#### Abstract

We prove some new characterizations of strongly continuous lattices using two new intrinsic topologies and a class of convergences. Lastly we show that the category of strongly continuous lattices and Scott continuous mappings is cartesian closed.

#### 1. Introduction and Preliminaries

The theory of continuous lattices was first introduced by Scott in 1972 (see [1]) and has been studied extensively by many people from various different fields due to its strong connections to computer science, general topology, algebra, and logics (see [2]). Later the continuous lattices were generalized to various other classes of ordered structures for different motivations, for example, the *L*-domains [3], generalized continuous lattices and hypercontinuous lattices [4], and the most general, continuous dcpos. In [5], in order to get more general results on the relationship between prime and pseudo-prime elements in a complete lattice, two new classes of lattices related to continuous lattices, namely, the semicontinuous lattices and the strongly continuous lattices, were introduced. The semicontinuous lattices were later generalized to *Z-*semicontinuous posets in [6]. Liu and Xie introduced several categories of semicontinuous lattice trying to construct a Cartesian closed category of semicontinuous lattices (see [7, 8]). In [9, 10], the first two authors studied other properties of semicontinuous lattices and proved a characterization of semicontinuous lattices in terms of the *S*-open subsets (to be defined later in this paper). Strongly continuous lattices form a proper subclass of continuous lattices. There is some fine connection between strong continuity and distributivity: every distributive continuous lattice is strongly continuous; every Noetherian (in particular, finite) strongly continuous lattice is distributive; a strongly continuous lattice in which the way-below relation is multiplicative is distributive. It has been know for a long time that the category *CONT* of continuous lattices and Scott continuous mappings is cartesian closed [2, Theorem ]. It is thus natural to ask if the subcategory of strongly continuous lattices and that of distributive continuous lattices are cartesian closed as well. In this paper we will give a positive answer to these questions. New characterizations of strongly continuous lattices based on some new topologies and convergence of nets are also obtained.

In Section 2, we introduce two new intrinsic topologies on complete lattices, which are then used to formulating a new characterization for strongly continuous lattices. In Section 3, we introduce the convergence of nets and show that a complete lattice is strongly continuous if and only if the convergence on the lattice is topological. In Section 4, we define several new types of mappings between complete lattices and study the relationship among *S-*continuous mappings, Scott continuous mappings, strongly continuous mappings, and semicontinuous mappings. In the last section, we consider the category *SCONT* of strongly continuous lattices and prove that *SCONT* is Cartesian closed. It is also pointed out that the subcategory *DCONT* of distributive continuous lattices is cartesian closed. For most of the basic definitions and results on continuous lattices we refer to the book [2].

Let be a poset. For any , define for some . We also write as . The sets and are defined dually.

A subset of is called an upper (lower) set if (). A subset of is directed provided it is nonempty, and every finite subset of has an upper bound in . An ideal of is a directed lower set.

Let . We say that is way-below , written , if for any directed subset with exists and , there is such that . Let and let . A complete lattice is called a continuous lattice if for every element . It is well known that for any complete lattice and , is an ideal.

*Definition 1 (see [11]). *An ideal of a lattice is called semiprime if for all , and imply .

We use to denote the family of all semiprime ideals of .

*Definition 2 (see [5]). *Let be a complete lattice. Define the relation on as follows: for any if for any semiprime ideal of , implies . For each , we write

For any , let and let .

*Definition 3 (see [5]). *A complete lattice is called semicontinuous, if for any ,
is called strongly continuous, if for any .

Theorem 4 (see [5]). *If is a semicontinuous lattice, then the relation has the interpolation property; that is, implies the existence of such that . *

Lemma 5 (see [5]). *Let be a complete lattice; then is a semiprime ideal for each . *

#### 2. Strong Continuity via Topology

For any complete lattice , is called Scott open if and only if and for any directed set , implies . All Scott open subsets of form a topology, called the Scott topology, denoted by [2]. One of the classic characterizations of continuous lattices is that a complete lattice is continuous if and only if is completely distributive. This result was later generalized to continuous posets [2]. In the current section we introduce two new intrinsic topologies on complete lattices and use them to establish some new characterizations for strongly continuous lattices.

