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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.2011-10-23
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    Start with sine or cosine and modify it slightly.2011-10-23
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    try a tangent function?2011-10-23
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    @Gary: You don’t want $\sin^{-1}$: it goes the wrong way.2011-10-23
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    @Brian, right. Thanks.2011-10-23
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    @Alice: That has the same problem as Gary’s suggestion, and the inverse tangent doesn’t pick up the endpoints.2011-10-23
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    Do I understand correctly that you are looking for a surjective function?2011-10-23
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    Notice that a bijection is not possible, though.2011-10-23
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    There is no continuous bijection from $(-\infty,\infty)$ to $[0,1]$, [see this thread](http://math.stackexchange.com/questions/42308/continuous-bijection-from-0-1-to-0-1/42310#42310) 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 $x \mapsto \tan{\pi(x-\frac{1}{2})}$.2011-10-23
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    $\dfrac1{1+e^x}$? $e^{-e^{-x}}$? (These are both bijections from $(-\infty,\infty)$ to $(0,1)$.)2015-05-27

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