AAA Abstract and Applied Analysis 1687-0409 1085-3375 Hindawi Publishing Corporation 942628 10.1155/2013/942628 942628 Research Article Some New Intrinsic Topologies on Complete Lattices and the Cartesian Closedness of the Category of Strongly Continuous Lattices 0000-0003-0842-9269 Wu Xiuhua 1 Li Qingguo 2 0000-0001-8446-3624 Zhao Dongsheng 3 Öziş Turgut 1 Institute of Mathematics and Physics Central South University of Forestry and Technology Changsha 410004 China csuft.edu.cn 2 College of Mathematics and Econometrics Hunan University Changsha 410082 China hnu.edu.cn 3 Mathematics and Mathematics Education National Institute of Education Nanyang Technological University 1 Nanyang Walt Singapore 637616 ntu.edu.sg 2013 12 3 2013 2013 02 12 2012 20 01 2013 2013 Copyright © 2013 Xiuhua Wu et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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 ) 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 ). Later the continuous lattices were generalized to various other classes of ordered structures for different motivations, for example, the L-domains , generalized continuous lattices and hypercontinuous lattices , and the most general, continuous dcpos. In , 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 . 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 II-2.12]. 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 lim-infS convergence of nets and show that a complete lattice is strongly continuous if and only if the lim-infS 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 .

Let P be a poset. For any AP, define A={xP:xy for some yA}. We also write {x} as x. The sets A and x are defined dually.

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

Let x,yP. We say that x is way-below y, written xy, if for any directed subset D with D exists and Dy, there is dD such that xd. Let x={aP:ax} and let x={aP:xa}. A complete lattice P is called a continuous lattice if for every element xP,x=x. It is well known that for any complete lattice P and xP, x is an ideal.

Definition 1 (see [<xref ref-type="bibr" rid="B11">11</xref>]).

An ideal I of a lattice P is called semiprime if for all x,y,zP, xyI and xzI imply x(yz)I.

We use Rd(P) to denote the family of all semiprime ideals of P.

Definition 2 (see [<xref ref-type="bibr" rid="B5">5</xref>]).

Let P be a complete lattice. Define the relation on P as follows: for any x,yP,xy if for any semiprime ideal I of P, yI implies xI. For each xP, we write (1)x={yP:yx},      x={yP:xy}.

For any AP, let A=xAx and let A=xAx.

Definition 3 (see [<xref ref-type="bibr" rid="B5">5</xref>]).

A complete lattice P is called semicontinuous, if for any xP, (2)xx.P  is called strongly continuous, if x=x for any xP.

Theorem 4 (see [<xref ref-type="bibr" rid="B5">5</xref>]).

If P is a semicontinuous lattice, then the relation has the interpolation property; that is, xy implies the existence of zP such that xzy.

Lemma 5 (see [<xref ref-type="bibr" rid="B5">5</xref>]).

Let P be a complete lattice; then x is a semiprime ideal for each xP.

2. Strong Continuity via Topology

For any complete lattice P, UP is called Scott open if and only if U=U and for any directed set DP, DU implies UD. All Scott open subsets of P form a topology, called the Scott topology, denoted by σ(P) . One of the classic characterizations of continuous lattices is that a complete lattice P is continuous if and only if (σ(P),) is completely distributive. This result was later generalized to continuous posets . 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 U of a complete lattice P is called S-open if and only if U=U and for any semiprime ideal I of P, IU implies UI.

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

Definition 7.

A subset U of a complete lattice is called T-open if and only if U=U.

We use τ(P) to denote the family of all T-open subsets of the complete lattice P.

Remark 8.

As xyz implies xz, it follows that every T-open set is an upper set. Furthermore, for any complete lattice P, τ(P) forms a topology on P. Obviously τ(P) contains the empty set and P and is closed under arbitrary union. Let U,Vτ(P). We show that UVτ(P). For any xUV, xU,xV, thus there exist mU,nV such that mx and nx. Therefore mnx. Note that mnUV, thus x(UV). Conversely, if x(UV), then there exists yUV such that yx. Since U=U and V=V, so xU and xV, thus xUV. Thus UV=(UV), which belongs to τ(P). Hence τ(P) forms a topology on P.

We call the topology τ(P) 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.

