Two cube edges in every direction force many vertices ($|\mathcal{F}|\ge \lceil 3n/2\rceil +1$)

For a family in the $n$-dimensional Boolean lattice, the loss of trace cardinality on deleting coordinate $i$ equals the number of pairs $S,S\cup \{i\}$ both in the family. If every direction has at least two such edges, downward coordinate compression preserves that property. The resulting down-set contains the empty set and all $n$ singletons. Each coordinate must also occur in a two-element member. Those pairs cover all $n$ coordinates and number at least $\lceil n/2\rceil$, giving the bound. For $n\ge 2$, the empty set, all singletons and a minimum collection of pairs covering every coordinate attain it.