Projection bands and atoms in pervasive pre-Riesz spaces

TL;DR

This paper explores the structure of projection bands and atoms in pre-Riesz spaces, establishing conditions for their extension from covers and linking atomic properties to the space being a vector lattice.

Contribution

It introduces new conditions for projection bands in pre-Riesz spaces and connects atomic elements to the pervasive property, extending the understanding of these structures.

Findings

Projection bands in pervasive spaces extend to vector lattice covers.

Atoms in pre-Riesz spaces are always discrete, and the converse holds if the space is pervasive.

In finite dimensions, pervasiveness is equivalent to being a vector lattice.

Abstract

In vector lattices, the concept of a projection band is a basic tool. We deal with projection bands in the more general setting of an Archimedean pre-Riesz space $X$. We relate them to projection bands in a vector lattice cover $Y$ of $X$. If $X$ is pervasive, then a projection band in $X$ extends to a projection band in $Y$, whereas the restriction of a projection band $B$ in $Y$ is not a projection band in $X$, in general. We give conditions under which the restriction of $B$ is a projection band in $X$. We introduce atoms and discrete elements in $X$ and show that every atom is discrete. The converse implication is true, provided $X$ is pervasive. In this setting, we link atoms in $X$ to atoms in $Y$. If $X$ contains an atom $a>0$, we show that the principal band generated by $a$ is a projection band. Using atoms in a finite dimensional Archimedean pre-Riesz space $X$, we establish that $X$ is pervasive if and only if it is a vector lattice.

Equations

$$\text{pos}A:=\left\{x\in X|\exists n\in \mathbb{N},{\lambda }_i\in {\mathbb{R}}_{>0},x_i\in A,i=1,\dots ,n\text{ with }x=\sum _{i=1}^n{\lambda }_i{x}_i\right\}.$$

$$(a+b)\wedge c=(a\vee b)\wedge c=(a\wedge c)\vee (b\wedge c)=(a\wedge c)+(b\wedge c).$$

$$x=\inf \left\{z\in D|x⩽z\right\},$$

$${\left\{x+y,-x-y\right\}}^u={\left\{x-y,-x+y\right\}}^u,$$

$$0⩽S(x_{\alpha })⩽T(x_{\alpha }).$$

$$X:=\left\{(x_k{)}_k\in {\ell }^{\infty }(\mathbb{Z})|L:=\underset{k\to \infty }{\lim }{x}_k\text{ exists and }\sum _{k=1}^{\infty }\frac{x_{-k}}{2^k}=L\right\}$$

$$B:=\left\{(x_k{)}_k\in X|x_k=0\text{ for }k⩽-2\right\}\text{ and }C:=\left\{(x_k{)}_k\in X|x_k=0\text{ for }k⩾0\right\}.$$

$$\sum _{k=1}^{\infty }\frac{c_{-k}}{2^k}⩾\frac{c_{-1}}{2^1}+\frac{c_{-2}}{2^2}+\frac{c_{-3}}{2^3}⩾\frac{1}{2}+\frac{1}{4}-\frac{2}{8}>0,$$

$$b_k:=0\text{ for }k\in {\mathbb{Z}}_{⩽2},$$

$$b_k:=x_k\text{ for }k\in {\mathbb{Z}}_{⩾0},$$

$$b_{-1}:=2\underset{k\to \infty }{\lim }{x}_k$$

$$\frac{{\tilde{b}}_{-1}}{2}=\sum _{k=1}^{\infty }\frac{{\tilde{b}}_{-k}}{2^k}=\underset{k\to \infty }{\lim }{\tilde{b}}_k=\underset{k\to \infty }{\lim }{x}_k,$$

$$y_1=\sup \left\{z\in i(B)|0⩽z⩽y_1\right\}.$$

$$\forall y_1\in \hat{B}\forall d\in B^{\text{d}}:y_1\perp i(d).$$

$${\tilde{y}}_1$$

$$=(y_1+y_2)\wedge {\tilde{y}}_1=y_1\wedge {\tilde{y}}_1+y_2\wedge {\tilde{y}}_1=y_1\wedge {\tilde{y}}_1,$$

$$y=y\wedge (y_1+y_2)=y\wedge y_1+y\wedge y_2=y_1\in \hat{B}.$$

$$X:=\mathop{\textnormal{span}}\big{(}\left\{x\in PA[-1,1]\hskip 2.84544pt\middle|\hskip 2.84544ptx(0)=0\right\}\cup\left\{q\right\}\big{)}$$

$$\hat{B}:={\mathcal{B}}_{q_1}=\left\{x\in Y|x([0,1])=\left\{0\right\}\right\}\text{ and }{\hat{B}}^{\text{d}}={\mathcal{B}}_{q_2}=\left\{x\in Y|x([-1,0])=\left\{0\right\}\right\}.$$

$$v_{1}=\scalebox{0.7}{$\left(\begin{array}[]{c}1\\ 0\\ 1\end{array}\right)$},\hskip 2.84544ptv_{2}=\scalebox{0.7}{$\left(\begin{array}[]{c}0\\ 1\\ 1\end{array}\right)$},\hskip 2.84544ptv_{3}=\scalebox{0.7}{$\left(\begin{array}[]{c}-1\\ 0\\ 1\end{array}\right)$},\hskip 2.84544ptv_{4}=\scalebox{0.7}{$\left(\begin{array}[]{c}0\\ -1\\ 1\end{array}\right)$},$$

$$f_{1}=\scalebox{0.7}{$\left(\begin{array}[]{c}-1\\ -1\\ 1\end{array}\right)$},\hskip 2.84544ptf_{2}=\scalebox{0.7}{$\left(\begin{array}[]{c}1\\ -1\\ 1\end{array}\right)$},\hskip 2.84544ptf_{3}=\scalebox{0.7}{$\left(\begin{array}[]{c}1\\ 1\\ 1\end{array}\right)$},\hskip 2.84544ptf_{4}=\scalebox{0.7}{$\left(\begin{array}[]{c}-1\\ 1\\ 1\end{array}\right)$}.$$

$$S_n:=\left\{z\in i(X)|0⩽z⩽y_n\right\}(n\in \left\{1,2\right\})$$

$$a_1=\sum _{k=2}^n{\lambda }_k{a}_k.$$

$$0<i(a_1)=i(a_1)\wedge \sum _{k=2}^n{\lambda }_{k}i(a_k)=\sum _{k=2}^n(i(a_1)\wedge {\lambda }_{k}i(a_k))=0.$$

$$[x-\mu a+(\mu -\lambda )a]\perp a,$$

$$0$$

$$=[(i(x)-\mu i(a))\wedge i(a)]+[\left((\mu -\lambda )i(a)\right)\wedge i(a)]$$

$$=[\left((\mu -\lambda )i(a)\right)\wedge i(a)]\ne 0,$$

$$X:=\mathop{\textnormal{span}}\big{(}C^{1}[0,1]\cup\left\{\mathds{1}_{\{x\}}\hskip 2.84544pt\middle|\hskip 2.84544ptx\in[0,1]\right\}\big{)}$$

$$Y:=\mathop{\textnormal{span}}\big{(}C[0,1]\cup\left\{\mathds{1}_{\{x\}}\hskip 2.84544pt\middle|\hskip 2.84544ptx\in[0,1]\right\}\big{)}.$$

$$|i(y)|=\sup \left\{z\in i(X)|0<z⩽|i(y)|\right\}.$$

$$x=\lambda a+(x-\lambda a)\text{ with }\lambda a\in {\left\{a\right\}}^{\text{dd}}\text{ and }(x-\lambda a)\in {\left\{a\right\}}^{\text{d}}.$$

$$x=x_a+x_d\text{ with }{x}_a\in {\left\{a\right\}}^{\text{dd}}={\mathcal{B}}_a,x_d\in {\left\{a\right\}}^{\text{d}},x_a\perp x_d.$$

Full text

Keywords: order projection, projection band, band projection, principal band, atom, discrete element, extremal vector, pervasive, weakly pervasive, Archimedean directed ordered vector space, pre-Riesz space, vector lattice cover

Mathematics Subject Classification (2010): 46A40, 06F20, 47B65

1 Introduction

Projection bands and atoms are both fundamental concepts in the vector lattice theory. For an Archimedean vector lattice $Y$, a band $B$ in $Y$ is called a projection band if $Y=B\oplus B^{\textnormal{d}}$, where $B^{\textnormal{d}}$ is the disjoint complement of $B$. The band $B$ is a projection band if and only if there exists an order projection onto $B$, i.e. a positive linear operator $P\colon Y\rightarrow Y$ with $P^{2}=P$ and $P(Y)=B$. Moreover, for every $y\in Y_{+}$ we have $0\leqslant P(y)\leqslant y$, see, e.g., [16, p. 133 ff.]. Projections bands can similarly be introduced in pre-Riesz spaces using the notion of a band in [12]. Pre-Riesz spaces are precisely those partially ordered vector spaces that can be order densely embedded into vector lattices, their so-called vector lattice covers; see [22]. In [4] it is shown that the relation between projection bands and order projections in pre-Riesz spaces is similar to the one in vector lattices.

