Completeness properties of transitive binary relational sets
aa r X i v : . [ m a t h . L O ] A p r COMPLETENESS PROPERTIES OF TRANSITIVE BINARY RELATIONALSETS
O. R. SayedDepartment of Mathematics, Faculty of Science, Assiut University, Assiut 71516, EGYPTo [email protected], o r [email protected]. H. SayedDepartment of Mathematics, Faculty of Science, New Valley University, [email protected] , nhs [email protected].
Abstract
The present paper is devoted to study some completeness properties of transitivebinary relational set, i.e., a set together with a transitive binary relation (so called t-set).
Abramsky and Jung [1] introduced a method to construct a canonical partially ordered setfrom a pre-ordered set and said: ”Many notions from the theory of order sets make sense evenif reflexivity fails”. Finally, they sum up these considerations with the slogan: ”Order theory isthe study of transitive relations”. Heckmann [3] introduced and studied the concepts of boundedcomplete poset, bounded complete domain, finitely complete poset, complete domain, finitarilycomplete poset, strongly compactly complete domain and compactly complete domain.This paper is organized as follows: In Section 2, some definitions and results concerningsome completeness properties of poset and domain were presented. In Section 3, boundedcomplete t-sets and bounded complete domain t-sets were introduced and studied. In Section4, finitely complete t-sets and complete domain t-sets were introduced and studied. In Section5, we extend the concept of finitary sets in transitive binary relational sets and then introducedand study the concept of finitarily complete t-sets. Finally, in Section 6, strongly compactlycomplete t-sets and compactly complete t-sets were introduced and studied.
For basic concepts of the poset, we refer to [5]. The concepts of directed subset, domain(directed complete poset), upper cone, upper closure etc., the reader is referred to [1, 3].
Definition 2.1
Let X be a nonempty set with a binary relation ” ” on X. The pair ( X, ) iscalled:(1) a partially ordered set (poset for short) [5] if ” ” is reflexive, antisymmetric and transi-tive; Corresponding author : O. R. Sayed.
Keywords and Phrases : Bounded complete poset; Bounded complete domain; Finitely complete poset; Fini-tarily complete poset; Strongly compactly complete domain. : 06A06, 06F30, 54E35, 54F05
2) a pre-ordered (quasi ordered) set [5] if ” ” is reflexive and transitive;(3) an equivalence set [5] if ” ” is reflexive, symmetric and transitive;(4) a continuous information system [4, 6] if ” ” is transitive and interpolative (if ∀ x, z ∈ X with x z there exists y ∈ X such that x y z ) ;(5) an abstract base [7] if ” ” is transitive and for every x ∈ X and every finite subset A ofX, if for every y ∈ A, y x , there exists z ∈ X such that y z x . Definition 2.2 (3) . Let ( X, ) be a domain. A subset A of X is called strongly compact iffor every O ∈ τ S with A ⊆ O there exists a finitary set F with A ⊆ F ⊆ O , where τ S is theScott-topology on X. Definition 3.1
A transitive binary relational set (t-set for short) is a pair ( X, ) , where X isa non-empty set and ” ” is a transitive binary relation on X. Example 3.1
Partially ordered sets, pre-ordered sets, equivalence sets, continuous informationsystem and abstract bases are t-sets.
