I couldn't find a bijective map from the open interval $(0,1)$ to $\mathbb{R}^2$. Is there any example?


Intuitively you can think of something like:

  1. Start with a number $x\in(0,1)$.
  2. Write it down as a decimal fraction, say $0.314159265...$
  3. Take all of the even decimal places and collect them into a new $y\in(0,1)$ (here, $0.1196...$), and all of the odd decimal places and collect them into $z\in(0,1)$ (here $0.34525...$).
  4. Return $(\tan(\pi y-\frac\pi2), \tan(\pi z-\frac\pi2))$.

Making this actually work is more complicated, though, because you need to handle some special cases to prevent $y$ or $z$ from being $0$ or $1$ (such as if $x=0.90909090...$) and to make sure your map is injective (since according to the above description, $x=0.250505050...$ and $x=0.159595959...$ would map to the same $y$ and $z$). Adjustments to deal with this can be written down explicitly, but are not particularly illuminating.

It is slicker to appeal to the Cantor-Bernstein theorem instead, and produce a bijection by combining one injection in each direction. It's easy to inject $(0,1)\to\mathbb R^2$. In the other direction, we can follow the above procedure in reverse:

  • Given $Y, Z$, write down the decimal expansions of $\frac{\arctan(Y)+\pi/2}\pi$ and $\frac{\arctan(Z)+\pi/2}{\pi}$, and interlace their digits to get the decimal expansion of a number in $(0,1)$.

This direction is injective, so Cantor-Bernstein produces a bijection.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.