Convergence structures and locally solid topologies on vector lattices of operators

For vector lattices $E$ and $F$, where $F$ is Dedekind complete and supplied with a locally solid topology, we introduce the corresponding locally solid absolute strong operator topology on the order bounded operators $\mathcal L_{\mathrm{ob}}(E,F)$ from $E$ into $F$. Using this, it follows that $\mathcal L_{\mathrm{ob}}(E,F)$ admits a Hausdorff uo-Lebesgue topology whenever $F$ does. For each of order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology, there are both a uniform and a strong convergence structure on $\mathcal L_{\mathrm {ob}}(E,F)$. Of the six conceivable inclusions within these three pairs, only one is generally valid. On the orthomorphisms of a Dedekind complete vector lattice, however, five are generally valid, and the sixth is valid for order bounded nets. The latter condition is redundant in the case of sequences of orthomorphisms on a Banach lattice, as a consequence of a uniform order boundedness principle for orthomorphisms that we establish. We also show that, in contrast to general order bounded operators, the orthomorphisms preserve not only order convergence of nets, but unbounded order convergence and -- when applicable -- convergence in the Hausdorff uo-Lebesgue topology as well.


INTRODUCTION AND OVERVIEW
Let X be a non-empty set. A convergence structure on X is a non-empty collection of pairs ((x α ) α∈A , x), where (x α ) α∈A is a net in X and x ∈ X , such that: (1) when ((x α ) α∈A , x) ∈ , then also ((x β ) β∈B , x) ∈ for every subnet (x β ) β∈B of (x α ) α∈A ; (2) when a net (x α ) α∈A in X is constant with value x, then ((x α ) α∈A , x) ∈ . One can easily vary on this definition. For example, one can allow only sequences. There does not appear to be a consensus in the literature about the notion of a convergence structure; [4] uses filters, for example. Ours is sufficient for our merely descriptive purposes, and close in spirit to what may be the first occurrence of such a definition in [12] for sequences. Although we shall not pursue this in the present paper, let us still mention that the inclusion of the subnet criterion in the definition makes it possible to introduce an associated topology on X in a natural way. Indeed, define a subset of S of X to be -closed when x ∈ S for all pairs ((x α ) α∈A , x) ∈ such that (x α ) α∈A ⊆ S. Then the collection of the complements of the -closed subsets of X is a topology on X .
The convergent nets in a topological space, together with their limits, are the archetypical example of a convergence structure. For a given convergence structure on a non-empty set X , however, it is not always possible to find a (obviously unique) topology τ on X such that the τ-convergent nets in X , together with their limits, are precisely the elements of . Such non-topological convergence structures arise naturally in the context of vector lattices. For example, the order convergent nets in a vector lattice, together with their order limits, form a convergence structure, but this convergence structure is topological if and only if the vector lattice is finite dimensional; see [8,Theorem 1] or [23,Theorem 8.36]. Likewise, the unbounded order convergent nets in a vector lattice, together with their unbounded order limits, form a convergence structure, but this convergence structure is topological if and only if the vector lattice is atomic; see [23,Theorem 6.54]. Topological or not, the order and unbounded order convergence structures, together with the (topological) structure for convergence in the Hausdorff uo-Lebesgue topology, when this exists, yield three natural and related convergence structures on a vector lattice to consider.
Suppose that E and F are vector lattices, where F is Dedekind complete. The above then yields three convergence structures on the vector lattice ob (E, F) of order bounded operators from E into F. On the other hand, there are also three convergence structures on ob (E, F) that are naturally derived from the three convergence structures on the vector lattice F. For example, one can consider all pairs ((T α ) α∈A , T ), where (T α ) α∈A is a net in ob (E, F) and T ∈ ob (E), such that (T α x) α∈A is order convergent to T x in F for all x ∈ E. These pairs also form a convergence structure on ob (E, F). Likewise, the pointwise unbounded order convergence in F and-when applicable-the pointwise convergence in the Hausdorff uo-Lebesgue topology on F both yield a convergence structure on ob (E, F). Motivated by the terminology for operators between Banach spaces, we shall speak of uniform and strong convergence structures on ob (E)-with the obvious meanings.
The present paper is primarily concerned with the possible inclusions between the uniform and strong convergence structure for each of order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology. Is it true that a uniformly order convergent net of order bounded operators is also strongly order convergent? Is the converse true? How is this for unbounded order convergence and, when applicable, convergence in the Hausdorff uo-Lebesgue topology? We consider these implications, six in all, for ob (E, F), but also for the orthomorphisms Orth(E) on a Dedekind complete vector lattice. 1 This special interest in Orth(E) stems 1 With six convergence structures under consideration, one can actually consider thirty nontrivial possible inclusions between them. With some more effort, one can determine for all of from representation theory. When a group acts as order automorphisms on a Dedekind complete vector lattice E, then the Boolean lattice of all invariant bands in E can be retrieved from the commutant of the group action in Orth(E). This commutant, therefore, plays the role of the von Neumann algebra which is the commutant of a unitary action of a group on a Hilbert space. It has been known long since that more than one topology on a von Neumann algebra is needed to understand it and its role in representation theory on Hilbert spaces, and the same holds true for the convergence structures as related to these commutants in an ordered context. Using these convergence structures, it is, for example, possible to obtain ordered versions of von Neumann's bicommutant theorem. We shall report separately on this. Apart from its intrinsic interest, the material on Orth(E) in the present paper is an ingredient for these next steps.
This paper is organised as follows.
Section 2 contains the basic notations, definitions, conventions, and references to earlier results.
In Section 3, we show how, given a vector lattice E, a Dedekind complete vector lattice F, and a (not necessarily Hausdorff) locally solid linear topology τ F on F, a locally solid linear topology can be introduced on ob (E, F) that deserves to be called the absolute strong operator topology that is generated by τ F . This is a preparation for Section 4, where we show that regular vector sublattices of ob (E, F) admit a Hausdorff uo-Lebesgue topology when F admits one.
For each of order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology, there are two conceivable implications between uniform and strong convergence of a net of order bounded operators. In Section 5, we show that only one of these six is generally valid. Section 9 will make it clear that the five failures are, perhaps, not as 'only to be expected' as one might think at first sight.
In Section 6, we review some material concerning orthomorphism and establish a few auxiliary result for use in the present paper and in future ones. It is shown here that a Dedekind complete vector lattice and its orthomorphisms have the same universal completion. Furthermore, a uniform order boundedness principle is established for sets of orthomorphisms. Section 7 briefly digresses from the main line of the paper. It is shown that orthomorphisms preserve not only the order convergence of nets, but also the unbounded order convergence and-when applicable-the convergence in the Hausdorff uo-Lebesgue topology. None of this is true for arbitrary order bounded operators.
In Section 8, we return to the main line, and we specialise the results in Sections 3 and 4 to the orthomorphisms. When restricted to Orth(E), the absolute strong operator topologies from Section 3 are simply strong operator topologies. these whether they are generally valid for the order bounded operators and for the orthomorphisms on a Dedekind complete vector lattice; see [10, Tables 3.1 and 3.2]. Section 9 on orthomorphisms is the companion of Section 5, but the results are quite in contrast. For each of order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology, both implications between uniform and strong convergence of a net of orthomorphisms are valid, with an order boundedness condition on the net being necessary only for order convergence. For sequences of orthomorphisms, this order boundedness condition is even redundant as a consequence of the uniform order boundedness principle for orthomorphisms from Section 6.

