The B-topology on S^∗-doubly quasicontinuous posets

Definition 1.1

[1] Let P be a poset and U ⊆ P . U is said to be B-open if and only if for every net ( x i ) i ∈ I → o x ∈ U , we have that x i ∈ U holds eventually.

Proposition 1.2

[7] Let P be a poset and U ⊆ P . Then U ∈ T P if and only if for any directed set D and filtered set F with sup D = inf F = x ∈ U , ↑ d 0 ∩ ↓ e 0 ⊆ U for some d 0 ∈ D and e 0 ∈ F .

Definition 1.3

[7] Let P be a poset and x , y , z ∈ P . We define y ≪ S x if for every directed subset D of P with sup D = x , there exists d ∈ D such that y ⩽ d ; dually, we define z ▹ S x if for every filtered subset F of P with inf F = x , there exists e ∈ F such that z ⩾ e .

Definition 1.4

[7] A poset P is said to be S -doubly continuous if for every x ∈ P , the sets ⇓ S x and ⇈ S x are directed and filtered, respectively, and sup ⇓ S x = x = inf ⇈ S x .

Definition 1.5

[7] An S -doubly continuous poset P is said to be S ∗ -doubly continuous if for every x ∈ P , y ∈ ⇓ S x and z ∈ ⇈ S x , there exist y 0 ∈ ⇓ S x and z 0 ∈ ⇈ S x such that ↑ y 0 ∩ ↓ z 0 ⊆ ⇑ S y ∩ ⇊ S z .

Proposition 1.6

[7] If P is a doubly continuous poset (see [15, Definition 2.5 ]), then P is S ∗ -doubly continuous.

Lemma 1.7

[8] Let P be a poset.

1. If ℳ is a directed subfamily of ℒ 0 ( P ) , then there exists a directed set D ⊆ ⋃ { M : M ∈ ℳ } such that D ∩ M ≠ ∅ for all M ∈ ℳ .

2. If N is a filtered subfamily of ℒ 0 ( P ) , then there exists a filtered set F ⊆ ⋃ { N : N ∈ N } such that F ∩ N ≠ ∅ for all N ∈ N .

Definition 2.1

Let P be a poset. A net ( x i ) i ∈ I in P is said to o s -converge to x ∈ P if there exists a directed family ℳ x ⊆ ℒ 0 ( ↓ x ) and a filtered family N x ⊆ ℒ 0 ( ↑ x ) such that

1. ⋂ { ↑ M : M ∈ ℳ x } = ↑ x and ⋂ { ↓ N : N ∈ N x } = ↓ x .

2. For every M ∈ ℳ x and N ∈ N x , x i ∈ ↑ M ∩ ↓ N holds eventually, i.e., there exists i 0 ∈ I such that x i ∈ ↑ M ∩ ↓ N for all i ≥ i 0 .

Proposition 2.2

Let P be a poset and ( x i ) i ∈ I a net in P . Then the o s -convergent point of ( x i ) i ∈ I is unique. That is, if ( x i ) i ∈ I → o s x ∈ P and ( x i ) i ∈ I → o s y ∈ P , then x = y .

Proof

Suppose that ( x i ) i ∈ I → o s x and ( x i ) i ∈ I → o s y . Then, by Definition 2.1, there exist directed families ℳ x ⊆ ℒ 0 ( ↓ x ) , ℳ y ⊆ ℒ 0 ( ↓ y ) and filtered families N x ⊆ ℒ 0 ( ↑ x ) , N y ⊆ ℒ 0 ( ↑ y ) such that

1. ⋂ { ↑ M x : M x ∈ ℳ x } = ↑ x and ⋂ { ↓ N x : N x ∈ N x } = ↓ x ;

2. ⋂ { ↑ M y : M y ∈ ℳ y } = ↑ y and ⋂ { ↓ N y : N y ∈ N y } = ↓ y ;

3. For any M x ∈ ℳ x and N x ∈ N x , x i ∈ ↑ M x ∩ ↓ N x holds eventually;

4. For any M y ∈ ℳ y and N y ∈ N y , x i ∈ ↑ M y ∩ ↓ N y holds eventually.

For every M x ∈ ℳ x , one can trivially check, by (3) and the finiteness of M x , that there exists m M x ∈ M x such that x i ∈ ↑ m M x holds frequently, that is, for every i ∈ I there exists j i ≥ i such that x j i ∈ ↑ m M x . Since x i ∈ ↓ N y holds eventually for every N y ∈ N y , there exists i N y ∈ I such that x i ∈ ↓ N y for all i ≥ i N y . Let M 0 = { m M x : M x ∈ ℳ x } . Then we have x j i N y ∈ ↑ m M x ∩ ↓ N y for every m M x ∈ M 0 and N y ∈ N y . This means M 0 ⊆ ⋂ { ↓ N y : N y ∈ N y } = ↓ y . Since x ∈ ⋂ { ↑ m M x : M x ∈ N x } ⊆ ⋂ { ↑ M x : M x ∈ N x } = ↑ x , we have sup M 0 = x . This implies x ⩽ y . Similarly, it can be proved that y ⩽ x . Thus, we conclude x = y .□

Proposition 2.3

Let P be a poset and ( x i ) i ∈ I a net in P . If ( x i ) i ∈ I → o x ∈ P , then ( x i ) i ∈ I → o s x .

