TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS SERGIO A. CELANI AND LEONARDO M. CABRER

Abstract. In this paper we will generalize the representation theory developed for finite Tarski algebras given in [7]. We will introduce the notion of Tarski space as a generalization of the notion of dense Tarski set, and we will prove that the category of Tarski algebras with semihomomorphisms is dually equivalent to the category of Tarski spaces with certain closed relations, called T -relations. By these results we will obtain that the algebraic category of Tarski algebras is dually equivalent to the category of Tarski spaces with certain partial functions. We will apply these results to give a topological characterization of the subalgebras.

1. Introduction and preliminaries The variety of Tarski algebras, also called Implication algebras, have been introduced and studied by J. C. Abbott in [2, 3], and by other authors (see [13],[9], [7]). This variety corresponds to the {→}-subreducts of Boolean algebras, and they are also the algebraic counterpart of the implication fragment of classical propositional logic (see [13]). On the other hand, it is known that if A is a Tarski algebra and X (A) is the set of all maximal filters (see page 2), also called irreducible deductive systems in the literature, then the map βA : A → P (X (A)) defined by βA (a) = {P ∈ X (A) : a ∈ P } is an injective homomorphism of Tarski algebras (see [7]). In other words, A is isomorphic to a Tarski subalgebra of the {→}-subreduct of the power set algebra P (X (A)). If A is a finite Boolean algebra, then the map βA is surjective, and thus A is isomorphic to P (X (A)). But if A is not a Boolean algebra, then the map βA is not surjective. In [7], the concept of Tarski sets was introduced with the aim of giving a representation theorem for finite Tarski algebras. A Tarski set is a pair hX, Ki where X is a non-empty set and K is a non-empty subset of the power set P (X). The fact that every finite Tarski algebra can be fully represented by an adequate Tarski set was showed in [7]. This result generalizes the Birkhoff’s representation for finite Boolean algebras. At the same time, in [1], a topological representation and a full duality between Tarski algebras and certain topological spaces, called implication spaces, were developed by Abad et al. The results of [1] are based on the fact that any Tarski algebra A can be represented as a union of a unique family of filters of a suitable Boolean algebra Bo(A). This duality appeals to the well known duality between Boolean algebras and Boolean spaces. In this paper we will develop other duality for Tarski algebras from a different point of view. We will introduce the notion of Tarski space as a generalization of the notion of dense Tarski set introduced in [7], and we will prove that there exists a dual equivalence between the category of Tarski algebras with semi-homomorphism and Tarski spaces with certain closed relations, called T -relations. This duality is a generalization of the results of Halmos and F. B. Wright on the duality between the category of Boolean algebras with semi-homomorphisms (1-preserving meethomomorphisms) and Boolean spaces with Boolean relations (see [11] and [15]). One of the advantages of our duality is that it is completely irrespective of the duality for the Boolean algebras. Moreover, the Boolean duality can be obtained as a consequence of this duality. The structure of the paper will be the following. In Section 2 we will recall some facts of Tarski algebras. In Section 3 we will introduce the Tarski spaces and we will prove that any Tarski algebra A is (isomorphic to) the dual Tarski algebra of some Tarski space, and conversely that for any Tarski space X there is a Tarski algebra A such that X is homeomorphic to dual Tarski space of A. 1991 Mathematics Subject Classification. 03G25, 06E15, 06F35. Key words and phrases. Tarski algebras, representation theorem, topological duality, subalgebras. 1

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SERGIO A. CELANI AND LEONARDO M. CABRER

In Section 4 we will define two categories whose objects are Tarski algebras. One of them will be the algebraic category of Tarski algebras with homomorphisms. The other one will be the category whose morphisms are the semi-homomorphisms between Tarski algebras. We will also define the category of Tarski spaces with T -relations, and its subcategory whose objects are the Tarski spaces and the morphisms are the T -partial functions. We will prove that there exists a correspondence between semi-homomorphisms (homomorphisms) and T -relations (T -partial functions). With these results we will obtain a duality; actually we are going to obtain two dualities. In section 5 we are going to prove a topological characterization of the subalgebras of a Tarski algebra. 2. Tarski algebras In this section we will recall the definitions and some basic properties of Tarski algebras. For detailed proofs of the results of this section see [2] and [3]. Definition 1. An algebra A = hA, →, 1i of type (2, 0) is a Tarski algebra if it satisfies the following identities: T1. 1 → a ≈ a, T2. a → a ≈ 1, T3. a → (b → c) ≈ (a → b) → (a → c) , T4. (a → b) → b ≈ (b → a) → a. We denote by T AR the variety of Tarski algebras. Given a Boolean algebra hB, ∨, ∧, ¬, 0, 1i, if we define the implication by a → b = ¬a ∨ b, then hB, →, 1i is a Tarski algebra. When there is no risk of misunderstanding given X a set and Y ⊆ X, we will note Y c instead of X − Y . Let X be a non-empty set. It is known that hP (X) , ⇒, Xi is a Tarski algebra, where U ⇒ V = U c ∪ V. Every subalgebra of hP (X) , ⇒, Xi will be called a Tarski algebra of sets. In a Tarski algebra A we can define an order relation ≤ by: a ≤ b if and only if a → b = 1. It is well known that hA, ≤i is a join-semilattice where the supremum of two elements a, b ∈ A is given by a ∨ b = (a → b) → b. Recall that a Tarski algebra can be defined as a Hilbert algebra hA, →, 1i satisfying the identity (a → b) → a ≈ a. This axiom is the characteristic identity for semisimple Hilbert algebras, so that the class of Tarski algebras is the class of semisimple Hilbert algebras. Let A be a Tarski algebra and let a1 , a2 , ..., an ∈ A, we define:  a1 → a if n = 1, (an , . . . , a1 ; a) = an → (an−1 , . . . , a1 ; a) if n > 1. A non-empty subset D of a Tarski algebra A is a called filter (or deductive system) if 1 ∈ D, and for any a, b ∈ A, if a, a → b ∈ D, then b ∈ D. The set of all filters of A is denoted by F i (A) . If X ⊆ A, then the filter generated by X, in symbols hXi, is the set hXi = {a ∈ A : ∃ {x1 , . . . , xn } ⊆ X such that (x1 , . . . , xn ; a) = 1} . In particular, the filter generated by {x}, is hxi = {a ∈ A : x → a = 1} = {a ∈ A : x ≤ a}. It is known (see [9], [10] or [5]) that F i (A) is an algebraic distributive lattice and its compact elements are Co (F i(A)) = {hXi : X is a finite subset of A} . A proper filter D A is maximal if and only if for any H ∈ F i (A) such that D ⊆ H, we have D = H or H = A. The set of all maximal filters is denoted by X (A). Theorem 2. Let A be a Tarski algebra. Then the following conditions are equivalent: (1) P ∈ X (A) . (2) P is prime, i.e., for all a, b ∈ A if a ∨ b ∈ P , then a ∈ P or b ∈ P. (3) For every a, b ∈ / P , there exists c ∈ / P such that a ≤ c and b ≤ c.

