I am trying to prove the proposition indicated in the title. Will it work if i define $h:(0,1)\to[0,1]$ as follows $$h(x) = \begin{cases}x\text{ if }x\in(0,\frac{1}{4})\cup(\frac{1}{2},\frac{1}{4})\cup(\frac{1}{2},1)\\0\text{ if }x = \frac{1}{4}\\1\text{ if }x = \frac{1}{2}\end{cases}$$
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1No. You have no $x\in(0,1)$ with $h(x)=\frac12$ – Hagen von Eitzen Oct 13 '18 at 20:21
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1Man, I gave you a function that gives you a very similar bijection. Use a technique similar. – Rushabh Mehta Oct 13 '18 at 20:23
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Hint: The following trick goes half the way, namely gives a bijection $(0,1)\to (0,1]$:
$$h(x)=\begin{cases}\frac1{\frac1x-1}&\text{if }\frac1x\in\Bbb N\\x&\text{otherwise}\end{cases} $$
Hagen von Eitzen
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