( 1 ) Let P=[0,1]{a,b}, where [0,1] is the unit interval; a and b are two distinct elements not in [0,1]. The partial order on P is defined by: 0<a<1,0<b<1; for x,y[0,1],xy if x is less than or equal to y according to the usual order on real numbers.

Obviously P is the unique semiprime ideal of P, so a (in fact, every upper set) is S-open. For each nN, let cn=1-(1/n). Then D={cn}nN is a directed set such that D=1>a, but for each n,cna. Thus a is not Scott open. Also, (a)=P, a is not T-open either.

( 2 ) Let P={0,a,b,c,1} be the five-element modular lattice. Then clearly {a,1} is Scott open. Again, in this example, P is the only semiprime ideal, so A=P for any nonempty set A. Hence {a,1} is not T-open.

Lemma 10.

( 1 ) If P is a semicontinuous lattice, then τ(P)σ(P)κ(P).

( 2 ) If P is strongly continuous, then τ(P)=σ(P)=κ(P).

Proof.

( 1 ) It only needs to show that τ(P)σ(P). Let Uτ(P). For any directed subset DP with DU, there exists aU such that aD. Since P is semicontinuous, aD{d:dD}=(dDd). Note that the union dDd is still a semiprime ideal, so there is dD such that ad. Thus daU. Therefore DU, hence Uσ(P). This proves τ(P)σ(P).

( 2 ) Now assume that P is strongly continuous. Then P is continuous. In a strongly continuous lattice, the relation and are the same by the Theorem 2.5 of . Also by  in a continuous lattice, every Scott open set A satisfies the condition A=A, it follows that every Scott open set of P is T-open. Now let U be S-open. For any directed set D with DU, let I=dDd. Then I is a semiprime ideal such that I=D. Thus, as U is S-open, IU. Hence DU. Therefore U is Scott open. By (1) we have τ(P)=σ(P)=κ(P).

Lemma 11.

For any complete lattice P, if τ(P)=κ(P) then P is semicontinuous.

Proof.

Suppose that P is not semicontinuous. Then there exists xP such that xx. Then xU=P(x) which is obviously S-open. Since τ(P)=κ(P), U is T-open. Thus there exists ux such that ux by Definition 7, a contradiction. Thus P must be semicontinuous.

Let P be a complete lattice. The long way-below relation on P is defined as follows: for any x,yP, xy if and only if for any nonempty subset BP, yB implies that xz for some zB. For each xP, we write (3)β(x)={yP:yx}.

Clearly xy implies xy. In , Raney proved that a complete lattice P is a completely distributive lattice if and only if x=β(x) for all xP.

It is well known that a complete lattice P is a continuous if and only if the topology σ(P) is a completely distributive lattice . If P is semicontinuous, κ(P) is generally not a completely distributive lattice. In Example 9(1), P is semicontinuous and κ(P) consists of all upper subsets of P. Then (κ(P),) is a complete lattice and the empty set is the largest element of (κ(P),), but β(). Thus κ(P) is not a completely distributive lattice.

Proposition 12.

Let (P,) be a complete lattice in which the relation satisfies the interpolation property. Then τ(P) is a completely distributive lattice.

Proof.

By Raney’s characterization of completely distributive lattices, we need to show that Eβ(E) hold for all Eτ(P).

For any xEτ(P), by the definition of τ(P), there exists yE such that xyE. Since the relation satisfies the interpolation property, yτ(P). Now we claim that yE. Let 𝒟τ(P) such that E𝒟. Since yE and 𝒟=𝒟, there exists U0𝒟 such that yU0. Therefore yU0. Hence yE. It now follows that E{y:yE}β(E). This completes the proof.

Lemma 13.

If P is a complete lattice such that τ(P)=κ(P), then P is a continuous lattice.

Proof.

If τ(P)=κ(P), then P is semicontinuous and τ(P) is a completely distributive lattice by Lemma 11 and Proposition 12. By Lemma 10, τ(P)=σ(P)=κ(P), thus (σ(P),) is a completely distributive lattice. It then follows from Theorem II-1.13 of  that P is continuous. This completes the proof.

Lemma 14.

Let P be a complete lattice. If τ(P)=κ(P), then P is strongly continuous.

Proof.

