Closure of set in $\mathbb{R}$ with standard metric

Question

Let $$ A = \left\{ \frac{mn}{m^2 + n^2 + 1} : m, n \in \mathbb{Z} \right\} \subseteq \mathbb{R}. $$ What is the closure of $A$ in $\mathbb{R}$ with standard metric?

So I know it contains $A$ itself but from there I don't know how to proceed. I may need just an hint, because I can't really see how this set is actually "made", meaning I can't really fully visualize it.

look's like pythagorean triplets : $(2mn,m^2+n^2,m^2-n^2)$... — Stéphane Jaouen, Feb 08 '24 at 18:14
$A=\left{\dfrac{q}{q^2+q+1},:,q\in \mathbb{Q}\right}\cup {0}.$ So the basic question is: which rational numbers do we get for $q+\frac{1}{q}+1$? — Marius S.L., Feb 08 '24 at 18:50
@MariusS.L. I'm not sure I fully understand how you got there — Folpo13, Feb 08 '24 at 19:55
@AnuragA I get $\frac{q}{q^2+1+1/n^2}$ what am I doing wrong? — Folpo13, Feb 08 '24 at 20:21

score 2 · Accepted Answer · answered Feb 08 '24 at 21:43

2

By Marius S.L.'s idea.

Given a rational number $q=\frac a b$, let $m=ak$ and $n=bk$ and then $$\lim_{k\to\infty}\frac{akbk}{a^2k^2+b^2k^2+1}=\frac{q}{q^2+1}.$$ Let $B=\{\frac q{q^2+1}| q\in\Bbb Q\}$.

Then $B\subset\bar A$ and so $\bar B\subset\bar A$. Clearly, $\bar B=[-\frac12,\frac12].$ Therefore, $[-\frac12,\frac12]\subset\bar A.$ Since $A\subset [-\frac12,\frac12]$, $\bar A=[-\frac12,\frac12]$.

answered Feb 08 '24 at 21:43

Bob Dobbs

10,988

1

I see now. Thank you very much – Folpo13 Feb 08 '24 at 21:59

Closure of set in $\mathbb{R}$ with standard metric

1 Answers1