0

I would like to map $\mathbb{R}$ onto $[0, 1)$. Is there a function similar to the sigmoid (shapewise) that is continuous and has range onto $[0, 1)$? Unfortunately the sigmoid doesn't include the $0$ i.e. $\sigma:\mathbb{R}\to(0, 1)$

Euler_Salter
  • 5,153
  • 1
    There is are surjective continuous functions from $\mathbb R$ onto $[0,1)$ and there are also bijective functions, but not a bijective continuous function. The problem is that all real numbers have neighbourhoods going above and below, but $0$ do not have a neighbourhood going below in $[0,1)$ – Henry Feb 10 '22 at 15:30
  • So for example $f(x) =1-e^{-x^2}$ is a surjective continuous function from $\mathbb R$ onto $[0,1)$ with $f(0)=0$ but $f(1)=f(-1) \approx 0.6321$ shows this is not bijective – Henry Feb 10 '22 at 15:36
  • 1
    You might be interested in the smoothstep functions. See also this post. – Ben Grossmann Feb 10 '22 at 19:01

0 Answers0