Birkhoff–Frink representation of algebraic closure systems

Remarks

Proof idea. For each finite set $B = \set{b_1, \dots, b_n} \subseteq A$ with $n$ distinct elements and each $c \in C(B)$, add an $n$-ary operation. It sends $(b_1, \dots, b_n)$ to $c$ and sends every other tuple to its first entry; for $n = 0$ it is the constant $c$. A closed set is closed under these operations because $C(B) \subseteq C(X)$ when $B \subseteq X$. A set closed under them contains $C(B)$ for each finite $B$ inside it, so it is closed because $C$ is algebraic.

Consequence: every algebraic lattice is isomorphic to the subalgebra lattice of some algebra. Indeed, an algebraic lattice is isomorphic to the lattice of closed sets of some algebraic closure operator.

The hypothesis that $C$ is algebraic is necessary. For any algebra with finitary operations, the union of a chain of subuniverses is a subuniverse. On $A = \N$, let $C(X) = X$ for finite $X$ and $C(X) = \N$ for infinite $X$. This is a closure operator, and it is not algebraic. The closed sets $\set{0, 2, \dots, 2k}$ form a chain whose union, the set of even numbers, is not closed. So no algebra on $\N$ has exactly the closed sets of $C$ as its subuniverses.

Going deeper

For a student

The theorem is the converse of a basic fact: in any algebra, the subuniverse operator $\operatorname{Sg}$ is an algebraic closure operator. Together they show that algebraic closure systems, subuniverse lattices and algebraic lattices describe the same objects. A similar but harder result for congruences is the theorem of Grätzer and Schmidt, which says that every algebraic lattice is a congruence lattice.