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Homedual of Stone representation theorem

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# 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 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 $x\neq y$, we want to show that $f(x)\neq f(y)$. 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 $U\neq V$, $f(x)\neq f(y)$. ∎

###### 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 $f^{{-1}}$ 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).

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). 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. ∎

## Mathematics Subject Classification

54D99*no label found*06E99

*no label found*03G05

*no label found*

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