Remark 3.1
One can deduce that any abstract base is a continuous information system (In-deed, let x ∈ X and { y } be a finite subset of X, where y ∈ X . If y x , then there exists z ∈ X such that y z x . Hence ” ” is interpolative.) but the converse need not be true as weillustrate by the following example: Example 3.2
Let X = { a, b, x } and = { ( a, a ) , ( b, b ) , ( a, x ) , ( b, x ) } . Then ” ” is transitiveand interpolative. Furthermore, if A = { a, b } , then ( X, ) is not an abstract base. Remark 3.2
Every pre-ordered set is an abstract base (Indeed, suppose that A is a finitesubset of a pre-ordered set X and such that for every x ∈ X and for every y ∈ A , y x . Then y x x .). The converse need not be true as we illustrate by the following example: Example 3.3
Let X = { a, b, c, d, e } and = { ( a, a ) } . Then ( X, ) is an abstract base. It isclear that ” ” is not reflexive. Hence ( X, ) is not a pre-ordered set. Definition 3.2
Let ( X, ) be a t-set and A ⊆ X .(1) The lower (resp. upper) bounded subset in X of A is denoted by lb ( A ) (resp. ub ( A ) ) anddefined as follows: lb ( A ) = { x ∈ X : ∀ y ∈ A, x y } (resp. ub ( A ) = { x ∈ X : ∀ y ∈ A, y x } ) .Each element in lb ( A ) (resp. ub ( A ) ) is called a lower (resp. an upper) bound of A.(2) The subset of least (resp. largest) elements of a subset A is denoted by le ( A ) (resp. la ( A ) )and defined as follows: le ( A ) = { x ∈ A : ∀ y ∈ A, x y } (resp. la ( A ) = { x ∈ A : ∀ y ∈ A, y x } ) . Each elementin le ( A ) (resp. la ( A )) is called a least (resp. a largest) element of A.
3) The infimum (resp. supremum) subset in X of A is denoted by inf( A ) (resp. sup( A ) ) anddefined as follows: inf( A ) = la ( lb ( A )) (resp. sup( A ) = le ( ub ( A )) ).Each element in inf( A ) (resp. sup( A ) ) is called an infimum (resp. a supremum) of A.(4) The lower (resp. upper) closure in X of A is denoted by ↓ ( A ) (resp. ↑ ( A ) ) and definedas follows: ↓ ( A ) = { x ∈ X : there exists y ∈ A, x y } (resp. ↑ ( A ) = { x ∈ X : there exists y ∈ A, y x } ) . Definition 3.3
Let ( X, ) be a t-set and A ⊆ X . Then A is called:(1) a directed subset if A = φ and for every distinct elements x, y in A, there exists z ∈ A ∩ ub ( { x , y } ) ;(2) an upper cone if there exists x ∈ A such that A = ↑ x . Definition 3.4
A t-set ( X, ) is called bounded complete if X is an upper cone and for every x, y ∈ X, ub ( { x, y } ) is empty or an upper cone. Theorem 3.1
For a t-set ( X, ) , the following statements are equivalent:(1) X is a bounded complete t-set;(2) le ( X ) = φ and for every x, y ∈ X with ub ( { x, y } ) = φ , sup( { x, y } ) = φ ;(3) If A is a finite bounded subset from above, then sup( A ) = φ ;(4) If A is a finite subset of X, then ub ( A ) is either φ or an upper cone. Proof. (1) ⇒ (2) : Since X is an upper cone, then there exists a ∈ X such that ↑ a = X .So, { a } ⊆ le ( X ). If ub ( { x, y } ) = φ , then ub ( { x, y } ) is an upper cone. Hence there exists b ∈ ub ( { x, y } ) such that ↑ b = ub ( { x, y } ). So, b ∈ le ( ub ( { x, y } )). Therefore sup( { x, y } ) = φ .