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

3

Let A be a Tarski algebra. A subset I of A is called an ideal of A if for every a, b ∈ A, if b ∈ I and a ≤ b, then a ∈ I, and for all a, b ∈ I, a ∨ b ∈ I. The following result can be found in [6] for Hilbert algebras. Theorem 3. Let A be a Tarski algebra. Let D ∈ F i (A) and let I be an ideal of A such that D ∩ I = ∅. Then there exists P ∈ X (A) such that D ⊆ P and P ∩ I = ∅. Theorem 4. Let A be a Tarski algebra. Then (1) For all a, b ∈ A, if a  b there exists P ∈ X (A) such that a ∈ P and b ∈ / P. (2) If P ∈ X (A) , then a → b ∈ / P if and only if a ∈ P and b ∈ / P. Theorem 5. Let A be a Tarski algebra. Then the map βA : A → P (X (A)) defined by βA (a) = {P ∈ X (A) : a ∈ P } is an injective homomorphism of Tarski algebras. 3. Topological representation In this section we will prove that the Tarski algebras can be represented by means of topological spaces with additional conditions. These results generalize the similar results given in [7] for the case of finite Tarski algebras. Definition 6. A Tarski space, or T -space for short, is a structure X = hX, K, τ i such that: (1) hX, τ i is a topological space where K is a basis of compact subsets for τ . (2) F or every A, B ∈ K, A ∩ B c ∈ K. (3) For every x, y ∈ X, if x 6= y, then there exists U ∈ K such that x ∈ / U and y ∈ U. (4) If F is a closed subset and K = {UTi : i ∈ I} is a dually directed subfamily of K such that F ∩ Ui 6= ∅ for all i ∈ I, then F ∩ {Ui : i ∈ I} = 6 ∅. Remark 7. By conditions (1), (2) and (3) of Definition 6 it is easy to see that every Tarski space is Hausdorff and totally disconnected. Given a T -space X = hX, K, τ i we can define two different Tarski subalgebras of hP (X) , ⇒, Xi. Let us consider the following subsets of P (X) : TK (X) = {W c ∪ S : W ∈ K and S ⊆ W } , and DK (X) = {U : U c ∈ K} . The set TK (X) is the Tarski algebra defined in [7]. Proposition 8. Let X = hX, K, τ i be a T -space. Then TK (X) = hTK (X) , ⇒, Xi and DK (X) = hDK (X) , ⇒, Xi are Tarski subalgebras of hP (X) , ⇒, Xi and DK (X) is a subalgebra of TK (X). Proof. By the results of [7], TK (X) is a subalgebra of hP (X) , ⇒, Xi. Moreover, if U ∈ DK (X), then U = W c for some W ∈ K. Since U = W c ∪ ∅, we get U ∈ TK (X) . Then DK (X) ⊆ TK (X). It is clear that X ∈ DK (X), because K is a basis of τ . If U, V ∈ DK (X), then U = W1c and V = W2c for some W1 , W2 ∈ K. By condition (2) of Definition 6, W1c ∩ W2 ∈ K. Then, c

U ⇒ V = U c ∪ V = W1 ∪ W2c = (W1c ∩ W2 ) ∈ DK (X) . Thus, DK (X) is a Tarski subalgebra of TK (X) .



Given a T -space X = hX, K, τ i , the Tarski algebra DK (X) will be called the dual Tarski algebra of the T -space X . Let us recall that a Boolean space is a topological space hX, τ i such that is compact and totally disconnected. A Boolean space can be characterized as a compact, Hausdorff topological space with a basis of open and closed (clopen) subsets. In the next result we see that the Boolean spaces are a particular class of T -spaces. Proposition 9. The Boolean spaces can be identified with the T -spaces X = hX, K, τ i such that X ∈ K.

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SERGIO A. CELANI AND LEONARDO M. CABRER

Proof. It is clear that any Boolean space is a T -space. Let X = hX, K, τ i be a T -space such that X ∈ K. Then hX, τ i is compact. By conditions (1), (2) and (3) of Definition 6, hX, τ i is Hausdorff and totally disconnected. Thus, hX, τ i is a Boolean space.  We note that if X = hX, K, τ i is a T -space such that X ∈ K, then ∅ ∈ DK (X). Thus, DK (X) is a Boolean algebra and it is the set of all clopen subsets of the topological space X . Now we will prove that if a T -space X = hX, K, τ i is finite, then TK (X) = DK (X). First, we will prove the following auxiliary result. Lemma 10. Let X = hX, K, τ i be a finite T -space. Then: (1) For each x ∈ X, {x} ∈ K. (2) If W ∈ K and S ⊆ W , then S ∈ K. Proof. (1). Let x ∈ X. Since K is a basis for τ , there exists W ∈ K such that x ∈ W . If W = {x}, then we have the result. If W − {x} = 6 ∅, since W is a finite set we can determine a finite subset {y1 , ..., yn } = W − {x}. By condition (3) of Definition 6, there are Uy1 , . . . , Uyn ∈ K such that x∈ / Uyi and yi ∈ Uyi , for 1 ≤ i ≤ n. Therefore we have {x} = W ∩ Uyc1 ∩ Uyc2 ∩ . . . ∩ Uycn . By condition (2) of Definition 6, {x} ∈ K. (2) Let W ∈ K and S ⊆ W . If S = {x} for some x ∈ W , then S ∈ K. If W ∩ S c = {y1 , ..., yn }, c c c c then S = W ∩ {y1 } ∩ ... ∩ {yn } . Since {yi } ∈ K for each 1 ≤ i ≤ n, S = W ∩ {y1 } ∩ ... ∩ {yn } ∈ K.  Corollary 11. If X = hX, K, τ i is a finite T -space, then TK (X) = DK (X). Proof. By Proposition 8, DK (X) is a subalgebra of TK (X). Let U ∈ TK (X). Then there exists c W ∈ K and S ⊆ W such that U = W c ∪ S = (W ∩ S c ) . By Lemma 10, S ∈ K, and by condition (2) of Definition 6, we have W ∩ S c ∈ K. Therefore, U ∈ DK (X).  Now we will provide a construction which shows that any Tarski algebra A is (isomorphic to) the dual Tarski algebra of some T -space, i.e., we will see that for any Tarski algebra A there exists a T -space X = hX, K, τ i such that A ∼ = DK (X). The first step is to analyze the structure of the maximal filters of a Tarski algebra. Let A ∈ T AR. Let us consider the set X (A) and let us consider the family of sets c

