TITLE: Are the error terms of the partial sums of inclusion-exclusion unimodal?
QUESTION [11 upvotes]: I often teach inclusion-exclusion: $$|A ∪ B| = |A| + |B| − |A ∩ B|$$
by suggesting that $|A∩B|$ is a correction factor for $|A|+|B|$.  Then I teach the three set version:
$$|A∪B∪C| = |A| + |B| + |C| − |A∩B| − |A∩C| − |B∩C| + |A∩B∩C|$$
again suggesting that $−(|A∩B|+|A∩C|+|B∩C|)$ and then $+|A∩B∩C|$ are correction factors.
Well, it occurred to me that maybe using more terms of the general inclusion-exclusion might not always be a good idea.
Indeed with five sets it is easy to find examples where $|A|+|B|+|C|+|D|+|E|$ is a better approximation to $|A∪B∪C∪D∪E|$ than is the second partial sum: $$|A|+|B|+|C|+|D|+|E| −\left({|A∩B|+|A∩C|+|A∩D|+|A∩E|+|B∩C|+ \atop |B∩D|+|B∩E|+|C∩D|+|C∩E|+|D∩E|}\right).$$
In other words, the "remainder term" in inclusion-inclusion is not necessarily monotonically non-increasing.
However, I haven't found an example where the remainder term is not unimodal.
Given a collection $\{A_i : 1 ≤ i ≤ k \}$ of subsets of $\{ 1, \ldots, n \}$, define the remainder terms
$$R_m = \left|\;\; \sum_{r=m}^k (-1)^r \sum_{I \in \Large\binom{[k]}{r}} \left| \bigcap \{ A_i : i \in I \}\right|\;\; \right|$$
Must $R$ be unimodal?

Is it always the case that there is some $j$ with $R_2 ≤ R_3 ≤ \cdots ≤ R_j ≥ R_{j+1} ≥ \cdots ≥ R_k$?

REPLY [9 votes]: Here's an explicit example of non-unimodality. We use $7$ sets: $$\eqalign{A_1&=A_2=\lbrace a,b,c,d,e,f,g,h,i,j,k\rbrace\cr A_3&=\lbrace g,h,i,j,k\rbrace\cr A_4&=A_5=A_6=A_7=\lbrace k\rbrace\cr}$$ Then the absolute remainders go $20,19,20,15,6,1,0$. 
This may be, in some sense, the minimal example. I don't think it can be done with fewer than $7$ sets, or fewer than $11$ elements all told (but my reasons involve considerable hand-waving and appeals to plausibility).