PRELIMINARIES
In this section, we collect a number of definitions, notations, conventions and earlier results.
All vector spaces are over the real numbers; all vector lattices are supposed to be Archimedean. We write E + for the positive cone of a vector lattice E. For a non-empty subset S of E, we let I S and B S denote the ideal of E and the band in E, respectively, that are generated by S; we write S ∨ for { s 1 ∨ · · · ∨ s n : s 1 , . . . , s n ∈ S, n ≥ 1 }.
Let E be a vector lattice, and let x ∈ E. We say that a net ( when there exists a net ( y β ) β∈B in E such that y β ↓ 0 and with the property that, for every β 0 ∈ B, there exists an α 0 ∈ A such that |x − x α | ≤ y β 0 whenever α in A is such that α ≥ α 0 . We explicitly include this definition to make clear that the index sets A and B need not be equal.
Let (x α ) α∈A be a net in a vector lattice E, and let x ∈ E. We say that (x α ) is Order convergence implies unbounded order convergence to the same limit. For order bounded nets, the two notions coincide.
Let E and F be vector lattices. The order bounded operators from E into F will be denoted by ob (E, F). We write E for ob (E, ). A linear operator T : E → F between two vector lattices E and F is order continuous when, for An order continuous linear operator between two vector lattices is automatically order bounded; see [3, Lemma 1.54], for example. The order continuous linear operators from E into F will be denoted by oc (E, F). We write E oc for oc (E, ). Let F be a vector sublattice of a vector lattice E. Then F is a regular vector sublattice of E when the inclusion map from F into E is order continuous. Ideals are regular vector sublattices. For a net in a regular vector sublattice F of E, its uo-convergence in F and in E are equivalent; see [14,Theorem 3.2].
When E is a vector space, a linear topology on E is a (not necessarily Hausdorff) topology that provides E with the structure of a topological vector space. When E is a vector lattice, a locally solid linear topology on E is a linear topology on E such that there exists a base of (not necessarily open) neighbourhoods of 0 that are solid subsets of E. For the general theory of locally solid linear topologies on vector lattices we refer to [2]. When E is a vector lattice, a locally solid additive topology on E is a topology that provides the additive group E with the structure of a (not necessarily Hausdorff) topological group, such that there exists a base of (not necessarily open) neighbourhoods of 0 that are solid subsets of E.
A topology τ on a vector lattice E is an o-Lebesgue topology when it is a (not necessarily Hausdorff) locally solid linear topology on E such that, for a net Let E be a vector lattice, let F be an ideal of E, and suppose that τ F is a (not necessarily Hausdorff) locally solid linear topology on F. Take a non-empty subset S of F. Then there exists a unique (possibly non-Hausdorff) locally solid [11,Theorem 3.1] for this, which extends earlier results in this vein in, e.g., [6] and [24]. This topology u S τ F is called the unbounded topology on E that is generated by τ F via S. Suppose that E admits a Hausdorff uo-Lebesgue topology τ E . The uniqueness of such a topology then implies that u E τ E = τ E . In the sequel we shall use this result from [6] and [24] a few times.
Finally, the characteristic function of a set S will be denoted by χ S , and the identity operator on a vector space will be denoted by I.

ABSOLUTE STRONG OPERATOR TOPOLOGIES ON ob (E, F)
Let E and F be vector lattices, where F is Dedekind complete. In this section, we start by showing how topologies can be introduced on vector sublattices of ob (E, F) that can be regarded as absolute strong operator topologies; see Corollary 3.5 and Remark 3.7, below. Once this is known to be possible, it is easy to relate this to o-Lebesgue topologies and uo-Lebesgue topologies on regular vector sublattices of ob (E, F). In particular, we shall see that every regular vector sublattice of ob (E, F) admits a (necessarily unique) Hausdorff uo-Lebesgue topology when F admits a Hausdorff o-Lebesgue topology; see Corollary 4.5, below.
When restricted to the orthomorphisms on a Dedekind complete vector lattice, the picture simplifies; see Section 8. In particular, the restrictions of absolute strong operator topologies are then simply strong operator topologies.
The construction in the proof of the following result is an adaptation of that in the proof of [11,Theorem 3.1]. The latter construction is carried out under minimal hypotheses and uses neighbourhood bases at zero as in [24, proof of Theorem 2.3] rather than Riesz pseudo-norms. Such an approach enables one to also understand various 'pathologies' in the literature from one central result; see [11,Example 3.10]. It is for this reason of maximum flexibility that we also choose such a neighbourhood approach here.  (a) for all T ∈ and s ∈ S, |ǫT ||s| is a (possibly non-Hausdorff ) linear topology on .
Proof. Suppose that τ F is a (not necessarily Hausdorff) locally solid additive topology on F.
It is clear from the required translation invariance of ASOT S τ F that it is unique, since the nets that are ASOT S τ F -convergent to zero are prescribed.
For its existence, we take a τ F -neighbourhood base {U λ } λ∈Λ of zero in F that consists of solid subsets of F. For x ∈ I S and λ ∈ Λ, we set The V λ,x 's are solid subsets of ob (E, F) since the U λ are solid subsets of F. Set We shall now verify that 0 satisfies the necessary and sufficient conditions in [17, Theorem 3 on p. 46] to be a base of neighbourhoods of zero for an additive topology on ob (E, F).
It is clear that There exists a µ ∈ Λ such that U µ + U µ ⊆ U λ , and it is easy to see that then An appeal to [17, Theorem 3 on p. 46] now yields that 0 is a base of neighbourhoods of zero for an additive topology on ob (E, F) that we shall denote by ASOT S τ F . It is a direct consequence of its definition that, for a net (T α ) α∈A Using the fact that τ F is a locally solid additive topology on F, it is routine to verify that the latter condition is equivalent to the condition that |T |x as well as to the condition that |T α ||s| τ F − → 0 for all s ∈ S. We turn to the statements in the parts (1)-(4).
We prove that part (4b) implies part (4a). Take T ∈ . Then ǫT ASOT S τ F | −−−−−−→ 0 as ǫ → 0 in . By construction, this implies that (and is, in fact, equivalent to) the fact that |ǫT ||s| Remark 3.2. It is clear from the convergence criteria for nets that the topologies ASOT S 1 τ F and ASOT S 2 τ F are equal when I S 1 = I S 2 . One could, therefore, work with ideals from the very start, but it seems worthwhile to keep track of a smaller set of presumably more manageable 'test vectors'. See also the comments preceding Theorem 4.3, below.
It is easy to see that then |T α |x τ F − → 0 uniformly on every order bounded subset of I S , so that then also T α x τ F − → 0 uniformly on every order bounded subset of I S .
Definition 3.4. The topology ASOT S τ F in Theorem 3.1 is called the absolute strong operator topology that is generated by τ F via S. We shall comment on this nomenclature in Remark 3.7, below.
The following result, which can also be obtained using Riesz pseudo-norms, is clear from Theorem 3.1.

