What is a metric for $\mathbb Q$ in the lower limit topology? A useful source of counterexamples in topology is $\mathbb R_\ell$, the set $\mathbb R$ together with the lower limit topology generated by half-open intervals of the form $[a,b)$. For example this space is separable, Lindelöf and first countable, but not second countable. So in particular, this implies that $\mathbb R_\ell$ is not metrizable.
Now Urysohn's metrization theorem says that any regular, second countable space is metrizable. Applying this to $\mathbb Q_\ell\subset \mathbb R_\ell$ gives the (for me rather counterintuitive) fact that there must exist a metric on $\mathbb Q$ inducing the lower limit topology. Which leads to the question:

What is a concrete example of a metric inducing the lower limit topology on $\mathbb Q_\ell$?

It certainly should be possible to go through the steps of the Urysohn metrization theorem to give a (semi-)concrete embedding of $\mathbb Q_\ell$ in $\mathbb R^\omega$ (for instance picking an enumeration of the rationals and constructing functions which separate points from closed sets, then look at the cartesian product of these maps) and then give the metric on $\mathbb Q_\ell$ in terms of this embedding.
But this is really not what I'm looking for, here. I'd be much more interested in a simple metric for $\mathbb Q_\ell$. 
Because "simple" is not well-defined, I will definitely also welcome answers which exhibit a metric, but which I would not consider to be simple. On the other hand, if you see a reason for why there may just be no "simple" metrics, please feel free to point this out as well.
Thanks!
 A: EDIT: This one should work out.
Here's another option: write your rationals as "mixed fractions," that is as their integer floor plus a fractional part and define 
$$d\left(a\frac{p}{q},b\frac{r}{s}\right)=\begin{cases}|a-b|,& a\neq b\\ d'\left(\frac{p}{q},\frac{r}{s}\right), &\textrm{otherwise} \end{cases}.$$
Define the distance between pure fractions separately. Assume WLOG that $\frac{p}{q}\leq \frac{r}{s}$. Then set
$$d'\left(\frac{p}{q},\frac{r}{s}\right)=\max\left(\left|\frac{p}{q}-\frac{r}{s}\right|,\frac{1}{m}\right), \frac{k}{m} \in \left(\frac{p}{q},\frac{r}{s}\right]$$
We see, for instance, that this gives $\left[\frac{1}{2},\frac{5}{6}\right)$ open, since the distance from anything smaller than $\frac{1}{2}$ to $\frac{1}{2}$ is $\frac{1}{2},$ but each distance from $\frac{1}{2}$ to something less than $\frac{5}{6}$ is no more than $\frac{1}{3}$. 
The symmetry and homogeneity axioms are immediate. Let's consider the triangle inequality. The only case in which it doesn't follow from that for the standard absolute metric is when we must show $d'\left(\frac{p}{q},\frac{r}{s}\right)+d'\left(\frac{r}{s},\frac{t}{u}\right)\geq d'\left(\frac{p}{q},\frac{t}{u}\right)$ and $d'\left(\frac{p}{q},\frac{t}{u}\right)=\frac{1}{m}$. But if the inequality failed, we'd have to have the interval $\left(\frac{p}{q},\frac{t}{u}\right]$ contain something with denominator $m$ bigger than any denominator in the intervals on the left-hand side-which is absurd, since depending on the ordering we have one of $ \left(\frac{p}{q},\frac{t}{u}\right]=\left(\frac{p}{q},\frac{r}{s}\right]\cup\left(\frac{r}{s},\frac{t}{u}\right], \left(\frac{p}{q},\frac{t}{u}\right]\subset \left(\frac{p}{q},\frac{r}{s}\right],$ or $\left(\frac{p}{q},\frac{t}{u}\right]\subset\left(\frac{r}{s},\frac{t}{u}\right]$.
So we have a metric. To get some half-open interval $[x,y)=\left[a\frac{p}{q},b\frac{r}{s}\right)$, take the union of all the $[m,m+1) \subset [a,b)$. Then for the least such $m$, construct $[x,m)$ by the infinite union $\bigcup_{i=0}^\infty\left[a\frac{p}{q+i},a\frac{p}{q+i}+\frac{1}{q+i+1}\right)$, getting all these intervals by the argument above about $\left[\frac{1}{2},\frac{2}{3}\right)$. Get $\left[b,b\frac{r}{s}\right)$ as the ball of radius $\frac{r}{s}$ around $b$, union the three pieces together, and we're done.