Atoms play an important role in the investigation of cones in vector lattices and finite dimensional ordered vector spaces, see, e.g., [2, Section 1.6] and [16, Section 26]. We call a strictly positive element $a$ of an ordered vector space $X$ an atom, if for every $x\in X$ with $0\leqslant x<a$ there is $\lambda\in\mathbb{R}$ with $x=\lambda a$. In an Archimedean vector lattice $Y$ a positive element $a$ is an atom if and only if $a$ is a discrete element, i.e. for every pair of disjoint elements $u,v\in Y$ with $0\leqslant u\leqslant a$ and $0\leqslant v\leqslant a$ it follows $u=0$ or $v=0$. The notion of an atom as well as the notion of a discrete element can be generalized to pre-Riesz spaces, since one can define disjointness in pre-Riesz spaces according to [12].

For the investigation of structures in pre-Riesz spaces, the approach to use vector lattice covers and the restriction and extention method described in [15, Section 2.8] turned out to be fruitful. The techniques used in the present paper follow the same spirit. We deal with the basic question how projection bands in a pre-Riesz space are related to projection bands in a corresponding vector lattice cover. Furthermore, we investigate under which conditions atoms and discrete elements in a pre-Riesz space coincide. We also study the problem how atoms in a pre-Riesz space and atoms in a corresponding vector lattice cover are linked.

In the vector lattice theory, the following statement is well-known, see [16, Theorem 26.4].

Theorem 1**.**

*Let $Y$ be an Archimedean vector lattice. If $a\in Y$ is an atom, then the principal band $\mathcal{B}_{a}$ generated by $a$ consists of all real multiples of $a$, and $\mathcal{B}_{a}$ is a projection band. *

Theorem 1 states that $Y$ admits the decomposition $Y=\mathcal{B}_{a}\oplus\mathcal{B}_{a}^{\textnormal{d}}$. We investigate under which conditions a similar statement is valid in pre-Riesz spaces. It turns out that pervasive pre-Riesz spaces play a crucial role.

The paper is organized as follows. In Section 2 all preliminaries are listed.

In Section 3 we deal with basic properties of band projections and projection bands in a pre-Riesz space $X$. We show that every band projection is order continuous. If $X$ is a vector lattice and $B,C\subseteq X$ are ideals with $X=B\oplus C$, then $B$ is a projection band, see [1, Theorem 1.41]. We give an example that this statement is not true in pre-Riesz spaces, even if $B$ and $C$ are bands. We give two different sets of conditions such that the statement is satisfied.

Section 4 is devoted to the restriction and extension of projection bands. If $X$ is pervasive, then a projection band in $X$ extends to a projection band in a vector lattice cover $Y$ of $X$; see Theorem 21. In Theorem 25 we give conditions such that the restriction of a projection band in $Y$ is a projection band in $X$. This implication is not true, in general.

In Section 5 we introduce atoms and discrete elements in ordered vector spaces. We show that in a pre-Riesz space every atom is a discrete element and that the converse is not true, in general. In Theorem 32 we first establish that in an Archimedean pervasive pre-Riesz space $X$ atoms and discrete elements coincide. Moreover, atoms in $X$ correspond to atoms in $Y$.

In Section 6 we deal with finite-dimensional spaces. Using the theory of atoms, we characterize finite-dimensional Archimedean pervasive pre-Riesz spaces. In Theorem 38 we show that these spaces are precisely the vector lattices.

In Section 7 we consider principal bands generated by atoms in an Archimedean pervasive pre-Riesz space $X$. We show that the ideal $\mathcal{I}_{a}$ generated by an atom $a\in X$ and the band $\mathcal{B}_{a}$ coincide. In Theorem 43 we generalize Theorem 1 and show that every Archimedean pervasive pre-Riesz space admits the decomposition $X=\mathcal{B}_{a}\oplus\mathcal{B}_{a}^{\textnormal{d}}$, provided $a\in X$ is an atom. We conclude that – similar to the vector lattice case – there exists an order projection onto the band $\mathcal{B}_{a}$.

2 Preliminaries

We list some basic terminology in partially ordered vector spaces. Let $X$ be a real vector space and let $X_{+}$ be a cone in $X$, that is, $X_{+}$ is a wedge ($x,y\in X_{+}$ and $\lambda,\mu\geqslant 0$ imply $\lambda x+\mu y\in X_{+}$) and $X_{+}\cap(-X_{+})=\left\{0\right\}$. In $X$ a partial order is defined by $x\leqslant y$ whenever $y-x\in X_{+}$. The space $(X,X_{+})$ (or, loosely, $X$) is then called a (partially) ordered vector space. For a linear subspace $D$ of $X$ we consider in $D$ the order induced from $X$, i.e. we set $D_{+}:=D\cap X_{+}$. A non-empty convex subset $F$ of a cone $X_{+}$ is called a face if $\alpha x+(1-\alpha)y\in F$ with $x,y\in X_{+}$ and $0<\alpha<1$ imply $x,y\in F$. For a subset $A$ of $X$ and $\lambda\in\mathbb{R}$ we use the notations $A+x:=\left\{a+x\hskip 2.84544pt\middle|\hskip 2.84544pta\in A\right\}$ and $\lambda A:=\left\{\lambda a\hskip 2.84544pt\middle|\hskip 2.84544pta\in A\right\}$. The positive-linear hull of $A\subseteq X$ is given by

[TABLE]

An ordered vector space $X$ is called Archimedean if for every $x,y\in X$ with $nx\leqslant y$ for every $n\in\mathbb{N}$ one has $x\leqslant 0$. Clearly, every subspace of an Archimedean ordered vector space is Archimedean. A subspace $D\subseteq X$ is called directed if for every $x,y\in D$ there is an element $z\in D$ such that $x\leqslant z$ and $y\leqslant z$. An ordered vector space $X$ is directed if and only if $X_{+}$ is generating in $X$, that is, $X=X_{+}-X_{+}$. The ordered vector space $X$ has the Riesz decomposition property (RDP) if for every $x_{1},x_{2},z\in X_{+}$ with $z\leqslant x_{1}+x_{2}$ there exist $z_{1},z_{2}\in X_{+}$ such that $z=z_{1}+z_{2}$ with $z_{1}\leqslant x_{1}$ and $z_{2}\leqslant x_{2}$. The space $X$ has the RDP if and only if for every $x_{1},x_{2},x_{3},x_{4}\in X$ with $x_{1},x_{2}\leqslant x_{3},x_{4}$ there exists $z\in X$ such that $x_{1},x_{2}\leqslant z\leqslant x_{3},x_{4}$. An element $u\in X_{+}$ is called an order unit, if for every $x\in X$ there is $\lambda\in\mathbb{R}_{>0}$ with $-\lambda u\leqslant x\leqslant\lambda u$. A net $(z_{\alpha})_{\alpha\in\mathbb{A}}$ in $X$ is decreasing, in symbols $z_{\alpha}\downarrow$, if for every $\alpha,\beta\in\mathbb{A}$ with $\alpha\leqslant\beta$ we have $z_{\alpha}\geqslant z_{\beta}$. We write $z_{\alpha}\downarrow 0$ if $z_{\alpha}\downarrow$ and $\inf\left\{z_{\alpha}\hskip 2.84544pt\middle|\hskip 2.84544pt\alpha\in\mathbb{A}\right\}=0$. A net $(x_{\alpha})_{\alpha}$ in $X$ order converges to $x\in X$, in symbols $x_{\alpha}\xrightarrow{o}x$, if there is a net $(z_{\alpha})_{\alpha}$ in $X$ with $z_{\alpha}\downarrow 0$ and $\pm(x-x_{\alpha})\leqslant z_{\alpha}$ for every $\alpha$.

For an ordered vector space $(X,X_{+})$, a linear operator $T\colon X\rightarrow X$ is called positive, if $T(X_{+})\subseteq X_{+}$. The operator $T$ is positive if and only if for every $x,z\in X$ the relation $x\leqslant z$ implies $T(x)\leqslant T(z)$. If $T$ is positive, we write $T\geqslant 0$. If $X$ is directed, then this introduces a partial order on the space $L(X)$ of linear operators on $X$. An operator $T\in L(X)$ is order continuous333There are several alternative definitions of order convergence of nets in partially ordered vector spaces. However, by [6, Theorem 4.4] an operator $T$ is order continuous if and only if $T$ is continuous with respect to any of these alternative notions. if for every net $(x_{\alpha})_{\alpha}$ in $X$ with $x_{\alpha}\xrightarrow{o}x\in X$ we have $T(x_{\alpha})\xrightarrow{o}T(x)$. We recall the following characterization from [9, Lemma 7].

Lemma 2**.**

Let $X$ and $Z$ be ordered vector spaces, $Z$ Archimedean, and $T\colon X\rightarrow Z$ a positive operator. Then the following are equivalent.

$T$* is order continuous.* 2. 2.

For every net $(x_{\alpha})_{\alpha}$ in $X$ with $x_{\alpha}\downarrow 0$ it follows $T(x_{\alpha})\downarrow 0$.

For standard notations in the case that $X$ is a vector lattice see [1]. We will use the following simple observation.

Lemma 3**.**

Let $X$ be a vector lattice.

Let $a,b,c\in X_{+}$ and $a\perp b$. Then $(a+b)\wedge c=a\wedge c+b\wedge c$. 2. 2.

Let $x\geqslant 0$ be such that $x=x_{1}+x_{2}$ with $x_{1}\perp x_{2}$. Then $x_{1},x_{2}\geqslant 0$.

