necessity of AC

Let X=⋃{αi∣i∈N},X = \bigcup\lbrace\alpha_i \mid i\in\mathbb{N}\rbrace, where each αi\alpha_i is a countable ordinal. My understanding is that in order to show that XX itself is a countable ordinal one must assume a weak form of the Axiom of Choice — namely, that the union of countably-many countable sets is countable. This confuses me, however. We know that, for each αi,\alpha_i, there is a bijection between it and N\mathbb{N}. Is the problem that we can’t “point” to these so as to then construct the bijection between XX and N\mathbb{N}? Are there no occasions when we know that such a bijection exists despite being unable to “point” to it — even in the context of ZF alone? Or is this precisely what we lack without AC? Thanks!

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A better title is certainly in order.
– Asaf Karagila
Oct 20 at 22:06

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2 Answers
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Yes, what you describe is correct. There are models of ZF in which the axiom of choice fails in such a way that ω1,\omega_1, which is the least uncountable ordinal, can be expressed as a union just as you wrote.

The issue is that although for each i∈Ni\in\mathbb{N} there is a bijection between αi\alpha_i and N,\mathbb{N}, there isn’t (in such a model) a function ff such that for each i<ω,i\lt\omega, f(i)f(i) is a bijection between αi\alpha_i and N.\mathbb{N}.      Got it -- thanks very much! – davidp Oct 21 at 1:03 If you know that there is a bijection between X=⋃{Xi∣i∈N}X=\bigcup\{X_i\mid i\in\Bbb N\} and N\Bbb N, then by restricting this bijection to each XiX_i, we can uniformly identify a bijection fi:Xi→Nf_i\colon X_i\to\Bbb N. Therefore, if you are in a situation where XX is uncountable, then it means exactly that we cannot uniformly choose bijections. (And indeed, Feferman and Levy showed that it is possible that the countable union of countable ordinals is in fact ω1\omega_1; and that even the real numbers can be written as a countable union of countable sets.)