Topological properties of the binary supremum function

We consider the binary supremum function sup:Z×Z→Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\sup :Z\times Z\rightarrow Z$$\end{document} on a sup semilattice Z and its topological properties with respect to the Scott topology and the product topology. It is well known that this function is continuous with respect to the Scott topology on Z×Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Z\times Z$$\end{document}. We show that it is open as well. Isbell has constructed several examples of complete lattices Z such that the binary supremum function on Z is discontinuous with respect to the product topology. Of course, in these cases the Scott topology on Z×Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Z\times Z$$\end{document} is strictly finer than the product topology. This raises the question whether there exists a complete lattice Z such that the Scott topology on Z×Z\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Z\times Z$$\end{document} is strictly finer than the product topology and such that the binary supremum function is continuous even with respect to the product topology. We construct such a lattice. Finally, by using any of the examples constructed by Isbell, we show the following result: Bounded completeness of a complete lattice Z is in general not inherited by the dcpo C(X, Z) of continuous functions from X to Z where X is a topological space and where on Z the Scott topology is considered. On the other hand, we show that bounded completeness of Z is inherited by C(X, Z) if the topology on X is the Scott topology of a partial order.


Introduction
In this article two examples in the theory of partial orders are constructed. The first example concerns the question whether the binary supremum function on a sup semilattice is continuous with respect to the product topology of the Scott topology on each B Peter Hertling peter.hertling@unibw.de 1 Fakultät für Informatik, Universität der Bundeswehr München, Werner-Heisenberg-Weg 39, 85577 Neubiberg, Germany copy of the sup semilattice. The second example is a counterexample concerning the question whether bounded completeness of a complete lattice is inherited by the dcpo of continuous functions from some topological space to the complete lattice.
The Scott topology on a partially ordered set (short: poset) (X , ≤) has turned out to be important in many areas of computer science. Given two posets (X , ≤ X ) and (Y , ≤ Y ) we have two natural topologies on the product poset (X × Y , ≤ X ×Y ): • The product topology of the Scott topology on (X , ≤ X ) and of the Scott topology on (Y , ≤ Y ), • the Scott topology on (X × Y , ≤ X ×Y ).
It is well known that the second topology on X × Y is always at least as fine as the first topology, and that there are examples where the second topology is strictly finer than the first topology. The best known example of this kind seems to consist of the Johnstone space (compare [12]) for (X , ≤) and the dcpo of its Scott open subsets for (Y , ≤); see [ [10,11].
If the supremum of any two elements of a poset (X , ≤) exists (in this case the poset is called a sup semilattice) then the binary supremum function sup (2) : X × X → X is well defined. Is it continuous? It is easy to see and well known that it is continuous if on X × X the finer topology of the two topologies discussed above, the Scott topology, is considered. Isbell [10,11] has constructed several examples of complete lattices such that the binary supremum function is discontinuous with respect to the product topology on X × X . Of course, in these cases the Scott topology on Z × Z is strictly finer than the product topology. This raises the question (posed by an anonymous referee of [8]) whether perhaps the binary supremum function sup (2) : X × X → X is always discontinuous with respect to the product topology on X × X if this is different from the Scott topology on X × X . We answer this question negatively by providing an example of a complete lattice X such that the Scott topology on X × X is strictly finer than the product topology but nevertheless the function sup (2) : X × X → X is continuous with respect to the product topology on X × X ; see Theorem 7.4.
Our second example concerns bounded completeness of the dcpo of continuous functions from a topological space to a bounded complete dcpo. A poset is called bounded complete if each bounded subset has a supremum. The following question arises: If (Z , ≤) is a bounded complete dcpo and X a topological space, is then the dcpo C(X , Z ) of continuous functions from X to Z (where on Z the Scott topology is considered) bounded complete as well? It turns out that the answer to this question is yes if the topology on X is the Scott topology of some partial order on X . One may now speculate whether the answer is yes even for arbitrary topological spaces. But it is the second main goal of this article to show that in general the answer to this question is no. In Sect. 8 we shall even construct a topological space X and a complete lattice Z such that the dcpo of continuous functions from X to Z is not bounded complete; see Theorem 8.5. This construction makes essential use of a complete lattice X such that the binary supremum function sup (2) : X × X → X is discontinuous with respect to the product topology, as constructed by Isbell [10,11] (Theorem 7.1). Note that in [13, Page 26] the closely related question is discussed whether the pointwise supremum of two continuous functions f , g : X → Z for a topological space X and a complete lattice Z is again continuous. The authors observe that this is true if the product topology on Z × Z is identical with the Scott topology on Z × Z . Concerning the general case, they write "But in general, we cannot conclude that f ∨ g or f ∧ g are Scott continuous". But they do not give an example where the pointwise supremum f ∨ g of f and g is not continuous. In Theorem 8.5, the main result of Sect. 8, such an example is constructed.
After introducing some notation and collecting several elementary notions concerning partial orders in the following two sections, in Sect. 4 we modify the Johnstone space in order to obtain a considerably simpler partial order which is even a complete lattice and has additional properties that will be needed in the proof of Theorem 7.4 in Sect. 7. In Sect. 5 we first remind the reader of some elementary algebraic properties of the binary supremum function sup (2) : X × X → X and then show that it is not only continuous with respect to the Scott topology on X × X but also an open mapping with respect to the Scott topology on X × X . In Sect. 6 we present an observation by Wei and Shouli [14] saying that for certain posets one can characterize the property "sobriety" by a condition on the binary supremum function. In Sect. 7, first we shortly speak about the constructions by Isbell [10,11] of complete lattices (X , ≤) such that the function sup (2) : X × X → X is not continuous with respect to the product topology on X × X . Then we construct a complete lattice (X , ≤) such that the function sup (2) : X × X → X is continuous with respect to the product topology on X × X although this topology is different from the Scott topology on X × X . Finally, in Sect. 8, first we show that for any poset X and any bounded complete dcpo Z the dcpo C(X , Z ) of continuous functions from X to Z is bounded complete as well. Then we construct a topological space X and a complete lattice Z such that the dcpo of continuous functions from X to Z is not bounded complete and such that there exist two continuous functions f , g : X → Z whose pointwise supremum is not continuous.

