ε\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varvec{\varepsilon }$$\end{document}-Shading Operator on Riesz Spaces and Order Continuity of Orthogonally Additive Operators

Given a Riesz space E and 0<e∈E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$0 < e \in E$$\end{document}, we introduce and study an order continuous orthogonally additive operator which is an ε\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\varepsilon $$\end{document}-approximation of the principal lateral band projection Qe\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Q_e$$\end{document} (the order discontinuous lattice homomorphism Qe:E→E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Q_e :E \rightarrow E$$\end{document} which assigns to any element x∈E\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x \in E$$\end{document} the maximal common fragment Qe(x)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Q_e(x)$$\end{document} of e and x). This gives a tool for constructing an order continuous orthogonally additive operator with given properties. Using it, we provide the first example of an order discontinuous orthogonally additive operator which is both uniformly-to-order continuous and horizontally-to-order continuous. Another result gives sufficient conditions on Riesz spaces E and F under which such an example does not exist. Our next main result asserts that, if E has the principal projection property and F is a Dedekind complete Riesz space then every order continuous regular orthogonally additive operator T:E→F\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$T :E \rightarrow F$$\end{document} has order continuous modulus |T|. Finally, we provide an example showing that the latter theorem is not true for E=C[0,1]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$E = C[0,1]$$\end{document} and some Dedekind complete F. The above results answer two problems posed in a recent paper by O. Fotiy, I. Krasikova, M. Pliev and the second named author.


Introduction
We use standard terminology and notation on Riesz spaces as in [2]. In the next section, we provide with all necessary information on the lateral order and orthogonally additive operators (OAOs, in short) on Riesz spaces. In the present section, we describe our main results.
Basic order continuity properties of OAOs essentially differ from that of linear operators. Let E, F be Riesz spaces. Below we provide assertions for linear operators which are false for OAOs.
(1) If E has the principal projection property and F is Dedekind complete then every horizontally-to-order continuous linear operator T : E → F is order continuous [5,Proposition 3.9]. For OAOs this is false: if 0 ≤ p ≤ ∞ then there exists a horizontally-to-order continuous orthogonally additive functional f : L p → R which is not order continuous (moreover, f is not uniformly-to-order continuous), see [3, Example 2.1]. (2) If F is Archimedean then every regular linear operator T : E → F is uniformly-to-order continuous [11,Proposition 4.6]. A typical example of a positive OAO which is not uniformly-to-order continuous is the principal lateral band projection Q e (see the next section), see also (1 for OAOs is much more involved for investigation. It was one of the main questions considered in [3]. Some partial results were obtained there.  [3,Corollary 3.5]. Let E be a Riesz space with the principal projection property and F a Dedekind complete Riesz space. Then for every order continuous operator T ∈ OA r (E, F ) the following assertions are equivalent: (1) |T | is order continuous; (2) T + is order continuous; (3) for every net (x α ) in E order convergent to some element x ∈ E, the following condition holds T + (x α ) − T + (x) So the following natural questions remained unsolved.
Problem 1 [3]. Under what assumptions on Riesz spaces E, F with F Dedekind complete every order bounded OAO T : E → F which is both horizontally-toorder continuous and uniformly-to-order continuous, is order continuous? Problem 2 [3]. Do there exist a Riesz space with the principal projection property E, a Dedekind complete Riesz space F and an order continuous OAO T : E → F such that |T | is not order continuous?
The principal lateral band projection Q e can serve as an "atomic" OAO in different constructions (as one of the summands) of an OAO with given properties. This technique was actively explored in [11] to prove the existence of OAOs with some pathological properties. However, Q e is order discontinuous, which makes impossible construction of an order continuous OAO with given properties.
Section 3 is devoted to construction and investigation of an order continuous operator Q ε e , which approximates Q e as ε → 0+. In Sect. 4 we provide the first example of an uniformly-to-order continuous and horizontally-to-order continuous OAO, acting from C[0, 1] to a Dedekind complete Riesz space, which is not order continuous. In the next section, we provide sufficient conditions, under which such an operator does not exist. One of the results asserts that, if E has the principal projection property and the horizontal Egorov property and F is Dedekind complete then every OAO T : E → F , which is both uniformly-to-order continuous and horizontally-to-order continuous, is order continuous. This gives a partial answer to Problem 1. In Sect. 6 we prove that, if E is a Riesz space with the principal projection property and F a Dedekind complete Riesz space then every order continuous regular OAO T : E → F has order continuous modulus. This gives a negative answer to Problem 2 for regular OAOs. In Sect. 7 we provide the first example of an order continuous order bounded OAO T : E → F between Riesz spaces E, F with F Dedekind complete with order discontinuous modulus |T |. This, in particular, shows that the assumption on E to have the principal projection property in the theorem from Sect. 6 cannot be removed. In the final part we pose the remaining open problems.

