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What is a continuous mapping of the real line $(-\infty, \infty)$ to the interval $[0, 1]$? I'm trying out logs and exponentials but they don't seem to work?

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Seems like $sin^{-1}$ composed with absolute value may work. – gary Oct 23 '11 at 12:14
Start with sine or cosine and modify it slightly. – Brian M. Scott Oct 23 '11 at 12:14
try a tangent function? – Alice Oct 23 '11 at 12:14
Notice that a bijection is not possible, though. – gary Oct 23 '11 at 12:28
There is no continuous bijection from $(-\infty,\infty)$ to $[0,1]$, see this thread where it is shown that there is no continuous bijection from $(0,1)$ to $[0,1]$ and notice that $x \mapsto \frac{1}{\pi}\arctan{x} + \frac{1}{2}$ maps $(-\infty,\infty)$ bijectively to $(0,1)$ with continuous inverse $x \mapsto \tan{\pi(x-\frac{1}{2})}$. – t.b. Oct 23 '11 at 12:40

I take it you want a continuous surjection from $\mathbb{R}$ to $[0,1]$. A simple example of such a function is $f(x)= \begin{cases} 0 & x<0 \\ x & 0 \le x \le 1 \\ 1 & x>1\end{cases}$

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AKA $\frac{|x|-|x-1|+1}{2}$ :-) – robjohn Oct 23 '11 at 12:38

$\frac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}}$ and $\frac{\tanh(x)+1}{2}$ both map $\mathbb{R}$ into $[0,1]$ but not onto. As Phira mentions $\frac{\sin(x)+1}{2}$ maps $\mathbb{R}$ onto $[0,1]$ but is not $1{-}1$. However, you won't find a $1{-}1$ and onto function whose inverse is continuous, since $[0,1]$ is compact and $\mathbb{R}$ is not.

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One can probably replace the hyperbolic tangent in the second example with the whole gamut of "sigmoidal functions": the arctangent, the logistic function, the error function... :) – J. M. Oct 23 '11 at 12:49
@J.M.: and even the first example :-) – robjohn Oct 23 '11 at 12:55

$\dfrac{\sin x +1}2$ is a continuous surjective function from the reals to this interval.

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All you need is a continuous function that maps $\mathbb{R}$ onto a closed interval; once you have that, adjusting the interval is fairly easy. (For instance, you can use a linear function.) The natural starting points are the sine and cosine functions.

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$\dfrac{1}{1+e^{-x}}$ is a one-to-one monotonically increasing map from $\mathbb R$ onto the open interval $(0,1)$.

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The question was about the closed interval, not the open one. – Julian Kuelshammer Nov 4 '12 at 12:25

As the question is formulated, any function that would map all number of the real line to a fixed number of $[0, 1]$ would do. A Lo's (down voted by Julian) answer is also a correct answer, since mapping to $(0,1)$, you map to $[0,1]$.

But what user18723 probably meant, but didn't specify, is to have more than just a continuous function. Perhaps the intent was to have a bijective function? But in that case, it's not possible. A good discussion on this situation can be found here.

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I am quite amazed that you know who's the downvoter. – Arctic Char May 27 '15 at 9:45
It's not true that $[0,1] \subset (0,1)$ – jkabrg May 27 '15 at 9:49

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