Some notations
Let N := {0, 1, 2, . . .} denote the set of natural numbers, that is, of non-negative integers. For k ∈ N and a total function f : N → N let For sets X , Y the projections π 1 : X × Y → X and π 2 : X × Y → Y are defined as usual by π 1 (x, y) := x and π 2 (x, y) := y, for (x, y) ∈ X × Y .

Basic notions: partial orders and topology
In this section, for the convenience of the reader, we collect several elementary notions concerning partial orders and topological spaces, in particular the definition of the Scott topology.
Let (Z , ≤) be a poset. For a subset S ⊆ Z we define ↑ S := {z ∈ Z : (∃y ∈ S)y ≤ z} and ↓ S : An element z ∈ Z is called a supremum or least upper bound of a subset S ⊆ Z if it is an upper bound of S and if for all upper bounds y of S one has z ≤ y. Obviously, if a subset S ⊆ Z has a supremum, then this supremum is unique. Then we denote it by sup ≤ (S) or simply by sup(S) if it is clear which relation ≤ is meant. An element z ∈ Z is called a lower bound of a subset S ⊆ Z if S ⊆ ↑{z}. An element z ∈ Z is called an infimum or greatest lower bound of a subset S ⊆ Z if it is a lower bound of S and if for all lower bounds y of S one has y ≤ z. Obviously, if a subset S ⊆ Z has an infimum, then this infimum is unique. Then we denote it by inf ≤ (S) or simply by inf(S). The poset (Z , ≤) is called a sup semilattice if sup({x, y}) exists for all x, y ∈ Z . The poset (Z , ≤) is called a complete lattice if sup(S) exists for all subsets S ⊆ Z . Note that then also inf(S) exists for all subsets S ⊆ Z . Indeed, it is well known-compare, e.g., [6, Proposition O-2.2(i)]and straightforward to check that if sup(S) exists for all subsets S ⊆ Z then for any subset T ⊆ Z , the supremum of the set of all lower bounds of T is an infimum of T . A subset S ⊆ Z is called directed if it is nonempty and for any two elements x, y ∈ S there exists an upper bound z ∈ S of the set {x, y}. The poset (Z , ≤) is called a dcpo if for any directed subset S ⊆ Z there exists a supremum of S in Z . It is clear that any complete lattice is a dcpo. In Sect. 6 we will consider posets that are at the same time a sup semilattice and a dcpo. Such posets can be characterized as follows. Proof "2 ⇒ 1": Trivial. "1 ⇒ 2": Let us assume that (Z , ≤) is a sup semilattice and a dcpo. Let S ⊆ Z be a nonempty subset of Z . We wish to show that sup(S) exists. The set is a directed subset of Z . Therefore, sup(D) exists. Obviously, an element z ∈ Z is an upper bound of D if, and only if, it is an upper bound of S. Hence, sup(D) is a least upper bound of S as well, that means, sup(S) exists and is equal to sup(D).
Next, we introduce the Scott topology.  is a T 0 topology on Z .
The topology described in this lemma is called the Scott topology of (Z , ≤). Its elements are called Scott open. Let us denote this topology by σ (Z , ≤). For a topological space X and a point x 0 ∈ X let cl({x 0 }) be the closure of the set {x 0 }, that is, the intersection of all closed sets containing x 0 (thus, cl({x 0 }) is the smallest closed set containing x 0 ). The following example of a Scott open set is useful for the proof of the assertion in Lemma 3.3 that the Scott topology is T 0 and will be useful later as well.

Products of partial orders
Let (X , ≤ X ) and (Y , ≤ Y ) be two posets. On the product Z := X × Y we define a binary relation ≤ X ×Y ⊆ Z × Z by for any x, x ∈ X and y, y ∈ Y .
If one, and the both, of these two conditions are satisfied then Proof It is obvious that ≤ X ×Y is a partial order. The second and the third assertion are straightforward to check as well. And the fourth assertion follows directly from the first and the third assertion.
For two posets (X , ≤ X ) and (Y , ≤ Y ), we have two natural topologies on the product X × Y : the Scott topology σ (X × Y , ≤ X ×Y ) and the product topology σ (X , ≤ X ) × σ (Y , ≤ Y ). What is the relation between these two topologies? It is well-known that the Scott topology is always at least as fine as the product topology. It is also well known that in general the converse of Lemma 4.2 is not true. Since in the following we are going to define a similar but simpler partial order (that is even a complete lattice) for comparison in the following example we define the Johnstone space [12] and sketch the well-known proof of Proposition 4.3. Then the pair (X , ≤ X ) is a dcpo. It is called the Johnstone space. Let Y be the set of all Scott open subsets of X , and let the binary relation ≤ Y on Y be set-theoretic inclusion. By Example 3.2 (Y , ≤ Y ) is a complete lattice, hence, a dcpo as well. One can check that the subset E ⊆ X × Y defined by is open in the Scott topology on (X × Y , ≤ X ×Y ) but not open in the product topology of the Scott topology on (X , ≤ X ) and the Scott topology on (Y , ≤ Y ).
For later use, we explicitly formulate one general step in the previous example. Proof Let (x, y) ∈ E and (x , y ) ∈ X × Y be a pair with (x, y) ≤ X ×Y (x , y ). Then x ≤ X x and y ≤ Y y . Furthermore, x ∈ y. As y is upwards closed, we obtain x ∈ y. And y ≤ Y y means y ⊆ y , hence, x ∈ y , hence, (x , y ) ∈ E. We have shown that E is ≤ X ×Y -upwards closed. Now let S ⊆ X × Y be a directed subset such that sup ≤ X ×Y (S) exists and is an element of E. According to Lemma 4.1(3) the suprema sup ≤ X (π 1 [S]) and sup ≤ Y (π 2 [s]) exist and satisfy (sup 2). We conclude that there is some y ∈ π 2 [S] with sup ≤ X (π 1 [S]) ∈ y. By Lemma 4.1 (2) the set π 1 [S] is a directed subset of X . As y is a Scott open subset of X we conclude that there exists some x ∈ π 1 [S] ∩ y. The fact x ∈ π 1 [S] implies that there exists some y ∈ Y with ( x, y) ∈ S. Analogously, the fact y ∈ π 2 [S] implies that there exists some x ∈ X with ( x, y) ∈ S. As S is directed there exists some (x, y) ∈ S with ( x, y) ≤ X ×Y (x, y) and ( x, y) ≤ X ×Y (x, y). Finally, similarly as in the proof above that E is upwards closed, from x ∈ y we conclude x ∈ y, hence, (x, y) ∈ E. This shows S ∩ E = ∅. We have shown that E is a Scott open subset of X × Y .
By a modification (which is mostly a simplification) of the definition of the Johnstone space we can prove the following slight improvement of Proposition 4.3. It will be used in the proof of Theorem 7.4 in Sect. 7.

