Phil 312: Intermediate Logic, Precept 11

Home , Isolated point, Stone space

Phil 312: Intermediate Logic, Precept 11. Alejandro Naranjo Sandoval, [email protected] 1879 Hall, 227 http://scholar.princeton.edu/ansandoval/classes/phi-312-intermediate-logic

1. Stone Duality: From Boolean Algebras to Stone Spaces.

• The main result we will prove this week is the Stone Duality theorem, i.e., the theorem that the categories Bool and Stone are duals of each other. Here I’ll review quickly this result.

• To show this theorem, one needs to show that, given a Boolean Algebra, one can construct a Stone Space corresponding to it, and vice versa. So let’s give ourselves a Boolean Algebra B. Let hom(B, 2) be the set of states of B (i.e., homomorphisms from B to 2). Now, given a ∈ B, some states of B might map it to 1, some might map it to 0. So, in particular, we can consider the set of states of B which map a to 1:

Ua = {φ ∈ hom(B, 2) | φ(a) = 1}.

• The family of set B = {Ua | a ∈ B} generates a topology on hom(B, 2) (i.e., let T be the family of (potentially arbitrary) union of elements of B. Make sure you see that T is a topology and B is a basis of T ).

• We deﬁne S(B) to be the set hom(B, 2) equipped with this topology T . As was proved in class, this topological space is a Stone Space, i.e., it is (i) totally separated, and (ii) compact.

2. Stone Duality: From Stone Spaces to Boolean Algebras.

• Now given a Stone Space X. (Suppose this is a set X with a topology T ). Then, deﬁne K(X) ⊆ T to be the set of clopen subsets of X. Note that K(X) is never empty, since X, ∅ ∈ K(X). So deﬁne 1K(X) := X, 0K(X) := ∅, and the Boolean operations to be the usual set-theoretic operations (make sure you see that the union and intersection of clopen sets is clopen, and that X \ U, for U clopen, is also clopen). Hence we get that K(X) is a Boolean algebra.

3. An application of Stone Duality.

• Example 1. A Boolean algebra has n atoms (recall: a ∈ B is an atom iff if b ≤ a is non-zero, then b = a) iff its corresponding Stone Space X has n isolated points (x ∈ X is an isolated point of X iff {x} is open in X). Proof. (⇒) Suppose a ∈ B is an atom. Hint: Consider f ∈ hom(B, 2) as defined in PSet6 by f(x) = 1 iff a ≤ x. The claim will be that {f} is open in S(B). (⇐) Let {x} be open in the Stone space X. Hint: last precept we proved that every

1 finite set in a Hausdorff space is closed. Also, we know that Stone spaces are Hausdorff.

In precept 8, I (almost) proved that the atoms of a theory T are in bijection with the atoms of its corresponding Boolean algebra. So now we have that these are also in bijection with the isolated points of the corresponding Stone space.

4. A brief word about the Cantor Set.

• Let’s define the Cantor set C first. Let I0 be the closed unit interval [0, 1] in the real line R. This set has two disconnected closed segments contained in it, namely 1 2 L = 0, 3 and R = 3 , 1 . Let I1 = L ∪ R. In turn, each of these segments will 1 2 1 contain two subsegments, i.e., L contains LL = 0, 9 and LR = 9 , 3 ; R contains 2 7 8 RL = 3 , 9 and RR = 9 , 1 . Let I2 = LL ∪ LR ∪ RL ∪ RR. We can continue this process ad infinitum, i.e., by deleting the open middle third of each subsegment, thus 1 2 1 creating two closed subsubsegments. (In other words, Ii+1 = 3 Ii ∪ 3 + 3 Ii ). So let \ C = Ii. i=0,1,2,...

• (Note that this defines C as a subset of R. For it to be a topological space, we equip it with a topology. This topology is the subspace topology, as defined in precept last time.) • We can prove various properties of the Cantor space. For example, it is uncountably infinite (optional: this is another application of a diagonal argument. For x ∈ C, the position of x in C can be represented by a string of Ls and Rs, depending on where x is in each of the two subsegments of each Ii. So suppose we have a list of all elements in C. Then we can form one y ∈ X which is not in the list by making it’s i-th letter R iff the i-th letter of the i-th element of the list is L). • Example 2. The Cantor Space is a Stone Space. Proof. Hint: (i) for compactness, you can assume that closed intervals (i.e., sets of the form [a, b]) of R are closed and compact. (ii) To prove that it is totally disconnected, use that two distinct x, y ∈ C differ in at least one letter, when they are represented as strings of Ls and Rs. • Example 3. (Optional). The Stone Space of a finitely axiomatizable theory T (with axioms p0, p1, . . . , pn) is a clopen subset of the Cantor set. m m+1 Proof. Consider a subsegment S = 3k , 3k ∩C of C. Then C \S is closed (since it is the union of segments of the same form). Hence S is open. So subsegments of that sort are clopen. This is the intuitive idea of the proof: a model of T corresponds to one of those subsegments of C. The truth of each propositional constant can be represented in the Cantor set as a choice of whether to “to go” right or left at each juncture, say, if true go left, otherwise go right. Only finitely many propositional constants are relevant to the truth of the axioms of T . So the models of T can be represented in the Cantor m m+1 set as a finite string of Ls and Rs, i.e., as a subsegment of the form 3k , 3k ∩ C, and we are done.

2 • Example 4. Let T be the theory with axioms p0 → pi, for i = 1, 2,.... Show that there is no finitely axiomatizable theory T 0 (in the same signature) that is equivalent to T . Proof. If T 0 is finitely axiomatizable, then by Example 3 it corresponds to a clopen subset of the Cantor space C. On the other hand, as you proved in PSet 1, p0 is an atom of T . So, by Example 1, there is an isolated point in the Stone space corresponding to T . But in the real numbers no open set can contain an isolated point (intuitively: any neighborhood of the isolated point will intersect with an open set in infinitely more points). So Mod(T ) is not open in C. So T 0 and T are not equivalent.

5. Warm-up for ﬁnal exercise: More applications of Stone Duality.

Try these at home, and let me know if you need some help. (Some hints are included).

• Example 5. A Stone space is separable (i.e. has a countable basis) iﬀ the corresponding Boolean algebra is countable. Proof. (⇐ ) Suppose B is a countable Boolean algebra. Hint: Straightforward. Use deﬁnition of S(B).

• Example 6. A Stone space is discrete iff the corresponding Boolean algebra is finite. Proof. (⇐ ) Given a finite Boolean algebra B. Hint: Since in the last precept we proved that finite sets in Hausdorff spaces are closed, and given that Stone spaces are Hausdorff, it suffices to prove that S(B) is finite.