International Mathematical Forum, Vol. 6, 2011, no. 19, 899 - 908

Relations and Upper Sets on Partially Ordered Sets

Yong Chan Kim

Department of Mathematics, Gangneung-Wonju National University Gangneung, Gangwondo 210-702, Korea [email protected]

Young Sun Kim

Department of Applied Mathematics, Pai Chai University Dae Jeon, 302-735, Korea [email protected]

Abstract

In this paper, we investigate the properties of various relations, op- erators and upper sets on partially ordered sets.

Mathematics Subject Classification: 03E72, 54A40, 54B10

Keywords: Relations, open (closed, upper) sets, isotone (antitone) maps

1 Introduction

Wille [7] introduced the formal concept lattices by allowing some uncertainty in data as examples as Galois, dual Galois, residuated and dual residuated connections. Formal concept analysis is an important mathematical tool for data analysis and knowledge processing [1-7]. Orlowska and Rewitzky [6] in- vestigated the algebraic structures of operators of Galois-style connections. In this paper, we investigate the properties of various relations, operators and upper sets on partially ordered sets. In particular, we study the relations and operation related to upper sets. Upper sets induce the closed and open sets. We investigate their functorial relations. Let X be a set. A pair (X, eX ) is called a partially order set (simply, poset)if eX ⊂ X × X is reflexive, transitive and anti-symmetric. If (X, eX )is −1 −1 a poset and we define a function (x, y) ∈ eX iff (y, x) ∈ eX , then (X, eX )is a poset. Let (X, eX ) and (Y,eY ) be posets and f : X → Y is an isotone map if (f(x1),f(x2)) ∈ eY for all (x1,x2) ∈ eX . Moreover, f is an antitone map if (f(x2),f(x1)) ∈ eY for all (x1,x2) ∈ eX . We can define a poset (P (X),eP (X)) 900 Yong Chan Kim and Young Sun Kim

where eP (X) ⊂ P (X) × P (X)as(A, B) ∈ eP (X) iff A ⊂ B for A, B ∈ P (X). V : P (X) → P (Y ) is an isotone map if (V (A),V(B)) ∈ eP (Y ) for all (A, B) ∈ −1 eP (X) and W : P (X) → P (Y ) is an antitone map if (W (A),W(B)) ∈ eP (Y ) for all (A, B) ∈ eP (X).

2 Relations and upper sets on partially ordered sets Definition 2.1 [6] Let R ⊂ X × Y be a relation. For each B ∈ P (Y ), we define operations [R], [[R]], R, R, [R]c, Rc : P (Y ) → P (X) as follows: [R](B)={x ∈ X | (∀y ∈ Y )((x, y) ∈ R → y ∈ B)}, [[R]](B)={x ∈ X | (∀y ∈ Y )(y ∈ B → (x, y) ∈ R)} R(B)={x ∈ X | (∃y ∈ Y )((x, y) ∈ R & y ∈ B)} R(B)={x ∈ X | (∃y ∈ Y )((x, y) ∈ Rc & y ∈ Bc)}. [R]c(B)={x ∈ X | (∀y ∈ Y )((x, y) ∈ R → y ∈ Bc)} Rc(B)={x ∈ X | (∃y ∈ Y )((x, y) ∈ R & y ∈ Bc)}. For each R−1 = {(y, x) ∈ Y ×X | (x, y) ∈ R}, we similarly define operations [R−1], [[R−1]], R−1, R−1, [R−1]c, R−1c : P (X) → P (Y ).

Definition 2.2 Let (X, eX ) be a poset. A set A ∈ P (X) is called an eX - upper set if (x ∈ A &(x, y) ∈ eX ) → y ∈ A for x, y ∈ X.