Proof*.*

1: Due to $a\wedge b=0$, by [1, Theorem 1.3(2)] we have $a+b=a\vee b$. Similarly, $(a\wedge c)\perp(b\wedge c)$ implies $(a\wedge c)+(b\wedge c)=(a\wedge c)\vee(b\wedge c)$. Thus, using [18, Proposition 1.1.2] in the second step of the following equation, we obtain

[TABLE]

2: Due to $|x_{1}|\wedge|x_{2}|=0$, by [1, Theorem 1.7(5)] we have $|x_{1}+x_{2}|=|x_{1}-x_{2}|$. Since $x$ is positive, we get $x=x_{1}+x_{2}=|x_{1}+x_{2}|$. By [1, Theorem 1.7(7)] we obtain $x=|x_{1}+x_{2}|\vee|x_{1}-x_{2}|=|x_{1}|+|x_{2}|$. It follows $x_{1}+x_{2}=x=|x_{1}|+|x_{2}|$ and therefore $x_{2}=(|x_{1}|-x_{1})+|x_{2}|\geqslant 0$. Analogously, $x_{1}\geqslant 0$. $\square$

The next proposition is from [24, Proposition III.10.1 b)].

Proposition 4**.**

*Let $Y$ be an Archimedean vector lattice, $y\in Y_{+}$ and $\Lambda\subseteq\mathbb{R}$ a bounded set. Then we have $\sup\left\{\lambda y\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\Lambda\right\}=(\sup\Lambda)y$ and $\inf\left\{\lambda y\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\Lambda\right\}=(\inf\Lambda)y$. *

Finite dimensional vector lattices are characterized as follows, see [15, Theorem 1.7.8].

Proposition 5**.**

Let $(X,K)$ be an $n$-dimensional Archimedean directed ordered vector space. Then the following statements are equivalent.

$(X,K)$* is a vector lattice.* 2. 2.

$(X,K)$* is order isomorphic to $(\mathbb{R}^{n},\mathbb{R}^{n}_{+})$.* 3. 3.

$(X,K)$* has the RDP.*

Recall that a vector lattice $X$ is Dedekind complete if every non-empty subset of $X$ which is bounded above has a supremum.

We say that a linear subspace $D$ of a vector lattice $X$ generates $X$ as a vector lattice if for every $x\in X$ there exist $a_{1},\ldots,a_{n},b_{1},\ldots,b_{m}\in D$ such that $x=\bigvee_{i=1}^{n}a_{i}-\bigvee_{j=1}^{m}b_{j}$. We call a linear subspace $D$ of an ordered vector space $X$ order dense in $X$ if for every $x\in X$ we have

[TABLE]

that is, the greatest lower bound of the set $\left\{z\in D\hskip 2.84544pt\middle|\hskip 2.84544ptx\leqslant z\right\}$ exists in $X$ and equals $x$, see [3, p. 360]. Recall that a linear map $i\colon X\rightarrow Y$, where $X$ and $Y$ are ordered vector spaces, is called bipositive if for every $x\in X$ one has $i(x)\geqslant 0$ if and only if $x\geqslant 0$. An embedding is a bipositive linear map, which implies injectivity. For an ordered vector space $X$, the following statements are equivalent, see [22, Corollaries 4.9-11 and Theorems 3.7, 4.13]:

There exist a vector lattice $Y$ and an embedding $i\colon X\rightarrow Y$ such that $i(X)$ is order dense in $Y$. 2. 2.

There exist a vector lattice $\tilde{Y}$ and an embedding $i\colon X\rightarrow\tilde{Y}$ such that $i(X)$ is order dense in $\tilde{Y}$ and generates $\tilde{Y}$ as a vector lattice.

If $X$ satisfies 1, then $X$ is called a pre-Riesz space, and $(Y,i)$ is called a vector lattice cover of $X$. For an intrinsic definition of pre-Riesz spaces see [22]. If $X$ is a subspace of $Y$ and $i$ is the inclusion map, we write briefly $Y$ for $(Y,i)$. As all spaces $\tilde{Y}$ in 2 are Riesz isomorphic, we call the pair $(\tilde{Y},i)$ the Riesz completion of $X$ and denote it by $X^{\varrho}$. The space $X^{\varrho}$ is the smallest vector lattice cover of $X$ in the sense that every vector lattice cover $Y$ of $X$ contains a Riesz subspace that is Riesz isomorphic to $X^{\varrho}$. For the following result see [17, Corollary 5] or [15, Proposition 1.6.2].

Proposition 6**.**

Let $X$ be a pre-Riesz space, $(Y,i)$ a vector lattice cover of $X$ and $S\subseteq X$ a non-empty subset.

If $\sup S$ exists in $X$, then $\sup i(S)$ exists in $Y$ and $\sup i(S)=i(\sup S)$. 2. 2.

If $\sup i(S)$ exists in $Y$ and $\sup i(S)\in i(X)$, then $\sup S$ exists in $X$ and $\sup i(S)=i(\sup S)$.

By [22, Theorem 17.1] every Archimedean directed ordered vector space is a pre-Riesz space. Moreover, every pre-Riesz space is directed. If $X$ is an Archimedean directed ordered vector space, then every vector lattice cover of $X$ is Archimedean. By [25, Chapter X.3] every Archimedean directed ordered vector space $X$ has a unique (up to isomorphism) Dedekind completion, which we denote by $X^{\delta}$. Clearly, $X^{\delta}$ is a vector lattice cover of $X$.

Let $X$ be a pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. By [22, Theorem 4.14] and [22, Theorem 4.5] the Dedekind completions $X^{\delta}$ and $Y^{\delta}$ are order and linearly isomorphic, i.e. we can identify $X^{\delta}=Y^{\delta}$. A subspace $D$ of $Y$ is called pervasive in $Y$, if for every $y\in Y_{+}$, $y\neq 0$, there exists $d\in D$ such that $0<d\leqslant y$. If $D$ is a pervasive and majorizing subspace of $Y$, then by [15, Proposition 2.8.5] $D$ is order dense in $Y$. By [15, Proposition 2.8.8] the pre-Riesz space $X$ is pervasive in $Y$ if and only if $X$ is pervasive in any vector lattice cover. Then $X$ is simply called pervasive. In the following result we give characterizations of pervasive pre-Riesz spaces. The characterization in 2 is from [23, Theorem 4.15 and Corollary 4.16] (a short proof can also be found in [17, Lemma 1]) and the one in 3 from [10, Proposition 6].

Proposition 7**.**

Let $X$ be an Archimedean pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. Then the following statements are equivalent.

$X$* is pervasive.* 2. 2.

For every $y\in Y_{+}$ with $y\neq 0$ we have $y=\sup\left\{x\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant x\leqslant y\right\}$. 3. 3.

For every $b\in X$ with $i(b)\vee 0>0$ there exists $x\in X$ such that $0<i(x)\leqslant i(b)\vee 0$.

Clearly, in the formula in (ii) we can replace the set $\left\{x\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0\leq x\leqslant y\right\}$ by the set $\left\{x\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0<x\leqslant y\right\}$. The pre-Riesz space $X$ is called fordable in $Y$ if for every $y\in Y$ there exists a set $S\subseteq X$ such that $\left\{y\right\}^{\textnormal{d}}=i(S)^{\textnormal{d}}$ in $Y$. By [15, Proposition 4.1.18] the space $X$ is fordable in $Y$ if and only if $X$ is fordable in any vector lattice cover of $X$. Then $X$ is simply called fordable. By [15, Proposition 4.1.15] every pervasive pre-Riesz space is fordable.

Next we define disjointness, bands and ideals in ordered vector spaces. For a subset $M\subseteq X$ denote the set of upper bounds of $M$ by $M^{u}:=\left\{x\in X\hskip 2.84544pt\middle|\hskip 2.84544pt\forall m\in M\colon m\leqslant x\right\}$. The elements $x,y\in X$ are called disjoint, in symbols $x\perp y$, if

[TABLE]

for motivation and details see [12]. If $X$ is a vector lattice, then this notion of disjointness coincides with the usual one, see [1, Theorem 1.4(4)]. The following result is established in [12, Proposition 2.1(ii)].

Proposition 8**.**

*Let $Y$ be a pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. Then $x\perp y$ holds in $X$ if and only if $i(x)\perp i(y)$ holds in $Y$. *

The subsequent result is obtained in [11, Theorem 22].

Theorem 9**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space, $a\in X$ and $S\subseteq X_{+}$ such that $\sup S$ exists in $X$. Then the relation $a\perp S$ implies $a\perp\sup S$. *

Let $X$ be an ordered vector space. The disjoint complement of a set $M\subseteq X$ is $M^{\textnormal{d}}:=\left\{x\in X\hskip 2.84544pt\middle|\hskip 2.84544pt\forall y\in M\colon x\perp y\right\}$. A linear subspace $B$ of $X$ is called a band in $X$ if $B^{\textnormal{dd}}=B$, see [12, Definition 5.4]. If $X$ is an Archimedean vector lattice, then this notion of a band coincides with the classical notion of a band in vector lattices (where a band is defined to be an order closed ideal). For every subset $M\subseteq X$ the disjoint complement $M^{\textnormal{d}}$ is a band, see [12, Proposition 5.5]. For an element $a\in X$, by $\mathcal{B}_{a}$ we denote the band generated by $a$, i.e. $\mathcal{B}_{a}:=\bigcap\left\{B\subseteq X\hskip 2.84544pt\middle|\hskip 2.84544ptB\textnormal{ is a band in }X\textnormal{ with }a\in B\right\}$. The band $\mathcal{B}_{a}$ is called a principal band. By [11, Lemma 4] we have $\mathcal{B}_{a}=\left\{a\right\}^{\textnormal{dd}}$.

