If a family of sets $\mathcal{F} \subseteq 2^E$ is union-closed and contains $\emptyset$, then $\mathcal{F}$ forms a lattice under the set-inclusion order. To see this, note that unions give the join operation, and that for $X, Y \in \mathcal{F}$, if we let $$Z = \bigcup \{ W | W \in \mathcal{F}\text{ and }W \le X, Y \},$$ then $Z \in \mathcal{F}$ is the meet of $X, Y$.

Now, I'd like to know whether a converse of this is true. Namely, if I am given a finite lattice $L$, can I find a union-closed (and containing $\emptyset$) family $\mathcal{F}$ over some set $E$, such that $L$ is isomorphic to $\mathcal{F}$?

*Background and discussion:*

I'm trying to characterize some class of lattices, and I have proven that they are always isomorphic to such an $\mathcal{F}$. I strongly suspect that they in fact are characterized by much stronger properties. But, I would like to know whether being isomorphic to such an $\mathcal{F}$ places any restrictions on the class of lattices whatsoever.

The question could be viewed as a generalization of Birkhoff's representation theorem, and also as a variant of this question: Is every finite poset a subset of a finite complemented distributive lattice?. However, the straightforward approach of that question, and generalizing Birkhoff-type ideas of join and meet irreducible, seems insufficient. Or, perhaps the result is false, but I haven't been able to find a counterexample $L$ or establish some property satisfies by this type of $\mathcal{F}$ but not by general lattices.

Thanks ahead of time!

3more comments