Lemma 2.3 Let (X, eX ) and (Y,eY ) be posets. For R ⊂ X × Y , A ∈ P (X),B ∈ P (Y ), we have the following properties: −1 −1 −1 −1 −1 −1 (1) (eX ◦ R) = R ◦ eX and (R ◦ eY ) = eY ◦ R . −1 c c (2) eX ◦ R ⊂ R iff eX ◦ R ⊂ R . −1 c c (3) R ◦ eY ⊂ R iff R ◦ eY ⊂ R . (4) R(B)=([R]c(B))c, R(B)=Rcc(B) = ([[R]](Bc))c , [[R]](B)= (Rcc(B))c =[Rc]c(B), (5) [R], R, [R−1], R−1 are isotone maps. (6) [[R]], R, [R]c, Rc are antitone maps. (7) A ⊂ [R](R−1(A)) and R−1([R](B)) ⊂ B. (8) R([R−1](A)) ⊂ A, B ⊂ [R−1](R(B)), R◦[R−1] ◦R = R and [R−1] ◦R◦[R−1]=[R−1]. (9) A ⊂ [[R]]([[R−1]](A)), B ⊂ [[R−1]]([[R]](B)), [[R]]◦[[R−1]]◦[[R]] = [[R]] and [[R−1]] ◦ [[R]] ◦ [[R−1]] = [[R−1]]. (10) A ⊂ [R]c([R−1]c(A)), B ⊂ [R−1]c([R]c(B)), [R]c[R−1]c[R]c =[R−1]c and [R−1]c[R]c[R−1]c =[R−1]c. (11) R(R−1(A)) ⊂ A, R−1(R(B)) ⊂ B, RR−1R = R and R−1RR−1 = R−1. Relations and upper sets on partially ordered sets 901

Proof. (1),(4),(5) and (6) are easily proved from their definitions. (2) It follows from

(x, z) ∈ eX &(z,y) ∈ R → (x, y) ∈ R x, y ∈ R → x, z ∈ e z,y ∈ R →⊥ ( ) (( ) X&( ) ) c c (x, y) ∈ R &(x, z) ∈ eX → (z,y) ∈ R

−1 (3) (w, y) ∈ R &(y, x) ∈ eX → (w, x) ∈ R iff (w, y) ∈ R &(x, y) ∈ c c eX → (w, x) ∈ R iff (w, x) ∈ R → (x, y) ∈ eX → (w, y) ∈ R iff (x, y) ∈ c c eX &(w, x) ∈ R → (w, y) ∈ R . −1 −1 (7) Let x ∈ [R](R (A)). Then (∃y ∈ Y )(y ∈R (A)&(x, y) ∈ R) ∃y ∈ Y ∀z ∈ X z ∈ A ∨ y, z ∈ R−1 x, y ∈ R implies ( ) ( )( ( ) )&( ) implies (∃y ∈ Y ) (x ∈ A ∨ (y, x) ∈ R−1)&(x, y) ∈ R implies x ∈ A. Hence A ⊂ [R](R−1(A)). Let y ∈R−1([R](B)). There exists w ∈ X such that (y, w) ∈ R−1 & w ∈ [R](B). Thus, (y, w) ∈ R−1 &((w, y) ∈ R → y ∈ B). So, y ∈ B. Hence R−1([R](B)) ⊂ B. (8) By (5)and (6), we prove it as a similar method in (7). −1 −1 (9) Let x ∈ [[R]][[R ]](A)). Then (∃y ∈ Y )(y ∈ [[R ]](A)&(x, y) ∈ R) ∃y ∈ Y ∀z ∈ X z ∈ A → y, z ∈ R−1 x, y ∈ R implies ( ) ( )( ( ) )&( ) implies (∃y ∈ Y ) (x ∈ A → (y, x) ∈ R−1)&(x, y) ∈ R implies x ∈ A. Thus, A ⊂ [[R]][[R−1]](A)). Similarly, B ⊂ [[R−1]]([[R]](B)). Other cases follows from (5) and (6). (10) From (4), (7) and (8), [R]c([R−1]c(A)) = [R]c(R−1(A))c =[R](R−1(A)) ⊃ A and [R−1]c([R]c(B)) = [R−1]c(R(B))c =[R−1](R(B)) ⊃ B. Other cases follows from (5) and (6). (11) Let x ∈R(R−1(A)). Then (∃y ∈ Y )((x, y) ∈ R & y ∈ R−1(A)) implies (∃y ∈ Y )((x, y) ∈ R &(∀z ∈ X)((y, z) ∈ R−1 ∨ z ∈ A)) implies (∃y ∈ Y )((x, y) ∈ R &((x, y) ∈ R ∨ x ∈ A)) implies x ∈ A. Thus, R(R−1(A)) ⊂ A. Other cases are similarly proved.

