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

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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.

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