Usual topology of $\mathbb{R}$ and filtration Liu asks in the book Algebraic Geometry and Arithmetic Curves on problem 1.3.1 whether the usual topology on $\mathbb R$ is defined by a subgroup filtration or not. 
I have some problems to understand what is a filtration. Liu defines that filtration is descending chain of subgroups $(G_n)_n$ of $G$ but can I just say that as a groups $\{0\}\subset \mathbb R$ is a descending chain so it is a filtration? Or as there are subgroups $A,B\subset \mathbb R$ with $A\not\subseteq B$ and $B\not\subseteq A$ then there is not a filtration?

The question is, probably, whether the usual topology on $\mathbb{R}$ is generated by a filtration by open subgroups. Given such a descending chain $G_n$ in a group $G$, we can take it as a neighborhood basis around the identity of $G$ for a topology. This completely describes a topology since then a neighborhood basis at $x\in G$ is given by $x G_n$. The fact that this requires open subgroups should immediately clue you in regarding the problem: what are the open subgroups of $\mathbb{R}$?
