In Chapter 10 of Atiyah-Macdonald's Introduction to Commutative Algebra they define, for a first-countable abelian topological group $G$, the completion $\hat G$ of $G$ to be the quotient group of all Cauchy sequences modulo null sequences. Then they define the notion of completeness by saying that $G$ is complete if the map $\phi:G\to\hat G$ mapping an element $g\in G$ to the constant sequence $(g)_n$ is an isomorphism. Being an isomorphism means:
- $\phi$ is surjective, this is, any Cauchy sequence in $G$ converges: indeed, given a Cauchy sequence $(g_n)_n$ in $G$, it is the image of some $g\in G$ via $\phi$, and it is easy to see that $g$ is a limit of $(g_n)_n$, so it converges.
- $\phi$ is injective, this is, $G$ is Hausdorff: since $\ker\phi$ is the closure of $\{0\}$, so points are closed if and only if $\ker\phi$ is 0.
The first point (namely that $\phi$ is surjective) is precisely the usual definition of completeness in metric spaces: all Cauchy sequences converge. The second point (that $\phi$ is injective) seems to be an independent statement. However, I have not been successful in finding an example of a group $G$ which is complete in the metric space sense but with $\phi$ not injective (i.e., not Hausdorff).
Can you give an example of a non-Hausdorff first countable abelian topological group which is complete in the sense that Cauchy sequences converge?
How about this for an example: $G = \mathbb{Z}/2\mathbb{Z}$ with the indiscrete topology. All sequences converge, in fact all sequences converge to the identity element and to the non-identity element. The only neighborhood of the identity is $G$ itself, so all sequences are Cauchy, and all sequences are null. It follows that $\hat G$ is the trivial group, and $\phi : G \to \hat G$ is not injective.