Why is the equivalent Cauchy sequence in a topological group well-defined? I was reading chapter 10 of Atiyah where I met the notion of 
equivalent Cauchy sequences for topological groups.
Atiyah does not explain the reason why equivalent Cauchy sequences indeed give a equivalence relation. I can manage to prove the reflexivity and symmetry of this relation. But I failed to work out the transitivity. 
Can anyon help me on this? Thank you.
FYI. A Cauchy sequence of a topological group $G$ is a sequence $(x_n)_{n \in \mathbb{N}}$ such that for any open neighbourhood $U$ of $0$ (the identity of $G$), there is $N \triangleq N(U)$ such that for any $n,m > N, x_n-x_m \in U$. 
And we say that two Cauchy sequences $(x_n)$ and $(y_n)$ are equivalent if $x_n-y_n$ converges to $0$, that is, for every open neighbourhood $U$ of $0$, there is $N$ such that for all $n > N, x_n-y_n \in U$.
I can see that for any open neighbourhood $U$ 0f $0$, if we can find an open neighbourhood $V$ of $0$ such that $V+V \subset U$, then the transivity will be proved. But, indeed how to explain the existence of such $V$?

The existence of such a $V$ follows from the definition of continuity of $+$ at $(0,0)$: 
There is a basic open neighbourhood $U_1 \times U_2$ that contains $(0,0)$ such that $+[U_1 \times U_2] = U_1 + U_2 \subseteq U$
Now use $V=U_1 \cap U_2$.
This works in any topological group.
