A new characterization of complete Heyting and co-Heyting algebras

We give a new order-theoretic characterization of a complete Heyting and co-Heyting algebra C. This result provides an unexpected relationship with the field of Nash equilibria, being based on the so-called Veinott ordering relation on subcomplete sublattices of C, which is crucially used in Topkis’ theorem for studying the order-theoretic stucture of Nash equilibria of supermodular games. Introduction Complete Heyting algebras — also called frames, while locales is used for complete co-Heyting algebras — play a fundamental role as algebraic model of intuitionistic logic and in pointless topology [6, 7]. To the best of our knowledge, no characterization of complete Heyting and co-Heyting algebras has been known. As reported in [1], a sufficient condition has been given in [4] while a necessary condition has been given by [3]. We give here an order-theoretic characterization of complete Heyting and co-Heyting algebras that puts forward an unexected relationship with Nash equilibria. Topkis’ theorem [9] is well known in the theory of supermodular games in mathematical economics. This result shows that the set of solutions of a supermodular game, i.e., its set of pure-strategy Nash equilibria, is nonempty and contains a greatest element and a least one [8]. Topkis’ theorem has been strengthned by [11], where it is proved that this set of Nash equilibria is indeed a complete lattice. These results rely on so-called Veinott’s ordering relation. Let 〈C,≤,∧,∨〉 be a complete lattice. Then, the relation ≤⊆ ℘(C)×℘(C) on subsets of C, according to Topkis [8], has been introduced by Veinott [9, 10]: for any S, T ∈ ℘(C), S ≤ T △ ⇐⇒ ∀s ∈ S.∀t ∈ T. s ∧ t ∈ S & s ∨ t ∈ T. This relation ≤ is always transitive and antisymmetric, while reflexivity S ≤ S holds if and only if S is a sublattice of C. If SL(C) denotes the set of nonempty subcomplete sublattices of C then 〈SL(C),≤〉 is therefore a poset. The proof of Topkis’ theorem is then based on the fixed points of a certain mapping defined on the poset 〈SL(C),≤〉. To the best of our knowledge, no result is available on the order-theoretic properties of the Veinott poset 〈SL(C),≤〉. When is this poset a lattice? And a complete lattice? Our efforts in investigating these questions led to the following main result: the Veinott poset SL(C) is a complete lattice if and only if C is a complete Heyting and co-Heyting algebra. This result therefore revealed an unexpected link between complete Heyting algebras and Nash equilibria of supermodular games. 1 Notation If 〈P,≤〉 is a poset and S ⊆ P then lb(S) denotes the set of lower bounds of S, i.e., lb(S) , {x ∈ P | ∀s ∈ S. x ≤ s}, while if x ∈ P then ↓ x , {y ∈ P | y ≤ x}. Let 〈C,≤,∧,∨〉 be a complete lattice. A nonempty subset S ⊆ C is a subcomplete sublattice of C if for all its nonempty subsets X ⊆ S, ∧X ∈ S and ∨X ∈ S, while S is merely a sublattice of C if this holds for all its nonempty and finite subsets X ⊆ S only. If S ⊆ C then the nonempty Moore closure of S is defined as M(S) , {∧X ∈ 1 C |X ⊆ S,X 6= ∅}. Let us observe that M is an upper closure operator on the poset 〈℘(C),⊆〉, meaning that: (1) S ⊆ T ⇒ M(S) ⊆ M(T ); (2) S ⊆ M(S); (3) M(M(S)) = M(S). C is a complete Heyting algebra (also called frame) if for any x ∈ C and Y ⊆ C, x ∧ ( ∨ Y ) = ∨ y∈Y x ∧ y, while it is a complete co-Heyting algebra if the dual equation x ∨ ( ∧ Y ) = ∧ y∈Y x ∨ y holds. Let us recall that these two notions are orthogonal, for example the complete lattice of open subsets of R ordered by ⊆ is a complete Heyting algebra, but not a complete co-Heyting algebra. C is (finitely) distributive if for any x, y, z ∈ C, x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). Let us define SL(C) , {S ⊆ C | S 6= ∅, S subcomplete sublattice of C}. Thus, if ≤ denotes the Veinott ordering defined in Section then 〈SL(C),≤〉 is a poset. 2 The Sufficient Condition To the best of our knowledge, no result is available on the order-theoretic properties of the Veinott poset 〈SL(C),≤〉. The following example shows that, in general, 〈SL(C),≤〉 is not a lattice. Example 2.1. Consider the nondistributive pentagon lattice N5, where, to use a compact notation, subsets of N5 are denoted by strings of letters. e d b c a Consider ed, abce ∈ SL(N5). It turns out that ↓ ed = {a, c, d, ab, ac, ad, cd, ed, acd, ade, cde, abde, acde, abcde} and ↓ abce = {a, ab, ac, abce}. Thus, {a, ab, ac} is the set of common lower bounds of ed and abce. However, the set {a, ab, ac} does not include a greatest element, since a ≤ ab and a ≤ ac while ab and ac are incomparable. Hence, ab and c are maximal lower bounds of ed and abce, so that 〈SL(N5),≤ 〉 is not a lattice. Indeed, the following result shows that if SL(C) turns out to be a lattice then C must necessarily be distributive. Lemma 2.2. If 〈SL(C),≤〉 is a lattice then C is distributive. Proof. By the basic characterization of distributive lattices, we know that C is not distributive iff either the pentagon N5 is a sublattice of C or the diamond M3 is a sublattice of C. We consider separately these two possibilities. (N5) Assume that N5, as depicted by the diagram in Example 2.1, is a sublattice of C. Following Example 2.1, we consider the sublattices ed, abce ∈ 〈SL(C),≤〉 and we prove that their meet does not exist. By Example 2.1, ab, ac ∈ lb({ed, abce}). Consider any X ∈ SL(C) such that X ∈ lb({ed, abce}). Assume that ab ≤ X . If x ∈ X then, by ab ≤ X , we have that b ∨ x ∈ X . Moreover, by X ≤ abce, b∨ x ∈ {a, b, c, e}. If b ∨ x = e then we would have that e ∈ X , and in turn, by X ≤ ed, d = e ∧ d ∈ X , so that, by X ≤ abce, we would get the contradiction d = d ∨ c ∈ {a, b, c, e}. Also, if b ∨ x = c then we would have that c ∈ X , and in turn, by ab ≤ X , e = b∧ c ∈ X , so that, as in the previous case, we would get the contradiction d = d ∨ c ∈ {a, b, c, e}. Thus, we necessarily have that b ∨ x ∈ {a, b}. On the one hand, if b ∨ x = b then x ≤ b so that, by ab ≤ X , x = b ∧ x ∈ {a, b}. On the other hand, if b ∨ x = a then x ≤ a so that, by ab ≤ X , x = a ∧ x ∈ {a, b}. Hence, X ⊆ {a, b}. Since X 6= ∅, suppose that a ∈ X . Then, by ab ≤ X , b = b ∨ a ∈ X . If, instead, b ∈ X then, by X ≤ abce, a = b ∧ a ∈ X . We have therefore shown that X = ab. An analogous argument shows that if ac ≤ X then X = ac. If the meet of ed and abce would exist, call it Z ∈ SL(C), from Z ∈ lb({ed, abce}) and ab, ac ≤ Z we would get the contradiction ab = Z = ac. (M3) Assume that the diamond M3, as depicted by the following diagram, is a sublattice of C.