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 . 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.


1 Answer 1


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 $A$.

  • $\begingroup$ Ok. I'm afraid you're right: I'll have to look the problem more closely. $\endgroup$ Aug 10, 2010 at 11:33

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