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.

You had the right idea, so let me spell out the argument with the fix I suggested in the comments:
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 $V \times V \subset V_1 \times V_2 \subset a^{-1}(U)$.
