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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31
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Lubell-Yamamoto-Meshalkin inequality for antichains in a Boolean lattice

Statement

Let AA be an nn-element set and let F⊆P(A)\mathcal F\subseteq\mathcal P(A) be an antichain. Then

∑S∈F1(n∣S∣)≤1.\sum_{S\in\mathcal F}\frac{1}{\binom n{|S|}}\le1.

Facts & Assumptions

Given: An nn-element set AA and an antichain F\mathcal F in its Boolean lattice.

[L1]

There are n!n! maximal chains in B(A)B(A), and a fixed kk-set belongs to exactly k!(n−k)!k!(n-k)! of them (The Boolean lattice on an nn-element set has n!n! maximal chains, and exactly k!(n−k)!k!(n-k)! contain a fixed kk-set).

[F1]

An antichain contains no two comparable distinct elements (Antichains, chain covers, and antichain covers of a poset).

[L2]

For 0≤k≤n0\le k\le n, (nk) k!(n−k)!=n!\binom nk\,k!(n-k)!=n! ((nk) k! (n−k)!=n!\binom{n}{k}\,k!\,(n-k)! = n! for k≤nk \le n; hence (nk) k!=nk‾\binom{n}{k}\,k! = n^{\underline{k}}, the quotient n!/(k!(n−k)!)n!/(k!(n-k)!) is a natural number, and (nk)=(nn−k)\binom{n}{k} = \binom{n}{n-k}).

[F2]

Finite sums may be indexed by an arbitrary finite set and reindexed without changing their value (The sum ∑i∈Sai\sum_{i \in S} a_i over a finite index set, and its product form).

Proof

technique · direct
1.1

Count pairs (S,C)(S,C) where S∈FS\in\mathcal F and CC is a maximal chain containing SS. By [L1], the number is ∑S∈F∣S∣!(n−∣S∣)!\sum_{S\in\mathcal F}|S|!(n-|S|)!.

givenL1F2
1.2

A maximal chain contains at most one member of F\mathcal F, because all members of a chain are comparable. Hence the number of pairs is at most the number n!n! of maximal chains.

givenF1L1
2.1

Combining steps 1.1 and 1.2 and dividing by the positive number n!n! gives ∑S∈F∣S∣!(n−∣S∣)!/n!≤1\sum_{S\in\mathcal F}|S|!(n-|S|)!/n!\le1.

step 1.1step 1.2algebra
3.1

By [L2], each summand in step 2.1 equals 1/(n∣S∣)1/\binom n{|S|}. Substitution yields the asserted LYM inequality.

step 2.1L2algebra∎

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