Similarity and Dominance.

Question

I have problem card($\mathbb N\times \mathbb N$) $=$ card($\mathbb N$).

So we have to show that $\mathbb N\times \mathbb N$ $\approx$ $\mathbb N$.
($A\approx$ B that mean there exists a bijection between A and B , where A and B be sets.)

We define a bijection that is $f(m,n)=2^{m-1}(2n-1)$ for all $m,n\in\mathbb N\times\mathbb N$

So i want to show that card ($\mathbb N\times \mathbb N\times\mathbb N$) $=$ card($\mathbb N$), i can't define a bijection between $\mathbb N\times \mathbb N\times\mathbb N$ and $\mathbb N$.

Please, hlep me.

Moreover , please hint me to show card ($\mathbb N^n$) $=$ card($\mathbb N$) for all $n\in \mathbb N$.

Answer 1

Hint: We can define a bijection $g:\Bbb N^3 \to \Bbb N^2$ by $$ g(p,q,r) = (f(p,q),r) $$ Note that $f \circ g: \Bbb N^3 \to \Bbb N$ is a bijection, since it is a composition of bijections.

In general, we can use this trick to get bijections between $\Bbb N^n$ and $\Bbb N^{n-1}$.