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Find a bijection from $ (3,4) $ to $ (7,8) - \{ 7+\frac{1}{2}, 7+\frac{1}{3}, 7+\frac{1}{4} ,\cdots \}$

Any ideas? I know it may be easy but I can't think anything.

Thank you!

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1 Answer 1

up vote 4 down vote accepted

Without loss of generality, suppose the first interval is also $(7,8)$. (you can just apply the function $x+4$ to go inbetween the two.) If we tried the function $f(x)=x$, that would give a bijection between $(7,8)$ and $(7,8)$. The problem is, how do we avoid the points $7+\frac{1}{2}$ etc..?

Consider the points of the form $7+\frac{n-1}{n}$ for $n=3,4,\dots$. Lets change the function $f(x)=x$ only on the numbers of the form $7+\frac{n-1}{n}$ or $7+\frac{1}{n}$. Send $7+\frac{1}{2}$ to $7+\frac{2}{3}$, and $7+\frac{2}{3}$ to $7+\frac{3}{4}$. Then send $7+\frac{1}{3}$ to $7+\frac{4}{5}$ and $7+\frac{3}{4}$ to $7+\frac{5}{6}$. Continuing in this way, send $7+\frac{1}{n}$ to $7+\frac{2n-2}{2n-1}$ and $7+\frac{n-1}{n}$ to $7+\frac{2n-3}{2n-2}$. This gives the bijection.

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Thank you for your time! –  passenger Feb 10 '12 at 11:55

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