TITLE: Closure of $\mathbb{Q}\times\mathbb{Q}$ in British Rail metric
QUESTION [5 upvotes]: I'm wondering what is the closure of $\mathbb{Q}\times\mathbb{Q}$ in $(\mathbb{R}^{2},d)$ where $d$ is British Rail metric:
$$ d(x,y) = \left\{ \begin{array}{lr} ||x-y|| & \text{if} \; \; x,y,0 \; \; \text{are collinear,}\\ || x || + ||y||& \;\;\;\; \text{otherwise.} \end{array} \right. $$
At this moment I'm thinking about set
$$\{(x,y)\in\mathbb{R}^{2}:\exists q\in\mathbb{Q} \quad qx=y\lor x=0\}$$
because I think it is a set of all points that lie on lines passing through $(0,0)$ with rational slope.
Is it correct answer to my question?

REPLY [2 votes]: Since this looks like a homework, I will only provide some hints. 
Given two points $p, q$ in $R^2$ I will say that the pair $(p,q)$ has type I if $p, q, 0$ are collinear and type II otherwise. Given any point $p\in R^2$ and a sequence $(q_n)$ in $R^2$ I will say that the sequence $(q_n)$ is of type I if all pairs $(p,q_n)$ are of type I. Similarly, for type  II. 
Now, fix $p$. Any sequence $q_n$  splits as (at most) two infinite subsequences, each subsequence has either type I or type II. Assuming that $(q_n)$ is of type II, what can you say about 
$$
\lim \inf_{n\to\infty} d(p, q_n) ? 
$$ 
For what points $p$ there is a sequence $(q_n)$ of type I consisting entirely of points in ${\mathbb Q}^2$?  
Once you have answered these two questions, you will obtain a description of the closure of ${\mathbb Q}^2$ (with respect to the metric $d$) as a union of certain lines in $R^2$. 
See also this, for Brian Scott's answer which will help.