Proposition 4.6
There exist complete lattices (X , ≤ X ) and (Y , ≤ Y ) such that the following three conditions are satisfied: 1. the Scott topology on (X × Y , ≤ X ×Y ) is strictly finer than the product topology on X × Y (of the Scott topology on (X , ≤ X ) and the Scott topology on (Y , ≤ Y )), 2. but the Scott topology on X × X is identical with the product topology on X × X, 3. and the Scott topology on Y × Y is identical with the product topology on Y × Y .
In the following example we shall define our modification (X , ≤ X ) of the Johnstone space. For (Y , ≤ Y ) we will then take the set of Scott open subsets of X and the subset relation. Then we formulate and prove several lemmas that collect the important properties of (X , ≤ X ) and (Y , ≤ Y ) and from which we then deduce Proposition 4.6.

Example 4.7
Our modification of the Johnstone space is defined as follows. Let where ⊥ and are different elements that are not elements of N × N. We define a binary relation ≤ X on X by for all x, y ∈ X . Compare Fig. 1.
The reader is invited to convince himself that this poset is indeed simpler than the Johnstone space by comparing the graphical representation in Fig 8 The construction in Example 4.7 was obtained by the author by slightly simplifying a similar construction in the preprint [8]. In the recent article [2] Chen, Kou, and Lyu arrive at the same construction. They also observe that it is a complete lattice and explicitly describe [2, Example 4.13] the lattice of its Scott open subsets in the same way as this is done in Lemma 4.9 (2) and 4.9 (3).
In the following lemma we show that (X , ≤ X ) is a complete lattice and give an explicit description of all Scott open subsets of X . 2. For any total function f : N → N the set Thus, sup(S) exists for every subset S ⊆ X . Hence, (X , ≤ X ) is a a complete lattice. 2. Let f : N → N be a total function. It is clear that U f is upwards closed. Let S ⊆ X be a directed subset such that sup(S) exists and sup(S) ∈ U f . We wish to show that there exists an element s ∈ S ∩ U f . This is clear if ∈ S because then sup(S) = ∈ U f , hence, ∈ S ∩ U f . So, let us assume that / ∈ S. Can there exist numbers i, j, p, q ∈ N with i = j and (i, p) ∈ S and ( j, q) ∈ S? No. As S is directed, in that case there would have to exist some s ∈ S with (i, p) ≤ X s and ( j, q) ≤ X s. But the only element s of X that satisfies (i, p) ≤ X s and ( j, q) ≤ X s is the element s = . Hence, we would get ∈ S. So, we can conclude that there Finally, let us consider the case that S is finite and S ⊆ {⊥} ∪ ({i} × N). As ⊥ is not an element of U f , the set S can neither be empty nor be equal to {⊥} because in both cases one would arrive at sup(S) = ⊥. Thus, we obtain sup(S) = (i, max{ p : (i, p) ∈ S}). But in this case we have not only So, we conclude that U = ∅ and U = X . According to Lemma 4.9 (3) there exists some function f : 11 It is instructive to discuss the previous lemma in the context of some further domain-theoretic notions. Let (Z , ≤) be a poset.
• For x, y ∈ Z we write x y and say that x is way-below y if for every directed set S ⊆ Z such that sup(S) exists and satisfies y ≤ sup(S) there exists an element In the following, as in Lemma 4.10, we consider the complete lattice (X , ≤ X ) defined in Example 4.7 and the complete lattice . Secondly, it is well known that the assertion proved in the previous lemma (that the set E := {(x, y) ∈ X × Y : x ∈ y} is not open with respect to the product topology on X × Y ) is equivalent to the assertion that (Y , ≤ Y ) is not continuous, thus, to the assertion that (X , ≤ X ) is not core-compact (indeed, the conditions (4) and (6)  Note also that, if (σ (X , ≤ X ), ⊆) were continuous then, due to [6,, this would imply that for any dcpo (Z , ≤ Z ) the product topology on X × Z would coincide with the Scott topology on X × Z . But, we have just seen that (σ (X , ≤ X ), ⊆) is not continuous. And, indeed, Lemmas 4.5 and 4.10 show that the product topology on X × Y (where again (Y , ≤ Y ) := (σ (X , ≤ X ), ⊆)) does not coincide with the Scott topology on X × Y . Nevertheless, the Scott topology on X × X coincides with the product topology on X × X .