Corollary 3.5. Let E and F be vector lattices, where F is Dedekind complete, and let τ F be a (not necessarily Hausdorff ) locally solid linear topology on F. Take a vector sublattice of ob (E, F) and a non-empty subset S of E.
There exists a unique additive topology ASOT S τ F on such that, for a net Let I S be the ideal of E that is generated by S. For a net (T α ) α∈A in , T α

on the group is, in fact, a locally solid linear topology on the vector lattice . When τ F is a Hausdorff topology on F, then ASOT S τ F is a Hausdorff topology on if and only if I S separates the points of .
Remark 3.6. Although in the sequel of this paper we shall mainly be interested in the nets that are convergent in a given topology, let us still remark that is possible to describe an explicit ASOT S τ F -neighbourhood base of zero in . Take a τ F -neighbourhood base {U λ } λ∈Λ of zero in F that consists of solid subsets of F. For λ ∈ Λ and x ∈ I S , set Remark 3.7. It is not difficult to see that ASOT S τ F is the weakest locally solid linear topology τ on such that, for every x ∈ I S , the map T → T x is a τ -τ F continuous map from into F. It is also the weakest linear topology τ ′ on such that, for every x ∈ I S , the map T → |T |x is a τ ′ -τ F continuous map from into F. The latter characterisation is our motivation for the name 'absolute strong operator topology'.
Take F = and S = E. Then ASOT E τ is what is commonly known as the absolute weak * -topology on E . There is an unfortunate clash of 'weak' and 'strong' here that appears to be unavoidable.

Remark 3.8.
For comparison with Remark 3.7, and to make clear the role of the local solidness of the topologies in the present section, we mention the following, which is an easy consequence of [1, Theorem 5.6], for example. Let E and F be vector spaces, where F is supplied with a (not necessarily) Hausdorff linear topology τ F . Take a linear subspace of the vector space of all linear maps from E into F, and take a non-empty subset S of E. Then there exists a unique (not necessarily Hausdorff) linear topology SOT S τ F on such that, for is Hausdorff if and only if S separates the points of . This strong operator topology SOT S τ F on that is generated by τ F via S, is the weakest linear topology τ on such that, for every s ∈ S, the map T → T x is τ -τ F -continuous.

LATTICES OF OPERATORS
To arrive at results concerning o-Lebesgue topologies and uo-Lebesgue topologies on regular vector sublattices of operators, we need a preparatory result for which we are not aware of a reference. Given its elementary nature, we refrain from any claim to originality. It will re-appear at several places in the sequel.

Lemma 4.1. Let E and F be vector lattices, where F is Dedekind complete, and let be a regular vector sublattice of
Proof. By the regularity of , we also have that T α o − → 0 in ob (E, F). Hence there exists a net (S β ) β∈B in ob (E, F) such that S β ↓ 0 in ob (E, F) and with the property that, for every β 0 ∈ B, there exists an α 0 ∈ A such that |T α | ≤ S β 0 for all α ∈ A such that α ≥ α 0 . We know from [3, Theorem 1.18], for example, that S β x ↓ 0 for all x ∈ E + . Since |T α x| ≤ |T α |x for x ∈ E + , it then follows We can now show that the o-Lebesgue property of a locally solid linear topology on the Dedekind complete codomain is inherited by the associated absolute strong operator topology on a regular vector sublattice of operators.  Proof. In view of Corollary 3.5, we merely need to show that, for a net (T α ) α∈A We conclude by showing that every regular vector sublattice of ob (E, F) admits a (necessarily unique) Hausdorff uo-Lebesgue topology when the Dedekind complete codomain F admits a Hausdorff o-Lebesgue topology. It is the unbounded topology that is associated with the members of a family of absolute strong operator topologies on the vector sublattice, with all members yielding the same result. Our most precise result in this direction is the following. The convergence criterion in part (2) is a 'minimal one' that is convenient when one wants to show that a net is convergent, whereas the criterion in part (3) maximally exploit the known convergence of a net. of , and a non-empty subset S of E. Then u ASOT S τ F is a uo-Lebesgue topology on . We let I S denote the ideal of E that is generated by S, and I the ideal of that is generated by . For a net (T α ) α∈A in , the following are equivalent: Suppose that τ F is actually a Hausdorff o-Lebesgue topology on F. Then the following are equivalent: Hausdorff uo-Lebesgue topology on ; (ii) I S separates the points of and I is order dense in .
In that case, the Hausdorff uo-Lebesgue topology u ASOT S τ F on is the restriction of the (necessarily unique) Hausdorff uo-Lebesgue topology on ob (E, F), i.e., of u ob (E,F ) ASOT E τ F , and the criteria in (1), (2), and (3) are also equivalent to: Proof. It is clear from Proposition 4.2 and [11, Proposition 4.1] that u ASOT S τ F is a uo-Lebesgue topology on . The two convergence criteria for nets follow from the combination of those in [11, Theorem 3.1] and in Corollary 3.5.
According to [11,Proposition 4.1], u ASOT S τ F is a Hausdorff topology on if and only if ASOT S τ F is a Hausdorff topology on and I is order dense in . An appeal to Proposition 4.2 then completes the proof of the necessary and sufficient conditions for u ASOT S τ F to be Hausdorff.
Suppose that τ F is actually also Hausdorff, that I S separates the points of , and that I is order dense in . From what we have already established, it is clear that u ob (E,F ) ASOT E τ F is a (necessarily unique) Hausdorff uo-Lebesgue topology on ob (E, F). Since the restriction of a Hausdorff uo-Lebesgue topology on a vector lattice to a regular vector sublattice is a (necessarily unique) Hausdorff uo-Lebesgue topology on the vector sublattice (see [24, Proposition 5.12]), the criterion in part (4) follows from that in part (3) As a consequence of the constructions of unbounded and absolute strong operator topologies, { V λ, T ,x : λ ∈ Λ, T ∈ I , x ∈ I S } is then a u ASOT S τ Fneighbourhood base of zero in .
The following consequence of Theorem 4.3 will be sufficient in many situations.

