Birkhoff–Frink theorem

Remarks

From 2 to 3: the subalgebra generated by a set is an algebraic closure operator, since each operation has finitely many arguments. From 3 to 1: in the lattice of closed sets, the compact elements are the closures of finite sets, and every closed set is the union of these. From 1 to 2: one equips the set $K$ of compact elements of $L$ with operations whose subalgebras are exactly the ideals of the join-semilattice $K$. The map $a \mapsto \set{k \in K : k \le a}$ identifies $L$ with these ideals. For example, the lattice of subgroups of a group and the lattice of ideals of a ring are algebraic. The lattice of closed subsets of $[0, 1]$ is complete but not algebraic, so it is not a subalgebra lattice.

Going deeper

For a student

The theorem explains why compactness is the right abstraction of "finitely generated": the compact elements of a subalgebra lattice are the finitely generated subalgebras. It is the first of two representation results: the Grätzer–Schmidt theorem shows that the same lattices are also exactly the congruence lattices of algebras. Together with the description of algebraic lattices as ideal lattices of join-semilattices with $0$, it gives several equivalent pictures of the same class.