The "Largeness" of Subsets of the Natural Numbers

Recently, I was thinking about the Erdős conjecture on arithmetic progressions, which says that, if, for a set $A\subseteq\mathbb{N}$, the sum $\sum_{a\in A}\frac{1}a$ diverges (i.e. "A is large"), then $A$ has arbitrarily long arithmetic progressions. My thoughts on this were that the condition of the sum of reciprocals diverges seems unnatural, so I wanted to study the condition of largeness further.

After a bit of work off of an order arising from a certain category, I came up with a conclusion that we can compare the largeness of sets as follows, for subsets $A$ and $B$ of $\mathbb{N}$:

Let $f_A$ and $f_B:\mathbb{N}\rightarrow\mathbb{N}$ such that $f_A(n)$ and $f_B(n)$ are the $n^{th}$ element of $A$ and $B$ respectively.

$A$ is larger than $B$ if the following set is small (i.e. sum of reciprocals converge) for some $k>0$: $$\left\{f_B(n):\frac{f_A(n)}{f_B(n)}>k \right\}$$

Intuitively, this says that $f_A$ and $f_B$ should be within a constant multiple of each other almost all the time, and if $f_A$ is consistently less than $f_B$, then $A$ must be larger than $B$ (i.e. it has more elements and so $f_A$ grows less quickly). This clearly forms a partial order; there are pathological examples which are incomparable, so it is not a total order - for instance: $$A=\mathbb{N}\cap\bigcup_{i=1}^{\infty}[(2i)!-(2i-1)!,2(2i)!)$$ $$B=\mathbb{N}\cap\bigcup_{i=1}^{\infty}[(2i+1)!-(2i)!,2(2i+1)!)$$ where both, containing infinitely many intervals $[x,2x)$, each having reciprocals sum to about $\log(2)$ are large; however, the ratio of $f_A$ to $f_B$ grows ever further away from $1$, so neither is larger than the other under my definition.

But those are awful functions; primarily, the issue is that $A$ is larger than the set of every other element of $A$ and similarly for $B$ - this contradicts intuition about sets. So, I want to consider only the sets for which $A$ and "every other element of $A$" are equally large and whether my order is a total order in that case. Thus, my question:

Suppose $f_A$ and $f_B$ are increasing functions $\mathbb{N}\rightarrow\mathbb{N}$ such that, for some constant $k$ the two sets: $$\left\{f_A(n):\frac{f_A(2n)}{f_A(n)}>k\right\}$$ $$\left\{f_B(n):\frac{f_B(2n)}{f_B(n)}>k\right\}$$ are small. Is it necessarily true that some $k_2$ can be chosen such that at least one of $$\left\{f_B(n):\frac{f_A(n)}{f_B(n)}>k_2 \right\}$$ $$\left\{f_A(n):\frac{f_B(n)}{f_A(n)}>k_2 \right\}$$ is small?

Your approach is interesting but I'm afraid we can construct a counterexample. Assume a monotonic natural number sequence $u_n, v_n$ are given. Let $$a_k=u_{n}k,\,(2^{n-1}\le k<2^n)\\b_k=v_{n}k, \,(2^{n-1}\le k<2^{n})$$ Then from $$\log2<\sum_{k=1}^{2^n-1}\frac{(-1)^{k-1}}{k}=\sum_{k=2^{n-1}}^{2^{n}-1}\frac1k< \sum_{k=2^{n-1}}^{2^n-1}\frac{1}{2^{n-1}}=1$$ we have $$\sum_{a_{2k}>2c_1a_k}\frac{1}{a_k}=\sum_{u_{n+1}>c_1u_{n}}\sum_{k=2^{n-1}}^{2^{n}-1}\frac{1}{u_{n}k}<\sum_{u_{n+1}>c_1u_{n}}\frac{1}{u_{n}}$$ Now for any $c_1>1$, $\displaystyle\sum_{u_{n+1}>c_1u_{n}}\frac{1}{u_{n}}$ converges since $$\sum_{u_{n+1}>c_1u_{n}}\frac{1}{u_{n}}<\frac{1}{u_1}\left(1+\frac{1}{c_1}+\frac{1}{c_1^2}+\cdots\right)=\frac{c_1}{(c_1-1)u_1}$$ So the sets $\displaystyle\left\{a_n:\frac{a_{2n}}{a_n}>c_1\right\},\,\left\{b_n:\frac{b_{2n}}{b_n}>c_1\right\}$ are small for any monotonic sequence $u_n$ and $v_n$. But $$\sum_{a_k>c_2b_k}\frac{1}{b_k}=\sum_{u_n>c_2v_n}\sum_{k=2^{n-1}}^{2^n-1}\frac{1}{kv_n}>\log2\sum_{u_n>c_2v_n}\frac{1}{v_n}$$ So setting $u_n$ and $v_n$ to be any uncomparable sets becomes a counterexample. Let $$A=\mathbb{N}\cap\bigcup_{i=1}^{\infty}[(2i)!-(2i-1)!,2(2i)!)\\B=\mathbb{N}\cap\bigcup_{i=1}^{\infty}[(2i+1)!-(2i)!,2(2i+1)!)$$ and $u_n=f_A(n)$, $v_n=f_B(n)$. Then $\displaystyle\sum_{u_n>c_2v_n}\frac{1}{v_n}$ diverges for any $c_2$ so $\{a_n:n\in\mathbb{N}\}$ and $\{b_n:n\in\mathbb{N}\}$ are not comparable but satisfy the conditions.