Let $M$ be a filtered module which is Hausdorff and complete with respect to the topology defined by the filtration. I want to show that if the sequence $\{x_n\}$ tends to $0$ the series $\sum x_n$ converges in $M$.
So we have to show that the sequence $s_n=\sum_{i=1}^{n}x_i$ converges in $M$. Since $M$ is complete enough to show that $\{s_n\}$ is a Cauchy sequence. Let $N_0$ denotes the set of filtration or the fundamental system of neighbourhood around $0$ and choose an arbitrary $\mathcal{U} \in N_0$, then there exists $k \in \mathbb{N}$ such that $x_n \in \mathcal{U}$ for all $n \geq k$, i.e., $s_n-s_{n-1} \in \mathcal{U}$ for all $n \geq k$. From here how do I complete the argument ? and also where exactly the Hausdorff is used, I know in a Hausdorff space if limit exists it exists uniquely. I need help, thanks.
You have to show that given $\mathcal{U}$, there exists $N$ such that for $m,n \ge N$ you have $s_m - s_n \in \mathcal{U}$. By assumption you can choose $N$ such that $s_{k+1} - s_k = x_{k+1} \in \mathcal{U}$ for $k \ge N$. We may assume that $m \ge n$. Then $$s_m - s_n = (s_m - s_{m-1}) + \dots + (s_{n+1} - s_n) \in \mathcal{U}.$$ Hausdorffness is not strictly necessary, but it is often included in the definition of completeness.