# Sum of Cauchy sequences is Cauchy in an Abelian Topological Group

Let $G$ be a topological abelian group and suppose $0$ has a countable fundamental system of neighborhoods. Let $(x_n),(y_n)$ be Cauchy sequences of $G$. Why is it true that $(x_n+y_n)$ is a Cauchy sequence?

I tried to generalize the case of real sequences: my problem is that if $U$ is a neighborhood of $0$, then i would need to use something like $\frac{1}{2} U$, but obviously this does not make sense.

I also looked at this relevant question Sum of Cauchy Sequences Cauchy? however it was not very helpful, since it refers to metric topological groups.

Thanks.

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Well, every first countable Hausdorff topological group has a compatible metric by the Birkhoff-Kakutani theorem... More seriously: replace $\frac{1}{2}U$ by an open set $V$ such that $0 \in V$ and $V+V \subset U$. To find such a set $V$ use continuity of addition. – t.b. Sep 5 '12 at 14:43
@t.b. Hahaha...Ok, i am gonna try this, looks interesting :-) – Manos Sep 5 '12 at 14:48
But now what about the title question, possibly non-abelian? Now you have to specify left-Cauchy or right-Cauchy ... and then what? – GEdgar Sep 5 '12 at 15:23
@GEdgar: Its abelian, i will fix it. – Manos Sep 5 '12 at 18:01

Suppose that for every $0$-neighborhood $U$ there is a $0$-neighborhood $V$ such that $V + V \subset U$. Since $(x_n)$, and $(y_n)$ are Cauchy, there is $N$ such that $x_n - x_m \in V$ and $y_n - y_m \in V$ for all $m,n \geq N$ and hence $(x_n+y_n)-(x_m+y_m) = (x_n-x_m)+(y_n-y_m) \in V+V \subset U$ for all $m,n \geq N$, showing that $(x_n+y_n)$ is Cauchy.
To see that our hypothesis is in fact true, we can argue as follows: Since the addition map $a\colon G \times G \to G, (g,h) \mapsto g+h$ is continuous, we know that $a^{-1}(U) \subset G \times G$ is open. Also, $(0,0) \in a^{-1}(U)$, so there are open $0$-neighborhoods $V_1, V_2 \subset G$ such that $V_1 \times V_2 \subset a^{-1}(U)$ by the definition of the product topology. Now $V = V_1 \cap V_2$ is a $0$-neighborhood such that $V + V \subset U$, since $V \times V \subset V_1 \times V_2 \subset a^{-1}(U)$.