Relational representations of algebraic lattices and their applications

Shuzhen Luo; Xiaoquan Xu

doi:10.1515/math-2022-0010

Open Access Published by De Gruyter Open Access April 13, 2022

Relational representations of algebraic lattices and their applications

Shuzhen Luo and Xiaoquan Xu

From the journal Open Mathematics

https://doi.org/10.1515/math-2022-0010

Abstract

Keywords: strongly regular relation; finitely strongly regular relation; generalized finitely strongly regular relation; Priestley spaces

MSC 2010: 06B35; 54H10; 06A11

1 Introduction

The regularity of binary relations was first characterized by Zareckiǐ. In [1], he proved the following remarkable result: a binary relation ρ on a set X is regular if and only if the complete lattice ( Φ ρ ( X ) , ⊆ ) is completely distributive, where Φ ρ ( X ) = { ρ ( A ) : A ⊆ X } , ρ ( A ) = { y ∈ X : ∃ a ∈ A with ( a , y ) ∈ ρ } . Further criteria for regularity were given by Markowsky [2] and Schein [3] (see also [4] and [5]). Motivated by the fundamental works of Zareckiǐ on regular relations, in [6,7,8], Xu and Liu successfully generalized the regular relation to finitely regular relation and generalized finitely regular relation, got the relational representations of hypercontinuous lattices and quasi-hypercontinuous lattices, and proved that a binary relation ρ is finitely regular if and only if the complete lattice ( Φ ρ ( X ) , ⊆ ) is a hypercontinuous lattice and ρ is generalized finitely regular if and only if ( Φ ρ ( X ) , ⊆ ) is a quasi-hypercontinuous lattice if and only if the interval topology on ( Φ ρ ( X ) , ⊆ ) is T 2 . Moreover, the discussion about the relation representation theory of lattices has also attracted a considerable deal of attention (see [9,10, 11,12,13, 14,15,16, 17,18]).

As a common generalization of continuous lattices, algebraic lattices play a very important role in domain theory. The so-called strongly algebraic [19], hyperalgebraic [20,21], and quasi-hyperalgebraic lattices [21,22] are among the most successful such generalizations, which now have been familiar objects as algebraic lattices in domain theory. However, the relation representations of them are still lacking. The aim of this paper is to introduce and investigate some binary relations in order to establish the theory of relation representations of algebraic lattices.

In this paper, we define the concepts of strongly regular relation, finitely strongly regular relation, and generalized finitely strongly regular relation, and get the relational representations of strongly algebraic, hyperalgebraic, and quasi-hyperalgebraic lattices. Meanwhile, several equivalent characterizations of them are obtained. Let ( X , δ , ≤ ) be a partially ordered topological space. ( X , δ , ≤ ) is called totally order-disconnected if whenever x ≰ y there is a clopen upper set U such that x ∈ U and y ∉ U . A compact totally order-disconnected space is called a Priestley space. The interest in these spaces is mainly due to the celebrated Priestley duality [23,24]. We also discuss this Priestley property about complete lattices with respect to the interval topology. In this paper, the main results are as follows: (1) a binary relation ρ : X ⇀ Y is strongly regular if and only if the complete lattice ( Φ ρ ( X ) , ⊆ ) is a strongly algebraic lattice; (2) a binary relation ρ : X ⇀ Y is finitely strongly regular if and only if ( Φ ρ ( X ) , ⊆ ) is a hyperalgebraic lattice if and only if the finite extension of ρ is strongly regular; and (3) a binary relation ρ : X ⇀ Y is generalized finitely strongly regular if and only if ( Φ ρ ( X ) , ⊆ ) is a quasi-hyperalgebraic lattice if and only if ( Φ ρ ( X ) , ⊆ ) equipped with the interval topology is a Priestley space.

2 Preliminaries

In this section, we recall some basic concepts needed in this paper; for other nonexplicitly stated elementary notions please refer to [6,20].

Let P be a poset with a partial order ≤ . For all x , y ∈ P , A ⊆ P , let ↑ x = { y ∈ P : x ≤ y } and ↑ A = ⋃ a ∈ A ↑ a ; ↓ x and ↓ A are defined dually. A is called the upper set if ↑ A = A and the lower set defined dually. The least upper bound of A in P is written as ∨ A or sup A ; similarly, the greatest lower bound of A is written as ∧ A or inf A . Define an order ≤ o p by x ≤ o p y ⇔ y ≤ x , and write P o p as short for ( P , ≤ o p ) , called the order dual of P . A poset L is called the complete lattice if for any subset A ⊆ L , ∨ A and ∧ A exist in L . For two complete lattices L 1 and L 2 , the symbol L 1 ≅ L 2 means that L 1 is order isomorphic to L 2 .

For a poset P , x ∈ P , A ⊆ P . P \ A = { x ∈ P : x ∉ A } . The topology generated by the collection of sets P ⧹ ↓ x (as a subbase) is called the upper topology and is denoted by υ ( P ) ; the lower topology ω ( P ) on P is defined dually. The topology θ ( P ) = υ ( P ) ∨ ω ( P ) is called the interval topology on P , i.e., the interval topology is the coarsest common refinement of the upper topology υ ( P ) and the lower topology ω ( P ) . For any set X , let X ( < ω ) = { F ⊆ X : F is nonempty and finite } . Set denotes the class of all sets. The class of all complete lattices is denoted by Com .

Let ( X , δ ) be a topology space and A ⊆ X . The interior of A in ( X , δ ) is denoted by int δ A . A subset U of X is called δ -clopen if U is open and closed in ( X , δ ) . The notation ( X , δ , ≤ ) is used to denote a set X endowed with a topology δ and a partially order ≤ . The space ( X , δ , ≤ ) is called totally order-disconnected, if given x , y ∈ X with x ≰ y , there exists a δ -clopen upper set U such that x ∈ U and y ∉ U . A compact totally order-disconnected space is called a Priestley space.

