Is there a bijective map from $(0,1)$ to $\mathbb{R}$?

Question

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

Answer 1

Here is a bijection from $(-\pi/2,\pi/2)$ and $\mathbb{R}$: $$ f(x)=\tan x. $$ You can play with this function and solve your problem.

Answer 2

Here is a nice one ${}{}{}{}{}{}{}{}{}$, can you find the equation?

Answer 3

$g(x)=\frac 1{1+e^x}$ gives a bijection from $\Bbb R$ to $(0,1)$, so take the inverse of this map.

Answer 4

A homeomorphism (continuous bijection with a continuous inverse) would be $f:(0,1)\to\Bbb R$ given by $$f(x)=\frac{2x-1}{x-x^2}.$$

Answer 5

For a less differentiable example, consider the bijection in the following picture,

In symbols, given $x \in (0,1)$ let $n$ be the largest natural number such that $1-\frac{1}{n}<x$, define $$y=\frac{x-n}{\frac{1}{n}-\frac{1}{n+1}}$$ to be the renormalized version of $x$ if the interval $(1-\frac{1}{n},1-\frac{1}{n+1}]$ is rescaled and shifted to map to $(0,1)$. Then we have the following bijection: $$f(x)=\begin{cases}\frac{n-1}{2}+y,& n \text{ odd} \\ -\frac{n-2}{2}-y,& n \text{ even}\end{cases}$$

Answer 6

Yes. let $f(x)=\tan((x-1/2)\pi)$. the domain is $(0,1)$ and range is $\mathbb{R}$

Answer 7

Yes, see above answers. There are even bijective maps between $(0,1)$ and $\mathbb{R}^n$. To see this, note that a bijection $\phi$ between $(0,1)$ and $(0,1)^2$ can be made in this way: Let $x= 0.b_1b_2\ldots$, with $b_j$ being the digits in a decimal expansion. Define $$\phi(x) = (0.b_1b_3b_5\ldots,0.b_2b_4b_6\ldots),$$ i.e., extract even and odd digits. For $\phi^{-1}(x_1,x_2)$, let $x_1 = 0.a_1a_2a_3\ldots$, and $x_2=b_1b_2b_3\ldots$. Then, $$ \phi^{-1}(x,y) = 0.a_1b_1a_2b_2\cdots$$ Some care has to be taken with identification between digital expansions like $0.199999\cdots$ and $0.20000\cdots$, but that is an exercise.

Having the bijection between $(0,1)$ and $(0,1)^2$, we can apply one of the other answers to create a bijection with $\mathbb{R}^2$.

The argument easily generalizes to $\mathbb{R}^n$.

Answer 8

The trigonometric function $\tan x$ is an invertible function from $(-\pi/2,\pi/2)$ to $\mathbb{R}$. Also to find an invertible function from $(0,1)$ to $(-\pi/2,\pi/2)$ find the equation of the straight line joining the points $(0,-\pi/2)$ and $(1,\pi/2)$. Now compose the two functions together. You can likewise find bijections between any two open intervals and any open interval and $\mathbb{R}$.

Answer 9

$x \mapsto \ln (- \ln x)$ with the inverse $y \mapsto e^{-e ^ {\ y}}$. It's also a $C ^ \infty$ diffeomorphism.