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单词 DualOfStoneRepresentationTheorem
释义

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.

Proof.

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

f(x):={UBxU}.

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.

Proof.

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.

Proof.

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.

Proof.

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.

Proof.

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

f(U)=M(V)={MB*VM}.

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.∎

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更新时间:2025/5/4 23:58:29