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  • How can one prove the existence of an order preserving bijection from $\mathbb{Q}$ to $\mathbb{Q}\backslash\lbrace{0}\rbrace$?

  • Can you give an example of such a bijection?

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    $\begingroup$ What is $\mathbb{Q}^*$? Nonzero rationals? $\endgroup$ – Batominovski Feb 16 '16 at 21:38
  • $\begingroup$ Yes. Is $\mathbb{Q}\backslash\lbrace{0}\rbrace$ better? $\endgroup$ – Albert Feb 16 '16 at 21:43
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    $\begingroup$ @Eliott: Yes, much. $\endgroup$ – hmakholm left over Monica Feb 16 '16 at 21:43
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    $\begingroup$ The quickest way to an existence proof is of course to know that all countable dense linear orders without first or last elements are isomorphic... $\endgroup$ – hmakholm left over Monica Feb 16 '16 at 21:46
  • $\begingroup$ You really ought to say something about what you've tried, but have a look at en.wikipedia.org/wiki/Dense_order. $\endgroup$ – Rob Arthan Feb 16 '16 at 21:46
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See the proof of theorem 3.7 here: http://www.math.wustl.edu/~freiwald/ch8.pdf

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Choose an irrational number $\alpha$.

Let $x_1, x_2, \ldots$ be a strictly increasing sequence of rational numbers that converge towards $\alpha$.

Let $y_1, y_2, \ldots$ be a strictly decreasing sequence of rational numbers that converge towards $\alpha$.

Then define $f:\mathbb Q\to\mathbb Q\setminus\{0\}$ as:

  • $f$ maps $(-\infty,x_1]$ to $(-\infty,-1]$ by subtracting $x_1+1$ from everything.
  • For every $n$, $f$ maps $[x_n,x_{n+1}]$ to $[-\frac1n,-\frac1{n+1}]$, by linear interpolation between the endpoints.
  • For every $n$, $f$ maps $[y_{n+1},y_n]$ to $[\frac1{n+1},\frac1n]$, by linear interpolation between the endpoints.
  • $f$ maps $[y_1,\infty)$ to $[1,\infty)$ by subtracting $y_1-1$ from everything.
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