Corollary 4.5. Let E and F be vector lattices, where F is Dedekind complete. Suppose that F admits a Hausdorff o-Lebesgue topology τ F .
Take a regular vector sublattice of ob (E, F). Then admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ . This topology equals u ASOT E τ F , and is also equal to the restriction to of the Hausdorff uo-Lebesgue topology For a net (T α ) α∈A in , the following are equivalent: Remark 4.6. There can, sometimes, be other ways to see that a given regular vector sublattice of ob (E, F) admits a Hausdorff uo-Lebesgue topology. For example, suppose that F oc separates the points of F. For x ∈ E and ϕ ∈ F oc , the map T → ϕ(T x) defines an order continuous linear functional on oc (E, F), and it is then clear that the order continuous dual of oc (E, F) separates the points of oc (E, F). Hence oc (E, F) can also be supplied with a Hausdorff uo-Lebesgue topology as in [11,Theorem 5.2] which, in view of its uniqueness, coincides with the one as supplied by Corollary 4.5.

COMPARING UNIFORM AND STRONG CONVERGENCE STRUCTURES ON
ob (E, F) Suppose that E and F are vector lattices, where F is Dedekind complete. As explained in Section 1, there exist a uniform and a strong convergence structure on ob (E, F) for each of order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology. In this section, we investigate what the inclusion relations are between the members of each of these three pairs. For example, is it true that the uniform (resp. strong) order convergence of a net of order bounded operators implies its strong (resp. uniform) order convergence to the same limit? We shall show that only one of the six conceivable implications is valid in general, and that the others are not even generally valid for uniformly bounded sequences of order continuous operators on Banach lattices. Whilst the failures of such general implications may, perhaps, not come as too big a surprise, the positive results for orthomorphisms (see Theorems 9.4,9.5,9.7, and 9.10, below) may serve to indicate that they are less evident than one would think at first sight.
For monotone nets in ob (E, F), however, the following result shows that then even all four (or six) notions of convergence in ob (E, F) coincide.
Proposition 5.1. Let E and F be vector lattices, where F is Dedekind complete, and let (T α ) α∈A be a monotone net in ob (E, F). The following are equivalent: Suppose that, in addition, F admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ F , so that ob (E, F) also admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ ob (E,F ) by Corollary 4.5. Then (1)-(4) are also equivalent to: Proof. We may suppose that T α ↓ and that x ∈ E + . For order bounded nets in a vector lattice, order convergence and unbounded order convergence are equivalent. Passing to an order bounded tail of (T α ) α∈A , we thus see that the parts (1) and (2) are equivalent. Similarly, the parts (3) and (4) (2) and (5) are equivalent, as are the parts (4) and (6).
We shall now give five examples to show that each of the remaining five conceivable implications between a corresponding uniform and strong convergence structures on ob (E, F) is not generally valid. In each of these examples, we can even take E = F to be a Banach lattice, and for the net (T α ) α∈A we can even take a uniformly bounded sequence (T n ) ∞ n=1 of order continuous operators on E. Example 5.2. We give an example of a uniformly bounded sequence (T n ) ∞ n=1 of positive order continuous operators on a Dedekind complete Banach lattice E with a strong order unit, such that T n because the sequence is not even order bounded in ob (E).
We choose ℓ ∞ ( ) for E = F. For n ≥ 1, we set T n := S n , where S is the right shift operator on E. The T n are evidently positive and of norm one. A moment's thought shows that they are order continuous. Furthermore, it is easy to see that We shall now show that { T n : n ≥ 1 } is not order bounded in ob (E). For this, we start by establishing that the T n are mutually disjoint. Let (e i ) ∞ i=1 be the standard sequence of unit vectors in E. Take m = n and i ≥ 1. Since e i is an atom, the Riesz-Kantorovich formula for the infimum of two operators shows that Hence (T m ∧ T n ) vanishes on the span of the e i . Since this span is order dense in E, and since T n ∧ T m ∈ oc (E), it follows that T n ∧ T m = 0.
We can now show that (T n ) ∞ n=1 is not order bounded in ob (E). Indeed, suppose that T ∈ ob (E) is a upper bound for all T n . Set e := ∞ i=1 e i . Then, for all N ≥ 1, This shows that Te cannot be an element of ℓ ∞ . We conclude from this contradiction that (T n ) ∞ n=1 is not order bounded in ob (E). Example 5.3. We give an example of a uniformly bounded sequence (T n ) ∞ n=1 of positive order continuous operators on a Dedekind complete Banach lattice E with a strong order unit, such that T n We choose ℓ ∞ ( ) for E = F. For n ≥ 1, we set T n := S n , where S is the right shift operator on E. Just as in Example 5.2, the T n are positive order continuous operators on E of norm one that are mutually disjoint. Since disjoint sequences in vector lattices are unbounded order convergent to zero (see [14, Corollary 3.6]), we have T n uo − → 0 in ob (E). On the other hand, if we let e be the two-sided sequence that is constant 1, then T n e = e for all n ≥ 1. Hence (T n e) ∞ n=1 is not unbounded order convergent to zero in E. For our next example, we require a preparatory lemma.
Since n is arbitrary, we see that (T S ∧ I)e = 0. Because 0 ≤ T S ∧ I ≤ I, T S ∧ I is order continuous. From the fact that the positive order continuous operator T S ∧ I vanishes on the weak order unit e of L p ([0, 1], , µ), we conclude that T S ∧ I = 0.
Example 5.5. We give an example of a uniformly bounded sequence (T n ) ∞ n=1 of order continuous operators on a separable reflexive Banach lattice E with a weak order unit, such that T n x . Let µ be the Lebesgue measure on the Borel σ-algebra of [0, 1], and let 1 ≤ p ≤ ∞. For E we choose L p ([0, 1], , µ), so that E is reflexive for 1 < p < ∞. For n ≥ 1, we let n be the sub-σ-algebra of that is generated by the intervals S n,i := [(i − 1)/2 n , i/2 n ] for i = 1, . . . , 2 n , and we let n : E → E be the corresponding conditional expectation. By On the other hand, it is not true that n τ ob (E) − −−− → I. To see this, we note that, by [5, Example 10.1.2], every n is a linear combination of operators as in Lemma 5.4. Hence n ⊥ I for all n. Since τ ob (E) is a locally solid linear topology, a possible τ ob (E) -limit of the n is also disjoint from I, hence cannot be I itself.
On setting T n := n − I for n ≥ 1, we have obtained a sequence of operators as desired.
Example 5.6. We give an example of a uniformly bounded sequence (T n ) ∞

n=1 of positive order continuous operators on a Dedekind complete Banach lattice E with a strong order unit that admits a Hausdorff uo-Lebesgue topology, such that
We choose E, the T n ∈ ob (E), and e ∈ E as in Example 5.3. There are several ways to see that E admits a Hausdorff uo-Lebesgue topology. This follows most easily from the fact that E is atomic (see [24,Lemma 7.4]) and also from [11, Theorem 6.3] in the context of measure spaces. By Corollary 4.5, ob (E) then also admits such a topology. Since we already know from Example 5.3 that T n uo − → 0, we also have that T n τ ob (E) − −−− → 0. On the other hand, the fact that T n e = e for n ≥ 1 evidently shows that (T n e) ∞ n=1 is not τ E -convergent to zero in E. Example 5.7. We note that Example 5.5 also gives an example of a uniformly bounded sequence (T n ) ∞ n=1 of order continuous operators on a separable reflexive Banach lattice E with a weak order unit that admits a Hausdorff uo-Lebesgue topology, such that T n x

