This is a question about completion in the sense of commutative algebra, i.e. inverse limit of quotients. Let $G$ be a topological abelian group, $G_0 \supseteq G_1 \supseteq \cdots$ and $H_0 \supseteq H_1 \supseteq \cdots$ two filtrations of subgroups of $G$. Each induces a topology by setting the filtration to be a fundamental system of neighbourhoods of $0$.
In a course I'm following, it was proved that if the two filtrations induce the same topology, then they induce the same completion as well. (This is clear, because the completion depends only on the topology.) But the converse, namely that if they induce the same completion then they induce the same topology, wasn't stated.
I've thought, asked and looked around for counterexamples but couldn't find any. What would be an example of two filtrations producing distinct topologies yet the same completion?
I have an example of a group $G$, and two filtrations $G = G_0 \supseteq G_1 \supseteq G_2 \dots$ and $G = H_0 \supseteq H_1 \supseteq H_2 \supseteq \dots$ such that the filtrations induce different topologies on $G$, but the completions of $G$ with respect to each topology are isomorphic as groups. The induced topologies on these completions are not equivalent, however, and so the completions are isomorphic as groups, but not as topological groups.
Consider the group $G = A[[x]]$, the additive group of formal power series with coefficients in some abelian group $A$. The completion of $G$ with respect to the $(x)$-adic topology on $G$ is isomorphic to $G$. (One way to see this is to note that $G$ is the completion of $A[x]$ with respect to the $(x)$-adic topology on $A[x]$.) In this case, the corresponding filtration is given by $G_n = (x^n) G$.
On the other hand, we could complete $G$ with respect to the discrete topology. This is induced by the filtration $H_0 = G$ and $H_n = 0$ for $n \geq 1$. (I only specify that $H_0 = G$ because this appears in the definition of a filtration in Atiyah-MacDonald, but this doesn't seem to be strictly necessary. $G$ will be a neighbourhood of $0$ even without this convention.) In this case, the completion of $G$ is again isomorphic to $G$ as a group. However, the induced topology is the discrete topology and not the $(x)$-adic topology.
More generally, if $G = G_0 \supseteq G_1 \supseteq G_2 \supseteq \dots$ is a filtration on $G$, then the induced topology on $\hat{G}$ is the discrete topology if $\bigcap_{i = 0}^{\infty} G_i = G_n$ for some $n$, because we then have that $\hat{G}_n = 0$. Thus if $\bigcap_{i = 0}^{\infty} G_i \neq G_n$ for each $n$ (for example if $G$ is Hausdorff and not discrete), then we can consider the group $\hat{G}$. Its completion with respect to the topology induced by the filtration $(\hat{G}_n)_{n = 0}^{\infty}$ is ismorphic to $\hat{G}$. On the other hand, its completion with respect to the discrete topology is also ismorphic as a group to $\hat{G}$, but with the discrete topology.