Topological concepts in partially ordered vector spaces

Abstract

In the context of partially ordered vector spaces one encounters different sorts of order convergence and order topologies. This article investigates these notions and their relations. In particular, we study and relate the order topology presented by Floyd, Vulikh and Dobbertin, the order bound topology studied by Namioka and the concept of order convergence given in the works of Abramovich, Sirotkin, Wolk and Vulikh. We prove that the considered topologies disagree for all infinite dimensional Archimedean vector lattices that contain order units. For reflexive Banach spaces equipped with ice cream cones we show that the order topology, the order bound topology and the norm topology agree and that order convergence is equivalent to norm convergence.

1 Introduction

In the study of partially ordered vector spaces one uses topological concepts like order convergence and order continuity as can be seen for example in [1,2,3,4,5]. In particular one encounters different types of order convergence, which lead to different types of order continuity as well as different types of order topology. The different types of order continuity are for example studied in [3] and [6] and we will focus in this paper on properties and relationships of different sorts of topologies that can be defined in partially ordered vector spaces. In particular we will investigate properties and relationships of the order topology [2, 7,8,9] and the order bound topology [5, 10]. In order to define the order topology, recall that we write $x_\alpha \downarrow 0$ whenever $(x_\alpha )_{\alpha \in A}$ is a decreasing net (i.e.  $\alpha \le \beta$ implies $x_\alpha \ge x_\beta$) which satisfies $\inf _{\alpha \in A}x_\alpha =0$.

Definition 1.1

A subset $M\subseteq X$ is called an order neighbourhood of $x\in X$, if for all nets $(\check{x}_\alpha )_{\alpha \in A}$ with $\check{x}_\alpha \downarrow 0$ there is $\alpha \in A$ such that $[x-x_\alpha ,x+x_\alpha ]\subseteq M$. A subset O of X is called order open if O is an order neighbourhood of each of its points. The order topology $\tau _o(X)$ (or simply $\tau _o$) is defined as the set of all order open subsets. Complements of order open sets are called order closed.

Following [5] we define the order bound topology for Archimedean partially ordered vector spaces that contain order units as follows.

Definition 1.2

An element $u\in X_+$ is called an order unit, whenever for all $x\in X$ there is $n\in {\mathbb {N}}$ with $x\le nu$. For $x\in X$ we define the Minkowski norm as $\Vert x\Vert _u:=\inf \{\lambda >0;\, \pm x\le \lambda u\}$. Recall that whenever X is Archimedean and u is an order unit, then $\Vert \cdot \Vert _u$ is indeed a norm on X. The corresponding topology $\tau _{ob}(X)$ (or simply $\tau _{ob}$) is independent from the choice of u and called the order bound topology.

It is natural to ask under which assumptions these topologies coincide. In Corollary 4.8 and Theorem 4.11 we provide the following characterization.

Theorem A

If X is an Archimedean partially ordered vector space (that contains order units) the following statements are equivalent.

• (i) $\tau _{o}=\tau _{ob}$.

• (ii) For all nets $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow x$ there holds $x_\alpha \overset{\tau _{ob}}{\rightarrow }x$.

• (iii) X contains non-empty, order open and order bounded sets.

• (iv) There exists $x\in X_+$ such that for any net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$ there exists $\alpha \in A$ with $x_\alpha \le x$.

We will use this characterization to show that for all finite dimensional Archimedean and directed partially ordered vector spaces there holds $\tau _o=\tau _{ob}$ (Example 6.1). As $\tau _{ob}$ is a linear topology we obtain in particular that this topology is the standard topology of X, i.e. the unique linear topology on X. In Example 6.2 we will furthermore show the following.

Example

For a reflexive Banach space equipped with an ice cream cone the order topology, the order bound topology and the norm topology agree and order convergence is equivalent to norm convergence.

However, for infinite dimensional Archimedean vector lattices X (with order units) we obtain $\tau _{ob}\ne \tau _{o}$ from the following result, which follows from Corollary 4.8 and Theorem 6.4.

Theorem B

An Archimedean vector lattice contains non-empty, order open and order bounded sets, if and only if it is finite dimensional.

Note that there exists a close relation between the order topology $\tau _o$ and various concepts of order convergence [3, 6]. It is thus natural to study the interplay of order convergence and the order bound topology. The following version of order convergence appears in [1,2,3, 6, 9].

Definition 1.3

Let $x \in X$ and let $(x_\alpha )_{\alpha \in A}$ be a net in X. We say that $(x_\alpha )_{\alpha \in A}$ order converges to x (and write $x_\alpha \xrightarrow {o} x$) whenever there exists a net $(\check{x}_\beta )_{\beta \in B}$ that satisfies $\check{x}_\beta \downarrow 0$ and such that for any $\beta \in B$ there exists $\alpha _\beta \in A$ with $\pm (x_\alpha -x) \le \check{x}_\beta$ for all $\alpha \ge \alpha _\beta$. In this case we call x an order limit of $(x_\alpha )_{\alpha \in A}$.

We prove the following in Theorem 3.1.

Theorem C

Let X be an Archimedean partially ordered vector space (that contains order units). Every $\tau _{ob}$-convergent net is also order convergent (to the same limit).

The reader might have noticed that we stressed the existence of order units in our statements. We did this as the order bound topology $\tau _{ob}$ can also be defined in case X does not contain order units (and is not Archimedean). In this case $\tau _{ob}$ can be defined as the linear topology generated by the zero neighbourhood base consisting of all convex and circled sets that absorb all order intervals [5, 10]. We will see in Theorem 3.1 that for the implication discussed in Theorem C ($\tau _{ob}$-convergence implies order convergence) the Archimedean property and the existence of order units are not only sufficient, but also necessary.

We end our introduction presenting the following remarkable parallelism between order units and elements x as considered in Theorem A(iv). In order to be more precise, we say that $x\in X_+$ is net catching, whenever for any net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$ there exists $\alpha \in A$ that satisfies $x_\alpha \le x$. In Corollary 3.2, Theorem 4.6 and Proposition 4.7 we show the following.

Theorem D

Let X be an Archimedean partially ordered vector space.

• (i) The $\tau _{ob}$-interior of $X_+$ is the set of all order units. Whenever there exist order units in X there holds $\tau _{o}\subseteq \tau _{ob}$.

• (ii) The $\tau _o$-interior of $X_+$ is the set of all net catching elements. Whenever there exist net catching elements in X there holds $\tau _{ob}\subseteq \tau _{o}$.

In particular, whenever $\tau _{o}=\tau _{ob}$ an element $x\in X$ is net catching, if and only if it is an order unit.

Combining the Theorems B and D we obtain in particular that for any infinite dimensional Archimedean vector lattice that contains order units the inclusion $\tau _{o}\subseteq \tau _{ob}$ is strict. So far, we do not know, whether for all Archimedean partially ordered vector spaces there holds $\tau _{o}\subseteq \tau _{ob}$. However, in Theorem 4.15 we prove that for Archimedean vector lattices all convex order open sets are $\tau _{ob}$-open.

The paper is organized as follows. In Sect. 2 we fix some notation. In Sect. 3 we will consider the interplay of order convergence and the Minkowski norm and present a proof of Theorem C. Sect. 4 is devoted to the interplay of order units, net catching elements, the order topology and the order bound topology. In particular, we will prove the Theorems A and D and present examples. This section ends with a brief investigation of convex order open sets. In order to present more interesting examples, such as the statement about ice cream cones of reflexive Banach spaces, we will provide a sufficient condition for the existence of net catching elements in Sect. 5. This technique is used in Sect. 6. In this section we furthermore prove Theorem B.

