Complete Boolean algebras • Every
finite Boolean algebra is complete. • The
algebra of subsets of a given set is a complete Boolean algebra. • 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 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.
Non-complete Boolean algebras • 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 algebra of all measurable subsets of a measure space is a ℵ1-complete Boolean algebra, but is not usually complete. • Another example of a Boolean algebra that is not complete is the Boolean algebra P(ω) of all sets of
natural numbers, quotiented out by the ideal
Fin of finite subsets. The resulting object, denoted P(ω)/Fin, consists of all
equivalence classes of sets of naturals, where the relevant
equivalence relation is that two sets of naturals are equivalent if their
symmetric difference is finite. The Boolean operations are defined analogously, for example, if
A and
B are two equivalence classes in P(ω)/Fin, we define A \land B to be the equivalence class of a \cap b, where
a and
b are some (any) elements of
A and
B respectively. :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==