THE SECOND ISOMORPHISM THEOREM ON ORDERED SET UNDER ANTIORDERS. Daniel A. Romano


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1 235 Kragujevac J. Math. 30 (2007) THE SECOND ISOMORPHISM THEOREM ON ORDERED SET UNDER ANTIORDERS Daniel A. Romano Department of Mathematics and Informatics, Banja Luka University, Mladena Stojanovića 2, Banja Luka, Bosnia and Herzegovina ( (Received October 09, 2006) Abstract. In this article we give two new characteristics of quasiantiorder relation on ordered set under antiorder. The new results in this article is cocalled the second isomorphism theorem on ordered sets under antiorders: Let (X, =,, α) be an ordered set under antiorder α, ρ and σ quasiantiorders on X such that σ ρ. Then the relation σ/ρ, defined by σ/ρ = {(x(ρ ρ 1 ), y(ρ ρ 1 ) X/(ρ ρ 1 ) X/(ρ ρ 1 ) : (x, y) σ}, is a quasiantiorder on X/(ρ ρ 1 ) and (X/(ρ ρ 1 ))/((σ/ρ) (σ/ρ) 1 ) = X/(σ σ 1 ) holds. Let A = {τ : τ is quasiantiorder on X such that τ σ}. Let B be the family of all quasiantiorder on X/q, where q = σ σ 1. We shall give connection between families A and B. For τ A, we define a relation ψ(τ) = {(aq, bq) X/q X/q : (a, b) τ}. The mapping ψ : A B is strongly extensional, injective and surjective mapping from A onto B and for τ, µ A we have τ µ if and only if ψ(τ) ψ(µ). Let (X, =, ) be a set in the sense of book [1], [2] and [6], where is a binary relation on X which satisfies the following properties: (x x), x y y x, x z x y y z, x y y = z x z
2 236 called apartness (A. Heyting). The apartness is tight (W. Ruitenburg) if (x y) x = y holds. Let Y be a subset of X and x X. The subset Y of X is strongly extensional in X if and only if y Y y x x Y ([3], [6]). If x X, we defined ([3]) x#y ( y Y )y x. Let f : (X, =, ) (Y, =, ) be a function. We say that it is: (a) f is strongly extensional if and only if ( a, b X)(f(a) f(b) a b); (b) f is an embedding if and only if ( a, b X)(a b f(a) f(b)). Let α X Y and β Y Z be relations. The filled product ([3]) of relations α and β is the relation β α = {(a, c) X Z : ( b Y )((a, b) α (b, c) β)}. A relation q X X is a coequality relation on X if and only if holds: q, q q 1, q q q. If q is a coequality relation on set (X, =, ), we can construct factorset (X/q, = q, q ) with aq = q bq (a, b)#q, aq q bq (a, b) q. M. Bozic and this author were first defining and studied notion of coequality relation in For more information on this relation readers can see in the paper [3]. A relation α on X is antiorder ([3], [4]) on X if and only if α, α α α, α α 1, α α 1 =. Antiorder relation on set with apartness the first defined in article [4]. Let f : (X, =,, α) (Y, =,, β) be a strongly extensional function of ordered sets under antiorders. f is called isotone if ( x, y S)((x, y) α (f(x), f(y)) β);
3 237 f is called reverse isotone if and only if ( x, y S)((f(x), f(y)) β (x, y) α). The strongly extensional mapping f is called an isomorphism if it is injective and embedding, onto, isotone and reverse isotone. X and Y called isomorphic, in symbol X = Y, if exists an isomorphism between them. As in [3] a relation τ X/ X is a quasiantiorder on X if and only if τ α( ), τ τ τ, τ τ 1 =. The first proposition gives some information on quasiantiorder: Lemma 1. If τ is a quasiantiorder on X, then the relation q = τ τ 1 is a coequality relation on X. Proof. (1) q = τ τ 1 1 = = (because 1 = ). (2) q = τ τ 1 = τ 1 τ = (τ τ 1 ) 1 = q 1. (3) q = τ τ 1 (τ τ) (τ 1 τ 1 ) (τ τ) (τ τ 1 ) (τ 1 τ) (τ 1 τ 1 ) = τ (τ τ 1 ) τ 1 (τ τ 1 ) = (τ τ 1 ) (τ τ 1 ) = q q. Now we give the second results. Let (X, =,, α, τ) be a ordered set under antiorder α and let τ be a quasiantiorder on X. Lemma 2. Let τ be a quasiantiorder relation on X, q = τ τ 1. Then the relation θ on X/q, defined by (aq, bq) θ (a, b) q, is an antiorder on X/q. The mapping π : X X/q strongly extensional surjective reverse isotone mapping and holds θ π = τ, in which case θ is equal to τ π 1. Proof. We shall check only linearity of the relation θ: Let aq and bq be arbitrary elements of set X/q such that aq bq. Then (a, b) q = τ τ 1. Thus (a, b) τ (b, a) τ. Therefore, we have (aq, bq) θ (bq, aq) θ. For the next proposition we need a lemma in which we will give explanation about two coequalities α and β on a set (X, =, ) such that β α.