2 Preliminaries

2.1 Nets and subnets

A mapping $x:A\rightarrow M$ is called a net, whenever A is a non-empty set equipped with a reflexive and transitive relation $\le$ that is directed, i.e. for $\alpha ,\alpha '\in A$ there exists $\alpha ''\in A$ with $\alpha \le \alpha ''$ and $\alpha '\le \alpha ''$. In this case we denote $(x_\alpha )_{\alpha \in A}$ with $x_\alpha =x(\alpha )$. We say that another net $(y_\beta )_{\beta \in B}$ is a subnet of $(x_\alpha )_{\alpha \in A}$, whenever there exists a function $h:B\rightarrow A$, such that (i) for any $\beta \le \beta '$ there holds $\alpha \le \alpha '$, (ii) for any $\alpha \in A$ exists $\beta \in B$ with $\alpha \le h(\beta )$, and (iii) for all $\beta \in B$ we have $y_\beta =x_{h(\beta )}$. See [11] for further details.

2.2 Partially ordered vector spaces

A partially ordered vector space is a real vector space X equipped with a reflexive, transitive and antisymmetric binary relation $\le$ such that for every $\lambda \in {\mathbb {R}}$ with $0\le \lambda$ and $x,y,z\in X$ with $x\le y$ one has that $x+z\le y+z$ and $\lambda x\le \lambda y$. We denote $X_+:=\{x\in X;\, 0\le x\}$ for the cone and call the elements of $X_+$ positive. For $x,y\in X$ we write $x<y$ if $x\le y$ and $x\ne y$. For $U,V\subseteq X$ we write $U\le V$ if for every $x\in U$ and $y\in V$ we have $x\le y$. For $x\in X$, we abbreviate $V\le \{x\}$ by $V\le x$, and $x\le V$ is defined similarly. For $x,y\in X$ the order interval is given by $[x,y]:=\{z\in X; \, x\le z\le y\}$. For a subset of X, the notions bounded above, bounded below, order bounded, upper (or lower) bound and infimum (or supremum) are defined as usual.

Consider a partially ordered vector space X. X is called directed if for every $x,y\in X$ there exists $z\in X$ with $x,y\le z$. Note that X is directed, if and only if $X_+$ is generating, i.e. $X=X_+-X_+$. We say that X is Archimedean if for every $x,y\in X$ with $nx\le y$ for all $n\in {\mathbb {N}}$ one has that $x\le 0$. We call X a vector lattice if for every non-empty finite subset of X the infimum and the supremum exist in X. X is called Dedekind complete whenever every non-empty set that is bounded from above has a supremum, and every non-empty set that is bounded from below has an infimum. Note that every Dedekind complete vector lattice is Archimedean. An element $u\in X_+$ is called an order unit, if for all $x\in X$ there is $n\in {\mathbb {N}}$ such that $x\le n u$. An element $x\in X_+$ is called net catching, if for all nets $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$ there exists $\alpha \in A$ such that $x_\alpha \le x$. We call $M\subseteq X$ order dense in X if for every $y\in X$ one has

$$\begin{aligned} \sup \{x\in M;\, x\le y\}=y=\inf \{x\in M;\, y\le x\}. \end{aligned}$$

A non-empty convex subset B of $X_+\setminus \{0\}$ is called a base (of $X_+$), if for each $x\in X_+\setminus \{0\}$ there are unique $b\in B$ and $\lambda \in {\mathbb {R}}_+$ such that $x=\lambda b$. Consider subsets $M,N\subseteq X$. We say that M is circled whenever for all $\lambda \in [-1,1]$ we have $\lambda M\subseteq M$. Furthermore, we say that M absorbs N, if there is $\mu >0$ with $\lambda N\subseteq M$ for all $\lambda \in [-\mu ,\mu ]$.

2.3 Embedding theorems

Let X and Y be partially ordered vector spaces. A linear mapping $f:X \rightarrow Y$ is said to be positive whenever all $x\in X_+$ satisfy $f(x)\ge 0$. We call a linear map $f:X \rightarrow Y$ order reflecting, whenever $f(x)\ge 0$ implies $x\ge 0$ for all $x\in X$. Note that every order reflecting mapping is injective. A linear order embedding is a positive and order reflecting map. A linear order dense embedding is a linear order embedding $f:X\rightarrow Y$ for which the image f(X) is order dense in Y. A linear order isomorphism is a surjective order embedding. For a reference of the following embedding theorem see [2, Theorem IV.11.1] or [12, Chapter V.3].

Theorem 2.1

If X is an Archimedean and directed partially ordered vector space, then there exists a Dedekind complete vector lattice $X^\delta$ and a linear order dense embedding $\iota :X \rightarrow X^\delta$. In this context $\iota :X\rightarrow X^\delta$ is called a Dedekind completion of X.

Recall that a topological space is called extremely disconnected whenever the closure of any open set is open. For reference of the following see [13, Theorem 45.5] or [2, Theorem V.2.2 and Theorem V.4.1].

Theorem 2.2

If X is a Dedekind complete vector lattice that contains order units, then there exists an extremely disconnected compact Hausdorff space $\Omega$ and a linear order isomorphism $\iota :X\rightarrow C(\Omega )$. Here $C(\Omega )$ is the vector lattice of all continuous function $f:\Omega \rightarrow {\mathbb {R}}$ equipped with the pointwise order.

2.4 Order convergence and the order topology

We next summarize some well known results that can be found in varying degrees of generality in [1,2,3, 6, 9] concerning the order topology and the order convergence as defined in the introduction.

Theorem 2.3

Let X be a partially ordered vector space.

• (i) A subset $M\subseteq X$ is order closed, if and only if the order limits of nets in M are contained in M.

• (ii) Any net $(x_\alpha )_{\alpha \in A}$ that satisfies $x_\alpha \downarrow x$ order converges to x.

• (iii) Any net that order converges to $x\in X$ also converges to x w.r.t. the order topology $\tau _o$.

• (iv) $X_+$ is order closed.

• (vi) Whenever $\tau _{o}$ is a linear topology, then X is Archimedean and directed.

• (v) $\tau _{o}({\mathbb {R}})$ is the standard topology of ${\mathbb {R}}$. Order convergence is equivalent to the standard notion of convergence of ${\mathbb {R}}$.

Proof

The statement (vi) follows from [6, Lemma 8.1] and (v) can easily be seen as the standard topology of the Dedekind complete lattice ${\mathbb {R}}$ is generated by the set of all order intervals. See [6, Theorem 3.14, Proposition 3.6, and Corollary 3.18 ] for the rest of the statements. $\square$

If X and Y are partially ordered vector spaces we call a positive map $f:X\rightarrow Y$ order continuous, whenever any net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$ satisfies $f(x_\alpha )\downarrow 0$. From [6, Theorem 4.4 and Corollary 4.5] we quote the following results.

Theorem 2.4

• (i) Every linear order dense embedding between partially ordered vector spaces is order continuous.

• (ii) For any positive map $f:X\rightarrow Y$ between partially ordered vector spaces the following statements are equivalent.

  • (a) f is order continuous.

  • (b) f is continuous with respect to $\tau _o(X)$ and $\tau _o(Y)$.

  • (c) For any order convergent net $(x_\alpha )_{\alpha \in A}$ in X with order limit $x\in X$ the net $(f(x_\alpha ))_{\alpha \in A}$ is order convergent with order limit f(x).

2.5 The order bound topology

The following definition generalizes the definition of the order bound topology presented in the introduction. See [5, Sect. 2.8] for details.

Definition 2.5

Let ${\mathcal {B}}_{ob}(X)$ (or simply ${\mathcal {B}}_{ob}$) be the set of all subsets $V \subseteq X$ such that V is convex, circled and absorbs all order intervals. Note that ${\mathcal {B}}_{ob}$ is a base at zero for a linear topology $\tau _{ob}(X)$ (or simply $\tau _{ob}$). This topology is called the order bound topology.