KA = {βA (a) : a ∈ A} . We note the following two facts: S c (1) X (A) = {βA (a) : a ∈ A}, and c c (2) For any a, b ∈ A and for any P ∈ X (A) such that P ∈ βA (a) ∩ βA (b) there exists c ∈ A c c c such that P ∈ βA (c) ⊆ βA (a) ∩ βA (b) Item (1) follows from the fact that any maximal filter is proper, and the second one follows from Theorem 2. Thus, the family KA is a basis for a topology τA on X (A). Let us denote by X (A) = hX (A) , KA , τA i the topological space associated with A, called the dual space of A. Now, for each D ∈ F i(A), let us consider the set ΦA (D) = {P ∈ X (A) : D ⊆ P } . Note that ΦA (D) =

\

{βA (a) : a ∈ D} ,

for each D ∈ F i(A). It is also easy to prove that if D = h{a1 , . . . , an }i, then ΦA (D) = βA (a1 ) ∩ . . . ∩ βA (an ). These facts will be used in the following proposition where we will conclude that in fact hX (A) , KA , τA i is a Tarski space for each Tarski algebra A. Proposition 12. Let A ∈ T AR and let X (A) be the dual space of A. Then:

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

5

(1) A subset U ⊆ X (A) is open in X (A) if and only if there exists a deductive system D of c A such that U = ΦA (D) . (2) A subset U ⊆ X (A) is compact and open in X (A) if and only if there exist a1 , . . . , an ∈ A c such that U = (ΦA (h{a1 , ..., an }i)) . (3) Every element of KA is a compact and open subset of X (A). c (4) For every a, b ∈ A, βA (a) ∩ βA (b) ∈ KA . (5) Let F be a closed subset in X (A), and let K = {Ui : iT∈ I} be a dually directed subfamily of KA such that F ∩ Ui 6= ∅ for any i ∈ I. Then, F ∩ {Ui : i ∈ I} = 6 ∅. Proof. (1) Let U be an open subset of X (A) . Since KA is a basis of the space X (A), U = S c {βA (a) : a ∈ B} for some B ⊆ A. Let us consider the filter D = hBi . It is evident that c U = ΦA (D) . T On the other hand it is easy to check that if D is a filter of A, then ΦA (D) = {βA (a) : a ∈ D} . c Thus, ΦA (D) is open in X (A) . c S (2) Letc U be a compact and open set in X (A). By item (1) above, U = ΦA (D) = {βA (a) : a ∈ D} for some filter D of A. Since U is compact, there exist {a1 , ..., an } ⊆ D such that c

c

U = βA (a1 ) ∪ ... ∪ βA (an ) = (βA (a1 ) ∩ ... ∩ βA (an ))

c

c

= (ΦA (h{a1 , ..., an }i)) . The other direction is easy and left to the reader. c c c (3) For every a ∈ A, βA (a) = ΦA (hai) . By assertion (2), we have that βA (a) is a compact and open subset of X (A) . (4) Let a, b ∈ A. By Proposition 4 item (2), we have that c

c

c

c

βA (a) ∩ βA (b) = (βA (a) ∪ βA (b)) = βA (a → b) ∈ KA . c

(5) Let F be a closed subset of X (A) and let K = {βA (a) } be a dually directed subfamily of c c KA such that F ∩ βA (a) 6= ∅ for any βA (a) ∈ K. Let us consider the set c

H = {a ∈ A : βA (a) ∈ K} and the decreasing set generated by H: (H] = {x ∈ A : x ≤ c for some c ∈ H} . We prove that (H] is an ideal of A. Let a, b ∈ (H] . Then there exist c1 , c2 ∈ H such that c c c a ≤ c1 and b ≤ c2 . Since βA (c1 ) , βA (c2 ) ∈ K, then there exists c ∈ H such that βA (c) ⊆ c c βA (c1 ) ∩ βA (c2 ) . So, a ≤ c and b ≤ c, thus (H] ∈ Id (A) . Since F is a closed subset, using assertion (1) above, we have F = ΦA (D) for some filter D. We prove that D ∩ (H] = ∅. c

Suppose that there exist a ∈ D and c ∈ H such that a ≤ c. By hypothesis, ΦA (D) ∩ βA (c) 6= ∅. So, there exists P ∈ X (A) such that D ⊆ P and c ∈ / P. But as a ∈ D, a ∈ P and consequently c ∈ P, which is a contradiction. Thus, D ∩ (H] = ∅. From Theorem 3,Tthere exists P ∈ X (A) c such that D ⊆ P and H ∩ P = ∅. Therefore, P ∈ ΦA (D) = F and P ∈ {βA (a) : i ∈ I} .  Theorem 13. Let A ∈ T AR. Then X (A) = hX (A) , KA , τA i is a T -space and the mapping βA : A → DKA (X (A)) is an isomorphism of Tarski algebras. Proof. It follows from Proposition 12 and the Definition of DKA (X (A)).



Let hX, τ i be a topological space. We will denote by C (X) (O(X)) the set of all closed subsets (open) of X. Let us denote by KO(X) the set of all compact and open subsets of X. Note that C (X) and O(X) are lattices under the inclusion relation, and that KO(X) is a join semilattice under the inclusion relation. Lemma 14. Let A ∈ T AR. Let X (A) be the dual space of A. Then:

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SERGIO A. CELANI AND LEONARDO M. CABRER

(1) The lattices F i(A) and C (X (A)) are dually isomorphic under the map ΦA : F i(A) → C (X (A)) defined by ΦA (D) = {P ∈ X (A) : D ⊆ P }. (2) The poset KO(X(A)) is isomorphic to the poset Co (F i(A)) of all compact elements of the algebraic lattice F i(A) under the map Ψ : Co (F i(A)) −→ KO(X(A)), defined by c

Ψ (D) = (Φ (D)) , for every D ∈ K (F i(A)). Proof. It follows from items (1) and (2) of Proposition 12.



Theorem 13 say that any Tarski algebra is isomorphic to the dual Tarski algebra DK (X) of some T -space X = hX, K, τ i. Now, the following question is see when a T -space is homeomorphic to the dual T -space of a Tarski algebra. In other words, given a T -space X = hX, K, τ i we will prove that there is a Tarski algebra A such that X is homeomorphic to X (A). Lemma 15. Let X = hX, K, τ i be a T -space. Then for each x ∈ X, the set HX (x) = {U ∈ DK (X) : x ∈ U } belongs to X (DK (X)) . Thus, the mapping HX : X → X (DK (X)) is well-defined. Proof. Let x ∈ X. It is clear that HX (x) is a filter of DK (X) . We will prove that it is prime. Let U, V ∈ DK (X) such that U, V ∈ / HX (x) , then x ∈ U c ∩ V c . As U c ∩ V c is open and K is a basis, c then there exists O ∈ K such that x ∈ Oc ⊆ U c ∩ V c . Thus O ∈ / HX (x) . By Theorem 2, HX (x) is a maximal filter of the Tarski algebra DK (X).  Theorem 16. Let X = hX, K, τ i be a T -space. Then the mapping HX : X → X (DK (X)) is an homeomorphism between the topological spaces X and X (DK (X)) . Proof. By condition (3) of Definition 6, it follows that HX is injective. We will prove that HX is onto. Let P ∈ X (DK (X)) . Let us consider the set H = {V c ∈ K : V ∈ / P } ⊆ K. The set F = case

