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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?

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  • $\begingroup$ 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. $\endgroup$ – Carl Mummert Jan 12 '16 at 5:57
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    $\begingroup$ 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. $\endgroup$ – BrianO Jan 12 '16 at 5:57
  • $\begingroup$ Agreed, it's a bit cagey and evidently confusing. $\endgroup$ – BrianO Jan 12 '16 at 6:05
  • $\begingroup$ @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. $\endgroup$ – Asaf Karagila Jan 12 '16 at 6:07
  • $\begingroup$ @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. $\endgroup$ – BrianO Jan 12 '16 at 6:23
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Based on the information you provide this is a mistake. You are correct in that it is consistent that there is a countable family of finite sets whose product is empty (namely, they do not have a choice function).

For example in Cohen's second model for the failure of choice there is such family of sets of size $2$.

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  • $\begingroup$ Is the statement true in a more concrete setting, for example if $S_n$ are non-empty subsets of $\mathbb{R}$? $\endgroup$ – Markus Jan 12 '16 at 6:04
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    $\begingroup$ Yes. There you can always choose from finite sets uniformly (even from uncountably many): since every finite set of reals has a minimum you can choose that one. $\endgroup$ – Asaf Karagila Jan 12 '16 at 6:06
  • $\begingroup$ 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? $\endgroup$ – Markus Jan 12 '16 at 6:15
  • $\begingroup$ You are correct. Choosing the finite subset requires choice. Because why not just choose a singleton, or just its element directly. $\endgroup$ – Asaf Karagila Jan 12 '16 at 6:23
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    $\begingroup$ 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 $\mathbb{R}$. $\endgroup$ – Markus Jan 12 '16 at 6:48

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