Theorem 2.4 Let (X, eX ) and (Y,eY ) be posets. Then the following prop- erties hold. c −1 (1) A is an eX -upper set iff A is an eX -upper set. −1 (2) If eX ◦ R ⊂ R for R ⊂ (X × Y ) and put Ry = {x ∈ X | (x, y) ∈ R} −1 c −1 −1 for each y ∈ Y , then (R )y is an eX -upper set and Ry is an eX -upper set. c In particular, for B ∈ P (Y ), [R](B), [R] (B) and R(B) are eX -upper sets. c c c c Moreover, for B ∈ P (Y ), R (B), R  (B) and [[R ]](B) are eX -upper sets. −1 −1 (3) If eX ◦ R ⊂ R, then Ry is an eX -upper set. In particular, for B ∈ c c P (Y ), R(B), R (B) and [[R]](B) are eX -upper sets. Moreover, [R ](B), c c c [R ] (B) and R (B) are eX -upper sets. 902 Yong Chan Kim and Young Sun Kim

(4) If R ◦ eY ⊂ R and put Rx = {w ∈ Y | (x, w) ∈ R}, then Rx is an eY - upper set. In particular, for A ∈ P (X), R−1(A), R−1c(A) and [[R−1]](A) c −1 c −1 c are eY -upper sets. Moreover, for A ∈ P (X), [(R ) ](A), [(R ) ] (A) and c −1 (R ) (A) are eY -upper sets. −1 c −1 (5) If R◦eY ⊂ R, then Rx is an eY -upper set and Rx is an eY -upper set. In −1 −1 c −1 particular, for A ∈ P (X), [R ](A), [R ] (A) and R (A) are eY -upper sets. Moreover, for A ∈ P (X), (Rc)−1(A), (Rc)−1c(A) and [[(Rc)−1]](A) are eY -upper sets. Proof. (1) It follows from:

x ∈ A &(x, z) ∈ eX → z ∈ A z ∈ A → x ∈ A x, z ∈ e →⊥ iff ( &( ) X) z ∈ A → x, z ∈ e → x ∈ A iff ( ) X iff (z ∈ A &(x, z) ∈ eX ) → x ∈ A −1 (2) Since (x, y) ∈ R &(z,x) ∈ eX → (z,y) ∈ R then x ∈ Ry &(x, z) ∈ −1 −1 −1 c eX → (z,y) ∈ R. So, Ry is an eX -upper set. Furthermore, (z,y) ∈ R → c z,y ∈ R x, y ∈ R z,x ∈ e z,y ∈ Rc → z,x ∈ e → ( ) ( ) &( ) X implies ( ) ( ) X c −1 c −1 c −1 c (x, y) ∈ R . Then z ∈ (Ry )  (z,x) ∈ eX → x ∈ (Ry ) . So, (Ry ) is an eX -upper set.

x ∈ [R](B)&(x, z) ∈ eX &(z,y) ∈ R → x ∈ [R](B)&(x, y) ∈ R → ((x, y) ∈ R → y ∈ B)&(x, y) ∈ R → y ∈ B. x ∈ R B x, z ∈ e → x, y ∈ R → y ∈ B Thus [ ]( )&( ) X (( ) ). Hence x ∈ [R](B)&(x, z) ∈ eX → z ∈ [R](B);i.e. [R](B)isaneX -upper set.

