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Are there any one-to-one functions of 2 variables? For each of the following prove or disprove whether there is a one-to-one function $f$ of 2 variables:

  1. $f$ is from $\Bbb{N}^2$ to $\Bbb{N}$
  2. $f$ is form $\Bbb Z^2$ to $\Bbb Z$
  3. $f$ is from $\Bbb R^2$ to $\Bbb R$
  4. $f$ is from $\Bbb Q^2$ to $\Bbb Q$
  5. Any other maps that may be more interesting like $\Bbb R^2$ to $\Bbb N$ or something.

I don't feel like thinking much right now because I'm quite tired, so I apologize if the question is obvious.

Edit: I must have been out of my mind earlier. By function I meant to say polynomial.

math_lover
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  • $f(m,n)= m$ is not one-to-one because $f(0,0) = f(0,1)$ – peterwhy Dec 23 '14 at 02:38
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    All of these exist. For 1,2,4 you can map the main set to $\mathbb{N}$ and then use the "diagonal" injection from $\mathbb{N}^2$ to $\mathbb{N}$. This basically proceeds by drawing the 2D array of $\mathbb{N}^2$ and traversing it along each diagonal. So it starts at $(1,1)$, then goes through the diagonal $(1,2)$ and $(2,1)$, then the diagonal $(3,1)$, $(2,2)$, $(1,3)$, etc. The idea is similar but slightly more complicated for 3. – Ian Dec 23 '14 at 02:39
  • @Strants: I've only ever seen "1-1" mean injective. I've seen bijective as "1-1 correspondence". – Sujaan Kunalan Dec 23 '14 at 03:11
  • @SujaanKunalan Fair point. I admit, I only remember hearing 1-to-1 meaning bijection when I was started to read about more advanced math in high school, and I may have misinterpreted. Comment retracted. –  Dec 23 '14 at 03:15
  • http://math.stackexchange.com/q/588468/73324 – vadim123 Dec 23 '14 at 05:10
  • https://en.wikipedia.org/wiki/Pairing_function#Cantor_pairing_function – vadim123 Dec 23 '14 at 05:12
  • http://math.stackexchange.com/q/1073961/73324 – vadim123 Dec 23 '14 at 05:13

1 Answers1

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Yes there are, since , e.g., $\mathbb R, \mathbb R^N; N>1$ have the same cardinality. This is equivalent to saying that there is not just an injection, but a bijection between them.

RikOsuave
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