# Any countable set $A$ with a dense total ordering, with neither a maximum nor a minimum, is isomorphic to the rationals.

I think I can do this first: index all elements of $A$ by the natural numbers, make a map $f$, first send $a_0$ to any rational number, call it $q_0$. Then inductively, depending on the ordering of $a_n$ with $a_0, \cdots, a_{n-1}$, make $q_n:=f(a_n)$ any rational so that the ordering of $q_1, \cdots, q_n$ preserves that $a_1, \cdots, a_n$, which I can always achieve since $\mathbb{Q}$ is dense. And if I ask myself what is $f(a_n)$ I can always find out in $n$ time. So $f$ should be well defined.

So $f(A)$ is a dense subset of $\mathbb{Q}$ without a maximum or a minimum. And I think now it suffices to show that any countable dense subset of $\mathbb{Q}$ without a maximum or a minimum is isomorphic to the whole of $\mathbb{Q}$. Any thoughts?

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how about choose the $q_n=p_i$ with minimun index satisfy the order . $\mathbb{Q}=\{p_i\}_{i<\omega}$ – Leitingok Nov 8 '12 at 15:51
For the back and forth method, please see this. – André Nicolas Nov 8 '12 at 16:03
Isomorphic as what? – Marc van Leeuwen Nov 8 '12 at 16:04
And your procedure does not ensure that $f(A)$ is dense (you could choose an interval with irrational end points and never choose any numbers in it). – Marc van Leeuwen Nov 8 '12 at 16:07
@Marc: It does ensure that $f[A]$ is a dense linear order, which is clearly the sense in which Zhang is using the term. – Brian M. Scott Nov 8 '12 at 16:27

Thoughts? Well, suppose things go well with your $f$, so that $f(A)$ is a dense countable set of rationals without endpoints (your construction seems underdescribed). Just why do you think the problem you now end up with is any easier than the problem you started with? It isn't (as far as I know).
Certainly, the standard 'back and forth' method due to Cantor which establishes that $f(A)$ with its natural order is isomorphic to $\mathbb{Q}$ can be used for an equally easy direct argument that $A$ is isomorphic to $\mathbb{Q}$ too. (And indeed, given the potential for accidental confusion in using two sets of rationals, it's probably better to stick to the original problem!)