Preliminary Information
Let E be a Riesz space and x, y ∈ E. We say that x is a fragment of y The set of all fragments of an element e ∈ E is denoted by F e . A disjoint sum in E is written using symbols , . So

The Lateral Order
The relation is a partial order on E, called the lateral order (see [6] for a systematic study of the lateral order). The supremum and infimum of a subset A ⊆ E with respect to the lateral order (if exists) is denoted by A and A respectively (for a two-point set A = {x, y} write x ∪ y and x ∩ y). A subset A ⊆ E is said to be laterally bounded if A ⊆ F e for some e ∈ E. Although any subset is laterally bounded from below by zero, a two-point set {x, y} ⊂ E may not have a lateral infimum x ∩ y (which is the maximal common fragment of x and y), see [6,Example 3.11]. A Riesz space E is said to have the intersection property provided every two-point subset of E has a lateral infimum. The principal projection property implies the intersection property [6,Theorem 3.13], however the converse is not true ( Every order ideal is a lateral ideal and every band is a lateral band. For every e ∈ E\{0} the set F e is a lateral band which is not an order ideal. Moreover, F e is both the minimal lateral ideal and minimal lateral band containing e. We say that F e is the principal lateral ideal and principal lateral band generated by e. The notion of a lateral ideal (lateral band) is so important for the study of orthogonally additive operators (order continuous orthogonally additive operators) as well as the order ideals (respectively, bands) are important for the study of order bounded (respectively, order continuous) linear operators [6,7]. Proposition 2.1 [6,11]. Let E be a vector lattice and e ∈ E. Then the following assertions hold.
(1) The set F e of all fragments of e is a Boolean algebra with zero 0, unit e with respect to the operations ∪ and ∩. Moreover,

Orthogonally Additive Operators
Let E be a Riesz space and X a real vector space. A function T : E → X is called an orthogonally additive operator (OAO in short) provided T (x + y) = T (x) + T (y) for any disjoint elements x, y ∈ E. Obviously, if T is an OAO then T (0) = 0. The set of all OAOs from E to X is a real vector space with respect to the natural linear operations.
Let E, F be vector lattices. An OAO T : E → F is said to be: • regular if T is a difference of two positive operators; • order bounded, or an abstract Uryson operator, if it maps order bounded subsets of E to order bounded subsets of F ; • laterally-to-order bounded (or a Popov operator, in terms of [9]) if the set The positivity of OAOs is completely different from that of linear operators, and the only linear operator which is positive in the sense of OAOs is zero. A positive OAO need not be order bounded. Indeed, every function T : R → R with T (0) = 0 is an OAO, and, obviously, not all such functions are order bounded. Obviously, every laterally non-expanding OAO preserves disjointness. The kernel of a positive OAO is a lateral ideal [6, Proposition 6.4] and every lateral ideal is a kernel of some positive OAO [7,Theorem 3.1].
Denote the sets of all positive, regular, order bounded and laterally-toorder bounded OAOs from E to F by OA + (E, F ), OA r (E, F ), U(E, F ) and P(E, F ) respectively. Observe that U(E, F ) is a vector subspace of P(E, F ) and the inclusion U(E, F ) ⊂ P(E, F ) is strict even for the one-dimensional case E = F = R ( [9]). We endow OA r (E, F ) with the order S ≤ T provided that T − S is a positive OAO, that is, Sx ≤ T x for all x ∈ E. Then OA r (E, F ) becomes an ordered vector space.