The order bound topology $\tau _{ob}$ is introduced and studied in [10]. For further details we recommend [5], where $\tau _{ob}$ is called ’order topology’. The following statements can be found in [5, Sect. 2.8].

Theorem 2.6

Let X be an Archimedean partially ordered vector space and let u be an order unit in X.

• (i) The norm $\Vert \cdot \Vert _u$ generates $\tau _{ob}$.

• (ii) The respective closed unit ball satisfies $\overline{B}_1^{\Vert \cdot \Vert _u}(0)=[-u,u]$.

3 Order convergence and the order bound topology

As presented above the link between the order convergence and the order topology was investigated in various works like for example [6, 9]. We thus focus next on the link between the order bound topology and the order convergence.

Theorem 3.1

Let X be a partially ordered vector space. The following statements are equivalent.

• (i) Any $\tau _{ob}$-convergent net is order convergent (to the same limit).

• (ii) X is Archimedean and contains order units.

From Theorem 2.3 we recall that order convergence implies convergence with respect to the order topology $\tau _o$. We thus obtain the following.

Corollary 3.2

If X is an Archimedean partially ordered vector space that contains order units, then the order bound topology is finer than the order topology, i.e. $\tau _o\subseteq \tau _{ob}$.

In order to show Theorem 3.1 we will need the following lemma.

Lemma 3.3

Let X be an Archimedean partially ordered vector space that contains order units. Then the set of order units U equipped with the reversed order from X is directed and satisfies $\inf U=0$ with respect to the order of X.

Proof

Let $u,v\in U$. There is $n \in {\mathbb {N}}$ such that $v \le n u$. As (1/n)v is also an order unit which satisfies $(1/n)v\le \{u,v\}$ we obtain U to be directed with respect to the reversed order.

Clearly $0\le U$ and we consider $z\in X$ with $z\le U$. For $u\in U$ and $n\in {\mathbb {N}}$ we know that also (1/n)u is an order unit. Thus there holds $nz\le u$ for all $n\in {\mathbb {N}}$ and we obtain $z\le 0$ as u is an order unit. This shows $\inf U=0$. $\square$

Proof of Theorem 3.1

To show that (i) implies (ii) we show first that X is Archimedean and consider $x,y\in X$ with $nx\le y$ for all $n\in {\mathbb {N}}$. Any $V\in {\mathcal {B}}_{ob}$ absorbs [y, y] there exists $\mu >0$ with $\lambda [y,y]\subseteq V$ for all $\lambda \in [-\mu ,\mu ]$. For $n\in {\mathbb {N}}$ with $n\ge 1/\mu$ we observe $(1/n)y\in (1/n)[y,y]\subseteq V$. Hence $((1/n) y)_{n\in {\mathbb {N}}}$ converges to 0 w.r.t. $\tau _{ob}$ and (i) implies that this net order converges to 0. Thus there exists a net $(\check{y}_\beta )_{\beta \in B}$ with $\check{y}_\beta \downarrow 0$ and such that for $\beta \in B$ there exists $n\in {\mathbb {N}}$ with $x\le (1/n)y \le \check{y}_\beta$. Hence $x\le 0$ and we obtain X to be Archimedean.

To show that X contains order units let $A:=\{(x,V);\,V\in {\mathcal {B}}_{ob},x\in V\}$ be ordered by $(x,V)\le (y,U)$ whenever $V\supseteq U$. Then A is easily seen to be directed and we define $x_{(x,V)}:=x$ for $(x,V)\in A$ in order to obtain a net $(x_\alpha )_{\alpha \in A}$. As ${\mathcal {B}}_{ob}$ is a base at zero for $\tau _{ob}$ we obtain $x_\alpha \overset{\tau _{ob}}{\rightarrow }0$ and (i) implies $x_\alpha \overset{o}{\rightarrow }0$. Thus there exists a net $(\check{x}_\beta )_{\beta \in B}$ with $\check{x}_\beta \downarrow 0$ and such that for any $\beta \in B$ there exists $\alpha _\beta \in A$ such that $\pm x_\alpha \le \check{x}_\beta$ for all $\alpha \ge \alpha _\beta$. We will now show that $\check{x}_\beta$ is an order unit for any $\beta \in B$.

Let $\beta \in B$ and consider $(y,V)=\alpha _\beta \in A$ as above. For $v\in V$ we have $(v,V)\ge (x,V)$ and hence $\pm v=\pm x_{(v,V)}\le \check{x}_\beta$. We observe $V\le \check{x}_\beta$. For $x\in X$ we consider the order interval [x, x] and obtain from the definition of ${\mathcal {B}}_{ob}$ that V absorbs [x, x]. Thus there exists $\mu >0$ with $\lambda x\le V$ for all $\lambda \in [-\mu ,\mu ]$. For $n\in {\mathbb {N}}$ with $n\ge 1/\mu$ we obtain $x\le nV\le n\check{x}_\beta$.

To show that (ii) implies (i) we consider a net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \overset{\tau _{ob}}{\rightarrow } x\in X$ but assume w.l.o.g. that $x=0$. Let B be the set of all order units, equipped with the reversed order from X. Define $\check{x}_\beta :=\beta$ for $\beta \in B$. By Lemma 3.3 the net $(\check{x}_\beta )_{\beta \in B}$ satisfies $\check{x}_\beta \downarrow 0$. For $\beta \in B$ we know that the norm $\Vert \cdot \Vert _{\beta }$ generates the order bound topology and thus $\Vert x_\alpha \Vert _\beta \rightarrow 0$. Thus there is $\alpha _\beta \in A$ such that $\inf \{\lambda >0;\,\pm x_\alpha \le \lambda \beta \}=\Vert x_\alpha \Vert _\beta \le \frac{1}{2}$ for $\alpha \ge \alpha _\beta$. We obtain $\pm x_\alpha \le \beta =\check{x}_\beta$ for every $\alpha \ge \alpha _\beta$. This shows $x_\alpha \xrightarrow {o} 0$. $\square$

Remark 3.4

Note that the net $(\check{x}_\beta )_{\beta \in B}$ constructed in the second part of the proof is independent from $(x_\alpha )_{\alpha \in A}$.

4 The interplay of the order topology and the order bound topology

4.1 About order units and net catching elements

We start the investigation of the interplay of the order topology and the order bound topology by relating net catching elements with order units.

Lemma 4.1

Let X be a partially ordered vector space.

• (i) If X contains net catching elements, then every order unit in X is net catching.

• (ii) If X is Archimedean and directed, then every net catching element of X is an order unit.

Proof

To show (i) let u be an order unit of X and consider a net catching element $x\in X$. There is $n \in {\mathbb {N}}$ such that $x\le n u$. it is straight forward to show that nu and hence u is net catching.

To show (ii) assume $x\in X$ to be net catching and consider $y\in X$. As X is directed there exists $z\in X_+$ such that $z\ge y$. The Archimedean property of X implies $({1}/{n})z \downarrow 0$. Thus since x is net catching there is $n \in {\mathbb {N}}$ such that $({1}/{n})z\le x$, i.e. $y\le z \le n x$.

Remark 4.2

Note that an element $x\in X$ is net catching, if and only if it is net catching with respect to the order restricted to the directed part $X_+-X_+$. As non-directed partially ordered vector spaces do not contain order units it is natural to assume directedness in (ii).

Example 4.3

Let $l_\infty$ be the Dedekind complete vector lattice of all bounded and real sequences, equipped with the pointwise order. Then $e:=(1)_{k \in {\mathbb {N}}}$ is an order unit of $l_\infty$ that is not net catching. Thus Lemma 4.1(i) implies that $l_\infty$ has no net catching elements.

