I'm completely lost, and I haven't been able to get anything yet. I looked at some that were similar to this problem, and still got pretty lost.
prove there is a bijection between $(0,1)$ and $[0,1)$
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real-analysis
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0One quick example: Map $1/2$ to $0$, then map $3/4$ to $1/2$, $7/8$ to $3/4$, $15/16$ to $7/8$, etc. Everything else gets mapped to itself. – 2017-02-06
1 Answers
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Consider a sequence $\left(\dfrac{1}{n+1}\right)$ and put $f\left(\dfrac{1}{2}\right)=0$, $f\left(\dfrac{1}{3}\right)=\dfrac{1}{2}$ and so on, i.e. $f\left(\dfrac{1}{n+1}\right)=\dfrac{1}{n}$ for $n=2,3\dots$. For the other $x$ put $f(x)=x$.