For two sets X and Y , we call ρ : X ⇀ Y a binary relation if ρ ⊆ X × Y . When X = Y , ρ is usually called a binary relation on X . Let ℬ ( X ) denote the set of all binary relations between X and Y , and Rel = ⋃ X ∈ Set ℬ ( X ) .

Definition 2.1

[6] Let ρ : X ⇀ Y , τ : Y ⇀ Z be two binary relations. Define

τ ∘ ρ = { ( x , z ) : ∃ y ∈ Y , ( x , y ) ∈ ρ , ( y , z ) ∈ τ } . The relation τ ∘ ρ : X ⇀ Z is called the composition of ρ and τ .
ρ − 1 = { ( y , x ) ∈ Y × X : ( x , y ) ∈ ρ } .
ρ ( A ) = { y ∈ Y : ∃ x ∈ A with ( x , y ) ∈ ρ } . We call it the image of A under a binary relation ρ . Instead of ρ ( { x } ) , we write ρ ( x ) for short.
Φ ρ ( X ) = { ρ ( A ) : A ⊆ X } .

Clearly, ( Φ ρ ( X ) , ⊆ ) is a complete lattice, in which the join operation ∨ is the set union operator ∪ , since ⋁ i ∈ I ρ ( A i ) = ⋃ i ∈ I ρ ( A i ) = ρ ( ⋃ i ∈ I A i ) for any A i ⊆ X . But the meet operation ∧ in ( Φ ρ ( X ) , ⊆ ) is not the set union operation ∩ in general. For instance, let X = { x 1 , x 2 } and Y = { y 1 , y 2 , y 3 } . Define a binary relation ρ : X ⇀ Y as follows: ρ ( x 1 ) = { y 1 , y 2 } and ρ ( x 2 ) = { y 2 , y 3 } . Then ρ ( x 1 ) ∩ ρ ( x 2 ) = { y 1 , y 2 } ∩ { y 2 , y 3 } = { y 2 } . But { y 2 } ∉ Φ ρ ( X ) since for any C ⊆ X , ρ ( C ) ≠ { y 2 } . Thus, { y 2 } is not the infimum of ρ ( x 1 ) and ρ ( x 2 ) in ( Φ ρ ( X ) , ⊆ ) . In fact, it is easy to see that ρ ( x 1 ) ∧ ρ ( x 2 ) = ∅ in ( Φ ρ ( X ) , ⊆ ) .

Definition 2.2

[6] Suppose p is a property about a complete lattice and s is a property about a binary relation. Let S = { ρ ∈ Rel : ρ has the property s } and P = { L ∈ Com : L has the property p } . We call the relation of s type is a representation of the complete lattices of p type, if the following two conditions are satisfied:

∀ ρ ⊆ X × Y , ρ ∈ S ⇔ ( Φ ρ ( X ) , ⊆ ) ∈ P ;
∀ L ∈ P , ∃ ρ : X ⇀ Y ∈ S such that L ≅ ( Φ ρ ( X ) , ⊆ ) .

In order to make sense of the s type relational representation of p type lattices, the properties p and s need to be explicitly defined.

Definition 2.3

[1] A binary relation ρ : X ⇀ Y is called regular, if there is a binary relation σ : Y ⇀ X such that ρ = ρ ∘ σ ∘ ρ .

In [1], Zareckiǐ gave the regular relation representation of completely distributive lattices. Furthermore, an intrinsic characterization of regular relations was obtained as follows.

Theorem 2.4

[6] For a binary relation ρ : X ⇀ X , the following two conditions are equivalent:

ρ is regular;
∀ ( x , y ) ∈ ρ , ∃ u , v ∈ X such that
1. ( x , v ) ∈ ρ , ( u , y ) ∈ ρ ;
2. For all s , t ∈ X , if ( s , v ) ∈ ρ , ( u , t ) ∈ ρ , then ( s , t ) ∈ ρ .

Definition 2.5

[6] For a binary relation ρ : X ⇀ Y , define a relation ρ ( < ω ) : X ( < ω ) ⇀ Y ( < ω ) , called the finite extension of ρ , by

∀ ( F , G ) ∈ X ( < ω ) × Y ( < ω ) , ( F , G ) ∈ ρ ( < ω ) ⇔ G ⊆ ρ ( F ) .

For a complete lattice L , x , y ∈ L . Define ⊲ on L by x ⊲ y ⇔ for any subset A ⊆ L , y ≤ ∨ A implies x ≤ a for some a ∈ A . Raney [14] proved that a complete lattice L is a completely distributive iff x = ∨ { y ∈ L : y ⊲ x } .

Definition 2.6

[19] A complete lattice L is called strongly algebraic, if ∀ x ∈ L , x = ∨ { y ∈ L : y ⊲ y ≤ x } .

Definition 2.7

[20, 22] Let L be a complete lattice and x , y ∈ L . We define y ≺ x if and only if x ∈ int υ ( L ) ↑ y . L is called hyperalgebraic, if for any x ∈ L , x = ∨ { y ∈ L : y ≺ y ≤ x } .

In [22], Yang proved that a complete lattice L is hyperalgebraic if and only if for any x ∈ L and U ∈ υ ( L ) with x ∈ U , there exists u ∈ L such that x ∈ int υ ( L ) ↑ u = ↑ u ⊆ U . As the generalization of hyperalgebraic lattices, Yang also gave the notion of quasi-hyperalgebraic lattices.

Definition 2.8

[22] Let L be a complete lattice. L is called quasi-hyperalgebraic, if for any x ∈ L and U ∈ υ ( L ) with x ∈ U , there exists F ∈ L ( < ω ) such that x ∈ int υ ( L ) ↑ F = ↑ F ⊆ U .

3 Strongly regular relation

In this section, we define a concept of strongly regular relation and get the strongly regular relation representation of strongly algebraic lattices. Meanwhile, we also obtain some equivalent characterizations of strongly regular relation.