x ∈R(B)&(x, z) ∈ eX c c ⇒ (∃y ∈ Y )((x, y) ∈ R & y ∈ B )&(x, z) ∈ eX ⇒ (∃y ∈ Y )((z,y) ∈ Rc & y ∈ Bc) ⇒ z ∈R(B). c c Since x ∈R (B)&(x, z) ∈ eX iff (∃y ∈ Y )((x, y) ∈ R & y ∈ −1 c c c B)&(x, z) ∈ eX ,byeX ◦R ⊂ R iff eX ◦R ⊂ R , then (z,y) ∈ R & y ∈ B;i.e. c c z ∈R (B). So, R (B)isaneX -upper set. Other case are similarly proved. −1 −1 (3) Since (x, z) ∈ eX &(z,y) ∈ R → (x, y) ∈ R, then Ry is an eX -upper set.

x ∈R(B)&(x, z) ∈ eX ⇒ (∃y ∈ Y )((x, y) ∈ R & y ∈ B)&(x, z) ∈ eX ⇒ (∃y ∈ Y )((z,y) ∈ R & y ∈ B) ⇒ z ∈R(B). Relations and upper sets on partially ordered sets 903

−1 c c Since eX ◦ R ⊂ R iff eX ◦ R ⊂ R , we have

c x ∈ [[R]](B)&(x, z) ∈ eX &(z,y) ∈ R c ⇒ (∃y ∈ Y )(y ∈ B → (x, y) ∈ R)&(x, z) ∈ eX &(z,y) ∈ R ⇒ (∃y ∈ Y )((x, y) ∈ Rc → y ∈ Bc)&(z,y) ∈ Rc) ⇒ y ∈ Bc.

c c Thus, (x ∈ [[R]](B)&(x, z) ∈ eX ) → ((z,y) ∈ R → y ∈ B . Hence, [[R]](B)isaneX -upper set. Other case are similarly proved. (4) Since (x, y) ∈ R &(y, w) ∈ eY → (x, w) ∈ R, Rx is an eY -upper set.

−1 y ∈R (A)&(y, w) ∈ eY −1 ⇒ (∃x ∈ X)((y, x) ∈ R & x ∈ A)&(y, w) ∈ eY ⇒ (∃x ∈ X)((w, x) ∈ R−1 & x ∈ A) ⇒ w ∈R(A).

Other cases are similarly proved as in (3). −1 −1 (5) Since (x, y) ∈ R &(y, w) ∈ eY → (x, w) ∈ R, Rx is an eY -upper set.

−1 y ∈ [R ](A)&(y, w) ∈ eY &(x, w) ∈ R ⇒ y ∈ [R−1](A)&(x, y) ∈ R ⇒ ((x, y) ∈ R → x ∈ A)&(x, y) ∈ R ⇒ x ∈ A. y ∈ R−1 A y, w ∈ e → x, w ∈ R → x ∈ A Thus [ ]( )&( ) Y (( ) ). −1 −1 Hence y ∈ [R ](A)&(y, w) ∈ eY → w ∈ [R ](A). Other cases are similarly proved as in (2).

In the above theorem, put R = eX . Since eX ◦ eX ⊂ eX , we obtain the following corollary.

Corollary 2.5 Let (X, eX ) be a poset. Then the following properties hold. −1 −1 c (1) Put (eX )z = {x ∈ X | (x, z) ∈ eX }, then (eX )z is an eX -upper set and −1 −1 c (eX )z is eX -upper set. For A ∈ P (X), [eX ](A), [eX ] (A) and eX (A) are c c c c eX -upper sets. Moreover, eX (A), eX  (A) and [[eX ]](A) are eX -upper set. (2) (eX )x = {z ∈ X | (x, z) ∈ eX } is an eX -upper set. For A ∈ P (X), −1 −1 c −1 −1 c eX (A), eX  (A) and [[eX ]](A) are eX -upper sets. Moreover, [(eX ) ](A), −1 c c −1 c [(eX ) ] (A) and (eX ) (A) are eX -upper sets.