The following notion of an ideal is introduced in [21, Definition 3.1]. A subset $M$ of an ordered vector space $X$ is called solid if for every $x\in X$ and $y\in M$ the relation $\left\{x,-x\right\}^{u}\supseteq\left\{y,-y\right\}^{u}$ implies $x\in M$. A solid subspace of $X$ is called an ideal. This notion of an ideal coincides with the classical definition, provided $X$ is a vector lattice. For an element $a\in X$, by $\mathcal{I}_{a}$ we denote the ideal generated by $a$, i.e. $\mathcal{I}_{a}:=\bigcap\left\{I\subseteq X\hskip 2.84544pt\middle|\hskip 2.84544ptI\textnormal{ is an ideal in }X\textnormal{ with }a\in I\right\}$, see also [11, Lemma 5]. The ideal $\mathcal{I}_{a}$ is called a principal ideal.

Next we discuss the restriction property and the extension property for ideals and bands. Let $X$ be a pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. For $S\subseteq Y$ we write $[S]i:=\left\{x\in X\hskip 2.84544pt\middle|\hskip 2.84544pti(x)\in S\right\}$. The pair $(L,M)\subseteq\mathcal{P}(X)\times\mathcal{P}(Y)$ is said to satisfy

the restriction property (R), if whenever $J\in M$, then $[J]i\in L$, and

• -

the extension property (E), if whenever $I\in L$, then there is $J\in M$ such that $I=[J]i$.

In [12] the properties (R) and (E) are investigated for ideals and bands. It is shown that the extension property (E) is satisfied for bands, i.e. for $L$ being the set of bands in $X$ and $M$ being the set of bands in $Y$. Moreover, the restriction property (R) is satisfied for ideals. In general, bands do not have (R) and ideals do not have (E). The appropriate sets $M$ and $L$ of directed ideals satisfy (E). If $X$ is fordable, then by [13, Proposition 2.5 and Theorem 2.6] we have (R) for bands. If for an ideal $I$ in $X$ and an ideal $J$ in $Y$ we have $I=[J]i$, then $J$ is called an extension ideal of $I$. An extension band $J$ for a band $I$ in $X$ is defined similarly. Extension ideals and bands are not unique, in general. If an ideal $I$ in $X$ has an extension ideal in $Y$, then $\hat{I}:=\bigcap\left\{J\subseteq Y\hskip 2.84544pt\middle|\hskip 2.84544ptJ\textnormal{ is an extension ideal of }I\right\}$ is the smallest extension ideal of $I$ in $Y$. For a band $B$ in $X$ the smallest extension band $\hat{B}$ of $B$ is defined similarly. From [11, Theorem 16] we obtain the following.

Lemma 10**.**

*Let $X$ be a pre-Riesz space and $I$ an ideal in $X$. If $I$ is directed, then $i(I)$ is majorizing in $\hat{I}$. *

By [8, Proposition 17 (a)] and its subsequent discussion the smallest extension band of $B$ is given by $\hat{B}=i(B)^{\textnormal{dd}}$ in $Y$. The following result can be found in [11, Theorem 31].

Theorem 11**.**

Let $X$ be an Archimedean pervasive pre-Riesz space with a vector lattice cover $(Y,i)$. Let $B\subseteq X$ be a band and $D\subseteq Y$ an extension band of $B$. Then the following statements are equivalent.

$i(B)$* is majorizing in $D$.* 2. 2.

$i(B)$* is order dense in $D$.*

*If 1 or 2 are satisfied, then $B$ is a pre-Riesz space and hence directed and $D$ is a vector lattice cover of $B$. Moreover, $B$ is pervasive. *

In the literature on vector lattices, atoms and discrete elements are defined in several different ways. We use the following notions. In a vector lattice $Y$, an element $a\in Y\backslash\{0\}$ is called an atom if for every $y\in Y$ with $|y|\leqslant|a|$ there is $\lambda\in\mathbb{R}$ such that $y=\lambda a$. An element $d\in Y\backslash\{0\}$ is called discrete if for every pair of disjoint elements $u,v\in Y$ with $0\leqslant u\leqslant|d|$ and $0\leqslant v\leqslant|d|$ it follows that $u=0$ or $v=0$, see [24, Definition III.13.1]. The following statements can be found in [1, Chapter 2, Exercises 5.(i)], [24, III.13.1 b)], [16, Lemma 26.2 (ii)].

Proposition 12**.**

Let $Y$ be an Archimedean vector lattice and $a\in Y_{+}\setminus\{0\}$. Then the following statements are equivalent.

The element $a$ is an atom. 2. 2.

The principal ideal $\mathcal{I}_{a}$ is one-dimensional. 3. 3.

The element $a$ is a discrete element.

Notice that if $Y$ is not Archimedean, then the implication from 3 to 1 is not true, in general. For further results see [24, III.13.1 a) to d)]. The following observation is established in [24, Proposition III.13.1 d)].

Proposition 13**.**

*Let $Y$ be an Archimedean vector lattice, $a\in Y_{+}$ an atom and $x\in Y$ with $0<a\leqslant x$. Then there exists a real number $\lambda>0$ such that $x-\lambda a\geqslant 0$ and $(x-\lambda a)\perp a$. *

3 Projection bands in pre-Riesz spaces

For a projection $P$ on an ordered vector space $X$, i.e. for a linear operator with $P^{2}=P$, there are two natural ways to relate $P$ with the order structure. On one hand, if $A\subseteq X$ is a linear subspace, a projection $P\colon X\rightarrow X$ with $0\leqslant P\leqslant I$ and $P(X)=A$ is called an order projection onto $A$. On the other hand, if $B\subseteq X$ is a band such that $X=B\oplus B^{\textnormal{d}}$, then $B$ is called a projection band, and the operator $P_{B}\colon X\rightarrow X$, $x\mapsto x_{1}$ is well-defined, where for every $x\in X$ we have the unique decomposition $x=x_{1}+x_{2}$ for $x_{1}\in B$ and $x_{2}\in B^{\textnormal{d}}$. The operator $P_{B}$ is called the band projection onto $B$. Order projections are introduced in [5] and considered in [4] in relation to band projections. In the following statement we collect the results from [4, Theorem 3.2 and Propositions 2.5, 2.3 and 3.1].

Theorem 14**.**

Let $X$ be a pre-Riesz space.

An operator $P\colon X\rightarrow X$ is an order projection if and only if it is a band projection. 2. 2.

Every projection band $B\subseteq X$ is directed. 3. 3.

A band $B\subseteq X$ is a projection band if and only if $B^{\textnormal{d}}$ is a projection band. 4. 4.

If two band projections $P$ and $Q$ have the same range, then $P=Q$.

Order projections in vector lattices are order continuous. To observe the same fact in Archimedean pre-Riesz spaces, we need the following result.

Proposition 15**.**

*Let $X$ be an Archimedean pre-Riesz space, $S,T\colon X\rightarrow X$ two operators with $0\leqslant S\leqslant T$ and $T$ order continuous. Then $S$ is order continuous. *

Proof*.*

As $P\geqslant 0$ and $X$ is Archimedean, by Lemma 2 it suffices to show that for every net $(x_{\alpha})_{\alpha}$ in $X_{+}$ with $x_{\alpha}\downarrow 0$ it follows $S(x_{\alpha})\downarrow 0$. Let $(x_{\alpha})_{\alpha}$ be a net in $X_{+}$ such that $x_{\alpha}\downarrow 0$. From $0\leqslant S\leqslant T$ for every $\alpha$ it follows

[TABLE]

As $T$ is order continuous, from $x_{\alpha}\downarrow 0$ it follows $T(x_{\alpha})\downarrow 0$. In particular, we have $\inf\left\{T(x_{\alpha})\hskip 2.84544pt\middle|\hskip 2.84544pt\alpha\in\mathcal{A}\right\}=0$. Due to $S\geqslant 0$ we obtain $S(x_{\alpha})\downarrow$. It is left to show that $\inf\left\{S(x_{\alpha})\hskip 2.84544pt\middle|\hskip 2.84544pt\alpha\in\mathcal{A}\right\}=0$. If $v$ is a lower bound of the set $M:=\left\{S(x_{\alpha})\hskip 2.84544pt\middle|\hskip 2.84544pt\alpha\in\mathcal{A}\right\}$, then $v$ is also a lower bound of the set $\left\{T(x_{\alpha})\hskip 2.84544pt\middle|\hskip 2.84544pt\alpha\in\mathcal{A}\right\}$, hence $v\leqslant 0$. As [math] is a lower bound of $M$, we obtain $0=\inf M$. $\square$

Since for every order projection $P$ we have $0\leqslant P\leqslant I$, where $I$ is the identity operator, from Proposition 15 we immediately obtain the following.