The Principal Lateral Band Projection Q e
A laterally non-expanding projection (that is, which is a lateral retract, is called a projection lateral band, and the lateral retraction of E onto A is called the lateral band projection of E onto A. Theorem 2.3 [11,Theorem 1.6]. Let E be a vector lattice with the intersection property. Then for every e ∈ E\{0} the function Q e : E → E defined by setting In Theorem 3.2 we give explicit formula (3.4) for x ∩ e in terms of lattice operations over x and e if x − e is a projective element of E.

The Intersection Property is a Lateral Analogue of the Principal Projection Property
where by E y we denote the minimal order ideal containing y. In this case the order projection P y of E onto E y is given (see [2,Theorem 1.47]) by (2.1) We need the following property of P y . By (3) of [2, Theorem 1.44], A Riesz space E is said to have the principal projection property provided every element of E is a projection element.
Let E be a vector lattice and x, y ∈ E. We say that x is laterally disjoint to y and write x †y if F x ∩ F y = {0}. Two subsets A and B of E are said to be laterally disjoint (write A †B) if x †y for every x ∈ A and y ∈ B. The laterally disjoint complement to a subset A of E is defined as follows: A † := {x ∈ E : (∀a ∈ A) x †a}. Note that x ⊥ y implies x †y for all x, y ∈ E and the converse is false. However, x †y implies x ⊥ y for every laterally bounded pair x, y ∈ E. An element e of a vector lattice E is called a laterally projection element provided E is decomposed into a nonlinear direct sum E = F e F † e , that is, every x ∈ E has a unique representation where Q e is the principal lateral band projection.

Different Types of Order Convergence and Order Continuity
A net (x α ) α∈A in a Riesz space E converges to a limit x ∈ E: • strongly order if there is a net (u α ) α∈A in E such that u α ↓ 0 and |x α − x| ≤ u α for some α 0 ∈ A and all α ≥ α 0 (write x α in this case we write x α e ⇒ x; • uniformly, provided (x α ) α∈A converges to x e-uniformly for some e ∈ E + ; in this case we write x α ⇒ x. Every strongly order convergent net weakly converges to the same limit, but the converse is false [1]. However, the strong and weak order convergence are equivalent if either E is Dedekind complete or the net (x α ) α∈A is monotone. In these two cases we write x α o −→ x. Note that one can equivalently replace the condition α∈A x α = x with x α o −→ x in the definition of horizontal convergence under the assumption x α x β for all α < β. The uniform convergence of a net implies the strong order convergence the net to the same limit.
The order continuity of operators we understand in the sense of strong order convergence. More precisely, let E, F be Riesz spaces. An OAO T : E → F is said to be: • order continuous provided for any x ∈ E and any net ( • horizontally-to-order continuous provided for any x ∈ E and any net • uniformly-to-order continuous provided for any x ∈ E and any net (

