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Let $G$ be the usual: a topological Abelian group with a topology induced by a countable neighbourhood basis $G_n$ of zero such that $G = G_1 \supset G_2 \supset \dots$. Let $\widehat{G}$ denote the completion of $G$.

Apparently, from $ \widehat{\widehat{G}} \cong \widehat{G}$

it follows that if $G$ is complete then it is Hausdorff. Maybe it's just a bit late but it's not obvious to me. So if $G$ is complete we have $G \hookrightarrow \widehat{G}$ so that $G$ is a subgroup of $\widehat{G}$.

We know that if $H$ is the intersections of all neighbourhoods of zero then $G$ is Hausdorff if and only if $H = \{0\}$. Does this help here? Thanks for your help.

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    @DavideGiraudo Sorry, that is indeed a very important point I forgot to mention: complete in the sense that every Cauchy sequence has its limit in there. Since we don't have a metric we define Cauchy to mean that for every neighbourhood $U$ of zero there is a $k_0$ such that for all i,j > k_0 we have $x_j - x_i \in U$.2012-07-29

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$H$ is a subgroup of $G$. If $x\in H$, then $\{nx,n\in\Bbb Z\}\subset H$. In particular, the sequence $\{nx\}_{n\geq 1}$ is Cauchy, and converge to some $y$. Since $\{(n+1)x\}$ converges to $y$, by continuity of the addition $x=0$.