 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/iii/paper-109/1/c/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Choose

k

distinct ranks whose

u (i)

values are the

k

largest, resolving any tie arbitrarily, and take the union of their full levels. Along a strict chain in a partially ordered set, cardinality strictly increases, so this set family is a k-Sperner family. Its weight is the sum of the selected rank weights because a full level of rank

i

has

(i n)

members. Thus the bound in the weighted theorem for k-Sperner families is always attained:

A = i \in I ⋃ X^{(i)}, ∣ I ∣ = k, w (A) = j = 1 \sum k u_{(j)} .

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