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I know this question has been asked and answered before, but I am working on my own through an analysis textbook and just wanted to check if the following construction would be appropriate:

Define $a|b=(a+b)/2$ to be the midpoint of interval $(a,b)$ and $a|b^n=(...(a\underbrace{|b)|b)...|b)}_{n\text{ times}}$ (note that $a|b^0=a$). Then define the bijection $f:[a,b)\rightarrow(a,b)$ as follows:$$f(x)=\begin{cases}a|b^{i+1}, & \text{if $x=a|b^i$, for $i=0,1,...$} \\x, & \text{otherwise} \end{cases}$$

So $a$ gets sent to the midpoint of $(a,b)$, which in turn gets sent to the midpoint of itself and $b$, and so on ad infinitum, while the rest of the elements stay fixed. I know this doesn't seem like something worthy of posting a question for, but I am working on my own and struggling with a lot of the exercises, so I am anxious for at least some of my solutions to be correct.

Hilbert Jr.
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  • Enough to get a bijection between $[0,1)$ and $(0,1).$ – little o Apr 19 '19 at 10:12
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    Yes, this is a standard trick: take a countable $A$ of subset of $X=[a,b)$ with $A={a_0=a,a_1,a_2,\ldots}$ and let $f(a_n)=a_{n+1}$ and $f(b)=b$ for $b\notin A$. – Angina Seng Apr 19 '19 at 10:12

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