By Lemma 11, P is semicontinuous. Assume that P is not strongly continuous. Then there exists bP such that b<b. Thus bPb which is obviously S-open. By Lemma 5 and Definition 6, there exists aP such that ab and aPb. By Lemma 13, a=aPbσ(P). Thus there exists xa such that xb. By Lemma 10, P is continuous and so xσ(P) for all xP. Again by Lemma 10 and τ(P)=κ(P),τ(P)=σ(P)=κ(P), so (x)=x. Thus ba(x)=x. And then xb. So xb, which contradicts the assumption on x. This completes the proof.

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

Theorem 15.

A complete lattice P is strongly continuous if and only if τ(P)=κ(P).

3. <inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M297"><mml:mi>lim</mml:mi><mml:mo /></mml:math></inline-formula><bold>-</bold><inline-formula><mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" id="M298"><mml:mrow><mml:msub><mml:mrow><mml:mi>inf</mml:mi><mml:mo /></mml:mrow><mml:mrow><mml:mi>S</mml:mi></mml:mrow></mml:msub></mml:mrow></mml:math></inline-formula> Convergence and Strongly Continuous Lattices

In , 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, lim-infS convergence, and show that a complete lattice is strongly continuous if and only if the lim-infS convergence on the lattice is topological.

Definition 16.

A net (xj)jJ in a complete lattice P is said to lim-infS converge to an element yP if there exists a semiprime ideal I of P such that

Iy, and

for any mI, xjm holds eventually (i.e., there exists kJ such that xjm for all jk).

In this case we write y  lim-infSxj.

For any complete lattice P, define (4)𝒫={((xj)jJ,x):(xj)jJ  isanetwithxlim-infS  xj((xj)jJ,x)}.The class 𝒫 is called topological if there is a topology T on P such that ((xj)jJ,x)𝒫 if and only if the net (xj)jJ converges to x with respect to the topology T.

As in usual cases, associated with 𝒫 is a family of sets, which is a topology on P: (5)𝒪(𝒫)={UP:U=Uandwhenever((xj)jJ,x)𝒫andxU,soeventuallyxjU((xj)jJ,x)}.

Proposition 17.

For any complete lattice P, 𝒪(𝒫)=κ(P).

Proof.

First, suppose U𝒪(𝒫). Let I be a semiprime ideal in P with IU. Consider the net (xd)dI with xd=d. Now for any aI,xda holds eventually. Thus ((xd)dI,I)𝒫. From the definition of 𝒪(𝒫) we conclude that the net (xd)dI must be eventually in U, and then there exists dI such that xwU for all wd, whence DU.

Conversely, suppose Uκ(P). For any ((xj)jJ,x)𝒫 such that xU, by the definition of 𝒫, we have xI for some semiprime ideal I and for each uI,xju holds eventually. Now IU, so by the definition of κ(P), there exists dU such that dI. Then there exists kI such that dxj for all jk. Thus xjU for all jk. Hence xjU holds eventually. Thus U𝒪(𝒫).

Lemma 18.

If P is a complete lattice and yintκx, then xy, where intκx denotes the interior of x with respect to the S-topology.

Proof.

Let P be a complete lattice and yintκx. For any semiprime ideal I with yI, we have Iintκx. Thus there exists d(intκx)I. Therefore xd and then xI. Thus xy.

Proposition 19.

Let P be a strongly continuous lattice. Then xlim-infS  xj if and only if the net (xj)jJ converges to the element x with respect to κ(P). In particular, the lim-infS convergence is topological.

Proof.

By Proposition 17, 𝒪(𝒫)=κ(P), so if xlim-infSxj then (xj)jJ converges to the element x with respect to κ(P). Conversely, suppose that we have a net (xj)jJ which converges to the element x with respect to κ(P). For each yx, we have xyκ(P) from the definition of κ(P). Thus there exists kI such that xjy for all jk, and then yxj for all jk. Since P is strongly continuous, x=x. By Lemma 5, x is a semiprime ideal. Therefore we have ((xj)jJ,x)𝒫. That is, x  lim-infSxj.

Lemma 20.

Let P be a complete lattice. If the lim-infS convergence is topological, then P is strongly continuous.

Proof.