Lemma 4.12
Let (X , ≤ X ) be the complete lattice defined in Example 4.7 (see Lemma 4.9 (1)). Then the Scott topology on X × X coincides with the product topology on X × X.
similarly for x 2 . So, we have to consider nine cases.
Case 1 x 1 = x 2 = ⊥. Then, as W is upwards closed, it must be equal to X × X . Hence, it is open in the product topology.
Case 2 x 1 = ⊥ and x 2 = . Similarly as in the proof of Lemma 4.9 (3) one shows that there exists a function f : Case 4 x 1 = and x 2 = ⊥. This case is symmetric to Case 2 and is treated in the same way.
Case 5 x 1 ∈ N × N and x 2 = ⊥. This case is symmetric to Case 3 and is treated in the same way.
We define the functions f and g step by step and, additionally, in parallel two increasing sequences of finite sets N 1 ⊆ N and N 2 ⊆ N (containing the numbers on which f and g have been defined already) as follows. In the first step we define Then we repeat the following infinitely often: Let us show that these definitions make sense and define functions f and g with the desired properties. By induction we show that after each step we have ((k 1 , f (k 1 )), (k 2 , g(k 2 ))) ∈ W , for all k 1 ∈ N 1 and all k 2 ∈ N 2 . It is clear that this is true after the first step, that is, for N 1 = {i 1 } and N 2 = {i 2 }. Let us assume by induction that this is true after some step. We show that it is true after the next step as well. Note that for every k 2 ∈ N 2 and j 1 : As W is upwards closed and N 2 is finite in every step of the definition of f , f ( j 1 ) is well-defined for j 1 = min(N\ N 1 ), and we obtain (( j 1 , f ( j 1 )), (k 2 , g(k 2 ))) ∈ W , for this j 1 and all k 2 ∈ N 2 . The same argument applies to the definition of g( j 2 ). This ends the proof by induction. Thus, by the steps above we have defined two functions f : N → N and g : N → N such that ((k 1 , f (k 1 )), (k 2 , g(k 2 ))) ∈ W , for all k 1 ∈ N and all k 2 ∈ N. This shows U f × U g ⊆ W . Finally, it is clear that This ends the treatment of Case 6.
Case 7 x 1 ∈ N × N and is an element of W . Hence, there exists some number p with (x 1 , (0, p)) ∈ W . By Case 6 there exist functions f : N → N and g : N → N with x 1 ∈ U f , (0, p) ∈ U g and U f × U g ⊆ W . As ∈ U g is clear, this ends the treatment of Case 7.
Case 8 x 1 = and x 2 ∈ N × N. This case is symmetric to Case 7 and is treated in the same way.
is an element of W . Hence, there exists some number p with ((0, p), ) ∈ W . By Case 7 there exist functions f : N → N and As ∈ U f is clear, this ends the treatment of Case 9.
In the following let (X , ≤ X ) be the complete lattice defined in Example 4.
Therefore, we can assume without loss of generality that dom(g) contains exactly one number. Let i 0 be this number, and let p 0 := g(i 0 ). Then It is clear that this set is upwards closed. Let S ⊆ Y be a directed subset with sup(S) ∈ V g . If sup(S) = X then we can conclude as above in the proof of the first assertion of this lemma that X ∈ S. So, in this case we have S ∩ V g = ∅. Let us assume that sup(S) = U f for some total function f : N → N with U f ∈ V g , hence, with f (i 0 ) ≤ p 0 , hence, with (i 0 , p 0 ) ∈ U f . As sup(S) is the union of all sets that are elements of S there must exist some U ∈ S with (i 0 , p 0 ) ∈ U . As U = X is impossible (this would imply sup(S) = X ) we conclude that there is some function h : N → N with U = U h and h(i 0 ) ≤ p 0 . We obtain U ∈ S ∩ V g , hence, S ∩ V g = ∅.
Actually, the Scott open subsets of Y identified in the previous lemma already form a base of the Scott topology of Y . Thus, the Scott topology of (Y , ≤ Y ) has a countable base. In the case U = ∅ we obtain Z = Y and can use V := Y . In the case U = X we can simply use V := {X }. In the following we consider the case U / ∈ {∅, X }. Then, according to Lemma 4.9 (3) there exists a function f : N → N with U = U f . We claim that there exists a number k ∈ N with V f k ⊆ Z. For the sake of a contradiction, let us assume that this is not the case. Then there exists a sequence V : N → Y such that, for all k ∈ N, V (k) ∈ V f k and V (k) / ∈ Z. For every k either V (k) = X or there exists a function g k : N → N with V (k) = U g k . In the first case we define g k := f and observe U g k ⊆ X = V (k) and U g k ∈ V f k . So, for every k ∈ N there exists a function g k : N → N with U g k ⊆ V (k) and U g k ∈ V f k . We are going to construct a sequence W : In order to construct such a sequence, we define a function g : N → N by , for all j. Furthermore, for all k, j ∈ N, g k ( j) ≤ g( j). Indeed, this is clear for k ≤ j by definition of g, and for j < k we observe that We observe that for all k, j ∈ N and A topological space (X , τ ) is called first-countable if for every point x ∈ X there exists a sequence B : N → τ of open sets such that, for all k ∈ N, x ∈ B(k) and, for any open subset U ⊆ X with x ∈ U there exists a k ∈ N with B(k) ⊆ U . Matthew de Brecht made the following observation, explained in detail in the blog of Jean Goubault-Larrecq [3]. [3]] Let (Z 1 , ≤ 1 ) and (Z 2 , ≤ 2 ) be posets such that the Scott spaces (Z 1 , σ (Z 1 , ≤ 1 )) and (Z 2 , σ (Z 2 , ≤ 2 )) are first-countable. Then the Scott topology on Z 1 × Z 2 coincides with the product topology on Z 1 × Z 2 . Hence, the Scott topology on X × Y is strictly finer than the product topology on X × Y . By Lemma 4.12 the Scott topology on X × X coincides with the product topology on X × X , and by Corollary 4.16 the Scott topology on Y × Y coincides with the product topology on Y × Y .

Elementary properties of the binary supremum function
Let X be a set. If ≤ is a partial order on X such that (X , ≤) is a sup semilattice then we define the binary supremum function sup (2) ≤ : X × X → X on X by for x, y ∈ X . If the relation ≤ is clear from the context then instead of sup ≤ we simply write sup (2) . For completeness sake let us mention the following well known algebraic characterisation of the binary supremum function. We call a function f : f (y, z)).
For a function f : X × X → X let us define a binary relation ≤ f ⊆ X × X by x ≤ f y : ⇐⇒ y = f (x, y), for x, y ∈ X . Lemma 5.1 Let X be a nonempty set.

