How to prove that the set of natural numbers $\mathbb{N}$ has the same size as $\mathbb{N\times N}$?

Question

How to prove that the set of natural numbers $\mathbb N$ has the same size as $\mathbb N \times \mathbb N$?

Definition: We know that two sets have the same size when there is an injection from one set to the other and the other way round.

I was thinking of the following proof:

$f: \mathbb N \to \mathbb N \times \mathbb N; f(n)=(n,1)$

we have that $f$ is an injective function because if we chose $j$ and $t$ in $\mathbb N$, then we will have $fj=j=t=ft$, hence the function is injective.

and we can take: $g: \mathbb N \times \mathbb N \to \mathbb N ; f(n,1)=n+1$

we have that $g$ is an injective function because if we chose $h$ and $k$ in $\mathbb N \times \mathbb N $, then we will have $fh=h=k=kt$, hence the function is injective.

Hence because of the definition above we conclude that the two sets have the same size.

To define a function $\mathbb{N}\to\mathbb{N}\times\mathbb{N}$, your output should have two coordinates; I don't see how $n+1$ qualifies. Similarly, a function $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$ must have two inputs, and again I don't see how $n+1$ qualifies. — Nick Peterson, Feb 03 '17 at 20:32
Should we refrain from voting to close since OP has provided his/her own argument, and retag as 'proof-strategy/vefification?' — A. Thomas Yerger, Feb 03 '17 at 20:44

score 1 · Answer 1 · answered Feb 03 '17 at 20:47

Your injective function is $\Bbb N\to\Bbb N\times\Bbb N$ is fine. If you're going to define an injective $g:\Bbb N\times\Bbb N\to\Bbb N$, you need to have two inputs, i.e. it should be of the form $g(n,m)=\;?$ . Right now, you are just writing $g(n,1)$ which does not give an output for every element of $\Bbb N\times\Bbb N$.

I'd recommend trying to show that $g(n,m)=2^n 3^m$ is injective.

score 0 · Accepted Answer · answered Feb 03 '17 at 20:48

I'd like to point out your function may be injective, but this is only because it is not actually defined on $\mathbb N \times \mathbb N$! It is in fact only defined on $\mathbb N$, a single copy, because it only accepts a single argument.

Now, you may try to write the related function $\mathbb N \times \mathbb N \to \mathbb N: (n,m) \mapsto n+m$, but this function would NOT be injective, as demonstrated by $(3,1) \mapsto 4$ and $(2,2) \mapsto 4$.

A natural injection is given by $(n,m) \mapsto 2^n3^m$. There are others in the comments. Since there is an obvious surjection from $\mathbb N \times \mathbb N \to \mathbb N$, we know that $|\mathbb N| \leq |\mathbb N \times \mathbb N|$ and also that $|\mathbb N \times \mathbb N| \leq |\mathbb N|$, since we have injections from each to the other. Then they have the same cardinality.

How to prove that the set of natural numbers $\mathbb{N}$ has the same size as $\mathbb{N\times N}$?

2 Answers2