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Heyting Duality Heyting Duality Vikraman Choudhury April, 2018 1 Heyting Duality Pairs of “equivalent” concepts or phenomena are ubiquitous in mathematics, and dualities relate such two different, even opposing concepts. Stone’s Representa- tion theorem (1936), due to Marshall Stone, states the duality between Boolean algebras and topological spaces. It shows that, for classical propositional logic, the Lindenbaum-Tarski algebra of a set of propositions is isomorphic to the clopen subsets of the set of its valuations, thereby exposing an algebraic viewpoint on logic. In this essay, we consider the case for intuitionistic propositional logic, that is, a duality result for Heyting algebras. It borrows heavily from an exposition of Stone and Heyting duality by van Schijndel and Landsman [2017]. 1.1 Preliminaries Definition (Lattice). A lattice is a poset which admits all finite meets and joins. Categorically, it is a (0, 1) − 푐푎푡푒푔표푟푦 (or a thin category) with all finite limits and finite colimits. Alternatively, a lattice is an algebraic structure in the signature (∧, ∨, 0, 1) that satisfies the following axioms. • ∧ and ∨ are each idempotent, commutative, and associative with respective identities 1 and 0. • the absorption laws, 푥 ∨ (푥 ∧ 푦) = 푥, and 푥 ∧ (푥 ∨ 푦) = 푥. Definition (Distributive lattice). A distributive lattice is a lattice in which ∧and ∨ distribute over each other, that is, the following distributivity axioms are satisfied. Categorically, this makes it a distributive category. 1 • 푥 ∨ (푦 ∧ 푧) = (푥 ∨ 푦) ∧ (푥 ∨ 푧) • 푥 ∧ (푦 ∨ 푧) = (푥 ∧ 푦) ∨ (푥 ∧ 푧) Definition (Complements in a lattice). A complement of an element 푥 of a lattice is an element 푦 such that, 푥 ∧ 푦 = 0 and 푥 ∨ 푦 = 1. Complements need not be unique. If it is unique, we denote the complement of 푥 by ¬푥. Definition (Boolean algebra). A Boolean algebra 픅 is a distributive lattice with unique complements. 1.2 Stone Duality We first state the Stone Representation Theorem, which establishes the duality of Boolean algebras and Stone spaces. Definition (Stone space). A topological space (푋, 휏) is totally disconnected if 휏 has a basis of clopen sets. A Stone space is a totally disconnected compact Hausdorff space. Definition (푃퐹). For 픏 a bounded distributive lattice, let 푃퐹(픏) be the set of prime filters. For 푎 ∈ 픏, we define 퐹 ∶ 픏 → 푃퐹(픏) 푎 ↦ 퐹푎 ≔ { 푃 ∈ 푃퐹(픏) ∣ 푎 ∈ 푃 } Note that 푃퐹(픏) is a also a poset in its own right, with respect to inclusion. It is easy to check that it is also a bounded distributive lattice. Definition (퐶푙). For 푋 a topological space, let 퐶푙(푋) be the set of clopen sets of 푋. Definition (퐶푈). For 푋 a topological space, let 퐶푈(푋) the set of clopen up-sets of 푋, ordered by inclusion. Lemma. The map 퐹 ∶ 픏 → 푃퐹(픏) is a homomorphism. It preserves joins, meets, 0, 1, and complements. Proof. 