If ≤ is a partial order on X such that (X , ≤) is a sup semilattice then the function f := sup
(2) ≤ is idempotent, commutative, and associative, and the relation ≤ f is identical with ≤.
2. If f : X × X → X is an idempotent, commutative, and associative function then the binary relation ≤ f on X is a partial order such that (X , ≤ f ) is a sup semilattice, and f is the binary supremum function of this sup semilattice.
This is well-known, and the proof is straightforward. We omit it. This lemma shows that the transition from a sup semilattice (X , ≤) (an order-theoretic object) to the pair (X , sup (2) ≤ ) consisting of a set X and a function with the three properties above (an algebraic object) on the one hand and the transition from a pair (X , f ) consisting of a set X and a function f with the three properties above (an algebraic object) to the sup semilattice (X , ≤ f ) (an order-theoretic object) on the other hand are inverse to each other. In this article we work with the order-theoretic description.
Let us have a first look at topological properties of the binary supremum function on a sup semilattice. Let (X , τ X ) and (Y , τ Y ) be topological spaces. A function f : Often the following fundamental characterization of Scott continuous functions between partial orders is useful.  The continuity of sup (2) is well known [13, p. 26]. It seems that the openness of sup (2) has not been stated so far.

f is Scott continuous, that is, continuous with respect to the Scott topology on X and the Scott topology on Y . 2. f is monotone
Proof For completeness sake, we prove the continuity of sup (2) . We use Lemma 5.2. It is obvious that sup (2) is monotone. Let S ⊆ X × X be a directed set such that sup(S) exists. We have to show that sup (2) (sup(S)) is a supremum of sup (2) [S]. For any (x, x ) ∈ S we have (x, x ) ≤ X ×X sup(S), hence, due to the monotonicity of sup (2) , also sup (2) (2) (sup(S)). This shows that sup (2) (sup(S)) is an upper bound of sup (2) [S]. Now let y ∈ X be an upper bound of sup (2) [S]. Then for all (x, x ) ∈ S we have sup (2) (x, x ) ≤ y, hence, x ≤ y and x ≤ y, hence, (x, x ) ≤ X ×X (y, y). This implies sup(S) ≤ X ×X (y, y) and, due to the monotonicity of sup (2) , sup (2) (sup(S)) ≤ sup (2) (y, y) = y. Thus, sup (2) (sup(S)) is indeed a supremum of sup (2) [S].
For the openness of sup (2) let us consider a Scott open subset U ⊆ X × X . We wish to show that sup (2) [U ] is a Scott open subset of X . First we show that sup (2) [U ] is upwards closed. This is clear if U is empty. Let us assume U = ∅, and let us consider some (x, x ) ∈ U and some y ∈ X with sup (2) (x, x ) ≤ y. Then x ≤ y and x ≤ y, hence, (x, x ) ≤ X ×X (y, y). As U is upwards closed we conclude (y, y) ∈ U , hence, y = sup (2) (y, y) ∈ sup (2) [U ]. We have shown that sup (2) [U ] is upwards closed. Now let us assume that S ⊆ X is a directed set such that sup(S) exists and is an element of sup (2) [U ]. Then the set P := {(s, s) ∈ X × X : s ∈ S} is a ≤ X ×X -directed set whose ≤ X ×X -supremum exists and is equal to (sup(S), sup(S)). Remember that sup(S) ∈ sup (2) [U ]. Let (x, x ) ∈ U be an element of U with sup(S) = sup (2) (x, x ). We conclude that x ≤ sup(S) and x ≤ sup(S), hence, (x, x ) ≤ X ×X (sup(S), sup(S)). As U is upwards closed we conclude (sup(S), sup(S)) ∈ U . Thus, sup ≤ X ×X (P) ∈ U . As U is Scott open we obtain P ∩ U = ∅. Thus, there exists some y ∈ S with (y, y) ∈ U . This implies y = sup (2) (y, y) ∈ sup (2) [U ]. Thus, S ∩ sup (2) [U ] = ∅. We have shown that sup (2) [U ] is Scott open.

Corollary 5.4 Let (X , ≤) be a sup semilattice. Then the function sup (2) : X × X → X is open with respect to the product topology on X × X of the Scott topologies on both copies of X (on the left hand side) and the Scott topology on X (on the right hand side).
Proof This follows from Proposition 5.3 and from Lemma 4.2.

Sobriety and the binary supremum function
In this section we present the observation by Wei and Shouli [14] that a slightly weaker property than continuity of the binary supremum function with respect to the product topology is equivalent to the property "sober", for suitable posets. The following definitions are copied from [6,, [7,Sect. 8.2]. Let X be a topological space. We call a subset C ⊆ X irreducible if it is nonempty and, for any closed subsets A, B ⊆ X the following implication is true: if C ⊆ A ∪ B then C ⊆ A or C ⊆ B. It is clear that the closure cl({x 0 }) of a point x 0 is irreducible, for any x 0 ∈ X . A topological space X (or its topology) is called sober if it is a T 0 space and for every irreducible closed subset C ⊆ X there exists a point x 0 ∈ X with C = cl({x 0 }). Note that the T 0 property of a topological space X implies for x 0 , We wish to analyze which posets that are dcpos and sup semilattices (compare Lemma 3.1) have a sober Scott topology. The following lemma is implicitly contained in the proof of [6, Proposition II-1.11]. Lemma 6.1 Let (X , ≤) be a dcpo and consider on X the Scott topology. For a closed subset C ⊆ X the following two conditions are equivalent.

C is directed. 2. There exists a point x 0 ∈ X with C = cl({x 0 }).
Proof Let x 0 ∈ X . On the one hand, in Example 3.4 we already mentioned that cl({x 0 }) = ↓{x 0 }. And it is clear that the set ↓{x 0 } is directed. On the other hand, if a set C is directed then sup(C) exists because we assume that (X , ≤) is a dcpo. And if C is closed and directed then sup(C) ∈ C. But in that case it is clear that C = ↓{sup(C)}.
It is well known that, if (X , ≤) is a dcpo and a sup semilattice and the function sup (2) : X × X → X is continuous with respect to the product topology on X , then X with the Scott topology is sober; see [5,, [6, Corollary II-1.12]. Wei and Shouli [14] observed that by weakening the continuity requirement on sup (2) one can obtain a condition that is equivalent to sobriety, under sufficiently strong further assumptions. The following theorem was essentially stated by them with slightly different assumptions. Theorem 6.2 Let (X , ≤) be a poset that is a dcpo and a sup semilattice. Then the following two conditions are equivalent.