A: Let $\nu:\mathbb{Q} \to \mathbb{N}$ be an enumeration of $\mathbb{Q}$. Then
$\displaystyle d(x,y):=\sum_{\min(x,y) < r \le \max(x,y)} 2^{-\nu(r)}$ will do.
A: To see how the metric from Alexander Shamov's post comes up while following the steps of Urysohn's theorem, the key is to observe that for any enumeration $\{q_n\}_{n \in \mathbb{N}}$ of the rationals $\mathbb{Q}$, the countable family of continuous function $$f_n(x) := \begin{cases} 1 & \text{ if } x \ge q_n \\ 0 & \text{ otherwise }\end{cases}$$ (continuous in $\displaystyle \mathbb{Q}_{\ell}$ topology) separates disjoint basic clopen neighborhoods. Hence instead of trying to embed it into $\mathbb{R}^\omega$ (as in the proof of Urysohn's lemma) we might as well do it in the more simpler Cantor space $\{0,1\}^{\omega}$.
The evaluation map $\displaystyle f: \mathbb{Q}_{\ell} \to \{0,1\}^{\omega}$ given by $\displaystyle f(x) = (f_n(x))_{n \in \mathbb{N}}$ is an imbedding into the Cantor space $\displaystyle \{0,1\}^{\omega}$ with product topology (as it separates points and closed sets).
Thus the pullback of the metric on Cantor space v.i.a. $f$-imbedding gives us the explicit metric for $\mathbb{Q}_{\ell}$: $$d(x,y) = \sum\limits_{n=1}^{\infty} \frac{|f_n(x) - f_n(y)|}{2^n} = \begin{cases} \sum\limits_{x < q_n \le y} \dfrac{1}{2^n} & \text{ when } x < y \\ 0 & \text{ when } x = y.\end{cases}$$
Interestingly, if we were to re-interpret the $\mathbb{Q}_\ell$ topology (containing the standard subspace topology of $\mathbb{Q}$ inherited from $\mathbb{R}$) as the minimal topology that makes the following function $$g(x) = \sum\limits_{q_n \le x} \frac{1}{2^n}$$ continuous, then Kevin Carlson's construction of the metric seems to be modeled on the discontinuities of the Thomae's function. In particular let us choose the separating family of functions $(f_r)_{r \in \mathbb{Q}}$ to be $$f_r(x) := \begin{cases} 1/q & \text{ if } x \ge r \\ 0 & \text{otherwise}
\end{cases}$$ where, $r = p/q \in \mathbb{Q}$ with $(p,q) = 1$. For a fixed $r \in \mathbb{Q}$ the metric $$d_r(x,y) := \begin{cases} |x-y| & \text{ if } x,y \ge r \text{ or } x,y \le r \\ \max\left(|x-y|, 1/q\right) & \text{ if } x < r \le y\end{cases}$$ precisely makes $f_r$ continuous. Modifying the above metric to incorporate jumps at all $r \in \mathbb{Q}$ which has denominator $q$ in irreducible representation followed by a factor $q$ scaling we may consider $$d_q^\prime(x,y) := \begin{cases} q|x-y| & \text{ if } x,y \in \left[\frac{p-1}{q},\frac{p}{q}\right) \text{ for some } p \in \mathbb{Z} \\ \max\left(q|x-y|, 1\right) & \text{ if } x < \frac{p}{q} \le y \end{cases}$$ which now precisely makes those $f_r$'s continuous s.t., denominator of $r \in \mathbb{Q}$ in irreducible representation is $q$. Then the product metric constructed from these $\{d_q'\}_{q \in \mathbb{N}}$ viz. $$d' := \sup_{q \in \mathbb{N}} \left(\frac{\max(d_q',1)}{q}\right)$$ gives a metric compatible to $\mathbb{Q}_\ell$ equivalent to Kevin Carlson's metric.
