complete Boolean algebra

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 $X$ such that the locale maps $X \to 1$ and $X \to X \times X$ (which in the category of frames are maps $0 \to X$ and $X + X \to X$) 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 $j$-indexed suprema and infima, have arities $j$ unbounded by any cardinal). It follows from general principles that the underlying-set functor $U: CompBoolAlg \to Set$ preserves and reflects limits and isomorphisms.

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

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

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

For instance around theorem 2.4 of

- Jaap van Oosten,
*Basic category theory*(pdf)

and

Revised on October 19, 2015 13:40:46
by Anonymous Coward
(76.99.53.22)