Theorem 2.6 Let (X, eX ) and (Y,eY ) be posets. For A ∈ P (X), we define operations C, I as follows:

I(X)={A ∈ P (X) | [eX ](A)=A}

C(X)={A ∈ P (X) |eX (A)=A} 904 Yong Chan Kim and Young Sun Kim

(A, B) ∈ eI(X) iff A ⊂ B ∀A, B ∈ I(X). Then we have the following properties: A e e A A e−1 Ac Ac e−1 A A (1) is an X -upper set iff [ X ]( )= iff [ X ]( )= iff X ( )= . (2) If Ai is an eX -upper set for all i ∈ Γ, then i∈Γ Ai and i∈Γ Ai are eX -upper sets. (3) If A ∈ I(A), then x ∈ A iff ((eX )x,A) ∈ eI(X). (4) [eX ]([eX ](A)) = [eX ](A) and eX (eX (A)) = eX (A). (5) (I(X), ∪, ∩, ∅,X) is a complete lattice. C X , ∪, ∩, ∅,X (6) ( ( ) ) is a complete lattice. (7) [eX ](A)= i{Ai | Ai ⊂ A, Ai : eX − upper set}. −1 (8) eX (A)= i{Ai | A ⊂ Ai,Ai : eX − upper set}. −1 ∗ c c c c (9) If eX ◦R ⊂ R , then (Ry ) , [R](B), [R] (B), R(B), R (B), R  (B), [[Rc]](B) ∈ I(X). −1 −1 c c c c (10) If eX ◦R ⊂ R, then Ry , R(B), R (B), [[R]](B), [R ](B), [R ] (B), Rc(B) ∈ I(X). −1 −1 c −1 c −1 c (11) If R◦eY ⊂ R, then Rx, R (A), [[R ]](A), [(R ) ](A), [(R ) ] (A), (Rc)−1(A) ∈ I(Y ). −1 c −1 −1 c −1 c −1 (12) If R ◦ eY ⊂ R, then Rx, [R ](A), [R ] (A), R (A), (R ) (A), (Rc)−1c(A), [[(Rc)−1]](A) ∈ I(Y ) −1 c −1 −1 (13) (eX )x, ((eX )x ) , [eX ](A), eX (A), eX (A),[[eX ]](A) ∈ I(X) for A ∈ P (X).

Proof. (1) Let (x ∈ A)&(x, y) ∈ eX → y ∈ A. Then (x ∈ A) → (x, y) ∈ eX → y ∈ A implies A ⊂ [eX ](A). Since eX is reflexive, [eX ](A) ⊂ A. Let [eX ](A)=A. Then (x ∈ A) → (x, y) ∈ eX → y ∈ A implies (x ∈ A)&(x, y) ∈ eX → y ∈ A. Hence A is an eX -upper set. Since (x ∈ A)&(x, y) ∈ eX → y ∈ A iff (y ∈ A) → ((x, y) ∈ eX → −1 c −1 x ∈ A)iff (y ∈ A)&(y, x) ∈ eX → x ∈ A iff A is an eX -upper set, thus −1 c c [eX ](A )=A . Moreover, we have: x ∈e−1 A ∃z ∈ X x, z ∈ e−1 z ∈ A →⊥ X ( )iff( )(( ) X & ) −1 iff (∀z ∈ X) ((x, z) ∈ eX → (z ∈ A →⊥) −1 c iff (∀z ∈ X)((x, z) ∈ eX → z ∈ A ) −1 c iff x ∈ [eX ](A ).

