Let $\Bbb{R}^{+}\_{0}$ be the set of non-negative real numbers and $\Bbb{R}^{+}$be the set of positive reals. Let us say that a function $f \colon \Bbb{R}^{+}\_{0} \to \Bbb{R}^{+}\_{0}$ is **eventually sublinear** if $\ \forall r \in \Bbb{R}^{+} \ \exists x_0 \in \Bbb{R}^{+} \colon \forall x \geq x_0, f(x) < rx$. Let $S$ be the set of non-decreasing, eventually sublinear functions (we require no continuity). We define an equivalence relation on $S$: $f \sim g$ if and only if $f$ and $g$ are eventually boundedly close, i.e., $\exists K>0\ \exists x_0 \in \Bbb{R}^{+} \colon \forall x \geq x_0, |f(x)-g(x)| < K$. Denote the set of equivalence classes of $S$ under $\sim$ by $\ T$.

Question 1: What is the cardinality of $\ T$?

Question 2: Does this cardinality change if we drop the requirement that the functions are non-decreasing? What if the co-domain becomes $\Bbb{R}$?

Looking at the set of pairwise non-equivalent sublinear functions $\{f(x) = r*ln(x) \mid r \in \Bbb{R}\}$ shows that $|T|$ must be at least as big as the cardinality of the reals, $\beth_1$. Also, $|T|$ can be no bigger than the cardinality of *all* functions from $\Bbb{R}$ to $\Bbb{R}$, which is $\beth_2$, the cardinality of the power set of the reals. So there seem to be three options:

Option 1: ZFC proves $|T| = \beth_1$.

Option 2: ZFC proves $|T| = \beth_2$.

Option 3: This question is independent from ZFC (but perhaps not independent from ZFC + Generalized CH?).

Different proof ideas involving explicit bounds or forcing are failing me. So does anyone know of an answer and a proof? Thanks.