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[TIFR GS-2013, Part D] Does there exist any bijection between $\mathbb R^2$ and the open interval $(0,1)$ ??

At the first glimpse, I thought about the function $f: \mathbb R^2 \to (0,1)$ defined by $f(x,y) = {0.2}^{x}{0.3}^{y}$. But then I realized that the preimage of any element in $(0,1)$ may not be unique. Here I am stuck with finding any example. Any help would be appreciated.

Anik Bhowmick
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The simple answer is yes because both sets have the cardinality of the continuum. You were not asked to supply a bijection. Coming up with an explicit one is hard, but proving one exists is not too bad. First we can biject $\Bbb R^2$ with $(0,1) \times (0,1)$. Then for a point $(x,y)$ in the unit square, express $x$ and $y$ in binary, using the terminating version if the number has two representations. Then form $z$ by alternating the digits of $x$ and $y$. This is not quite surjective because numbers like $0.1y1y1y1y1y1y\ldots$ where the $y$s are a mix of $0$s and $1$s have no preimage, but it is injective. Now note you can inject $(0,1)$ into the square and use Schroeder-Bernstein to assert a bijection.

Ross Millikan
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  • Thanks a ton, but I was wondering if we can construct such function. – Anik Bhowmick Sep 21 '18 at 04:42
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    The above constructs the function in a number of steps. Schroeder-Bernstein is a constructive proof in that it defines the function explicitly. That doesn't mean there is a nice formula for it but if you choose the bijection between $\Bbb R^2$ and $(0,1)^2$ you can (in theory) follow through the proof and find the image of any point $(x,y)$ or the inverse image. I don't think there is a bijection with a nice formula. – Ross Millikan Sep 21 '18 at 04:46