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I know there exists one, but I want to find the explicit form of some bijective function between $(0,1)$ and $(0,1)\times(0,1)$.

Michael Hardy
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David Malagón
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  • Look here: http://wstein.org/edu/Spring2002/21b/handouts/Math21b3.pdf

    By the way, would there be some function which construction (Or finding) does not rely on some power series expansions with some basis?

    (I edit this since I´d said decimal expansions, and there are binary, too)

     – LeviathanTheEsper Feb 21 '16 at 03:51
  • @LeviathanTheEsper Sure, you have all those that rely on binary expansion. – Clement C. Feb 21 '16 at 03:52
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    More seriously: see this question (in particular for the detailed construction involving binary expansion, with the fun subtleties that arise), for the similar question of $[0,1]^2 \to [0,1]$. – Clement C. Feb 21 '16 at 03:58
  • Very interesting, thanks. – LeviathanTheEsper Feb 21 '16 at 04:01
  • What is (0,1)x (0,1)? A single point? A set of four elements? – fleablood Feb 21 '16 at 07:56
  • @fleablood: $(0,1)\times(0,1):={(x,y)\mid 0<x<1,0<y<1}\subset\mathbb{R}^2$ – Xaver Feb 21 '16 at 13:01
  • Oh, for the love of... okay, I have no idea why the absolute obvious, clear, and straightforward meaning escaped me. I got caught up on (0,1) being an ordered pair ... or being the set {0,1}. D'oh! – fleablood Feb 21 '16 at 18:12

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