Core Universal Algebra

Subuniverses, Subalgebras and the Generation Operator

A subuniverse is a closed subset; a subalgebra is a subuniverse carrying the induced operations. Keeping the two apart makes the empty case and the lattice structure behave.

Engineering · Mathematics4 min readKV-MATH-0210

Learning objectives

Distinguish subuniverse from subalgebra and explain why both terms are needed.
Compute Sg(X) by intersection and by iterated term application.
Prove the two descriptions agree.
Show that Sg is a finitary closure operator.
Identify finitely generated subuniverses as the compact elements.

01Subuniverse and subalgebra

A subuniverse of A is a subset B closed under every operation of A: for each n-ary f and all b₁,…,bₙ ∈ B, the value f(b₁,…,bₙ) lies in B. A subalgebra is an algebra whose universe is a subuniverse of A and whose operations are the restrictions.

Key resultThe empty set is a subuniverse, never a subalgebra

When the type has no constants, ∅ satisfies the closure condition vacuously and so is a subuniverse. It is not a subalgebra because algebras have non-empty universes. Maintaining the distinction lets Sub(A) be a complete lattice with ∅ as its bottom, without forcing an empty algebra into existence.

Notation: Sub(A) for the set of subuniverses, and Sub(A) for that set regarded as a lattice under inclusion. The source distinguishes them typographically, and the distinction is the usual one between a carrier and a structure on it.

02The generation operator from above

An arbitrary intersection of subuniverses is a subuniverse, so the subuniverses form a closure system and the smallest subuniverse containing a set X exists.

Sg(X) = ⋂ { B : B is a subuniverse of A and X ⊆ B }

Well defined because A itself is a subuniverse containing X, so the family is non-empty.

This description is clean but non-constructive. It tells you Sg(X) exists without telling you what is in it, and for computation the description from below is required.

03The generation operator from below

ProcedureComputing Sg(X) by iterated closure

in: A, X → out: Sg(X) as an ascending union

input: algebra A of type F, subset X ⊆ A
E(Y) := Y ∪ { f(y₁,…,yₙ) : f ∈ F n-ary, y₁,…,yₙ ∈ Y }
X₀ := X
X_{k+1} := E(X_k)
Sg(X) = ⋃_{k ≥ 0} X_k
terminates in finitely many rounds iff the union stabilises

Correctness: the union is closed because any operation applied to finitely many elements of the union has all its arguments in some single X_k, hence its value in X_{k+1}. Minimality is immediate by induction. Caveat: the union need not stabilise at any finite stage, though each element enters at some finite stage.

The step that makes this work is that operations are finitary. An n-ary operation applied to arguments drawn from an ascending chain has all n of them in one member of the chain, because n is finite. For an infinitary operation the argument fails outright, and the ascending union would not be closed.

04Sg is a finitary closure operator

Extensivity, monotonicity and idempotency are immediate from the intersection description. Finitariness follows from the construction from below.

Every element has a finite witness
If a ∈ Sg(X) then a ∈ X_k for some finite k, and a is produced by finitely many operation applications, each with finitely many arguments.
Collect the leaves
Tracing the construction back gives a finite subset Y ⊆ X from which a is built.
Conclude finitariness
So a ∈ Sg(Y) for finite Y ⊆ X, giving Sg(X) = ⋃ { Sg(Y) : Y ⊆ X finite }.
Read off algebraicity
By the closure-operator correspondence, Sub(A) is an algebraic lattice and its compact elements are the finitely generated subuniverses.

05Generating sets and minimality

A set X generates A when Sg(X) = A. An algebra is finitely generated when some finite set generates it. Minimal generating sets need not have equal cardinality, unlike bases of vector spaces — the exchange property that makes dimension well defined is special, not general.

CautionGenerating sets are not bases

In a general algebra two minimal generating sets can have different sizes, and an independent set need not extend to a generating set. Vector spaces, and more generally algebras with a suitable exchange property, are the exception. The Irredundant Basis Theorem quantifies exactly how badly this can fail, and is the subject of the next page.

Frequently asked

Is the union of two subuniverses a subuniverse?

Generally no. Closure fails as soon as an operation takes one argument from each piece. The join in Sub(A) is Sg of the union, not the union itself — the same asymmetry between easy meets and hard joins seen in congruence lattices.

Does Sg(∅) always make sense?

Yes. If the type has constants, Sg(∅) is the subuniverse they generate and is non-empty. If it has none, Sg(∅) = ∅. Either way the operator is defined, which is why permitting ∅ as a subuniverse is convenient.

How does Sub(A) relate to Con A?

Both are algebraic lattices arising from finitary closure operators on A and on A × A respectively, and both have the finitely generated objects as compact elements. They are not otherwise related: knowing Sub(A) tells you little about Con A. The classification programme in universal algebra concerns Con A almost exclusively.

Sources and further reading

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.