Proof

For $n,k\in {\mathbb {N}}$ let $e^{(n)}_k:=0$, if $k<n$ and $e^{(n)}_k:=2$ otherwise. This defines a sequence $(e^{(n)})_{n \in {\mathbb {N}}}$ in $l_\infty$ such that $e^{(n)}\downarrow 0$. Since for no $n \in {\mathbb {N}}$ we have $e^{(n)}\le e$ we obtain e to be not net catching. $\square$

There also exist net catching elements, which are not order units.

Example 4.4

Equip $l_\infty$ with the lexicographic order, i.e. with $(x_n)_{n \in {\mathbb {N}}} \le (y_n)_{n \in {\mathbb {N}}}$ whenever $(x_n)_{n \in {\mathbb {N}}}=(y_n)_{n \in {\mathbb {N}}}$ or whenever there exists $n \in {\mathbb {N}}$ such that $x_n < y_n$ and $x_m=y_m$ for all $m<n$. $l_\infty$ equipped with $\le$ is a (non-Archimedean) vector lattice that contains order units and furthermore net catching elements which are not order units.

Proof

Note that the lexicographic order is total, i.e. that there holds $x\le y$ or $y\le x$ for all $x,y\in l_\infty$. One easily obtains that $(u_n)_{n\in {\mathbb {N}}}\in l_\infty$ is an order unit with respect to this ordering, if and only if $u_1>0$. Consider $x=(x_n)_{n\in {\mathbb {N}}}\in l_\infty$ with $x_1=0$ and $x_n:=1$ for $n\ge 2$, which is obviously no order unit. To show that x is net catching let $(x_\alpha )_{\alpha \in A}$ be a net in $l_\infty$ with $x_\alpha \downarrow 0$. As $0<x$ and as $\le$ is total we observe that there exists $\alpha \in A$ such that $x_\alpha \le x$. $\square$

4.2 A parallelism between order units and net catching elements

We next investigate the interplay of order units with the order bound topology, and of net catching elements with the order topology respectively.

Lemma 4.5

Let X be a partially ordered vector space and $x\in X$.

• (i) x is an order unit, if and only if $[-x,x]\in {\mathcal {B}}_{ob}$.

• (ii) x is net catching, if and only if $[-x,x]$ is an order neighbourhood of 0.

Proof

’(i)’ Let us assume that x is an order unit. Since $[-x,x]$ is clearly convex and circled it remains to show that $[-x,x]$ absorbs order intervals. Consider an order interval $[y,z]\subseteq X$. As x is an order unit there is $n\in {\mathbb {N}}$ with $-y,z\le nx$ and we obtain $(1/n)[y,z]\subseteq [-x,x]$. As $[-x,x]$ is circled this shows that $[-x,x]$ absorbs [y, z].

To show the converse assume $[-x,x]\in {\mathcal {B}}_{ob}$ and let $y\in X$. As $[-x,x]$ absorbs [y, y] there is $\mu >0$ with $\lambda [y,y]\subseteq [-x,x]$ for all $\lambda \in [-\mu ,\mu ]$. For $n\in {\mathbb {N}}$ with $n\ge 1/\lambda$ we obtain $y\le nx$, which proves x to be an order unit.

’(ii)’ Assume that x is net catching. For a net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$ there exists $\alpha \in A$ with $x_\alpha \le x$. Thus $[-x_\alpha ,x_\alpha ]\subseteq [-x,x]$ and we obtain $[-x,x]$ to be an order neighbourhood. Similarly one shows that whenever $[-x,x]$ is an order neighbourhood, then x is net catching. $\square$

In analogy with Corollary 3.2 we next show the following.

Theorem 4.6

If there are net catching elements in X, then the order topology is finer than the order bound topology, i.e. $\tau _{ob} \subseteq \tau _o$.

Proof

We show that every $V\in {\mathcal {B}}_{ob}$ is an order neighbourhood of 0. Let x be a net catching element. Since V absorbs all order intervals there is $\lambda >0$ such that $[-\lambda x,\lambda x]\subseteq V$. Note that $\lambda x$ is net catching. Thus Lemma 4.5 shows $[-\lambda x,\lambda x]\subseteq V$ to be an order neighbourhood of 0. $\square$

Proposition 4.7

Let X be a partially ordered vector space.

• (i) The $\tau _{ob}$-interior of $X_+$ is the set of all order units of X.

• (ii) The $\tau _{o}$-interior of $X_+$ is the set of all net catching elements of X.

Proof

’(i)’ Let u be contained in the $\tau _{ob}$-interior of $X_+$ and let $x\in X$. There exists $V\in {\mathcal {B}}_{ob}$ such that $V+u\subseteq X_+$. As V absorbs all order intervals we argue as above to find $n\in {\mathbb {N}}$ with $[-x,-x]\subseteq nV$ and observe $nu-x\le nV+nu\subseteq nX_+=X_+$, i.e. $x\le nu$.

To show the converse we will show that the set U of all order units is order open. Indeed, whenever $u\in U$, then $(1/2) u\in U$. Thus by Lemma 4.5 we obtain $[(-1/2)u,(1/2)u]\in {\mathcal {B}}_{ob}$. As $u+[(-1/2)u,(1/2)u]=[(1/2)u,(3/2)u]$ consists only of order units we obtain that U is a $\tau _{ob}$-neighbourhood of u.

’(ii)’ Let x be contained in the $\tau _o$-interior $X_+^{o_{\tau _o}}$ of $X_+$. To show that x is net catching let $(x_\alpha )_{\alpha \in A}$ be a net in $X_+$ with $x_\alpha \downarrow 0$. Since $X_+^{o_{\tau _o}}$ is order open it is an order neighbourhood of x. Thus there is $\alpha \in A$ such that $[x-x_\alpha ,x+x_\alpha ]\subseteq X_+^{o_{\tau _o}}\subseteq X_+$ and hence $x_\alpha \le x$. The converse can be shown similarly to the corresponding implication of (i).

Corollary 4.8

Let X be a partially ordered vector space.

• (i) X contains order units, if and only if there exist non-empty $\tau _{ob}$-open subsets of X that are order bounded.

• (ii) X contains net catching elements, if and only if there exist non-empty order open subsets of X that are order bounded.

Proof

If the $\tau _{ob}$-interior $X_+^{o_{\left( \tau _{ob}\right) }}$ is non-empty, then $((2x)-X_+^{o_{\left( \tau _{ob}\right) }})\cap X_+^{o_{\left( \tau _{ob}\right) }}$ is a non-empty, $\tau _{ob}$-open and order bounded set for any $x\in X_+^{o_{\left( \tau _{ob}\right) }}$. Conversely, if O is a non-empty, $\tau _{ob}$-open and order bounded set, then there exist $x,y\in X$ with $-x\le O\ni y$ and $x+y\in O+x\subseteq X_+$ is an interior point of $X_+$ with respect to $\tau _{ob}$. This shows (i) and (ii) can be obtained by considering $\tau _o$ instead of $\tau _{ob}$. $\square$

4.3 Examples

Example 4.9

Recall from Example 4.3 that $l_\infty$ equipped with the pointwise order is an Archimedean vector lattice that contains order units but no net catching elements. From Corollary 3.2 we know that $\tau _{o}\subseteq \tau _{ob}$. This inclusion is strict, as we know from Proposition 4.7 that the interior of $X_+$ with respect to $\tau _{o}$ is empty while the interior of $X_+$ with respect to $\tau _{ob}$ contains elements.

Example 4.10

Recall from Example 4.4 that $l_\infty$ equipped with the lexicographic order is a vector lattice that contains net catching elements that are not order units. From Theorem 4.6 we obtain $\tau _{ob}\subseteq \tau _{o}$. This inclusion is strict as the interior of $X_+$ with respect to $\tau _o$ is not the same as the interior of $X_+$ with respect to $\tau _{ob}$.

4.4 Characterizations of $\tau _o=\tau _{ob}$

From our investigations we conclude the following characterization.