ORTHOMORPHISMS
In this section, we review some material concerning orthomorphism and establish a few auxiliary result for use in the present paper and in future ones. Let E be a vector lattice. We recall from [3, Definition 2.41] that an operator on E is called an orthomorphism when it is a band preserving order bounded operator. An orthomorphism is evidently disjointness preserving, it is order continuous (see [

3, Theorem 2.44]), and its kernel is a band (see [3, Theorem 2.48]). We denote by Orth(E) the collection of all orthomorphism on E.
Even when E is not Dedekind complete, the supremum and infimum of two orthomorphisms S and T in E always exists in ob (E). In fact, we have Let E be a vector lattice, let T ∈ ob (E), and let λ ≥ 0. Using [3, Theorem 2.40], it is not difficult to see that the following are equivalent: (1) −λI ≤ T ≤ λI; (2) |T | exists in ob (E), and |T | ≤ λI; The set of all such T is a unital subalgebra (E) of Orth(E) consisting of ideal preserving order bounded operators on E. It is called the ideal centre of E.
Let E be a vector lattice, and define the stabiliser of E, denoted by (E), as the set of linear operators on E that are ideal preserving. It is not required that these operators be order bounded, but this is nevertheless always the case. In fact, (E) is a unital subalgebra of Orth(E) for every vector lattice E (see [25, Proposition 2.6]), so that we have the chain For every Banach lattice E, Orth(E) is a unital Banach subalgebra of the bounded linear operators on E in the operator norm. This follows easily from the facts that bands are closed and that a band preserving operator on a Banach lattice is automatically order bounded; see [3,Theorem 4.76].
Let E be a Banach lattice. Since the identity operator is an order unit of Orth(E), we can introduce the order unit norm · I with respect to I on Orth(E) by setting T I := inf{ λ ≥ 0 : |T | ≤ λI } for T ∈ Orth(E). Then T = T I for all T ∈ Orth(E); see [25, Proposition 4.1]. Since we already know that Orth(E) is complete in the operator norm, it follows that Orth(E), when supplied with · = · I , is a unital Banach lattice algebra that is also an AM-space. When E is a Dedekind complete Banach lattice, then evidently T = T I = |T | I = |T | = T r for T ∈ Orth(E).

Hence Orth(E) is then also a unital Banach lattice subalgebra of the Banach lattice algebra of all order bounded operators on E in the regular norm.
Let E be Banach lattice. It is clear from the above that (Orth(E), · ) = (Orth(E), · I ) is a unital Banach f -algebra in which its identity element is also a (positive) order unit. The following result is, therefore, applicable with = Orth(E) and e = I. It shows, in particular, that Orth(E) is isometrically Banach lattice algebra isomorphic to a C(K)-space. Both its statement and its proof improve on the ones in [ Proof. Since ( , · I ) is an AM-space with order unit e, there exist a compact Hausdorff space K ′ and an isometric surjective lattice homomorphism ψ ′ : → C(K ′ ) such that ψ ′ (e) = 1; see [21, Theorem 2.1.3] for this result of Kakutani's, for example. Via this isomorphism, the f -algebra multiplication on C(K ′ ) provides the vector lattice with a multiplication that makes into an f -algebra with e as its positive multiplicative identity element. Such a multiplication is, however, unique; see [3, Theorem 2.58]. Hence ψ ′ also preserves multiplication, and we conclude that ψ ′ : → C(K ′ ) is an isometric surjective Banach lattice algebra isomorphism.
We now turn to . It is clear that · e is Banach subalgebra of . After moving to the C(K ′ )-model for that we have obtained, [13,Lemma 4.48] shows that · e is also a vector sublattice of . Hence · e is a Banach fsubalgebra of . When e ∈ · e , we can then apply the first part of the proof to · e , and obtain a compact Hausdorff space K and an isometric surjective Banach lattice algebra isomorphism ψ : We now proceed to show that E and Orth(E) have isomorphic universal completions. We start with a preparatory lemma.

Proposition 6.2. Let E be a Dedekind complete vector lattice, and let x ∈ E. Let I x be the principal ideal of E that is generated by x, let B x be the principal band
in E that is generated by x, let P x : E → B x be the corresponding order projection, and let P x be the principal ideal of ob (E) that is generated by P x . For T ∈ P x , set ψ x (T ) := T |x|. Then ψ x (T ) ∈ I x , and: (1) the map ψ x : P x → I x is a surjective vector lattice isomorphism such that ψ x (P x ) = |x|; (2) P x = P x (E).
Proof. Take T ∈ P x . There exists a λ ≥ 0 such that |T | ≤ λP x , and this implies that |T y| ≤ λP x | y| for all y ∈ E. This shows that T |x| ∈ I x , so that ψ x maps P x into I x ; it also shows that T (B d x ) = {0}. Suppose that T |x| = 0. Since the kernel of T is a band in E, this implies that T vanishes on B x . We already know that it vanishes on B d x . Hence T = 0, and we conclude that ψ x is injective. We show that ψ x is surjective. Let y ∈ I x . Take a λ > 0 such that 0 ≤ | y/λ| ≤ |x|. An inspection of the proof of [3,Theorem 2.49] shows that there exists a T ∈ (E) with T |x| = y/λ. Since λT P x ∈ P x and (λT P x )|x| = y, we see that ψ x is surjective. Finally, it is clear from equation (1) that ψ x is a vector lattice homomorphism. This completes the proof of part (1).
We turn to part (2). It is clear that P x ⊇ P x (E). Take T ∈ P x ⊆ (E). Then also P x T ∈ P x . Since ψ x (T ) = ψ x (P x T ), the injectivity of ψ x on P x implies that T = P x T ∈ P x (E).
The first part of Proposition 6.2 is used in the proof of our next result.

