complete Boolean algebra

Complete boolean algebras


A complete Boolean algebra is a complete lattice that is also a Boolean algebra. Since lattice homomorphisms of Boolean algebras automatically preserves the Boolean structure, the complete Boolean algebras form a full subcategory CompBoolAlg of CompLat.


Assuming excluded middle, complete atomic Boolean algebras are (up to isomorphism) precisely power sets. In fact, taking power sets defines a fully faithful functor from the opposite category of Set to Comp Bool Alg whose essential image consists of the complete atomic boolean algebras. See at Set – Properties – Opposite category. These abstract representations of power sets are important enough to have their own abbreviation: ‘CABA’.

This property of CABAs is not applicable in constructive mathematics, where power sets are rarely boolean algebras. However, we can use discrete locales instead (or rather, their corresponding frames). That is, define a CABA to be (not a complete atomic boolean algebra but) a frame XX such that the locale maps X1X \to 1 and XX×XX \to X \times X (which in the category of frames are maps 0X0 \to X and X+XXX + X \to X) are open (as locale maps). Then it should be (I will check) a classical theorem that CABAs and complete atomic boolean algebras are the same, and a constructive theorem that CABAs and power sets are the same (in the same functorial manner as above).


Complete Boolean algebras are the models of an algebraic theory (in which the operations, notably jj-indexed suprema and infima, have arities jj unbounded by any cardinal). It follows from general principles that the underlying-set functor U:CompBoolAlgSetU: CompBoolAlg \to Set preserves and reflects limits and isomorphisms.

However, this functor UU is not monadic; in fact, it does not even possess a left adjoint. Indeed, while the free complete Boolean algebra on a finite set XX exists and coincides with the free Boolean algebra on XX (it is finite, being isomorphic to the double power set P(PX)P(P X)), we have

Theorem (Gaifman-Hales; Solovay)

There is no free complete Boolean algebra on countably many generators.

As a consequence, CompBoolAlgCompBoolAlg is not cocomplete (otherwise there would exist a countable coproduct of copies of P(P1)P(P 1), which is ruled out by the previous theorem).


For instance around theorem 2.4 of


Revised on January 23, 2013 09:43:46 by Todd Trimble (