# Let $L$ be a lattice. Is there a union closed family of subsets of $[n]$ ordered by inclusion that correspond to $L$.

Let $$\mathcal{C}$$ be a collection of subsets of $$[n]$$ that is closed under taking unions (including taking the union of the empty set which is the empty set). Then $$(\mathcal{C},\subseteq)$$ is a lattice. But is the converse true? In other words, given any lattice is there a union closed family of subsets of $$[n]$$ (for some $$n$$) whose members obey the order of the lattice?

I looked at all the lattices with up to 6 points and haven't found a counter-example but I can't come up with any algorithm that would prove the statement.

P.S. I do know that the converse is true for distributive lattices. Also, I am only considering finite lattices.

• Are you only considering finite lattices? May 24, 2022 at 18:49
• Yes. Thank you. Here, all lattices are finite. May 24, 2022 at 18:50

If $$L$$ is a lattice and $$x\in L$$, define $$A_x = \{z\in L\;|\;z\not\geq x\}$$. The function $$A\colon L\to {\mathcal P}(L)\colon x\mapsto A_x$$ is a join-embedding of $$(L,\vee)$$ into $$({\mathcal P}(L),\cup)$$. The collection $$\{A_x\;|\;x\in L\}$$ is therefore a union-closed set of subsets of $$L$$ which, when ordered by inclusion, forms a lattice isomorphic to $$L$$.