ε-Shading Operator
The principal lateral band projection Q e (see Theorem 2.3) gives an important tool for constructing examples of OAOs defined on a Riesz space with the intersection property. However, Q e is not order continuous (because In the present section, we construct a "blurring" version of Q e , which mainly has similar properties and is order continuous. Another superiority of this version is that it acts in an arbitrary Riesz space.
Let E be a Riesz space, 0 < e ∈ E and 0 < ε < 1. Define a map Q ε e : E → E by setting The map Q ε e defined by (3.1) will be called the ε-shading operator generated by e. Lemma 3.1. Let E be a Riesz space, 0 < e ∈ E and 0 < ε < 1. Then for every Proof. Remark that, by the well known formula [2, Theorem 1.7] it is enough to prove one of the equalities. For every x ∈ E, using theorems 1.3 and 1.8 of [2], we obtain The following theorem collects main properties of the ε-shading operator.
is a positive disjointness preserving order continuous OAO possessing the following properties.
If, moreover, e and e are projection elements of E then Q ε e Q ε e = Q ε e . For the proof, we need the following lemma.
is a positive disjointness preserving order continuous OAO possessing the following properties: Proof. First fix any x, y ∈ E + with x ⊥ y and prove Now we show that Indeed, on the one hand, On the other hand, Indeed, Taking into account that x + y = x ∨ y, we obtain where Observe that w 1 ≤ x and w 1 ≤ v. Hence Analogously, w 2 ≤ y, w 2 ≤ v and hence Taking into account that 0 ≤ w 1 ≤ x and 0 ≤ w 2 ≤ y, we obtain that w 1 ⊥ w 2 and hence, w 1 ∨ w 2 = w 1 + w 2 = Q(x) + Q(y). By (3.8), one gets (3.5).
To prove (3.5) for the general case of x, y ∈ E with x ⊥ y, by the above, it is enough to prove that (3.9) Fix any x ∈ E. We need two claims and the following known elementary fact (see item (2) (3.10) Now we obtain Proof of Claim 2. We have (3.13), this yields that w = 0. Now we continue the proof of Lemma 3.3. By (3.3) and Claim 1, and analogously, Q(x + ) ≤ x + . Hence, Thus, by Claim 2, and (3.9) is proved. So, Q is an OAO. Item (i) is already proved by (3.9) and (3.14). Q preserves disjointness by (i). The order continuity of Q follows from (ii). So, it remains to prove (ii). Observe that, for every a, b, c, d ∈ E one has Hence, for every x, y ∈ E we obtain Proof of Theorem 3.2. By Lemma 3.3, Q ε e is a positive order continuous OAO and (ii) holds true. ( On the other hand, Finally, (3.15), (3.16) and (3.17) together give (iv) follows from (3.1).
(vi) Assume x − e is a projection element. Then by (2.1) and Lemma 3.1 (3.18) Show that z = e ∩ x. By (2.2), e = P x−e e z, which implies that z e. Then Observe that for every n ∈ N the relation t e implies t = Q 1/n e (t) by (iii), and the relation t (vii) Given any x ∈ E, one has On the other hand, 0 ≤ w ≤ εe . Hence, 0 ≤ w ≤ e ∧ e = 0, which implies w = 0. Analogously, εe ∧ (e − x) = εe ∧ (e − x). By that Analogously, e ∧ 1 ε |x − e| = e − Q ε e (x). Hence and the first part of (vii) is proved. Assume now that e and e are projection elements. By (2) of Theorem 2.2, Analogously, Q ε e (z) = 0 which confirms by (3.19) that Q ε e ⊥ Q ε e .

A Uniformly-to-Order Continuous and Horizontally-to-Order Continuous OAO, Which is Not Order Continuous
In the present section, we provide the first example of such an operator. Moreover, it is a functional, that is, with values in R. It is defined on C[0, 1] possessing the intersection property, but failing to have the principal projection property.
Observe that e n ∈ C[0, 1] and e n+1 (t) < e n (t) for all t ∈ [0, 1] and n ∈ N. Choose numbers ε n > 0 so that Then define a functional f : where Q εn en is the ε n -shading operator generated by e n . To show that the functional is well defined by (4.2), set Prove that f is uniformly-to-order continuous. Since the uniformly convergence in C[0, 1] is equivalent to the norm-convergence, it is enough to show that f is norm-to-order continuous. By Theorem 3.2 (ii), every function is norm-to-order continuous. By (4.3), Since (e n ) ∞ n=1 has no norm-convergent subsequence and ε n → 0, the sequence (B n ) ∞ n=1 is locally finite with respect to the norm. Therefore, f is norm-toorder continuous as locally finite sum of norm-to-order continuous functions.
Prove that f is horizontally-to-order continuous. Let x ∈ C[0, 1] and

Sufficient Conditions of the Order Continuity of Horizontally-to-Order and Uniformly-to-Order Continuous Operators
In this section, we provide sufficient conditions on Riesz spaces E, F , under which every horizontally-to-order continuous and uniformly-to-order continuous OAO is order continuous, giving a partial answer to Problem 1. Let F be a Riesz space. By D(F ) we denote the set of all Riesz spaces E such that every OAO T : E → F which is both horizontally-to-order continuous and uniformly-to-order continuous, is order continuous.
Next is our main result of the section.

Theorem 5.1. Let E be a Riesz space such that for every net
For the proof of Theorem 5.1 we need some lemmas.

