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This question already has an answer here:

I know a continuous bijection from $[0,1)$ to $\mathbb R$ cannot exists but what happens if we lift the restriction of continuous $?$

Can there exists a bijection , not necessarily continuous from $[0,1)$ to $\mathbb R ?$

$(0,1)$ is bijective with $\mathbb R.$ Although I doubt that would be any useful here.

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marked as duplicate by user228113, Vlad, Watson, zz20s, JMP Apr 27 '16 at 13:18

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    $\begingroup$ $(0,1)$ being bijective to $\Bbb R$ is very useful if you know how to construct a bijection from $[0, 1)$ to $(0,1)$. $\endgroup$ – Arthur Apr 27 '16 at 8:30
  • $\begingroup$ @Arthur : I don't. help please? $\endgroup$ – user118494 Apr 27 '16 at 8:32
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    $\begingroup$ You can construct a bijection between $[0,1)$ and $(0,1)$ in the same way you construct a bijection between $\Bbb N$ and $\Bbb N\setminus\{0\}$. $\endgroup$ – user228113 Apr 27 '16 at 8:33
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    $\begingroup$ Possible duplicate of How to define a bijection between $(0,1)$ and $(0,1]$?. And this, and this. $\endgroup$ – user228113 Apr 27 '16 at 8:35
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    $\begingroup$ If you instead of writing the formula for that bijection, think about what it does: It takes $0$, and moves it to $1$. To make "room" for that, it has to take $1$ and move it to $2$. And so on. In the same way, for the bijection from $[0,1)$ to $(0,1)$ you take $0$ and move it somewhere, say $1/2$. To make "room" for that, you take $1/2$ and move it somewhere, say $1/3$. And so on. All other numbers you leave untouched. $\endgroup$ – Arthur Apr 27 '16 at 8:38
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Consider the function

$f:[0,1) \to (0,1)$

$f(x):= \begin{cases} \frac{1}{2} & \text{ if } x = 0 \\ \frac{1}{2^{n+1}} & \text{ if }x=\frac{1}{2^n} \text{ for some } n\in \mathbb N \\ x & \text{ else} \end{cases}$

This is obviously a bijection.

Then consider $g(x) = \tan(\pi (x-\frac{1}{2}))$ where $g:(0,1) \to \mathbb R$. Then $g$ is obviously a bijection too. Then

$$h(x) := (g \circ f)(x)$$ is your desired bijection between $[0,1)$ and $\mathbb R$.

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  • $\begingroup$ Damn dollar signs always intruding. $\endgroup$ – B. Pasternak Apr 27 '16 at 8:43
  • $\begingroup$ @B.Pasternak Haha Americanization everywhere=) Lets petition a change from $ to . $\endgroup$ – flawr Apr 27 '16 at 8:44
  • $\begingroup$ Count me in. Europe might be done soon anyways! $\endgroup$ – B. Pasternak Apr 27 '16 at 8:45
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here is a way to do it :

Step one : take the set $\{0;1/2;3/4;7/8;15/16;...;1-1/2^n;...\}$ and map it (in increasing order, for example) to $\{0;1;2;3;4;...\}$

Step two : make a bijection between the remaining sets, that is make a bijection between $(0;1/2)$ and $(-\infty;0)$, a bijection between $(1/2;3/4)$ and $(0;1)$, a bijection between $(3/4;7/8)$ and $(1;2)$, and so on.

And it's done.

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