Definition 3.1

A binary relation ρ : X ⇀ Y is called strongly regular, if there is a binary relation σ ⊆ ρ − 1 : Y ⇀ X such that ρ = ρ ∘ σ ∘ ρ .

By the definition of strongly regular relation, it is easy to obtain the following conclusion.

Remark 3.2

If a binary relation ρ : X ⇀ Y is strongly regular, then ρ is regular.
If ρ : X ⇀ Y is strongly regular, then ρ − 1 : Y ⇀ X is also.
Let X , Y be two sets and f a function from X to Y . Then f is strongly regular. In fact, define a binary relation τ : Y ⇀ X as ( y , x ) ∈ τ ⇔ y = f ( x ) . Then τ ⊆ f − 1 and f ∘ τ ∘ f = f .
Let X be a set and P ( X ) a power set of X . Then the binary relation ∈ : X ⇀ P ( X ) , x ∈ A is strongly regular. In fact, define a binary relation σ : P ( X ) ⇀ X as ( A , x ) ∈ σ ⇔ A = { x } . Then σ ⊆ ∈ − 1 and ∈ ∘ σ ∘ ∈ = ∈ .

Lemma 3.3

Let ρ : X ⇀ Y be a binary relation and M ∈ Φ ρ ( X ) . Then for any y ∈ M , ⋀ y ∈ N ∈ Φ ρ ( X ) N ⊲ M . Furthermore, if z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , then ⋀ z ∈ N ∈ Φ ρ ( X ) N ⊲ ⋀ z ∈ N ∈ Φ ρ ( X ) N .

Proof

Let y ∈ M and M ⊆ ⋃ i ∈ I ρ ( A i ) . Then ∃ i ∈ I such that y ∈ ρ ( A i ) , which implies ⋀ y ∈ N ∈ Φ ρ ( X ) N ⊆ ρ ( A i ) . Thus, ⋀ y ∈ N ∈ Φ ρ ( X ) N ⊲ M . Therefore, if z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , then ⋀ z ∈ N ∈ Φ ρ ( X ) N ⊲ ⋀ z ∈ N ∈ Φ ρ ( X ) N .□

Now we give the strongly regular relation representation of strongly algebraic lattices.

Theorem 3.4

For a binary relation ρ : X ⇀ Y , the following three conditions are equivalent:

ρ is strongly regular.
∀ ( x , y ) ∈ X × Y with ( x , y ) ∈ ρ , ∃ ( u , v ) ∈ X × Y such that
1. ( u , v ) ∈ ρ ;
2. ( x , v ) ∈ ρ , ( u , y ) ∈ ρ ;
3. ∀ ( s , t ) ∈ X × Y , if ( s , v ) ∈ ρ and ( u , t ) ∈ ρ , then ( s , t ) ∈ ρ .
( Φ ρ ( X ) , ⊆ ) is a strongly algebraic lattice, i.e., ∀ A ⊆ X , ρ ( A ) = ⋃ { G ∈ Φ ρ ( X ) : G ⊲ G ⊆ ρ ( A ) } .
∀ M ∈ Φ ρ ( X ) , y ∈ M , ∃ z ∈ M such that z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , y ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N .
∀ M ∈ Φ ρ ( X ) , M = ⋃ v ∈ M ⋀ v ∈ N ∈ Φ ρ ( X ) N , and ∀ y ∈ M , ∃ z ∈ M such that z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , y ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N .

Proof

(1) ⇒ (2) Let ρ be a strongly regular relation. Then there exists σ ⊆ ρ − 1 such that ρ = ρ ∘ σ ∘ ρ . ∀ ( x , y ) ∈ X × Y with ( x , y ) ∈ ρ , ∃ ( v , u ) ∈ Y × X such that ( x , v ) ∈ ρ , ( v , u ) ∈ σ ⊆ ρ − 1 , ( u , y ) ∈ ρ , so conditions (i) and (ii) hold. ∀ ( s , t ) ∈ X × Y , if ( s , v ) ∈ ρ and ( u , t ) ∈ ρ , by ( v , u ) ∈ σ and ρ = ρ ∘ σ ∘ ρ , we have ( s , t ) ∈ ρ .

(2) ⇒ (3) ∀ N = ρ ( A ) ∈ Φ ρ ( X ) , y ∈ N , ∃ x ∈ A such that ( x , y ) ∈ ρ . By (2), there is a ( u , v ) ∈ X × Y such that

( u , v ) ∈ ρ ;
( x , v ) ∈ ρ , ( u , y ) ∈ ρ ;
∀ ( s , t ) ∈ X × Y , if ( s , v ) ∈ ρ and ( u , t ) ∈ ρ , then ( s , t ) ∈ ρ .

Thus, y ∈ ρ ( u ) ⊆ ρ ( x ) . In fact, ∀ t ∈ ρ ( u ) , since ( x , v ) ∈ ρ , ( u , t ) ∈ ρ and (iii), we have that ( x , t ) ∈ ρ , i.e., t ∈ ρ ( x ) .

Now we only need to show ρ ( u ) ⊲ ρ ( u ) . Let ρ ( u ) ⊆ ⋃ i ∈ I ρ ( A i ) . Then ∃ i ∈ I such that v ∈ ρ ( A i ) since ( u , v ) ∈ ρ , which implies that there exists x i ∈ A i such that v ∈ ρ ( x i ) , i.e., ( x i , v ) ∈ ρ . For any t ∈ ρ ( u ) , i.e., ( u , t ) ∈ ρ , by (iii) ( x i , t ) ∈ ρ , that is, t ∈ ρ ( x i ) , so ρ ( u ) ⊆ ρ ( x i ) ⊆ ρ ( A i ) . Hence, ρ ( u ) ⊲ ρ ( u ) . Therefore, ρ ( A ) = ⋃ { G ∈ Φ ρ ( X ) : G ⊲ G ⊆ ρ ( A ) } .