(2) ⇒ (3) : Now φ is a finite bounded set from above since ub ( φ ) = X = φ . Since le ( X ) = le ( ub ( φ )) = φ , then sup( φ ) = φ . Let A be a non-empty finite bounded subset fromabove. If A = { z } and ub ( { z } ) = φ , then sup( A ) = φ . Suppose A = { x , x , x , ..., x n } and ub ( A ) = φ . Now A , = { x , x } and ub ( A , ) = φ , then sup( A , ) = φ . Take u , ∈ sup( A , )and consider A , , = { u , , x } . Then sup( A , , ) = φ because ub ( A , , ) = φ . We can proceeduntil consider the set B = { u , ,...,n − , x n } . Since ub ( B ) = φ , then sup( B ) = φ . Now, for every l ∈ sup( B ), l ∈ ub ( A ). Since m ∈ ub ( A ), one can deduce that l m . Then l ∈ sup( A ). So,sup( A ) = φ .(3) ⇒ (4) : Let A be a finite subset of X . If A is not bounded from above, then ub ( A ) = φ .Let A be finite bounded subset from above. Then sup( A ) = le ( ub ( A )) = φ . Then there exists x ∈ ub ( A ) such that ↑ x = ub ( A ).(4) ⇒ (1) : Now, X = ub ( φ ) and so X is an upper cone. Since for every x, y ∈ X , { x, y } isfinite. Then ub ( { x, y } ) = φ or ub ( { x, y } ) is an upper cone. Lemma 3.1
For a t-set ( X, ) , the following statements are equivalent:
1) If A is bounded subset from above, then sup( A ) = φ ;(2) If A is a non-empty subset of X, then inf( A ) = φ . Proof. (1) ⇒ (2): Let A be non-empty subset of X and B = lb ( A ). Now, ub ( B ) ⊇ A = φ .Then sup( B ) = φ . Let x ∈ sup( B ). Now, x ∈ ub ( B ). Then for every a ∈ A , x a . Then x ∈ la ( lb ( A )) = inf( A ). Hence inf( A ) = φ .(2) ⇒ (1): Suppose that A is a bounded subset of X from above and B = ub ( A ) = φ . Theninf( B ) = φ . Let x ∈ inf( B ). Since A ⊆ lb ( B ) and x ∈ inf( B ), then for every a ∈ A , a x .Thus x ∈ le ( ub ( A )) = sup( A ). Therefore sup( A ) = φ . Definition 3.5
A t-set ( X, ) is called a bounded complete domain if it is bounded completeand domain. Theorem 3.2
For a domain t-set ( X, ) , the following statements are equivalent:(1) X is a bounded complete t-set;(2) le ( X ) = φ and ∀ x, y ∈ X with ub ( { x, y } ) = φ , sup( { x, y } ) = φ ;(3) If A is a finite bounded subset from above, sup( A ) = φ ;(4) If A is a finite subset of X, ub ( A ) is either φ or an upper cone;(5) If A is bounded subset from above, sup( A ) = φ ;(6) If A is a non-empty subset of X, inf( A ) = φ . Proof.
From Theorem 3.1 and Lemma 3.1, it rests to prove that (3) and (5) are equivalent.(3) ⇒ (5): Let A be a bounded subset of X from above and D = { x : x is a fixed elementof sup( F ) for every finite subset F of A } . Since sup( φ ) = φ and for every y ∈ sup( F ∪ F ) , y ∈ ub (sup( F ) ∪ sup( F )), where F and F are finite subsets of A , then D is directed. Thussup( D ) = φ . Now, for every l ∈ sup( D ), l ∈ ub ( A ). Suppose that z ∈ ub ( A ). Then for all m ∈ A, m z so that z ∈ ub ( A ). Thus l z so that l ∈ sup( A ). Hence sup( D ) ⊆ sup( A ).Therefore sup( A ) = φ .(5) ⇒ (3): Obvious. Definition 4.1
A t-set ( X, ) is called finitely complete if X is an upper cone, and for all x, y ∈ X, ub ( { x, y } ) is an upper cone. One can easily deduce that any finitely complete t-set is a bounded complete t-set.
