Let $B\_i^p$ be a family of sets, where $p\in \mathbb{N}$ and $i \in I$, $I$ being a directed set, and such that, for every $i$, we have a descending chain of inclusions


$$
\dots \supset B\_i^{p-1} \supset B\_i^p \supset B\_i^{p+1} \supset \dots
$$


Question: is the following implication true?


$$
\bigcap\_p B\_i^p = \emptyset, \ \text{for all} \ i \quad \Longrightarrow \quad \bigcap\_p \varinjlim\_i B\_i^p = \emptyset \ .
$$


Since $\bigcap\_{}$ is a limit, this seems a problem of an interchange of limits and filtered colimits and indeed there is a universal map


$$
\varphi: \varinjlim\_i \bigcap\_p B\_i^p \ \longrightarrow \ \bigcap\_p \varinjlim\_i B\_i^p
$$


If $\varphi$ was a bijection, then my implication would be true with no doubts, 
but, since the intersection is not finite, I can not say that $\varphi$ is a bijection. Nevertheless, could my implication still be true, without $\varphi$ being a bijection?


The reason behind my question is the following: let $(A\_i, F\_i)$ be a directed family of filtered sets (or abelian groups, or modules; in fact, in my problem they are cochain complexes). Since filtered colimits (direct limits) are exact, you can define a filtration on the colimit like this:


$$
F^p\varinjlim\_i A\_i = \varinjlim\_iF\_i^pA\_i \ .
$$


Now assume all the filtrations $F\_i$ are Hausdorff; that is, $\bigcap\_p F\_i^pA\_i = 0$ for all $i$. Is it then necessarily true that the filtration $F$ on $\varinjlim\_iA\_i$ is Hausdorff too?


This question is a sequel to my previous question [Convergence of right half-plane spectral sequence bounded on the right](https://mathoverflow.net/questions/32762/convergence-of-right-half-plane-spectral-sequence-bounded-on-the-right) . Despite Tilman's counterexemple to my guess there, I think I've managed almost to prove it because my spectral sequences are right half-plane and this is the final detail I need.



Colimit of intersections
Unfortunately, there seems to be a counterexample: Let $B^p\_i$ be the interval $(p-i,\infty)$ in the integers $\mathbb{Z}$, for natural numbers $p$ and $i$. In this case, we have $B^{p+1}\_i\subset B^p\_i$, and for any fixed $i$, we have $\bigcap\_p B^p\_i=0$, since $p$ runs out to infinty. But for fixed $p$, the limit of $B^p\_i$ as $i$ increases is all of $\mathbb{Z}$. Thus, the intersection of these limits is also $\mathbb{Z}$, which is not empty.


A simpler counterexample, using the same idea: let $B^p\_i=A$ for some fixed non-empty $A$, if $p\lt i$, otherwise $B^p\_i=$ emptyset. In this case, for fixed $i$ we have $\bigcap\_p B^p\_i=0$, since eventually $p$ exceeds $i$, but for fixed $p$, the $B^p\_i$ are eventually equal to $A$ as $i$ increases, and so have limit $A$.