Proposition 6.3. Let E be a Dedekind complete vector lattice. Then there exist an order dense ideal I of E and an order dense ideal of Orth(E) such that I and are isomorphic vector lattices.
, and the vector lattice isomorphism ψ x α : P x α → I x α be as in Proposition 6.2.
Since the x α 's are mutually disjoint, it is clear that the ideal α∈A I x α of E is, in fact, an internal direct sum α∈A I x α . Since the disjoint system is maximal, α∈A I x α is an order dense ideal of E. It follows easily from equation (1) that the P x α are also mutually disjoint. They even form a maximal disjoint system in Orth(E). To see this, suppose that T ∈ Orth(E) is such that |T | ∧ P x α = 0 for all α ∈ A. Then (|T |x α ) ∧ x α = (|T | ∧ P x α )x α = 0 for all α ∈ A. Since |T | is band preserving, this implies that |T |x α = 0 for all α ∈ A. The fact that the kernel of |T | is a band in E then yields that |T | = 0. Just as for E, we now conclude that the ideal α∈A P x α of Orth(E) is an internal direct sum α∈A P x α that is order dense in Orth(E). Since α∈A ψ x α : α∈A P x α → α∈A I x α is a vector lattice isomorphism by Proposition 6.2, the proof is complete.
It is generally true that a vector lattice and an order dense vector sublattice of it have isomorphic universal completions; see [2, Theorems 7.21 and 7.23]. Proposition 6.3 therefore implies the following.

Corollary 6.4. Let E be a Dedekind complete vector lattice. Then the universal completions of E and of Orth(E) are isomorphic vector lattices.
The previous result enables us to relate the countable sup property of E to that of Orth(E). We recall that vector lattice E has the countable sup property when, for every non-empty subset S of E that has a supremum in E, there exists an at most countable subset of S that has the same supremum in E as S. In parts of the literature, such as in [20] and [26], E is then said to be order separable. We also recall that a subset of a vector lattice is said to be an order basis when the band that it generates is the whole vector lattice. We shall now establish a uniform order boundedness principle for orthomorphisms. It will be needed in the proof of Theorem 9.5, below. Proposition 6.6. Let E be a Dedekind complete vector lattice, and let { T α : α ∈ A } be a non-empty subset of Orth(E). The following are equivalent: (1) { T α : α ∈ A } is an order bounded subset of ob (E); (2) for each x ∈ E, { T α x : α ∈ A } is an order bounded subset of E.
Before proceeding with the proof, we remark that, since Orth(E) is a projection band in ob (E), the order boundedness of the net could equivalently have been required in Orth(E).
Proof. It is trivial that part (1) implies part (2). We now show the converse. Take an x ∈ E + . The hypothesis in part (2), together with equation (2), shows that { |T α |x : α ∈ A } is an order bounded subset of E. Hence the same is true for Then the same is true for { |T α | : α ∈ A }, as desired. Proposition 6.6 fails for nets of general order bounded operators. It can, in fact, already fail for a sequence of order continuous operators on a Banach lattice, as is shown by the following example. Example 6.7. Let E := ℓ ∞ ( ), and let (e i ) ∞ i=1 be the standard unit vectors in E. Let S ∈ oc (E) be the right shift, and set T n := S n for n ≥ 1. It is easy to see that (T n x) ∞ n=1 is order bounded in E for all x ∈ E. We shall show, however, that (T n ) ∞ n=1 is not order bounded in ob (E). To see this, we first note that T m ⊥ T n for m, n ≥ 1 with m = n. Indeed, for all i ≥ 1, we have 0 ≤ (T m ∧ T m )e i ≤ T m (e i ) ∧ T n (e i ) = e m+i ∧ e n+i = 0. Hence (T m ∧ T n )x = 0 for all x ∈ I, where I is the ideal of E that is spanned by { e i : i ≥ 1 }. Since I is order dense in E and T n ∧ T m ∈ oc (E), it follows that T n ∧ T m = 0 for all m, n ≥ 1 with m = n. Suppose that T is an upper bounded of (T n ) ∞ n=1 in ob (E). Set e := ∞ i=1 e i . Using the disjointness of the T n , we have for all n ≥ 1, which is impossible. So (T n ) ∞ n=1 is not order bounded in ob (E). As a side result, we note the following consequence of Proposition 6.6. It is an ordered analogue of the familiar result for a sequence of bounded operators on a Banach space. Corollary 6.8. Let E be a Dedekind complete vector lattice, and let (T n ) ∞ n=1 be a sequence in Orth(E). Suppose that the sequence (T n x) ∞ n=1 is order convergent in E for all x ∈ E. Then { T n : n ≥ 1 } is an order bounded subset of ob (E). For x ∈ E, define T : E → E by setting Then T ∈ Orth(E).
Proof. Using Proposition 6.6, it is clear that T is a linear and order bounded operator on E. Since each of the T n is a band preserving operator, the same is true for T . Hence T is an orthomorphism on E.
We conclude by giving some estimates for orthomorphisms that will be used in the sequel. As a preparation, we need the following extension of [3, Exercise 1.3.7]. Lemma 6.9. Let E be a vector lattice with the principal projection property. Take x, y ∈ E. For λ ∈ , let P λ denote the order projection in E onto the band generated by (x − λ y) + . Then λP λ y ≤ P λ x. When x, y ∈ E + and λ ≥ 0, then x ≤ λ y + P λ x.
Proof. The first inequality follows from the fact that For the second inequality, we note that x − λ y ≤ (x − λ y) + = P λ (x − λ y) + for all x, y, and λ. When x, y ∈ E + and λ ≥ 0, then (x − λ y) + ≤ x + = x, so that Proposition 6.10. Let E be a Dedekind complete vector lattice, and letT∈Orth(E) + . For λ > 0, let P λ be the order projection in Orth(E) onto the band generated by (T − λI) + in Orth(E). There exists a unique order projection P λ in E such that P λ (S) = P λ S for all S ∈ Orth(E). Furthermore: (1) λP λ ≤ P λ T ≤ T ; (2) T ≤ λI + P λ T ; (3) (P λ T x) ∧ y ≤ 1 λ T y for all x, y ∈ E + . Proof. Since 0 ≤ P λ ≤ I Orth(E) , it follows from [3, Theorem 2.62] that there exists a unique P λ ∈ Orth(E) with 0 ≤ P λ ≤ I such that P λ (S) = P λ S for all S ∈ Orth(E). The fact that P λ is idempotent implies that P λ is also idempotent. Hence P λ is an order projection.
The inequalities in the parts (1) and (2) are then a consequence of those in Lemma 6.9. For part (3), we note that (P λ T x) ∧ y is in the image of the projection P λ . Since order projections are vector lattice homomorphisms, we have, using part (1) in the final step, that We shall have use for the following corollary, which has some appeal of its own.

