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Let $M$ be a complete $R$-module with respect to the filteration $\{M_n\}.$ Also let $N$ be a closed submodule of $M.$ Then how can I show that $M/N$ is complete with respect to the induced filteration, i.e., $\{(M/N)_n=(N+M_n)/N\}.$

We should show that $M/N$ is Hausdorff and every Cauchy sequence in $M/N$ converges. Since $N= \cap_{n=1}^{\infty}(N+M_n)$ clearly $M/N$ is Hausdorff. But How can I show that every Cauchy sequence in $M/N$ is convergent ? I need some help. Thanks.

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A Cauchy sequence in $M/N$ has the form $\{x_n+N\}$, where the $x_n\in M$ satisfy relations $x_{n+1}-x_n\in N+M_{s(n)}$ with $s(n)\to \infty$. Though $\{x_n\}$ might not be a Cauchy sequence in $M$ we can a define a Cauchy sequence $\{y_n\}$ in $M$ such that $x_n\equiv y_n\mod N$. The procedure is as follows. We set $y_1=x_1$. Assume that we have defined $y_n$ so that $y_n\equiv x_n\mod N$. Note that $x_{n+1}-x_n-u\in M_{s(n)}$ for some $u\in N$. We thus define $y_{n+1}=x_{n+1}-(x_n-y_n)-u$. By construction we have $y_{n+1}\equiv x_{n+1}\mod N$ and $y_{n+1}-y_n=x_{n+1}-x_n-u\in M_{s(n)}$. It follows that $\{y_n\}$ is a Cauchy sequence in $M$ and has therefore a limit, say $y\in M$. It is now easy to see that $y+N$ is the limit of $\{x_n+N\}=\{y_n+N\}$.

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