Let $\mathcal{A}$ be a finite family of sets. For $\mathcal{A}' \subseteq \mathcal{A}$, define $\cup \mathcal{A}'=\bigcup_{A \in \mathcal{A}'}A$. Let $U(\mathcal{A})=\{\cup \mathcal{A}' : \mathcal{A}' \subseteq \mathcal{A}\}$, considered as a poset ordered by inclusion.
$U(\mathcal{A})$ is a lattice if every pair $x,y \in L$ has a unique largest common lower bound, called their $\textbf{meet}$, written $x \wedge y$, and also has a unique smallest common upper bound, called their $\textbf{join}$ and written $x \vee y$. That is to say, $\forall z \in U(\mathcal{A})$
$$z \leq x \text{ and } z \leq y \rightarrow z \leq x \wedge y$$
$$z \geq x \text{ and } z \geq y \rightarrow z \geq x \vee y$$
With a bit of help from you all, I've managed to show that $U(\mathcal{A})$ is a lattice.
I'm now pondering if every finite lattice is of this form? It seems like it should be true that corresponding to any lattice we can construct another lattice with just sets, but I do not know how to prove this. Also, I wouldn't be suprised if it wasn't actually true, as things like this that seems to be oviously true often are not.
I would appreciate some insight on this one!! Thanks