There exist $\{a_{n}\},\{b_{n}\}$ such $\lim_{n\to\infty}\frac{a_{n}}{b_{n}}=c？$ Let $ A$ and $ B$ are two infinite subsets of the natural numbers $\mathbb{N}$,  such that
$$A\cap B=\emptyset \qquad A\cup B=\mathbb N$$ 
Question: is it true that for every natural $c>0$, 
 there exist two increasing sequences $\{a_{n}\},\{b_{n}\}$ such that 
$\{a_{n}\}\subseteq  A,\{b_{n}\}\subseteq  B$ and 
$$\lim_{n\to\infty}\dfrac{a_{n}}{b_{n}}=c?$$
My try: maybe I guess  there doesn't exist such this condition $\{a_{n}\},\{b_{n}\}$
Thank you for you help
 A: UPDATE 12/02/2013 The first version of this answer was incorrect
(it used a false version of Ramsey’s theorem, for ordered tuples instead
of unordered tuples). The corrected version below does not use Ramsey’s
theorem. 

The answer to your question is YES.
The case $c=1$ is simpler than the others and has been explained in 
Harald's comment below.
Also, the problem stays the same if we replace $A$ by $B$, $B$ by
$A$, and $c$ by $\frac{1}{c}$.
So we may assume without loss of generality that $c > 1$.  
It will suffice to show the following : for any $\varepsilon >0$ and any
integer $m$, there are integers $a\in A,b\in B$ with $a\geq m,b\geq m$, such that
$$
c-\varepsilon \leq \frac{a}{b} \leq c+\varepsilon \tag{1}
$$
So suppose that this is false for some $\varepsilon >0$ and some
$m$. We then have, for any $(a,b)\in A\times B$ such that $a,b\geq m$,
$$
\frac{a}{b} < c-\varepsilon \text{ or }
c+\varepsilon < \frac{a}{b} \tag{2}
$$
We may assume that $c-\varepsilon > 1$ (this will be true for small
enough $\varepsilon$).
Let $M_1=\frac{c^2-\varepsilon^2+c}{\varepsilon(c-\varepsilon)}$,
$M_2=\frac{2}{\varepsilon}$, $M_3=\frac{c+\varepsilon+2}{c+\varepsilon-1}$,
$M_4=\frac{c+c^2-\varepsilon^2}{\varepsilon(c-\varepsilon)}$
and $M={\sf max}(m,M_1,M_2,M_3,M_4)$.
Lemma 1 If $B$ contains an integer interval 
$I=\big[\big|x,y\big|\big]=\lbrace x,x+1,x+2,\ldots ,y \rbrace$ with
$x\geq M$. Then $B$ also contains the interval
$J=\big[\big| \ 1+\lceil(c-\varepsilon)x\rceil,
\lfloor(c+\varepsilon)(y-1)\rfloor \ \ \big|\big]$.
Proof of lemma 1  Let $w\in J$. Consider the numbers
$$
\alpha=\frac{w}{c+\varepsilon}+1, \beta=\frac{w}{c-\varepsilon}-1
$$
We have
$$
\begin{array}{lcl}
\beta-\alpha &=& \frac{2}{c^2-\varepsilon^2}(w-\frac{c^2-\varepsilon^2}{\varepsilon})
\geq \frac{2}{c^2-\varepsilon^2} ((c-\varepsilon)x+1-\frac{c^2-\varepsilon^2}{\varepsilon}) \geq 0 \ (\text{because } x\geq M_1) \\
y-\alpha &=& \frac{(c+\varepsilon)(y-1)-w}{c+\varepsilon} \geq 0 \\
\beta-x &=& \frac{w-(c-\varepsilon)x-1}{c-\varepsilon} \geq 0
\end{array}
$$
We deduce ${\sf max}(x,\alpha) \leq {\sf min}(y,\beta)$ and hence
${\sf max}(x,\lceil\frac{w}{c+\varepsilon}\rceil) 
\leq {\sf min}(y,\lfloor\frac{w}{c+\varepsilon}\rfloor)$. Let $b$ be any integer
between those two integers. Then $b\in[|x,y|]$ so $b\in B$, but at the same time
$b\in [|\frac{w}{c+\varepsilon},\frac{w}{c-\varepsilon}|]$, so 
$c-\varepsilon \leq \frac{w}{b} \leq c+\varepsilon$. By (2) we see that $w$ cannot be
in $A$, so it must be in $B$.
Lemma 2 If $B$ contains an integer interval 
$I=\big[\big|x,y\big|\big]=\lbrace x,x+1,x+2,\ldots ,y \rbrace$ with
$x\geq M$ and $y\geq (c-\frac{\varepsilon}{2})x$, then $y+1\in B$.
Proof of lemma 2 We have $y+1\geq (c-\frac{\varepsilon}{2})x+1 \geq (c-\varepsilon)x+2$
(because $x\geq M_2$). So $y+1\geq 1+\lceil(c-\varepsilon)x\rceil$. 
Also, we have $y+1 \leq (c+\varepsilon)(y-1)-1$ (because $y \geq x \geq M_3$).
In the end we have $y+1\in J$, so lemma 1 applies.
Lemma 3 $B$ cannot contain an   integer interval 
$I=\big[\big|x,y\big|\big]=\lbrace x,x+1,x+2,\ldots ,y \rbrace$ with
$x\geq M$ and $y\geq (c-\frac{\varepsilon}{2})x$.
Proof of lemma 3 If it did, we could iterate lemma 2 to deduce
that B contains every integer $\geq x$, contradicting the infinitude of 
$A$.
Lemma 4 The interval $J=[|x_J,y_J|]$ defined in lemma 1 satisfies
$x_J \geq x$ and 
$\frac{y_J}{x_J} \geq (1+\frac{\varepsilon}{c-\varepsilon})\frac{y}{x}$.
Proof of lemma 4 We have $(c+\varepsilon)(x-1)-1 \geq x$ because of
$x\geq M_3$, so $x_J \geq x$.
Next, we have $1 \geq 
\frac{c^2-\varepsilon^2}{\varepsilon(c-\varepsilon)x-c}$ because of
$x\geq M_1$. We deduce $y\geq x \geq \frac{c^2-\varepsilon^2}{\varepsilon(c-\varepsilon)x-c}$,
so $(\varepsilon(c-\varepsilon)x-c)y \geq (c^2-\varepsilon^2)x$, which is equivalent
to $\frac{(c+\varepsilon)(y-1)}{1+(c-\varepsilon)x} \geq (1+\frac{\varepsilon}{c-\varepsilon})\frac{y}{x}$, which
implies  $\frac{y_J}{x_J} \geq (1+\frac{\varepsilon}{c-\varepsilon})\frac{y}{x}$.
Conclusion of proof Since $B$ is infinite, there is a $x_0\in B$
such that $x_0 \geq M$. Let $y_0=x_0$. Iterating lemmata 1 and 4,
we find for each $n$ an integer interval $I=[|x_n,y_n|]$ with
$x_n \geq M$ and $\frac{y_n}{x_n} \geq 
\bigg(1+\frac{\varepsilon}{c-\varepsilon}\bigg)^n$. 
We then have $\frac{y_n}{x_n} > c-\frac{\varepsilon}{2}$ for large enough
$n$, contradicting
lemma 3.
