Best way to have bijection between $\mathbb N$ and $n$ infinite sets of $\mathbb N\times \mathbb N\times...\times \mathbb N$? I tried to search for a way to have bijection between $n$ indexed infinite sets of $\mathbb N\times \mathbb N\times...\times \mathbb N$,  and integers where each set is an arbitrary multiple of $\mathbb N$, but found nothing.
What is the best formula to accomplish the goal?
 A: Consider the first $n$ primes sorted in the usual order then there is a bijection between elements of $\mathbb{N}^n$ and powers of these primes. So for say, $n=3$ we would have the primes $2,3,5$ and we'd map $(x,y,z) \rightarrow 2^x3^y5^z$ and this is a bijection by construction. Since this is a subset of $\mathbb{N}$ we can write them in the usual order which would be something like $2,3,4,5,6,8,9,10,12,15,16,18,20,24,...$ for $n=3$. From here the bijection to the naturals is obvious and so the composition of these two functions is also a bijection giving us a bijection between $\mathbb{N}^n$ and $\mathbb{N}$.
A: Consume the elements of the $n$ sets in a round-robin way.
You can even handle a countable infinity of sets by increasing the number of sets on each round.
A: Iterate Cantor's pairing function.  See "Pairing function - Wikipedia" https://en.m.wikipedia.org/wiki/Pairing_function.
Geometrically, the resulting "tuple function" is utterly trivial: put the numbers in an array and wind your way back and forth starting from the upper left hand corner.
