Lattice Theory Foundations
Algebraic Lattices and Compact Elements
Compact elements, algebraic lattices, and the theorem that the subuniverse and congruence lattices of any algebra are algebraic — the structural fact that finitary arity buys.
Learning objectives
- Define compact element and algebraic lattice
- Prove that finitely generated subuniverses are exactly the compact elements
- State the Grätzer–Schmidt representation theorem and its significance
Compactness
An element c of a complete lattice L is compact if whenever c ≤ ⋁X for some subset X, there is a finite Y ⊆ X with c ≤ ⋁Y.
The name is borrowed from topology and the analogy is exact: the compact elements of the lattice of open sets of a topological space are precisely the compact open sets.
A complete lattice in which every element is the join of the compact elements below it.
Why Sub(A) and Con(A) are algebraic
In Sub(A), the compact elements are exactly the finitely generated subuniverses — those of the form Sg(X) for finite X.
The argument in one direction: if Sg(X) with X finite lies below a join of subuniverses, then each of the finitely many elements of X lies in the join, and each was produced from finitely many of the joined subuniverses — so finitely many suffice overall.
The reason the argument closes is that every basic operation takes finitely many arguments. An element of Sg(X) is built by a finite term from finitely many generators, so only finite information is ever required. Drop finitary arity and the theorem fails.
Since every subuniverse is the join of the finitely generated subuniverses it contains, Sub(A) is algebraic. The same argument, applied to principal congruences Θ(a, b), shows Con(A) is algebraic with the finitely generated congruences as its compact elements.
The representation theorem
Every algebraic lattice is isomorphic to Con(A) for some algebra A. Conversely, every congruence lattice is algebraic.
This is a complete answer to the question of which lattices are congruence lattices, and it is a strong statement in both directions. It says the constraint “is a congruence lattice” is exactly the constraint “is algebraic” — no more and no less.
The corresponding question for finite algebras — which finite lattices are congruence lattices of finite algebras — is the finite lattice representation problem, and it remains unresolved. Every finite lattice is known to be the congruence lattice of some algebra; whether a finite algebra always suffices is not known.
Algebraic lattices in the wild
| Lattice | Compact elements |
|---|---|
| Sub(A) | Finitely generated subuniverses |
| Con(A) | Finitely generated congruences |
| Subgroups of a group | Finitely generated subgroups |
| Ideals of a ring | Finitely generated ideals |
| Su(A), the power set | Finite subsets |
| Closed sets of an algebraic closure operator | Closures of finite sets |
The pattern is uniform: algebraic lattices are exactly the lattices of closed sets of algebraic closure operators — those where membership in a closure is always witnessed by a finite subset. That correspondence is the subject of the next page.
Frequently asked questions
Is every complete lattice algebraic?
No. The unit interval of real numbers under the usual order is complete, but its only compact element is 0, so it is very far from algebraic.
Why is the finite representation problem hard?
Because the Grätzer–Schmidt construction produces an infinite algebra even from a finite lattice, and no method is known for cutting it down to a finite one in general. The problem connects to questions in finite group theory, which is part of why it has resisted attack.
Source. S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, The Millennium Edition — a corrected re-typesetting of Springer-Verlag Graduate Texts in Mathematics 78 (1981). Section I.4, book pages 19-20.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates and reorganises mathematical results; it is not a reproduction of the source text.