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dual space of a Boolean algebra∗ CWoo† 2013-03-22 3:20:27 Let B be a Boolean algebra, and B ∗ the set of all maximal ideals of B. In this entry, we will equip B ∗ with a topology so it is a Boolean space. Definition. For any a ∈ B, define M (a) := {M ∈ B ∗ | a ∈ / M }, and B := {M (a) | a ∈ B}. It is know that in a Boolean algebra, maximal ideals and prime ideals coincide. From this entry, we have the three following properties concerning M (a): M (a)∩M (b) = M (a∧b), M (a)∪M (b) = M (a∨b), B ∗ −M (a) = M (a0 ). Furthermore, if M (a) = M (b), then a = b. From these properties, we see that M (0) = ∅ and M (1) = B ∗ . As a result, we see that Proposition 1. B ∗ is a topological space, whose topology T is generated by the basis B. Proof. ∅ and B ∗ are both open, as they are M (0) and M (1) respectively. Also, the intersection of open sets M (a) and M (b) is again open, since it is M (a ∧ b). We may in fact treat B as a subbasis for T , since finite intersections of elements of B remain in B. Proposition 2. Each member of B is closed, hence T is generated by a basis of clopen sets. In other words, B ∗ is zero-dimensional. Proof. Each M (a) is open, by definition, and closed, since it is the complement of the open set M (a0 ). Proposition 3. B ∗ is Hausdorff. Proof. If M, N ∈ B ∗ such that M 6= N , then there is some a ∈ B such that a ∈ M and a ∈ / N . This means that N ∈ M (a) and M ∈ / M (a), which means that M ∈ B ∗ − M (a) = M (a0 ). Since M (a) and M (a0 ) are open and disjoint, with N ∈ M (a) and M ∈ M (a0 ), we see that B ∗ is Hausdorff. ∗ hDualSpaceOfABooleanAlgebrai created: h2013-03-2i by: hCWooi version: h42043i Privacy setting: h1i hDefinitioni h06E05i h03G05i h06B20i h03G10i h06E20i † This text is available under the Creative Commons Attribution/Share-Alike License 3.0. You can reuse this document or portions thereof only if you do so under terms that are compatible with the CC-BY-SA license. 1 Now, based on a topological fact, every zero-dimensional Hausdorff space is totally disconnected. Hence B ∗ is totally disconnected. Proposition 4. B ∗ is compact. Proof. Suppose {Ui | i ∈ I} is a collection of open sets whose union is B ∗ . Since each Ui is a union of elements of B, we might as well assume that B ∗ is covered by elements of B. In other words, we may assume that each Ui is some M (ai ) ∈ B. Let J be the ideal generated by the set {ai | i ∈ I}. If J 6= B, then J can be extended to / M (ai ), so S a maximal ideal M . Since each ai ∈ M , we see that M ∈ that M ∈ / {M (ai ) | i ∈ I} = B ∗ , which is a contradiction. Therefore, J = B. In particular, 1 ∈ J, which means that 1 can be expressed as the join of a finite number of the ai ’s: _ 1 = {ai | i ∈ K}, where K is a finite subset of J. As a result, we have [ _ {M (ai ) | i ∈ K} = M ( {ai | i ∈ K}) = M (1) = B ∗ . So B ∗ has a finite subcover, and hence is compact. Collecting the last three results, we see that B ∗ is a Boolean space. Remark. It can be shown that B is isomorphic to the Boolean algebra of clopen sets in B ∗ . This is the famous Stone representation theorem. 2