Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 109 / 4 / ii

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- Solution ii

Solution

Fix

A \in F

and consider the traces

T = {A \cap B : B \in F ∖ {A}} \subseteq P (A) .

For distinct

B, C \in F ∖ {A}

, the hypothesis applied in both orders says

A \cap B \neq \subseteq C, A \cap C \neq \subseteq B .

Equivalently, neither trace contains the other. Thus the traces are distinct and form an antichain in the Boolean lattice on the

2 r

points of

A

. By Sperner theorem,

∣ F ∣ - 1 = ∣ T ∣ \leq (r 2 r),

which is the required bound.

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