# function from N\mathbb{N} to N×N×N\mathbb{N}\times\mathbb{N}\times\mathbb{N}

I want to prove that the size (cardinality) of $\mathbb{N}$ is the same as the size of the set of all ordered triples of natural numbers ; just to be more clear that $|\mathbb{N}|=|\mathbb{N}\times\mathbb{N}\times\mathbb{N}|$.

Here is one way: I assume $0\in{\mathbb N}$. Given a triple $(a,b,c)$, assign to it the number $2^a(2(2^b(2c+1)-1)+1)-1$. The same idea shows that ${\mathbb N}$ has the same size as ${\mathbb N}^k$ for any positive integer $k$.
There are of course many other ways. The idea I am using above is to first find a bijection between ${\mathbb N}\times{\mathbb N}$ and ${\mathbb N}$, and then use it to create the one you want. The bijection I used is $(a,b)\mapsto 2^a(2b+1)-1$. Another popular choice uses Cantor’s pairing function, $\displaystyle (a,b)\mapsto \frac{(a+b)(a+b+1)}2 +b$. As a technical comment, there is actually an advantage to Cantor’s choice, in that it is a polynomial, and it is easier to define (in a first-order way) in the usual structure of the natural numbers $({\mathbb N},+,\times,0,1,<)$. The choice I used requires exponentiation, and although this can also be defined in a first-order way, it is significantly less trivial.