I thought I could use $x = \tan(\pi x/2)$, but this only works for $x: = (0,1)$. How should I address the 0? I saw somewhere else's to find a function for $x = 1/n$ (eg ($f(x) = 1/(n+1)$), but I did not understand this.
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Hi there, I suppose you mean $f(x)=\tan(\pi x/2)$, if so, notice that $f(0)=0$. – Cristian Baeza Sep 05 '20 at 20:19
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@CristianBaeza $0$ isn't positive. – Stephen Montgomery-Smith Sep 05 '20 at 20:31
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A common idea is to pick out a tiny (countable) subset of the domain and codomain and "let them deal with the troublesome endpoint", while making the rest of the function something simple. The many answers to this question illustrate that idea in several ways. – Greg Martin Sep 05 '20 at 20:41
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1Those were different days. A question like that wouldn't attract about 500 likes today. Aram2 and I will be lucky to get 3 likes total. But I do notice that none of their answers are the same as mine. https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1 – Stephen Montgomery-Smith Sep 05 '20 at 20:52
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ups!! my bad... – Cristian Baeza Sep 05 '20 at 21:06
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Write the positive numbers and $[0,1)$ as $$(0,\infty) = \bigcup_{n=0}^\infty (n,n+1], \quad [0,1) = \bigcup_{n=0}^\infty [1-\tfrac1{n+1},1-\tfrac1{n+2}) .$$ Let $f$ restricted to $(n,n+1]$ be the linear function that takes $n$ to $1-\frac1{n+2}$, and $n+1$ to $1-\frac1{n+1}$.
Stephen Montgomery-Smith
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