# 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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1

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

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