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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?

zinsek
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1 Answers1

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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.

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Moishe Kohan
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  • To be honest it's my own extension of homework (homework required to show that $\mathbb{Q}\times\mathbb{Q}$ is not dense in that metric space). And basicly to answer "For what points $p$..." I can repeat my answer from original post. I think that $p$ are points from lines with rational slope that pass through origin. This is based on my conclusion from homework that for point $x=(\sqrt[]{2},4)$ there is no such $q_{n}$ that its limit would be $x$. – zinsek Dec 06 '16 at 20:15
  • @zinsek OK, if you can prove this guess (and this is not even a calculus problem), you are half-way through the solution. – Moishe Kohan Dec 06 '16 at 20:50
  • Ok, so $x$ is from line with slope $2\sqrt[]{2}$, for for every point $(a,b)\in\mathbb{R}^{2}$ if $a$ is rational then $b$ cannot be. Based on that, any point from $\mathbb{Q}^{2}$ cannot belong to this line apart from $(0,0)$. So every $q_{n}$ is type $II$ with $(x,q_{n})$ so $x$ cannot be limit of $q_{n}$ because every point from $q_{n}$ is at least $d(0,x) = 3\sqrt[]{2}$ away from $x$. That's a bit informal but I think it's clear enough. So now I would need to show that for any point from set mentioned in original post exists $q_{n}$ that converges to that point? – zinsek Dec 06 '16 at 21:51
  • @zinsek: Very good! The second part is simple calculus since you are using the usual norm on rational lines. – Moishe Kohan Dec 06 '16 at 21:56