Theorem 4.1
For a t-set ( X, ) , the following statements are equivalent:(1) X is a finitely complete t-set;(2) X has a least element and for all x, y ∈ X , sup( { x, y } ) = φ ;(3) If A is a finite subset of X, then sup( A ) = φ ;
4) If A is a finite subset of X, then ub ( A ) is an upper cone. Proof. (1) ⇒ (2): Since X is an upper cone, then there exists a ∈ X such that ↑ a = X . So, a ∈ le ( X ). Suppose that x, y ∈ X . Then there exists z ∈ ub ( { x, y } ) such that ↑ z = ub ( { x, y } ).Therefore z ∈ sup( { x, y } ).(2) ⇒ (3): First, the empty set is finite. Since there exists x ∈ le ( X ), then there exists x ∈ sup( φ ). Suppose that A = { z } . Now, we have that sup( { z } ) = sup( { z, z } ) = φ . Let A = { x , x , x , ..., x n } , i.e. A is a finite set. Now, A , = { x , x } , then there exists u , ∈ sup( A , ). Put A , , = { u , , x } so that there exists u , , ∈ sup( A , , ). We can proceed untilconsider the set B = { u , ,...,n − , x n } so that there exists l ∈ sup( B ). Then l ∈ ub ( A ). Let m ∈ ub ( A ). One can deduce that l m . Therefore l ∈ sup( A ).(3) ⇒ (4) : Let A be a finite set. Then sup( A ) = φ . Thus, there exists l ∈ le ( ub ( A )) sothat ↑ l = ub ( A ). Therefore ub ( A ) is an upper cone.(4) ⇒ (1) : Since φ is finite and ub ( φ ) = X , then X is an upper cone. Since the set { x, y } is finite for every x, y ∈ X , then ub ( { x, y } ) is an upper cone. Definition 4.2 ( X, ) is called a complete domain t-set if it is finitely complete t-set anddomain t-set. Theorem 4.2
For a t-set ( X, ) , the following statements are equivalent:(1) X is a complete domain t-set;(2) X is a bounded complete domain t-set with la ( X ) = φ ;(3) If A is a subset of X, inf( A ) = φ ;(4) If A is a subset of X, sup( A ) = φ ;(5) If A is a finite subset of X or a directed subset of X, sup( A ) = φ . Proof. (1) ⇒ (2): Any complete domain t-set is bounded complete domain. Now, since forall x, y ∈ X, ub ( { x, y } ) is an upper cone, then ub ( { x, y } ) = φ . Hence X is directed. Therefore la ( X ) = sup( X ) = φ .(2) ⇒ (3): Let A be a subset of X . First, if A = φ , then X = lb ( φ ). Since le ( X ) = φ , thenthere exists l ∈ inf( φ ). Second, if A = φ , then from Theorem 3.2(6), inf( A ) = φ .(3) ⇒ (4): Let A be a subset of X . Since inf( φ ) = φ , then there exists l ∈ la ( X ) so thatevery subset of X is bounded from above. From Lemma 3.1, sup( A ) = φ ;(4) ⇒ (5): Obvious.(5) ⇒ (1): Since for every directed subset A of X , then sup( A ) = φ . Hence X is a domaint-set. From Theorem 4.1 (3), X is a finitely complete t-set. Definition 5.1
Let ( X, ) be a t-set. A subset A of X is called finitary if there exists a finitesubset F of A with A ⊆↑ ( F ) . Proposition 5.1
Let ( X, ) be a t-set and { A j : j ∈ { , , ..., n }} be a family of finitary subsetsof X. Then S nj =1 A j is a finitary subset. roof. Since for every j ∈ { , , ..., n } there exists a finite subset K j such that K j ⊆ A j ⊆↑ ( K j ) , then S nj =1 K j ⊆ S nj =1 A j ⊆ S nj =1 ↑ ( K j ) ⊆↑ ( S nj =1 K j ). Since S nj =1 K j is finite, then it isclear that S nj =1 A j is finitary. Definition 5.2
A t-set ( X, ) is called finitarily complete if X is finitary, ∀ x, y ∈ X, ub ( { x, y } ) is finitary. Theorem 5.1