퐹0 = ∅ because no prime filter contains 0. 퐹1 = 푃퐹(픏) because every prime filter contains 1. 퐹푎 ∪ 퐹푏 ⊆ 퐹푎∨푏 because filters are up-sets and 푎 ∈ 푃 or 푏 ∈ 푃 implies 푎 ∨ 푏 ∈ 푃. Moreover, 퐹푎∨푏 ⊆ 퐹푎 ∪ 퐹푏 because 푃 ∈ 퐹푎∨푏 is prime, hence 푎 ∨ 푏 ∈ 푃 implies 푎 ∈ 푃 or 푏 ∈ 푃. The case for meets follows dually because filters are closed under finite meets, and are up-sets. 2 ′ If 픏 has complements, we want to prove that ¬퐹푎 = 퐹푎 = 퐹¬푎. Since 푃 ∈ 푃퐹(픏) is proper, it cannot contain both 푎 and ¬푎. Since 푃 is prime, and hence an ultrafilter (using the ultrafilter lemma), it contains either 푎 or ¬푎. This means that the set of prime filters containing ¬푎 is precisely the set of prime filters not containing 푎. Definition (Topology τ on 푃퐹(픏)). We equip 푃퐹(픏) with a topology 휏 as follows. ′ For 푎, 푏 ∈ 픏, let 푆 ≔ { 퐹푎 ∣ 푎 ∈ 픏 } ∪ { 퐹푏 ∣ 푏 ∈ 픏 } be a subbasis. Then 푇 ≔ ′ { 퐹푎 ∩ 퐹푏 ∣ 푎, 푏 ∈ 픏 } is a basis of 휏. Lemma. If 픏 is a distributive lattice, then 푃퐹(픏) is a Stone space. ′ Proof. Let 퐴 be in 푇, the basis of 휏, there are 푎, 푏 ∈ 픏 such that 퐴 = 퐹푎 ∩ 퐹푏. Since ′ ′ ′ ′ 퐴 is in the basis of 휏, 퐴 is open. Now, 퐴 = (퐹푎 ∩ 퐹푏) = 퐹푎 ∪ 퐹푏 by de Morgan’s ′ ′ laws. Since 퐹푎 and 퐹푏 are in the basis, their union 퐴 is open, so 퐴 is closed. Let 푃, 푄 be prime filters with 푃 ≠ 푄. Then either 푃 ⊈ 푄 or 푄 ⊈ 푃. If 푃 ⊈ 푄, ′ there is an 푎 ∈ 픏 with 푎 ∈ 푃 and 푎 ∉ 푄, so 푃 ∈ 퐹푎 and 푄 ∈ 퐹푎. Since 푃 and 푄 are separated by disjoint open sets, 푃퐹(픏) is Hausdorff. To show that 푃퐹(픏) is compact, we need to show that any open cover 풰 ⋃ ′ of 푃퐹(픏) has a finite subcover. If 푈 ∈ 풰, then 푈 = 푎,푏∈퐴 { 퐹푎 ∩ 퐹푏 }, for some 퐴 ⊆ 픏. Since 풰 is a cover, for fixed 퐴1, 퐴2 ⊆ 퐴, ′ ′ 푃퐹(픏) ⊆ ﾌ { 퐹푎 ∩ 퐹푏 } ⊆ ﾌ { 퐹푎 } ∪ ﾌ { 퐹푏 } 푎,푏∈퐴 푎∈퐴1 푎∈퐴2 which is also an open cover of 푃퐹(픏). Hence, ⋂ 퐹 ⊆ ⋃ 퐹 . 푏∈퐴2 푏 푎∈퐴1 푎 Let 퐼 be the ideal generated by 푎 ∈ 퐴1 and 퐺 the filter generated by 푏 ∈ 퐴2.A proof by contradiction and the use of the prime ideal theorem shows that 퐺∩퐼 ≠ ∅. If 푥 ∈ 퐺 ∩ 퐼, 푥 ≥ 푏∗ ≔ 푏1 ∧ … ∧ 푏푛 for some 푏푖 ∈ 퐴2. Also, 푥 ≤ 푎∗ = 푎1 ∨ … ∨ 푎푚 for some 푎 ∈ 퐴 . Because 퐹 s are prime filters, 퐹 ∩…∩퐹 ⊆ 퐹 ⊆ 퐹 ⊆ 퐹 ∪…∪퐹 . 푖 1 푖 푏1 푏푛 푏∗ 푎∗ 푎1 푎푚 This gives a covering of 푃퐹(픏) which makes it compact. Lemma. If 푋 is a Stone space, then 퐶푙(푋) is a Boolean algebra. Proof. This follows by a tedious checking of the axioms of a Boolean algebra, using the fact that clopenness is preserved under intersection, union and complement. We now claim that 푃퐹 and 퐶푙 establish an equivalence, which proves the Stone representation theorem. 