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Find a bijection between the set of real numbers and the interval $(−1, 1) ≡ \{\,x ∈ \Bbb R\mid − 1 < x < 1\,\}$.

Hi am I trying to revise for an an exam and I came across this question which I can not figure out. I tried looking online but I kept just finding graphs. Thank You

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You have to find bijection map

$g:\mathbb{R} \to \left( { - 1,1} \right)$

Then, first write the any bijection map from $\mathbb{R}$ to any open interval.

And then make its range (co-domain) by some adjustment to required open interval.

Example:

$f\left( x \right) = {\tan ^{ - 1}}x$

then,

$\begin{gathered} Dom\left( f \right) = \mathbb{R} \\ Range\left( f \right) = \left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \\ \end{gathered} $

then, your job is to arrange your range (co-domain) as $\left( { - 1,1} \right)$

then, for this multiply your function f(x) by $\frac{2}{\pi }$,

$\frac{2}{\pi }f\left( x \right) = \frac{2}{\pi }{\tan ^{ - 1}}x$

then take,

$\begin{gathered} g\left( x \right) = \frac{2}{\pi }f\left( x \right) \\ = \frac{2}{\pi }{\tan ^{ - 1}}x \\ \end{gathered} $

And then,

$\begin{gathered} Dom\left( g \right) = \mathbb{R} \\ Range\left( g \right) = \frac{2}{\pi }\left( { - \frac{\pi }{2},\frac{\pi }{2}} \right) \\ = \left( { - 1,1} \right) \\ \end{gathered} $

Hence, the required bijection map $g:\mathbb{R} \to \left( { - 1,1} \right)$ is,

$g\left( x \right) = \frac{2}{\pi }{\tan ^{ - 1}}x$

I hope this will fulfill you required bijection map.

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How about $x\mapsto \frac x{|x|+1}$?

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  • $\begingroup$ Where did you get that function from? Thanks $\endgroup$ – Emma May 13 '16 at 11:16
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Use $tan^-{1}x$ function to define a bijection from $R$ to the interval$(-\pi/2,\pi/2)$ and then use linear map to construct explicit bijection

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  • $\begingroup$ I will try that. Thank you $\endgroup$ – Emma May 13 '16 at 11:27

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