This is a strengthened dual version of this answered question. Here we are adding to that question requirement $3$ below.
Let $\mathcal{F}$ be a family of $n$ finite sets. In this case, the family can be regarded as a multiset, since it is allowed to contain multiple instances of the same set. Let $U(\mathcal{F})$ be the union of all sets in $\mathcal{F}$.
We require that:
- $\mathcal{F}$ is union-closed ($\mathcal{F}$ must contain at least one instance of $A \cup B$ for any $A,B \in \mathcal{F}$);
- every element in $U(\mathcal{F})$ belongs to at most $\lfloor (n+1)/2 \rfloor$ sets of $\mathcal{F}$ (every set is counted with its multiplicity);
- if $A \in \mathcal{F}$ has multiplicity $> 1$ then there must exist at least $\lceil (n+1)/2 \rceil$ members of $\mathcal{F}$, counted with their multiplicity, which are (not necessarily proper) subsets of $A$.
Is it possible to prove or disprove that, for any possible choice of $\mathcal{F}$, there exist two disjoint sets in $\mathcal{F}$?