The Scott topology σ (X , ≤) is sober. 2. For every Scott open subset W
X whose complement is irreducible, the set (sup (2) ) −1 [W ] ⊆ X × X is open in the product topology on X × X.
Proof "(1) ⇒ (2)": Let (X , σ (X , ≤)) be sober. Let W X be a proper Scott open subset of X whose complement C := X \ W is irreducible. Let x, y ∈ X be two elements with sup (2) (x, y) =: z ∈ W . We wish to show that there are Scott open subsets U , V ⊆ X with x ∈ U , y ∈ V and sup (2) [U × V ] ⊆ W . As C is irreducible and X is sober there exists a point c ∈ X with C = cl({c}). If x and y would satisfy x ≤ c and y ≤ c then their supremum z would satisfy z ≤ c as well, that is, z ∈ ↓{c} = cl({c}) = C. Thus, we observe that x c or y c. Without loss of generality we assume x c, hence, x / ∈ ↓{c} = cl({c}) = C, hence, x ∈ W . But now the open sets U := W and V := X satisfy x ∈ U , y ∈ V and sup (2) [U ×V ] = sup (2) [W × X ] ⊆ W . "(2) ⇒ (1)": Let us assume that (2) is true. Let C ⊆ X be a closed and irreducible subset of X . According to Lemma 6.1 it is sufficient to show that C is directed. Let us consider two elements x, y ∈ C. We claim that the supremum z := sup (2) (x, y) is an element of C. Suppose that this is not the case. Then z ∈ W := X \C. By Condition (2) there exist Scott open subsets U , V ⊆ X with x ∈ U , y ∈ V and sup (2) [U × V ] ⊆ W . As U and V are upwards closed we obtain sup (2) [U × V ] = U ∩ V . Hence, we have U ∩ V ⊆ W . For the closed sets A := X \ U and B := X \ V that means C ⊆ A ∪ B. As C is irrreducible this implies C ⊆ A or C ⊆ B, hence, U ⊆ W or V ⊆ W . We conclude x ∈ W or y ∈ W , in contradiction to x, y ∈ C.

Continuity or discontinuity of the supremum function with respect to the product topology on a complete lattice
Let (X , ≤) be a sup semilattice. Proposition 5.3 says that the binary supremum function is continuous with respect to the Scott topology on X × X . This leads to the question whether it is necessarily even continuous with respect to the product topology on X × X as well. This question was answered negatively by Isbell.
Theorem 7.1 (Isbell [10,11]) There exists a complete lattice (Z , ≤) such that the binary supremum function sup (2) : Z × Z → Z is not continuous with respect to the product topology on Z × Z of the Scott topology on each copy of Z (on the left hand side) and the Scott topology on Z (on the right hand side).
In fact, in [10] Isbell constructed a complete lattice such that this lattice with the Scott topology is not sober. It follows from [5, Corollary II. [1][2][3][4][5][6][7][8][9][10][11][12] as well as [6, Corollary II-1.12] or Theorem 6.2 that the binary supremum function on this lattice is not continuous with respect to the product topology. This result by Isbell leads to the question (posed by Hoffmann, according to [11]) whether there exists a sober complete lattice such that the binary supremum function is discontinuous with respect to the product topology. This question was answered positively by Isbell, who constructed such a lattice [11].
The observation described in the following lemma was made by Isbell in [11] for complete lattices. This fact was rediscovered by the author [8, Proof of Theorem 4.2], who at the time of writing [8] was not aware of the articles [10,11] by Isbell.