Proof

Straightforward by the definitions of o -convergence and o s -convergence.□

Example 2.4

Let P = { x } ∪ { a 0 , a 1 , a 2 , a 3 , … , a n , … } ∪ { b 0 , b 1 , b 2 , b 3 , … , b n , … } . The partially order ⩽ on P is defined by setting

1. a 0 ⩽ a 1 ⩽ a 2 ⩽ a 3 ⩽ … , a n ⩽ ⋯ ⩽ x ;

2. b 0 ⩽ b 1 ⩽ b 2 ⩽ b 3 ⩽ … , b n ⩽ ⋯ ⩽ x .

Let x 2 k = a k and x 2 k + 1 = b k for every k ∈ N , where N denotes the set of all natural numbers. Then it is easy to verify that the net ( x i ) i ∈ N o s -converges to x but not o -converge to x .

Proposition 2.5

Given a poset P and x , y , z ∈ P . Let a directed family ℳ x ⊆ ℒ 0 ( ↓ x ) and a filtered family N x ⊆ ℒ 0 ( ↑ x ) satisfy ⋂ { ↑ M : M ∈ ℳ x } = ↑ x and ⋂ { ↑ N : N ∈ N x } = ↓ x .

1. If y ≪ S x , then M 0 ⊆ ↑ y for some M 0 ∈ ℳ x .

2. If z ▹ S x , then N 0 ⊆ ↓ z for some N 0 ∈ N x .

Proof

(1): Suppose that M ⊈ ↑ y for every M ∈ ℳ x . Then it is easy to verify that { M \ ↑ y : M ∈ ℳ x } ⊆ ℒ 0 ( ↓ x ) is a directed family such that ⋂ { ↑ ( M \ ↑ y ) : M ∈ ℳ x } = ↑ x . By Rudin lemma, there exists a directed D ⊆ ⋃ { M \ ↑ y : M ∈ ℳ x } such that D ∩ ( M \ ↑ y ) ≠ ∅ for all M ∈ ℳ x . This implies

Thus, we have sup D = x . It follows from the definition of ≪ S that there exists d 0 ∈ D such that d 0 ∈ ↑ y . This contradicts that d 0 ∈ D ⊆ ⋃ { M \ ↑ y : M ∈ ℳ x } .

The proof of (2) can be completed similarly.□

Proposition 2.6

Let P be a S -doubly continuous poset and x ∈ P . Then a net

Proof

Straightforward by Proposition 2.5, the definitions of o -convergence and o s -convergence.□

Proposition 2.7

Let P be a poset and x ∈ P . Suppose that ℳ x ⊆ ℒ 0 ( ↓ x ) is a directed family and N x ⊆ ℒ 0 ( ↑ x ) a filtered family with ⋂ { ↑ M : M ∈ ℳ x } = ↑ x and ⋂ { ↓ N : N ∈ N x } = ↓ x . If we

1. set ℐ 0 = { ↑ M ∩ ↓ N : M ∈ ℳ x & N ∈ N x } and ℐ ℳ x N x = { ( i , I ) ∈ P × ℐ 0 : i ∈ I } ;

2. define the directed preorder ≦ on ℐ ℳ x N x by

   ( ∀ ( i 1 , I 1 ) , ( i 2 , I 2 ) ∈ ℐ ℳ x N x ) ( i 1 , I 1 ) ≦ ( i 2 , I 2 ) ⇔ I 2 ⊆ I 1 ;

3. let x ( i , I ) = i for all ( i , I ) ∈ ℐ ℳ x N x ,

then the net ( x ( i , I ) ) ( i , I ) ∈ ℐ ℳ x N x → o s x .

Proof

The proof can be easily completed by checking that the directed family ℳ x and the filtered family N x satisfy Conditions (Q1) and (Q2) of Definition 2.1.□

Lemma 3.1

Let P be a poset and U ∈ T P . Then for every x ∈ U , every directed subfamily ℳ x ⊆ ℒ 0 ( ↓ x ) and every filtered subfamily N x ⊆ ℒ 0 ( ↑ x ) such that ⋂ { ↑ M : M ∈ ℳ x } = ↑ x and ⋂ { ↓ N : N ∈ N x } = ↓ x , we have x ∈ ↑ M 0 ∩ ↓ N 0 ⊆ U for some M 0 ∈ ℳ x and N 0 ∈ N x .