(3) ⇒ (4) Let M = ρ ( A ) ∈ Φ ρ ( X ) and y ∈ M . Since ( Φ ρ ( X ) , ⊆ ) is a strongly algebraic lattice, ∃ G ∈ Φ ρ ( X ) such that y ∈ G ⊲ G ⊆ M . Let G = ρ ( B ) = ⋃ b ∈ B ρ ( b ) . By the definition of ⊲ , ∃ b ∈ B such that G = ρ ( b ) and ρ ( b ) ⊈ ⋃ ( Φ ρ ( X ) \ ↑ ρ ( b ) ) = ⋃ { H ∈ Φ ρ ( X ) : ρ ( b ) ⊈ H } . Thus, ∃ z ∈ ρ ( b ) \ ⋃ ( Φ ρ ( X ) \ ↑ ρ ( b ) ) .

Now we show that ρ ( b ) = ⋀ z ∈ N ∈ Φ ρ ( X ) N . Let N ∈ Φ ρ ( X ) with z ∈ N . Assume that ρ ( b ) ⊈ N , then N ∈ Φ ρ ( X ) \ ↑ ρ ( b ) . Thus, z ∈ N ⊆ ⋃ ( Φ ρ ( X ) \ ↑ ρ ( b ) ) , which contradicts z ∉ ⋃ ( Φ ρ ( X ) \ ↑ ρ ( b ) ) . Thus, ρ ( b ) ⊆ N , which implies ρ ( b ) ⊆ ⋀ z ∈ N ∈ Φ ρ ( X ) N . The converse inclusion is always true, since z ∈ G = ρ ( b ) . Hence, M = ⋃ v ∈ M ⋀ v ∈ N ∈ Φ ρ ( X ) N , and for any y ∈ M , there is a z ∈ M such that z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , y ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N .

(4) ⇒ (5) Obviously.

(5) ⇒ (1) Define a relation σ : Y ⇀ X by

( y , x ) ∈ σ ⇔ y ∈ ρ ( x ) = ⋀ y ∈ N ∈ Φ ρ ( X ) N .

Obviously σ ⊆ ρ − 1 .

Next we will prove that ρ = ρ ∘ σ ∘ ρ . ∀ ( x , y ) ∈ ρ , y ∈ ρ ( x ) , by (5), ρ ( x ) = ⋃ v ∈ M ⋀ v ∈ N ∈ Φ ρ ( X ) N , and ∃ z ∈ ρ ( x ) such that z , y ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N . By Lemma 3.3, ⋀ z ∈ N ∈ Φ ρ ( X ) N ⊲ ⋀ z ∈ N ∈ Φ ρ ( X ) N . Let ⋀ z ∈ N ∈ Φ ρ ( X ) N = ρ ( C ) = ⋃ c ∈ C ρ ( c ) . Then ∃ c ∈ C such that ⋀ z ∈ N ∈ Φ ρ ( X ) N = ρ ( c ) . Thus, z ∈ ρ ( c ) = ⋀ z ∈ N ∈ Φ ρ ( X ) N . Follows from the definition of σ , it is that ( z , c ) ∈ σ . Note that ( x , z ) ∈ ρ and ( c , y ) ∈ ρ , we have that ( x , y ) ∈ ρ ∘ σ ∘ ρ . Hence, ρ ⊆ ρ ∘ σ ∘ ρ . ∀ ( x , y ) ∈ ρ ∘ σ ∘ ρ , then ∃ ( p , q ) ∈ X × Y such that ( x , q ) ∈ ρ , ( q , p ) ∈ σ and ( p , y ) ∈ ρ . Since ( q , p ) ∈ σ , q ∈ ρ ( p ) = ⋀ q ∈ N ∈ Φ ρ ( X ) N . Thus, y ∈ ρ ( p ) = ⋀ q ∈ N ∈ Φ ρ ( X ) N ⊆ ρ ( x ) , i.e., ( x , y ) ∈ ρ . Therefore, ρ is strongly regular.□

Theorem 3.5

Let L be a complete lattice. Then the following conditions are equivalent:

L is a strongly algebraic lattice, i.e., ∀ x ∈ L , x = ∨ { z ∈ L : z ◃ z ≤ x } .
There is a strongly regular relation ρ : X ⇀ X such that L ≅ ( Φ ρ ( X ) , ⊆ ) .
The relation ≰ on L is strongly regular, i.e., ∀ x , y ∈ L with x ≰ y , ∃ u , v ∈ L such that
1. u ≰ v ;
2. u ≰ y , x ≰ v ;
3. ∀ s , t ∈ L , if u ≰ t , s ≰ v , then s ≰ t .
∀ x , y ∈ L with x ≰ y , ∃ u , v ∈ L such that
1. u ≰ v ;
2. u ≰ y , x ≰ v ;
3. ∀ z ∈ L , u ≤ z or z ≤ v .
∀ x , y ∈ L with x ≰ y , ∃ u , v ∈ L such that
1. x ∉ ↓ v , y ∉ ↑ u ;
2. ↓ v ∪ ↑ u = L , ↓ v ∩ ↑ u = ∅ .

Proof

(1) ⇒ (2) Let X = L . Define a binary relation ρ : X ⇀ X , ( x , y ) ∈ ρ ⇔ y ◃ y ≤ x and a relation σ : X ⇀ X , ( u , v ) ∈ σ ⇔ u ◃ v and v = u . Obviously, σ ⊆ ρ − 1 . We claim that ρ = ρ ∘ σ ∘ ρ . In fact, for any ( x , y ) ∈ ρ , i.e., y ◃ y ≤ x . Since L is strongly algebraic, ∃ z ∈ L such that y ◃ y ≤ z ◃ z ≤ x . By the definition of ρ and σ , we have ( x , z ) ∈ ρ , ( z , z ) ∈ σ , and ( z , y ) ∈ ρ . Hence, ( x , y ) ∈ ρ ∘ σ ∘ ρ . For any ( x , y ) ∈ ρ ∘ σ ∘ ρ , ∃ u ∈ L such that ( x , u ) ∈ ρ , ( u , u ) ∈ σ , and ( u , y ) ∈ ρ . Then y ◃ y ≤ u ◃ u ≤ x , which implies y ◃ y ≤ x . So ( x , y ) ∈ ρ . All the above show that ρ = ρ ∘ σ ∘ ρ , i.e., ρ is strongly regular.