*Definition 6. *A subset of a complete lattice is called *S-*open if and only if and for any semiprime ideal of , implies .

We use to denote the family of all *-*open subsets of ; it is easy to verify that forms a topology on , called the -topology.

*Definition 7. *A subset of a complete lattice is called *T*-open if and only if .

We use to denote the family of all *T*-open subsets of the complete lattice .

*Remark 8. *As implies , it follows that every *T-*open set is an upper set. Furthermore, for any complete lattice , forms a topology on . Obviously contains the empty set and and is closed under arbitrary union. Let . We show that . For any , , thus there exist such that and . Therefore . Note that , thus . Conversely, if , then there exists such that . Since and , so and , thus . Thus , which belongs to . Hence forms a topology on .

We call the topology the *T*-topology. Obviously every Scott open set is *S*-open. The reverse conclusion does not need to be true. There is not any inclusion relation between the *T*-topology and the Scott topology applying to all complete lattices.

*Example 9. * Let , where is the unit interval; and are two distinct elements not in . The partial order on is defined by: ; for if is less than or equal to according to the usual order on real numbers.

Obviously is the unique semiprime ideal of , so (in fact, every upper set) is *S*-open. For each , let . Then is a directed set such that , but for each . Thus is not Scott open. Also, , is not *T*-open either.

Let be the five-element modular lattice. Then clearly is Scott open. Again, in this example, is the only semiprime ideal, so for any nonempty set . Hence is not *T*-open.

Lemma 10. * If is a semicontinuous lattice, then .** If is strongly continuous, then . *

*Proof. * It only needs to show that . Let . For any directed subset with , there exists such that . Since is semicontinuous, . Note that the union is still a semiprime ideal, so there is such that . Thus . Therefore , hence . This proves .

Now assume that is strongly continuous. Then is continuous. In a strongly continuous lattice, the relation and are the same by the Theorem 2.5 of [5]. Also by [2] in a continuous lattice, every Scott open set satisfies the condition , it follows that every Scott open set of is *T-*open. Now let be *S-*open. For any directed set with , let . Then is a semiprime ideal such that . Thus, as is *S-*open, . Hence . Therefore is Scott open. By we have .

Lemma 11. *For any complete lattice , if then is semicontinuous. *

*Proof. *Suppose that is not semicontinuous. Then there exists such that . Then which is obviously *S-*open. Since , is *T*-open. Thus there exists such that by Definition 7, a contradiction. Thus must be semicontinuous.

Let be a complete lattice. The long way-below relation on is defined as follows: for any , if and only if for any nonempty subset , implies that for some . For each , we write

Clearly implies . In [12], Raney proved that a complete lattice is a completely distributive lattice if and only if for all .

It is well known that a complete lattice is a continuous if and only if the topology is a completely distributive lattice [2]. If is semicontinuous, is generally not a completely distributive lattice. In Example 9, is semicontinuous and consists of all upper subsets of . Then is a complete lattice and the empty set is the largest element of , but . Thus is not a completely distributive lattice.

Proposition 12. *Let be a complete lattice in which the relation satisfies the interpolation property. Then is a completely distributive lattice. *

*Proof. *By Raney’s characterization of completely distributive lattices, we need to show that hold for all .

For any , by the definition of , there exists such that . Since the relation satisfies the interpolation property, . Now we claim that . Let such that . Since and , there exists such that . Therefore . Hence . It now follows that . This completes the proof.

Lemma 13. *If is a complete lattice such that , then is a continuous lattice. *

*Proof. *If , then is semicontinuous and is a completely distributive lattice by Lemma 11 and Proposition 12. By Lemma 10, , thus is a completely distributive lattice. It then follows from Theorem of [2] that is continuous. This completes the proof.

Lemma 14. *Let be a complete lattice. If , then is strongly continuous. *

*Proof. *By Lemma 11, is semicontinuous. Assume that is not strongly continuous. Then there exists such that . Thus which is obviously *S-*open. By Lemma 5 and Definition 6, there exists such that and . By Lemma 13, . Thus there exists such that . By Lemma 10, is continuous and so for all . Again by Lemma 10 and , so . Thus . And then . So , which contradicts the assumption on . This completes the proof.

