The statements $f(n) = O(n^{\epsilon})$ for all $\epsilon > 0$ and $f(n) = n^{o(1)}$. Consider the statements
\begin{align}
\tag{A}
f(n) &= O(n^{\epsilon}) \text{ for all } \epsilon > 0 \\
\tag{B}
f(n) &= n^{o(1)}
\end{align}
Questions:


*

*It's clear that (B) implies (A). Does (A) imply (B)?

*If the answer to the first question is no, is there a more brief way to write (A)?
 A: Consider the statements
\begin{align}
\tag{$A$} f(n) &= O(n^{\epsilon}) \text{ for all } \epsilon > 0 \\
\tag{$B$} f(n) &= n^{o(1)} \\
\tag{$C$} f(n) &= o(n^{\epsilon}) \text{ for all } \epsilon > 0
\end{align}
These statements are equivalent to each other.
To see this, we first rewrite the statements by unwrapping their definitions and taking logarithms.
\begin{align}
\tag{$A$} \forall \epsilon > 0 \,\, \exists c > 0 \,\, \exists N \,\, \forall n \geq N : \,\, \frac{\log f(n)}{\log n} &\leq \epsilon + \frac{\log c}{\log n} \\
\tag{$B$} \forall \epsilon > 0 \,\, \exists N \,\, \forall n \geq N : \,\, \frac{\log f(n)}{\log n} &\leq \epsilon \\
\tag{$C$} \forall \epsilon > 0 \,\, \forall \eta > 0 \,\, \exists N \,\, \forall n \geq N : \,\, \frac{\log f(n)}{\log n} &\leq \epsilon + \frac{\log \eta}{\log n}
\end{align}
Now we show that $(A) \Leftrightarrow (B) \Leftrightarrow (C)$ by proving (i.e., sketching the proofs of) all the pairwise implications.
$(A) \Rightarrow (B)$: Let $\epsilon^{\prime} > 0$ be given. Put $\epsilon = \epsilon^{\prime}/2$. Write down $(A)$ with this choice of $\epsilon$. Take $N^{\prime} \geq N$ large enough that 
$$
\frac{\log c}{\log n} \leq \frac{\epsilon^{\prime}}{2} \text{ for all } n \geq N^{\prime}.
$$ 
This gives (a primed version of) $(B)$.
$(B) \Rightarrow (A)$: $(B)$ implies $(A)$ with $c=1$
$(A) \Rightarrow (C)$: Let $\epsilon^{\prime} > 0$ and $\eta^{\prime} > 0$ be given. Put $\epsilon = \epsilon^{\prime}/2$. Write down $(A)$ with this choice of $\epsilon$. Take $N^{\prime} \geq N$ large enough that 
$$
\frac{\log c}{\log n} \leq \frac{\epsilon^{\prime}}{4} \leq \frac{\epsilon^{\prime}}{2} + \frac{\log \eta^{\prime}}{\log n} \text{ for all } n \geq N^{\prime}.
$$
This gives (a primed version of) $(C)$.
$(C) \Rightarrow (A)$: Clear.
$(B) \Rightarrow (C)$: Let $\epsilon^{\prime} > 0$ and $\eta^{\prime} > 0$ be given. Put $\epsilon = \epsilon^{\prime}/2$. Write down $(B)$ with this choice of $\epsilon$. Take $N^{\prime} \geq N$ large enough that 
$$
0 \leq \frac{\epsilon^{\prime}}{2} + \frac{\log \eta^{\prime}}{\log n} \text{ for all } n \geq N^{\prime}.
$$
This gives (a primed version of) $(C)$.
$(C) \Rightarrow (B)$: Write down $(C)$ with $\eta=1$.
