Talk:Countable Union of Countable Sets is Countable/Proof 2

Why specifically countably infinite sets? The theorem is as easy to prove, but a little stronger when just countable is required. --Lord_Farin 12:42, 19 January 2012 (EST)
 * No reason. I think there was a contributor some years back who argued that "countable" should always mean "countably infinite" rather than "countably infinite or finite" so an effort must have been made to write the proofs so they distinguished between the two. Can't remember the details now. You're quite correct. Feel free to upgrade this and any other similar. --prime mover 16:34, 19 January 2012 (EST)
 * My lecture notes on axiomatic set theory require the Axiom of Choice for this theorem. They state '...without AC we cannot prove that $\R$ is not the union of countably many countable sets.' That sounds like they are quite certain that AC is necessary (well, of course, Axiom of Countable Choice would suffice, but I accept AC anyway, so no big deal there; I only pursue that it is known when AC is made use of). --Lord_Farin 18:11, 19 January 2012 (EST)
 * Wah! I'm seriously unhappy with that. Countable sets can be well-ordered without the need for AoC, surely? I'm going to have to think about this ... acceptance of AoC or not, it would be good for PW to be able to distinguish rigorously what needs AoC and what doesn't, and this statement is difficult. I'm to bed, I'll catch up on this later. --prime mover 18:16, 19 January 2012 (EST)