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 $B^{\mathrm{*}}$, 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 $f:X\to B^{*}$ by

$$f(x):=\{U\in B\mid x\notin U\}.$$

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 $x\in X$. To see this, first note that if $U,V\in f(x)$, then so is $U\cup V\in f(x)$, and if $W$ is any clopen set of $X$, then $U\cap W\in f(x)$ too. Finally, suppose that $U\cap V\in f(x)$. Then $x\in X-(U\cap V)=(X-U)\cup (X-V)$, which means that $x\notin U$ or $x\notin V$, which is the same as saying that $U\in f(x)$ or $V\in f(x)$. Hence $f(x)$ is a prime ideal, or a maximal ideal, since $B$ is Boolean. ∎

Lemma 2.

$f$ is injective.

Proof.

Suppose $x\ne y$, we want to show that $f(x)\ne f(y)$. Since $X$ is Hausdorff, there are disjoint open sets $U,V$ such that $x\in U$ and $y\in V$. 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 $U\in f(y)$ and $V\in f(x)$. Since $U\ne V$, $f(x)\ne f(y)$. ∎

Lemma 3.

$f$ is surjective.

Proof.

Pick any maximal ideal $I$ of $B^{*}$. We want to find an $x\in X$ such that $f(x)=I$. If no such $x$ exists, then for every $x\in X$, there is some clopen set $U\in I$ such that $x\in U$. This implies that $\bigcup I=X$. Since $X$ is compact, $X=\bigcup J$ for some finite set $J\subseteq I$. Since $I$ is an ideal, and $X$ is a finite join of elements of $I$, we see that $X\in I$. 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 $f^{\mathrm{-}\mathrm{1}}$ 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 $V\in B$ (so that $V$ is clopen in $X$) such that

$$f(U)=M(V)=\{M\in B^{*}\mid V\notin M\}.$$

This is because every clopen set in $B^{*}$ has the form $M(V)$ for some $V\in B^{*}$ (see the lemma in this entry (http://planetmath.org/StoneRepresentationTheorem)). Now, $f(U)=\{f(x)\mid x\in U\}=\{f(x)\mid U\notin f(x)\}=\{M\mid U\notin M\}$, 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