# Complete Boolean algebra

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In mathematics, a complete Boolean algebra is a Boolean algebra in which every subset has a supremum (least upper bound). Complete Boolean algebras are used to construct Boolean-valued models of set theory in the theory of forcing. Every Boolean algebra A has an essentially unique completion, which is a complete Boolean algebra containing A such that every element is the supremum of some subset of A. As a partially ordered set, this completion of A is the Dedekind-MacNeille completion.

More generally, if κ is a cardinal then a Boolean algebra is called κ-complete if every subset of cardinality less than κ has a supremum.

## Examples

• Every finite Boolean algebra is complete.
• The regular open sets of any topological space form a complete Boolean algebra. This example is of particular importance because every forcing poset can be considered as a topological space (a base for the topology consisting of sets that are the set of all elements less than or equal to a given element). The corresponding regular open algebra can be used to form Boolean-valued models which are then equivalent to generic extensions by the given forcing poset.
• The algebra of all measurable subsets of a σ-finite measure space, modulo null sets, is a complete Boolean algebra. When the measure space is the unit interval with the σ-algebra of Lebesgue measurable sets, the Boolean algebra is called the random algebra.
• The algebra of all measurable subsets of a measure space is a ℵ1-complete Boolean algebra, but is not usually complete.
• The algebra of all subsets of an infinite set that are finite or have finite complement is a Boolean algebra but is not complete.
• The Boolean algebra of all Baire sets modulo meager sets in a topological space with a countable base is complete; when the topological space is the real numbers the algebra is sometimes called the Cantor algebra.
Now let a0, a1,... be pairwise disjoint infinite sets of naturals, and let A0, A1,... be their corresponding equivalence classes in P(ω)/Fin . Then given any upper bound X of A0, A1,... in P(ω)/Fin, we can find a lesser upper bound, by removing from a representative for X one element of each an. Therefore the An have no supremum.

## Properties of complete Boolean algebras

• Sikorski's extension theorem states that

if A is a subalgebra of a Boolean algebra B, then any homomorphism from A to a complete Boolean algebra C can be extended to a morphism from B to C.

• Every subset of a complete Boolean algebra has a supremum, by definition; it follows that every subset also has an infimum (greatest lower bound).
• For a complete boolean algebra both infinite distributive laws hold.
• For a complete boolean algebra infinite de-Morgan's laws hold.

## The completion of a Boolean algebra

The completion of a Boolean algebra can be defined in several equivalent ways:

• The completion of A is (up to isomorphism) the unique complete Boolean algebra B containing A such that A is dense in B; this means that for every nonzero element of B there is a smaller non-zero element of A.
• The completion of A is (up to isomorphism) the unique complete Boolean algebra B containing A such that every element of B is the supremum of some subset of A.

The completion of a Boolean algebra A can be constructed in several ways:

• The completion is the Boolean algebra of regular open sets in the Stone space of prime ideals of A. Each element x of A corresponds to the open set of prime ideals not containing x (which open and closed, and therefore regular).
• The completion is the Boolean algebra of regular cuts of A. Here a cut is a subset U of A+ (the non-zero elements of A) such that if q is in U and pq then p is in U, and is called regular if whenever p is not in U there is some rp such that U has no elements ≤r. Each element p of A corresponds to the cut of elements ≤p.

If A is a metric space and B its completion then any isometry from A to a complete metric space C can be extended to a unique isometry from B to C. The analogous statement for complete Boolean algebras is not true: a homomorphism from a Boolean algebra A to a complete Boolean algebra C cannot necessarily be extended to a (supremum preserving) homomorphism of complete Boolean algebras from the completion B of A to C. (By Sikorski's extension theorem it can be extended to a homomorphism of Boolean algebras from B to C, but this will not in general be a homomorphism of complete Boolean algebras; in other words, it need not preserve suprema.)

## Free κ-complete Boolean algebras

Unless the Axiom of Choice is relaxed,[1] free complete boolean algebras generated by a set do not exist (unless the set is finite). More precisely, for any cardinal κ, there is a complete Boolean algebra of cardinality 2κ greater than κ that is generated as a complete Boolean algebra by a countable subset; for example the Boolean algebra of regular open sets in the product space κω, where κ has the discrete topology. A countable generating set consists of all sets am,n for m, n integers, consisting of the elements x∈κω such that x(m)<x(n). (This boolean algebra is called a collapsing algebra, because forcing with it collapses the cardinal κ onto ω.)

In particular the forgetful functor from complete Boolean algebras to sets has no left adjoint, even though it is continuous and the category of Boolean algebras is small-complete. This shows that the "solution set condition" in Freyd's adjoint functor theorem is necessary.

Given a set X, one can form the free Boolean algebra A generated by this set and then take its completion B. However B is not a "free" complete Boolean algebra generated by X (unless X is finite or AC is omitted), because a function from X to a free Boolean algebra C cannot in general be extended to a (supremum-preserving) morphism of Boolean algebras from B to C.

On the other hand, for any fixed cardinal κ, there is a free (or universal) κ-complete Boolean algebra generated by any given set.

## References

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