It is well-known that the set of rational numbers $\mathrm Q$ is dense; that is, given any two rationals, say $r\ne s,$ then there exist infinitely many rationals $r_i$ such that $r<r_i<s$ for every integer $i.$
Obviously, this property depends on the usual ordering of the rationals; this can also be seen by using Cantor's famous ordering in order to show the equinumerousity of the integers $\mathrm Z$ and $\mathrm Q,$ beginning $$\frac 01,\frac 11,\frac 12,\frac 21,\frac 13,\frac 31,\cdots.$$ Clearly, each rational is in this sequence, so that there is no rational between each consecutive pair; the first two, say.
My question is about the possibility of a reverse process. Is there a way to order the integers so that they become dense? Clearly one will have to design an ordering different from the usual one to make this happen, were it possible. But I don't even have a hint where to start, or whether it is in fact possible.
Thank you.
Use your favorite bijection $b:\mathbb Z\to\mathbb Q$ to define an ordering $\prec$ of $\mathbb Z$ by setting $m\prec n$ if and only if $b(m)<b(n)$. This ordering $\prec$ of $\mathbb Z$ is dense because the standard ordering $<$ of $\mathbb Q$ is dense.