−1 c c −1 Hence [eX ](A )=A iff eX (A)=A. A e i ∈ (2) Let i be an X -upper set for each Γ. From (1), we only show that e−1 A A x ∈e−1 A z ∈ A x, z ∈ X ( i∈Γ i)= i∈Γ i from: X ( i∈Γ i)iff ( i∈Γ i)&( ) e−1 ∃j ∈ z ∈ A z,x ∈ e ∃j ∈ x ∈ A x ∈ A X iff ( Γ)( j &( ) X )iff( Γ)( j)iff i∈Γ i. Similarly, i∈Γ Ai is an eX -upper set. Relations and upper sets on partially ordered sets 905

(3) Let A ∈ I(Y ), by (1), x ∈ A &(x, z) ∈ eX → z ∈ A implies x ∈ A → ((x, z) ∈ eX → z ∈ A) and ((eX )x,A) ∈ eI(X) implies x ∈ A. Hence x ∈ A iff ((eX )x,A) ∈ eI(X). (4) [eX ](A)isaneX -upper set from Corollary 2.5 (1). Moreover, eX (eX (A)) = −1 −1 −1 −1 eX ([eX ](A)) = [eX ]([eX ](A)) = [eX ](A). (5) and (6) are easily proved from (1) and (2). (7) Since [eX ](A) ⊂ A and [eX ](A)isaneX -upper set, [eX ](A) ⊂ i{Ai | Ai ⊂ A, Ai : eX − upper set}. c c c Since [eX ](A)=(eX (A )) , let x ∈ [eX ](A). Then x ∈eX (A ). So, c c c there exists z ∈ X such that (x, z) ∈ eX & z ∈ A ⊂ (∪iAi) . Since Ai is an −1 c −1 c eX -upper set, ∩iAi is an eX -upper set. Hence x ∈ (∪iAi) . (8) By Theorem 2.4 (1), we have

c c eX (A)=([eX ](A )) c c = i{Ai | Ai ⊂ A ,Ai : eX − upper set} c c c c −1 = i{Ai | A ⊂ Ai ,Ai : eX − upper set}

(9)-(13) are easily proved from Theorem 2.4 and Corollary 2.5.

Definition 2.7 Let (X, eX ) be a poset. A set A ∈ I(X) (resp. A ∈ C(X)) is called an open set (resp. closed set).

Theorem 2.8 Let (X, eX ) and (Y,eY ) be posets. Then the following prop- erties: −1 (1) If f :(X, eX ) → (Y,eY ) is an isotone map, then f (B) is an eX -upper set for each eY -upper set B, (2) If f :(X, eX ) → (Y,eY ) is an isotone map, then f(eX (A)) ⊂eY (f(A)). −1 −1 (3) f(eX (A)) ⊂eY (f(A)) iff f ([eY ](B)) ⊂ [eX ](f (B)), for each A ∈ P (X),B ∈ P (Y ). (4) If R([R−1](D(X))) ⊂ D(X) and E(Y ) ⊂ [R−1](R(E(Y ))) for each D(X) ⊂ P (X),E(Y ) ⊂ P (Y ), then R(eP (Y )(E(Y ))) ⊂eP (X)(R(E(Y ))). (5) If R−1([R](E(Y ))) ⊂ E(Y ) and D(X) ⊂ [R](R−1(D(X))) for each −1 −1 D(X) ⊂ P (X),E(Y ) ⊂ P (Y ), then R (eP (X)(D(X))) ⊂eP (Y )(R (D(X))), [R]([eP (Y )](E(Y ))) ⊂ [eP (X)]([R](E(X))). (6) If V (W (E(Y ))) ⊂ E(Y ) and D(X) ⊂ W (V (D(X))) for each D(X) ⊂ P (X),E(Y ) ⊂ P (Y ) and (V,W) ∈{([[R−1]], [[R]]), ([R−1]c, [R]c)}, then −1 −1 V (eP (X)(D(X))) ⊂eP (Y )(V (D(X))), W ([eP (Y )](E(Y ))) ⊂ [eP (X)](W (E(X))). (7) If R−1(R(E(Y ))) ⊂ E(Y ) and D(X) ⊂R(R−1(D(X))) −1 for each D(X) ⊂ P (X),E(Y ) ⊂ P (Y ), then R (eP (X)−1 (D(X))) ⊂ −1 −1 eP (Y )(R (D(X))), R([eP (Y )](E(Y ))) ⊂ [eP (X)](R(E(X))). 906 Yong Chan Kim and Young Sun Kim