From Lemmas 10 and 14, we obtain the following new characterization of strongly continuous lattices.

Theorem 15. *A complete lattice is strongly continuous if and only if . *

#### 3. **-** Convergence and Strongly Continuous Lattices

In [2], the lim-inf convergence is introduced, and it is proved that a complete lattice is continuous if and only if the lim-inf convergence on the lattice is topological. This result was later generalized to continuous dcpos and continuous posets (see [2, 13]). Now we introduce a similar convergence, convergence, and show that a complete lattice is strongly continuous if and only if the convergence on the lattice is topological.

*Definition 16. *A net in a complete lattice is said to converge to an element if there exists a semiprime ideal of such that(1), and(2)for any , holds eventually (i.e., there exists such that for all ).

In this case we write .

For any complete lattice , define The class is called topological if there is a topology on such that if and only if the net converges to with respect to the topology .

As in usual cases, associated with is a family of sets, which is a topology on :

Proposition 17. *For any complete lattice , . *

*Proof. *First, suppose . Let be a semiprime ideal in with . Consider the net with . Now for any holds eventually. Thus . From the definition of we conclude that the net must be eventually in , and then there exists such that for all , whence .

Conversely, suppose . For any such that , by the definition of , we have for some semiprime ideal and for each holds eventually. Now , so by the definition of , there exists such that . Then there exists such that for all . Thus for all . Hence holds eventually. Thus .

Lemma 18. *If is a complete lattice and , then , where denotes the interior of with respect to the S-topology. *

*Proof. *Let be a complete lattice and . For any semiprime ideal with , we have . Thus there exists . Therefore and then . Thus .

Proposition 19. *Let be a strongly continuous lattice. Then if and only if the net converges to the element with respect to . In particular, the convergence is topological. *

*Proof. *By Proposition 17, , so if then converges to the element with respect to . Conversely, suppose that we have a net which converges to the element with respect to . For each , we have from the definition of . Thus there exists such that for all , and then for all . Since is strongly continuous, . By Lemma 5, is a semiprime ideal. Therefore we have . That is, .

Lemma 20. *Let be a complete lattice. If the convergence is topological, then is strongly continuous. *

*Proof. *By Proposition 17, the topology arising from convergence is the *S-*topology. Thus if the convergence is topological, we must have that if and only if the net converges to the element with respect to . For any , let , where consists of all *S-*open sets containing , and define an order on to be the lexicographic order on the first two coordinates. That is, if and only if is a proper subset of or and . Let for each . Then it is easy to verify that the net converges to the element with respect to . Thus , and we conclude that there exists a semiprime ideal such that and holds eventually for each . Let , then there exists such implies . In particular, we have for all . Thus . It follows that . Furthermore, . By Lemma 18, and then . Therefore . Since is a semiprime ideal with a supremum greater than or equal to , it follows that . Hence and so . All these show that is strongly continuous.

What we now have proved is the following characterization of strongly continuous lattices.

Theorem 21. *Let be a complete lattice, then the following statements are equivalent.*(1)* is strongly continuous.*(2)*The convergence is the convergence for the S-topology; that is, for all and all nets in if and only if the net converges to the element with respect to .*(3)

*The convergence is topological.*

#### 4. Continuous Mappings

In this section, we will investigate the relations among semicontinuous mappings, strongly semicontinuous mappings, Scott continuous mappings, and *S-*continuous mappings. Firstly recall from [2] that a mapping is said to be Scott continuous, if is continuous from topological spaces to . It is known that is Scott continuous if and only if preserves all directed suprema (see [2]). Now we give some new basic definitions.

*Definition 22. *Let be complete lattices. An order preserving mapping is called a semicontinuous mapping, if preserves suprema of semiprime ideals; is called a strongly semicontinuous mapping, if is semicontinuous and for any , . A mapping is called -continuous if it is continuous from topological spaces to .

Lemma 23. *Let be complete lattices. If is S-continuous, then is order-preserving. *

*Proof. *Let in . Suppose that ; then the -open set contains . Thus is a *S-*open neighborhood of not containing . But then as is an upper set, a contradiction. Thus implies .

