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Question :
Propose a function $f:(0,1) \to \mathbb R$ such that $f$ is continous on $(0,1)$ and $range(f)=\mathbb R$.

My thoughts :

I've proved that $(0,1)$ has the same cardinality as $\mathbb R$. But in that proof, i never proposed the function which takes something from $(0,1)$ and gives something in $\mathbb R$. I just proved that it exists. That's the problem ... I have no idea about how a function's range can span the whole $\mathbb R$.

Any solutions or even good hints would be nice from you.

Thanks in advance.

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    $\begingroup$ start with $\tan$ and do some modifications $\endgroup$ – user302982 Nov 28 '16 at 14:27
  • $\begingroup$ $\ln$ maps $(0,\infty)\longleftrightarrow \Bbb R$. So, pick something like $\ln\frac{x}{1-x}$. $\endgroup$ – user228113 Nov 28 '16 at 14:31
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Hint: the function $$\tan : \left( -\tfrac{\pi}{2},\tfrac{\pi}{2} \right) \to \mathbb{R} : x \mapsto \tan x$$ is a bijection.

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You need a pole at $0$ and a pole at $1$. The simplest function providing these poles with the correct orientation is $$ -\frac1x+\frac1{1-x} $$

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