Lemma 5.2. Let E be a Dedekind complete Riesz space and A an upper bounded subset of
Proof. For every x ∈ A we set A x = {y ∈ A : x + y ∈ A} and u x = sup A x . Assume that v > 0. Since E is an Archimedean Riesz space, there exists a number n ∈ N such that n 2 v ≤ u 0 = sup A. Then w 0 := ( n 2 v − u 0 ) + > 0. Let x 0 = 0. Now we construct finite sequences (x k ) n k=1 and (w k ) n k=1 of elements x k ∈ A and w k ∈ E + such that for every k = 1, . . . , n Then w 1 ∈ B w0 and w 1 > 0. Moreover, Since w 1 ∈ B w0 and w 0 ∈ B v , one has w 1 ∈ B v and P 1 (v) > 0. Hence, taking into account that (by the choice of x 1 ) x 0 + x 1 ∈ A, we obtain It is clear that w 2 ∈ B w1 and w 2 > 0. Moreover, To complete the construction of (x k ) n k=1 and (w k ) n k=1 , it remains to repeat the reasoning n − 2 times. Now we consider the element By (2), x ∈ A. On the one hand But on the other hand we have that

Lemma 5.3. Let E be a Riesz space and F a Dedekind complete Riesz space,
T : E → F a function and x 0 ∈ E. Then the following conditions are equivalent: Proof. (i) ⇒ (ii). It is enough to prove the first equality only, because the second equality coincide with the first one for −T in place of T . If A has a maximal element then the claim of the lemma is obvious. Assume A has no maximal element. Fix any net (u α ) α∈A in E with u α ↓ 0. Endow the set with the following partial order: (α , β ) ≤ (α , β ) if and only if either α < α or α = α and β ≤ β . Obviously, B is a directed set. Now consider the net x (α,β) (α,β)∈B defined by setting x (α,β) = x 0 + β for all (α, β) ∈ B. Set also u (α,β) := u α for all (α, β) ∈ B. Since u α ↓ 0, we have u (α,β) ↓ 0 as well. Since By the order continuity of T at x 0 , choose a net t (α,β) (α,β)∈B in F and (α 0 , β 0 ) ∈ B so that t (α,β) ↓ 0 and Hence, Since t (γ,0) ↓ 0, the latter inequality implies the first equality in (5.1).
Proof of Theorem 5.1. Let F be a Dedekind complete Riesz space and T : E → F be an OAO which is both horizontally-to-order continuous and uniformlyto-order continuous. Fix any x 0 ∈ E and show that T is order continuous at We choose a net (E i ) i∈I of projection bands E i of E to satisfy (1)-(3). For every i ∈ I we set T i = T | Ei . Clearly, every T i is an OAO which is both horizontally-to-order continuous and uniformly-to-order continuous. If the directed set I has a maximal element i 0 then by (1)-(3) one has E = E i0 , T = T i0 and v = 0.
Let I have no maximal element. For every α ∈ A and i ∈ I we set By the orthogonal additivity of T , Proof. Since Z α ⊆ Y α , one has w α ≤ v α . It remains to show that y ≤ w α for every y ∈ Y α . Let y ∈ Y α be an arbitrary point and x ∈ E such that It follows from the horizontally-to-order continuity of T that Therefore, Proof. By (2) and [2, Theorem 1.46], for every y ∈ E and j > i we have that P i (y) P j (y) and hence P j (y) = P i (y) P j (y) − P j (y) . By the orthogonal additivity of T , Let y ∈ Y α,i and z ∈ Z α,i be arbitrary elements. Then there exist x 1 , x 2 ∈ E and j > i such that and T is orthogonally additive, for every x , x ∈ E. Therefore, Consider the element x = x 2 −P i (x 2 ) +P i (x 1 ) and observe that P i (x) = P i (x 1 ) and Therefore, in particular,

