Stone duality expresses an anti-equivalence of categories between the category of boolean algebras and the category of boolean spaces. A *boolean space* is a compact totally disconnected topological space, and the category of boolean spaces is the full subcategory of the category of topological spaces. Boolean spaces are also known as Stone spaces.

Under this anti-equivalence, a boolean space *S* is mapped to the boolean algebra of clopen subsets of *S*. Conversely, given a boolean algebra *B*, the associated space *S* is the space of all ultrafilters on *B*, or equivalently, the set of all boolean-algebra homomorphisms from *B* to the two-element boolean algebra {0,1}. The space *S* is topologized by taking as basic clopen sets the sets of the form

- $ \{\mathcal{U} \in S : b \in \mathcal{U}\} \subset S $

for *b* an element of *B*.

This functor sends surjections to injections and vice versa. In fact, in both categories, the monomorphisms are exactly the injections and the epimorphisms are exactly the surjections.

## Significance in Model TheoryEdit

If *T* is a theory, the space of completions of *T* is naturally the Stone space associated to the boolean algebra *B* of "sentences mod *T*". Here *B* is the set of all sentences in the language of *T*, modulo the equivalence relation of being equivalent in all models of *T*. From this point of view, ultraproducts correspond to ultralimits in the Stone space. Theories extending *T* correspond to closed subsets of the Stone space, with complete theories corresponding to singletons.

If *M* is a model and *A* is a subset of *M*, then one can consider the boolean algebra of *A*-definable subsets of *M ^{n}*. The corresponding boolean space under Stone duality is exactly the Stone space $ S_n(A) $ of (complete)

*n*-types over

*A*. This is because a complete

*n*-type over

*A*can be thought of as a homomorphism of boolean algebras from the

*A*-definable subsets of

*M*to the two-element boolean algebra {0,1}.

^{n}Similarly, the space of quantifier-free types over *A* is the Stone space associated with the boolean algebra of quantifier-free definable sets. If *B* denotes the boolean algebra of all *A*-definable subsets of *M ^{n}*, and

*B*denotes the subalgebra of

_{qfree}*A*-definable quantifier-free subsets, then there is an obvious inclusion

- $ B_{qfree} \hookrightarrow B $

which is a surjection if and only if quantifier-elimination holds in *M* after naming the elements of *A*.
Under Stone duality, we see that quantifier elimination is equivalent to the *injectivity* of the map

- $ S(A) \to S_{qfree}(A) $

of type spaces.

In particular, quantifier elimination holds as long as each complete quantifier-free type determines a unique type. This can be proven directly by compactness, but the proof using Stone spaces is arguably more intuitive.

Stone spaces of types also play an important role in Stability Theory, where a number of ranks are defined as Cantor-Bendixson rank on various type spaces.

## Proof of Stone dualityEdit

Let **B** be the contravariant functor from boolean spaces to boolean algebras sending a boolean space *S* to the boolean algebra of clopen subsets of *S*. Given a morphism *f: S -> T* of boolean spaces, the induced map **B**(*T*) -> **B**(*S*) sends a clopen subset *C* ⊆ *T* to its preimage *f ^{-1}(C) ⊆ S*. This is a well-defined contravariant functor.

To show an anti-equivalence of categories, it suffices to show that **B** is fully faithful and essentially surjective.

**Faithfulness:** Suppose *f* and *g* are two distinct maps between boolean spaces *S* and *T*. Suppose for the sake of contradiction that *f* and *g* induce the same map **B**(*T*) -> **B**(*S*). Let *p* be a point in *S* such that *f(p) ≠ g(p)*. Because *T* is a boolean space, there is a clopen subset *C ⊆ T* containing *f(p)* but not *g(p)*. Then * p ∈ f ^{-1}(C) \ g^{-1}(C)*. As

*f*, we conclude that

^{-1}(C) ≠ g^{-1}(C)**B**(

*f*) ≠

**B**(

*g*). So

**B**is faithful.

*Claim:* Let *U* be an ultrafilter on **B**(*T*). Then there is a unique point *q ∈ T* such that *U* is exactly the clopen subsets of *T* containing *q*.

*Proof:* Let *C* be the intersection of all the elements of *U*. Then *C* is closed. Any finite intersection of elements of *U* is also in *U*, hence non-empty. By compactness of *T*, it follows that *C* is non-empty. We claim that it is a singleton. If not, let *q*_{1} and *q*_{2} be two distinct points in *C*. By total disconnectedness of *T*, there is a clopen set *D* containing *q*_{1} but not *q*_{2}. Then either *D* or its complement is an element of *U*. But neither *D* nor its complement contains both *q*_{1} and *q*_{2}, hence neither contains *C*, contradicting the fact that *C* is the intersection of all elements in *U*. Therefore *C* is a singleton {*q*}. Now any clopen set not containing *q* is not in *U*, and therefore its complement must be in *U*. So any clopen set containing *q* is in *U*. This completes the proof of the claim.