T

{U : U ∈ P } is closed in X and F ∩ V c 6= ∅ for V c ∈ H, because in the opposite Vc ⊆

[

{U c : U ∈ P } ,

and since V c is compact, V c ⊆ U0c ∪ U1c ∪ ... ∪ Unc for some U0 , U1 , ..., Un ∈ F, i.e., X = U0c ∪ U1c ∪ ... ∪ Unc ∪ V = (U0 , . . . , Un ; V ) . It follows that V ∈ P , which is a contradiction. By condition (4) of Definition 6 we have that T F ∩ {U c : U ∈ / P} = 6 ∅. Then there exists \ \ x∈ {V : V ∈ P } ∩ {U c : U ∈ / P}, this implies that P = HX (x) . Thus, HX is onto. Now we will prove that HX is continuous. By Proposition 12, given an open set U of X (DK (X)) there exists a filter D of DK (X) such that U c = ΦDK (X) (D) . Let \ V = {O : O ∈ D} .

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

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−1 Then, V is closed in X. It is easy to see that V c = HX (U ) . Thus, HX is continuous. To see that HX is an open map consider W ∈ K. Since HX is a biyection, we have that:

HX (W ) = {HX (x) : x ∈ W } = DK (X) − {HX (x) : x ∈ W c } c = DK (X) − βDK (X) (W c ) = βDK (X) (W c ) , and the result follows.

 4. Topological Duality

In the previous section we have seen that Tarski algebras are related to T -spaces. In this section we will consider two categories whose objects are Tarski algebras and we will prove that they are dually equivalent to two other categories whose objects are T -spaces. For detailed definitions and results on categories and dualities see [4, Chapter 1.3]. First let us define the arrows of the categories of Tarski algebras. Definition 17. Let A, B ∈ T AR. A mapping h : A → B is a semi-homomorphism if for every a, b ∈ A : (1) h (a → b) ≤ h (a) → h (b) , (2) h (1) = 1. A homomorphism between two Tarski algebras A and B is a semi-homomorphism h such that h (a) → h (b) ≤ h (a → b) , for all a, b ∈ A. Note that a semi-homomorphism h : A → B is monotonic, i.e., if a ≤ b, then h (a) ≤ h (b), for every a, b ∈ A. We will denote by ST (A, B) (HT (A, B)) the set of all semi-homomorphisms (homomorphisms) from A into B. We note also that if A and B are Boolean algebras, then a semi-homomorphism can be defined as a function h : A → B satisfying the conditions h (1) = 1 and h (a ∧ b) = h(a) ∧ h(b), for every a, b ∈ A. At this point, we will define the following two algebraic categories whose objects are Tarski algebras: ST = Tarski algebras + semi-homomorphisms HT = Tarski algebras + homomorphisms It is clear that HT is a subcategory of ST . We will prove that these categories are dually equivalent to the following categories (precise definitions are given below): TR TF

= Tarski spaces + T -relations = Tarski spaces + T -partial functions,

respectively. These dualities are a generalization of the results of Halmos and F. B. Wright on the duality between the category of Boolean algebras with semi-homomorphisms, and Boolean spaces with Boolean relations (see [11] and [15]), and the duality between Boolean algebras with homomorphisms and Boolean spaces with continuous functions [12]). The following example is crucial in the next results. Example 18. Let X1 and X2 be two sets. Let R ⊆ X1 × X2 be a binary relation. For each x ∈ X1 , let R (x) = {y : (x, y) ∈ R} . Define the mapping hR : P (X2 ) → P (X1 ) by hR (U ) = {x ∈ X1 : R (x) ⊆ U } . Then it is easy to prove that hR ∈ ST (P (X2 ) , P (X1 )) . Definition 19. Let X1 and X2 be two T -spaces. We will say that a subset R ⊆ X1 × X2 is a T -relation if: (1) hR (U ) = {x ∈ X1 : R (x) ⊆ U } ∈ DK1 (X1 ), for every U ∈ DK2 (X2 ) . (2) R (x) is a closed subset of X2 , for all x ∈ X1 . A T -partial function between X1 and X2 is partial map f : X1 → X2 such that f −1 (U ) ∈ DK1 (X1 ) for each U ∈ DK2 (X2 ) .

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SERGIO A. CELANI AND LEONARDO M. CABRER

Remark 20. Note that if a T -partial function f : X1 → X2 is a function, i.e., its domain is X1 , then f will be a continuous function. But the reverse is not true, i.e., there exist continuous functions between T -spaces that are not T -partial functions, because it is not necessary for a continuos function f to satisfy the condition f −1 (U ) ∈ DK1 (X1 ) for each U ∈ DK2 (X2 ) . Let X1 and X2 be two Tarski spaces. We will denote by T R (X1 , X2 ) the set of all Tarski relations between X1 and X2 . We will denote by T F (X1 , X2 ) the set of all T -partial function f from X1 into X2 . Remark 21. It is clear that any f ∈ T F (X1 , X2 ) is also a T -relation. Then we can say that T F (X1 , X2 ) ⊆ T R (X1 , X2 ). We also note that hf (U ) = f −1 (U ), for each U ∈ DK2 (X2 ). Lemma 22. Let X1 and X2 be two T -spaces. Let R ∈ T R (X1 , X2 ). For each closed subset C of X1 , R (C) = {y : (x, y) ∈ R, for some x ∈ C } is a closed subset of X2 . Proof. Suppose that C is a closed subset of X1 . We will prove that R(C) is a closed subset of X2 by showing that for any y ∈ / R(C) there exists U ∈ K2 such that R (C) ⊆ U c and y ∈ U . So, let y ∈ / R (C). For every x ∈ C, y ∈ / R (x) . Since R (x) is closed, there exists Ux ∈ K2 , for c c each x ∈ U such that y ∈ Ux and R (x) ⊆ (Ux ) , i.e., x ∈ hR ((Ux ) ). Then we have that [ c C⊆ {hT (Uxc ) : x ∈ C, y ∈ Ux and x ∈ hR ((Ux ) ) } . So, F∩

\

c

c

{hR (Uxc ) : x ∈ C, y ∈ Ux and x ∈ hR ((Ux ) ) } = ∅

y∈U ∈K2

Now consider the set c

c

H = {hR (Uxc ) : x ∈ C, y ∈ Ux and x ∈ hR ((Ux ) ) } Since R is a T -relation, H ⊆ K1 , and it is easy to prove that H is a dually directed family. Then c by condition (5) of Definition 6, there exists hR (U c ) ∈ H such that c