Theorem 4.11

Let X be an Archimedean partially ordered vector space that contains order units. Then the following statements are equivalent.

• (i) $\tau _o=\tau _{ob}$.

• (ii) For all nets $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow x$ there holds $x_\alpha \overset{\tau _{ob}}{\rightarrow }x$.

• (iii) X contains net catching elements.

Proof

Recall from Theorem 2.3 that $x_\alpha \downarrow x$ implies $x_\alpha \overset{\tau _o}{\rightarrow } x$. Thus (i) implies (ii).

To show that (ii) implies (iii) consider an order unit $u\in X_+$. Let $(x_\alpha )_{\alpha \in A}$ be a net in X with $x_\alpha \downarrow 0$. From (ii) we know $x_\alpha \rightarrow 0$ with respect to $\tau _{ob}$. Now recall from Lemma 4.5 that $[-u,u]\in {\mathcal {B}}_{ob}$. Thus there is $\alpha \in A$ with $x_\alpha \in [-u,u]$ and we observe $x_\alpha \le u$.

To show that (iii) implies (i) recall from Corollary 4.8 that the existence of order units implies $\tau _{o}\subseteq \tau _{ob}$. If X contains net catching elements we furthermore know from Theorem 4.6$\tau _{o}\supseteq \tau _{ob}$ and obtain (i). $\square$

Having the parallelism between order units and net catching elements in mind it is natural to ask, how $\tau _o=\tau _{ob}$ can be characterized under the assumption of the existence of net catching elements.

Theorem 4.12

Let X be a partially ordered vector space that contains net catching elements. Then the following statements are equivalent.

• (i) $\tau _o=\tau _{ob}$.

• (ii) $\tau _o$ is a linear topology.

• (iii) X is Archimedean and directed.

• (iv) X is Archimedean and contains order units.

In particular, if one of these statements is valid, then each net catching element $x\in X$ is an order unit of X and $\Vert \cdot \Vert _x$ is a norm that generates the order topology $\tau _o$.

Proof

As the order bound topology is a linear topology we obtain that (i) implies (ii). From Theorem 2.3 we furthermore obtain that (ii) implies (iii). If X is Archimedean and directed, then any net catching element is an order unit by Lemma 4.1. Thus (iii) implies (iv). From Theorem 4.11 we furthermore obtain that (iv) implies (i). $\square$

Recall that order convergence implies convergence with respect to $\tau _o$. From Theorem 4.12 and Theorem 3.1 we thus obtain the following corollary.

Corollary 4.13

If X is Archimedean and directed and contains net catching elements, then the following statements are equivalent for any net $(x_\alpha )_{\alpha \in A}$ in X.

$$\begin{aligned} \mathrm{(i)} x_\alpha \overset{o}{\rightarrow }0. \quad \mathrm{(ii)} x_\alpha \overset{\tau _o}{\rightarrow }0. \quad \mathrm{(iii)} x_\alpha \overset{\tau _{ob}}{\rightarrow }0. \end{aligned}$$

Remark 4.14

In [9, Corollary 4.11] a different proof for the equivalence of (i) and (ii) in Corollary 4.13 is given. Note that the existence of net catching elements is not stated explicitly in [9] but used in the proof. Indeed, it is used that the set of all non-empty, order bounded and order open sets is not-empty, a property equivalent to the existence of net catching elements as shown in Corollary 4.8. See [6, Example 8.3] for an example of an Archimedean vector lattice for which $\tau _o$-convergence does not imply order convergence.

4.5 About convex order open sets

So far we do not now, whether for Archimedean and directed partially ordered vector spaces (without order units) any order open set is $\tau _{ob}$-open. For the following partial result in this direction recall that a partially ordered vector space X is said to satisfy the (Riesz) decomposition property whenever for $x,u,v\in X_+$ with $x\le u+v$ there exist $u',v'\in X_+$ with $x=u'+v'$, $u'\le u$ and $v'\le v$. Any vector lattice satisfies the decomposition property. See [5] for further details on this notion.

Theorem 4.15

If X is an Archimedean and directed partially ordered vector space that satisfies the decomposition property, then any convex order open set is $\tau _{ob}$-open.

The proof of Theorem 4.15 is split into several lemmas about the local structure of order neighbourhoods. For the rest of this section we assume that X is a partially ordered vector space.

Lemma 4.16

Any order neighbourhood of 0 contains a circled order neighbourhood of 0.

Proof

Let $U\subseteq X$ be an order neighbourhood of 0. Define $I(U):=\{x\in X_+;\, [-x,x]\subseteq U\}$. The set $V:=\bigcup _{x\in I(U)}[-x,x]$ is easily seen to be circled and to be contained in U. To show that V is an order neighbourhood of 0 let $(x_\alpha )_{\alpha \in A}$ be a net in X with $x_\alpha \downarrow 0$. As U is an order neighbourhood of 0 there is $\alpha \in A$ with $[-x_\alpha ,x_\alpha ]\subseteq U$. Thus $x_\alpha \in I(U)$ and we obtain $[-x_\alpha ,x_\alpha ]\subseteq V$. $\square$

Remark 4.17

The construction of the set V is given in [9] and it is shown there that V is an order neighbourhood of 0.

Lemma 4.18

If X satisfies the decomposition property, then any convex order neighbourhood of 0 contains a convex and circled order neighbourhood of 0.

Proof

Let U be a convex order neighbourhood of 0 and define V as in the proof of Lemma 4.16. It remains to show that V is convex and we consider $x,y\in V$ and $\lambda \in (0,1)$. There are $x',y'\in X_+$ with $x\in [-x',x']\subseteq U$ and $y\in [-y',y']\subseteq U$. Denoting $z':=\lambda x'+(1-\lambda )y'$ we observe $\lambda x+(1-\lambda )y\in [-z',z']$.

To show $z'\in I(U)$ we consider $\tilde{z}\in [-z',z']$. There holds $0\le \tilde{z}+z'\le 2z'=2\lambda x'+2(1-\lambda )y'$ and from the decomposition property we observe that there exist $u,v\in X_+$ with $\tilde{z}+z'=u+v$, $u\le 2\lambda x'$ and $v\le 2(1-\lambda )y'$. We compute

$$\begin{aligned} \tilde{x}:=\frac{1}{\lambda }(u-\lambda x')\in \frac{1}{\lambda }[-\lambda x',\lambda x'] =[-x',x']\subseteq U \end{aligned}$$

and similarly $\tilde{y}:=(1/(1-\lambda ))(v-(1-\lambda )y')\in U$. As U is convex we observe

$$\begin{aligned} \tilde{z} =u+v-z' =u+v-(\lambda x'+(1-\lambda )y') =\lambda \tilde{x}+(1-\lambda )\tilde{y} \in U. \end{aligned}$$

This shows $[-z',z']\subseteq U$, i.e. $z'\in I(U)$. We conclude $\lambda x+(1-\lambda )y\in [-z',z']\subseteq V$.

$\square$

Lemma 4.19

If X is Archimedean and directed, then any order neighbourhood of 0 absorbs all order intervals.

Proof

Let U be an order neighbourhood of 0. By Lemma 4.16U contains a circled order neighbourhood V of 0. Let [x, y] be a non-empty order interval in X. Since X is directed there is $w\in X$ with $w\ge -x$ and $w\ge y$. From the Archimedean property we observe $(1/n)w\downarrow 0$. Now recall that V is an order neighbourhood of 0. We thus obtain the existence of $n\in {\mathbb {N}}$ with $(1/n)[x,y]\subseteq [-(1/n)w,(1/n)w]\subseteq V$. For $\lambda \in [-1/n,1/n]$ we use that V is circled to see $\lambda [x,y]\subseteq V\subseteq U$, which proves U to absorb [x, y]. $\square$

Proof of Theorem 4.15

Let O be a convex and order open set and consider $x\in O$. Then $O-x$ is a convex order neighbourhood of 0. From Lemma 4.18 we know that $O-x$ contains a convex and circled order neighbourhood V of 0. By Lemma 4.19 we know that V absorbs all order intervals and hence $V\in {\mathcal {B}}_{ob}$. This proves O to be $\tau _{ob}$-open. $\square$

We finish this section with an Example of an Archimedean vector lattice that contains a circled order neighbourhood that does not contain convex sets that absorb order intervals.

