CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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A symmetric chain decomposition gives a second proof of Sperner's bound

Statement

If $A$ has $n$ elements, every antichain in $B(A)$ has cardinality at most $\binom n{\lfloor n/2\rfloor}$ .

Facts & Assumptions

Given: An $n$ -element set $A$ and an antichain $\mathcal F$ in $B(A)$ .

[L1]

The Boolean lattice has a partition into symmetric chains (Every finite Boolean lattice has a symmetric chain decomposition).

[F1]

An antichain contains at most one element from any chain (Antichains, chain covers, and antichain covers of a poset).

[F2]

Rank $\lfloor n/2\rfloor$ consists of the $\lfloor n/2\rfloor$ -subsets of $A$ and has cardinality $\binom n{\lfloor n/2\rfloor}$ (The Boolean lattice of subsets of a finite set and its rank levels, The set $[A]^{k}$ of $k$ -element subsets and the binomial coefficient $\binom{n}{k} := \lvert [n]^{k}\rvert$ ).

Proof

technique · direct

1.1

Fix the symmetric saturated-chain decomposition supplied by [L1]. Every chain in it meets rank $\lfloor n/2\rfloor$ exactly once, because its consecutive ranks run from some $r\le\lfloor n/2\rfloor$ through $n-r\ge\lfloor n/2\rfloor$ .

givenL1

2.1

Consequently the number of chains in the decomposition equals the cardinality of rank $\lfloor n/2\rfloor$ , hence equals $\binom n{\lfloor n/2\rfloor}$ .

step 1.1F2

3.1

By [F1], the antichain $\mathcal F$ contains at most one member from each chain. Its cardinality is therefore at most the number in step 2.1, which is Sperner's bound.

step 2.1F1∎

Depends on

Used by

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Sources

M. Keller and W. T. Trotter, Applied Combinatorics, §6.2 (standard reference, not scraped)