Given a field $F$ and a modulus $|\cdot|:F \to [0,+\infty)$, then for any $c>0$, does it always exist $x \in F \backslash \{0\}$ s.t. $|x|<c$? Or is there any special field s.t. we can construct a modulus where there is no $x \in F \backslash \{ 0\}$ satisfying $|x|<c$ for some $c>0$.
Modulus is defined as the following,
The reason for asking this question is the following problem,
where a norm is defined as the following,
Yes, as long as $\lvert \cdot \rvert$ is nontrivial (i.e. there is some $x$ such that $\lvert x\rvert\notin \{0,1\}$).
This follows from multiplicativity: if $\lvert x\rvert<1$, then $\lvert x^n\rvert=\lvert x\rvert^n\to 0$ as $n\to \infty$, while if $\lvert x\rvert >1$, then $1=\lvert 1\rvert=\lvert xx^{-1}\rvert=\lvert x\rvert\lvert x^{-1}\rvert$, so $\lvert x^{-1}\rvert<1$ and you can work as in the case of $\lvert x\rvert<1$.
Put it differently: apart from $0$, $\lvert\cdot\rvert$ is in particular a homomorphism from the multplicative group of the field to $({\bf R}_{>0},\cdot)$. In particular, its range is a subgroup of positive reals, and any nontrival subgroup of positive reals is dense near $0$.
On the other hand, on any field you can put trivial modulus, in which case there is no $x\neq 0$ with $\lvert x\rvert<c$ for $c\leq 1$.