Now we only need to verify L ≅ ( Φ ρ ( X ) ⊆ ) . ∀ ρ ( A ) ∈ Φ ρ ( X ) , ρ ( A ) = { y ∈ L : ∃ a ∈ A , ( a , y ) ∈ ρ } = { y ∈ L : ∃ a ∈ A , y ◃ y ≤ a } = ⋃ a ∈ A { y ∈ L : y ◃ y ≤ a } = { y ∈ L : y ◃ y ≤ ∨ A } . Define two functions f : L → ( Φ ρ ( X ) , ⊆ ) , f ( x ) = ρ ( x ) = { y ∈ L : y ◃ y ≤ x } and g : ( Φ ρ ( X ) , ⊆ ) → L , g ( ρ ( A ) ) = ∨ ρ ( A ) = ∨ A . Since L is strongly algebraic, it is easy to see that f ∘ g = i d Φ ρ ( X ) and g ∘ f = i d L . Hence, L ≅ ( Φ ρ ( X ) , ⊆ ) .

(2) ⇒ (1) By Theorem 3.4.

(1) ⇒ (3) ∀ x , y ∈ L with x ≰ y . Since L is a strongly algebraic lattice, there is a u ∈ L such that u ◃ u ≤ x with u ≰ y . Let v = sup ( L \ ↑ u ) . Then it is easy to see that u ≰ v , u ≰ y , and x ≰ v . ∀ s , t ∈ L , if u ≰ t , s ≰ v , then t ∈ L \ ↑ u . Thus, t ≤ v which implies s ≰ t (otherwise, s ≤ v , a contradiction). This proof shows that ≰ is strongly regular.

(3) ⇒ (4) ∀ x , y ∈ L with x ≰ y . By (4), ∃ u , v ∈ L such that

u ≰ v ;
u ≰ y , x ≰ v ;
∀ s , t ∈ L , if u ≰ t , s ≰ v , then s ≰ t .

∀ z ∈ L , let s = t = z . Then u ≤ z or z ≤ v . Thus, condition (5) holds.

(4) ⇔ (5) ⇔ (1) see [25].□

From the above theorem, we can obtain the following corollary.

Corollary 3.6

Let L be a complete lattice. If L is strongly algebraic, then L o p is strongly algebraic.

4 Finitely strongly regular relation

In this section, we define a concept of finitely strongly regular relation and obtain the finitely strongly regular relation representation of hyperalgebraic lattices, prove that ρ is finitely strongly regular if and only if ( Φ ρ ( X ) , ⊆ ) is a hyperalgebraic lattice if and only if the finite extension ρ ( < ω ) of ρ is strongly regular.

Definition 4.1

A binary relation ρ : X ⇀ Y is called finitely strongly regular, if ∀ ( x , y ) ∈ X × Y with ( x , y ) ∈ ρ , ∃ { u 1 , u 2 , … , u n } ∈ X ( < ω ) and { v 1 , v 2 , … , v m } ∈ Y ( < ω ) such that

{ v 1 , v 2 , … , v m } ⊆ ρ ( { u 1 , u 2 , … , u n } ) ;
y ∈ ρ ( { u 1 , u 2 , … , u n } ) , and ( x , v j ) ∈ ρ ( j = 1 , 2 , … , m ) ;
∀ { s 1 , s 2 , … , s m } ∈ X ( < ω ) , t ∈ Y , if t ∈ ρ ( { u 1 , u 2 , … , u n } ) , ( s j , v j ) ∈ ρ ( j = 1 , 2 , … , m ) , then ∃ k ∈ { 1 , 2 , … , m } such that ( s k , t ) ∈ ρ .

Remark 4.2

It is easy to observe that if the relation ρ is strongly regular, then ρ is finitely strongly regular.
Condition (iii) of Definition 4.1 is equivalent to the following condition:
( iii ′ ) ∀ S ∈ X ( < ω ) , t ∈ Y , if t ∈ ρ ( { u 1 , u 2 , … , u n } ) , { v 1 , v 2 , … , v m } ⊆ ρ ( S ) , then t ∈ ρ ( S ) .
Let ρ : X ⇀ Y be a binary relation satisfying conditions (i) and (iii) of Definition 4.1. Then for any S ⊆ X ,
{ v 1 , v 2 , … , v m } ⊆ ρ ( S ) ⇔ ρ ( { u 1 , u 2 , … , u n } ) ⊆ ρ ( S ) .

Let U = { u 1 , u 2 , … , u n } , V = { v 1 , v 2 , … , v m } . Then U ∈ X ( < ω ) , V ∈ Y ( < ω ) . By Definition 4.1, it is easy to obtain the following lemma.

Lemma 4.3

Let ρ : X ⇀ Y be a binary relation. Then the following conditions are equivalent:

ρ : X ⇀ Y is finitely strongly regular.
∀ ( x , y ) ∈ X × Y with ( x , y ) ∈ ρ , ∃ ( U , V ) ∈ X ( < ω ) × Y ( < ω ) such that
1. V ⊆ ρ ( U ) ;
2. y ∈ ρ ( U ) , V ⊆ ρ ( x ) ;
3. ∀ S ∈ X ( < ω ) , t ∈ Y , if t ∈ ρ ( U ) , V ⊆ ρ ( S ) , then t ∈ ρ ( S ) .

Now we give the finitely strongly regular relation representation of hyperalgebraic lattices.