3 Theorem (Stone Representation Theorem). The map 푓 ∶ 픅 → 퐶푙(푃퐹(픅)) given by 푏 ↦ 퐹푏 is an isomorphism of Boolean algebras. The map 푔 ∶ 푋 → 푃퐹(퐶푙(푋)) given by 푥 ↦ 퐹푈 ≔ { 푈 ∈ 퐶푙(푋) ∣ 푥 ∈ 푈 } is a homeomorphism. Proof. First, we notice that 퐶푙(푃퐹(픅)) is a Boolean algebra by the above lemma, hence a subalgebra of the power set of 푃퐹(픅). Since 퐹푏 is an element of the ′ subbase 푆 of 휏, it is open. Since (퐹푏) is also an element, 퐹푏 is closed, hence clopen. To show that 푓 is injective, consider 푎 ≠ 푏. We want to show that 퐹푎 ≠ 퐹푏. Either 푎 ≰ 푏 or 푏 ≰ 푎. Using 푎 ≰ 푏, let 퐺 =↑ 푎 be the filter generated by 푎 and 퐼 =↓ 푏 be the ideal generated by 푏, so that 퐺 ∩ 퐼 = ∅. Then there is a prime filter 푃 with 퐺 ⊆ 푃 and 푃 ∩ 퐼 = ∅. Hence, 푎 ∈ 푃 and 푏 ∉ 푃, so 푃 ∈ 퐹푎 and 푃 ∈ 퐹푏, so that 퐹푎 ≠ 퐹푏. To show that 푓 is surjective, consider 퐶 a clopen subset of 푃퐹(픅), we need to show that 퐶 is in the image of 푓. Since 퐶 is open, for 퐵1, 퐵2 ⊆ 픅, ′ 퐶 = ﾌ { 퐹푏 ∩ 퐹푐 } = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푏 ∈ 푃, 푐 ∉ 푃 } 푏∈퐵1,푐∈퐵2 푏∈퐵1,푐∈퐵2 Since 푃 is also an ultrafilter, 푐 ∉ 푃 implies 푐′ ∈ 푃, and since 푃 is closed under meets, 푏 ∧ 푐′ ∈ 푃. Hence, ′ 퐶 = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푏 ∧ 푐 ∈ 푃 } = ﾌ { 푃 ∈ 푃퐹(픅) ∣ 푑 ∈ 푃 } = ﾌ 퐹푑 푏∈퐵1,푐∈퐵2 푑∈퐵3 푑∈퐵3 Since 퐹푑 is open for each 푑, { 퐹푑 ∣ 푑 ∈ 퐵3 } is an open cover of 퐶. Since 퐶 is closed and 푃퐹(픅) is compact, 퐶 is also compact. Hence, 퐶 = 퐹 ∪ … ∪ 퐹 = 퐹 푑1 푑푛 푑1∨…∨푑푛 which is in the image of 푓. To see that 푔 is well-defined, for any 푥 ∈ 푋, 푔(푥) should be a prime filter of 퐶푙(푋). Let 푉 ∈ 퐶푙(푋) with 푉 ∈ 푔(푥) and 푉 ⊆ 푊. Then 푥 ∈ 푊 and 푊 ∈ 푔(푥), hence 푔(푥) is an up-set of 퐶푙(푋). If 푈, 푉 ∈ 푔(푥), then 푥 ∈ 푈 and 푥 ∈ 푉, so 푥 ∈ 푈 ∩ 푉. Thus 푈 ∩ 푉 ∈ 푔(푥) and 푔(푥) is a filter. Suppose 푈 ∪ 푉 ∈ 푔(푥), then 푥 ∈ 푈 ∪ 푉, so 푥 ∈ 푈 or 푥 ∈ 푉, hence 푔(푥) is a prime filter. Let 휏1 be the topology of 푃퐹(퐶푙(푋)). To check that 푔 is continuous, it suffices to show that the inverse image of every set in the basis of 휏1 is open. If 퐴 is an element in the basis of 휏1, since 퐹 is a homomorphism, ′ { 퐹푈 ∩ 퐹푉 ∣ 푈, 푉 ∈ 퐶푙(푋) } = { 퐹푈∩푉′ ∣ 푈, 푉 ∈ 퐶푙(푋) } So 퐴 = 퐹푊 for some 푊 ∈ 퐶푙(푋). −1 푔 (퐴) = { 푥 ∈ 푋 ∣ 푔(푥) ∈ 퐹푊 } = { 푥 ∈ 푋 ∣ 푊 ∈ 푔(푥) } = 푊 4 Since 푊 is a clopen subset of 푋, the inverse image of 퐴 is open for every 퐴 in the basis of 휏1. Thus, 푔 is continuous. To show that 푔 is injective, let 푥, 푦 ∈ 푋 with 푔(푥) = 푔(푦). Every 푎, 푏 ∈ 푋 with 푎 ∉ 푏 can be separated by open sets, because 푋 is Hausdorff. But 푋 also has a basis of clopen sets, thus we can separate 푎, 푏 by clopen sets. Hence, ﾎ 푔(푥) ≔ ﾎ { 푈 ∈ 퐶푙(푋) ∣ 푥 ∈ 푈 } ⊆ { 푥 } If 푔(푥) = 푔(푦), then ⋂ 푔(푥) = ⋂ 푔(푦), hence { 푥 } = { 푦 }, which implies 푥 = 푦. Lastly, we check that 푔 is surjective.
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