I am trying to form functions from $[0,1)$ to different sets. first one was continuous surjection from $[0,1)$ to $R$. little modification of sinusoidal exponential function will satisfy the requirement. but for bijection from $[0,1)$ to $R$, I am not able to come up with any function. Is it possible? bijection doesn't have to be continuous.
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I would say that this is not an exact duplicate. The sets are different and it is not a bijection that is asked for. – JezuzStardust Nov 17 '16 at 09:50
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@JezuzStardust Yes, it is, because a bijection from $(0,1) \to \mathbb{R}$ is trivial. We have dozens of these sorts of questions on this site; we don't need more. – MathematicsStudent1122 Nov 17 '16 at 09:51
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Ok, you are right. Sorry. – JezuzStardust Nov 17 '16 at 09:53
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ok. so if I map $[0,1)$ to $(0,1)$ as $0$ to $1/2$, $1/n$ to $1/(n+1)$, where n is natural numbers greater than 1, and identity map for other numbers, i would have a bijection from $[0,1)$ to $(0,1)$ and using arc tan function I could complete the mapping. right? – jnyan Nov 17 '16 at 10:54