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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?

$$\bigcap_p B_i^p = 0, \emptyset, \ \text{for all} \ i \quad \Longrightarrow \quad \bigcap_p \varinjlim_i B_i^p = 0 \emptyset \ .$$

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 http://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.

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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?

$$\bigcap_p B_i^p = 0, \ \text{for all} \ i \quad \Longrightarrow \quad \bigcap_p \varinjlim_i B_i^p = 0$$

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 $\varphi$ were 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 http://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.

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