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I've realized that there are many results in set theory rely on the theorem that states "the union of countably many countable sets is countable". However, all of the proofs of this theorem I've found so far, including ones on SE, rely on the axiom of choice.

So, I'm wonder without AC is it still true or false or neither provable nor refutable?

Tran Khanh
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  • Related? Duplicate? https://math.stackexchange.com/q/1935724/42969, https://math.stackexchange.com/q/1289709/42969. – Martin R Nov 01 '23 at 08:32
  • @MartinR, correct me if I'm wrong but that's a different question asking for why it's needed in the proof right? Here I'm asking if it can be proven/rejected (or either) without AC. – Tran Khanh Nov 01 '23 at 08:38
  • You can do it with countable choice, but even that is slightly stronger in $\mathsf{ZF}$ than the desired result. – J.G. Nov 01 '23 at 08:45
  • If you are asking "is it needed", then the answer is one word. Yes. If that is a satisfactory answer, then surely "Yes, and here's why" or even "Yes, here's why, so in these cases you don't need it" is even better. – Asaf Karagila Nov 01 '23 at 09:06

1 Answers1

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It's independent of ZF; more specifically it's known that it's consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, see here.

However, most of the time in practice a slightly different statement can be used, namely that a counted union of counted sets is counted, where a "counted set" is a set equipped with a bijection to $\mathbb{N}$ (rather than just a set such that there exists such a bijection). This saves you from having to choose bijections and you can just use the given bijections to explicitly construct the desired bijection; this is provable in ZF.

Qiaochu Yuan
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  • How would the outline of the proof look, then? I could construct such a bijection if the sets in question are disjoint, but what if they aren't? Then the usual proof only gives us a surjection, and from there I would use that the existence of surjections in both directions imply the existence of a bijection, but that requires choice. How does one sidestep this issue? – Vercassivelaunos Nov 01 '23 at 08:36
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    @Vercassivelaunos: whatever argument you know that explicitly constructs a bijection in the disjoint case (which reduces to showing that $\mathbb{N} \times \mathbb{N}$ is countable), perform the same construction but just skip any elements that have already appeared. – Qiaochu Yuan Nov 01 '23 at 08:42