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单词 DualSpaceOfABooleanAlgebra
释义

dual space of a Boolean algebra


Let B be a Boolean algebraMathworldPlanetmath, and B* the set of all maximal idealsMathworldPlanetmathPlanetmath of B. In this entry, we will equip B* with a topologyMathworldPlanetmath so it is a Boolean space.

Definition. For any aB, define M(a):={MB*aM}, and :={M(a)aB}.

It is know that in a Boolean algebra, maximal ideals and prime idealsMathworldPlanetmathPlanetmath coincide. From this entry (http://planetmath.org/RepresentingABooleanLatticeByFieldOfSets), we have the three following properties concerning M(a):

M(a)M(b)=M(ab),M(a)M(b)=M(ab),B*-M(a)=M(a).

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 intersectionMathworldPlanetmathPlanetmath of open sets M(a) and M(b) is again open, since it is M(ab). ∎

We may in fact treat as a subbasis for 𝒯, since finite intersections of elements of remain in .

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 complementPlanetmathPlanetmath of the open set M(a). ∎

Proposition 3.

B* is HausdorffPlanetmathPlanetmath.

Proof.

If M,NB* such that MN, then there is some aB such that aM and aN. This means that NM(a) and MM(a), which means that MB*-M(a)=M(a). Since M(a) and M(a) are open and disjoint, with NM(a) and MM(a), we see that B* is Hausdorff. ∎

Now, based on a topological fact, every zero-dimensional Hausdorff space is totally disconnected. Hence B* is totally disconnected.

Proposition 4.

B* is compactPlanetmathPlanetmath.

Proof.

Suppose {UiiI} is a collectionMathworldPlanetmath of open sets whose union is B*. Since each Ui is a union of elements of , we might as well assume that B* is covered by elements of . In other words, we may assume that each Ui is some M(ai).

Let J be the ideal generated by the set {aiiI}. If JB, then J can be extended to a maximal ideal M. Since each aiM, we see that MM(ai), so that M{M(ai)iI}=B*, which is a contradictionMathworldPlanetmathPlanetmath. Therefore, J=B. In particular, 1J, which means that 1 can be expressed as the join of a finite number of the ai’s:

1={aiiK},

where K is a finite subset of J. As a result, we have

{M(ai)iK}=M({aiiK})=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.

Titledual spacePlanetmathPlanetmath of a Boolean algebra
Canonical nameDualSpaceOfABooleanAlgebra
Date of creation2013-03-22 19:08:35
Last modified on2013-03-22 19:08:35
OwnerCWoo (3771)
Last modified byCWoo (3771)
Numerical id6
AuthorCWoo (3771)
Entry typeDefinition
Classificationmsc 06E05
Classificationmsc 03G05
Classificationmsc 06B20
Classificationmsc 03G10
Classificationmsc 06E20
Related topicStoneRepresentationTheorem
Related topicMHStonesRepresentationTheorem

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