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$$\mathbb N\times\mathbb N \text{ is countable}.$$

Is there any way to prove it using induction? without fundamental theorem of arithmetic

Michael Hardy
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Meer
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  • You mean the Cartesian product? – Vim May 03 '16 at 03:30
  • yes $$\Bbb N × \Bbb N$$ – Meer May 03 '16 at 03:32
  • Using induction...? I am not quite sure I understand. Do you want an explicit bijection between $\mathbb{N}$ and $\mathbb{N}\times \mathbb{N}$? – Fimpellizzeri May 03 '16 at 03:33
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    Yes, for example: "The set is countable because there are proofs that show that, and if it wasn't countable then they wouldn't work. The other proofs work, and therefore the set is countable. QED" – Asaf Karagila May 03 '16 at 03:35
  • yea proving its a bijection using induction – Meer May 03 '16 at 03:35
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    @AsafKaragila You should reference where this technique comes from: http://smbc-comics.com/index.php?id=1099 :) – Luiz Cordeiro May 03 '16 at 03:36
  • This is a duplicate http://math.stackexchange.com/questions/54158/the-cartesian-product-mathbbn-times-mathbbn-is-countable – MathematicsStudent1122 May 03 '16 at 03:37
  • Perhaps if you have an bijection N-->N , that would be the initial step for induction, then you can extend to N-->NxN ? ( and then NxNxN, etc, that would be novel) – user247608 May 03 '16 at 03:38
  • @user247608: Your induction idea works in showing that $\mathbb{N}\times \mathbb{N}\times \cdot \times \mathbb{N}$ ($n$ of them) is countable. The induction step from $n=k$ to $n=k+1$, where $k\ge 2$, is easy. Unfortunately, going from $1$ to $2$ takes some work. – André Nicolas May 03 '16 at 03:50
  • Yeah, I'll have to think about that - of course, this is not the easy or obvious way, but it could be interesting if I have time to think about it. As far as I know, induction is generally used on finite sets, so this is a very non-obvious problem to consider! – user247608 May 03 '16 at 04:41
  • how about $f: N \times N -> N$ given by f((m,n))= m + $\sum_{i=0}^{m+n-2} i$ – Meer May 03 '16 at 05:08

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I assume that $N$ means $\Bbb N$ and $*$ means Cartesian product. So you are actually requesting a proof that $\Bbb N^2$ is countable.

There is indeed no need for an appeal to induction, since we already have many very apparent ways to enumerate them all, like $(0,0),(1,0),(0,1),(2,0),(1,1),(0,2),(3,0),(2,1),(1,2),(0,3),\cdots$ in which every element appears exactly once.

PS: then comes the part where induction comes into play :). Starting from here, you should be able to show that for any $p\in\Bbb N^+$, $\Bbb N^p$ is countable, and, in the same spirit, that the Cartesian product of a finite collection of countable sets is again countable.

Vim
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