For every $r >0$, exists a sequences $a_n\in A$, $b_n \in B$ such that $\displaystyle\lim_{n\to \infty}\frac{a_n}{b_n} = r$ Let $A,B$ be two sets such that
$A\cup B = \mathbb{N}, A\cap B = \emptyset$ with $|A| = |B| = \infty$
Then for every $r \in \mathbb{R}>0$, exists sequences such that
$a_n\in A$, $b_n \in B$
such that
$\lim_{n\to \infty}\frac{a_n}{b_n} = r$
I do not know how to construct this two sequences. My ideas was using the density of the rationals in the reals and construct a sequence or using the cantor series but I can not get anything. 
 A: First it suffices to show the case of $r\geq 1$. Then it suffices to show that for any $r\geq 1$, 
$$\inf_{a\in A}\inf_{b\in B}|\frac{a}{b}-r|=0.$$
Suppose not. Then there exists $r>c>0$ such that for all $a\in A,b\in B$, $|\frac{a}{b}-r|>c$, which implies $a\notin[b(r-c),b(r+c)]$ for all $a\in A$. This implies $[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This further implies that for all $n\in\mathbb{N}$, $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$.
Now for $n$ large enough, $(r+c)^n>(r-c)^{n+1}$, thus the fact that $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$ for all $n\in\mathbb{N}$ implies that $B$ contains all numbers sufficiently large, which contradicts $A\cap B=\emptyset$ and $|A|=\infty$.  Thus the result is proved. 
$\bf{Edit}$: The above argument has an error in the following statement:
"...$[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This further implies that for all $n\in\mathbb{N}$, $[b(r-c)^n,b(r+c)^n]\cap\mathbb{Z}\subset B$."
Here is a modification. We have $[b(r-c),b(r+c)]\cap \mathbb{Z}\subset B$ for all $b\in B$. This implies that $[(b(r-c)+1)(r-c),(b(r+c)-1)(r+c)]\cap\mathbb{Z}\subset B$." Inductively, we get $[b(r-c)^n+(r-c)^{n-1}+\cdots+(r-c), b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)]
\cap\mathbb{Z}\subset B$ for all $n\in\mathbb{N}$.
But note that $b(r-c)^{n+1}+(r-c)^{n}+\cdots+(r-c)<b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)$ for $n$ large enough, say $n\geq N$. Thus the intervals $[b(r-c)^n+(r-c)^{n-1}+\cdots+(r-c), b(r+c)^{n}-(r+c)^{n-1}-\cdots-(r+c)]$ for $n\geq N$ connect to form $[b(r-c)^N+(r-c)^{N-1}+\cdots+(r-c), \infty)$, and 
$B\supset[b(r-c)^N+(r-c)^{N-1}+\cdots+(r-c), \infty)\cap \mathbb{N}$. 
