{S:S\in\mathcal{S}\}$, the author writes the following:

It is straightforward that $\prod\{S:S\in\mathcal{S}\}\neq\emptyset$ whenever $\mathcal{S}$ is finite and $S\neq\emptyset$ for any $S\in\mathcal{S}$. The same can be shown if $\mathcal{S}$ is countable, see Exercise 2.

The "Exercise 2" asks to prove that $\prod_n S_n$ of non-empty sets is non-empty, without invoking Zorn's lemma.

Is it indeed that straightforward to show the claim for countable $\mathcal{S}$? I was under the impression that while the full strength of AC is not needed, one still needs the weaker Axiom of Countable Choice to make that claim. Can it be proved in ZF alone?

• I think this indeed a small mistake in an overall very nice textbook. At the very least, there is no clue in the text about the solution the author may have had in mind. Jan 12, 2016 at 5:57
• You can prove this "Exercise 2" without invoking Zorn's Lemma (equivalent to full AC), but you need at least countable AC to prove the statement — in fact it's equivalent to that. Jan 12, 2016 at 5:57
• Agreed, it's a bit cagey and evidently confusing. Jan 12, 2016 at 6:05
• @Brian: Actually this concerns a countable family of finite sets. So it's much weaker than countable choice. It is equivalent to König's lemma about trees. Jan 12, 2016 at 6:07
• @AsafKaragila, ah and oops, I totally overlooked that the sets $S\in\mathcal{S}$ remain finite. Yes, a good bit weaker than full-blown countable choice, and equivalent, as you note, to "every infinite finitely-branching tree has an infinite branch". In any case, not a theorem of ZF. Jan 12, 2016 at 6:23

For example in Cohen's second model for the failure of choice there is such family of sets of size $2$.
• Is the statement true in a more concrete setting, for example if $S_n$ are non-empty subsets of $\mathbb{R}$? Jan 12, 2016 at 6:04
• So pick $F_n\subset S_n$ finite, and since $\Pi F_n$ is non-empty, so is $\Pi S_n$? This would work then for any subsets of an well-ordered set, right? Edit: This can be right. Doesn't "picking" finite subsets $F_n$ already involve some choice? Jan 12, 2016 at 6:15
• I misunderstood your statement, sorry for the confusion. So the statement is true when $S_n$ are finite subsets of $\mathbb{R}$, but when they are infinite one still may need some form of choice. I had the impression you claimed the statement is true for any non-empty subsets of $\mathbb{R}$. Jan 12, 2016 at 6:48