How to neatly organize all real increasing functions? Let $I$ be the set of increasing functions $\mathbb{R}\longrightarrow\mathbb{R}$ modulo asymptotic behavior, i.e.
\begin{equation}
f\sim g \qquad\Longleftrightarrow\qquad \lim_{x\rightarrow+\infty}\frac{f(x)}{g(x)}=1
\end{equation}
It would be super cool if one could "index" each function $f\in I$ with a real number $\lambda_f$ such that
\begin{equation}
\lambda_f < \lambda_g \qquad\Longleftrightarrow\qquad \lim_{x\rightarrow+\infty}\frac{f(x)}{g(x)}=0
\end{equation}
If this could be done, one would have for each increasing function a unique index indicating how rapidly the function is increasing; vice versa for each real number one could determine a unique function that has a particular way of increasing to $+\infty$. It would be a "universal increasing function classifier". For instance I imagine that all polynomial functions would be indexed by a bounded interval on the reals.
My question: does such a bijection exist? Can it be constructed?
 A: Can't be done. The order type of the reals is not the same as the partial order on non-decreasing functions implicit above.
Since $\Bbb R$ is order-isomorphic to $(0,\infty)$ the space $X$ is order-isomorphic to $Y$, the space of non-decreasing functions from $(0,\infty)$ to itself. The construction I have in mind comes out simpler with $Y$ in place of $X$.


Exercise If $f_0,f_1,\dots\in Y$ there exists  $g\in Y$ such that $\lim_{x\to\infty}f_n(x)/g(x)=0$ for every $n$.


(Now if your $\lambda_f$ thing exists, a sequence $f_n$ with $\lambda_{f_n}>n$ leads to a contradiction.)
Hint:$$g(x)=x\max_{n\le x}f_n(x).$$
Note I'm assuming here that the map $f\mapsto\lambda_f$ is a bijection. This seems to follow from the "vice versa" in the question. In any case, if you don't want to assume that, do this:
Say $A=\sup_f\lambda_f$. First note that there does not exist $f$ with $\lambda_f=A$, because for any $f$ there exists $g$ with $\lambda_g>\lambda_f$. Now choose $f_n$ so $\lambda_{f_n}\to A$; define $g$ as above, and $\lambda_g>\lambda_{f_n}$ leads to a contradiction.
