dual of Stone representation theorem
The Stone representation theorem characterizes a Boolean algebra as a field of sets in a topological space
. 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 homeomorphic to B*, the dual space
(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 operations of intersection
, union, and complement
, is a Boolean algebra.
Next, define a function f:X→B* by
f(x):={U∈B∣x∉U}. |
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 ideal in B* for any x∈X. To see this, first note that if U,V∈f(x), then so is U∪V∈f(x), and if W is any clopen set of X, then U∩W∈f(x) too. Finally, suppose that U∩V∈f(x). Then x∈X-(U∩V)=(X-U)∪(X-V), which means that x∉U or x∉V, which is the same as saying that U∈f(x) or V∈f(x). Hence f(x) is a prime ideal, or a maximal ideal
, since B is Boolean. ∎
Lemma 2.
f is injective.
Proof.
Suppose x≠y, we want to show that f(x)≠f(y). Since X is Hausdorff, there are disjoint open sets U,V such that x∈U and y∈V. 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 U∈f(y) and V∈f(x). Since U≠V, f(x)≠f(y). ∎
Lemma 3.
f is surjective.
Proof.
Pick any maximal ideal I of B*. We want to find an x∈X such that f(x)=I. If no such x exists, then for every x∈X, there is some clopen set U∈I such that x∈U. This implies that ⋃I=X. Since X is compact, X=⋃J for some finite set
J⊆I. Since I is an ideal, and X is a finite join of elements of I, we see that X∈I. But this would mean that I=B*, contradicting the fact that I is a maximal, hence a proper ideal
of B*. ∎
Lemma 4.
f and f-1 are continuous.
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 V∈B (so that V is clopen in X) such that
f(U)=M(V)={M∈B*∣V∉M}. |
This is because every clopen set in B* has the form M(V) for some V∈B* (see the lemma in this entry (http://planetmath.org/StoneRepresentationTheorem)). Now, f(U)={f(x)∣x∈U}={f(x)∣U∉f(x)}={M∣U∉M}, the last equality is based on the fact that f is a bijection. Thus by setting V=U completes 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 |