Let $G$ be an abelian group and assume that there are some subgroups $H_i$ For $i\in\mathbb Z$ with the following properties:
- Each $H_i$ is a complete topological group
- $H_i\supset H_{i+1}$
- $G=\bigcup_{i\in\mathbb Z} H_i$ and $0=\bigcap_{i\in\mathbb Z} H_i$
We now endow $G$ with the direct limit topology: a subset is open iff each intersection with $H_i$ is open.
Question: Is it true that each open neighborhood of $0$ in $G$ must contain some $H_i$? Intuitively, since all $H_i$ shrink to $0$ this should be true, but I don't have a formal proof
Here's a sketch of the proof. Note that I am not very familiar with complete topological groups, so I will use a few fact which hold for complete metric spaces and hope they generalise to this context.
Suppose by absurd there exists $U$ open neighbourhood of $0$ such that $\forall i \in \mathbb{Z}$ $H_i \not \subset U$. Then $H_i \setminus U$ is a non-empty closed subset of $H_i$ thus we can take $h_i \in H_i \setminus U$. By the condition $\bigcap_i H_i =0$ you should get $\lim_{i \in \mathbb{N}} h_{-i}=0$. But $H_0 \setminus U$ is a closed subset of $H_0$ complete, hence itself complete. Therefore the limit of the sequence $0$ should also be in $H_0 \setminus U$ and you have the absurd.
The fact that closed subsets of complete spaces are complete and that limit of $h_{-i}$ must be zero are trivial for metric spaces but I think you should be able to adapt the proof even for complete groups.
What makes me unsure about my proof is that I needed the completeness only of one of the groups $H_i$ while you state that all the $H_i$ have this property. Anyway I hope that this would help in some way.