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Say you have a set $A_i$ for $i$ in the natural numbers $\mathbb{N}$, and that is a countable set. Then for all natural numbers $n$, the union of those sets is countable.

I must prove this by induction, and I do realize that to do that, I must show that there is an injection( or surjection ) between the first two sets. Everything should follow accordingly afterwards, but I had the idea to utilize Cantor's Diagonal Argument in the inductive proof, but I am not sure how to go about defining injections for the argument in a way that shows that the sets are countable.

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    $\begingroup$ So, you're trying to prove that a countable union of countable sets is countable? While that's true, I don't think proving it by induction is appropriate, as (unless I'm missing some really creative idea) induction will only ever show that finite unions of countable sets are countable. $\endgroup$ – Theo Bendit Oct 25 '18 at 2:25
  • $\begingroup$ Why would you use the diagonal argument? That's a way to prove something is not countable. $\endgroup$ – Matt Samuel Oct 25 '18 at 2:27
  • $\begingroup$ Also (again, unless I'm missing something really creative), Cantor's diagonal argument will only ever show a set is strictly larger than a given cardinality, not equal to a given cardinality. You'll need to construct the maps explicitly. $\endgroup$ – Theo Bendit Oct 25 '18 at 2:28
  • $\begingroup$ For aall natural numbers $n$ the union of WHAT sets is countable? All of the sets $A_i$? No, must be something to do with $n$, I guess. The union of all the sets $A_i$ for $i\gt n$ is countable? It would help a lot if we know what sets you were trying to prove countable. $\endgroup$ – bof Oct 25 '18 at 9:25
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Are you talking about a finite union of countable sets? If so, for the induction hypothesis write $$ \cup_{i=1}^n A_i=(\cup_{i=1}^{n-1}A_i)\cup A_n. $$ Now both sets on the right hand side are countable, so there is a surjection from $\mathbb{N}$ to each of them. Can you turn this into a surjection from $\mathbb{N}$ to their union, possibly by splitting $\mathbb{N}$ into $\mathbb{N}=\{2k:k\in \mathbb{N}\}\cup \{2k+1:k\in \mathbb{N}\}$?

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