Claim 6. For every fixed
Proof. Since P i (u α ) ⇒ 0 and T i is uniformly-to-order continuous at x 0 , it follows from Lemma 5.3 thatṽ α ↓ 0, wherẽ Proof. Fix any i ∈ I. By Claims 4, 5 and 6, w α, Proof. Fix any α ∈ A. It is enough to show that the set B = Z α satisfies the following assumption of Lemma 5.2 sup{z ∈ B : y + z ∈ B} ≥ v for every y ∈ B. Indeed, let y ∈ Z α be an arbitrary element. We choose i ∈ I so that y ∈ Y α,i . By Claims 5 and 7, By Claim 8, the theorem is proved. To obtain an important consequence of Theorem 5.1, we need more information. Following [10], we say that a Riesz space E has the horizontal Egorov property provided for every f ∈ E, every e ∈ E + and every net (f α ) α∈A in E with f α o −→ f and |f α | + |f | ≤ e for all α ∈ A, there exists a net (e i ) i∈I of fragments of e with e i h −→ e such that for every i ∈ I, |f − f α | ∧ e i e ⇒ α 0. In this definition, the order convergence f α o −→ f could be understood in both weak and strong senses, which coincide if E is Dedekind complete. Since we are going to apply the above property to monotone nets, for which the weak order convergence is equivalent to the strong one, it does not matter for us, what kind of order convergence we consider in the above definition.
A Boolean algebra B satisfies the weak distributive law, if whenever (Π n ) n∈N is a sequence of partitions of B, there is a partition Π of B such that every element of Π is finitely covered by each of Π n , that is, Recall that a partition of B is a maximal disjoint subset of B, or, equivalently, a disjoint subset A of B such that A = 1, where 1 is unity of B.
We say that a Riesz space E satisfies the weak distributive law provided for every e ∈ E + \{0} the Boolean algebra F e of fragments of e satisfies the weak distributive law. Likewise, E is said to be measurable provided for every e ∈ E + \{0} the Boolean algebra F e is measurable [4]. It is well known that every measurable Boolean algebra satisfies the weak distributive law (see [10] for details). Hence, every measurable Riesz space satisfies the weak distributive law, e.g., every Riesz space of equivalence classes of measurable functions on a measure space (Ω, Σ, μ). The above information yields the following consequences of Theorem 5.1.

Corollary 5.5. Let E be a Riesz space with the principal projection property and the horizontal Egorov property. Then E ∈ D(F ) for every Dedekind complete
Proof. We prove that E satisfies the assumptions of Theorem 5.1.
(1) Let (u α ) α∈A be a net in E with u α ↓ 0. With no loss of generality, we may and do assume that the net (u α ) α∈A is order bounded by some e ∈ E + . By the horizontal Egorov property of E, choose a net (e i ) i∈I of fragments of e with e i h −→ e such that u α ∧ e i e ⇒ α 0 for all i ∈ I. Given any y ∈ E, by E y we denote the principal band generated by y, and by P y the order projection of E onto E y . The desired net of bands (F i ) i∈I we define by setting (1) For every n ∈ N and i ∈ I one has Thus, for every α ∈ A and i ∈ I (2) is obvious.
(3) For every x ∈ E one has by [ Given any nonempty set Ω, by c 00 (Ω) we denote the Riesz space of all functions f : Ω → R with finite support, endowed with the natural order. Clearly, for every finite set Ω the uniform convergence in c 00 (Ω) is equivalent to the order convergence. Thus, c 00 (Ω) ∈ D(F ) for every Riesz space F .

