Does the following bijection work:

Take any point $(x,y) \in (0,1) \times (0,1).$ Each real number $r \in (0,1)$ may be represented by an infinitely-long decimal expansion (0.235, for example, is the same as 0.234999999...). Take the real numbers $x,y \in \mathbb{R}$ and interlace their decimal expansions to produce a unique real number $r' \in (0,1).$ The number $r'$ being unique, the mapping is 1-1. Given a real number $r' \in (0,1),$ one can unlace the decimal expansion of that number according to the pattern set by the mapping and produce the real numbers $x$ and $y$, and therefore arrive at a unique point $(x,y) \in (0,1) \times (0,1).$

Does a bijection exist?

  • $\begingroup$ Not quite. But if $C$ is the middle-thirds Cantor set you can do something just like this to give a bijection between $C^2$ and $C$. $\endgroup$ – David C. Ullrich May 31 '18 at 17:51

Not quite. The correspondence between infinite decimals and elements of $(0,1)$ is not itself a bijection because some numbers have more than one decimal expansion.

You solve that by restricting to the expansion with infinitely many non-zero digits, fine. But now the mapping from $(0,1)^2$ to $(0,1)$ is not surjective. For example there are no $x$ and $y$ that give $r'=0.1101010101...$. Because that would require $x=0.10000...$ and $y=0.111...$. But $x=0.1000...=0.09999...$, so the image of that $(x,y)$ is actually $0.0191919...\ne0.11010101...$.

  • $\begingroup$ Hey David. Thanks for you response! I made two corrections: one the book only required that the mapping be injective. Moreover, I did try to provide an example of a bijection at the bottom of the post. Would you mind glancing over it? It is two or three sentences long! Thank you! $\endgroup$ – Rafael Vergnaud May 31 '18 at 19:31
  • $\begingroup$ That map defined by those last few sentences is obviously not injective. If the problem is to show that there's a bijection from $(0,1)^2$ onto $(0,1)$ I think the easiest thing is this: Say $C$ is the Cantor set. Show that there is a bijection from $[0,1]$ onto $C$, and uuse a ternary digit interlacing to give a bijection between $C^2$ and $C$. $\endgroup$ – David C. Ullrich May 31 '18 at 19:38

You have an injection $(0,1)^2\to(0,1)$ -- but it is not surjective, because there is nothing that maps to, for example, $$ \frac{1}{99} = 0.0101010101010\ldots $$

De-interlating the digits of this would produce $\langle 0,\frac19\rangle$, but that is not in $(0,1)^2$.

However, an injection is really all you need, because it is easy to find an injection in the other direction, and then the Schröder-Bernstein theorem does the work of stitching them together to a single bijection for you.

  • $\begingroup$ Thank you for catching my mistake! $\endgroup$ – Rafael Vergnaud May 31 '18 at 17:42
  • $\begingroup$ The map isn't even well defined, in fact. $\endgroup$ – Noah Schweber May 31 '18 at 17:43
  • 2
    $\begingroup$ @NoahSchweber: I think the OP's specification of "an infinitely-long decimal expansion" was meant to say that when you have a choice you choose the representation that ends in infinite nines rather than the one that ends in infinite zeroes. $\endgroup$ – Henning Makholm May 31 '18 at 17:45
  • $\begingroup$ Ah, I didn't read it that way. @RafaelVergnaud Was that what you intended? If so, I'll delete my answer. $\endgroup$ – Noah Schweber May 31 '18 at 17:48
  • $\begingroup$ Hey Noah. I did not see your comment earlier! Sorry for the later response Infinite nines! $\endgroup$ – Rafael Vergnaud May 31 '18 at 19:26

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