Theorem 4.4

For a binary relation ρ : X ⇀ Y , then the following conditions are equivalent:

ρ is finitely strongly regular.
The finite extension ρ ( < ω ) of ρ : X ( < ω ) ⇀ Y ( < ω ) is strongly regular.
∀ ( F , G ) ∈ X ( < ω ) × Y ( < ω ) , G ⊆ ρ ( F ) ⇒ ∃ ( U , V ) ∈ X ( < ω ) × Y ( < ω ) such that
1. V ⊆ ρ ( U ) ;
2. G ⊆ ρ ( U ) , V ⊆ ρ ( F ) ;
3. ∀ ( S , T ) ∈ X ( < ω ) × Y ( < ω ) , if V ⊆ ρ ( S ) and T ⊆ ρ ( U ) , then T ⊆ ρ ( S ) .
( { ⋃ F ∈ ℱ ρ ( F ) ( < ω ) : ℱ ⊆ X ( < ω ) } , ⊆ ) is a strongly algebraic lattice.
( Φ ρ ( X ) , ⊆ ) is a hyperalgebraic lattice.

Proof

(1) ⇒ (2) Define a relation δ : Y ( < ω ) ⇀ X ( < ω ) by

( G , F ) ∈ δ ⇔ G ⊆ ρ ( F ) , and ∀ ( S , T ) ∈ X ( < ω ) × Y ( < ω ) , if G ⊆ ρ ( S ) , F ∩ ρ − 1 ( T ) ≠ ∅ , then T ∩ ρ ( S ) ≠ ∅ .

Then

δ ⊆ ( ρ ( < ω ) ) − 1 .
ρ ( < ω ) ∘ δ ∘ ρ ( < ω ) ⊆ ρ ( < ω ) .

For any ( H , W ) ∈ ρ ( < ω ) ∘ δ ∘ ρ ( < ω ) , there exists ( G , F ) ∈ Y ( < ω ) × X ( < ω ) such that ( H , G ) ∈ ρ ( < ω ) , ( G , F ) ∈ δ , and ( F , W ) ∈ ρ ( < ω ) , that is, G ⊆ ρ ( H ) , ( G , F ) ∈ δ , and W ⊆ ρ ( F ) . ∀ w ∈ W , let S = H and T = { w } . Then G ⊆ ρ ( S ) and F ∩ ρ − 1 ( T ) ≠ ∅ . Then T ∩ ρ ( S ) ≠ ∅ since ( G , F ) ∈ δ . Hence, w ∈ ρ ( S ) , it follows that W ⊆ ρ ( H ) , i.e., ( H , W ) ∈ ρ ( < ω ) .

ρ ( < ω ) ⊆ ρ ( < ω ) ∘ δ ∘ ρ ( < ω ) .

For any ( H , W ) ∈ ρ ( < ω ) and w ∈ W , ∃ h ( w ) ∈ H such that ( h ( w ) , w ) ∈ ρ . Since ρ is finitely strongly regular, by Lemma 4.3, ∃ U ( w ) ∈ X ( < ω ) and V ( w ) ∈ Y ( < ω ) satisfy the following conditions:

V ( w ) ⊆ ρ ( U ( w ) ) ;
w ∈ ρ ( U ( w ) ) , V ( w ) ⊆ ρ ( h ( w ) ) ;
∀ S ∈ X ( < ω ) , t ∈ Y , if t ∈ ρ ( U ( w ) ) , V ( w ) ⊆ ρ ( S ) , then t ∈ ρ ( S ) .

Let F = ⋃ w ∈ W U ( w ) , G = ⋃ w ∈ W V ( w ) . Then ( F , G ) ∈ X ( < ω ) × Y ( < ω ) , G ⊆ ρ ( F ) , G ⊆ ⋃ w ∈ W ρ ( h w ) ⊆ ρ ( H ) , and W ⊆ ρ ( F ) . By Definition 2.5, ( H , G ) ∈ ρ ( < ω ) and ( F , W ) ∈ ρ ( < ω ) . Now we have to verify ( G , F ) ∈ δ . Clearly, ( G , F ) ∈ ρ ( < ω ) . ∀ ( S , T ) ∈ X ( < ω ) × Y ( < ω ) , if G ⊆ ρ ( S ) , F ∩ ρ − 1 ( T ) ≠ ∅ , then there exist w 0 ∈ W and t 0 ∈ T such that t 0 ∈ ρ ( U ( w 0 ) ) , and ∀ w ∈ W , V ( w ) ⊆ ρ ( S ) . Thus, V ( w 0 ) ⊆ ρ ( S ) . By 3 ∘ , t 0 ∈ ρ ( S ) , which implies T ∩ ρ ( S ) ≠ ∅ . Thus, ( G , F ) ∈ δ . Hence, ( H , W ) ∈ ρ ( < ω ) ∘ δ ∘ ρ ( < ω ) .

By (a), (b), and (c), we have that δ ⊆ ( ρ ( < ω ) ) − 1 and ρ ( < ω ) = ρ ( < ω ) ∘ δ ∘ ρ ( < ω ) , i.e., ρ ( < ω ) is strongly regular.

(2) ⇒ (3) By Theorem 3.4 and Definition 2.5.

(2) ⇔ (4) Let σ = ρ ( < ω ) . Then by Theorem 3.4, σ is strongly regular ⇔ ( Φ σ ( X ( < ω ) ) , ⊆ ) = ( { ⋃ F ∈ ℱ ρ ( < ω ) ( F ) : ℱ ⊆ X ( < ω ) } , ⊆ ) = ( { ⋃ F ∈ ℱ ρ ( F ) ( < ω ) : ℱ ⊆ X ( < ω ) } , ⊆ ) is a strongly algebraic lattice.

