Explicit map from countable subset of $[0,1]\subset\mathbb{R}$ to $[0,1]\subset\mathbb{Q}$ Let $A\subsetneq[0,1]$ be some countable set of real numbers. Since the rationals are dense in the reals and since all countable linear orders are embeddable into $\mathbb{Q}$, it seems to me that there has to be a map $f:A\rightarrow[0,1]_\mathbb{Q}$ with $[0,1]_\mathbb{Q}$ being the rational interval such that for any $a\in A$:
$$a\leq a'\text{ iff }f(a)\leq f(a')$$
Is it possible to construct $f$ explicitly, without resorting to the axiom of choice?

$f$ does not have to be onto.
 A: From my limited understanding of set theory, you can do this with induction without using any variant of the axiom of choice at all.
Since $A$ is countable, you are assuming there exists some enumeration of its elements $\{a_n\}_{n\in\mathbb{N}}$.  Start with the empty map mapping the empty subset of $A$ to an empty subset of $\mathbb{Q}\cap [0,1]$.  The induction step is assuming you already have a mapping of $a_1,...,a_k$ to some rationals in this interval $b_1, ..., b_k$ that preserves the linear order.  To perform the induction step you need to choose a rational to send $a_{k+1}$ to.  Ok.  There's a unique pair $a_s, a_t$ such that $s, t \le k$, $a_s$ and $a_t$ are adjacent in the linear order and $a_{k+1}$ is between them.  So Let $b_{k+1} = (b_s+b_t)/2$.  You need to handle the case where $a_{k+1}$ is to the left, or right of all $\{a_1, ..., a_k\}$ and make sure you don't go beyond the interval $[0,1]$.  I'll let you complete the detail, but the point is there is no use of the axiom of choice because you're giving an algorithm how to choose $b_{k+1}$.
To finish the proof, you take the union of all finite partial maps you constructed this way to get a complete mapping of $\{a_n\}_{n\in\mathbb{N}}$ to $\{b_n\}_{n\in\mathbb{N}}$.
A: I think it is not possible, in general, since $0$ and $1$ are rational numbers in $[0,1]$ which are smaller (resp. larger then all others). you will not have this numbers for $A$ in general.