Let ( X, ) be a t-set. Then the following statements are equivalent:(1) X is finitarily complete;(2) X is finitary and if A and B are finitary upper sets, then A ∩ B is finitary;(3) If A , ..., A n are finitary subsets of X, then T nj =1 A j is finitary;(4) If B is finite subset of X, then ub ( B ) is finitary. Proof. (1) ⇒ (2): If X is finitarily complete, then X is finitary. If A is finitary upper set,then there exists a finite set F ⊆ A such that A ⊆↑ ( F ) and ↑ ( A ) ⊆ A . Hence A = ↑ ( F )and if B is finitary upper set, then there exists a finite set F ⊆ B such that B ⊆↑ ( F ) and ↑ ( B ) ⊆ B . Hence B = ↑ ( F ). Thus A ∩ B = ↑ ( F ) ∩ ↑ ( F ) = ( S a ∈ F ( ↑ a )) T ( S b ∈ F ( ↑ b )) = S a ∈ F ,b ∈ F ( ↑ a ∩ ↑ b ). So, A ∩ B is a finite union of finitary sets. Therefore A ∩ B is finitary.(2) ⇒ (3): By indication. The empty intersection is X .(3) ⇒ (4): If B is finite, then ub ( B ) = S e ∈ B ( ↑ e ) upper cones are finitary. This is finiteintersection of finitary sets.(4) ⇒ (1): X is the set of upper bounds of φ , and ↑ x ∩ ↑ y is the set of upper bounds of { x, y } . Theorem 5.2 (1) Every finite pre-ordered set ( X, ) is finitarily complete;(2) Every bounded complete ( X, ) pre-ordered set is finitarily complete. Proof. (1) Since X is finite and ub ( { x, y } ) for each x, y ∈ X is finite also, then one can easily deducethat X is finitarily complete.(2) Since X is upper cone, then X = ↑ { x } for some x ∈ X . Hence X is finitary. Also, onecan deduce that ub ( { x, y } ) for each x, y ∈ X is a finitary subset of X because ub ( { x, y } )is empty or upper cone. Therefore, X is finitarily complete.The following two examples illustrate that the concepts of bounded completeness and finite-ness are independent notions for t-sets (moreover for posets). Example 5.1
Let X = { a, b, c, d } and = { ( a, a ) , ( b, b ) , ( c, c ) , ( d, d ) } . Then ( X, ) is finiteposet but not bounded complete. Example 5.2
Let N = { , , , ... } and be the usual partially ordered relation on N. Then ( N, ) is bounded complete but not finite. Strongly compactly complete t-sets
Definition 6.1
A triple ( X, , τ ) is called a topological t-set, where ( X, ) is a t-set and ( X, τ ) is a topological space. Definition 6.2
Let ( X, , τ ) be a topological t-set. A subset A of X is called strongly compactif for all O ∈ τ such that A ⊆ O , there exists a finitary subset F of X such that A ⊆ F ⊆ O . Theorem 6.1
Let ( X, , τ ) be a topological t-set such that each member of τ is an upper subset.If a subset A of X is strongly compact, then A is compact. Proof.
Let ℜ be an open cover of A , i.e. A ⊆ S B ∈ℜ B and ℜ ⊆ τ . Put S B ∈ℜ B = G .Then A ⊆ G ∈ τ . Since A is strongly compact, then there exists a finitary subset K of G such that A ⊆ K ⊆ G so that there exists a finite subset F of K such that K ⊆↑ ( F ).Then for every x ∈ F there exists B x ∈ ℜ such that x ∈ B x . So, F ⊆ S x ∈ F B x . Hence A ⊆ K ⊆↑ ( F ) ⊆↑ ( S x ∈ F B x ) = S x ∈ F B x . Therefore, A is compact. Corollary 6.1 (1) If A is strongly compact subset of X with respect to the topological t-set ( X, , τ Alx ) , then A is compact, where τ Alx is the Alexandroff topology induced by ” ” ;(2) If A is a strongly compact subset of X with respect to the topological t-set ( X, , τ s ∗ ) thenA is compact, where τ S ∗ is the Scott*-topology induced by ” ” . Theorem 6.2
Let ( X, ) be a t-set and { A j : j ∈ { , , ..., n }} be a family of finitary subsetsof X. Then S nj =1 A j is a finitary subset. Proof.