Proof

Suppose not, that is, for every M ∈ ℳ x and every N ∈ N x , we have x ∈ ↑ M ∩ ↓ N ⊈ U . Now, we show

1. For every M = { m 1 , m 2 , … , m k M } ∈ ℳ x , the set

   M K = { m ∈ M : ( ∀ N ∈ N x ) ↑ m ∩ ↓ N ⊈ U } ≠ ∅ .

   Suppose M K = ∅ . Then, for every i ∈ { 1 , 2 , … , k M } , there exists N i ∈ N x such that ↑ m i ∩ ↓ N i ⊆ U . Since N x is a filtered subfamily of ℒ 0 ( ↓ x ) , we can take N 00 ∈ N x such that N 00 ⊆ ↓ N 1 ∩ ↓ N 2 ∩ … ∩ ↓ N k M . This implies ↑ M ∩ ↓ N 00 ⊆ U , a contradiction to the hypothesis. Thus, we have M K ≠ ∅ .

2. The family ℳ x K = { M K : M ∈ ℳ x } is also directed.

   Let ( M 1 ) K , ( M 2 ) K ∈ ℳ x K . According to the directness of ℳ x , it follows that there exists M 00 ∈ ℳ x such that M 00 ⊆ ↑ M 1 ∩ ↑ M 2 . To show the directness of ℳ x K , it suffices to prove ( M 00 ) K ⊆ ↑ ( M 1 ) K ∩ ↑ ( M 2 ) K . Suppose ( M 00 ) K ⊈ ↑ ( M 1 ) K . Then, we have m 0 ∈ ↑ m 1 ′ for some m 0 ∈ ( M 00 ) K and m 1 ′ ∈ M 1 \ ( M 1 ) K . Since m 1 ′ ∈ M 1 \ ( M 1 ) K , there exists N 1 ∈ N x such that ↑ m 1 ′ ∩ ↓ N 1 ⊆ U , which implies

   ↑ m 0 ∩ ↓ N 1 ⊆ ↑ m 1 ′ ∩ ↓ N 1 ⊆ U .

   This contradicts the fact that m 0 ∈ ( M 00 ) K . Thus, we have ( M 00 ) K ⊆ ↑ ( M 1 ) K . A similar verification can show ( M 00 ) K ⊆ ↑ ( M 2 ) K . Therefore, we have ( M 00 ) K ⊆ ↑ ( M 1 ) K ∩ ↑ ( M 2 ) K .

3. The existence of directed set D satisfying

   1. ( ∀ M ∈ ℳ x ) D ∩ M K ≠ ∅ ;

   2. D ⊆ ⋃ { M K : M ∈ ℳ x } ;

   3. sup D = x ;

   4. ( ∀ d ∈ D , ∀ N ∈ N x ) ↑ d ∩ ↓ N ⊈ U .

   By Rudin lemma and the directness of ℳ x K , we can conclude the existence of a directed set D satisfying (1) and (2). Since

   ↑ x ⊆ ⋂ { ↑ d : d ∈ D } ( by (2) and ℳ x K ⊆ ℒ 0 ( ↓ x ) )

   ⊆ ⋂ { ↑ M K : M ∈ ℳ x } (by (1) and (2)) ⊆ ⋂ { ↑ M : M ∈ ℳ x } ( by ( ∀ M ∈ ℳ x ) M K ⊆ M ) = ↑ x ,

   x is the supremum of the directed set D , i.e., sup D = x . Now, it is easy to verify, by (2) and the definition of ℳ x K , that the directed set D satisfies (4).

4. For every N ∈ N x , the set

   N K = { n ∈ N : ( ∀ d ∈ D ) ↑ d ∩ ↓ n ⊈ U } ≠ ∅ .

   The verification is similar to that of (i).

5. The family N x K = { N K : N ∈ N x } is filtered.

   The verification is similar to that of (ii).

6. The existence of a filtered set F satisfying

   1. ( ∀ N ∈ N x ) F ∩ N K ≠ ∅ ;

   2. F ⊆ ⋃ { N K : N ∈ N x } ;

   3. inf F = x ;

   4. ( ∀ d ∈ D , ∀ e ∈ F ) ↑ d ∩ ↓ e ⊈ U .

   The verification is similar to that of (iii).

