Bourbaki

Exercise 5, § 5

Tag 03MH, permanent. Cite it as [Bourbaki, Tag 03MH].

Theory of Sets, III, 235

\* (a) Let E be a distributive lattice and ff be a mapping of E into a commutative semigroup M (written additively) such that

f(x)+f(y)=f(sup⁡(x,y))+f(inf⁡(x,y))f(x) + f(y) = f(\sup (x, y)) + f(\inf (x, y))

for all xx, yy in E. Show that for each finite subset I of E we have

f(sup⁡(I))+∑2n≤Card⁡(I)(∑H⊂I, Card⁡(H)=2nf(inf⁡(H)))=∑2n+1≤Card⁡(I)(∑H⊂I, Card⁡(H)=2n+1f(inf⁡(H))) f(\sup (I))+ \sum_{2n\leq \operatorname{Card}(I)} \left( \sum_{\substack{H\subset I,\ \operatorname{Card}(H)=2n}} f(\inf (H)) \right) = \sum_{2n+1\leq \operatorname{Card}(I)} \left( \sum_{\substack{H\subset I,\ \operatorname{Card}(H)=2n+1}} f(\inf (H)) \right)

(By induction on Card (I).) \*

(b) In particular, let A be a set, let (Bi)i∈I(B_i)_{i\in I} be a finite family of finite subsets of A, and let B be the union of the BiB_i. For each subset H of I, put BH=⋂i∈HBiB_H=\bigcap_{i\in H}B_i. Show that

Card⁡(B)+∑2n≤Card⁡(I)(∑Card⁡(H)=2nCard⁡(BH))=∑2n+1≤Card⁡(I)(∑Card⁡(H)=2n+1Card⁡(BH)). \operatorname{Card}(B)+ \sum_{2n\leq \operatorname{Card}(I)} \left( \sum_{\operatorname{Card}(H)=2n} \operatorname{Card}(B_H) \right) = \sum_{2n+1\leq \operatorname{Card}(I)} \left( \sum_{\operatorname{Card}(H)=2n+1} \operatorname{Card}(B_H) \right).
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