Let $G$ be a topological abelian group. Recall that a Cauchy sequence $(x_n)$ in $G$ is defined to be a sequence such that for any neighborhood $U$ of $0$, there exists an integer $N$ with $x_n-x_m\in U$ for all $n,m\geq N$. And a Cauchy sequence $(x_n)$ is said to be convergent to $0$ if for any neighborhood $U$ of $0$, almost all members of the sequence $x_n$ lie in $U$.
Question: Now let $(x_n)$ be a Cauchy sequence which does not converge to $0$, is there a neighborhood $U$ of $0$ such that there exists an integer $N$ with $x_n\not\in U$ for all $n\geq N$?
By definition, we can only obtain that there are infinitely many elements of the $(x_n)$ outside $U$. Maybe I missed something basic, I can't find a way to prove it is right.
Thanks!
Suppose there were no such $U$. Then given any neighbourhood $U$ of $0$, choose another $0$-neighbourhood $V$ with $V + V \subseteq U$. There is an $N$ such that $x_n - x_m \in V$ for all $n, m \ge N$ and by assumption there is an $N' \ge N$ such that $x_{N'} \in V$. This gives for $n \ge N'$: $$ x_n = (x_n - x_{N'}) + x_{N'} \in V + V \subseteq U$$ As $U$ was arbitrary, $x_n \to 0$. Contradiction.