Shapes of initial segements of $\mathbb Q$ Two sets of points in a metric space have the same shape if some bijection between them is a scalar multiple of an isometry.

$\Big[$ By that I mean that for a particular bijection $f$ there is some scalar $c$ such that for
  $\qquad$ all points $x,y$ in the metric space, $d(f(x),f(y)) = cd(x,y).$ In metric spaces
  $\qquad$ in general, that does not mean there is actually some isometry of which $f$ is a
  $\qquad$ scalar multiple, so I chose somewhat unfortunate nomeclature. But in the space
  $\qquad$ $\mathbb Q,$ that is not problematic. $\Big]$ 

An initial segment of the real line with the usual metric has only one of two shapes: it is shaped either like an open interval $(-\infty,c)$ or like a half-open interval $(-\infty,c].$ But initial segments of the rationals come in many shapes: $(-\infty,\sqrt 2)\cap\mathbb Q$ is not shaped like $(-\infty,2)\cap\mathbb Q$ and neither of those is shaped like $(-\infty,\pi)\cap\mathbb Q.$ Describing one of these shapes would require an infinite amount of information. Can one somehow present increasingly larger finite amounts of this information? For example, could each initial segment be assigned a sequence of integers in such a way that too such initial segments have the same such sequence if and only if they have the same shape?
 A: First, let's observe that if $I$ and $J$ are initial segments of $\mathbb{Q}$ which contain maximum elements $a$ and $b$, respectively, then $I$ and $J$ have the same shape, since they are conjugate by the isometry $x\mapsto x-a+b$. So it suffices to consider initial segments of $\mathbb{Q}$ with no maximal element. Every such initial segment has the form $I_r = (-\infty,r)\cap \mathbb{Q}$ for $r\in \mathbb{R}$. 
Now we're considering bijections $f\colon I_r \to I_s$ such that there exists $c$ with $d(f(x),f(y)) = cd(x,y)$ for all $x,y\in I_r$. Such a map must be a rational affine map (of the form $x\mapsto cy + d$ with $c,d\in \mathbb{Q}$), so such a bijection exists iff there are rational $c\neq 0$ and $d$ such that $cr + d = s$. 
So the initial segments of $\mathbb{Q}$ modulo the "same shape" equivalence relation is essentially the same as $\mathbb{R}$ modulo the action of the group of invertible rational affine maps.
As Carl notes in the comments, the set of equivalence classes and the set $\mathbb{N}^\mathbb{N}$ both have size continuum, so they can be put in bijection, enabling you to code an equivalence class by a sequence of natural numbers. But Carl calls this "cheating" - indeed, it's not at all clear how to effectively or definably associate a sequence of natural numbers to an equivalence class. 
These kinds of considerations lead us naturally to the study of Borel equivalence relations. The notion of Borel reduction of Borel equivalence relation is a very flexible formalization of the intuitive idea of assigning invariants to equivalence classes in a "reasonable" way.
In particular, a lot is known about countable Borel equivalence relations (where every equivalence class is countable). Such an equivalence relation is called smooth if it is Borel reducible to the equality relation on $2^\mathbb{N}$ (or equivalently $\mathbb{N}^\mathbb{N}$). That is to say, the smooth Borel equivalence relations are the ones whose classes can be classified by a sequence of natural numbers in a "reasonable" way. 
Here is a very nice introduction to the subject of countable Borel equivalence relations, by Konstantin Slutsky: http://homepages.math.uic.edu/~kslutsky/papers/cber.pdf 
In Proposition 2.1.1, Slutsky shows that the tail equivalence relation $E_t$ on $2^\mathbb{N}$ is not smooth. The definition is that $x E_t y$ iff there exist $k_1,k_2\in \mathbb{N}$ such that $x(k_1+n) = y(k_2+n)$ for all $n\in \mathbb{N}$. This equivalence relation is very similar to the one induced on $\mathbb{R}$ by the group of invertible rational affine maps: thinking of a binary sequence as the base $2$ representation of a real number, changing finitely many bits of the sequence corresponds to adding a dyadic rational, and shifting the sequence left or right corresponds to multiplying by a dyadic rational. This is imprecise, but it makes me suspect that a very similar argument to the proof of 2.1.1 would suffice to show that the "same shape" equivalence relation is not smooth.
