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It is well known that $\mathbb{N}$ is bijective with $\mathbb{N}\times\mathbb{N}$, there are many possible such bijections as described in this site.

What I am curious, what are some bijections that generalizes well to higher dimensions, i.e. generalizes to bijection from $\mathbb{N}$ to $\mathbb{N}\times\dots\times\mathbb{N}$?

For instance the "Cantor tuple function" https://en.wikipedia.org/wiki/Pairing_function generalizes to higher dimensions.

Are there any others?

Thanks for any references.

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    One way is to recursively build one from the bijection $\Bbb N\to\Bbb N\times \Bbb N$, and make a bijection $\Bbb N\times (\Bbb N)\to \Bbb N\times (\Bbb N\times \Bbb N)$, and so on, always splitting the last $\Bbb N$ into two. – 2017-02-07

4 Answers 4

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One can do by showing injections from $\mathbb{N} \to \mathbb{N} \times \mathbb{N}.. \mathbb{N} $ and also on the reverse side. The $ \Rightarrow$ direction is obvious just map it to any one of the coordinates and rest as 1's.

For the other one , let $ p_1,p_2,..p_n$ be n primes. Let ther other map be $ (k_1,k_2,..k_n) \to p_1^{k_1}p_2^{k_2}..p_n^{k_n}$. This is also injective.

Hence there is a bijection between the two sets.

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An explicit bijection is given by $(a_1,a_2,...,a_k) \rightarrow \sum\limits_{s=1}^{k}{\binom{\sum\limits_{r=1}^{s}{a_r}+s-1}{s}}$. You can get more bijections by composing with bijections $\mathbb{N} \rightarrow \mathbb{N}$.

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I can give you an explicit bijection between $\mathbb{N}$ and $\displaystyle\bigcup_{n\in \mathbb{N}} \mathbb{N}^n$ (where $\mathbb{N}^n$ denotes the cartesiok product iterated $n$ times). To each $m\in \mathbb{N}$ you associate its prime decomposition $p_1 ^{\alpha_1}....p_k^{\alpha_k}$ (where the $p_i$'s are the prime numbers in order, $\alpha_i$ being $0$ if $p_i$ doesn't divide $m$, and $\alpha_k\neq 0$), and then you associate the finite sequence $(\alpha_1,....,\alpha_k)$. You can easily check that it's a bijection, and it should give you some ideas for bijections $\mathbb{N}\to \mathbb{N}^n$

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Another construction goes as follows: $(a_1,\ldots,a_n)$ is sent to the product $2^{a_1}3^{a_2}\cdots p_{n-1}^{a_{n-1}}\cdot b_{a_n}$ where $p_i$ is the $i$-th prime number, and $b_i$ is the $i$-th natural number not divisible by any of the primes $2,3,\ldots,p_{n-1}$.

This is a generalisation of $(a,b)\mapsto 2^a(2b+1)$.