Relational representations of algebraic lattices and their applications

Definition 2.1

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

1. 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 .

2. ρ − 1 = { ( y , x ) ∈ Y × X : ( x , y ) ∈ ρ } .

3. 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.

4. Φ ρ ( X ) = { ρ ( A ) : A ⊆ 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:

1. ∀ ρ ⊆ X × Y , ρ ∈ S ⇔ ( Φ ρ ( X ) , ⊆ ) ∈ P ;

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

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 .

Theorem 2.4

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

1. ρ is regular;

2. ∀ ( 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

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 } .

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 .

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 .

Remark 3.2

1. If a binary relation ρ : X ⇀ Y is strongly regular, then ρ is regular.

2. If ρ : X ⇀ Y is strongly regular, then ρ − 1 : Y ⇀ X is also.

3. 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 .

4. 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 .□

Theorem 3.4

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

1. ρ is strongly regular.

2. ∀ ( 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 .

3. ( Φ ρ ( X ) , ⊆ ) is a strongly algebraic lattice, i.e., ∀ A ⊆ X , ρ ( A ) = ⋃ { G ∈ Φ ρ ( X ) : G ⊲ G ⊆ ρ ( A ) } .

4. ∀ M ∈ Φ ρ ( X ) , y ∈ M , ∃ z ∈ M such that z ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N , y ∈ ⋀ z ∈ N ∈ Φ ρ ( X ) N .

5. ∀ 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

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 .

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

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:

1. L is a strongly algebraic lattice, i.e., ∀ x ∈ L , x = ∨ { z ∈ L : z ◃ z ≤ x } .

2. There is a strongly regular relation ρ : X ⇀ X such that L ≅ ( Φ ρ ( X ) , ⊆ ) .

3. 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 .

4. ∀ 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 .

5. ∀ 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

1. u ≰ v ;

2. u ≰ y , x ≰ v ;

3. ∀ 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].□

Corollary 3.6

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

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

1. { v 1 , v 2 , … , v m } ⊆ ρ ( { u 1 , u 2 , … , u n } ) ;

2. y ∈ ρ ( { u 1 , u 2 , … , u n } ) , and ( x , v j ) ∈ ρ ( j = 1 , 2 , … , m ) ;

3. ∀ { 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

1. It is easy to observe that if the relation ρ is strongly regular, then ρ is finitely strongly regular.

2. 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 ) .

3. 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 ) .

Lemma 4.3

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

1. ρ : X ⇀ Y is finitely strongly regular.

2. ∀ ( 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 ) .

Theorem 4.4

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

1. ρ is finitely strongly regular.

2. The finite extension ρ ( < ω ) of ρ : X ( < ω ) ⇀ Y ( < ω ) is strongly regular.

3. ∀ ( 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 ) .

4. ( { ⋃ F ∈ ℱ ρ ( F ) ( < ω ) : ℱ ⊆ X ( < ω ) } , ⊆ ) is a strongly algebraic lattice.

5. ( Φ ρ ( X ) , ⊆ ) is a hyperalgebraic lattice.