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Assume $M$ is $A$ module, where $A$ is a commutative ring with unity. Assume $\{M_i,i\in I\}$ is a set of some submodules of $M$, where $I$ is a directed set( this means there is a partial order relation $\le$ on $I$, and whenever $i,j\in I$,there is a $k\in I$ satisfying $i\le k$ and $j\le k$ ). And $i,j\in I,i\le j$ implies $M_j\subset M_i$. The linear topology on $M$ (induced by $\{M_i,i\in I\}$), is the topology generated by $\{x+M_i,x\in M,i\in I\}$. Now let $\hat{M}$ be the completion of $M$ (induced by $\{M_i,i\in I\}$). To get the topology on $\hat{M}$, first we let $M/M_i$ be a discrete space, then take product topology on $\prod_iM/M_i$, finally take the subspace topology, as $\hat{M}$ is a subset of $\prod_iM/M_i$.

Now take $M_i^*$ be the kernel of project map $\hat{M}\to M/M_i$, i want to prove that the topology of $\hat{M}$ coincides with the linear topology of $\hat{M}$ induced by $\{M_i^*,i\in I\}$. And i'm stuck in one step: take a non-empty base element $U$ in the original topology of $\hat{M}$ , this should be the form of $U=\hat{M}\cap\left(\prod_{i\in I_0}U_i\times\prod_{i\in I,\not\in I_0}M/M_i\right)$,where $I_0$ is a finite subset of $I$, and $U_i$ is any subset of $M/M_i$(since $M/M_i$ has the discrete topology). Take any $x\in U$, i need to find one $M_i^*$ satisfying $x+M_i^*\subset U$. My lecture note says use the condition $I$ being a directed set, we can find $i\in I$ that $i>i_0,\forall i_0\in I_0$, this $i$ will do the work. Can anyone tell me why $x+M_i^*\subset U$?

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If $i > i_{0}$, then there is a natural map $M/M_{i} \to M/M_{i_{0}}$ compatible with the projections from $\hat{M}$ so the image of $M_{i}^{\ast}$ in $M/M_{i_{0}}$ is $0$, in particular the image of $x+M_{i}^{\ast}$ in $M/M_{i_{0}}$ is the same as the image of $x$.

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