Corollary 6.11. Let E be a Dedekind complete vector lattice, and let T ∈ Orth(E) + . Then
for all x, y ∈ E + and λ > 0.
Proof. For λ > 0, we let P λ be the order projection in Orth(E) onto the band generated by (T − λI) + in Orth(E). According to Proposition 6.10, there exists a unique order projection P λ in E such that P λ (S) = P λ S for all S ∈ Orth(E). By applying part (2) of Proposition 6.10 in the first step and its part (3) in the third, we have, for x, y ∈ E + ,

CONTINUITY PROPERTIES OF ORTHOMORPHISMS
Orthomorphisms preserve order convergence of nets. In this short section, we show that they also preserve unbounded order convergence and-when applicable-convergence in the Hausdorff uo-Lebesgue topology. Before doing so, let us note that this is in contrast to the case of general order bounded operators. Surely, there exist order bounded operators that are not order continuous. For the remaining two convergence structures, we consider ℓ 1 with its standard basis (e n ) ∞ n=1 . It follows from [14, Corollary 3.6] that e n uo − → 0. There are several ways to see that ℓ 1 admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ ℓ 1 . This follows from the fact that its norm is order continuous (see [24, p. 993]), from the fact that it is atomic (see [24,Lemma 7.4]), and from a result in the context of measure spaces (see [11,Theorem 6.3]). The latter two results also show that τ ℓ 1 is the topology of coordinatewise convergence. In particular, e n τ ℓ 1 −→ 0 which is, of course, also a consequence of the fact that e n uo − → 0. Define T : ℓ 1 → ℓ 1 by setting T x := ∞ n=1 x n e 1 for x = ∞ n=1 x n e n ∈ ℓ 1 . Since Te n = e 1 for all n ≥ 1, the order continuous positive operator T on ℓ 1 preserves neither uo-convergence nor τ ℓ 1 -convergence of sequences in ℓ 1 .

Proposition 7.1. Let E be a Dedekind complete vector lattice, and let T ∈ Orth(E).
Proof. Using equation (2), one easily sees that we may suppose that T and the x α 's are positive. Let B T (E) denote the band in E that is generated by T (E). Take a y ∈ T (E) + . Since a positive orthomorphism is a lattice homomorphism, there exists an x ∈ E + such that y = T x. Using the fact that x α uo − → 0 in E, the order continuity of T then implies that For the case of a Hausdorff uo-Lebesgue topology, we need the following preparatory result that has some independent interest. Lemma 9.9 is of the same flavour.

Proposition 7.2. Let E be a Dedekind complete vector lattice that admits a (not necessarily Hausdorff ) locally solid linear topology τ E , and let T ∈ Orth(E). Sup-
Proof. As in the proof of Proposition 7.1, we may suppose that T and the x α are positive. For n ≥ 1, we let P n be the order projection in Orth(E) onto the band generated by (T − nI) + in Orth(E) again, so that again there exists a unique order projection P n in E such that P n (S) = P n S for all S ∈ Orth(E). Fix e ∈ E + . Take a solid τ E -neighbourhood U of 0 in E, and choose a τ E -neighbourhood V of 0 such that V +V ⊆ U. Take an n 0 ≥ 1 such that Te/n 0 ∈ V . As x α τ E − → 0, there exists an α 0 ∈ A such that n 0 x α ∈ V for all α ≥ α 0 . By applying Corollary 6.11 in the first step, we have, for all α ≥ α 0 , The solidness of V then implies that (T x α ) ∧ e ∈ U for all α ≥ α 0 . Since U and e were arbitrary, we conclude that T α x Since the unbounded topology u E τ E that is generated by a Hausdorff uo-Lebesgue topology τ E equals τ E again, the following is now clear.

Corollary 7.3. Let E be a Dedekind complete vector lattice that admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ E , and let T ∈ Orth(E).
Suppose

TOPOLOGIES ON Orth(E)
Let E be a Dedekind complete vector lattice, and suppose that τ E is a (not necessarily Hausdorff) locally solid additive topology on E. Take a non-empty subset S of E. According to Theorem 3.1, there exists a unique additive topology ASOT S τ E on ob (E) such that, for a net (T α ) α∈A in ob (E), T α and only if |T α ||s| τ E − → 0 for all s ∈ S. When (T α ) α∈A ⊆ Orth(E), equation (2) and the local solidness of τ E imply that this convergence criterion is also equivalent to the one that T α s τ E − → 0 for all s ∈ S. Hence on subsets of Orth(E), an absolute strong operator topology that is generated by a locally solid additive topology on E coincides with the corresponding strong operator topology. In order to remind ourselves of the connection with the topology on the enveloping vector lattice ob (E) of Orth(E), we shall keep writing ASOT S τ F when considering the restriction of this topology to subsets of Orth(E), rather than switch to, e.g., SOT S τ F .
The above observation can be used in several results in Section 3. For the ease of reference, we include the following consequence of Corollary 3.5. For a Dedekind complete vector lattice E, Orth(E), being a band in ob (E), is a regular vector sublattice of ob (E). A regular vector sublattice of Orth(E) is, therefore, also a regular vector sublattice of ob (E), and Proposition 4.2 then shows how o-Lebesgue topologies on can be obtained from an o-Lebesgue topology on E as (absolute) strong operator topologies. In particular, this makes the fact that part (1) of Proposition 8.2 implies its part (2) more concrete. The fact that part (1) implies part (2) is made more concrete as a special case of the following consequence of Theorem 4.3.

Theorem 8.3. Let E be a Dedekind complete vector lattice. Suppose that E admits
an o-Lebesgue topology τ E . Take a regular vector sublattice of Orth(E), a nonempty subset of , and a non-empty subset S of E. Then u ASOT S τ E is a uo-Lebesgue topology on . We let I S denote the ideal of E that is generated by S, and I the ideal of that is generated by . For a net (T α ) α∈A in , the following are equivalent: Suppose that τ E is actually a Hausdorff o-Lebesgue topology on E. Then the following are equivalent: (a) u ASOT S τ E is a (necessarily unique) Hausdorff uo-Lebesgue topology on ; (b) I S separates the points of and I is order dense in . In that case, the Hausdorff uo-Lebesgue topology u ASOT S τ E on is the restriction of the (necessarily unique) Hausdorff uo-Lebesgue topology on ob (E, F), i.e., of u ob (E,F ) ASOT E τ E , and the criteria in (1), (2), and (3) are also equivalent to: − → 0 for all T ∈ ob (E) and x ∈ E.