Corollary 16**.**

*Let $X$ be an Archimedean pre-Riesz space and $P$ an order projection. Then $P$ is order continuous. *

Recall that for two ideals $B,C$ in a vector lattice $X$ the relation $X=B\oplus C$ implies that $B$ is a projection band and $C=B^{\textnormal{d}}$. The two subsequent Theorems 17 and 18 generalize this result to pre-Riesz spaces.

Theorem 17**.**

*Let $X$ be a pervasive pre-Riesz space and $B,D\subseteq X$ two ideals such that $X=B\oplus D$. Then $B$ is a projection band and $X=B\oplus B^{\textnormal{d}}$. *

Proof*.*

Let $B,D\subseteq X$ be two ideals with $X=B\oplus D$. Assume, on the contrary, that $B\not\perp D$. Then there exist two elements $b\in B$ and $d\in D$ such that $b\not\perp d$. For a vector lattice cover $(Y,i)$ of $X$, by Proposition 8 it follows $i(b)\not\perp i(d)$. We obtain $0<|i(b)|\wedge|i(d)|$. Since $X$ is pervasive, there exists $x\in X$ with $0<i(x)\leqslant|i(b)|\wedge|i(d)|$. We show $x\in B$. Indeed, $i(x)\leqslant|i(b)|$ is equivalent to $\left\{i(x)\right\}^{u}\supseteq\left\{i(b),-i(b)\right\}^{u}$. Taking the intersection of these sets of upper bounds with $X$ we obtain due to $x>0$ that $\left\{x,-x\right\}^{u}=\left\{x\right\}^{u}\supseteq\left\{b,-b\right\}^{u}$ in $X$. It follows $x\in B$. Similarly, we obtain $x\in D$. From the uniqueness of the decomposition $X=B\oplus D$ we have $B\cap D=\left\{0\right\}$, which implies $x=0$, a contradiction. We thus obtain $B\perp D$. It follows $D\subseteq B^{\textnormal{d}}$.

To establish $B^{\textnormal{d}}\subseteq D$, let $x\in B^{\textnormal{d}}$. From $X=B\oplus D$ it follows $x=b+d$ with $b\in B$ and $d\in D\subseteq B^{\textnormal{d}}$. Due to $x,d\in B^{\textnormal{d}}$ we obtain $b=x-d\in B^{\textnormal{d}}$. This yields $b\in B\cap B^{\textnormal{d}}=\left\{0\right\}$ and therefore $x=d\in D$. This establishes $B^{\textnormal{d}}\subseteq D$. We conclude $B^{\textnormal{d}}=D$. $\square$

We show a result similar to Theorem 17, where the ideals are directed and the condition on the underlying space is weaker. A pre-Riesz space $X$ is called weakly pervasive if for every $b,d\in X$ with $b,d>0$ and $b\not\perp d$ there exists $x\in X$ such that $0<x\leqslant b,d$, see [10, Definition 8 and Lemma 9]. There it is also shown that every pervasive pre-Riesz space is weakly pervasive.

Theorem 18**.**

*Let $X$ be a weakly pervasive pre-Riesz space and $B,D\subseteq X$ two directed ideals such that $X=B\oplus D$. Then $B$ is a projection band and $X=B\oplus B^{\textnormal{d}}$. *

Proof*.*

Let first $b\in B_{+}$. We show for every $d\in D_{+}$ that $b\perp d$. Let $d\in D_{+}$. Assume, on the contrary, that $b\not\perp d$. Since $X$ is weakly pervasive, there exists $\tilde{x}\in X_{+}$ with $0<\tilde{x}\leqslant b,d$. Due to $B,D$ being ideals we obtain $\tilde{x}\in B,D$. From the uniqueness of the decomposition $X=B\oplus D$ we have $B\cap D=\left\{0\right\}$, which is a contradiction to $\tilde{x}>0$. It follows $b\perp d$. Since $X$ is a pre-Riesz space, disjoint complements are linear subspaces of $X$. Since $D$ is directed, it follows $b\perp D$. The directedness of $B$ implies $B\perp D$. We conclude $D\subseteq B^{\textnormal{d}}$.

To establish $B^{\textnormal{d}}\subseteq D$, let $x\in B^{\textnormal{d}}$. From $X=B\oplus D$ it follows $x=b+d$ with $b\in B$ and $d\in D\subseteq B^{\textnormal{d}}$. Due to $x,d\in B^{\textnormal{d}}$ we obtain $b=x-d\in B^{\textnormal{d}}$. This yields $b\in B\cap B^{\textnormal{d}}=\left\{0\right\}$ and therefore $x=d\in D$. This establishes $B^{\textnormal{d}}\subseteq D$. We conclude $B^{\textnormal{d}}=D$. $\square$

The following example demonstrates that, in general, Theorem 17 is not true if $X$ is not pervasive, and Theorem 18 is not true if one of the ideals is not directed.

Example 19**.**

A weakly pervasive pre-Riesz space $X$ with two bands $B,C\subseteq X$ such that $X=B\oplus C$ and $C\not=B^{\textnormal{d}}$.

Let $\ell^{\infty}(\mathbb{Z})$ be the vector space of bounded sequences on $\mathbb{Z}$ endowed with the pointwise order. In [14, Example 5.2] it is shown that Y:=\left\{(y_{k})_{k}\in\ell^{\infty}(\mathbb{Z})\hskip 2.84544pt\middle|\hskip 2.84544pt\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}y_{k}\textnormal{ exists}\right\} is a vector lattice and its linear subspace

[TABLE]

is a pre-Riesz space and order dense in $Y$. Thus by Proposition 8 disjointness is pointwise in $X$. In [10, Example 10] it is established that $X$ is weakly pervasive. Consider the following subspaces of $X$:

[TABLE]

We first show that $B$ is a directed band. Let $a=(a_{k})_{k}$, $b=(b_{k})_{k}\in B$. For $k\leqslant-2$ we have $a_{k}=0$, which implies \raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}a_{k}=\sum_{k=1}^{\infty}\frac{a_{-k}}{2^{k}}=\frac{1}{2}a_{-1}. Similarly, \frac{1}{2}b_{-1}=\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}b_{k}. Define a bounded sequence $c=(c_{k})_{k}$ by $c_{k}:=\frac{1}{2}\max\left\{a_{k},b_{k}\right\}$ for every $k\in\mathbb{Z}$. Then \raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}c_{k}=\frac{1}{2}\max\left\{a_{-1},b_{-1}\right\}=\frac{1}{2}c_{-1}. Moreover, for $k\in\mathbb{Z}_{\leqslant-2}$ we have $c_{k}=0$. It follows \sum_{k=1}^{\infty}\frac{c_{-k}}{2^{k}}=\frac{1}{2}c_{-1}=\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}c_{k}, i.e. $c\in X$. Clearly, $c\in B$ and $a,b\leqslant c$. Thus $B$ is directed. We show that $B$ is a band. For every $n\in\mathbb{N}_{\geqslant 1}$ define a sequence $x^{(n)}=(x^{(n)}_{k})_{k}\in X$ with $x^{(n)}_{k}:=0$ for $k\in\mathbb{Z}\backslash\left\{-n,-(n+1)\right\}$ and $x^{(n)}_{-n}:=1$, $x^{(n)}_{-(n+1)}:=-2$. Then for every $n\in\mathbb{N}_{\geqslant 1}$ we have \sum_{k=1}^{\infty}\frac{x^{(n)}_{-k}}{2^{k}}=\frac{x^{(n)}_{-n}}{2^{n}}+\frac{x^{(n)}_{-(n+1)}}{2^{n+1}}=0=\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}x^{(n)}_{k}. That is, $x^{(n)}\in X$. Moreover, for every $n\in\mathbb{N}_{\geqslant 2}$ we have $x^{(n)}\perp B$. For the set $S:=\left\{x^{(n)}\hskip 2.84544pt\middle|\hskip 2.84544ptn\in\mathbb{N}_{\geqslant 2}\right\}\subseteq X$ we then have $S^{\textnormal{d}}=B$. It follows that $B$ is a band.

Next we establish that $C$ is a non-directed band. To show that $C$ is not directed, consider the two elements $x^{(1)},x^{(2)}\in C$. Assume that there exists $c=(c_{k})_{k}\in C$ with $x^{(1)},x^{(2)}\leqslant c$. Then $c_{-1},c_{-2}\geqslant 1$ and $c_{-3}\geqslant-2$. Moreover $c_{k}\geqslant 0$ for every $k\in\mathbb{Z}_{\leqslant-3}$ and due to $c\in C$ we have $c_{k}=0$ for $k\in\mathbb{Z}_{\geqslant 0}$. It follows

[TABLE]

from which we obtain \raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}c_{k}>0, a contradiction to $c_{k}=0$ for $k\in\mathbb{Z}_{\geqslant 0}$. Thus $C$ is not directed. To see that $C$ is a band, for every $n\in\mathbb{N}_{\geqslant 0}$ define a sequence $z^{(n)}=(z^{(n)}_{k})_{k}$ by $z^{(n)}_{n}:=1$ and $z^{(n)}_{k}:=0$ for $k\neq n$. Clearly, for every $n\in\mathbb{N}_{\geqslant 0}$ we have $z^{(n)}\in X$. For the set $T:=\left\{z^{(n)}\hskip 2.84544pt\middle|\hskip 2.84544ptn\in\mathbb{N}_{\geqslant 0}\right\}\subseteq X$ we obtain $T^{\textnormal{d}}=C$. It follows that $C$ is a band.