Example 4.20

Let $l_0$ be the space of all sequences $x=(x_n)_{n\in {\mathbb {N}}}$ in ${\mathbb {R}}$ with finite support ${\text {supp}}(x)=\{n\in {\mathbb {N}};x_n\ne 0\}$. $l_0$ is an Archimedean vector lattice equipped with the pointwise order. Let M be the set of all x, for which there exists $F\subseteq {\mathbb {N}}$ finite with ${\text {supp}}(x)\subseteq F$ and $|x_n|\le 1/|F|$ for $n\in F$. Then M is a circled order neighbourhood of 0 that does not contain a convex set which absorbs all order intervals. In particular, we obtain from Lemma 4.19 that M does not contain a convex order neighbourhood.

Proof

Clearly M is circled. To show that M is an order neighbourhood let $(x^{(\alpha )})_{\alpha \in A}$ be a net in $l_0$ with $x^{(\alpha )}\downarrow 0$. Considering $\alpha _0\in A$ we obtain the existence of $N\in {\mathbb {N}}$ with ${\text {supp}}(x^{(\alpha )})\subseteq {\text {supp}}(x^{(\alpha _0)})\subseteq \{1,\dots , N\}$ for all $\alpha \ge \alpha _0$. Let $\alpha \ge \alpha _0$ such that $x^{(\alpha )}_k\le 1/N$ for $k=1,\cdots , N$ to obtain $[-x^{(\alpha )},x^{(\alpha )}]\subseteq M$.

Assume now that there exists a convex subset $C\subseteq M$ that absorbs all order intervals and let $x\in C$ and $N\in {\mathbb {N}}$. Choose $F\subseteq {\mathbb {N}}$ disjoint from ${\text {supp}}(x)$ with $|F|=N$. Let $y\in l_0$ be defined by $y_n:=1$ for $n\in F$ and $y_n=0$ for $n\notin F$. As C absorbs order intervals there exists $\lambda >0$ such that $\lambda y\in C$. Since C is convex we obtain $z:=1/2x+1/2y\in C\subseteq M$. Thus for $n\in {\text {supp}}(x)$ we compute $0\le (1/2)|x_n|=|(1/2)x_n+(1/2)y_n|\le 1/|{\text {supp}}(z)|\le 1/N$. As $N\in {\mathbb {N}}$ was arbitrary this shows $x=0$ and we have shown $C=\{0\}$, a contradiction as C is supposed to absorb order intervals. $\square$

5 A sufficient condition for net catching elements

In order to be able to present more examples of partially ordered vector spaces that contain net catching elements we will prove the following in this section.

Theorem 5.1

Let X be a partially ordered vector space and B be a base of $X_+$. If there exists a linear topology $\tau$ on X such that B is $\tau$-compact and such that $X_+$ is $\tau$-closed, then all upper bounds of B are net catching.

To prove this theorem it will be convenient to consider the following correspondence between strictly positive linear functionals and bases. Recall that a linear functional $f:X\rightarrow {\mathbb {R}}$ is called strictly positive, whenever we have $0<f(x)$ for all non-zero $x\in X_+$. For a strictly positive linear functional f the set $B_f:=\{x \in X_+;\, f(x)=1\}$ is a base of $X_+$. Furthermore, for any base B of $X_+$ there exists a unique strictly positive linear functional $f:X\rightarrow {\mathbb {R}}$ such that $B=B_f$. Thus there is a one-to-one correspondence between strictly positive linear functionals and bases of $X_+$. For reference see [5]. We show next that order continuity translates into being order closed in this correspondence.

Proposition 5.2

Let $f:X \rightarrow {\mathbb {R}}$ be a strictly positive linear functional. Then $B_f$ is order closed if and only if f is order continuous.

Proof

Assume that f is order continuous and note that f is positive. From Sect. 2.4 we thus know that f is continuous with respect to $\tau _o(X)$ and the standard topology of ${\mathbb {R}}$. As $X_+$ is always order closed we observe that $B=X_+ \cap f^{-1}(\{1\})$ is order closed.

To show the converse assume $B_f$ to be order closed and let $(x_\alpha )_{\alpha \in A}$ be a net in X with $x_\alpha \downarrow 0$. Clearly $(f(x_\alpha ))_{\alpha \in A}$ is a net in ${\mathbb {R}}_+$ such that $f(x_\alpha ) \downarrow$. To get a contradiction assume that there does not hold $f(x_\alpha )\downarrow 0$. Then there exists $\lambda >0$ with $f(x_\alpha )\ge \lambda$ for all $\alpha \in A$. We define $y_\alpha :=({1}/{f(x_\alpha )})x_\alpha$ for all $\alpha \in A$ to obtain a net $(y_\alpha )_{\alpha \in A}$ in $B_f$. For $\alpha \in A$ we observe

$\pm y_\alpha \le (1/\lambda )x_\alpha$, which proves $y_\alpha \xrightarrow {o}0$. Since $B_f$ is order closed we obtain that $0\in B_f$, i.e. $f(0)=1$, a contradiction. $\square$

The following proposition gives a sufficient condition on X to have an order closed base.

Lemma 5.3

Let B be a base of $X_+$. If there exists a linear topology $\tau$ on X such that B is $\tau$-compact and such that $X_+$ is $\tau$-closed, then B is order closed.

Proof

Let $f:X\rightarrow {\mathbb {R}}$ be the unique strictly positive linear functional on X such that $B=\{x\in X_+ ;\, f(x)=1\}$. By Proposition 5.2 it suffices to show that f is order continuous. Let $(x_\alpha )_{\alpha \in A}$ be a net in X such that $x_\alpha \downarrow 0$. Since f is positive we obtain $(f(x_\alpha ))_{\alpha \in A}$ to be a net in ${\mathbb {R}}_+$ that satisfies $f(x_\alpha )\downarrow$. To obtain a contradiction we assume that $f(x_\alpha )\downarrow 0$ is not satisfied. As above we obtain that $0<\lambda :=\inf _{\alpha \in A} f(x_\alpha )$ and define $y_\alpha :=(1/f(x_\alpha ))x_\alpha$ in order to obtain a net $(y_\alpha )_{\alpha \in A}$ in the $\tau$-compact set B. Thus there exists a subnet $(z_\gamma )_{\gamma \in C}$ of $(y_\alpha )_{\alpha \in A}$ that converges with respect to $\tau$ to some $z\in B$. Let us fix a map $h:C \rightarrow A$ as in the definition of a subnet.

Then $(f(x_{h(\gamma )}))_{\gamma \in C}$ is a subnet of $(f(x_\alpha ))_{\alpha \in A}$ we observe $f(x_{h(\gamma )})\downarrow \lambda$. Thus $(f(x_{h(\gamma )}))_{\gamma \in C}$ converges to $\lambda$ with respect to the standard topology of ${\mathbb {R}}$. As $\tau$ is a linear topology we obtain

$$\begin{aligned} x_{h(\gamma )}=f(x_{h(\gamma )})y_{h(\gamma )}=f(x_{h(\gamma )})z_\gamma \xrightarrow {\tau } \lambda z. \end{aligned}$$

Furthermore, we have $x_{h(\gamma )}\downarrow$ and that $X_+$ is $\tau$-closed. A standard argument [5, Lemma 2.3] therefore yields $x_{h(\gamma )}\downarrow \lambda z$. Thus $x_{h(\gamma )}\downarrow 0$ allows to observe $\lambda z=0$. Since $z\in B$ this implies $\lambda =0$, a contradiction. $\square$

We obtain Theorem 5.1 by combining the following Lemma with Lemma 5.3 above.