COMPARING UNIFORM AND STRONG CONVERGENCE STRUCTURES ON Orth(E)
Let E and F be vector lattices, where F is Dedekind complete, and let (T α ) α∈A be a net in ob (E, F). In Section 5, we studied the relation between uniform and strong convergence of (T α ) α∈A for order convergence, unbounded order convergence, and-when applicable-convergence in the Hausdorff uo-Lebesgue topology. In the present section, we consider the case where (T α ) α∈A is actually contained in Orth(E). As we shall see, the relation between uniform and strong convergence is now much more symmetrical than in the general case of Section 5; see Theorem 9.4 (and Theorem 9.5), Theorem 9.7, and Theorem 9.10, below.
These positive results might, perhaps, lead one to wonder whether some of the three uniform convergence structures under consideration might actually even be identical for the orthomorphisms. This, however, is not the case. There Using χ [(k−1)2 −n ,k2 −n ] for n ≥ 1 and k = 1, . . . , 2 n , one easily finds a sequence that is convergent to zero in measure, but that is not convergent in any point of [0, 1].
We now start with uniform and strong order convergence for nets of orthomorphisms. For this, we need a few preparatory results. The first one is on general order continuous operators. Lemma 9.1. Let E be a Dedekind complete vector lattice, let (T α ) α∈A be a decreasing net in oc (E) + , and let F be an order dense vector sublattice of E. The following are equivalent: Proof. We need to show only that part (1) implies part (2). Let T ∈ ob (E) be such that T α ↓ T in ob (E). Then T ∈ oc (E) + . The hypothesis under part (1) and [3,Theorem 1.18] imply that T x = 0 for all x ∈ F + . Since F is order dense in E and T is order continuous, it now follows from [3, Theorem 1.34] that T = 0. Using [3, Theorem 1.18] once more, we conclude that T α x ↓ 0 in E for all x ∈ E + , and the statement in part (2) follows. Proposition 9.2. Let E be a Dedekind complete vector lattice, let (T α ) α∈A be a decreasing net in Orth(E) + , and let S be a non-empty subset of E. The following are equivalent: In particular, if E has a positive weak order unit e, then T α Proof. We need to show only that part (1) implies part (2). Take y ∈ I + S . There exist s 1 , . . . , s n ∈ S and λ 1 , . . . , and the assumption then implies that T α y ↓ 0 in E. Since orthomorphisms preserve bands, we have T α y ∈ B S for all α ∈ A, and the fact that B S is an ideal of E now shows that T α y ↓ 0 in B S . It follows that T α y o − → 0 in B S for all y ∈ I S . Since the restriction of each T α to the regular vector sublattice B S of E is again order continuous, and since I S is an order dense vector sublattice of the vector lattice B S , Lemma 9.1 implies that T α y o − → 0 in B S for all y ∈ B S . The fact that B S is a regular vector sublattice of E then yields that T α y o − → 0 in E for all y ∈ B S .

Lemma 9.3. Let E be a Dedekind complete vector lattice, and let
be a subset of Orth(E) that is bounded above in ob (E). Then, for x ∈ E + , Proof. Using [2, Theorem 1.67.(b)] in the second step, we see that, for x ∈ E + , By equation (1), this equals We can now establish our main result regarding uniform and strong order convergence for nets of orthomorphisms. As for Proposition 6.6, the order boundedness of the net could equivalently have been required in Orth(E).
Since Lemma 9.3 shows that T α |s| = β≥α |T β ||s| for α ∈ A and s ∈ S, we see that T α |s| ↓ 0 in E for all s ∈ S. Proposition 9.2 then yields that T α x o − → 0 for all x ∈ B |S| = E. Using that T α ↓, it follows that T α ↓ 0 in ob (E). Since |T α | ≤ T α for α ∈ A, we see that |T α | o − → 0 in ob (E), as required.
In view of Lemma 4.1, the most substantial part of Theorem 9.4 is the fact that the parts (3) and (4) imply the parts (1) and (2). For this to hold in general, the assumption that (T α ) α∈A be order bounded is actually necessary. To see this, let Γ be an uncountable set that is supplied with the counting measure, and consider E := ℓ p (Γ ) for 1 ≤ p < ∞. Set A := { (n, S) : n ≥ 1, S ⊂ Γ is at most countably infinite } and, for (n 1 , S 1 ), (n 2 , S 2 ) ∈ A, say that (n 1 , S 2 ) ≤ (n 2 , S 2 ) when n 1 ≤ n 2 and S 1 ⊆ S 2 . For (n, S) ∈ A, define T (n,S) ∈ (E) = Orth(E) by setting T (n,S) x := nχ Γ \S x for all x : Γ → in E. Take an x ∈ E. Then the net (T (n,S) x) (n,S)∈A has a tail (T (n,S) x) (n,S)≥(1,supp x) that is identically zero. Hence T (n,S) x o − → 0 in E for all x ∈ E. We claim that (T (n,S) ) (n,S)∈A is not order convergent in Orth(E), and not even in ob (E). For this, it is sufficient to show that it does not have any tail that is order bounded in ob (E). Suppose, to the contrary, that there exist an n 0 ≥ 1, an at most countably infinite subset S 0 of Γ , and a T ∈ ob (E) such that T (n,A) ≤ T for all (n, A) ∈ A with n ≥ n 0 and A ⊇ A 0 . As Γ is uncountable, we can choose an x 0 ∈ Γ \ A 0 ; we let e x 0 denote the corresponding atom in E. Then, in particular, T (n,A 0 ) e x 0 ≤ Te x 0 for all n ≥ n 0 . Hence Te x 0 ≥ ne x 0 for all n ≥ n 0 , which is impossible.
Using Theorem 9.4 and Corollary 6.8, the following is easily established. In contrast to Theorem 9.4, there is no order boundedness in the hypotheses because this is taken care of by Corollary 6.8. Remark 9.6. Even for Banach lattices with order continuous norms, the condition that I S = E in Theorem 9.5, cannot be relaxed to B S = E as in Theorem 9.4. To see this, we choose E := c 0 and set e := n≥1 e i /i 2 , where (e i ) ∞ i=1 is the standard unit basis of E. It is clear that B e = E. For n ≥ 1, there exists a unique T n ∈ Orth(E) such that, for i ≥ 1, T n e i = ne i when i = n, and T n e i = 0 when i = n. It is clear that T n e o − → 0 in E. However, a consideration of T n i≥1 e i /i for n ≥ 1 shows that (T n ) ∞ n=1 fails to be order bounded in Orth(E), hence cannot be order convergent in Orth(E).
We continue our comparison of uniform and strong convergence structures on the orthomorphisms by considering unbounded order convergence. In that case, the result is as follows. Proof. Since Orth(E) is a regular vector sublattice of ob (E), the equivalence of the parts (1) and (2) is clear from [14,Theorem 3.2] We prove that part (2) implies part (4). Suppose that T α uo − → 0 in ob (E), so that, in particular, |T α | ∧ I o − → 0 in ob (E). Take x ∈ E. Using equation (1)  Since the net (|T α ||x|) α∈A is contained in the band B |x| , it now follows from [11,Proposition 7.4] that |T α ||x| uo − → 0 in E. As |T α ||x| = |T α x|, we conclude that T α x uo − → 0 in E. It is clear that part (4) implies part (3). We prove that part (3) implies part (2). Suppose that T α s uo − → 0 in E for all s ∈ S, so that also |T α ||s| = |T α s| uo − → 0 in E for s ∈ S. Using equation (1)  We now consider uniform and strong convergence of nets of orthomorphisms for the Hausdorff uo-Lebesgue topology. Let E be a Dedekind complete vector lattice. Suppose that E admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ E . We recall from Theorem 8.3 that Orth(E) then also admits a (necessarily unique) Hausdorff uo-Lebesgue topology τ Orth(E) , and that this topology equals u Orth(E) ASOT E τ E . Furthermore, for a net (T α ) α∈A in Orth(E),