4 238 Lemma 3. ([5]) Let α and β be coequality relations on a set X with apartness such that β α. Then the relation β/α on X/α, defined by β/α = {(xα, yα) X/q X/q : (x, y) β}, is an coequality relation on X/α and there exists the strongly extensional and embedding bijection f : (X/α)/(β/α) (X/β). So, at the end of this article, we are in position to give a description of connection between quasiantiorder ρ and quasiantiorder σ of set X with apartness such that σ ρ. Theorem 1. Let (X, =,, α) be a set, ρ and σ quasiantiorders on X such that σ ρ. Then the relation σ/ρ defined by σ/ρ = {(x(ρ ρ 1 ), y(ρ ρ 1 )) X/(ρ ρ 1 ) X/(ρ ρ 1 ) : (x, y) σ}, is a quasiantiorder on X/(ρ ρ 1 ) and (X/(ρ ρ 1 )/((σ/ρ) (σ/ρ) 1 )) = X/(σ σ 1 ) holds. Proof. (1) Put q = (ρ ρ 1 ), and construct the factorset (X/q, = q, q ). If a and b are elements of X, then (aq, bq) σ/ρ (a, b) σ (a, b) ρ (because σ ρ) (a, b) q aq q bq. (aq, cq) σ/ρ (a, c) σ ( b X)((a, b) σ (b, c) σ) ( bq X/q)((aq, bq) X/q (bq, cq) q). Suppose that (σ/ρ) (σ/ρ) 1, and let (aq, bq) (σ/ρ) (σ/ρ) 1. Then (aq, bq) σ/ρ (bq, aq) σ/ρ, and thus (a, b) σ (b, a) σ, i. e. (a, b)
5 239 σ σ 1. It is impossible. So, must be (σ/ρ) (σ/ρ) 1 =. Therefore, the relation σ/ρ is a quasiantiorder on set X/q. (2) From σ ρ it follows ρ ρ 1 σ σ 1. By lemma 1, there exist following coequality relations: q = ρ ρ 1 on the set (X, =, ), r = σ σ 1 on the set (X, =, ), r/q = (σ/ρ) (σ/ρ) 1 on factorset X/r, and, by Lemma 3, there exists the strongly extensional and embedding bijective function ϕ : (X/(ρ ρ 1 ))/((σ/ρ) (σ/ρ) 1 ) X/(σ σ 1 ). (3) By Lemma 2, on factorset X/r with r = σ σ 1 there exists the antiorder θ(σ) such that (a(σ σ 1 ), b(σ σ 1 )) θ(σ) (a, b) σ, and on factorset X/q with q = ρ ρ 1 there exists the antiorder θ(ρ) such that (a(ρ ρ 1 ), b(ρ ρ 1 )) θ(ρ) (a, b) ρ. Also, on the factorset X/((σ/ρ) (σ/ρ) 1 ) there exists the antiorder θ(σ/ρ) such that (a((σ/ρ) (σ/ρ) 1 ), b((σ/ρ) (σ/ρ) 1 )) θ(σ/ρ) (ar, br) θ(σ). It is easily to verify that the mapping ϕ : (X/(ρ ρ 1 ))/((σ/ρ) (σ/ρ) 1 ) X/(σ σ 1 ) is isotone and reverse isotone strongly extensional embedding bijection. At the end of this consideration, we shall give the following proposition:
6 240 Theorem 2. Let (X, =,, α, σ) be an ordered set under an antiorder α, σ a quasiantiorder on X such that σ σ 1 =. Let A = {τ : τ is quasiantiorder on X such that τ σ}. Let B be the family of all quasiantiorder on X/q, where q = σ σ 1. For τ A, we define a relation ψ(τ) = {(aq, bq) X/q X/q : (a, b) τ}. The mapping ψ : A B is strongly extenstional, injective and surjective mapping from A onto B and for τ, µ A we have τ µ if and only if ψ(τ) ψ(µ). Proof. (1) f is