Topological (abelian) groups with a neighbourhood basis of $0$ consisting of subgroups.

Question

Let $G$ be a topological abelian group (written additively).

Suppose that $\{G_\alpha\}_{\alpha\in A}$ is a neighbourhood basis of $0$ consisting of subgroups.

I want to prove that $U\subseteq G$ is open if and only if it is a union of some cosets of some of the $G_\alpha.$

Does it suffice to note that $\{g+G_\alpha \, : \, g \in G,\; \alpha \in A\}$ is a base for the topology on $G?$

Many thanks!

Answer 1

Yes, that's right. And it doesn't depend on the neighborhood basis consisting of groups. If $\{G_\alpha\}_{\alpha\in A}$ is any base for the neighborhoods of zero, then $\{g+G_\alpha \, : \, g \in G,\; \alpha \in A\}$ is a base for the topology.

Answer 2

Since translations are homeomorphisms, $\{x+G_\alpha\}_{\alpha\in A}$ is a basis of neighborhoods of $x$.

If $U$ is an open set, for every $x\in U$ there is $\alpha(x)\in A$ such that $x+G_{\alpha(x)}\subseteq U$. Thus $U$ is a union of cosets of the requested type.

The converse follows from the fact that a subgroup $H$ which is a neighborhood of $0$ is open. Indeed, if $x\in H$, then $H=x+H$ is a neighborhood of $x$.