Let $M$ be an abelian group. A quasi-seminorm on $M$ is a function $|\cdot|:M\to \mathbb{R}_{\geq 0}$ s.t.
- $|0|=0$;
- symmetry:$|-f|=|f|$;
- generalized triangle inequality: $\exists K\geq1,\forall f,g\in M,|f+g|\leq K(|f|+|g|)$.
It is a seminorm if we can take $K=1$.
Denote the open ball $B(f,r):=\{g\in M:|f-g|<r\}$. Denote $\mathcal{B}(f):=\{B(f,r):r>0\}$, it is a filter base. Denote $\mathcal{U}(f)$ by the filter generated by $\mathcal{B}(f)$.
Now there are two ways to generate a topology on $M$:
- the topology $\tau_1$ generated by $\bigcup_{f\in M}\mathcal{B}(f)$, i.e. the smallest topology s.t. every open ball $B(f,r)$ is open.
- the collection $(\mathcal{U}(f))_{f\in M}$ forms a neighborhood system and thus there exists a unique topology $\tau_2$ on $M$ s.t. the neighborhood system of $\tau_2$ is $(\mathcal{U}(f))_{f\in M}$.
The problem is: are they the same topology? If not, which one should we choose if I expect to have a topological group? Is there any other reason that we should prefer one over another? Maybe $|\cdot|:M\to \mathbb{R}_{\geq 0}$ should be continuous?
In particular, $U\in \tau_1$ iff $U$ is an arbitrary union of finite intersections of open balls; and $U\in \tau_2$ iff $\forall f\in U,U\in \mathcal{U}(f)$ iff $\forall f\in U,\exists r>0,B(f,r)\subset U$.
If $|\cdot|$ is a seminorm, i.e. we have triangle inequality, then it can be shown $U\in \tau_1$ iff $U$ is a union of open balls and thus the neighborhood system of $\tau_1$ agrees with $\tau_2$, so $\tau_1=\tau_2$. I doubt that it can be done with generalized triangle inequaltiy.
I kind of believe that $(M,\tau_2)$ is always a topological group, because the topology on a topological abelian group is uniquely determined by a neighborhood base around $0$, in this case $\mathcal{B}(0)$ satisfies $\forall B(0,r)\in \mathcal{B}(0),\exists B(0,s)\in \mathcal{B}(0),B(0,s)+B(0,s)\subset B(0,r)$ and then we can construct a group topology identical to $\tau_2$.
I am not sure if $(M,\tau_1)$ is a topological group. The concept of seminormed abelian group appear in an introduction course of Berkovich space, I personally consider quasi-seminorm to make sure that $|\cdot|^t$ is also a quasi-seminorm for all $t>0$.