We show $B\not\perp C$. To that end, let $b=(b_{k})_{k}$ be defined for $k\in\mathbb{Z}_{\leqslant-2}$ by $b_{k}:=0$, for $k\in\mathbb{Z}_{\geqslant 0}$ by $b_{k}=\frac{1}{2}$ and $b_{-1}:=1$. Then $b\in B$. In $Y$ we have $i(b)\not\perp i(x^{(1)})$. By Proposition 8 for the sequence $x^{(1)}\in C$ and $b\in C$ we have $b\not\perp x^{(1)}$. It follows $B\not\perp C$.

Next we show $X=B\oplus C$. Let $x=(x_{k})_{k}\in X$. Define $b=(b_{k})_{k},c=(c_{k})_{k}\in X$ by

[TABLE]

Then $b\in B$ and $c\in C$. Clearly, for $k\in\mathbb{Z}\backslash\left\{-1\right\}$ we have $x_{k}=b_{k}+c_{k}$. Moreover, x_{-1}=2\frac{x_{-1}}{2}=2\big{(}\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}x_{k}-\sum_{k=2}^{\infty}\frac{x_{-k}}{2^{k}}\big{)}=b_{-1}+c_{-1}. That is, $x=b+c$. We show that this decomposition is unique. Let $\tilde{b}=(\tilde{b}_{k})_{k}\in B$ and $\tilde{c}=(\tilde{c}_{k})_{k}\in C$ be such that $x=\tilde{b}+\tilde{c}$. From $b,\tilde{b}\in B$ and $c,\tilde{c}\in C$ it follows for $k\in\mathbb{Z}\backslash\left\{-1\right\}$ that $b_{k}=\tilde{b}_{k}$ and $c_{k}=\tilde{c}_{k}$. It is left to show $b_{-1}=\tilde{b}_{-1}$ and $c_{-1}=\tilde{c}_{-1}$. As $\tilde{b}\in B$, we have $\sum_{k=1}^{\infty}\frac{\tilde{b}_{-k}}{2^{k}}=\frac{\tilde{b}_{-1}}{2}$. Since for $k\in\mathbb{Z}_{\geqslant 0}$ we have $x_{k}=\tilde{b}_{k}+\tilde{c}_{k}=\tilde{b}_{k}$, we obtain

[TABLE]

from which it follows \tilde{b}_{-1}=2\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}x_{k}=b_{-1}. Moreover, as $\tilde{c}\in C$, we have 0=\raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}\tilde{c}_{k}=\sum_{k=1}^{\infty}\frac{\tilde{c}_{-k}}{2^{k}}=\frac{\tilde{c}_{-1}}{2}+\sum_{k=2}^{\infty}\frac{\tilde{c}_{-k}}{2^{k}}. As $\tilde{c}_{k}=x_{k}$ for $k\in\mathbb{Z}_{\leqslant-2}$, we get $\tilde{c}_{-1}=-2\sum_{k=2}^{\infty}\frac{\tilde{c}_{-k}}{2^{k}}=-2\sum_{k=2}^{\infty}\frac{x_{-k}}{2^{k}}=c_{-1}$. We conclude $b=\tilde{b}$ and $c=\tilde{c}$.

Finally, we show that $X$ is not pervasive. To that end, we use the characterization of pervasiveness in Proposition 7 3. Consider the sequence $x^{(1)}\in X$ defined above. The sequence $i(x^{(1)})\vee 0\in Y$ is zero in every coordinate, except in $(i(x^{(1)})\vee 0)_{-1}=1$. Thus $i(x^{(1)})\vee 0>0$. Assume, on the contrary, that $X$ is pervasive, that is, there is $x\in X$ with $0<i(x)\leqslant i(x^{(1)})\vee 0$. Then for every $k\in\mathbb{Z}\backslash\left\{-1\right\}$ we have $x_{k}=0$ and it follows \raisebox{2.15277pt}{\scalebox{0.8}{\displaystyle\lim_{k\to\infty};}}x_{k}=0=\sum_{k=1}^{\infty}\frac{x_{-k}}{2^{k}}=\frac{x_{-1}}{2}. We obtain $x=0$, a contradiction. We conclude that $X$ is not pervasive.

4 Extension and restriction of projection bands

In view of the embedding technique for pre-Riesz spaces, the question arises how projection bands in a pre-Riesz space $X$ are related to projection bands in a vector lattice cover $Y$ of $X$. The restriction and extension properties for bands in pervasive pre-Riesz spaces suggest that there projection bands are linked in a similar way. We show that indeed the extension property for projection bands is satisfied in Archimedean pervasive pre-Riesz spaces, whereas for the restriction property stronger assumptions are needed.

The next statement is a technical result that we use to establish Theorem 21.

Lemma 20**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. Let $B\subseteq X$ be a band such that $X=B\oplus B^{\textnormal{d}}$. Moreover, let $\hat{B}$ be the smallest extension ideal of $B$ and $A$ the smallest extension ideal of $B^{\textnormal{d}}$ in $Y$. Then $A\subseteq\hat{B}^{\textnormal{d}}$ and $\hat{B}\subseteq A^{\textnormal{d}}$. *

Proof*.*

The bands $B$ and $B^{\textnormal{d}}$ are ideals and by Theorem 142 directed. Therefore in $Y$ there exist the smallest extension ideals $\hat{B}$ of $B$ and $A$ of $B^{\textnormal{d}}$.

Let $y_{1}\in\hat{B}_{+}$. Due to $X$ being Archimedean and pervasive, by Proposition 7 we have $y_{1}=\sup\left\{z\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant z\leqslant y_{1}\right\}$. Since $B$ is a directed ideal, by Lemma 10 the subspace $i(B)$ is majorizing in $\hat{B}$. Thus there exists $b\in B$ with $0\leqslant y_{1}\leqslant i(b)$. For every $x\in X$ with $0\leqslant i(x)\leqslant y_{1}$, due to $0\leqslant x\leqslant b$ it follows $x\in B$. This yields

[TABLE]

Since $B$ is a band in $X$, for every $x\in B$ and $d\in B^{\textnormal{d}}$ we have $x\perp d$, which implies $\left\{z\in i(B)\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant z\leqslant y_{1}\right\}\perp i(d)$. Applying Theorem 9 in the vector lattice $Y$, for every $d\in B^{\textnormal{d}}$ we obtain $y_{1}\perp i(d)$. From the fact that $\hat{B}$ is directed we conclude

[TABLE]

As the roles of $B$ and $B^{\textnormal{d}}$ are interchangeable, similarly to (2) for every $y_{2}\in A_{+}$ we obtain $y_{2}=\sup\left\{z\in i(B^{\textnormal{d}})\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant z\leqslant y_{2}\right\}$. Due to (3), for every $y_{1}\in\hat{B}$ we have $y_{1}\perp\left\{z\in i(B^{\textnormal{d}})\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant z\leqslant y_{2}\right\}$. By Theorem 9 it follows $y_{1}\perp y_{2}$. Since $A$ is directed we conclude that for every $y_{1}\in\hat{B}$ and every $y_{2}\in A$ we have $y_{1}\perp y_{2}$. It follows $A\subseteq\hat{B}^{\textnormal{d}}$ and $\hat{B}\subseteq A^{\textnormal{d}}$. $\square$

Theorem 21**.**

Let $X$ be an Archimedean pervasive pre-Riesz space and $B\subseteq X$ a band such that $X=B\oplus B^{\textnormal{d}}$. Let $(Y,i)$ be a vector lattice cover of $X$ and $\hat{B}$ the smallest extension ideal of $B$ in $Y$. Then we have the following.

$Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$, 2. 2.

$\hat{B}$* equals the smallest extension band of $B$,* 3. 3.

$i(B)$* is majorizing in $\hat{B}$ and $i(B^{\textnormal{d}})$ is majorizing in $\hat{B}^{\textnormal{d}}$,* 4. 4.

$\hat{B}^{\textnormal{d}}=\widehat{B^{\textnormal{d}}}$, 5. 5.

$\hat{B}$* coincides with every extension band and every extension ideal of $B$,* 6. 6.

$\hat{B}$* and $\hat{B}^{\textnormal{d}}$ are vector lattice covers of $B$ and $B^{\textnormal{d}}$, respectively, and both $B$ and $B^{\textnormal{d}}$ are pervasive.*

Proof*.*

1: To establish $Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$, let first $y\in Y_{+}$. Since $i(X)$ is majorizing in $Y$, there exists $x\in X$ such that $0\leqslant y\leqslant i(x)$. Due to $X=B\oplus B^{\textnormal{d}}$ there exist $x_{1}\in B$ and $x_{2}\in B^{\textnormal{d}}$ such that $x=x_{1}+x_{2}$. The vector lattice $Y$ has the RDP, therefore $0\leqslant y\leqslant i(x)=i(x_{1})+i(x_{2})$ implies that there are $y_{1},y_{2}\in Y_{+}$ with $0\leqslant y_{1}\leqslant i(x_{1})$ and $0\leqslant y_{2}\leqslant i(x_{2})$ such that $y=y_{1}+y_{2}$. From $i(x_{1})\in i(B)\subseteq\hat{B}$ and $0\leqslant y_{1}\leqslant i(x_{1})$ we obtain $y_{1}\in\hat{B}$. Let $A$ be as in Lemma 20. Similarly, due to $i(x_{2})\in A$, with Lemma 20 we have $y_{2}\in A\subseteq\hat{B}^{\textnormal{d}}$. Since $\hat{B}\perp\hat{B}^{\textnormal{d}}$, it follows that $y=y_{1}+y_{2}$ is a disjoint decomposition.

