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Consider the union $$ \bigcup_{n=1}^\infty \mathbb{Z}^{2n}. $$ Is the union countable? Each of the $\mathbb{Z}^{2n}$ is countable, and if I remember correctly the countable union of countable sets is countable but I wanted to verify this fact.

user918212
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    The set is countable since you can enumerate it as follows : For $k=0,1,2,\cdots$ determine all sequences with length $k$ or smaller and entries with an absolute value not larger than $k$. Each finite sequence has a finite length and a finite maximum absolute value hence it will eventually appear. – Peter Sep 04 '23 at 15:53
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    You remembered it correctly. A countable union of countable sets is countable. – PinkRabbit Sep 04 '23 at 15:53
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    Technically, you need a weak form of the axiom of choice in order to establish the countable union of countable sets being countable. – Dan Rust Sep 04 '23 at 15:53
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    But here you don't. You can make everything explicit. – Anne Bauval Sep 04 '23 at 16:08
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    It does not make any difference but is there any reason why you have $\mathbb Z^{2n}$ rather than $\mathbb Z^{n}$? – Henry Sep 04 '23 at 16:13
  • @Henry no it doesn't I do not know why I bothered adding that. – user918212 Sep 04 '23 at 16:15
  • See also https://math.stackexchange.com/q/200389/399263 which is related. – zwim Sep 04 '23 at 18:31

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Your argument is correct, as mentioned in the comments (using choice) a countable union of countable sets is countable. But if for each of the sets $X_i$ we a known bijection $f_i:\mathbb{N}\to X_i$.Then we can construct a surjection $f:\mathbb{N} \to \bigcup_{i=1}^\infty X_i$. It goes like this:

First, get a bijection $g:\mathbb{N}\to \mathbb{N} \times \mathbb{N}$, and define $$f(n)=f_{\pi_1(g(n))}(\pi_2(g(n)))$$ Now any element $x^* \in X_i$ is the image of some $c$ under $f_i$ so that $k=g^{-1}(i,c)$ is such that $f(k)=x^*$, hence $f$ is surjective.

For your question, since the $X_i=\mathbb{Z}^{2n}$ are pairwise disjoint, this function $f$ is also injective as you can check, so therefore it's also bijective.

Fernando Chu
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