0

Say I have a set S of an arbitrary infinite size N, and each member of S is a countable set. Is the union always at most size N? I know that Tarski has a theorem that AxA is equinumerous to A for any infinite set A, and that this is equivalent to the axiom of choice, and this seems like it could prove what I want, as the union of N countable sets should inject to SxS. Is this valid? And is there a way to prove this without the axiom of choice?

Tsvi Benschar
  • 101
  • 2
    Yes, this is correct, and no, there is no way to prove it without choice. For instance, without choice it is consistent that the reals are a countable union of countable sets. (Also note that without choice it is consistent that there is an infinite set with no countably infinite subset.) – spaceisdarkgreen Nov 08 '21 at 05:25
  • Both those examples only require very weak choice principles to refute, though. I'm not sure offhand how much choice is needed to prove your statement. – spaceisdarkgreen Nov 08 '21 at 05:35
  • @spaceisdarkgreen is absolutely right. However, there's a crucial subtlety: if the elements of $S$ are uniformly countable, that is if a set $(f_s)_{s\in S}$ exists with $f_s$ an injection from $s$ to $\omega$, then $\bigcup S$ is again countable. So in practice the countable unions of countable sets we actually care about can be proved to be countable with no reliance on choice at all. – Noah Schweber Nov 08 '21 at 05:37
  • In addition to the linked duplicate, see also this question (and a bunch of others - this topic comes up a lot on this site). – Noah Schweber Nov 08 '21 at 05:41
  • @NoahSchweber (With regard to the closing.) The question isn't about countable unions of countable sets though, it's about arbitrary unions of countable sets. – spaceisdarkgreen Nov 08 '21 at 05:48
  • @spaceisdarkgreen Oh crud! To the poster: I'm sorry, I read your "N" as "$\mathbb{N}$." I've reopened. (Apparently I have unilateral closing and reopening powers, I did not know that.) That said, I believe this is still a duplicate ... – Noah Schweber Nov 08 '21 at 05:49
  • As @NoahSchweber said I was talking more generally than countable unions, I was proving an exercise that separable metric spaces have a countable base, and realized that if you define a more general notion of separability to involve any cardinality N then the same proof can show that there is a base of at most size N assuming that the union of N countable sets is at most size N. All good though, it was a confusing wording. – Tsvi Benschar Nov 08 '21 at 06:31
  • @TsviBenschar This is a good example of what Noah is talking about with his first comment (and also of why you should generally ask about the problem at hand rather than the problem you've decided is the crux of the problem at hand). No choice is needed to show separable implies a countable base for metric spaces, even though choice is required to prove a countable union of countable sets is countable. (For your general problem I think some choice is needed in general, but it is a different question than the question you asked since we have enumerations of the countable sets.) – spaceisdarkgreen Nov 08 '21 at 07:03
  • @spaceisdarkgreen I was more interested in the general question but fair point – Tsvi Benschar Nov 08 '21 at 07:49

0 Answers0