Exhaustive filtration implies separated and complete filtration

Its union is $k$, so it is exhaustive. Its intersection is also $k\mathrm{\ne }0$, so it is not separated. Every quotient $k\mathrm{/}{F}_{p}k$ is zero, and the inverse system therefore has zero limit: a cone into zero objects has exactly the unique zero map into the zero object. The completion map $k\to 0$ kills the nonzero class of $1$ and is not an isomorphism. Thus the same exhaustive filtration fails both asserted conclusions.

Separately filter $S$ by $F_{-m}S=T_{m}S$ for $m\ge 0$ and $F_{p}S=S$ for $p>0$. This is exhaustive since $F_{0}S=S$. If a sequence lies in every tail, its coordinate $j$ is zero by taking $m=j+1$, so the filtration is separated. Its quotients are $k^m$ with truncation maps, and their limit is $P$ by [F2]. The completion map misses the constant-one sequence, hence is not onto. This second example shows that even adding separatedness to exhaustiveness does not force completeness. The index $m=0$ gives the zero quotient by the whole group; the zero group itself would satisfy all three properties and is not a refuting witness. All maps and sequences used are explicit and require no AC.