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In any topological group it is easy to show open subgroups are closed.

Prove that the converse holds in the following situation:

Let G be an abelian group, and let $$G=G_0 \supset G_1 \supset G_2\supset \dots$$ be a chain of subgroups. We turn $G$ into a topological group by defining the following topology: $U \subset G$ is open if and only if for every $x \in U$ there is some $n \in \mathbb{N}$ such that $x+G_n \subset U$.

I want to show that every closed subgroup of $G$ is open.

Possible useful facts:

  • The open subsets of $G$ are precisely unions of cosets of the $G_i$'s.
  • A subgroup $K \subset G$ is open if and only if there is some $n \in \mathbb{N}$ with $G_n \subset K$.
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This is false.

Consider $G=\mathbb{Z}$, with $G_n=p^n\mathbb{Z}$. Then $\{0\}$ is a closed subgroup which is not open.

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