 Minimal-member bound for a Razborov-closed family

For $r\ge 2$, an $r$-closed set family has at most $(r-1{)}^k$ inclusion-minimal members of size $k$. This is stronger than the factorial estimate obtained from the Erdős–Rado sunflower lemma alone.

Here is the induction on $r$. For $r=2$, two distinct minimal members force their proper intersection, a contradiction, so at most one minimal member exists. For $r>2$, choose a minimal $k$-set $Z$ and group the minimal members $X$ by $W=X\cap Z$. The remainder $X\setminus Z$ is minimal in the family of sets $B$ outside $Z$ for which $B\cup W$ belongs to the original family.

That remainder family is $(r-1)$-closed: adding $Z$ as the final witness turns each $(r-1)$-witness forcing step into an original $r$-witness step. Induction bounds the group for $W$ by $(r-2{)}^{k-|W|}$; the group $W=Z$ contains only $Z$. Therefore the binomial theorem givesThe case $k=0$ has at most the single empty set. The forcing argument explains why the stronger bound uses the full closure rule rather than only a sunflower prohibition.^[ref]