Exercise 20, § 6
General Topology, I
Bourbaki marks this one as harder than the rest.
A lattice is said to be a Boolean algebra if it is distributive, has a least element , and a greatest element and if, for each , there is an element such that and ; such an element is said to be a complement of in .
a) Let be the injection of a Boolean algebra into the set of subsets of the set of prime prefilters on , defined in Exercise 18 d). Show that if is a complement of in , then in ; deduce that an element has only one complement , that the complement of is , and that the complement of is ; give direct proofs of these statements. The ideals of the Boolean ring corresponding to , other than , are the coprefilters on .
b) Show that, in a Boolean algebra , every prime prefilter is maximal (remark that, for each , every prime prefilter must contain either or its complement).
c) Conversely, let be a distributive lattice which has a least element and a greatest element , and is such that every prime prefilter on is maximal. Show that is a Boolean algebra. (For an element , consider the coprefilter (Exercise 17) consisting of those such that ; if has no complement, there is a maximal coprefilter containing and ; remark that the complement of in is a prime prefilter [Exercise 17 b) and 18 b)] and use Exercise 17 f)).
d) Show that if is a topological space then the set of subsets of which are both open and closed is a Boolean algebra if ordered by inclusion. If we take to be the topological space associated with the Fréchet filter (no. 5), show that the Boolean algebra thus obtained is countably infinite and therefore cannot be isomorphic to the ordered set of all subsets of a set .
Solution
No solution has been written yet.