By Proposition 17, the topology arising from lim-infS convergence is the S-topology. Thus if the lim-infS convergence is topological, we must have that xlim-infSxj if and only if the net (xj)jJ converges to the element x with respect to κ(P). For any xP, let J={(U,n,a)N(x)××P:aU}, where N(x) consists of all S-open sets containing x, and define an order on J to be the lexicographic order on the first two coordinates. That is, (U,m,a)(V,n,b) if and only if V is a proper subset of U or U=V and mn. Let xj=a for each j=(U,n,a)J. Then it is easy to verify that the net (xj)jJ converges to the element x with respect to κ(P). Thus xlim-infSxj, and we conclude that there exists a semiprime ideal I such that xI and xju holds eventually for each uI. Let dI, then there exists k=(U,m,a)I such (V,n,b)=jk implies dxj=b. In particular, we have (U,m+1,b)(U,m,a)=k for all bU. Thus xUd. It follows that Ix. Furthermore, xintκd. By Lemma 18, dx and then Ix. Therefore x=Ix. Since I is a semiprime ideal with a supremum greater than or equal to x, it follows that xI. Hence I=x and so x=x. All these show that P is strongly continuous.

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

Theorem 21.

Let P be a complete lattice, then the following statements are equivalent.

P is strongly continuous.

The lim-infS convergence is the convergence for the S-topology; that is, for all xP and all nets (xj)jJ in Pxlim-infSxj if and only if the net (xj)jJ converges to the element x with respect to κ(P).

The lim-infS 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  that a mapping f:LM is said to be Scott continuous, if f is continuous from topological spaces (L,σ(L)) to (M,σ(M)). It is known that f is Scott continuous if and only if f preserves all directed suprema (see ). Now we give some new basic definitions.

Definition 22.

Let L,M be complete lattices. An order preserving mapping f:LM is called a semicontinuous mapping, if f preserves suprema of semiprime ideals; f is called a strongly semicontinuous mapping, if f is semicontinuous and for any IRd(L), f(I)Rd(M). A mapping f:LM is called S-continuous if it is continuous from topological spaces (L,κ(L)) to (M,κ(M)).

Lemma 23.

Let L,M be complete lattices. If f:LM is S-continuous, then f is order-preserving.

Proof.

Let xy in L. Suppose that f(x)f(y); then the S-open set V=Mf(y) contains f(x). Thus U=f-1(V) is a S-open neighborhood of x not containing y. But then xy as U is an upper set, a contradiction. Thus xy implies f(x)f(y).

Lemma 24.

Let P be a complete lattice and A=AP. Then A is S-closed if and only if IA holds for any IRd(P) with IA.

Lemma 25.

Let L,M be complete lattices. If f:LM is a strongly semicontinuous mapping, then f is S-continuous.

Proof.

Let A be an S-closed subset of M. First, f-1(A) is a lower set because f is order preserving and A is a lower set. For arbitrary semiprime ideal If-1(A), we have f(I)A. Since f is strongly semicontinuous, f(I)=f(I)=f(I) and f(I)Rd(M). By Lemma 24, f(I)=f(I)A. Therefore If-1(A). Again by Lemma 24, f-1(A) is S-closed. Thus f is S-continuous. This proves our result.

Proposition 26.

Let f:LM be an order preserving mapping from a strongly continuous lattice L to a complete lattice M. Then f is Scott continuous if and only if f is semicontinuous.

Proof.

Necessity. Obvious.

Sufficiency. Let D be arbitrary directed subset of L. Since f is order-preserving, f(D)f(D). Suppose that f(D)f(D). Then f(D)U=M(f(D))σ(M). Thus D=Df-1(U). Since L is a strongly continuous lattice, by Theorem  2.5 in , there exists a semiprime ideal ID such that I=Df-1(U). Hence f(I)=f(I)Uσ(M). It then follows that f(I)U. Thus there exist tU and dD such that tf(d). Therefore f(d)U. That is, f(d)f(D), which contradicts f(D)f(D). Hence f(D)=f(D) holds for all directed set D; thus f is Scott continuous.

Proposition 27.

Let f:LM be a map from a strongly continuous lattice L to a distributive lattice M. Then the following conditions are equivalent:

f is Scott continuous;

f is strongly semicontinuous;

f is S-continuous;

f is semicontinuous.

Proof.

( 1 ) ( 2 ) . Let f be Scott continuous. Then f is order preserving and preserves the supremum of all semiprime ideals. Let IRd(L). Then f(I) is an ideal of M. Since M is distributive, f(I) is the semiprime ideal of M.

