Is every positive real number the limit of a sequence of ratios obtained from a double partition of $ \mathbb{N} $? Let $ \{ A,B \} $ be a partition of $ \mathbb{N} $ into two infinite subsets. For every $ r \in \mathbb{R}_{> 0} $, can we find an increasing sequence $ (a_{n})_{n \in \mathbb{N}} $ in $ A $ and an increasing sequence $ (b_{n})_{n \in \mathbb{N}} $ in $ B $ such that $ \displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{b_{n}} = r $?
 A: For $r=1$, note that there are infinitely many pairs of adjacent elements of $A$ and $B$, whose ratio tends to $1$.
For $r\ne1$, without loss of generality let $r\gt1$. Form sequences $a_n$, $b_n$ such that $|a_n/b_n-r|\lt\frac rn$, and for the sake of contradiction assume that at some point this is no longer possible. That is, for some $k,n\in\mathbb N$ there is no pair $a_n\in A$ and $b_n\in B$ such that $a_n,b_n\ge k$ and $|a_n/b_n-r|\lt\frac rn$.
Pick some $b\in B$ with $b\ge k$. If necessary, increase $n$ to ensure $r(1-\frac1n)\gt1$. Then none of the natural numbers between $br(1-\frac1n)$ and $br(1+\frac1n)$ is in $A$. Therefore they are in $B$, and we can apply the same argument to them. Thus for all $m\in\mathbb N$ none of the numbers between $br^m(1-\frac1n)^m$ and $br^m(1+\frac1n)^m$ is in $A$. But at some point $r^{m+1}(1+\frac1n)^{m+1}\gt r^m(1-\frac1n)^m$, so the intervals overlap and cover the rest of $\mathbb N$ from that point on, contradicting the fact that $A$ is infinite.
The contradiction shows that contrary to the assumption the sequence can be extended indefinitely.
(I glossed over some rounding issues, but I think it's clear enough that by going to sufficiently high numbers we can ensure that the intervals grow by a constant factor even when restricted to the natural numbers they contain.)
A: Many thanks go to Barto and Joriki for their valuable insights. What follows is simply a reorganization of the solutions of these two gentlemen. All credit, therefore, goes to them and to them only.
Let $ r \in \mathbb{R}_{>1} $, and choose an $ \epsilon \in (0,r - 1) $.
Claim For every $ N \in \mathbb{N} $, there exists $ (a,b) \in (A \cap \mathbb{N}_{>N}) \times (B \cap \mathbb{N}_{>N}) $ such that $ \displaystyle \left| r - \frac{a}{b} \right| < \epsilon $.
Proof of claim By way of contradiction, assume that the claim is false. Then there exists an $ N \in \mathbb{N} $ such that the inequality $ \displaystyle \left| r - \frac{a}{b} \right| < \epsilon $ has no solution for $ (a,b) \in (A \cap \mathbb{N}_{>N}) \times (B \cap \mathbb{N}_{>N}) $. As the inequality $ \displaystyle \left| r - \frac{a}{b} \right| < \epsilon $ is equivalent to $ b(r - \epsilon) < a < b(r + \epsilon) $, it follows that $ (b(r - \epsilon),b(r + \epsilon)) \cap \mathbb{N} \subseteq B $ for all $ b \in B \cap \mathbb{N}_{>N} $.
Now, the gist of Barto's and Joriki's observation is that for a fixed $ b \in B \cap \mathbb{N}_{>N} $, the following proposition is true:
\begin{equation}
\forall k \in \mathbb{N}: \quad (b(r - \epsilon)^{k},b(r + \epsilon)^{k}) \cap \mathbb{N} \subseteq B.
\end{equation}
We prove this by induction. When $ k = 1 $, the proposition is true by the previous paragraph. Next, suppose that the proposition is true for $ k = l $. Pick any $ n \in (b(r - \epsilon)^{l},b(r + \epsilon)^{l}) \cap \mathbb{N} $. As $ n \in B $ by the induction hypothesis, and as $ n > b(r - \epsilon)^{l} > N \cdot 1 = N $, we see that $ n \in B \cap \mathbb{N}_{>N} $. We may thus apply the argument in the previous paragraph to deduce that $ (n(r - \epsilon),n(r + \epsilon)) \cap \mathbb{N} \subseteq B $. However, this is true for all $ n \in (b(r - \epsilon)^{l},b(r + \epsilon)^{l}) \cap \mathbb{N} $, so we can conclude that $ (b(r - \epsilon)^{l+1},b(r + \epsilon)^{l+1}) \cap \mathbb{N} \subseteq B $. The proposition is therefore true for $ k = l + 1 $, and by induction, it is true for all $ k \in \mathbb{N} $.
Finally, observe that the collection of intervals $ \{ (b(r - \epsilon)^{k},b(r + \epsilon)^{k}) \}_{k \in \mathbb{N}} $ must cover $ \mathbb{R} $ from some point on. To prove this, it suffices to show that the left-endpoint of the interval $ (b(r - \epsilon)^{k+1},b(r + \epsilon)^{k+1}) $ is less than the right-endpoint of the interval $ (b(r - \epsilon)^{k},b(r + \epsilon)^{k}) $ for all $ k \in \mathbb{N} $ large enough. However, this is true because
\begin{equation}
\lim_{k \rightarrow \infty} \frac{b(r - \epsilon)^{k+1}}{b(r + \epsilon)^{k}} = \lim_{k \rightarrow \infty} \left( \frac{r - \epsilon}{r + \epsilon} \right)^{k} (r - \epsilon) = 0.
\end{equation}
We thus see that $ (b(r - \epsilon)^{k},\infty) \cap \mathbb{N} \subseteq B $ for all $ k \in \mathbb{N} $ large enough, which contradicts the hypothesis that $ A $ is infinite. The assumption is therefore false, and the claim is proven. Q.E.D.
The claim is now sufficient to prove the case when $ r > 1 $ or $ 0 < r < 1 $. The case $ r = 1 $ is settled by taking pairs of adjacent numbers in $ A $ and $ B $, as Joriki has mentioned.