Proof (1) Let B be an eY -upper set. Since f is an isotone map, x ∈ −1 f (B)&(x, z) ∈ eX implies f(x) ∈ B &(f(x),f(z)) ∈ eY implies f(x) ∈ B. −1 Hence f (B)isaneX -upper set. (2)

eY (f(A)) −1 = i{Bi | f(A) ⊂ Bi,Bi : eY − upper set} −1 −1 = i{Bi | A ⊂ f (Bi),Bi : eY − upper set} −1 −1 −1 −1 ⊃ i{f(f (Bi)) | A ⊂ f (Bi),f (Bi):eX − upper set} −1 −1 −1 −1 ⊃ f( i{f (Bi) | A ⊂ f (Bi),f (Bi):eX − upper set} ⊃ f(eX (A))

−1 −1 −1 (3) (⇒) Put A = f (B). f(eX (f (B)) ⊂eY (f(f (B))) ⊂eY (B) −1 −1 −1 c c −1 c c implies eX (f (B)) ⊂ f (eY (B)). ([eX ](f (B ))) ⊂ (f ([eY ](B ))) −1 c −1 c implies [eX ](f (B )) ⊃ f ([eY ](B )). −1 −1 −1 c c −1 c c (⇐) Let [eX ](f (B)) ⊃ f ([eY ](B)). Then (eX (f (B ))) ⊃ (f (eY (B ))) −1 c −1 c −1 c c implies eX (f (B )) ⊂ f (eY (B )). Put f (B )=A. Then B ⊂ −1 c −1 f(f (B )) = f(A). So, eX (A) ⊂ f (eY (f(A))). Hence f(eX (A)) ⊂ eY (f(A)). (4-5) Since R, R−1 and [R], [R−1] are isotone maps, by the conditions, we easily prove as a similar method in (3). −1 −1 c −1 c (6-7) Since [[R ]], [R ] :(P (X),eP (X)) → (P (Y ),eP (Y ))[[R]], [R] : −1 −1 −1 (P (Y ),eP (Y )) → (P (X),eP (X)), R  :(P (X),eP (X)) → (P (Y ),eP (Y )) and −1 R :(P (Y ),eP (Y )) → (P (X),eP (X)) are isotone maps, by the conditions, we easily prove as a similar method in (3).

Example 2.9 Let (X = {a, b, c, d},eX) and (Y = {x, y, z},eY ) be a poset with

eX = {(a, a), (a, b), (a, c), (a, d), (b, b), (b, d), (c, c), (c, d), (d, d)}.

eY = {(x, x), (x, y), (x, z), (y, y), (z,y), (z,z)}. Put a relation R ⊂ X × Y as

R = {(a, x), (b, x), (b, y), (b, z), (c, x), (c, z), (d, x), (d, y), (d, z)}.

Then (eX )a = {a, b, c, d}, (eX )b = {b, d}, (eX )c = {c, d}, (eX)d = {d} are eX - upper sets. Put A = {a, b}. Then a ∈ A &(a, d) ∈ eX but d ∈ A. Hence −1 −1 −1 A is not an eX -upper set. Since eX ◦ R = R, Rx = {a, b, c, d},Ry = −1 −1 c c {b, d},Rz = {b, c, d} are eX -upper sets. Since R ◦ eY = R, Ra = {y, z},Rb = c c ∅,Rc = {y},Rd = ∅ are eY -upper sets. Since R ◦ eY ⊂ R, Ra = {x} is not an eY -upper sets because x ∈ Ra &(x, y) ∈ eY , but y ∈ Ra. Relations and upper sets on partially ordered sets 907