C ∩ hR (U c ) = ∅, i.e., C ⊆ hT (U c ). Thus there exists U ∈ K2 such that R (C) ⊆ U c and y ∈ U . It follows that R (C) is closed in X2 .  Let X1 , X2 and X3 three sets. Let R ⊆ X1 × X2 and S ⊆ X2 × X3 two relations. The composition of the relations R and S is the relation defined as S ◦ R = {(x, z) : ∃y ∈ X2 (x, y) ∈ R and (y, z) ∈ S} . The following is a technical result needed to affirm that T R and T F are in fact categories. Theorem 23. Let X1 , X2 and X3 be three T -spaces. Let R ∈ T R (X1 , X2 ) and S ∈ T R (X2 , X3 ). Then: (1) (2) (3) (4)

IX1 = {(x, x) : x ∈ X1 } ∈ T F (X1 , X1 ) ⊆ T R (X1 , X1 ). R ◦ IX1 = R = IX2 ◦ R. S ◦ R ∈ T R (X1 , X3 ). If f ∈ T F (X1 , X2 ) and g ∈ T F (X2 , X3 ), then g ◦ f ∈ T F (X1 , X3 ).

Proof. Items (1), (2) and (4) are easy and left to the reader. (3) First we have to prove that hS◦R (U ) ∈ DK1 (X1 ), for every U ∈ DK3 (X3 ) . Let U ∈ DK3 (X3 ) . Then: hR◦S (U ) = {x ∈ X1 : (S ◦ R) (x) ⊆ U } = {x ∈ X1 : S (R (x)) ⊆ U } = {x ∈ X1 : R (x) ⊆ hS (U )} = hR (hS (U )) . Thus the result follows. We need to prove that for every x ∈ X1 , (R ◦ S) (x) is a closed subset of X3 . But this result follows from Lemma 22 

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

9

By the previous results we get that T R, the Tarski spaces with T -relations as arrows, and T F, the Tarski spaces with T -partial functions as arrows, are categories. Now we will determine the isomorphisms in the category T R. This result will be useful to prove that T R and ST are dually equivalent. Lemma 24. Let X1 and X2 be two T -spaces. Let R ∈ T R (X1 , X2 ). Then the following propositions are equivalent: (1) R is an isomorphism in the category T R. (2) R is a homeomorphism and R−1 (U ) ∈ K1 if and only if U ∈ K2 . Where R−1 (U ) = {x ∈ X1 : ∃y ∈ U, (x, y) ∈ R}. Proof. If R is an isomorphism in the category T R there exists S ∈ T R (X2 , X1 ) such that R ◦ S = IX2 and S ◦ R = IX1 . Then it is easy to see that R and S are bijective maps. By Theorem 22 we have that R and S are closed functions, then S is an homeomorphism. By Remark 21 and since R ∈ T R (X1 , X2 ) and S ∈ T R (X2 , X1 ), we have that U ∈ K2 if and only if R−1 (U ) ∈ K1 . The converse is easy and left to the reader.  In Section 3 we already study the relation between Tarski algebras and Tarski spaces. Now we will complete the duality studying the correspondence between semi-homomorphism and T relations. Let A, B ∈ T AR. Let h ∈ ST (A, B). Let us define a binary relation Rh ⊆ X (B) × X (A) by: (P, Q) ∈ Rh ⇔ h−1 (P ) ⊆ Q, −1 where h (P ) = {a ∈ A : h (a) ∈ P }. In the following Proposition items (1), (2) are necesary to conclude in (3) that Rh is a T -relation for each semi-homomorphism h. Items (4), (5) are some technical results needed to show that there is a contravariant functor from the category ST into T R. Item (6) will be needed later to see that ST and T R are dually aquivalent. Proposition 25. Let h ∈ ST (A, B). Then: (1) h−1 (P ) ∈ F i (A) for each P ∈ X (B). If h ∈ HT (A, B), then h−1 (P ) ∈ X (B) ∪ {B}, for each P ∈ X (A). (2) For all P ∈ X (B) and for all a ∈ A, h (a) ∈ / P if and only if there exists Q ∈ X (A) such that (P, Q) ∈ Rh and a ∈ / Q. (3) The relation Rh is a T -relation between X (B) and X (A). (4) If h ∈ HT (A, B), then Rh ∈ T F (X (B) , X (A)). (5) If C is a Tarski algebra and k ∈ ST (B, C), then Rk◦h = Rh ◦ Rk . (6) The mapping hRh : DKA (X (A)) → DKB (X (B)) satisfies βB ◦ h = hRh ◦ βA . Proof. (1) Since h(1) = 1, 1 ∈ h−1 (P ). Let a, a → b ∈ h−1 (P ). Then h (a) , h(a → b) ∈ P . Since h is a semi-homomorphism, h(a) → h(b) ∈ P , and taking into account that P is a filter, we get h(b) ∈ P , i.e. b ∈ h−1 (P ). Therefore, h−1 (P ) is a filter of B. Suppose that h is a homomorphism. Then h(a ∨ b) = h ((a → b) → b) = h(a) ∨ h(b), −1

for every a, b ∈ A. Assume that h (P ) 6= B, and let a, b ∈ B such that a ∨ b ∈ h−1 (P ). Then h(a ∨ b) = h (a) ∨ h (b) ∈ P . As P is prime, a ∈ h−1 (P ) or b ∈ h−1 (P ). So, h−1 (P ) is prime. Again by Theorem 2 we get h−1 (Q) ∈ X (A). (2) Let P ∈ X (B) and h (a) ∈ / P. Since h is a semi-homomorphism and a ∈ / h−1 (P ) , h−1 (P ) −1 is a proper filter of A. So, there exists Q ∈ X (A) such that h (P ) ⊆ Q and a ∈ / Q. Conversely, if there exists Q ∈ X (A) such that (P, Q) ∈ Rh and a ∈ / Q, then clearly h (a) ∈ / P. (3) By item (2) it follows that hRh (βA (a)) = βB (h(a)) ,