welldefined function: Let τ A. Then ψ(τ) is a quasiantiorder on X/q by Lemma 3. Let τ, µ A with τ = µ. If (aq, bq) ψ(τ) then (a, b) τ = µ, so (aq, bq) ψ(µ). Similarly, ψ(µ) ψ(τ). Therefore, ψ(µ) = q ψ(τ), i. e. the mapping ψ is a function. (2) ψ is an injection. Let τ, µ A, ψ(τ) = q ψ(µ). Let (a, b) τ. Since (aq, bq) ψ(τ) = q ψ(µ), we have (a, b) µ. Similarly, we conclude µ τ. So, µ = τ. (3) ψ is strongly extensional. Let τ, µ A, ψ(τ) q ψ(µ) i. e. let there exists an element (aq, bq) ψ(τ) and (aq, bq)#ψ(µ). Then, (a, b) τ. Let (x, y) be an arbitrary element of µ. Then, (xq, yq) ψ(µ) and (xq, yq) (aq, bq). It means xq q aq yq q bq, i. e. (x, a) q (y, b) q. Therefore, from x a y b we have (a, b) τ and (a, b) (x, y) µ. Thus we have τ µ. Similarly, from (aq, bq)#ψ(τ) and (aq, bq) ψ(µ) we conclude τ µ. (4) ψ is onto. Let M B. We define a relation µ on X as follows: µ = {(x, y) X X : (xq, yq) M}. µ is a quasiantiorder. In fact:
7 241 (I) Let (x, y) µ. Since (xq, yq) M q on X/q, we conclude that xq q yq, i. e. (x, y) q = σ σ 1. Hence (x, y) σ or (y, x) σ. Therefore, we have x y. Let (x, z) µ i. e. let (xq, zq) M. Then (xq, yq) M or (yq, zq) M for arbitrary yq X/q by cotransitivity of M. Thus (x, y) µ or (y, z) µ. Therefore, the relation µ is a quasiantiorder relation on X. Let M M 1 =. Suppose that (a, b) µ µ 1. Then (aq, bq) M M 1 =, which is impossible. So, µ µ 1 =. (II) ψ(µ) = M. Indeed: (xq, yq) ψ(µ) (x, y) µ (xq, yq) M. (III) µ σ. In the matter of fact, we have sequence (a, b) µ (aq, bq) ψ(µ) = M (π(a), π(b)) ψ(µ) = M (π : X X/q is strongly extensional function) (a, b) π 1 (ψ(µ)) = π 1 (M) (by π 1 (M) q = Coker(π) = σ σ 1 ) By Corollary the mapping π : X X/q is a strongly extensional reverse isotone surjective mapping. Therefore, we have π 1 (M) σ. (5) Let τ, µ A. We have τ µ if and only if ψ(τ) ψ(µ). Indeed: Let τ µ and (xq, yq) ψ(τ). Since (x, y) τ µ, we have (xq, yq) ψ(µ). Opposite, let ψ(τ) ψ(µ) and (x, y) τ. Since (xq, yq) ψ(τ) ψ(µ), we conclude that (x, y) µ. References [1] E. Bishop, Foundation of Constructive Analysis; McGrawHill, NY (1967). [2] R. Mines, F. Richman, W. Ruitenburg, A Course of Constructive Algebra, SpringerVerlag, New York (1988). [3] D. A. Romano, Coequality Relation, a Survey, Bull. Soc. Math. Banja Luka, 3 (1996), 1 36
8 242 [4] D. A. Romano,Semivaluation on Heyting Field, Kragujevac Journal of Mathematics, 20 (1998), [5] D. A. Romano, A Theorem on Subdirect Product of Semigroups with Apartness, Filomat, 14 (2000), 1 8 [6] A. S. Troelstra, D. van Dalen, Constructivism in Mathematics, An Introduction, Volume II, NorthHolland, Amsterdam (1988).
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