Furthermore, I(X)={∅,X,{d}, {b, d}, {c, d}, {b, c, d}} and I(Y )={∅,Y,{y}, {y, z}}. ⎧ ⎪ X if x ∈ B, ⎨⎪ c c {b, c, d} if B ∈{{y, z}, {z}}, R(B)=R (B )= ⎪ {b, d} if B = {y}, ⎩⎪ ∅ if B = ∅. ⎧ ⎨⎪ X if B ∈{Y,{y, z}} [Rc](B)=[[R]](Bc)= {b, c, d} if B ∈{{y}, {x, y}} ⎩⎪ {b, d} otherwise. ⎧ ⎨⎪ X if B ⊂{x} [[R]](B)=[Rc]c(B)= {b, c, d} if B ∈{{x, z}, {z}} ⎩⎪ {b, d} otherwise. ⎧ ⎪ Y if A = X, ⎨⎪ −1 {y, z} if A = {b, c, d}, [R ](A)= ⎪ {y} if A ∈{{b, d}, {a, b, d}} ⎩⎪ ∅ otherwise ⎧ ⎨⎪ X if B = Y, R−1(A)= {y, z} if A ∈{{b, c, d}, {c, d}, {b, c}, {b, d}{d}, {b}, ∅} ⎩⎪ ∅ otherwise. ⎧ ⎨⎪ {y, z} if A = X, (Rc)−1(A)= {y} if A ∈{{b, c, d}, {c, d}} ⎩⎪ ∅ if A ∈{∅, {d}, {b, d}} Let E(Y )={{x, y}, {y}} be given. Then

E(Y ) ⊂ [R−1](R(E(Y ))) = {Y,{y}} eP (Y )(E(Y )) = {A | (A, B) ∈ eP (Y ),B ∈ E(Y )} = {∅, {x}, {y}, {x, y}} R(eP (Y )(E(Y ))) = {∅,X,{b, d}} R(E(Y )) = {X, {b, d}} eP (X)(R(E(Y ))) = P (X).

{∅,X,{b, d}} = R(eP (Y )(E(Y ))) ⊂eP (X)(R(E(Y ))) = P (X).

Let D1(X)={{b, d}, {c, d}},D2(X)={{a, b, d},Y} be given. Then

−1 R([R ](D1(X))) = {∅, {b, d}} ⊂ D1(X) [eP (X)](D2(X)) = {A | (A, B) ∈ eP (X) → B ∈ D(X)} = D2(X) −1 −1 [R ](D2(X)) = {Y,{y}} [eP (Y )]([R ](D2(X))) = {Y }.

−1 −1 {Y,{y}} =[R ]([eP (X)](D2(X))) ⊂ [eP (Y )]([R ](D(X))) = {Y }. 908 Yong Chan Kim and Young Sun Kim

References

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[2] G. Georgescu, A. Popescue, Non-dual fuzzy connections, Arch. Math. Log., 43 (2004), 1009-1039.

[3] J. J¨arvinen, M. Kondo, J. Kortelainen, Logics from Galois connections, Int. J. Approx. Reasoning, 49 (2008), 595-606.

[4] Y.C. Kim, J.W. Park, Join preserving maps and various concepts, Int.J. Contemp. Math. Sciences, 5 (5) (2010), 243-251.

[5] J.M. Ko, Y.C. Kim, Antitone Galois connections and formal concepts, Int. J. Fuzzy Logic and Intelligent Systems, 10(2) (2010), 107-112.

[6] Ewa. Orlowska, I. Rewitzky, Algebras for Galois-style connections and their discrete duality, Fuzzy Sets and Systems, 161 (2010), 1325-1342.

[7] R. Wille, Restructuring lattice theory; an approach based on hierarchies of concept, in: 1. Rival(Ed.), Ordered Sets, Reidel, Dordrecht, Boston, 1982.

Received: November, 2010