# Does my function exist?

Is it possible to define a bijection $f:\mathbb{R}\setminus\mathbb{R}^{-}\rightarrow[0,1)$ such that $f$ is continuously differentiable on its entire domain?

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Please tell us what the strange ${\mathbb R}\setminus{\mathbb R}^-$ means: ${\mathbb R}_{>0}$ or ${\mathbb R}_{\geq0}$ or something else? – Christian Blatter Mar 11 '11 at 16:28
It means the set of all nonnegative real numbers. (see: en.wikipedia.org/wiki/Set_subtraction#Relative_complement) – user8116 Mar 11 '11 at 17:16

Yes. For example, $f(x)=1-\frac{1}{1+x}$.
If your domain is the reals greater than or equal to zero, $1-\exp(-x)$ seems to fill the bill.
Let us consider $f\,:\,\left[0;+\infty\right[\rightarrow\left[0;1\right[$ defined as: $$f = \frac{2}{\pi} \arctan$$