I was wondering if $\bigcup_{i \in \mathbb{N}} \mathbb{N}^i$ were countable or not, where $\mathbb{N}^{i}$ means the direct product of $i$ copies of $\mathbb{N}$. I may have read that this is countable, but I think I have never seen a proof of the countability (or uncountability). The fact is that I know a countable union of countable set is countable, and I know how to prove that by using the axion of countable choice. The problem is that $\prod_{1}^{\infty} \mathbb{N}$ is not countable, and I don't know how to get rid of the problem.
Is $\bigcup_{i \in \mathbb{N}} \mathbb{N}^i$ countable?
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elementary-set-theory
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1$\mathbb{N}^k$ is countable for all $k$ so you have a countable union of countable sets. The infinite product of $\mathbb{N}$ with itself doesn't appear in the union – 2017-02-08
2 Answers
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For all $i \in \mathbb{N}$, $\mathbb{N^i}$ is countable (finite product of countable sets). So $\bigcup_{i \in \mathbb{N}} \mathbb{N}^i$ is a countable union of countable sets, hence countable.
$\prod_{i\in \mathbb{N}} \mathbb{N}$ is not needed here, you just have finite products.
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You don't need to get rid of a problem you don't have. As it is already stated in the comments, $\Bbb N^i$ is countable for every $i\in \Bbb N$, so you have a countable union of countable sets which is again countable.
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0Yeah but the fact that it goes for every $n$... I also thought that it is countable for every $n$ and so it was trivial, but it's the infinite that makes me cautios, I never know how to behave in these situations when infinite is involved – 2017-02-08