dual of Stone representation theorem

The Stone representation theorem characterizes a Boolean algebraMathworldPlanetmath as a field of sets in a topological spaceMathworldPlanetmath. There is also a dual to this famous theorem that characterizes a Boolean space as a topological space constructed from a Boolean algebra.

Theorem 1.

Let X be a Boolean space. Then there is a Boolean algebra B such that X is homeomorphicMathworldPlanetmath to B*, the dual spacePlanetmathPlanetmath (http://planetmath.org/DualSpaceOfABooleanAlgebra) of B.


The choice for B is clear: it is the set of clopen sets in X which, via the set theoretic operationsMathworldPlanetmath of intersectionMathworldPlanetmathPlanetmath, union, and complementPlanetmathPlanetmath, is a Boolean algebra.

Next, define a function f:XB* by


Our ultimate goal is to prove that f is the desired homeomorphism. We break down the proof of this into several stages:

Lemma 1.

f is well-defined.


The key is to show that f(x) is a prime idealMathworldPlanetmathPlanetmathPlanetmath in B* for any xX. To see this, first note that if U,Vf(x), then so is UVf(x), and if W is any clopen set of X, then UWf(x) too. Finally, suppose that UVf(x). Then xX-(UV)=(X-U)(X-V), which means that xU or xV, which is the same as saying that Uf(x) or Vf(x). Hence f(x) is a prime ideal, or a maximal idealMathworldPlanetmath, since B is Boolean. ∎

Lemma 2.

f is injectivePlanetmathPlanetmath.


Suppose xy, we want to show that f(x)f(y). Since X is HausdorffPlanetmathPlanetmath, there are disjoint open sets U,V such that xU and yV. Since X is also totally disconnected, U and V are unions of clopen sets. Hence we may as well assume that U,V clopen. This then implies that Uf(y) and Vf(x). Since UV, f(x)f(y). ∎

Lemma 3.

f is surjectivePlanetmathPlanetmath.


Pick any maximal ideal I of B*. We want to find an xX such that f(x)=I. If no such x exists, then for every xX, there is some clopen set UI such that xU. This implies that I=X. Since X is compactPlanetmathPlanetmath, X=J for some finite setMathworldPlanetmath JI. Since I is an ideal, and X is a finite join of elements of I, we see that XI. But this would mean that I=B*, contradicting the fact that I is a maximal, hence a proper idealMathworldPlanetmath of B*. ∎

Lemma 4.

f and f-1 are continuousPlanetmathPlanetmath.


We use a fact about continuous functions between two Boolean spaces:

a bijection is a homeomorphism iff it maps clopen sets to clopen sets (proof here (http://planetmath.org/HomeomorphismBetweenBooleanSpaces)).

So suppose that U is clopen in X, we want to prove that f(U) is clopen in B*. In other words, there is an element VB (so that V is clopen in X) such that


This is because every clopen set in B* has the form M(V) for some VB* (see the lemma in this entry (http://planetmath.org/StoneRepresentationTheorem)). Now, f(U)={f(x)xU}={f(x)Uf(x)}={MUM}, the last equality is based on the fact that f is a bijection. Thus by setting V=U completesPlanetmathPlanetmathPlanetmathPlanetmathPlanetmathPlanetmath the proof of the lemma. ∎

Therefore, f is a homemorphism, and the proof of theorem is complete. ∎

Title dual of Stone representation theorem
Canonical name DualOfStoneRepresentationTheorem
Date of creation 2013-03-22 19:08:38
Last modified on 2013-03-22 19:08:38
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 12
Author CWoo (3771)
Entry type Theorem
Classification msc 54D99
Classification msc 06E99
Classification msc 03G05
Related topic BooleanSpace
Related topic HomeomorphismBetweenBooleanSpaces
Defines dual space