( 2 ) ( 3 ) follows from Lemma 25.

( 3 ) ( 4 ) . Let f:LM be S-continuous. By Lemma 23, f is order-preserving. Let I be any semiprime ideal of L. Thus f(I)f(I). Assume that f(I)<f(I). Then f(I)U=M(f(I)) which is obviously S-open. Hence If-1(U)κ(L). Therefore If-1(U) and then there exists dI such that f(d)U. That is, f(d)f(I) a contradiction.

( 1 ) ( 4 ) follows from Proposition 26.

Corollary 28.

Let L,M be strongly continuous lattice. Then f is Scott continuous if and only if f is semicontinuous.

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

Proposition 29.

Let L be a semicontinuous lattice and let M be a complete lattice. If there is a surjective S-continuous mapping f:LM that preserves the relation , then M 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 D and E we will use 0E to denote the least element of E, [DE] to denote the set of all order-preserving maps from D to E, [D,E] to denote the set of all Scott continuous mappings from D to E, and [DE] to denote the set of all semicontinuous mappings from D to E. Obviously, [D,E][DE][DE], and they are all posets with respect to the pointwise order. In particular, [D,E] is a complete lattice. It is well known that the category CONT is Cartesian closed [2, Theorem II-2.12]. 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 D and E, the set [DE] is closed under taking supremum in [DE]; thus it is a complete lattice.

Proof.

Let (x)=0E for all xD. It is easy to show that is the bottom element of [DE].

Let [DE] and h(x)=ff(x) for any xD. Then h:DE is the supremum of in [D,E]. Now for semiprime ideal I in D, we have (6)h(I)=xIh(x)=xIff(x)=ff(I)=ff(I)=h(I). Thus =h[DE].

Let D and E be two complete lattices. For every xD and yE, define the interpolating step function [xy]:DE by (7)[xy](a)={y,xa,0E,otherwise.

Lemma 31.

Let D be a complete lattice for which the way-below relation satisfies the interpolation property. Then for all xD and yE, [xy] is Scott continuous.

Proof.

Clearly [xy] is order preserving. Let I be any ideal of D. If xI, then [x](I)=y. Since the way below relation satisfies the interpolation property, there exists cD such that xcI. Thus cI. Therefore [xy](I)=y=[xy](I). If x  I, then xz for all zI. Hence [xy](I)=[xy](I)=0E. Therefore [xy] 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 D={0,1}{xi:i=1,,n,}{xij:i,j=1,,n,} and E={0E,1E} with 0E<1E. The order on D is given by (see also Figure 1)

0x1 for all xD;

for each n, xnxn+1;

for each xi and each n, xi,nxi,n+1xi.

Then x11 in D, but x1  xi for i=2,,n,. Thus the way-below relation on D does not satisfy the interpolation property. Consider the mapping [x11E]. Let I=D{1}. Then I = 1. [x11E](I) = 1E, but [x11E](I) = 0E[x11E](I). Thus [x11E] is not Scott continuous.

Lemma 33.

Let D be a complete lattice for which the way-below relation satisfies the interpolation property and f[D,E]. Then for all xD and yE, yf(x) implies [xy]f.

Proof.

Let be any ideal of [D,E] with f. Then f(z)gg(z) for all zD. Then yf(x)gg(x). Since {g(x):g} is directed, there exists g* such that yg*(x). Now we claim that [xy]g*.

For arbitrary aD,xa implies xa and [xy](a)=yg*(x)g*(a). If x  a, then [xy](a)=0Eg*(a). Therefore [xy](a)g*(a) for all aD. That is, [xy]g*. This proves our claim. Hence [xy]. Therefore [xy]f.

The following proposition can be found in .

Proposition 34 (see [<xref ref-type="bibr" rid="B2">2</xref>]).

Let D and E be continuous lattices. Then for each f[D,E], f={[xy]:yf(x)}; hence [D,E] 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 D and E be continuous lattices and f,g[D,E]. Then gf holds in [D,E] if and only if there exist xiD,yiE, for i=1,2,,n, such that (8)yif(xi),gi=1n[xiyi].

Corollary 36.

If D and E are continuous lattices and gf holds in [D,E], then g(a)f(a) holds for all aD.

Proof.