Lemma 24. *Let be a complete lattice and . Then is S-closed if and only if holds for any with . *

Lemma 25. *Let be complete lattices. If is a strongly semicontinuous mapping, then is S-continuous. *

*Proof. *Let be an *S-*closed subset of . First, is a lower set because is order preserving and is a lower set. For arbitrary semiprime ideal , we have . Since is strongly semicontinuous, and . By Lemma 24, . Therefore . Again by Lemma 24, is *S-*closed. Thus is *S-*continuous. This proves our result.

Proposition 26. *Let be an order preserving mapping from a strongly continuous lattice to a complete lattice . Then is Scott continuous if and only if is semicontinuous. *

*Proof. **Necessity*. Obvious.*Sufficiency*. Let be arbitrary directed subset of . Since is order-preserving, . Suppose that . Then . Thus . Since is a strongly continuous lattice, by Theorem 2.5 in [5], there exists a semiprime ideal such that . Hence . It then follows that . Thus there exist and such that . Therefore . That is, , which contradicts . Hence holds for all directed set ; thus is Scott continuous.

Proposition 27. *Let be a map from a strongly continuous lattice to a distributive lattice . Then the following conditions are equivalent: *(1) is Scott continuous;(2) is strongly semicontinuous;(3) is *S-*continuous;(4) is semicontinuous.

*Proof. *. Let be Scott continuous. Then is order preserving and preserves the supremum of all semiprime ideals. Let . Then is an ideal of . Since is distributive, is the semiprime ideal of .

follows from Lemma 25.

. Let be *S-*continuous. By Lemma 23, is order-preserving. Let be any semiprime ideal of . Thus . Assume that . Then which is obviously *S-*open. Hence . Therefore and then there exists such that . That is, a contradiction.

follows from Proposition 26.

Corollary 28. *Let be strongly continuous lattice. Then is Scott continuous if and only if is semicontinuous. *

The following proposition follows directly from Lemma 23 and Theorem 1 of [7].

Proposition 29. *Let be a semicontinuous lattice and let be a complete lattice. If there is a surjective -continuous mapping that preserves the relation , then is semicontinuous. *

#### 5. The Cartesian Closedness of the Category of Strongly Continuous Lattices

Let *SCONT* denote the category of all strongly continuous lattices and semicontinuous mappings between them. By Corollary 28, the morphisms of *SCONT* are the Scott continuous mappings. Thus *SCONT* is a full subcategory of the category *CONT* of continuous lattices and Scott continuous mappings. Given two complete lattices and we will use to denote the least element of , to denote the set of all order-preserving maps from to , to denote the set of all Scott continuous mappings from to , and to denote the set of all semicontinuous mappings from to . Obviously, , and they are all posets with respect to the pointwise order. In particular, is a complete lattice. It is well known that the category *CONT* is Cartesian closed [2, Theorem ]. Thus it is natural to ask whether *SCONT* is cartesian closed as well. In this section we will give a positive answer to this question.

Lemma 30. *For any two complete lattices and , the set is closed under taking supremum in ; thus it is a complete lattice. *

*Proof. *Let for all . It is easy to show that is the bottom element of .

Let and for any . Then is the supremum of in . Now for semiprime ideal in , we have
Thus .

Let and be two complete lattices. For every and , define the interpolating step function by

Lemma 31. *Let be a complete lattice for which the way-below relation satisfies the interpolation property. Then for all and , is Scott continuous. *

*Proof. *Clearly is order preserving. Let be any ideal of . If , then . Since the way below relation satisfies the interpolation property, there exists such that . Thus . Therefore . If , then for all . Hence . Therefore preserves supremum of arbitrary ideal. So it is Scott continuous.

The following example illustrates that the assumption that the way below relation satisfies the interpolation property in Lemma 31 is necessary.

*Example 32. *Let and with . The order on is given by (see also Figure 1)(i) for all ;(ii)for each , ;(iii)for each and each , .

Then in , but for . Thus the way-below relation on does not satisfy the interpolation property. Consider the mapping . Let . Then = . = , but = . Thus is not Scott continuous.