Lemma 5.4

If B is an order closed base of $X_+$, then all upper bounds of B are net catching.

Proof

Let $x\in X$ be such that $B\le x$. To show that x is net catching consider a net $(x_\alpha )_{\alpha \in A}$ with $x_\alpha \downarrow 0$. Furthermore, consider a strictly positive functional $f:X \rightarrow {\mathbb {R}}$ such that $B=\{x \in X_+;\, f(x)=1\}$. From Proposition 5.2 we know that f is order continuous and thus satisfies $f(x_\alpha )\downarrow 0$. in particular, there is $\alpha \in A$ with $f(x_\alpha )\le 1$. Whenever $f(x_\alpha )=0$, then the strict positivity of f implies $x_\alpha =0\le x$. Whenever $f(x_\alpha )>0$, then $(1/f(x_\alpha ))x_\alpha \in B$ and we obtain $x_\alpha \le (1/f(x_\alpha ))x_\alpha \le x$. $\square$

6 Examples

6.1 Finite dimensional spaces

Example 6.1

For a finite dimensional, Archimedean and directed partially ordered vector space X the topology $\tau _o=\tau _{ob}$ is the standard topology of X, i.e. the unique linear topology on X. A net is order convergent, if and only if it is convergent with respect to the standard topology. The interior of $X_+$ with respect to the discussed topologies is non-empty and consists of the net catching elements which are precisely the order units.

Proof

Let us denote $\tau$ for the standard topology of X. As X is Archimedean we know that $X_+$ is $\tau$-closed and that $X_+$ has a $\tau$-closed base. See [5, Chapter 3] for reference.

From X being directed we furthermore obtain that X contains order units [5, Lemma 2.5 and Lemma 3.2]. By Theorem 4.11 it thus suffices to show that X contains net catching elements in order to obtain $\tau _o=\tau _{ob}$. Note that $\tau _{ob}$ is a linear topology and thus the standard topology of X. The statements about the order convergence and the interior of the cone follow from Proposition 4.7 and Corollary 4.13.

It remains to show that X contains net catching elements. Let u be an order unit of X and consider a norm $\Vert \cdot \Vert$ on X such that the closed unit ball $\overline{B}_1(0)$ is the convex hull of a finite set $F\subseteq X$. Since B is compact with respect to $\Vert \cdot \Vert$ there is $\kappa >0$ such that $B\subseteq \overline{B}_\kappa (0)$. Since u is an order unit there is $n\in {\mathbb {N}}$ such that $\{\kappa x;\, x\in F\}\le n u$ and we observe $B\subseteq \overline{B}_\kappa (0)\le n u$. From Theorem 5.1 we obtain that nu is net catching. $\square$

6.2 Reflexive Banach spaces with ice cream cones

Example 6.2

Let $(X,\Vert \cdot \Vert )$ be a reflexive Banach space. Let f be a linear functional on X such that $\sup \{{|f(x)|}/{\Vert x\Vert };\, x \in X\}=1$ and $\varepsilon \in (0,1)$. The ice cream cone with parameters f and $\varepsilon$ is the cone

$$\begin{aligned} K_{f,\varepsilon }:=\{x \in X;\,f(x) \ge \varepsilon \Vert x\Vert \}. \end{aligned}$$

A reflexive Banach space X equipped with an ice cream cone becomes an Archimedean and directed partially ordered vector space and $K_{f,\varepsilon }$ contains $\Vert \cdot \Vert$-interior points [5, Theorem 2.52 and Lemma 2.3]. We prove next that in this case $\tau _{o}=\tau _{ob}$ is the norm topology of X. A net is order convergent, if and only if it is $\Vert \cdot \Vert$-convergent. A point in X is a $\Vert \cdot \Vert$-interior point of the cone, iff it is an order unit, iff it is net catching.

Proof

Note that the statements about order convergence and the interior of the cone follow from the first statement, Proposition 4.7 and Corollary 4.13. We show first that X (equipped with $K_{f,\varepsilon }$) contains net catching elements. Recall from [5, Theorem 2.52] that f is strictly positive and that $K_{f,\varepsilon }$ is $\Vert \cdot \Vert$-closed.

In order to apply Theorem 5.1 we consider the cone base $B:=\{x\in K_{f,\varepsilon };\, f(x)=1\}$ and the weak topology

$\tau$ of X. As X is a reflexive Banach space we know from the Banach-Alaoglu theorem that the closed ball $\overline{B}_\varepsilon ^{\Vert \cdot \Vert }(0)$ is $\tau$-compact. From $\sup \{{|f(x)|}/{\Vert x\Vert };\, x \in X\}=1$ we obtain that f is continuous with respect to $\Vert \cdot \Vert$. Thus f is continuous with respect to $\tau$ and $B=\overline{B}_\varepsilon ^{\Vert \cdot \Vert }(0)\cap f^{-1}(\{1\})$ is $\tau$-compact. Note furthermore that $K_{f,\varepsilon }$ is $\tau$-closed as a $\Vert \cdot \Vert$-closed and convex set. To show that B has upper bounds we consider a $\Vert \cdot \Vert$-interior point x of $K_{f,\varepsilon }$. There exists $\kappa >0$ such that $\overline{B}^{\Vert \cdot \Vert }_{\kappa }(x)\subseteq K_{f,\varepsilon }$. We define $y:=(1/\kappa )x$. For $b\in B=\overline{B}_\varepsilon ^{\Vert \cdot \Vert }(0)\cap f^{-1}(\{1\})$ we compute

$$\begin{aligned} \Vert x-\kappa (y-b)\Vert =\Vert \kappa b\Vert =\kappa \Vert b\Vert \le \kappa \end{aligned}$$

and obtain $\kappa (y-b)\in K_{f,\varepsilon }$. Since $K_{f,\varepsilon }$ is a cone we obtain $y-b\in K_{f,\varepsilon }$, i.e. $b\le y$. Thus y is an upper bound of B and Theorem 5.1 yields that y is a net catching element. In particular, we observe that $x=\kappa y$ is net catching.

Recall that X is Archimedean and directed. Thus by Theorem 4.12 there holds $\tau _o=\tau _{ob}$ and x is an order unit in X. It remains to show that $\Vert \cdot \Vert$ and $\Vert \cdot \Vert _x$ are equivalent norms, which we will do by showing

$$\begin{aligned} \overline{B}_\kappa ^{\Vert \cdot \Vert }(0) \subseteq \overline{B}_1^{\Vert \cdot \Vert _x}(0) \subseteq \overline{B}_\lambda ^{\Vert \cdot \Vert }(0), \end{aligned}$$

where we denote $\lambda :=(2/\varepsilon )f(x)+\Vert x\Vert$. For $z\in \overline{B}_\kappa ^{\Vert \cdot \Vert }(0)$ we obtain $x+z\in \overline{B}_\kappa ^{\Vert \cdot \Vert }(x)\subseteq K_{f,\varepsilon }$, i.e. $-x\le z$ and similarly $z\le x$. This shows $\overline{B}_\kappa ^{\Vert \cdot \Vert }(0)\subseteq [-x,x]=\overline{B}_1^{\Vert \cdot \Vert _x}(0)$. To show the second inclusion we consider $z\in \overline{B}_1^{\Vert \cdot \Vert _x}(0)=[-x,x]$. As f is positive we know $f(z)\le f(x)$. From $z+x\in K_{f,\varepsilon }$ we thus observe

$$\begin{aligned} \varepsilon \Vert z+x\Vert \le f(z+x)=f(z)+f(x)\le 2f(x). \end{aligned}$$

Computing $\Vert z\Vert \le \Vert z+x\Vert +\Vert x\Vert \le \lambda$ we obtain $z\in \overline{B}_\lambda ^{\Vert \cdot \Vert }(0)$. $\square$

6.3 Spaces of continuous functions on extremely disconnected compact hausdorff spaces

Example 6.3

Let $\Omega$ be an extremely disconnected, compact and infinite Hausdorff space. Consider the Dedekind complete vector lattice $C(\Omega )$ of all continuous maps $f:\Omega \rightarrow {\mathbb {R}}$ equipped with the pointwise order. Any element of $C(\Omega )$ with values in $(0,\infty )$ is an order unit as any $f\in C(\Omega )$ is bounded. Nevertheless, $C(\Omega )$ contains no net catching elements. Thus by Corollary 3.2 and Theorem 4.11 the inclusion $\tau _{o}(C(\Omega ))\subseteq \tau _{ob}(C(\Omega ))$ is valid but strict.

