Order structure of asymptotics Consider the set $A = \{\theta(f) \mid f : \mathbb{R} \rightarrow \mathbb{R},\;f\,\text{non-decreasing}\}$ where $\theta(f)$ denotes the set of functions which are asymptotically within a constant factor of $f$. Define an order on $A$ by setting $\theta(f) \le \theta(g)$ if $f \in O(g)$.
What is the order structure of $A$? For that matter, what is the cardinality?
$\mathbb{R}$ can be embedded in $A$ by sending $\alpha$ to $x\mapsto x^\alpha$, but $A$ has a bottom element, so they are certainly not isomorphic.

NB: The following is for an earlier version of the question in which the functions are not necessarily monotone.
Here are two things you can say:


*

*$\langle \mathcal{P}([0, 1)), \subseteq \rangle$ can be embedded into $A$. In particular, $A$ has cardinality $2^\mathfrak{c}$.

*$A$ is dense. That is, whenever $f < g$ there is $h$ such that $f < h < g$.


There are probably a lot more things you can say as well. For instance, I suspect that it's possible to embed $\omega_2$ into $A$.
Proof of 1
For $X \subseteq [0, 1)$ define for each $n \in \mathbb{Z}$ and each $x \in [0, 1)$,
$$f_X(n + x) = \begin{cases} 0 \qquad x \notin X \\ n \qquad x \in X \end{cases}$$
Note that this defines a total function on $\mathbb{R}$ for each $X$. We clearly have that $X \subseteq Y$ implies $f_X \leq f_Y$. For the converse, suppose that $f_X \leq f_Y$. Then there is some $M$ and some $x_0$ such that for $x > x_0$, $f_X(x) \leq M f_Y(x)$. Choose $n > x_0$ such that $n \in \mathbb{N}$, and let $x \in X$. Then $f_Y(n + x)$ can't be zero, because then $M f_Y(n + x) = 0$ and that would contradict $f_X(n + x) \leq M f_Y(n + x)$.
Proof of 2
Suppose that $f < g$. Since $g \nleq f$, we know that for every $n$ there is some $x_n \in \mathbb{R}$ such that $x_n > n$ and $g(x_n) > n f(x_n)$. Define $h$ as follows:
$$h(x) = \begin{cases} f(x) \qquad \exists n \text{ such that } x = x_{2n} \\
g(x) \qquad \text{otherwise} \end{cases}$$
Then we clearly have $f \leq h \leq g$. Suppose that $g \leq h$. Then there would be some $N$ such that whenever $x > N$, $g(x) \leq N h(x)$. However, we know that $x_{2N} > N$ and $g(x_{2N}) > 2N f(x_{2N}) = 2N h(x_{2N})$, giving a contradiction. Hence, $h < g$ and a similar argument using $2N + 1$ instead of $2N$ shows that $f < h$.
