# Inductive Proof of a countable set Cartesian product [duplicate]

Possible Duplicate:
Proving $\mathbb{N}^k$ is countable

I would like to prove that if S is countable then for any positive integer n the set $S^n$ (the n-fold Cartesian product of S with itself) is countable using mathematical induction.

I think I should initialize it at n=0 but I don't know where to go from there.

Thanks so much for the help

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## marked as duplicate by Asaf Karagila, Martin Sleziak, Chris Eagle, t.b., J. M.Aug 18 '12 at 1:29

This question was marked as an exact duplicate of an existing question.

The result is trivial for $n=0$ and $n=1$, so your first step should be to prove it for $n=2$. Then you can use that result in your induction step to go from countability of $S^n$ to countability of $S^{n+1}$, since $S^{n+1}$ clearly admits a bijection with $S^n\times S$, a product of two countable sets. – Brian M. Scott May 1 '12 at 23:08
Thanks Brian. How do you prove the countability of a cartesian product of two countable sets ? – BrainOverfl0w May 1 '12 at 23:11
There are many ways; one is discussed in considerable detail here. – Brian M. Scott May 1 '12 at 23:17
I made it with your help. Many thanks ! – BrainOverfl0w May 1 '12 at 23:39

Brian gave me some excellent advise and I found a way to do it. Showing that the cartesian $S^n \times S$