k-Sperner family

A $k$-Sperner family contains no inclusion chain of length $k+1$. Every maximal chain in a Boolean lattice therefore meets it in at most $k$ members, so its Lubell mass is at most $k$.

Let $1\le k\le n+1$. Give each subset of $[n]$ a positive weight $w(i)$ depending only on its size $i$, and define the total set family weight and full-level weights byThe largest weight of a k-Sperner family is the sum of the $k$ largest numbers $u_0,\dots ,u_n$. Every maximizing set family is a union of $k$ full rank levels whose weights are largest; ties permit any choice among the tied levels.

Indeed, let $a_i=|\mathcal{F}\cap (\genfrac{}{}{0ex}{}{[n]}{i})|/(\genfrac{}{}{0ex}{}{n}{i})$ be the occupied proportion of level $i$. Then $0\le a_i\le 1$, ${\sum }_i{a}_i\le k$ by the Lubell mass bound, and $W(\mathcal{F})={\sum }_i{u}_i{a}_i$. Allocating this mass to the largest $u_i$ proves the upper bound, attained by the corresponding full levels. Since every $u_i$ is positive, equality forces ${\sum }_i{a}_i=k$. Partition $\mathcal{F}$ into $k$ antichains according to the longest chain ending at each member. Each part has Lubell mass at most one, so equality forces every part to have Lubell mass one. Equality in the LYM inequality makes each part a full rank level.

If $k>n+1$, every set family is a k-Sperner family, and the unique maximum with positive weights is the entire Boolean lattice. Thus the general maximum uses the $\min \{k,n+1\}$ largest level weights, rather than asking for more distinct levels than exist.

If $t$ is the $k$th largest rank weight $u_i$, exactly $h$ levels have weight greater than $t$, and exactly $s$ have weight $t$, then there are $\left(\genfrac{}{}{0ex}{}{s}{k-h}\right)$ maximizing families. For fixed $n,k$, the possible counts are exactly $\left(\genfrac{}{}{0ex}{}{s}{j}\right)$ with $1\le j\le k$ and $j\le s\le n+1-k+j$.

Among $k$-Sperner subfamilies of $\mathcal{P}([n])$, the maximum cardinality is the sum of the $k$ largest binomial coefficients $\left(\genfrac{}{}{0ex}{}{n}{j}\right)$. The union of the corresponding levels attains the bound. For the upper bound, the Lubell mass is at most $k$, and assigning its available mass to the levels with largest binomial coefficients maximizes cardinality.