If $\{M_\lambda\}_{\lambda\in\Lambda}$ is a collection of submodules of $M$, which defines a linear topology on $M$, then why is each $M_\lambda$ both open and closed in $M$?
I understand that $M_\lambda$ is open since $x+M_\lambda \subset M_\lambda$, where the inclusion "$\subset$" comes from the fact that $M_\lambda$ is an abelian group, and so it's cosets $x+M_\lambda$ are subgroups of $M_\lambda$, correct?
Also, why is $M_\lambda$ closed? If we have that $M_\lambda$ is open, then so is $x+M_\lambda$, and also the union $\cap_\lambda x+M_\lambda$. We would then have that $M_\lambda$ is closed if $\cap_\lambda x+M_\lambda$ equals the complement $M-M_\lambda$, but I don't see why it does that?
All the submodules $M_{\lambda}$ are open simply by the definition of the linear topology on $M$: To have a linear topology defined by the system of submodules $\{M_{\lambda}\}_{\lambda}$ means that you proclaim $\{M_{\lambda}\}_{\lambda}$ to be system of open neighbourhoods of $0$.
To show that each $M_{\lambda}$ is closed, note that each coset $x+M_{\lambda}$ is open: this is because the translation map $y\mapsto x+y$ is necessarily a homeomorphism on $M$ (this follows from axioms of a topological module), so it takes the open set $M_{\lambda}$ to an open set $x+M_{\lambda}$. Consequently, $\bigcup_{x \notin M_{\lambda}}x+M_{\lambda}$ is open, and $M_{\lambda}$ is its complement. Thus, $M_{\lambda}$ is closed.