= Solution
{c}
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 $\binom ni$ members. Thus the bound in the <weighted theorem for k-Sperner families> is always attained:
$$
\boxed{\mathcal A=\bigcup_{i\in I}X^{(i)},\qquad |I|=k,\qquad w(\mathcal A)=\sum_{j=1}^k u_{(j)}.}
$$
Back to article page