Since for every j ∈ { , , ..., n } there exists a finite subset K j such that K j ⊆ A j ⊆↑ ( K j ), then S nj =1 K j ⊆ S nj =1 A j ⊆ S nj =1 ↑ ( K j ) ⊆↑ ( S nj =1 K j ). Since S nj =1 K j is finite, then it isclear that S nj =1 A j is finitary. Theorem 6.3
Let ( X, , τ ) be a topological t-set and { A j : j ∈ { , , ...n }} be a family ofstrongly compact subsets. Then S nj =1 A j is strongly compact subset. Proof.
Suppose O ∈ τ such that S nj =1 A j ⊆ O . Then for all j ∈ J there exists a finitary subset B j such that A j ⊆ B j ⊆ O so that S nj =1 A j ⊆ S nj =1 B j ⊆ O . From Theorem 6.2, S nj =1 B j isfinitary. Hence S nj =1 A j is strongly compact. Definition 6.3
Let ( X, , τ ) be a topological t-set. X is called strongly compactly complete t-setif X is strongly compact and for every x, y ∈ X, ub ( { x, y } ) is strongly compact. Theorem 6.4
Let ( X, , τ ) be a topological t-set. Consider the following statements:(1) X is strongly compactly complete;(2) X is finitary and the intersection of two finitary upper sets is strongly compact. Then:(A) (1) ⇒ (2).(B) If ” ” is reflexive, then (2) ⇒ (1). roof. (A) Since X is strongly compact and open, then there exists a finitary set B of X such that X ⊆ B ⊆ X . So, X is finitary. Suppose A and B be two finitary upper sets. Then there arefinite sets E and F such that A ⊆↑ ( E ) ⊆↑ ( A ) ⊆ A and B ⊆↑ ( F ) ⊆↑ ( B ) ⊆ B . Hence, wehave A = ↑ ( E ) and B = ↑ ( F ). Now, A ∩ B = ↑ ( E ) ∩ ↑ ( F ) = S e ∈ E,f ∈ F ( ↑ e ∩ ↑ f ). Therefore,from Theorem 6.2, A ∩ B is strongly compact.(B) Since X is finitary and the only open set containing X is X itself, then X is stronglycompact. Let x, y ∈ X . Since ” ” is reflexive, then for every x ∈ X, ↑ { x } is finitary andsince ” ” is transitive, then ↑ x is upper set. Hence ↑ { x }∩ ↑ { y } is strongly compact. Definition 6.4
Let ( X, , τ ) be a topological t-set. X is called compactly complete if X iscompact and for all x, y ∈ X, ub ( { x, y } ) is compact. Theorem 6.5
Let ( X, , τ ) be a topological t-set. Consider the following statements:(1) X is compactly complete;(3) X is finitary and the intersection of two finitary upper sets is compact. Then:(A) If τ has the property F, then (1) ⇒ (2) ;(B) If ” ” is reflexive, and each member of τ is an upper set then (2) ⇒ (1) . Proof. (A) since τ has the property F , then X is finitary. Let U and V be two finitary uppersets. Then there are finite sets E and M such that U ⊆↑ ( E ) ⊆↑ ( U ) ⊆ U and V ⊆↑ ( M ) ⊆↑ ( V ) ⊆ V . So, U = ↑ ( E ) and V = ↑ ( M ). Now, U ∩ V = ↑ ( E ) ∩ ↑ ( M ) = S e ∈ E,m ∈ M ( ↑ e ∩ ↑ m ).Therefore U ∩ V is compact because a finite union of compact subsets is compact.(B) Since X is finitary, then X is strongly compact. From Theorem 6.1, X is compact. Since” ” is reflexive, then for all x ∈ X, x ∈↑ x . Thus, we have ↑ x is finitary. Furthermore, forall x ∈ X, ↑ x is an upper set. Therefore, ↑ x ∩ ↑ y is compact for all x, y ∈ X . Conclusion:
Theorems 3.1, 3.2, 4.1, 4.2, 5.1, 6.1, 6.2, 6.3, 6.4, Proposition 5.1, Lemma5.1 and Corollary 6.1 can be obtained if we replace the condition of t-set by a pre-ordered set(resp. abstract base, continuous information system)
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