Can it be proved in ZF that the product $\prod_{\alpha < \kappa} S_\alpha$ is nonempty given$\{ S_\alpha \}_{\alpha < \kappa}$ a family of nonempty set and $\kappa > 0$ is an ordinal?
It seems possible with transfinite induction, yet I cannot seems to show the limit case holds for the predicate $\varphi(\alpha) := \exists f (f \text{ is a choice function for} \{ S_\beta \}_{\beta < \alpha})$
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Asaf Karagila
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I realize that this might be a duplicate of https://math.stackexchange.com/questions/7065/transfinite-induction-and-the-axiom-of-choice, let me know if it is. – Asaf Karagila Jul 04 '19 at 21:59
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@AsafKaragila . The possible duplicate asks for an explanation of why the case $\kappa=\omega$ cannot be proven in ZF by induction, as the proposer felt that it could. This Q merely asks whether it is provable for $\kappa \in On$ and notes that (s)he cannot see how to. – DanielWainfleet Jul 22 '19 at 18:09
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An extreme example : It has been shown to be consistent with $ZF$ that there exists ${S_n:n\in \omega}$ where each $S_n$ has exactly $2$ members, and $\prod_{n\in \omega}S_n=\emptyset.$ – DanielWainfleet Jul 22 '19 at 18:17
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@Daniel: But the reason is the same reason, at the end of the day. I also didn't close it as a duplicate, just said it might be one. – Asaf Karagila Jul 22 '19 at 18:25
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No, of course not. Not even in the case that $\kappa=\omega$, as that would be exactly countable choice which is known to be unprovable.
Moreover, even if you assume that for any ordinal $\kappa$ the product of $\kappa$ non-empty sets is non-empty, you still cannot prove the axiom of choice in general. In fact, you cannot even prove that every uncountable cardinal is comparable with $\aleph_1$.
Asaf Karagila
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Does the principle in your second paragraph have a standard name? – Alex Kruckman Jul 05 '19 at 03:34
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Well ordered choice? Denoted by $\sf AC_{WO}$ sometimes. (Some authors use superscrip, it can get confusing without reading the definitions.) – Asaf Karagila Jul 05 '19 at 06:33
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The transfinite induction you're probably imagining fails at every limit ordinal.
I suppose you're thinking "then take the union of the functions we have so far", but which of them? Since there are many sequences for each shorter length, in order for "take the union" to work you need to be able to point to just one of each length in a way that fits together -- but that is exactly what $\alpha$-choice which you're just trying to prove is for. So this argument actually begs the question.
Even though we think of the induction as a process that happens in sequence, that's not how it works formally. In the induction step for $\alpha$ all the induction hypothesis lets you assume is that there exist one or more sequences of every length shorter than $\alpha$ -- you don't have a particular selection of them already singled out as "the ones we chose in previous steps". There are no "previous steps", because the induction step in transfinite induction must be proved just once but for an arbitrary $\alpha$.
This may be easier to wrap your mind around if you remember how to prove transfinite induction from the fact that ordinals are well-ordered. In this proof you're not actually doing anything step by step -- it's an indirect proof where we say, once and for all, something like this:
Suppose the desired property fails for at least one ordinal. Then bla bla well-ordered bla bla, and therefore there is a smallest $\alpha$ that the property fails for. Then apply the induction step just once, for that $\alpha$, and conclude that the property doesn't actually fail there after all -- a contradiction.
This proof never applies the induction step to all the ordinals before the supposed first point where the property fails. So we shouldn't assume when we apply it that it "has already run", so to say, for the smaller ordinals. We just know that its conclusion is, somehow, true for each of them independently.
hmakholm left over Monica
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