Lemma 33. *Let be a complete lattice for which the way-below relation satisfies the interpolation property and . Then for all and , implies . *

*Proof. *Let be any ideal of with . Then for all . Then . Since is directed, there exists such that . Now we claim that .

For arbitrary implies and . If , then . Therefore for all . That is, . This proves our claim. Hence . Therefore .

The following proposition can be found in [2].

Proposition 34 (see [2]). *Let and be continuous lattices. Then for each , ; hence is a continuous lattice. *

The preceding proposition yields a characterization of the way-below relation on function spaces via interpolating step functions.

Corollary 35. *Let and be continuous lattices and . Then holds in if and only if there exist , for , such that
*

Corollary 36. *If and are continuous lattices and holds in , then holds for all . *

*Proof. *With the notation of the previous corollary, we have that
Let . For any , if , then . Hence . If , then . Thus . Therefore . This completes our proof.

Lemma 37. *Let be continuous and let be strongly continuous. Then for any , is a semiprime ideal of . *

*Proof. *Obvious is an ideal because is a complete lattice. Let , . By Proposition 34, is continuous, so the relation satisfies the interpolation property. Thus there exists such that and . For arbitrary , by Corollary 36, , . Since is strongly continuous, and . Thus . And then . Since is arbitrary, . And follows . Thus is a semiprime ideal of . This completes our proof.

Theorem 38. *If is a continuous lattice and is a strongly continuous lattice, then is strongly continuous. *

*Proof. *By Proposition 34, is a continuous lattice and for all , so is a semicontinuous lattice. Let with . By Lemma 37, is an semiprime ideal, so , that is, . By Theorem 2.5 of [5], is a strongly continuous lattice. This completes our proof.

For cartesian closedness we adopt the elementary definition in [14].

*Definition 39 (see [14]). *A category **K** is called a cartesian category if it satisfies the following conditions.(i)There is a terminal object.(ii)Each pair of objects and of **K** has a product with projections and .(iii)Each pair of objects and of **K** has an exponentiation , that is, an object and an arrow eval: with the property that for any , there is a unique arrow such that the composite
is .

Lemma 40. *The cartesian product of two strongly continuous lattices is strongly continuous. *

*Proof. *Let be strongly continuous lattices. Then are continuous lattices. By Proposition of [2], the product of and is continuous. It is easy to verify that the projections , are Scott continuous. Let in . Since are strongly continuous lattices, , and . Since both and are semiprime ideals, it follows that is the semiprime ideal of . Hence . Thus is strongly continuous by the Theorem 2.5 of [5]. By Corollary 28, are semicontinuous. This completes our proof.

Theorem 41. *The category SCONT is cartesian closed in which the exponential is .*

*Proof. *Note that the morphisms in *SCONT* are the Scott continuous mappings. Clearly the singleton set is a strongly continuous lattice and serves as a terminal object in *SCONT*. By Lemma 40, the cartesian product of two strongly continuous lattices is again a strongly continuous lattice.

By Theorem 38, is a strongly continuous lattice. Define eval: by
Then eval is a Scott continuous mapping. Let be a strongly continuous lattice, let be a semicontinuous (or Scott continuous) mapping, and define by . For any directed set and , because is directed. Thus . So . Thus is Scott continuous. And eval = eval . So eval. Clearly is the unique morphism satisfying the condition. By Definition 39, the category *SCONT* is cartesian closed. The proof is completed.

*Remark 42. *By [5], every distributive continuous lattice is strongly continuous. One can verify straightforwardly that for any two distributive continuous lattices and , the lattice of Scott continuous mappings from to is distributive and thus is a distributive continuous lattice. Hence one can prove, by a similar argument as for *SCONT*, that the fully subcategory *DCONT* of *SCONT* consisting of all distributive continuous lattices is cartesian closed.

#### Acknowledgments

This work has been supported by the National Natural Science Foundation of China (no. 11071061, no. 11201490, and no. 11226151), the National Basic Research Program of China (no. 2011CB311808), the Natural Science Foundation of Hunan Province under Grant no. 10JJ2001, and Science and Technology Plan Projects of Hunan Province no. 2012FJ3145.