A 'missing' condition in a question in Homological Algebra With very little knowledge of Homological Algebra, apparently the statement that 'Assuming some $mild$ conditions, if we have a filtered chain complex with trivial homology for its associated graded complex, then the complex is also acyclic' makes not-so-much sense to me. What are the conditions that we need to achieve such? Some little ideas or proof will be very helpful as well.
 A: Suppose $X_{\bullet}$ is a chain complex with decreasing filtration $\ldots\supset X^n_\bullet\supset X^{n+1}_\bullet\supset\ldots$ by subcomplexes $X^n_\bullet$. The question is to understand when the acyclicity of all quotients $X^n_\bullet / X^{n+1}_\bullet$ implies the acyclicity of $X_\bullet$.
A sufficient ('mild') condition is that in each degree $k$, the filtration $\{X^n_k\}_n$ of $X_k$ is effectively bounded, meaning that $X^n_k = X_k$ for $n\ll 0$ and $X^n_k=0$ for $n\gg 0$. For the proof, note first that since acyclicity concerns one degree a time, you may in fact assume that the entire filtration $\{X^n_\bullet\}_n$ is effectively bounded, i.e. finite. In this case, the claim follows from iteratively applying the long exact homology sequence.
Given the effectively bounded case, we can also tackle the case of a filtration which is exhaustive and degree-wise bounded below, i.e. for any $k$ we have $X^n_k=0$ for $n\gg 0$ and $X_k = \bigcup_n X^n_k$. Namely, in this case $X_\bullet$ is the directed limit of $X^n_\bullet$ as $n\to -\infty$, and all $X^n_\bullet$ carry an effectively bounded filtration with acyclic quotients. Hence all $X^n_\bullet$ are acyclic by what we have already seen, and so is $X_\bullet$, if filtered colimits are exact in the underlying abelian category (e.g. if it's a module category).
Remark: At this point, note that if we're not assuming that the filtration is exhaustive and, dually, separating, i.e. $\bigcap_n X_k^n=0$, there'll be no chance for the statement to be true - to see that, just pick any constant filtration.
What about bounded above or even unbounded complexes? Given a separating, bounded-above filtration, we might try to dualize the argument, but that would require exact directed limits, and such are rare in practice: Module categories, and more generally Grothendieck categories, never  have exact filtered limits (except for the zero category).
Finally, let's see an example. A very simple one is the chain complex $0\to {\mathbb Z}^{(\mathbb N)}\to {\mathbb Z}^{\mathbb N}\to 0$, i.e. you have an infinite sum of copies of ${\mathbb Z}$ embedded into the infinite product. As ${\mathbb Z}^{(\mathbb N)}\subsetneq {\mathbb Z}^{\mathbb N}$, the homology of this chain complex is not zero. On the other hand, consider the filtration by the subcomplexes $0\to {\mathbb Z}^{({\mathbb N}_{\geq n})}\to {\mathbb Z}^{{\mathbb N}_{\geq n}}\to 0$ where the first $n$ entries of the elements of ${\mathbb Z}^{(\mathbb N)}$ resp. ${\mathbb Z}^{\mathbb N}$ must vanish (as abstract complexes, these are all isomorphic to the original complex). The filtration quotients are all isomorphic to $0\to {\mathbb Z}\xrightarrow{1} {\mathbb Z}\to 0$, hence acyclic. 