Lemma 7.2
Let (X , ≤ X ) and (Y , ≤ Y ) be sup semilattices such that ⊥ X := inf(X ) and ⊥ Y := inf(Y ) exist and such that the product topology on X × Y is different from the Scott topology on X × Y . Then Z := X × Y with ≤ X ×Y is a sup semilattice such that the binary supremum function sup (2) : Z × Z → Z is discontinuous with respect to the product topology on Z .
For completeness sake we give the proof. First, we formulate and prove another simple lemma.
of E under the function sup (2) is not open in the product topology on Z × Z (of the Scott topologies on each copy of Z ). For the sake of a contradiction, let us assume that D is open in the product topology on Z × Z . We are going to show that this assumption would imply that E is open in the product topology on Z = X × Y . Let us fix some element (x (0) , y (0) ) ∈ E. It is sufficient to show that the assumption that D is open in the product topology on Z × Z implies that there exist a Scott open subset F ⊆ X and a Scott open subset G ⊆ Y with x (0) ∈ F, with y (0) ∈ G and with F ×G ⊆ E. How do we arrive at such sets? Note that sup (2) ((x (0) , ⊥ Y ), (⊥ X , y (0) )) = (x (0) , y (0) ) ∈ E, hence, Our assumption that D is open in the product topology on Z × Z implies that there exist two Scott open subsets D 1 , y (0) ) ∈ D 2 , and with D 1 × D 2 ⊆ D. By Lemma 7.3 the sets are Scott open subsets of X resp. of Y . It is clear that In view of Theorem 7.1 one might even suspect that the binary supremum function is necessarily discontinuous with respect to the product topology if the product topology is different from the Scott topology. The question whether this is the case or not was posed by an anonymous referee of [8]. The following theorem shows that this is not the case. The binary supremum function can be continuous with respect to the product topology even if the product topology is strictly weaker than the Scott topology.
Theorem 7.4 There exists a complete lattice (Z , ≤) such that the product topology on Z × Z of the Scott topology on each copy of Z is different from the Scott topology on Z × Z , and the binary supremum function sup (2) : Z × Z → Z is still continuous with respect to the product topology on Z × Z of the Scott topology on each copy of Z (on the left hand side) and the Scott topology on Z (on the right hand side).
Proof Let (X , ≤ X ) and (Y , ≤ Y ) be complete lattices as in Proposition 4.6. We will construct a suitable Z as the direct sum of X and Y except that we add an additional 638 P. Hertling   Fig. 2 The complete lattice Z in the proof of Theorem 7.4, constructed out of two complete lattices X and Y smallest and an additional largest element in order to make Z a complete lattice. Formally, this can be done as follows. Let where ⊥ and are different elements that are not elements of (X × {0}) ∪ (Y × {1}), and let the binary relation ≤ Z on Z be defined by z ≤ Z z : ⇐⇒ z = ⊥ or z = or (∃x, x ∈ X )(z = (x, 0) and z = (x , 0) and x ≤ X x ) or (∃y, y ∈ Y )(z = (y, 1) and z = (y , 1) and y ≤ Y y ), for all z, z ∈ Z .
We come to the correctness proof. In fact, for notational simplicity, in the following we will identify X with X × {0} and Y with Y × {1}. In other words, we will assume that ⊥ = and that the three sets X , Y , and {⊥, } are pairwise disjoint. Then the pair (Z , ≤ Z ) consisting of a set Z and a binary relation ≤ Z on Z can be defined as follows.
and z ≤ Z z : ⇐⇒ z = ⊥ or z = or (z ∈ X and z ∈ X and z ≤ X z ) or (z ∈ Y and z ∈ Y and z ≤ Y z ), for all z, z ∈ Z . See Fig. 2.
It is straightforward to check that (Z , ≤ Z ) is a partial order. And for any subset S ⊆ Z the supremum of S exists: Hence, (Z , ≤ Z ) is a complete lattice. Next, we show that the Scott topology on Z × Z is strictly finer than the product topology on Z × Z (that is, than the product topology on Z × Z of the Scott topology on each copy of Z ). Let E ⊆ X × Y be a Scott open subset of X × Y that is not open with respect to the product topology on X × Y . We claim that the set is a Scott open subset of Z × Z and not open with respect to the product topology on and, furthermore, z 1 ∈ X ⇒ x ≤ X z 1 , and, similarly, z 2 ∈ Y ⇒ y ≤ Y z 2 . If z 1 = then (z 1 , z 2 ) ∈ (Y ∪ { }) × Z ⊆ G, and similarly, if z 2 = then (z 1 , z 2 ) ∈ Z × (X ∪ { }) ⊆ G. So, let us assume that z 1 ∈ X and z 2 ∈ Y . Then x ≤ X z 1 and y ≤ Y z 2 , hence (x, y) ≤ X ×Y (z 1 , z 2 ). As E is ≤ X ×Y -upwards closed we conclude (z 1 , z 2 ) ∈ E ⊆ G. We have shown that G is ≤ Z ×Z -upwards closed.
Next, let S ⊆ Z × Z be a directed set such that sup ≤ Z ×Z (S) is an element of G (note that sup Z ×Z (S) exists because Z × Z is a complete lattice by Lemma 4.1 (4)). We wish to show that S ∩ G = ∅. For the sake of a contradiction, let us assume that We still have to show that G is not open with respect to the product topology on Z × Z . For the sake of a contradiction, let us assume that G is open with respect to the product topology on Z × Z . We are going to show that this implies that the subset E of X × Y is open with respect to the product topology on X × Y . Let us consider an arbitrary element (x, y) of E. We wish to show that there exist Scott open subsets U ⊆ X and V ⊆ Y with x ∈ U , y ∈ V , and U × V ⊆ E. We make use of the fact that (x, y) ∈ E implies (x, y) ∈ G. As by assumption G is open with respect to the product topology on Z × Z there exist Scott open subsets U ⊆ Z and V ⊆ Z with (x, y) ∈ U × V and U × V ⊆ G. Then the set U := U ∩ X is a Scott open subset of X , and the set V Hence, G is indeed not open with respect to the product topology on Z × Z .
Finally, we have to show that the function sup (2) : Z × Z → Z is continuous with respect to the product topology on Z × Z . Let W ⊆ Z be a Scott open subset. It is sufficient to show that (sup (2) ) −1 [W ] is open in the product topology on Z × Z . We can assume that W is not empty. Let (z 1 , z 2 ) ∈ Z × Z be a pair with sup (2) It is sufficient to show that there exist Scott open subsets W 1 , W 2 ⊆ Z with z 1 ∈ W 1 , z 2 ∈ W 2 and W 1 × W 2 ⊆ (sup (2) ) −1 [W ]. We distinguish several cases for (z 1 , z 2 ).
Case 2 z 2 = ⊥. This case is symmetric to Case 1 and is treated in the same way. Case 4 z 2 = . This case is symmetric to Case 3 and is treated in the same way.
Case 6 z 1 ∈ Y and z 2 ∈ X . This case is symmetric to Case 5 and is treated in the same way.
Case 7 z 1 ∈ X and z 2 ∈ X . By Proposition 5.3 the function sup (2) : Z × Z → Z is continuous with respect to the Scott topology on Z × Z . Hence, (sup (2) is a Scott open subset of Z × Z . We claim that the set W := (sup (2) is a Scott open subset of X × X (that means, Scott open with respect to ≤ X ×X ). This follows directly from the fact that the restriction of ≤ Z ×Z to X × X is identical with ≤ X ×X . By the second condition in Proposition 4.6 the Scott topology on X × X is identical with the product topology. Hence, the subset W of X × X is open in the product topology. Furthermore (z 1 , z 2 ) ∈ W . Hence, there exist ≤ X -Scott open subsets U 1 , U 2 ⊆ X with z 1 ∈ U 1 , z 2 ∈ U 2 and U 1 × U 2 ⊆ W . The sets W 1 := U 1 ∪ { } and W 2 := U 2 ∪ { } are ≤ Z -Scott open subsets of Z and satisfy z 1 ∈ W 1 , z 2 ∈ W 2 , and W 1 × W 2 ⊆ (sup (2) Case 8 z 1 ∈ Y and z 2 ∈ Y . This case is treated in the same way as Case 7, using the third condition in Proposition 4.6 instead of the second condition.

