A new characterization of complete Heyting and co-Heyting algebras

نویسنده

  • Francesco Ranzato
چکیده

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.

برای دانلود متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

ثبت نام

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

منابع مشابه

Dually quasi-De Morgan Stone semi-Heyting algebras II. Regularity

This paper is the second of a two part series. In this Part, we prove, using the description of simples obtained in Part I, that the variety $mathbf{RDQDStSH_1}$ of regular dually quasi-De Morgan Stone semi-Heyting algebras of level 1 is the join of the variety generated by the twenty 3-element $mathbf{RDQDStSH_1}$-chains and the variety of dually quasi-De Morgan Boolean semi-Heyting algebras--...

متن کامل

Dually quasi-De Morgan Stone semi-Heyting algebras I. Regularity

This paper is the first of a two part series. In this paper, we first prove that the variety of dually quasi-De Morgan Stone semi-Heyting algebras of level 1 satisfies the strongly blended $lor$-De Morgan law introduced in cite{Sa12}. Then, using this result and the results of cite{Sa12}, we prove our main result which gives an explicit description of simple algebras(=subdirectly irreducibles) ...

متن کامل

On Heyting algebras and dual BCK-algebras

A Heyting algebra is a distributive lattice with implication and a dual $BCK$-algebra is an algebraic system having as models logical systems equipped with implication. The aim of this paper is to investigate the relation of Heyting algebras between dual $BCK$-algebras. We define notions of $i$-invariant and $m$-invariant on dual $BCK$-semilattices and prove that a Heyting semilattice is equiva...

متن کامل

Finitely Presented Heyting Algebras

In this paper we study the structure of finitely presented Heyting algebras. Using algebraic techniques (as opposed to techniques from proof-theory) we show that every such Heyting algebra is in fact coHeyting, improving on a result of Ghilardi who showed that Heyting algebras free on a finite set of generators are co-Heyting. Along the way we give a new and simple proof of the finite model pro...

متن کامل

Injective and Projective Heyting Algebras^) by Raymond Balbes and Alfred Horn

The determination of the injective and projective members of a category is usually a challenging problem and adds to knowledge of the category. In this paper we consider these questions for the category of Heyting algebras. There has been a lack of uniformity in terminology in recent years. In [6] Heyting algebras are referred to as pseudo-Boolean algebras, and in [1] they are called Brouwerian...

متن کامل

ذخیره در منابع من


  با ذخیره ی این منبع در منابع من، دسترسی به آن را برای استفاده های بعدی آسان تر کنید

برای دانلود متن کامل این مقاله و بیش از 32 میلیون مقاله دیگر ابتدا ثبت نام کنید

ثبت نام

اگر عضو سایت هستید لطفا وارد حساب کاربری خود شوید

عنوان ژورنال:
  • Logical Methods in Computer Science

دوره 13  شماره 

صفحات  -

تاریخ انتشار 2017