10

SERGIO A. CELANI AND LEONARDO M. CABRER

for every a ∈ A. Thus, hRh (βA (a)) ∈ DKA (X(A)). Also, from (1) we get \ Rh (P ) = {βA (a) : h (a) ∈ P } . Thus, Rh (P ) is a closed subset of X(B), for all P ∈ X(B). (4) It follows directly from the second part of assertion (1).  −1 (5) Let (P, Q) ∈ Rk◦h . Then (k ◦ h) (P ) ⊆ Q, i.e., h−1 k −1 (P ) ⊆ Q. Let us consider the set h (Qc ) = {h (q) : q ∈ / Q} . It is easy to see that (k (Qc )] = {d : d ≤ k (q) for some q ∈ / Q} is an ideal. We will prove that k −1 (P ) ∩ (h (Qc )] = ∅. If we suppose the contrary, there exist k (a) ∈ P and q ∈ / Q such that a ≤ h (q). Then, k (a) ≤ k (h (q)), because k is monotonic. So, k (h (q)) ∈ P , i.e., q ∈ h−1 k −1 (P ) ⊆ Q, which is a contradiction. Thus, there exists D ∈ X (B) such that k −1 (P ) ⊆ D and h−1 (D) ⊆ Q. Then (P, Q) ∈ Rh ◦ Rk . The proof of that Rh ◦ Rk ⊆ Rk◦h is easy and it is left to the reader. (6) It follows from (2).  Remark 26. Given a Tarski algebra A, if we consider IdA : A → A where IdA (a) = a for every a ∈ A, then clearly RIdA = {(P, P ) : P ∈ X (A)} = IX(A) . By Theorem 13, Proposition 25 and the previous Remark we can define a contravariant functor: X : ST −→ T R by X (A) = hX (A) , KA , τA i if A is a Tarski algebra and by X (h) = Rh if h is a semi-homomorphism between Tarski algebras. By Item (4) of Proposition 25 the restriction of X to HT , denoted by X |HT , is a contravariant functor from HT into T F. Now, to complete the duality, we will prove that there exist a contravariant functor from T R into ST . Theorem 27. Let X1 ,X2 and X3 be three T -spaces. Let R ∈ T R (X1 , X2 ) and T ∈ T R (X2 , X3 ). Then: (1) Consider the map hR : DK2 (X2 ) → DK1 (X1 ) (defined in Definition 19). Then hR ∈ ST (DK2 (X2 ) , DK1 (X1 )). (2) hR◦T = hR ◦ hT . (3) For any (x, y) ∈ X1 × X2 , (x, y) ∈ R if and only if (HX1 (x) , HX2 (y)) ∈ RhR . (4) RhR ◦ HX1 = HX2 ◦ R. (5) hR ∈ HT (DK2 (X2 ) , DK1 (X1 )) if and only if R ∈ T F (X1 , X2 ). Proof. (1) By definition of T -relation, hR (U ) ∈ DK1 (X1 ). So, hR is well-defined. We will prove that hR (U ⇒ V ) ⊆ hR (U ) ⇒ hR (V ) for any U, V ∈ DK2 (X2 ). Let x ∈ hR (U ⇒ V ). So c R (x) ⊆ U ⇒ V = U c ∪ V. If x ∈ / (hR (U )) , then R (x) ⊆ U. It follows that R (x) = R (x) ∩ U ⊆ V , and consequently x ∈ hR (V ) . Then we have that c

x ∈ (hR (U )) ∪ hR (V ) = hR (U ) ⇒ hR (V ) . It is clear that hR (X2 ) = X1 . Therefore hR is a semi-homomorphism. (2) It follows from the proof of Item (3) of Theorem 23. (3) Let (x, y) ∈ R. Suppose that (HX1 (x) , HX2 (y)) ∈ / RhR , i.e., −1 hR (HX1 (x)) * HX2 (y) .

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

11

Then there exists U ∈ DK1 (X1 ) such that x ∈ hR (U ) and y ∈ / U . It follows that R (x) ⊆ U , but as y ∈ R (x), y ∈ U , which is a contradiction. Suppose that (x, y) ∈ X1 × X2 is such that (x, y) ∈ / R. Since R (x) is closed and K2 is a basis for the topology τ2 , there exists U ∈ K2 such that R (x) ⊆ U c and y ∈ U . It follows that c U c ∈ h−1 / HX2 (y), i.e., (HX1 (x) , HX2 (y)) ∈ / RhR . R (HX1 (x)) and U ∈ (4) Let (x, y) ∈ RhR ◦ HX1 . There exist z ∈ X (DK1 (X1 )) such that (x, z) ∈ HX1 and (z, y) ∈ RhR . Since HX1 and HX2 are bijective functions (Theorem 16), HX1 (x) = z and there exists t ∈ X2 such that y = HX2 (t) . Thus (HX1 (x) , HX2 (t)) ∈ RhR and by the assertion (3), we have that (x, t) ∈ R. Therefore (x, y) ∈ HX2 ◦ R. The converse follows in a similar way. (5) First suppose that hR is a homomorphism of Tarski algebras. To prove that R is a partial map, we have to prove that if x ∈ X1 and y, z ∈ X2 such that (x, y) ∈ R and (x, z) ∈ R, then y = z. By (3), we have that (HX1 (x) , HX2 (y)) ∈ RhR and (HX1 (x) , HX2 (z)) ∈ RhR , i.e., −1 h−1 R (HX1 (x)) ⊆ HX2 (y) and hR (HX1 (x)) ⊆ HX2 (z) .

By Proposition 25, h−1 R (HX1 (x)) ∈ X (DK2 (X2 )) ∪ {DK2 (X2 )} . c If h−1 R (HX1 (x)) = DK2 (X2 ), then for every U ∈ K2 we have that hR (U ) ∈ HX1 (x). Thus x ∈ hR (U ), i.e., R (x) ⊆ U . Since ∅ ∈ K2 , R (x) = ∅, which is a contradiction with the fact that −1 y, z ∈ R (x). Then we have that h−1 R (HX1 (x)) ∈ X (DK2 (X2 )). As hR (HX1 (x)) is maximal,

h−1 R (HX1 (x)) = HX2 (y) = HX2 (z) . By the injectivity of HX2 we conclude that y = z. For the converse suppose that R ∈ T F (X1 , X2 ). Then hR (U ) = {x ∈ X : R (x) ∈ U } = R−1 (U ) . We already have that hR is a semi-homomorphism. Suppose that there exist U, V ∈ DK2 (X2 ) such c that hR (U hR (V ) * hR (U ⇒ V ) . Then there exists x ∈ X1 such that x ∈ hR (U ) ∪ hR (V ) = c) ⇒ −1 −1 −1 c R (U ) ∪R (V ), and x ∈ / R (U ∪ V ) . So R (x) ∈ U and R (x) ∈ / V , and thus x ∈ / R−1 (V ).  c So x ∈ R−1 (U ) , which is a contradiction. We conclude that hR is an homomorphism of Tarski algebras.  By Theorems 8 and 27 we can define a contravariant functor: D : T R −→ ST by D (hX, K, τ i) = DK (X) if hX, K, τ i is a Tarski space and by D (R) = hR if R is a T -relation. By Item (2) of Theorem 27, the restriction of D to T F, denoted by D |T F , is a contravariant functor from T F into HT . By Theorem 16, item (4) of Theorem 27, and Lemma 24 we have that HX (defined for every Tarski space X in Lemma 15) is a natural equivalence between the identity functor of T R and X ◦ D. By Theorems 13 and item (6) of Proposition 25, we have that βA is a natural equivalence between the identity functor of ST and the composition functor D ◦ X. We can summarize these results in the following Theorem and its Corollary. Theorem 28. The contravariant functors X and D define a dual equivalence between the category of Tarski algebras with semi-homomorphisms and the category of Tarski spaces with T -relations. Corollary 29. The contravariant functors X |T F and D |HT define a dual equivalence between the algebraic category of Tarski algebras with homomorphisms and the category of Tarski spaces with T -partial functions.