(3) ⇒ (5) Let L = ( Φ ρ ( X ) , ⊆ ) . For any M = ρ ( A ) ∈ L , y ∈ M , there is a x ∈ A such that y ∈ ρ ( x ) , by (3), there exists ( U , V ) ∈ X ( < ω ) × Y ( < ω ) such that

V ⊆ ρ ( U ) ;
y ∈ ρ ( U ) , V ⊆ ρ ( x ) ;
∀ ( S , T ) ∈ X ( < ω ) × Y ( < ω ) , if V ⊆ ρ ( S ) and T ⊆ ρ ( U ) , then T ⊆ ρ ( S ) .

Let V = { v 1 , v 2 , … , v m } and N j = ⋃ { N ∈ L : v j ∉ N } ( j = 1 , 2 , … , m ) ( N j may be an empty set). Now we have to verify ρ ( U ) ≺ ρ ( U ) ⊆ M . Since V ⊆ ρ ( x ) ⊆ ρ ( A ) = M and V ⊆ ρ ( U ) , we have M ∈ L \ ↓ { N 1 , N 2 , … , N m } and ρ ( U ) ∈ L \ ↓ { N 1 , N 2 , … , N m } . ∀ N = ρ ( B ) ∈ L \ ↓ { N 1 , N 2 , … , N m } , V ⊆ N = ρ ( B ) . Let S = B , T = ρ ( U ) , by the condition (iii), ρ ( U ) ⊆ ρ ( S ) = ρ ( B ) = N . Thus, N ∈ ↑ ρ ( U ) . This proves that ρ ( U ) ∈ L \ ↓ { N 1 , N 2 , … , N m } ⊆ ↑ ρ ( U ) . Thus, M ∈ int υ ( L ) ↑ ρ ( U ) = ↑ ρ ( U ) with y ∈ ρ ( U ) . Hence, M = ⋃ { G ∈ L : G ≺ G ⊆ M } = ⋁ { G ∈ L : G ≺ G ⊆ M } , i.e., L = ( Φ ρ ( X ) , ⊆ ) is a hyperalgebraic lattice.

(5) ⇒ (1) Let L = ( Φ ρ ( X ) , ⊆ ) . ∀ ( x , y ) ∈ ρ , i.e., y ∈ ρ ( x ) , since L is a hyperalgebraic lattice, there is a N = ρ ( A ) ∈ L such that y ∈ N ≺ N ⊆ ρ ( x ) . Since N ≺ N , ∃ M 1 , M 2 , … , M m ∈ L ( < ω ) such that y ∈ ρ ( x ) ∈ ↑ N = L \ ↓ { M 1 , M 2 , … , M m } . ∀ j ∈ { 1 , 2 , … , m } , by N ⊈ M j , ∃ v j ∈ N but v j ∉ M j . It follows that there is a u j ∈ A such that v j ∈ ρ ( u j ) . Also, since y ∈ ρ ( A ) , ∃ u ∈ A such that y ∈ ρ ( u ) .

Let U = { u , u 1 , u 2 , … , u m } ∈ X ( < ω ) , V = { v 1 , v 2 , … , v m } ∈ Y ( < ω ) . Then it is easy to see that ( U , V ) satisfies the following conditions:

V ⊆ ρ ( U ) ;
y ∈ ρ ( U ) , V ⊆ ρ ( x ) .

Now we check that ( U , V ) satisfies condition (c) of Lemma 4.3. ∀ S ∈ X ( < ω ) , t ∈ Y , if V ⊆ ρ ( S ) , t ∈ ρ ( U ) , then ∀ j ∈ { 1 , 2 , … , m } , ∃ s j ∈ S such that v j ∈ ρ ( s j ) . Let S 0 = { s 1 , s 2 , … , s m } . Then S 0 ⊆ S and ρ ( S 0 ) = ⋃ j = 1 m ρ ( s j ) ∈ L \ ↓ { M 1 , M 2 , … , M m } = ↑ N . Note that U ⊆ A , thus ρ ( U ) ⊆ ρ ( A ) = N ⊆ ρ ( S 0 ) . Since t ∈ ρ ( U ) , ∃ k ∈ { 1 , 2 , … , m } such that t ∈ ρ ( s k ) ⊆ ρ ( S 0 ) ⊆ ρ ( S ) . Therefore, ρ is finitely strongly regular.□

Theorem 4.5

Let L be a complete lattice. Then the following conditions are equivalent:

L is a hyperalgebraic lattice, i.e., ∀ x ∈ L , x = ∨ { y ∈ L : y ≺ y ≤ x } .
∀ x , y ∈ L with x ≰ y , ∃ u ∈ L , and { v 1 , v 2 , … , v m } ∈ L ( < ω ) such that
1. u ≰ v j ( j = 1 , 2 , … , m ) ;
2. u ≰ y , x ≰ v j ( j = 1 , 2 , … , m ) ;
3. ∀ { s 1 , s 2 , … , s m } ∈ L ( < ω ) and t ∈ L , if u ≰ t , s j ≰ v j ( j = 1 , 2 , … , m ) , then ∃ k ∈ { 1 , 2 , … , m } such that s k ≰ t .
∀ x , y ∈ L with x ≰ y , ∃ u ∈ L , and { v 1 , v 2 , … , v m } ∈ L ( < ω ) such that
1. u ≰ v j ( j = 1 , 2 , … , m ) ;
2. u ≰ y , x ≰ v i ( i = 1 , 2 , … , m ) ;
3. ∀ z ∈ L , u ≤ z or z ≤ v k for some k ∈ { 1 , 2 , … , m } .
∀ x , y ∈ L with x ≰ y , ∃ u ∈ L , and F ∈ L ( < ω ) such that
1. x ∉ ↓ F , y ∉ ↑ u ;
2. ↓ F ∪ ↑ u = L , ↓ F ∩ ↑ u = ∅ .
∀ x , y ∈ L with x ≰ y , ∃ u ∈ L such that x ∈ int υ ( L ) ↑ u = ↑ u ⊆ L \ ↓ y .
∀ x , y ∈ L with x ≰ y , ∃ υ ( L ) -closed subset C and u ∈ L such that
1. x ∉ C , y ∉ ↑ u ;
2. C ∪ ↑ u = L , C ∩ ↑ u = ∅ .
The relation ≰ on L is finitely strongly regular.
The finite extension ≰ ( < ω ) of ≰ : L ( < ω ) ⇀ L ( < ω ) is strongly regular.
( { ⋃ F ∈ ℱ ( L \ ↑ ∨ F ) ( < ω ) : ℱ ⊆ L ( < ω ) } , ⊆ ) is a strongly algebraic lattice.
There is a finitely strongly regular relation ρ : X ⇀ X such that L ≅ ( Φ ρ ( X ) , ⊆ ) .

