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
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.