12

SERGIO A. CELANI AND LEONARDO M. CABRER

5. Subalgebras In section 3 we have proved that every Tarski algebra is isomorphic to a subalgebra of a Tarski algebra of subsets of a topological space. This subalgebra is obtained by means of a particular basis for the topology of the dual space of a Tarski algebra. This idea will allow us to determine the lattice of subalgebras of a Tarski algebra by means of subsets of this particular basis with an additional property. We will call the lattice of subalgebras of a Tarski algebra A, Sub (A). It is known that in the case of Boolean algebras the lattice of subalgebras of an algebra B is isomorphic to the lattice of certain equivalence relations of its dual Stone space (see [12]). Given a subalgebra C of B the associated relation RC in X (B) is defined by: (R)

(P, Q) ∈ RC iff P ∩ C = Q ∩ C.

In the following example we will see that this representation of subalgebra by means of binary relations cannot be applied to the variety of Tarski algebras. Example 30. Consider the Boolean algebra with four elements B = {0, a, b, 1} 1 s @ @ @s a

b s @ @ @s 0 Fig.1 Now consider B = hB, →, 1i the Tarski algebra where the implication is defined by x → y = ¬x ∨ y. It is easy to see that X (B) = {{a, 1} , {b, 1}}. Now consider the Tarski subalgebras of B whose universes are B1 = B and B2 = {a, b, 1} . It follows that for every P, Q ∈ X (B), P ∩ Bi = Q ∩ Bi if and only if P = Q, for each i ∈ {1, 2}. Thus RB 1 = RB 2 . Definition 31. Let hX, K, τ i be a T-space . A subset ∅ 6= L ⊆ K will be called a T -basic set if the following condition is held: (TB)

for any U, V ∈ L, U ∩ V c ∈ L.

Given a T -space X = hX, K, τ i we will denote T B (X ) = {L ⊆ K : L is a T -basic set} . Proposition 32. Let X be a T -space and let L ∈ T B (X ). Then the following propositions hold: (1) L ∪ {X} is a basis for a topology in X. (2) DL (X) = {U c : U ∈ L} is a universe of a subalgebra of DK (X). (3) hT B (X) , ⊆i is lattice. Proof. (1) Since L is non empty, let U ∈ L. By condition (TB), U ∩U c = ∅ ∈ L. Now let U, V ∈ L, c by condition (TB) we have that U ∩ V c ∈ L. Thus U ∩ (U ∩ V c ) = U ∩ (U c ∪ V ) = U ∩ V ∈ L. Therefore the result follows. (2) By Item (1). ∅ ∈ L, then X ∈ DL (X). If W, Z ∈ DL (X), then c

W ⇒ Z = W c ∪ Z = (W ∩ Z c ) . As W c , Z c ∈ L, by condition (TB) W ∩ Z c ∈ L. Thus W ⇒ Z ∈ DL (X) and the result follows. (3) It follows from the fact that arbitrary intersections of T -basic sets are T -basic sets and because K is the greatest T -basic set of hX, K, τ i . 

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

13

We will denote DL (X) to the subalgebra hDL (X) , ⇒, Xi of DK (X). Given a Tarski algebra A, we define c

T (B) = {(βA (b)) : b ∈ B} for each B ∈ Sub (A) . Proposition 33. Let A be a Tarski algebra. Then the map T : Sub (A) −→ T B (X (A)) is an order preserving function. Proof. Let B ∈ Sub (A). Clearly T (B) ⊆ KA . If U, V ∈ T (B) there exist a, b ∈ B such that c

U = (βA (a)) V = (βA (b)) c

c c

c

c

Then U ∩ V c = (βA (a)) ∩ βA (b) = (βA (b) ∪ βA (a)) = (βA (b) ⇒ βA (a)) . c c Since βA is a homomorphism of Tarski Algebras, (βA (b) ⇒ βA (a)) = (βA (b → a)) ∈ T (B). Thus T (B) is a T -basic set of X (A) . The proof that T is an order preserving function is easy and left to the reader.  Let A be a Tarski algebra and let L ∈ T B (X (A)). We consider the set c

S (L) = {a ∈ A : (βA (a)) ∈ L} We will prove that S (L) ∈ Sub (A). Consider a, b ∈ S (L). Since L is a T -basic set, and c c c (βA (a)) , (βA (b)) ∈ L, (βA (b)) ∩ (βA (a)) ∈ L. But c c

c

(βA (b)) ∩ (βA (a)) = (βA (b) ∪ (βA (a)) ) = (βA (a) ⇒ βA (b))

c

c

= (βA (a → b)) . Thus a → b ∈ S (L). Proposition 34. Let A be a Tarski algebra. Then the lattice of subalgebras of A is isomorphic to the lattice of T -basic subsets of KA . c

Proof. Consider B ∈ Sub (A). Thus a ∈ S (T (B)) iff (βA (a)) ∈ T (B) iff there exists b ∈ B c c such that (βA (a)) = (βA (b)) iff a = b ∈ B. Thus S (T (B)) = B. Consider L ∈ T B ((X (A))) . Thus U ∈ T (S (L)) iff there exists a ∈ S (L) such that U = c (βA (a)) iff U ∈ L.  Now we are going see that the relation associated to a Boolean subalgebra of a Boolean algebra, given in (R), adapted to Tarski algebras, plays an important role in the characterization of the Tarski space of the subalgebra obtained by a T -basic set. S Given a T -space X = hX, K, τ i and a L ∈ T B (X ) , consider YL = L ⊆ X, and let τL be the L topology generated by L in YL . We define a binary relation ∼ over YL by: L

x ∼ y if and only if for every U ∈ L, x ∈ U iff y ∈ U . Proposition 35. Given a T -space hX, K, τ i and a L ∈ T B (X ). Let ρL : YL −→ (YL ) / L ∼ D E be the natural projection and let (YL ) / L , τL / L be the quotient topological space given by ρL and ∼ ∼ the topology τL .   If we define L0 ⊆ P Y / L by L0 = {ρL (U ) : U ∈ L} then: ∼ D E XL = (YL ) / L , L0 , τL / L is a T -space ∼

and it is isomorphic in T R to X (DL (X)).



