Lattice Theory Foundations
Complete Lattices and Algebraic Lattices
Completeness is not equational, so complete lattices are not a variety. Yet every congruence lattice is complete, and algebraic besides. That tension organises the chapter.
- Define completeness and show that arbitrary meets imply arbitrary joins.
- Define compact elements and algebraic lattices.
- Explain why algebraicity is the residue of finitary operations.
- State the correspondence with finitary closure operators.
- Recognise Sub(A) and Con A as algebraic lattices.
01Completeness and the one-sided criterion
A lattice is complete when every subset — including the empty set and infinite subsets — has both a supremum and an infimum. The empty-set clause forces the existence of a greatest and a least element, since sup ∅ is the least element and inf ∅ is the greatest.
If every subset of a poset has an infimum, then every subset has a supremum as well: the supremum of X is the infimum of the set of upper bounds of X. So completeness need only be checked on one side. This is used constantly, because closure-system arguments naturally produce arbitrary intersections and hence arbitrary meets.
The proof is short and worth carrying. Let U be the set of upper bounds of X, and let p = inf U, which exists by hypothesis. Every element of X is below every element of U, so every element of X is a lower bound of U, hence below inf U = p. So p is an upper bound of X, so p ∈ U, so p is the least element of U — that is, sup X.
02Compact elements
An element c of a complete lattice is compact when c ≤ sup X implies c ≤ sup Y for some finite Y ⊆ X. The name is borrowed from topology and the analogy is exact: compactness is the property that a cover can be reduced to a finite subcover.
In the lattice of subuniverses of an algebra, the compact elements are exactly the finitely generated subuniverses. That identification is the substance behind the abstract definition, and it is why compactness is the right notion to isolate.
03Algebraic lattices
A complete lattice is algebraic when every element is the supremum of the compact elements below it. Equivalently, the compact elements are join-dense.
The unit interval [0, 1] under the usual order is a complete lattice. Its only compact element is 0, because any positive number is the supremum of the numbers strictly below it with no finite subset sufficing. So [0, 1] is complete and very far from algebraic, and it is not the congruence lattice of any algebra.
04The representation theorem
- input: complete lattice L
- compute K = { c ∈ L : c is compact }
- for each a ∈ L:
- check a = ⋁ { c ∈ K : c ≤ a }
- if the identity holds for every a: L is algebraic
- then L ≅ the lattice of closed sets of a finitary closure operator
- and L ≅ Sub(A) for some algebra A
The correspondence with closure operators runs in both directions and is developed on its own page in this stream. Its practical value is that it converts questions about generated substructures into lattice-theoretic questions and back.
05The congruence lattice representation problem
Every congruence lattice is algebraic. The converse question — is every algebraic lattice a congruence lattice? — was answered affirmatively by Grätzer and Schmidt: every algebraic lattice is isomorphic to Con A for some algebra A.
Whether every finite lattice is the congruence lattice of a finite algebra is a different and much harder question, and the general representation theorem does not settle it. This is one of the places where the 1981 source and the present state of the field diverge, and the Research Frontier stream addresses it.
Frequently asked
Why isn't the class of complete lattices a variety?
Because completeness cannot be expressed by identities. Terms are finite, so an equation constrains only finitely many elements at a time, whereas completeness asserts the existence of suprema for arbitrary subsets. Concretely, the class of complete lattices is not closed under subalgebras: a sublattice of a complete lattice need not be complete.
Are the compact elements of Con A the finitely generated congruences?
Yes — the compact elements of Con A are precisely the congruences generated by finitely many pairs, and in particular the principal congruences Θ(a, b) are compact. This is the congruence-lattice analogue of finitely generated subuniverses being the compact elements of Sub(A).
Does algebraicity constrain the lattice much?
It rules out a great deal — [0, 1] and other continuous structures — but by Birkhoff–Frink and Grätzer–Schmidt it is the only constraint for arbitrary algebras. So in the infinite setting, algebraicity is exactly the right characterisation and nothing further can be said. The interest shifts entirely to restricted settings, such as finite algebras or algebras in a fixed variety.
- S. Burris and H. P. Sankappanavar, A Course in Universal Algebra, Millennium Edition (a corrected re-typesetting of Springer GTM 78, 1981).
- G. Grätzer, Universal Algebra, 2nd edition, Springer.
- R. McKenzie, G. McNulty and W. Taylor, Algebras, Lattices, Varieties, Volume I.
Original KEVOS® explanatory article. Written from the topic map of the cited works; no text is reproduced from them.