Proof

( 1 ) ⇔ ( 4 ) ⇔ ( 5 ) See Lemma 2.2 of [22].

(1) ⇒ (2) ∀ x , y ∈ L with x ≰ y . Since L is a hyperalgebraic lattice, there exists u ∈ L such that u ≺ u ≤ x with u ≰ y . By the definition of ≺ , we have that x ∈ int υ ( L ) ↑ u = ↑ u ⊆ L \ ↓ y . Hence, there exists a finite set F = { v 1 , v 2 , … , v m } ⊆ L such that u ∈ L \ ↓ F ⊆ int υ ( L ) ↑ u , which implies x ∈ L \ ↓ F = ↑ u ⊆ L \ ↓ y . So u ≰ v j ( j = 1 , 2 , … , m ) , u ≰ y , and x ≰ v j ( j = 1 , 2 , … , m ) .

Now we only need to show that u and F = { v 1 , v 2 , … , v m } satisfy condition (iii) of (2). Let { s 1 , s 2 , … , s m } ∈ L ( < ω ) and t ∈ L with u ≰ t , s j ≰ v j for all j = 1 , 2 , … , m . Assume that for any j ∈ { 1 , 2 , … , m } , s j ≤ t , then t ∈ L \ ↓ F ; otherwise, there exists a j 0 ∈ { 1 , 2 , … , m } such that t ≤ v j 0 , and then s j 0 ≤ v j 0 , a contradiction. This implies t ∈ L \ ↓ F ⊆ ↑ u , a contradiction. Hence, there exists k ∈ { 1 , 2 , … , m } such that s k ≰ t .

(2) ⇒ (3) We only need to show condition (iii) of (3). ∀ z ∈ L . Let t = z and s j = z ( j = 1 , 2 , … , m ) . By condition (iii) of (2), u ≤ t = z or z ≤ v k for some k ∈ { 1 , 2 , … , m } .

(3) ⇒ (1) ∀ x ∈ L , if y ∈ L with x ≰ y , by (3), ∃ u ∈ L and { v 1 , v 2 , … , v m } ∈ L ( < ω ) such that

u ≰ v j ( j = 1 , 2 , … , m ) ;
u ≰ y , x ≰ v i ( i = 1 , 2 , … , m ) ;
∀ z ∈ L , u ≤ z , or z ≤ v k for some k ∈ { 1 , 2 , … , m } .

Let F = { v 1 , v 2 , … , v m } ∈ L ( < ω ) . Then it is easy to see that x ∈ ↑ u = L \ ↓ F . Thus, u ≺ u ≤ x and u ≰ y . Hence, x = ⋁ { y ∈ L : y ≺ y ≤ x } , i.e., L is a hyperalgebraic lattice.

(5) ⇒ (6) Let C = L \ int υ ( L ) ↑ u . Then C is υ ( L ) -closed and satisfies condition (6) with u .

(6) ⇒ (5) ∀ x , y ∈ L with x ≰ y , by (6), ∃ υ ( L ) -closed subset C and u ∈ L such that

x ∉ C , y ∉ ↑ u ;
C ∪ ↑ u = L , C ∩ ↑ u = ∅ .

By condition (b), ↑ u = L \ C ∈ υ ( L ) , and it follows from condition (a) that x ∈ L \ C = int υ ( L ) ↑ u = ↑ u ⊆ L \ ↓ y .

(2) ⇒ (7) ∀ x , y ∈ L with x ≰ y , by (2), there exist u ∈ L and { v 1 , v 2 , … , v m } ∈ L ( < ω ) such that

u ≰ v j ( j = 1 , 2 , … , m ) ;
u ≰ y , x ≰ v j ( j = 1 , 2 , … , m ) ;
∀ { s 1 , s 2 , … , s m } ∈ L ( < ω ) and t ∈ L , if u ≰ t , s j ≰ v j ( j = 1 , 2 , … , m ) , then ∃ k ∈ { 1 , 2 , … , m } such that s k ≰ t .

Let u i = u ( i = 1 , 2 … , n ) . Then it is easy to see that { u 1 , u 2 , … , u n } and { v 1 , v 2 , … , v m } satisfy condition (7).

(7) ⇔ (8) ⇔ (9) By Theorem 4.4.

(7) ⇒ (10) Let X = L and ρ = ≰ : X ⇀ X . Then L ≅ ( Φ ≰ ( X ) , ⊆ ) .

(10) ⇒ (1) Using Theorem 4.4.□

5 Generalized finitely strongly regular relation

In this section, we define a concept of generalized finitely strongly regular relation and get the generalized finitely strongly regular representation of quasi-hyperalgebraic lattices, prove that ρ is generalized finitely strongly regular if and only if ( Φ ρ ( X ) , ⊆ ) is a quasi-hyperalgebraic lattice if and only if ( Φ ρ ( X ) , ⊆ ) with respect to the interval topology is a Priestley space. Meanwhile, some equivalent characterizations of quasi-hyperalgebraic lattices are obtained.

Definition 5.1

A binary relation ρ : X ⇀ Y is called generalized finitely strongly regular, if ∀ ( x , y ) ∈ ρ , there exist { F 1 , F