For $f,g:\Bbb R_+\to\Bbb R$, let's define $f<g$ to mean $\exists x_0,\forall x>x_0,f(x)<g(x)$.
Observations about $<$:
- It's finer than the ordering provided by O-notation. For example, $x<x+\frac 1x<x+1$.
- If $m:\Bbb R\to\Bbb R$ is increasing, then $m\circ f<m\circ g\iff f<g$. (In contrast, O-notation distinguishes $\exp(x)$ from $\exp(2x)$ but not $x$ from $2x$.)
- If $m:\Bbb R\to\Bbb R$ is increasing and surjective, then $f\mapsto m\circ f$ is an order isomorphism of $<$.
- For any functions $a<b$, the interval $\{f:a<f<b\}$ is order-isomorphic to the whole space of functions (via squeezing $\Bbb R$ into $(a(x),b(x))$ for each $x$). So this order has a lot of symmetry and self-similarity.
- $<$ is not a total order. In fact, for any $f$, we can find an interval of mutually-comparable functions that are all incomparable with $f$, such as $\{g_\alpha:\alpha\in[-1,1]\}$ where $g_\alpha(x)=f(x)+\sin(x)+\alpha$.
To understand the space of limiting behaviors captured by $<$, I'd like to describe a maximal chain of it. Beyond maximality, I feel like I should forbid the chain from having "coarse" segments that can be replaced by a larger set of mutually-comparable functions to obtain a "finer" chain. That is, our chain $C$ should have the property that if $(C\setminus X)\cup Y$ is also a chain, then $|X\cap C|\ge|Y\setminus C|$. This implies maximality, so let's call such a chain "finely maximal". (Does this condition seem appropriate? I'm not confident that cardinality is the right tool for the job.)
Then my questions are:
- Is there a finely maximal chain of $<$?
- Are all finely maximal chains order-isomorphic to each other?
- Is there a canonical choice of finely maximal chain? For example, modulo exchanging functions that are eventually identically equal, is there exactly one finely maximal chain of $<$ containing the $0$ function?
- (If yes to any of the above) Is there a simple description of this order structure?
