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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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    You’ve basically just replaced the original problem with a new instance of itself. The trick is to make $f$ surjective by well-ordering $\Bbb Q$ as $\{q_n:n\in\Bbb N\}$ and letting $f(a_n)$ be the *first* rational in this ordering that is in the right place relative to the $f(a_k)$ with k.2012-11-08

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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!)