We show that this decomposition is unique. Let there be another disjoint decomposition $y=\tilde{y}_{1}+\tilde{y}_{2}$ with $\tilde{y}_{1}\in\hat{B}$ and $\tilde{y}_{2}\in\hat{B}^{\textnormal{d}}$. Due to the disjointness of the decompositions, with Lemma 32 we have $\tilde{y}_{1},\tilde{y}_{2}\geqslant 0$. Lemma 31 then yields

[TABLE]

i.e. $\tilde{y}_{1}\leqslant y_{1}$. Similarly, $y_{1}\leqslant\tilde{y}_{1}$. We conclude $y_{1}=\tilde{y}_{1}$. Analogously, we obtain $y_{2}=\tilde{y}_{2}$, i.e. the disjoint decomposition is unique.

Let $y\in Y$ be arbitrary. We have $y=y^{+}-y^{-}$ with unique disjoint decompositions $y^{+}=y^{+}_{1}+y^{+}_{2}$ and $y^{-}=y^{-}_{1}+y^{-}_{2}$, where $y^{+}_{1},y^{-}_{1}\in\hat{B}$ and $y^{+}_{2},y^{-}_{2}\in\hat{B}^{\textnormal{d}}$. Due to $y^{+}\perp y^{-}$ it follows that $y=(y^{+}_{1}-y^{-}_{1})+(y^{+}_{2}-y^{-}_{2})$ is a unique decomposition of $y$ with $(y^{+}_{1}-y^{-}_{1})\in\hat{B}$ and $(y^{+}_{2}-y^{-}_{2})\in\hat{B}^{\textnormal{d}}$. We conclude $Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$.

2: First we establish that $\hat{B}$ is a band. We need to show $\hat{B}^{\textnormal{dd}}\subseteq\hat{B}$. Let $y\in\hat{B}^{\textnormal{dd}}$ and consider first the case $y\geqslant 0$. By 1 there is a decomposition $y=y_{1}+y_{2}$ with $y_{1}\in\hat{B}$ and $y_{2}\in\hat{B}^{\textnormal{d}}$. Lemma 32 yields $y_{1},y_{2}\geqslant 0$. Therefore, by Lemma 31 and due to $y\in\hat{B}^{\textnormal{dd}}$, we obtain

[TABLE]

For an arbitrary $y\in\hat{B}^{\textnormal{dd}}$ with $y=y^{+}-y^{-}$ it similarly follows $y^{+},-y^{-}\in\hat{B}$ and therefore $y\in\hat{B}$. This implies that $\hat{B}$ is a band. Since $\hat{B}$ is the smallest extension ideal of $B$ and every extension band $D$ of $B$ is an ideal, it follows $\hat{B}\subseteq D$, i.e. $\hat{B}$ is the smallest extension band of $B$.

3: To establish that $i(B)$ is majorizing in $\hat{B}$ it suffices to consider positive elements. Let $y\in\hat{B}_{+}$. Since $i(X)$ is majorizing in $Y$ and $X=B\oplus B^{\textnormal{d}}$, there exists an $x\in i(X)$ with $y\leqslant x=x_{1}+x_{2}$, where $x_{1}\in i(B)$ and $x_{2}\in i(B^{\textnormal{d}})$. As $Y$ has the RDP, there exist $y_{1},y_{2}\in Y$ with $y=y_{1}+y_{2}$ such that $0\leqslant y_{1}\leqslant x_{1}$ and $0\leqslant y_{2}\leqslant x_{2}$. From $0\leqslant y_{1}\leqslant x_{1}\in i(B)\subseteq\hat{B}$ it follows $y_{1}\in\hat{B}$. Then $y_{1}\in i(B)\subseteq\hat{B}$. The uniqueness of the decomposition established in 1 yields $y=y_{1}$ and $y_{2}=0$. From $0\leqslant y=y_{1}\leqslant x_{1}\in i(B)$ we conclude that $i(B)$ is majorizing in $\hat{B}$.

To establish that $i(B^{\textnormal{d}})$ is majorizing in $\hat{B}^{\textnormal{d}}$, let $y\in(\hat{B}^{\textnormal{d}})_{+}$. Similarly to the previous case there exist $x_{1},x_{2}\in i(X)$ and $y_{1},y_{2}\in Y$ with $y=y_{1}+y_{2}$ such that $0\leqslant y_{1}\leqslant x_{1}\in i(B)$ and $0\leqslant y_{2}\leqslant x_{2}\in i(B^{\textnormal{d}})$. The uniqueness of the decomposition leads to $y_{1}=0$ and $y=y_{2}\leqslant x_{2}\in i(B^{\textnormal{d}})$, i.e. $i(B^{\textnormal{d}})$ is majorizing in $\hat{B}^{\textnormal{d}}$.

4: To establish the inclusion $\widehat{B^{\textnormal{d}}}\subseteq\hat{B}^{\textnormal{d}}$, let $A$ be the smallest extension ideal of $B^{\textnormal{d}}$ in $Y$. Applying 2 to $B^{\textnormal{d}}$ we obtain that the ideal $A$ equals the smallest extension band of $B^{\textnormal{d}}$, i.e. $\widehat{B^{\textnormal{d}}}=A$. Due to Lemma 20 it follows $\widehat{B^{\textnormal{d}}}=A\subseteq\hat{B}^{\textnormal{d}}$.

To prove the inclusion $\hat{B}^{\textnormal{d}}\subseteq\widehat{B^{\textnormal{d}}}$, we show that $\widehat{B^{\textnormal{d}}}\backslash\hat{B}^{\textnormal{d}}=\varnothing$. Let, on the contrary, $y\in\widehat{B^{\textnormal{d}}}\backslash\hat{B}^{\textnormal{d}}$ with $y\neq 0$. We can assume that $y>0$, otherwise consider one of the two elements $y^{+},y^{-}$. As $y\notin\hat{B}^{\textnormal{d}}$, we have $y\not\perp\hat{B}$. That is, there exists a positive element $z\in\hat{B}$ such that $z\wedge y>0$. Since $X$ is pervasive, there exists $x\in X_{+}$ with $0<i(x)\leqslant z\wedge y$. The relation $0<i(x)\leqslant y\in\widehat{B^{\textnormal{d}}}$ implies $i(x)\in\widehat{B^{\textnormal{d}}}$ and thus $x\in[\widehat{B^{\textnormal{d}}}]i=B^{\textnormal{d}}$. On the other hand, the relation $0<i(x)\leqslant z\in\hat{B}$ implies $i(x)\in\hat{B}$ and thus $x\in[\hat{B}]i=B$, a contradiction. We conclude $\hat{B}^{\textnormal{d}}\subseteq\widehat{B^{\textnormal{d}}}$.

5: First we show that every extension ideal of $B$ is contained in $\hat{B}$. Let $I\subseteq Y$ be an extension ideal of $B$. Let $y\in I_{+}$. From 1 it follows $y=y_{1}+y_{2}$ for $y_{1}\in\hat{B}$ and $y_{2}\in\widehat{B^{\textnormal{d}}}$. Due to $X$ being pervasive, by Proposition 7 we have $y_{2}=\sup\left\{x\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0\leqslant x\leqslant y_{2}\right\}=\sup i(\left[[0,y_{2}]\right]i)$. From $y_{2}\in\widehat{B^{\textnormal{d}}}$ it follows $[0,y_{2}]\subseteq\widehat{B^{\textnormal{d}}}$ and thus we obtain the first inclusion $\left[[0,y_{2}]\right]i\subseteq B^{\textnormal{d}}$. Moreover, due to $y_{1}\perp y_{2}$, Lemma 3 2 yields $y_{1},y_{2}\geqslant 0$. From $0\leqslant y_{2}\leqslant y\in I$ we obtain $[0,y_{2}]\in I$. Since $I$ is an extension ideal of $B$, we obtain the second inclusion $\left[[0,y_{2}]\right]i\subseteq[I]i=B$. The two inclusions together yield $\left[[0,y_{2}]\right]i\subseteq B^{\textnormal{d}}\cap B=\left\{0\right\}$. This leads to $y_{2}=\sup i(\left[[0,y_{2}]\right]i)=0$. We conclude that $y=y_{1}\in\hat{B}$, i.e. $I\subseteq\hat{B}$.

Since for every extension ideal $I$ of $B$ we have $I\subseteq\hat{B}$ and due to $\hat{B}$ being the smallest extension ideal of $B$, it follows $I=\hat{B}$. That is, $\hat{B}$ coincides with every extension ideal of $B$.

We show that $\hat{B}$ equals every extension band of $B$. Let $C$ be an extension band of $B$. In particular, $C$ is an extension ideal of $B$, which implies $C\subseteq\hat{B}$. On the other hand, by 2 the ideal $\hat{B}$ is the smallest extension band of $B$, so $\hat{B}\subseteq C$. That is, $\hat{B}=C$.