When Does the Order Continuity of T Imply that of |T |?
The following theorem gives a negative answer to Problem 2. Moreover, in the next section we provide an example which demonstrates that the assumption on E to have the principal projection property in Theorem 6.1 cannot be removed, or even replaced with the intersection property. For the proof, we need the following lemma. Lemma 6.2. Let E be a Riesz space, x, y ∈ E and u x. If u is a projection element then there exists v y such that and prove that v possesses the desired properties. The relation v y follows from (iii) of [2,Theorem 1.44].
Since x − u ⊥ nu for all n ∈ N, one has Observe that |v| = P u |y|. Since |y − v| ∧ |y| = |y − v|, we obtain Finally, the latter two conclusions give The following simple example shows that Lemma 6.2 is false without the assumption on u to be a projection element. Let E = C[0, 1], x(t) = |t − 1/2| for all t ∈ [0, 1], u(t) = 1/2 − t if t ≤ 1/2 and u(t) = 0, if t > 1/2, and y(t) = 1 for all t ∈ [0, 1]. Then y has two fragments 0 and y, none of which satisfies the requirements.
Proof of Theorem 6.1. Let x ∈ E be an arbitrary point and (x α ) α∈A a net in E with x α s.o −→ x in E. Let (u α ) α∈A be a net in E such that u α ↓ 0 and |x α − x| ≤ u α for some α 0 ∈ A and all α ≥ α 0 . To prove the order continuity of |T |, by Theorem 1.3 it is enough to prove that In the case where A has a maximal element the proof is obvious. So assume that A has no maximal element. For every α ∈ A we set z α = x α − x and and endow the set B = α∈A B α with the lexicographic partial order ≤, that is (α, z) ≤ (α , z ) ⇔ ((α < α ) or (α = α and z z ).
For every α ∈ A we set w α = v (α,0) . Then w α ≥ w α for all α, α ∈ A with α < α . Since A has no maximal element, for every α ∈ A there exists α ∈ A with α < α , and for every z ∈ F zα we have that Thus, w α ↓ 0.
Let α 2 > α 1 be a fixed index and α ≥ α 2 an arbitrary index. Let s x α be an arbitrary fragment. By Lemma 6.2, there exists t s x such that Thus, T (s) ≤ T (t s ) + w α for every s x α . Hence, which implies (6.1).
The following proposition shows that, the order continuity of T at zero implies the order continuity of |T | at zero without any assumption on E. Proposition 6.3. Let E and F be Riesz spaces with F Dedekind complete and T ∈ OA r (E, F ). If T is order continuous at 0 then |T | is order continuous at 0.
In the case where A has a maximal element the proof is obvious. So assume that A has no maximal element. Let (u α ) α∈A be a net in E such that |x α | ≤ u α for all α ≥ α 0 , where α 0 ∈ A and u α ↓ 0. For every α ∈ A we set and endow the set B = α∈A B α with the lexicographic partial order ≤, that is Clearly, (B, ≤) is a directed set.
For any β = (α, x) ∈ B we set x β = x and show that the net (x β ) β∈B strongly order converges to 0 in E. For every β = (α, x) ∈ B we set u β := u α . Since u α ↓ 0, one has u β ↓ 0. Moreover, It follows from the order continuity of T at 0 that there exists a net For every α ∈ A we set w α = v (α,0) . Observe that w α ≥ w α for every α, α ∈ A with α < α . Since A has no maximal element, for every α ∈ A there exists α ∈ A with α < α , and for every Thus, w α ↓ 0.
Let α 2 be a fixed index with α 2 > α 1 and α ≥ α 2 be an arbitrary index. Then for every x ∈ F xα . Therefore, for every x ∈ F xα , by Theorem 2.2, (3) Remark 6.4. As the above proof shows, the claim of Proposition 6.3 is valid if we replace the strong order convergence of nets in E with the weak order convergence.

An Order Continuous OAO with Discontinuous Modulus
In this section, we provide the first example of an order continuous abstract Uryson operator T ∈ U (E, F ) between Riesz spaces E, F with F Dedekind complete such that the modulus |T | is not order continuous.
Given Riesz spaces E, F, G with F ⊆ G and a mapping f : E → F , by f G we denote the same mapping f G : E → G. By C we denote the Dedekind completion of C[0, 1] in the sense that C[0, 1] is a majorizing order dense Riesz subspace of C. By O(E, F )  To prove Theorem 7.2, we need some lemmas and preliminaries.
For every x ∈ C[0, 1] we set The following two propositions are obvious.
Proposition 7.4. Let x, y ∈ C[0, 1] with x ⊥ y. Then (1) x l ⊥ y l and (x + y) l = x l + y l ; (2) x r ⊥ y r and (x + y) r = x r + y r ; (3) (x + y) c = x c + y c , moreover, x c = 0 or y c = 0. The condition x 1 ⊥ x 2 means that A 1 ∩ A 2 = ∅. Since ϕ is strictly monotone, Therefore, T x 1 ⊥ T x 2 . To prove (3) Proof of Theorem 7.1. Set S := T C , where T is defined in item (2) of Theorem 7.2. The order boundedness of S, as well as the order continuity of S and order discontinuity of |S| follows from Theorem 7.2 and Lemma 7.6.

Open Problems
The following problems still remain unsolved. Data Availability Statement Data sharing not applicable to this article as no datasets were generated or analyzed during the preparation of the paper.

Conflict of interest The authors have not disclosed any competing interests.
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