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I'm having trouble envisioning a bijective relationship that maps $(0,1)$ to $(0,1]$.

My professor gave the hint that it can be expressed as a piece-wise function $f(x)$ comprising of two cases: _ if $x=$ __ and $x$ otherwise.

It could be $1$ if $x=\frac{1}{2}$ and $x$ if $0<x<\frac{1}{2}$ or $\frac{1}{2}<x<1$.

However, $f$ would not be a bijection then, because there would not be an $x$ value for $y=\frac{1}{2}$.

Joel Christophel
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2 Answers2

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Hint: You've got the right idea with $1=f(\frac12)$. Now make $\frac12=f(\frac14)$, $\frac14=f(\frac18)$, etc. So $f(x)=2x$ for some values of $x$, and $f(x)=x$ otherwise.

bof
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    I understand your first sentence. But I don't understand "$f(x) = 2x$ for some values of $x$." Isn't "some values of $x$" ambiguous? – Joel Christophel Oct 08 '14 at 05:12
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    Your professor's hint "$f(x)=$_ if $x=$_ and $x$ otherwise" is ambiguous. I filled in the first blank for you, and hinted at how you should fill in the other blank. Did you want somebody to write a complete solution that you can copy verbatim and hand in? – bof Oct 08 '14 at 05:27
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    Of course I wanted that. But it is fine that you did not provide that. I just thought you were attempting to provide it. Simple miscommunication :) – Joel Christophel Oct 08 '14 at 05:35
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    Of course you wanted a complete solution. How silly of me to think you wanted the least possible hint to get you unstuck so you could figure out the rest yourself. 8-( – bof Oct 08 '14 at 05:41
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    Funny indeed. So I'm feeling the missing predicate is $x=\frac{1}{n}$ for $n\in\mathbb{N}$ because when $x = \frac{1}{2}$, $y$ should be $2(\frac{1}{2}) = 1$, etc. – Joel Christophel Oct 08 '14 at 05:43
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    Revision to the above: $2x$ if $x=\frac{1}{n}\wedge2|n\wedge n\in\mathbb{N}$. Will you confirm if I get it right? – Joel Christophel Oct 08 '14 at 06:00
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    ${\frac12,\frac14,\frac18,\frac1{16},\dots}={\frac1{2^n}:n\in\mathbb N}$. (I thought $\frac12,\frac14,\frac18$ would be enough of a hint.) Of course there are a zillion other ways to do this, e.g., see the other answer. – bof Oct 08 '14 at 06:43
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    Yeah, but $\frac{1}{n}\wedge2|n\wedge n\in\mathbb{N}\Leftrightarrow\frac{1}{2^{n}}:n\in\mathbb{N}$, silly – Joel Christophel Oct 08 '14 at 13:59
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    ${\frac12,\frac14,\frac16,\frac18,\frac1{10},\frac1{12},\dots} \ne {\frac12,\frac14,\frac18,\frac1{16},\frac1{32},\dots}$ – bof Oct 08 '14 at 15:07
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The trick is 'shift' a countably infinite subset of $(0,1)$. Consider

$$ f(x) = \left\{\begin{array}{cc} \frac{1}{n-1} & \textrm{ if } x = \frac{1}{n} \textrm{ and } n \in \mathbb{N} \\ x & \textrm{ otherwise } \end{array}\right. $$

this map is bijective.

Bruce Zheng
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