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I am not a mathematician. Let $\mathbb {Z}$ be a positive integer set. I need to know whether there exist a bijection from $\mathbb {Z}^3$ to $\mathbb {Z}$, what might be a possible mapping?

I know that bijection exists from $\mathbb {R}^3$ to $\mathbb {R}$.

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kaka
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    Yes. First consider a bijection between $\mathbb{Z}$ and $\mathbb{N}$, say $f:\mathbb{Z}\to\mathbb{N}$ (http://math.stackexchange.com/questions/965460/what-is-an-example-of-function-f-bbbn-to-bbbz-that-is-a-bijection). Then consider a bijection between $\mathbb{N}$ and $\mathbb{N}^2$, say $g:\mathbb{N}^2\to\mathbb{N}$. Then compose these appropriately to get a bijection between $\mathbb{Z}^2$ and $\mathbb{Z}$, and then make some more compositions to make a bijection between $\mathbb{Z}^3$ and $\mathbb{Z}$. – Luiz Cordeiro May 04 '16 at 00:40
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    $f:(a,b)\mapsto \binom{a+b+1}{2}+\binom{a}{1}$ is a bijective map between $\mathbb{N}^2$ and $\mathbb{N}$, as well as $f:(a,b,c)\mapsto \binom{a+b+c+2}{3}+\binom{a+b+1}{2}+\binom{a}{1}$ is a bijective map between $\mathbb{N}^3$ and $\mathbb{N}$. There is little to suitably modify. – Jack D'Aurizio May 04 '16 at 00:48
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    Anyway, consider that https://en.wikipedia.org/wiki/Schr%C3%B6der%E2%80%93Bernstein_theorem gives a shortcut. It is trivial that and injective map $g:\mathbb{Z}\mapsto\mathbb{Z}^3$ exists, a little less trivial that an injective map $h:\mathbb{Z}^3\mapsto\mathbb{Z}$ exists, too, but given that we are done. – Jack D'Aurizio May 04 '16 at 01:02

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You have an injective map $Z\rightarrow N$ defined by $f(n)=2^n, n>0$ and $f(n)=3^{-n}, n\leq 0$, this induces an injective map $g:Z^3\rightarrow N$ defined by $g(a,b,c)=2^{f(a)}3^{f(b)}5^{f(c)}$. The image of $g$ is in bijection with $N$, now take a bijection between $N$ and $Z$ and compose with $g$.

Tsemo Aristide
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