This is related to Serre, Local Algebra, Chpt 2, Sec 3.
"If $M$ is a filtered $A-$module, we write $\hat{M}$ for its hausdorff completion, this is an $\hat{A}-$module, isomorphic to $\lim M/M_n$(inverse limit)."
$A$ has decreasing filtration given by ideals $I_i$. $M$ has decreasing filtration given by submodule $M_i$. $I_iM_i\subset M_i$ is required.
$\textbf{Q:}$ Is $\hat{M}$ necessarily hausdorff? I will note the following. $Ker(M\to\lim M/M_n)=\cap_iM_i$. In particular, I have $M/\cap_iM_i\subset\lim M/M_n$. If $M$'s image is dense, I would hope that $\hat{M}$ is hausdorff. It seems pulling back neighborhood of $0$ around $\hat{M}$, I get neighborhood of $0\in M/\cap_iM_i$. However, I can always shrink. So I get intersection of neighborhood of $0\in\hat{M}$ necessarily $0$ as they are closed. Is this correct? If not, what is the mistake and can I also have a counter example? The naming "hausdorff completion" is confusing to me as it implies this completion is hausdorff irregardless $M$ hausdorff or not.
The answer is yes; this is from Strooker's Homological questions in local algebra.