What is the cardinality of $O(f(x))$?

Question

I was wondering what the cardinality of

$$O(f(x))=\left\{g:\mathbb R\to\mathbb R\left|\lim_{x\to\infty}\sup\frac{|g(x)|}{f(x)}<\infty \right.\right\}$$ is.

I know that if $g(x)=cf(x)$ for some $c\in\mathbb R$ then $g(x)\in O(f(x))$, hence $\{c\in\mathbb R|cf(x)\}\subset O(f(x))$. As $|\mathbb R|=\beth_1$, this implies $|\{c\in\mathbb R|cf(x)\}|=\beth_1$. At the same time, $O(f(x))\subset\{g:\mathbb R\to\mathbb R\}$ and $|\{g:\mathbb R\to\mathbb R\}|=\beth_2$.

The implication is that

$$\beth_1\leq|O(f(x))|\leq\beth_2$$

However, I'm not sure whether $|O(f(x))|=\beth_1$ or $|O(f(x))|=\beth_2$. Could anybody figure out what the cardinality of $O(f(x))$ is?

Answer 1

Take any subset $A\subset [0,1]$ and define $f_A(x):=\chi_A(x)+f(x)\chi_{[0,1]^c}(x)$.

Since $A\subset[0,1]$, it is clear taht $f_A\in O(f(x))$ and $|\{f_A:\, A\subset [0,1]\}=|\cal P[0,1]|=\beth_2$

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For the sake of completenes $$\chi_A(x)=\begin{cases}1&x\in A\\0&x\notin A\end{cases}$$