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

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.

• yes ${}{}{}{}{}$ Commented Apr 13 at 22:00
• @mathworker21 care to elaborate? Commented Apr 13 at 22:11
• Given $x\in L$, let $S_x=\{y\in L\,:\,y\not\geq x\}$. Commented Apr 13 at 23:40
• @RichardStanley wow! Very simple - thank you. If you'd like to copy your comment into an answer (just to 'check the box' etc) that would be excellent Commented Apr 14 at 6:26
• @ClayThomas: sorry, I decided these comments actually made sense as an answer to the related question mathoverflow.net/questions/450680. By the way now you can see that all Stanley did for the answer for this question is to take complements for the answer to that one (changing an intersection-closed family to a union-closed family). Commented Apr 14 at 17:53

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