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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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    how about choose the $q_n=p_i$ with minimun index satisfy the order . $\mathbb{Q}=\{p_i\}_{i<\omega}$2012-11-08
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    For the back and forth method, please [see this.](http://en.wikipedia.org/wiki/Back-and-forth_method)2012-11-08
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    Isomorphic as what?2012-11-08
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    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).2012-11-08
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    @Marc: It does ensure that $f[A]$ is a dense linear order, which is clearly the sense in which Zhang is using the term.2012-11-08
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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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