Order preserving bijection from $\mathbb{Q}$ to $\mathbb{Q}\backslash\lbrace{0}\rbrace$

Question

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?

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... — hmakholm left over Monica, Feb 16 '16 at 21:46
You really ought to say something about what you've tried, but have a look at en.wikipedia.org/wiki/Dense_order. — Rob Arthan, Feb 16 '16 at 21:46

Igor Rivin · Accepted Answer · 2016-02-16 21:53:33Z

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answered Feb 16 '16 at 21:53

Igor Rivin

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hmakholm left over Monica · Accepted Answer · 2016-02-16 21:52:03Z

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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