topology of completion of a module

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 $$x+M_i^*\subset U$$?

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 $$x$$.