To summarize what we have proved, we picked a directed set D and a filtered set F such that sup D = inf F = x ∈ U , and ↑ d ∩ ↓ e ⊈ U for every d ∈ D and e ∈ F . By Proposition 1.2, U is not a B-open set. This contradicts to the assumption U ∈ T P . Therefore, the proof is complete.□

Theorem 3.2

Let P be a poset. Then U ∈ T P if and only if for every net ( x i ) i ∈ I → o s x ∈ U , x i ∈ U holds eventually.

Proof

( ⇒ ): Let U ∈ T P and a net ( x i ) i ∈ I → o s x ∈ U . Then, by Definition 2.1, there exist a directed subfamily ℳ x ⊆ ℒ 0 ( ↓ x ) and a filtered subfamily N x ⊆ ℒ 0 ( ↑ x ) such that

1. ⋂ { ↑ M : M ∈ ℳ x } = ↑ x and ⋂ { ↓ N : N ∈ N x } = ↓ x ;

2. For every M ∈ ℳ x and N ∈ N x , x i ∈ ↑ M ∩ ↓ N eventually.

By Lemma 3.1, we have x ∈ ↑ M 0 ∩ ↓ N 0 ⊆ U for some M 0 ∈ ℳ x and N 0 ∈ N x . This implies that x i ∈ ↑ M 0 ∩ ↓ N 0 ⊆ U holds eventually.

( ⇐ ): By Proposition 2.3 and Definition 1.1.□

Definition 4.1

Let P be a poset and A , M , N ⊆ P .

1. We say that M S 0 -approximates A below, in symbol M ≪ S 0 A , if for every directed subset D of P , sup D ∈ A implies d ∈ ↑ M for some d ∈ D . We simply write M ≪ S 0 x for M ≪ S 0 { x } and y ≪ S 0 A for { y } ≪ S 0 A .

2. Dually, we say that N S 0 -approximates A above, in symbol N ▹ S 0 A , if for every filtered subset F of P , inf F ∈ A implies e ∈ ↓ N for some e ∈ F . We simply write x ▹ S 0 A for { x } ▹ S 0 A and N ▹ S 0 x for N ▹ S 0 { x } .

Proposition 4.2

Let P be a poset and A , B , M , N ⊆ P . Then

1. M ≪ S 0 A ⇔ ( ∀ a ∈ A ) M ≪ S 0 a , and N ▹ S 0 A ⇔ ( ∀ a ∈ A ) N ▹ S 0 a ;

2. B ≤ M ≪ S 0 A ⇒ B ≪ S 0 A , and B ≥ o p N ▹ S 0 A ⇒ B ▹ S 0 A ;

3. M ≪ S 0 A ⇒ M ≤ A , and N ▹ S 0 A ⇒ N ≥ o p A ;

4. { x } ≪ S 0 { y } and { z } ▹ S 0 { y } in the sense of Definition 4.1 are equivalent to x ≪ S y and z ▹ S y in the sense of Definition 1.3, respectively.

Proposition 4.3

Let P be a poset, x ∈ P and A , B ⊆ P . For every directed family ℳ x ⊆ ℒ 0 ( ↓ x ) and every filtered family N x ⊆ ℒ 0 ( ↑ x ) ,

1. if A ≪ S 0 x and ⋂ { ↑ M : M ∈ ℳ x } = ↑ x , then M 0 ⊆ ↑ A for some M 0 ∈ ℳ x ;

2. if B ▹ S 0 x and ⋂ { ↓ N : N ∈ N x } = ↓ x , then N 0 ⊆ ↓ B for some N 0 ∈ N x .

Proof

(1): Suppose M ⊈ ↑ A for every M ∈ ℳ x . Then { M \ ↑ A : M ∈ ℳ x } is a directed subfamily of ℒ 0 ( ↓ x ) and ⋂ { ↑ ( M \ ↑ A ) : M ∈ ℳ x } = ↑ x . By Rudin Lemma, there exists a directed set D such that D ⊆ ⋃ { M \ ↑ A : M ∈ ℳ x } and D ∩ ( M \ ↑ A ) ≠ ∅ for every M ∈ ℳ x . Thus, we have

This means sup D = x and d ∉ ↑ A for all d ∈ D . By the definition of ≪ S 0 , we conclude that A ≪̸ S 0 x , which contradicts the assumption A ≪ S 0 x .

(2) It can be similarly proved.□

Proposition 4.4

Let P be a poset and A , B , H ⊆ P . Then

1. A ≪ S 0 H , if and only if for every net ( x i ) i ∈ I → o s x ∈ H , x i ∈ ↑ A holds eventually;

2. B ▹ S 0 H , if and only if for every net ( x i ) i ∈ I → o s x ∈ H , x i ∈ ↓ B holds eventually.