**Fullness:** Suppose *S* and *T* are two boolean spaces, and *f* is a homomorphism of boolean algebras from **B**(*T*) to **B**(*S*). We will construct the corresponding map *g* from *S* to *T*. Given a point *p* in *S*, consider the ultrafilter *U_p* on **B**(*S*) containing exactly the clopen sets containing *p*. Then *f ^{-1}(U_p)* is an ultrafilter on

**B**(

*T*). By the claim, there is a unique point

*g(p) ∈ T*such that

*f*is exactly the clopen neighborhoods of

^{-1}(U_p)*g(p)*. We have just defined a map

*g*:

*S*->

*T*with the property that if

*C*is a clopen subset of

*T*, then

- $ g(p) \in C \iff C \in f^{-1}(U_p) \iff f(C) \in U_p \iff p \in f(C) $

In particular, *g ^{-1}(C) = f(C)*, so pre-images of clopen sets under

*g*are still clopen. Because

*T*is a boolean space, its clopen subsets form a basis, and therefore

*g*is continuous. Also, as

*g*, we see that

^{-1}(C) = f(C)**B**(

*g*) is exactly

*f*.

**Essential surjectivity:** Given a boolean algebra *B*, we need to construct a boolean space *S* such that *B* is isomorphic to **B**(*S*). Let *S* be the set of all ultrafilters on *B*.
For each *b* ∈ *B* let

- $ D(b) = \{U \in S : b \in U\} $.

Note that if *U* is an ultrafilter on *B*, and *b _{1}* and

*b*are two elements of

_{2}*B*, then

- $ b_1 \vee b_2 \in U \iff (b_1 \in U) \vee (b_2 \in U) $
- $ b_1 \wedge b_2 \in U \iff (b_1 \in U) \wedge (b_2 \in U) $
- $ \neg b_1 \in U \iff \neg(b_1 \in U) $

by definition of "ultrafilter." In terms of the *D*(*b*)'s, this means

- $ D(b_1 \vee b_2) = D(b_1) \cup D(b_2) $
- $ D(b_1 \wedge b_2) = D(b_1) \cap D(b_2) $
- $ D(\neg b_1) = S \setminus D(b_1) $

Because the class of *D*(*b*)'s is closed under intersections, it is the basis for some topology on *S.*

We claim that this topology makes *S* a boolean space.
For total disconnectedness, let *U*_{1} and *U*_{2} be two distinct ultrafilters on *B*. Then some element *b* is in *U*_{1} but not *U*_{2}, or vice versa. Thus *D*(*b*) contains exactly one of *U*_{1} and *U*_{2}. But *D*(*b*) is a basic open and its complement is *D*(¬*b*), another basic open. Consequently, *D*(*b*) is a clopen set separating *U*_{1} and *U*_{2}. Therefore *S* is totally disconnected.

For compactness, it suffices to show that any cover of *S* by basic opens has a finite subcover. Suppose that

- $ \{D(b_i) : i \in I\} $

is a covering of *S*. If there is a finite subset *I _{0} ⊆ I* such that

- $ \bigvee_{i \in I_0} b_i = 1 $,

then a finite subcover exists, because

- $ \bigcup_{i \in I_0} D(b_i) = D\left(\bigvee_{i \in I_0} b_i\right) = D(1) = S $.

If *not*, then $ \{b_i : i \in I\} $ generates a proper ideal in *B*. In particular, some maximal ideal contains all the *b _{i}*. The complement of this is an ultrafilter

*U*such that

*b*is not in

_{i}*U*for every

*i*, or equivalently,

- $ U \notin \bigcap_{i \in I} D(b_i) $

This contradicts the fact that the *D*(*b _{i}*) were supposed to cover the entirety of

*S*.

Consequently *S* is truly a boolean space. Moreover, there is a natural homomorphism *D* of boolean algebras from *B* to **B**(*S*) sending *b* ∈ *B* to *D(b)*. The identities listed above ensure that this is a homomorphism of boolean algebras. We claim that *D* is an isomorphism. For injectivity, suppose that *b* and *c* are two distinct elements of *B*. Let *d* be the symmetric difference

- $ d = (b \wedge \neg c) \vee (c \wedge \neg b) \in B $

Then *b ≠ c* ensures *d ≠ 0*. Consequently, *d* generates a proper filter. Let *U* be a maximal filter (an ultrafilter) containing *d*. Then exactly one of *b ∈ U* and *c ∈ U* holds. Therefore, *U* is in exactly one of *D(b)* and *D(c)*. In particular, *D(b) ≠ D(c)*. So *D* is an injective map.

For surjectivity of *D*, let *C* be a clopen subset of *S*. Then *C* is a closed subset of a compact Hausdorff space, and is therefore compact. For each point *p* in *C*, there is a *basic* clopen neighborhood of *p* contained in *C*. Such neighborhoods cover *C*, so by compactness of *C*, we can write *C* as a finite union of basic clopens. But, by the identities listed above, any union of *D*(*b*)'s is itself a *D*(*b*). So *C* must be in the image of *D*.

This shows that *D* is an isomorphism of boolean algebras between *B* and **B**(*S*). This completes the proof that **B** is an essentially surjective functor, as well as the proof of Stone duality. QED