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Does there exist a function from $(0,1)$ to $(0,1]$ both one-to-one and onto, not necessarily continuous?

I couldn't think of any. Any help would be appreciated!

Thanks,

Hendrix
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scsnm
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    https://math.stackexchange.com/questions/1965089/find-a-bijection-between-2-intervals, https://math.stackexchange.com/questions/160738/how-to-define-a-bijection-between-0-1-and-0-1 – Hans Lundmark Aug 22 '19 at 14:18

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Well, take the series $(\frac{1}{n})_{n\geq 1}$ and cut this series out of the set $(0,1] = (\frac{1}{n})_{n\geq 1} \cup A$, where the union is disjoint.

Then define the mapping $f:(0,1]\rightarrow (0,1)$ as follows: $$f(x) =\left\{\begin{array}{ll} \frac{1}{n+1} & \mbox{if } x=\frac{1}{n}\\ x & \mbox{if } x\in A. \end{array} \right.$$ So the mapping maps the elements of $A$ to themselves, and the elements $1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots$ to $\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots$. This mapping is bijective.

Wuestenfux
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