Proof

The statement about the order units is standard and it remains to show that $C(\Omega )$ does not contain net catching elements. By Lemma 4.1 this can be achieved by showing that the order unit $u\in C(\Omega )$ defined by $u(\omega ):=1/2$ for all $\omega \in \Omega$ is not net catching.

As $\Omega$ is infinite and compact there exists $\delta \in \Omega$ such that $\{\delta \}$ is not open. Let A be the set of all open and closed subsets $M\subseteq \Omega$ that contain $\delta$. We order A by reversed inclusion and obtain a directed set. For $M \in A$ we denote the charachteristic function of M by $\chi _{M}$ and obtain $(\chi _M)_{M\in A}$ to be a decreasing net in $C(\Omega )_+$. As for no $M\in A$ we have $\chi _M\le u$ it remains to show that there holds $\chi _M\downarrow 0$.

For this it suffices to show that any lower bound x of $\{\chi _M;\, M\in A\}$ satisfies $x\le 0$. Consider $\omega \in \Omega \setminus \{\delta \}$. Since $\Omega$ is Hausdorff there are disjoint and open sets $U,O\subseteq \Omega$ with $\delta \in U$ and $\omega \in O$. Let M be the closure of U and note that M is open by the extreme disconnectedness of $\Omega$. From $\delta \in M$ we observe that we have $M \in A$. As $\omega \notin M$ we compute $x(\omega )\le \chi _M(\omega )=0$. If $x(\delta )>0$, then the continuity of x implies $\{\delta \}=x^{-1}((0,\infty ))$ to be open, a contradiction. We observe $x\le 0$, which proves $\chi _M\downarrow 0$. $\square$

6.4 Archimedean vector lattices

In Example 6.3 we have seen that an infinite dimensional Archimedean vector lattice might not contain net catching elements. We show next that no infinite dimensional Archimedean vector lattice contains net catching elements.

Theorem 6.4

An Archimedean vector lattice contains net catching elements, if and only if it is finite dimensional.

Recall from Corollary 3.2 that whenever an Archimedean vector lattice contains order units, then $\tau _o\subseteq \tau _{ob}$. As a direct consequence of Proposition 4.7 and Theorem 6.4 we thus observe the following.

Corollary 6.5

For an Archimedean vector lattice that contains order units the inclusion $\tau _{o}\subseteq \tau _{ob}$ is valid but strict.

Remark 6.6

Note that Example 4.10 provides an infinite dimensional vector lattice that contains net catching elements and for which the inclusion $\tau _{ob}\subseteq \tau _o$ is strict. Clearly this vector lattice is not Archimedean. In Example  6.2 we have seen that there are infinite dimensional Archimedean partially ordered vector spaces that contain net catching elements and satisfy $\tau _{o}= \tau _{ob}$. Clearly such partially ordered vector spaces are no vector lattices.

We will prove Theorem 6.4 by embedding X into a space of the form discussed in Example 6.3. We thus investigate next how net catching elements behave under embedding.

Lemma 6.7

Let X be an Archimedean vector lattice and Y be a partially ordered vector space. Let $\iota :X\rightarrow Y$ be a linear order dense embedding. An element $x\in X$ is net catching in X, if and only if $\iota (x)$ is net catching in Y.

Proof

Assume that x is net catching w.r.t. X and consider a net $(y_\alpha )_{\alpha \in A}$ with $y_\alpha \downarrow 0$. Let B be the set of all $\beta \in X_+$ for which there exists $\alpha \in A$ with $\iota (\beta )\ge y_\alpha$. Using that $\iota$ preserves infima it is straight forward to show that for $\beta ,\beta '\in B$ also the infimum $\beta \wedge \beta '$ is contained in B. Thus B equipped with the reversed order of X is directed. We define $x_\beta :=\beta$ for $\beta \in B$ to obtain a net $(x_\beta )_{\beta \in B}$ in X that satisfies $x_\beta \downarrow$.

We next show that this net satisfies $x_\beta \downarrow 0$. Obviously, we have $0\le x_\beta$ for all $\beta \in B$. Let $z\in X$ with $z\le x_\beta$ for all $\beta \in B$ and consider $\alpha \in A$. For $y\in \iota (X)$ with $y\ge y_\alpha$ there exists $\beta \in X$ with $\iota (\beta )=y$ and we observe $\beta \in B$. Thus $\iota (z)\le \iota (y_\beta )=\iota (\beta )=y$. As $\iota (X)$ is order dense in Y we obtain

$$\begin{aligned} \iota (z)\le \inf \{y\in \iota (X);\, y\ge y_\alpha \}=y_\alpha . \end{aligned}$$

From the arbitrary choice of $\alpha \in A$ and $y_\alpha \downarrow 0$ we get $\iota (z)\le 0$. As $\iota$ is order reflecting we observe $z\le 0$, which proves $x_\beta \downarrow 0$.

Now recall that x is net catching. We therefore obtain the existence of $\beta \in B$ with $\beta =x_\beta \le x$. From the definition of B we thus get the existence of $\alpha \in A$ with $x_\alpha \le \iota (\beta )\le \iota (x)$ and observe $\iota (x)$ to be net catching in Y.

To show the converse assume that $\iota (x)$ is net catching and consider a net $(x_\alpha )_{\alpha \in A}$ in X with $x_\alpha \downarrow 0$. As any linear order dense embedding is order continuous we obtain $\iota (x_\alpha )\downarrow 0$. Thus there exists $\alpha \in A$ with $\iota (x_\alpha )\le \iota (x)$. Since $\iota$ is order reflecting we obtain x to be net catching. $\square$

Remark 6.8

Note that there exist infinite dimensional Archimedean partially ordered vector spaces X with net catching elements (Example 6.2). Considering the Dedekind completion $X^\delta$ and a linear order dense embedding $\iota :X\rightarrow X^\delta$ we obtain from Theorem 6.4 that $\iota (x)$ is not net catching for all net catching elements $x\in X$.

Proof of Theorem 6.4

Let us assume that X is finite dimensional. Then X contains net catching elements as presented in Example 6.1.

To show the converse assume that X is an Archimedean vector lattice that contains net catching elements. As X is Archimedean and directed we obtain from Lemma 4.1(ii) that X contains an order unit u. Let $\iota :X\rightarrow X^\delta$ be a Dedekind completion of X. Then $X^\delta$ contains the order unit $\iota (u)$ and by Theorem 2.2 we assume w.l.o.g. that $X^\delta$ is of the form $C(\Omega )$ for a compact and extremely disconnected Hausdorff space $\Omega$. As $\iota$ is a linear order dense embedding we obtain from Lemma 6.7 that $C(\Omega )$ contains net catching elements and our considerations in Example 6.3 imply $\Omega$ to be finite. Thus $C(\Omega )$ is finite dimensional. As any linear order embedding is injective we obtain that X is finite dimensional as well. $\square$