14

SERGIO A. CELANI AND LEONARDO M. CABRER

D E Proof. First we have to prove that XL = (YL ) / L , L0 , τL / L is a Tarski space. ∼



We will prove that L0 is a basis for τL / L Clearly if x ∈ Y, there exists U ∈ L such that ∼. D E S x ∈ U , then ρL (x) ∈ ρL (U ) ∈ L0 . Thus Y / L = L0 . If W is an open set in Y / L , τL / L , ∼ ∼ ∼ S −1 −1 (ρL ) (W ) ∈ τL Thus there exists H ⊆ L such that (ρL ) (W ) = U . Since ρL is an onto U ∈H S map, W = ρL (U ). We conclude that L0 is a basis for τL / L . ∼

U ∈H

Since every U ∈ L is compact, every ρL (U ) is compact. Thus L0 satisfies condition (1) of Definition 6. −1 −1 Note that if U ∈ L, then U = (ρL ) (ρL (U )) . Clearly U ⊆ (ρL ) (ρL (U )) . For the converse L −1 let x ∈ (ρL ) (ρL (U )). Then there exists y ∈ U such that ρL (y) = ρL (x) , i.e., x ∼ y. Thus −1 x ∈ U . Therefore Z ∈ L0 if and only if (ρL ) (Z) ∈ L. −1 −1 To prove that L0 satisfies condition (2)  of Definition  6, consider W, Z ∈ L’. Then (ρL ) (W ),(ρL ) (Z) ∈ L. By condition (TB), (ρL )

−1

−1

c

(W ) ∩ (ρL ) (Z) ∈ L. Since ρL is an onto map we have that  c −1 −1 −1 (ρL ) (W ∩ Z c ) = (ρL ) (W ) ∩ (ρL ) (Z) . Thus W ∩ Z c ∈ L0 . L

Condition (3) of Definition 6 follows directly from the definition of ∼. Condition (4) follows from the fact that L ⊆ K. Now we will prove that XL is isomorphic to X (DL (X)) in the category T R. In order to do so, let us consider f : Y / L −→ X (DL (X)) ∼

L

defined by f (ρL (x)) = {V ∈ DL (X) : x ∈ V }. By definition of ∼, if ρL (x) = ρL (y) clearly f (ρL (x)) = f (ρL (y)). The fact that f (ρL (x)) ∈ X (DL (X)) follows in the same way that Lemma 15 and the proof that f is an homeomorphism is analogous c to the proof of Theorem 16. If U ∈ KDL (X) , there exists V ∈ L such that U = βDL (X) (V c ) . We will prove that f −1 (U ) = ρL (V ) ∈ L0 . ρL (x) ∈ f −1 (U ) iff f (ρL (x)) ∈ U iff f (ρL (x)) ∈ / βDL (X) (V c ) iff V c ∈ / f (ρL (x)) iff x ∈ V . −1 0 Thus f (U ) = ρL (V ) ∈ L . Clearly, if f −1 (U ) ∈ L0 then U ∈ KDL (X) . Thus by Lemma 24 XL is isomorphic in T R to X (DL (X)).  6. Conclusions and future works In this paper we have developed a topological duality for two categories based on Tarski algebras with semi-homomorphisms and homomorphisms as arrows. To achieved that duality we define the notion of Tarski space and T -relation. We have described the lattice of subalgebras of a Tarski algebras by means of T -basics sets. The reason why we analized the category of Tarski algebras as objects and semi-homomorphisms as arrows, is because we are interested in the study of modal Tarski algebras. A modal Tarski algebra can be defined as a pair hA, i where A is a Tarski algebra and  is a semi-homomorphism from A into itself (see [7]). So, in the same way that Tarski algebras correspond to the {→}subreducts of Boolean algebras, the modal Tarski algebras correspond to the {→, }-subreducts of modal algebras, and they are also the algebraic counterpart of the {→, }-fragment of classical modal propositional logic. As a consequence of these remarks we have that the topological duality developed in this article for the category of Tarski algebras with semi-homomorphisms can be applied to give a duality between Tarski modal algebras and Tarski spaces endowed with a T relation. This duality and some applications will appear in a future paper. On the other hand, following the ideas presented in this work, we think that we are going to be able to give a full duality for Hilbert algebras which extends the one given by the authors in [8] for finite Hilbert algebras.

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS

15

Acknowledgement. We thank one of the referees for his observations and suggestions which have contributed to improve this paper. References [1] M. Abad, J. P. D´ıaz Varela and A. Torrens, Topological representation for implication algebras, Algebra Univers. 52 (2004) 39–48. [2] J. C. Abbott, Implicational algebras, Bull. Math, R. S.. Roumaine, 11 (1967), pp. 3-23. [3] J. C. Abbott, Semi-Boolean algebras, Mat. Vesnik 19 (1967), pp. 177–198. [4] R. Balbes and Ph. Dwinger, Distributive Lattices, University of Missouri Press, (1974). [5] D. Busneag, A note on deductive systems of a Hilbert algebra. Kobe J. Math. 2 (1985), pp. 29-35. [6] S. A. Celani, A note on homomorphism of Hilbert algebras. Int. Journal of Math. and Mathematical Sc. Vol. 29, No. 1 (2002), pp. 55-61. [7] S. A. Celani, Modal Tarski algebras. Reports on Math. Logic, No. 39 (2005), pp.113-126. [8] S. A. Celani and L.M. Cabrer, Duality for finite Hilbert algebras. Discrete Mathematics, 305 (2005), pp. 74-99 [9] I. Chajda, P. Halaˇ s and J. Zedn´ık, Filters and annihilators in Implication algebras, Acta Univ. Palacki. Olomuc., Fac. rer. nat., Mathematica 37 (1998), pp.41-45. [10] A. Diego, Sur les alg` ebres de Hilbert. Coll´ ection de Logique Math` ematique, serie A, 21, Gouthier-Villars, Paris (1966). [11] P. R. Halmos, Algebraic Logic, Chelsea Pub. Co., New York, (1962). [12] S. Koppelberg, General theory of Boolean algebras, in: Handbook of Boolean Algebras vol 1. Edited by Donald Monk and Robert Bonnet, North-Holland (1989). [13] H. Rasiowa, An algebraic approach to non-classical logics, North-Holland, Studies in Logic and the Foundations of Mathematics, Vol. 78, Amsterdam (1974). [14] G. Sambin and V. Vaccaro, Topology and duality in modal logic. Annals of Pure and Applied Logic 37 (1988), pp. 249-296. [15] F. B. Wright, Some remarks on Boolean duality. Portugaliae Math. 16 (1957), pp. 109-117. E-mail address: [email protected] E-mail address: [email protected] ´ ticas, Facultad de Ciencias Exactas, Univ. Nac. del Centro, CONICET and Departamento de Matema Pinto 399, 7000 Tandil, Argentina

TOPOLOGICAL DUALITY FOR TARSKI ALGEBRAS 1 ...

In Section 4 we will define two categories whose objects are Tarski algebras. One of ...... It follows that for every P, Q ∈ X (B), P ∩ Bi = Q ∩ Bi if and only if P = Q, ...

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