Failure of bounded completeness for the directed complete partial order of Scott continuous functions
Let us consider some poset (Z , ≤) and an arbitrary topological space X . We call a function f : X → Z Scott continuous if it is continuous with respect to the given topology on X and the Scott topology on Z . Let C(X , Z ) denote the set of all Scott continuous functions f : X → Z . It is clear that the binary relation ≤ C defined on X by , is a partial order, that is, (C(X , Z ), ≤ C ) is a poset.  Proof Let (Z , ≤) be a poset. We have already seen that then (C(X , Z ), ≤ C ) is a poset as well.
1. Let ∈ Z be an upper bound for Z . The constant function f : X → Z with f (x) := for all x ∈ X is Scott continuous, thus, an element of C(X , Z ), and it is an upper bound for C(X , Z ). 2. Let ⊥ ∈ Z be a least element of Z . The constant function f : X → Z with f (x) := ⊥ for all x ∈ X is Scott continuous, thus, an element of C(X , Z ), and it is a lower bound for C(X , Z ).
The following question arises: If (Z , ≤) is bounded complete, is then C(X , Z ) bounded complete as well? This is true if the topology considered on X is the Scott topology induced by a partial order relation on X . Proof Let F ⊆ C(X , Z ) be a ≤ C -bounded set. Then for each x ∈ X the set Z ) is a ≤ C -upper bound of F then, for every x ∈ X , the element h(x) ∈ Z is a ≤ Z -upper bound of the set F(x)). Since we assume (Z , ≤ Z ) to be bounded complete, we can define a function g : X → Z by g(x) := sup(F(x)).
First, we show that this function g is Scott continuous. We use the characterization in Lemma 5.2. First we show that g is monotone. Let us consider x, y ∈ X with x ≤ X y. As each function f ∈ F is monotone, we have for each function f ∈ F f (x) ≤ Z f (y) ≤ Z g(y), hence, g(x) ≤ Z g(y). Thus, g is monotone. Now, let S ⊆ X be a directed set such that sup(S) exists. For every s ∈ S we have g(s) ≤ Z g(sup(S)) because g is monotone, hence, g(sup(S)) is an upper bound of g [S]. Let z 0 ∈ Z be an arbitrary upper bound of g [S]. We wish to show g(sup(S)) ≤ Z z 0 . For every f ∈ F and every s ∈ S, f (s) ≤ Z g(s) ≤ Z z 0 .
As every f ∈ F is Scott continuous, for every f ∈ F, sup( f [S]) exists and we obtain f (sup(S)) = sup( f [S]) ≤ Z z 0 .
This completes the proof of our claim that g is Scott continuous.
It is obvious that g is an upper bound of F. If h ∈ C(X , Z ) is any upper bound of F then for all x ∈ X and all f ∈ F, f (x) ≤ Z h(x), hence, g(x) = sup{ f (x) : f ∈ F} ≤ Z h(x), hence, g ≤ C h. We have shown that g is the least upper bound of F.
One may now speculate whether for an arbitrary topological space X and an arbitrary bounded complete dcpo Z (with least element) the dcpo C(X , Z ) is bounded complete as well. For example, Edalat [4, p. 502] claims that this is the case by writing: "For any topological space Z and any bounded complete dcpo D with bottom ⊥, let Z → D be the bounded complete dcpo of Scott continuous functions from Z to D." But we are going to show that in general this is not the case; see Theorem 8.5 below.

Remark 8.4
The claim by Edalat, cited above, is a crucial step in his argument in [4] why the notion of the L-derivative, introduced by him in [4], is well-defined. As this claim is incorrect the reader might now be worried whether the L-derivative is welldefined at all. But indeed it is. One can replace the cited claim by a different argument; compare [9]. It is clear that Property (a) in Theorem 8.5 implies Property (b). The converse in not so obvious, but true as well, as the following lemma shows. It will be used in the proof of Theorem 8.5.

Lemma 8.6
For any topological space X , any bounded sup semilattice (Z , ≤ Z ), and any two functions f , g ∈ C(X , Z ) the following two conditions are equivalent. 1. The set { f , g} has a ≤ C -supremum in C(X , Z ). Proof Fix some functions f , g ∈ C(X , Z ), and let h : X → Z be the pointwise supremum of f and g.
"2 ⇒ 1": If h is Scott continuous and r ∈ C(X , Z ) any function with f ≤ C r and g ≤ C r then h ≤ C r . Thus, in this case h is a supremum of { f , g} in C(X , Z ). complete (every subset has a supremum), and it has a least element. For X we choose Z × Z with the product topology of the Scott topologies on each copy of Z . By Proposition 8.1 C(X , Z ) is a dcpo and by Lemma 8.2 it is bounded and has a least element. Once we have shown that there exist two functions f , g ∈ C(X , Z ) that do not have a supremum in C(X , Z ) we have shown that C(X , Z ) is not bounded complete.
In fact, for f and g we can take the projection functions π 1 , π 2 : X = Z × Z → Z defined by π i (z 1 , z 2 ) := z i for i = 1, 2 and all z 1 , z 2 ∈ Z . For i = 1, 2 the projection function π i is continuous with respect to the product topology on X = Z × Z , thus, π i ∈ C(X , Z ). Note that the function sup (2) is the pointwise supremum of π 1 and π 2 . As we have chosen Z to be a complete lattice as in Theorem 7.1 and X to be Z × Z with the product topology, the function sup (2) is not an element of C(X , Z ), that is, sup (2) is not continuous. Thus, the pointwise supremum of the continuous functions π 1 : X → Z and π 2 : X → Z is not continuous. And, according to Lemma 8.6, the set {π 1 , π 2 } does not have a ≤ C -supremum in C(X , Z ). This ends the proof of Theorem 8.5.