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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
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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.
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