Cardinality of growth rates

Question

$Let f=o(g)$ stand for $\lim_{x\to\infty} f(x)/g(x)=0$. It is simple to prove the following fact.

Proposition. Let $f_0,f_1:\mathbb{R}\to(0,\infty)$ be such that $f_0=o(f_1)$. Then there is a family of functions $(f_t)_{t\in[0,1]}$ such that $f_s=o(f_t)$ if $0\leq s<t\leq1$.

Proof. Consider $f_t=f_0^{1-t}f_1^t$.

I wonder if the following conjecture is also true.

Conjecture. Let $f_0,f_1:\mathbb{R}\to(0,\infty)$ be such that $f_0=o(f_1)$. Then there is a bounded totally ordered set $I$ and a family of functions $(f_i)_{i\in I}$ such that

1. $|I|>\mathfrak{c}$,
2. for every $i,j\in I$ with $i<j$ one has $|\{k\in I: i<k<j\}|=|I|$,
3. $f_{\min I} = f_0$, $f_{\max I}=f_1$,
4. for every $i,j\in I$ with $i<j$ one has $f_i=o(f_j)$.

Answer 1

Let me rather define $f=o(g)$ as $\forall \varepsilon > 0, \exists x_0, \forall x \geq x_0, |f(x)| \leq \varepsilon |g(x)|$.

Let $f_0,f_1$ be such that $f_0 = o(f_1)$. Let us assume that we have a family $(f_i)_{i \in I}$ satisfying the properties $1,3,4$.

I am going to show that $\exists x_0, \forall x \geq x_0, f_0(x)= 0$ (and conversely, if the latter holds, just take $f_i = f_0$ for $i < \mathrm{max} \ I$). If this is not the case, then there exists a sequence $(x_n)_n$ tending to $+\infty$ such that for eqch $n$, $f_0(x_n) \neq 0$.

Let us consider the map $I \rightarrow \mathbb{R}^{\mathbb{N}}$ sending $i$ to $(f_i(x_n))_n$. Since $\mathbb{R}^{\mathbb{N}}$ has the cardinality of the continuum, this map is not injective, so we can find $i < j$ such that $\forall n, f_i(x_n) = f_j(x_n)$. Since $f_i = o(f_j)$, this implies $f_i(x_n) = 0$ for $n$ large enough. Since $f_0 = o(f_i)$, this implies $f_0(x_n)= 0$ for $n$ large enough, a contradiction.