Logical Methods in Computer Science Vol. 13(3:25)2017, pp. 1–11 Submitted Apr. 15, 2015 www.lmcs-online.org Published Sep. 14, 2017 A NEW CHARACTERIZATION OF COMPLETE HEYTING AND CO-HEYTING ALGEBRAS FRANCESCO RANZATO Dipartimento di Matematica, University of Padova, Italy e-mail address: [email protected] ABSTRACT. 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 [Johnstone 1982, Johnstone 1983]. To the best of our knowledge, no characterization of complete Heyting and co-Heyting algebras has been known. As reported in [Balbes and Dwinger 1974], a sufficient condition has been given in [Funayama 1959] while a necessary condition has been given by [Chang and Horn 1962]. 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 [Topkis 1998] 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 [Topkis 1978]. Topkis’ theorem has been strengthned by [Zhou 1994], 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 (also called strong set relation). Let hC; ≤; ^; _i be a complete lattice. Then, the relation ≤v⊆ }(C) × }(C) on subsets of C, according to Topkis [Topkis 1978], has been introduced by Veinott [Topkis 1998, Veinott 1989]: for any S; T 2 }(C), 4 S ≤v T () 8s 2 S:8t 2 T: s ^ t 2 S & s _ t 2 T: This relation ≤v is always transitive and antisymmetric, while reflexivity S ≤v 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 hSL(C); ≤vi is therefore a poset. The proof of Topkis’ theorem is then based on the fixed points of a certain mapping defined on the poset hSL(C); ≤vi. Key words and phrases: Complete Heyting algebra, Veinott ordering. lLOGICAL METHODS c F. Ranzato IN COMPUTER SCIENCE DOI:10.23638/LMCS-13(3:25)2017 CC Creative Commons 2 F. RANZATO To the best of our knowledge, no result is available on the order-theoretic properties of the Veinott poset hSL(C); ≤vi. 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 finding therefore reveals an unexpected link between complete Heyting algebras and Nash equilibria of supermodular games. This characterization of the Veinott relation ≤v could be exploited for generalizing a recent approach based on abstract interpretation for approximating the Nash equilibria of supermodular games introduced by [Ranzato 2016]. 1. NOTATION If hP; ≤i is a poset and S ⊆ P then lb(S) denotes the set of lower bounds of S, i.e., lb(S) , fx 2 P j 8s 2 S: x ≤ sg, while if x 2 P then #x , fy 2 P j y ≤ xg. Let hC; ≤; ^; _i be a complete lattice. A nonempty subset S ⊆ C is a subcomplete sublattice of C if for all its nonempty subsets X ⊆ S, ^X 2 S and _X 2 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) , f^X 2 C j X ⊆ S; X 6= ?g. Let us observe that M is an upper closure operator on the poset h}(C); ⊆i, meaning that: (1) S ⊆ T )M∗(S) ⊆ M∗(T ); (2) S ⊆ M∗(S); (3) M∗(M∗(S)) = M∗(S). We define SL(C) , fS ⊆ C j S 6= ?;S subcomplete sublattice of Cg: Thus, if ≤v denotes the Veinott ordering defined in Section then hSL(C); ≤vi is a poset. C is a complete Heyting algebra (also called frame) if for any x 2 C and Y ⊆ C, x ^ (W Y ) = W y2Y x ^ y, while it is a complete co-Heyting algebra (also called locale) if the dual equation V V x _ ( Y ) = y2Y 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 2 C, x ^ (y _ z) = (x ^ y) _ (x ^ z). Let us also recall that C is completely distributive if for any family fxj;k j j 2 J; k 2 K(j)g ⊆ C, we have that ^ _ _ ^ xj;k = xj;f(j) j2J k2K(j) f2J K j2J where J and, for any j 2 J, K(j) are sets of indices and J K , ff : J ![j2J K(j) j 8j 2 J: f(j) 2 K(j)g denotes the set of choice functions. It turns out that the class of completely distributive complete lattices is strictly contained in the class of complete Heyting and co-Heyting algebras. Clearly, any completely distribuitive lattice is a complete Heyting and co-Heyting algebra. On the other hand, this containment turns out to be strict, as shown by the following counterexample. Example 1.1. Let us recall that a subset S ⊆ [0; 1] of real numbers is a regular open set if S is open and S coincides with the interior of the closure of S. For example, (1=3; 2=3) and (0; 1=3) [ (2=3; 1) are both regular open sets, while (1=3; 2=3) [ (2=3; 1) is open but not regular. Let us consider C = hfS ⊆ [0; 1] j S is a regular open setg; ⊆i. It is known that C is a complete Boolean algebra (see e.g. [Vladimirov 2002, Theorem 12, Section 2.5]). As a consequence, C is a complete Heyting and co-Heyting algebra (see e.g. [Vladimirov 2002, Theorem 3, Section 0.2.3]). Recall that an element a 2 C in a complete lattice C is an atom if a is different from the least element ?C of C and for any x 2 C, if ?C < x ≤ a then x = a, while C is atomic if for any x 2 C r f?C g there exists an atom a 2 C such that a ≤ x. It turns out that C does not have atoms: in fact, any regular open set S 2 C is a union of open sets, namely, S = [fU ⊆ [0; 1] j U is open;U ⊆ Sg A NEW CHARACTERIZATION OF COMPLETE HEYTING AND CO-HEYTING ALGEBRAS 3 (see e.g. [Vladimirov 2002, Section 2.5]), so that no S 2 C r f?g can be an atom of C. In turn, this implies that C is a complete Boolean algebra which is not atomic. It known that a complete Boolean algebra is completely distributive if and only if it is atomic (see [Koppelberg 1989, Theorem 14.5, Chapter 5]). Hence, since C is not atomic, we obtain that C is not completely distributive. 2. THE SUFFICIENT CONDITION To the best of our knowledge, no result is available on the order-theoretic properties of the Veinott poset hSL(C); ≤vi. The following example shows that, in general, hSL(C); ≤vi 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 2 SL(N5). It turns out that #ed = fa; c; d; ab; ac; ad; cd; ed; acd; ade; cde; abde; acde; abcdeg and #abce = fa; ab; ac; abceg. Thus, fa; ab; acg is the set of common lower bounds of ed and abce. However, the set fa; ab; acg does not include a greatest element, since a ≤v ab and a ≤v ac while ab and ac are incomparable. Hence, ab and c are maximal lower bounds of ed and v abce, so that hSL(N5); ≤ i 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 hSL(C); ≤vi 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 2 hSL(C); ≤vi and we prove that their meet does not exist. By Example 2.1, ab; ac 2 lb(fed; abceg).