Is every finite lattice isomorphic to a union-closed family of sets containing $\emptyset$?

Question

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!

Answer 1

I am converting my comment into an answer at the request of the proposer. Given $x\in L$, let $S_x=\{y\in L\, :\, y\not\geq x\}$.

Answer 2

Finite complete lattices and sup-preserving functions form a very nice category C; the category of models for the theory with one nullary operation (say 0) and a commutative, associative, idempotent binary operation (say +) having 0 as unit. This theory is commutative, so C has homs and tensors, making it a monoidal closed category with unit object (say I) the free model on one generator. Moreover, Hom( - , I) is a duality. Hom(X,I) is just the poset X turned upside down, and Hom(f,I) is the right adjoint of the homomorphism f if we think of the objects of C as categories and the maps of C as colimit-preserving functors. If P(S) denotes the Boolean algebra of subsets of S, as an object of C, then P(S) is the free model on S. Every model X is a quotient of some P(S), and hence, by duality, a subobject of some P(S'), which answers the question in the affirmative. A convenient set of generators for a model X is the set of those elements x for which the subset { a | 0 <= a <= x} is linearly ordered. So X embeds as a poset in P(S') where S' = { x | { x <= a <= 1} is linearly ordered} where 1 is the top element of X. X is distributive if for any x, y and z with z <= x+y there exist a and b with a <= x, b <= y and z = a+b (i.e. X has pullbacks of sup-diagrams).

All this pother is saying no more than Richard Stanley's more compact comment. A question: are semisimplicial resolutions of objects of C by free objects finite? In other words, given a finite sup-lattice X is there an essentially finite semi-simplicial sup-lattice, with free members, whose sup-lattice of connected components is isomorphic to X? My guess is yes.