TITLE: countably infinite union of countably infinite sets is countable
QUESTION [6 upvotes]: How do you prove that any collection of sets $\{X_n : n \in \mathbb{N}\}$ such that for every $n \in \mathbb{N}$ the set $X_n$ is equinumerous to the set of natural numbers, then the union of all these sets, $\bigcup_{i\in \mathbb{N}}$ $X_i$ is also equinumerous to the set of natural numbers? (By equinumerous I mean there exists a one-to-one and onto function $f:X_n \to \mathbb{N}$.)
Is this statement false as it stands?

REPLY [2 votes]: If you want to show that the countable union of countable subsets is countable, you can use Cantor-Schroeder-Bernstein (I don't think it uses AC --even in summer :) ), and set up injections between $\mathbb N$ and $\mathbb N \times \mathbb N $, and the other-way-around , by generalizing this:
take any two primes , say 2,3, and map : $(a,b)\rightarrow 2^a3^b$ ( you can see that, to generalize to a product of k-copies of $\mathbb N$, just take k different primes; if you want an actually countably-infinite product, this is maybe more delicate), and an injection in the opposite direction is given by , e.g., n->(n,0,0,...). 
And, BTW, any choice of injections in CSBernstein allows to construct an actual bijection. 
EDIT: I think it is not too hard to show the map (a,b)->$2^a3^b$ is an injection; if we had $2^a3^b=2^{a'}3^{b'}$, it would follow that $2^{a-a'}3^{b-b'}=1$; by simple divisibility arguments, each of the factors on the left-hand side would have to divide 1; it then follows that a-a'=0 and b-b'=0, i.e., a=a', b=b'.
EDIT#2 : Please see some of the caveats in the comments section about concluding that the union of countables is countable.