1

I am looking for a one-to-one function which maps (0,1)^2 to R. It is preferable that the function doesn't involve trig functions. I have tried several mappings like $\ln(\frac{x_2}{1-x_1}),$ but they are not one-to-one. The challenge for me is the one-to-one requirement.

I have read Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$. I like the idea there, but I need to use this function to do further calculation, so it has to be in explicit form. Is it possible to find such a function?

I appreciate any ideas and comments.

Community
  • 1
DEW
  • 89
  • What do you mean use this to do further calculation? Any such function will be discontinuous because you are mixing dimensions. I think the functions you find in the link are the best you will do. – Ross Millikan Jul 24 '16 at 01:22
  • Thank you @RossMillikan. Further calculation means summation/subtraction/derivative, etc. Did you mean there is no such functions can be continuous? – DEW Jul 24 '16 at 02:00

1 Answers1

0

Work with the decimal representation ( we take the finite one if there is ambiguity).

We map $(0.a_1a_2\dots,0.b_1b_2\dots )$ to $0.a_1b_1a_2b_2\dots$

I think this was Cantor's idea.

Asinomás
  • 105,651
  • Thank you @Carry on Smiling. This is also the idea in the link I mentioned. I was just curious if there was other ways to avoid decimals. – DEW Jul 24 '16 at 01:02
  • @Piquito: No, the first goes to $0.11231123\ldots$ and the second to $0.11321132\ldots$. You need to worry about the terminating decimals, but all others work fine. – Ross Millikan Jul 24 '16 at 01:16
  • This does not have a point that maps to $0.09090909090909\ldots $ – Ross Millikan Jul 24 '16 at 01:18
  • Oh God! I understood $a_ib_i$ as the product , not by the position. I apologize. – Piquito Jul 24 '16 at 01:33
  • @RossMillikan: I think (0.00000....,0.99999......)=(0,1) applies. (See what says Carry in cases of ambiguity). Am I wrong? – Piquito Jul 24 '16 at 02:10
  • @Piquito the problem is that the point I mentioned in $\Bbb R$ should go to $(0.0000\ldots, 0.99999\ldots)$ in $\Bbb R^2$, but we said to consider the terminating version in $\Bbb R^2$. Meanwhile $(0.0000\ldots, 1.0000\ldots )$ goes to $1.00000\ldots$ In the answer linked by DEW this problem is solved, but it needs to be solved. – Ross Millikan Jul 24 '16 at 03:42
  • @RossMillikan: I understand. Thank you. – Piquito Jul 24 '16 at 13:10