Topology on abelian groups

Take $(G,+)$ an abelian additive group. I was looking for a proof or a counterexample of the following statement

For every family of subgroups $G_n, n \in \Bbb N$ such that $G_0=G$ and $G_{n+1} \subset G_n$ for all $n$, there exists a topology on $G$ such that $G_n$ is a neighbourhood base for the neutral element and $(G,+)$ is a topological group.

Solution 1:

Yes, there always is such a topology, as is well-known. Go via uniformities (which is the standard way to look at topologies on a group): let $D_n = \{(x,y) \in G \times G \mid x-y \in G_n\}$ which is a translation invariant family of entourages which is nested and so it's a filter base and we take the uniformity generated by this filter base, call it $\mathcal{U}$. It's well-known that this induces a topology on $G$ which is topological group (precisely as the $D_n$ are translation invariant) and which has the $G_n$ as a local base at $0$ (easy to see). It will be a separated uniformity (or equivalently a $T_1$ group) iff $\bigcap_n G_n = \{0\}$ and by the Birkhoff-Kakutani theorem in that latter case $G$ is metrisable in the uniform topology.

For a possible reference see (here, e.g.) or any good book on topological groups (Bourbaki is one).