With the notation of the previous corollary, we have that (9)gi=1n{[xiyi]:yif(xi)}. Let aD. For any i, if xia, then [xiyi](a)=yif(xi)f(a). Hence [xiyi](a)f(a). If xi  a, then [xiyi](a)=0Ef(a). Thus g(a)i=1n{[xiyi](a):yif(xi)}f(a). Therefore g(a)f(a). This completes our proof.

Lemma 37.

Let D be continuous and let E be strongly continuous. Then for any f[D,E], f={g[D,E]:gf} is a semiprime ideal of [D,E].

Proof.

Obvious f is an ideal because [D,E] is a complete lattice. Let hg1, hg2f. By Proposition 34, [D,E] is continuous, so the relation satisfies the interpolation property. Thus there exists f*f such that hg1f*f and hg2f*f. For arbitrary aD, by Corollary 36, (hg1)(a)=h(a)g1(a)f*(a), (hg2)(a)=h(a)g2(a)f*(a). Since E is strongly continuous, h(a)g1(a)f*(a) and h(a)g2(a)f*(a). Thus h(a)(g1(a)g2(a))f*(a). And then (h(g1g2))(a)=h(a)(g1(a)g2(a))f*(a). Since a is arbitrary, h(g1g2)f*f. And follows h(g1g2)f. Thus f is a semiprime ideal of [D,E]. This completes our proof.

Theorem 38.

If D is a continuous lattice and E is a strongly continuous lattice, then [D,E] is strongly continuous.

Proof.

By Proposition 34, [D,E] is a continuous lattice and f=f for all f[D,E], so [D,E] is a semicontinuous lattice. Let f,g[D,E] with gf=f. By Lemma 37, f is an semiprime ideal, so gf, that is, gf. By Theorem 2.5 of , [D,E] is a strongly continuous lattice. This completes our proof.

For cartesian closedness we adopt the elementary definition in .

Definition 39 (see [<xref ref-type="bibr" rid="B14">14</xref>]).

A category K is called a cartesian category if it satisfies the following conditions.

There is a terminal object.

Each pair of objects D and E of K has a product D×E with projections p1:D×ED and p2:D×EE.

Each pair of objects D and E of K has an exponentiation ED, that is, an object ED and an arrow eval: ED×DE with the property that for any f:X×DE, there is a unique arrow λf:XED such that the composite (10)X×Dλf×DED×DevalE is f.

Lemma 40.

The cartesian product of two strongly continuous lattices is strongly continuous.

Proof.

Let D,E be strongly continuous lattices. Then D,E are continuous lattices. By Proposition I-2.1 of , the product D×E of D and E is continuous. It is easy to verify that the projections p1:D×ED, p2:D×EE are Scott continuous. Let (a,b)(c,d) in D×E. Since D,E are strongly continuous lattices, cRd(D), dRd(E) and (c,d)=(c,d)=(c×d). Since both c and d are semiprime ideals, it follows that c×d is the semiprime ideal of D×E. Hence (a,b)(c,d). Thus D×E is strongly continuous by the Theorem 2.5 of . By Corollary 28, p1,p2 are semicontinuous. This completes our proof.

Theorem 41.

The category SCONT is cartesian closed in which the exponential ED is [D,E].

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, [D,E] is a strongly continuous lattice. Define eval: ED×DE by (11)eval(h,a)=h(a). Then eval is a Scott continuous mapping. Let X be a strongly continuous lattice, let f:X×DE be a semicontinuous (or Scott continuous) mapping, and define λf:XED by λf(x)(a)=f(x,a). For any directed set IX and aD, λf(I)(a)=f(I,a)=f(I×{a})={f(x,a):xI} because I×{a} is directed. Thus λf(I)(a)={λf(x,a):xI}=xIλf(x)(a). So λf(I)=λf(I). Thus λf is Scott continuous. And eval  (λf×idD)(x,a) = eval   (f(x),a)=f(x)(a)=f(x,a). So eval  (λf×idD)=f. Clearly λf is the unique morphism satisfying the condition. By Definition 39, the category SCONT is cartesian closed. The proof is completed.

Remark 42.

By , every distributive continuous lattice is strongly continuous. One can verify straightforwardly that for any two distributive continuous lattices S and T, the lattice [S,T] of Scott continuous mappings from S to T 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.

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