# Every linearly-ordered real-parametrized family of asymptotic classes is nowhere dense?

Upon writing this post, I had the following natural conjecture: $\def\nn{\mathbb{N}} \def\rr{\mathbb{R}}$

Take any family $\{ C_r : r\in\rr \}$ of functions from $\nn$ to $\rr$ where $0 < C_s(n) \ll C_t(n)$ as $n \to \infty$ for every $s,t \in \rr$ such that $s < t$. Then for every $r \in \rr$ there is some function $D_r$ from $\nn$ to $\rr$ such that for every $ε \in \rr^+$ we have $C_r(n) \ll D_r(n) \ll C_{r+ε}(n)$ as $n \to \infty$. (By considering the reciprocal family this immediately implies the other side as well.)

Intuitively, I claim that there will always be some asymptotic class that falls between the cracks of any linearly-ordered real-parametrized family of asymptotic classes. I think that this is equivalent to the stronger claim of nowhere-denseness, but I am not sure.

For example:

• If $C_r(n) = n^r$ for every $r \in \rr$ and $r \in \nn$, then $D_r$ where $D_r(n) = n^r·\ln(n)$ for every $n \in \nn$ provides a suitable witness, since $n^0 \ll \ln(n) \ll n^ε$ as $n \to \infty$ for every $ε \in \rr^+$.

• If $C_r(n) = r^n$ for every $r \in \rr$ and $r \in \nn$, then $D_r$ where $D_r(n) = n^r·n$ for every $n \in \nn$ provides a suitable witness, since $r^0 \ll n \ll (1+\frac{ε}{r})^n$ as $n \to \infty$ for every $ε \in \rr^+$.

It is easy to show that $C$ has strict upper and lower bounds since $C_{-n}(n) \ll C_r(n) \ll C_n(n)$ as $n \to \infty$ for every $r \in \rr$. But I cannot find a general way to construct 'in-between' functions. I know that $C_{r+\frac1n}(n) \ll C_{r+ε}(n)$ as $n \to \infty$ for every $r \in \rr$ and $ε \in \rr^+$, but it is possible that $C_{r+\frac1n}(n) \sim C_r(n)$, as is indeed the case in both the above examples.

Is my conjecture true? If so, it suffices to prove that $C_0(n) \ll D_0(n) \ll C_ε(n)$ as $n \to \infty$ for some function $D_0$ from $\nn$ to $\rr$, since the general claim follows by translation. If not, it suffices to prove that for some family $C$ there is no such $D_0$, again due to translation.

• You'll want to look at Gordon Fisher's 1981 paper The infinite and infinitesimal quantities of du Bois-Reymond and their reception and Hausdorff gaps (see this survey). Jan 16, 2018 at 17:56
• @DaveL.Renfro: If you have an answer to my question, could you kindly post an answer that doesn't rely on links or articles hidden behind a great paywall? Thank you! Jan 17, 2018 at 2:02
• I've been too busy at work lately (non-academic) to spend time trying to dig into this topic, but maybe the following will help. First, even if you can't visit a library to access the Gordon Fisher paper (my copy is a photocopy I made back in the late 1980s; I don't have online access either), the title of the paper itself provides words and phrases you can use in searches for relevant work and for books/papers that cite Fisher's paper. Second, Math Origins: Orders of Growth at the MAA website should be useful. Jan 23, 2018 at 11:11
• @DaveL.Renfro: Thanks for that! I'm too busy these few weeks to have time for mathematical research, but I'll try to get to them eventually. I took a look at the last link you provided, and I already know everything on that page, I think. Jan 23, 2018 at 14:32

You can diagonalize to find a sequence asymptotically bigger than $$C_{r}$$ but asymptotically smaller than $$C_{r+\epsilon}$$ for any $$\epsilon>0$$.
Fix $$r$$. Define $$d_r(n)$$ to be the greatest positive integer such that $$C_{r+1/j}(n')/C_r(n')\ge d_r(n)$$ for all positive integer $$j \le d_r(n)$$ and $$n'\ge n$$, or $$d_r(n)=1$$ if none exists. Take $$D_r(n)=C_r(n)d_r(n)$$.
For any positive integer $$j$$, either $$j \le d_r(n)$$ or $$d_r(n) \le j$$, so by construction $$D_r(n) \le C_{r+1/j}(n)$$ or $$D_r(n) \le j·C_r(n)$$ as $$n \to \infty$$. Therefore $$D_r\ll C_{r+\epsilon}$$ for any $$\epsilon>0.$$
I claim that $$d_r(n)\to\infty$$ as $$n\to\infty$$. For any fixed positive integer $$\Delta$$, from the conjunction of $$C_r\ll C_{r+1/j}$$ for positive integer $$j \le \Delta$$, there is a large enough $$n$$ such that for all $$j\le\Delta$$ and $$n'\ge n$$ we have $$C_{r+1/j}(n')/C_r(n')\ge\Delta$$. For this $$n$$ we have $$d_r(n) \ge \Delta$$. And $$d_r$$ is clearly non-decreasing. So $$d_r$$ tends to $$\infty$$. Therefore $$C_r\ll D_r$$.