discrete mathematics - Prove that the set of $mathbb{N} times mathbb{N} times mathbb{N}$ is countable.

I have seen proofs of $\mathbb{N}\times\mathbb{N}$. The proof by list where you list each element in a pair of elements and then counting them diagonally is the most convincing. I have two thoughts in mind:

1. Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.


2. Show that $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ has the same cardinality as $\mathbb{N}\times\mathbb{N}$, and $\mathbb{N}\times\mathbb{N}$ has the same cardinality as $\mathbb{N}$.

Generally, I'm having difficulty going from $\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}$, since it easy to prove the bijection from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$.

Answer

If you already know that $\Bbb{N\times N}$ is countable, fix a bijection $f\colon\Bbb{N\times N\to N}$. Now consider $g\colon\Bbb{N\times N\times N\to N\times N}$ defined as: $$g(n,m,k)=(n,f(m,k)),$$ or better yet $h\colon\Bbb{N\times N\times N\to N}$ defined as $$h(n,m,k)=f(n,f(m,k))$$ and cut out the middle-man.

As my freshman discrete mathematics professor used to tell us, go home and convince yourself this is correct.