6: As was established in 2, due to $B$ being a directed ideal, the subspace $i(B)$ is majorizing in the band $\hat{B}$. Thus by Theorem 11 the band $\hat{B}$ is a vector lattice cover of $B$ and $B$ is pervasive. As the roles of $B$ and $B^{\textnormal{d}}$ are interchangeable, a similar statement follows for the band $B^{\textnormal{d}}$ and its extension band $\hat{B}^{\textnormal{d}}$. $\square$

Note that the assumptions of Theorem 21 do not imply that $X$ is a vector lattice, see Example 41 and Theorem 43 below.

From Theorem 211 and 2 we obtain the following.

Corollary 22**.**

*Archimedean pervasive pre-Riesz spaces have the extension property (E) for projection bands. *

In the next example we see that the converse of Theorem 211 is not true, in general.

Example 23**.**

A restriction of a projection band in a vector lattice cover need not be a projection band in the corresponding pre-Riesz space, even if the pre-Riesz space is a vector lattice.

We call a function $x\colon[0,1]\rightarrow\mathbb{R}$ piecewise right continuous, if there exists an $n\in\mathbb{N}$ and $t_{0},\ldots,t_{n}\in[0,1]$ with $0=t_{0}<t_{1}<\ldots<t_{n}=1$ such that $x$ is continuous on $]t_{k},t_{k+1}[$ and continuous from the right in $t_{k}$ for every $k\in\left\{0,n-1\right\}$. Let $Y$ be the vector lattice of bounded piecewise right continuous functions on the interval $[0,1]$. The vector lattice $X:=C([0,1])$ of continuous functions is order dense in $Y$. That is, we can view $(Y,i)$ as a vector lattice cover of $X$, where $i$ is the identity embedding map. For the functions $y_{1}:=\mathds{1}_{\left[0,\tfrac{1}{2}\right[}$ and $y_{2}:=\mathds{1}_{\left[\tfrac{1}{2},1\right]}$ we have $y_{1},y_{2}\in Y$. Consider the band $\hat{B}:=\mathcal{B}_{y_{1}}=\left\{y\in Y\hskip 2.84544pt\middle|\hskip 2.84544pty(\left[\tfrac{1}{2},1\right])=0\right\}$ generated by $y_{1}$ in $Y$. Clearly, $Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$ and $y_{2}\in\hat{B}^{\textnormal{d}}$. Since $X$ is a vector lattice, $X$ is pervasive. Thus we have the restriction property for bands. It follows that $B:=\hat{B}\cap X$ and $B^{\textnormal{d}}=\hat{B}^{\textnormal{d}}\cap X$ are bands in $X$. However, we have $X\neq B\oplus B^{\textnormal{d}}$. Indeed, for the element $x:=y_{1}+y_{2}=\mathds{1}_{[0,1]}\in X$ there do not exist two continuous functions $x_{1}\in B$ and $x_{2}\in B^{\textnormal{d}}$ such that $x=x_{1}+x_{2}$.

Notice that $B$ is not majorizing in $\hat{B}$. Clearly, the vector lattice $X$ has the RDP.

The following example shows that in the setting of Theorem 211 the converse is not true, even if $B$ is majorizing in $\hat{B}$.

Example 24**.**

A restriction $B$ of a projection band $\hat{B}$ in a vector lattice cover need not be a projection band in the corresponding pre-Riesz space, even if $i(B)$ is majorizing in $\hat{B}$.

Let $PA[-1,1]$ be the vector lattice of continuous piecewise affine functions on the interval $[-1,1]$ and define $q(t):=t^{2}$ for every $t\in[-1,1]$. Let

[TABLE]

be endowed with pointwise order. Then $X$ is directed and Archimedean and therefore a pre-Riesz space. Consider the sublattice $Y:=\left\{y\in C[-1,1]\hskip 2.84544pt\middle|\hskip 2.84544pt\exists\hskip 2.84544ptx\in X\colon|y|\leqslant x\right\}$ of $C[-1,1]$. It is immediate that $X$ is pervasive in $Y$, since for every positive $y\in Y$ there is a non-zero function $x\in PA[-1,1]$ with $x(0)=0$ such that $0<x\leqslant y$. Moreover, $X$ is majorizing in $Y$. It follows that $Y$ is a vector lattice cover of $X$. We show that $X$ does not have the RDP. Indeed, define two functions $x_{1},x_{2}$ on $[-1,1]$ by $x_{1}(t):=-t$ and $x_{2}(t):=0$ for $t\in[-1,0]$ and $x_{1}(t):=0$ and $x_{2}(t):=t$ for $t\in[0,1]$. Then $x_{1},x_{2}\in X_{+}$ and $q\leqslant x_{1}+x_{2}$. However, there exist no elements $z_{1},z_{2}\in X_{+}$ with $q=z_{1}+z_{2}$ such that $0\leqslant z_{1}\leqslant x_{1}$ and $0\leqslant z_{2}\leqslant x_{2}$.

Define two functions $q_{1},q_{2}\in Y$ by $q_{1}(t):=t^{2}$ and $q_{2}(t):=0$ for $t\in[-1,0]$ and by $q_{1}(t):=0$ and $q_{2}(t):=t^{2}$ for $t\in[0,1]$. Then in $Y$ we have

[TABLE]

It is immediate that $Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$. Since $X$ is pervasive, it has the band restriction property. It follows that $B:=[\hat{B}]i$ is a band in $X$. Notice that $[\hat{B}^{\textnormal{d}}]i=B^{\textnormal{d}}$.

We have $X\neq B\oplus B^{\textnormal{d}}$. Indeed, for $q\in X$ there do not exist $z_{1}\in B$ and $z_{2}\in B^{\textnormal{d}}$ such that $q=z_{1}+z_{2}$.

Notice that in this example, $B$ is majorizing in $\hat{B}$.

In Examples 23 and 24 we saw that the conditions in the following theorem can not be omitted.

Theorem 25**.**

*Let $X$ be an Archimedean fordable pre-Riesz space with RDP and $(Y,i)$ a vector lattice cover of $X$. Let $\hat{B}$ be a band in $Y$ such that $Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$ and let $B:=[\hat{B}]i$. Moreover, let $i(B)$ be majorizing in $\hat{B}$ and $i(B^{\textnormal{d}})$ be majorizing in $\hat{B}^{\textnormal{d}}$. Then $X=B\oplus B^{\textnormal{d}}$. *

Proof*.*

Since $X$ is fordable, $X$ has the restriction property for bands. Thus $[\hat{B}]i=B$ is a band in $X$. Let $x\in X_{+}$. Due to $i(x)\in Y=\hat{B}\oplus\hat{B}^{\textnormal{d}}$ there exist $y_{1}\in\hat{B}$ and $y_{2}\in\hat{B}^{\textnormal{d}}$ such that $i(x)=y_{1}+y_{2}$. By Lemma 3 2 we have $y_{1},y_{2}\geqslant 0$. Since $i(B)$ is majorizing in $\hat{B}$ and $i(B^{\textnormal{d}})$ is majorizing in $\hat{B}^{\textnormal{d}}$, there exist $b_{1}\in B$ and $b_{2}\in B^{\textnormal{d}}$ such that $i(x)\leqslant i(b_{1})+i(b_{2})$. That is, we have $x\leqslant b_{1}+b_{2}$ and $b_{1},b_{2}\geqslant 0$. Since $X$ has the RDP, there exist $x_{1},x_{2}\in X$ with $x=x_{1}+x_{2}$ such that $0\leqslant x_{1}\leqslant b_{1}$ and $0\leqslant x_{2}\leqslant b_{2}$. Since $B$ and $B^{\textnormal{d}}$ are ideals in $X$, it follows $x_{1}\in B$ and $x_{2}\in B^{\textnormal{d}}$.

We show the uniqueness of this decomposition. Let $x\in X_{+}$ be such that $x=x_{1}+x_{2}=\tilde{x}_{1}+\tilde{x}_{2}$, where $x_{1},\tilde{x}_{1},x_{2},\tilde{x}_{2}\in X_{+}$ with $x_{1},\tilde{x}_{1}\in B$ and $x_{2},\tilde{x}_{2}\in B^{\textnormal{d}}$. Then $x_{1}-\tilde{x}_{1}\in B$ and $x_{1}-\tilde{x}_{1}=\tilde{x}_{2}-x_{2}\in B^{\textnormal{d}}$. Due to $B\cap B^{\textnormal{d}}=\left\{0\right\}$ it follows $x_{1}-\tilde{x}_{1}=0=\tilde{x}_{2}-x_{2}$, i.e. $x_{1}=\tilde{x}_{1}$ and $x_{2}=\tilde{x}_{2}$.

Now let $x\in X$. As $X$ is directed, there exist $x^{(1)},x^{(2)}\in X_{+}$ such that $x=x^{(1)}+x^{(2)}$. Since by the first part of the proof both $x^{(1)}$ and $x^{(2)}$ have disjoint decompositions $x^{(1)}=x^{(1)}_{1}+x^{(1)}_{2}$ and $x^{(2)}=x^{(2)}_{1}+x^{(2)}_{2}$, it follows $x=(x^{(1)}_{1}+x^{(2)}_{1})-(x^{(1)}_{2}+x^{(2)}_{2})$, where $x^{(1)}_{1}+x^{(2)}_{1}\in B$ and $x^{(1)}_{2}+x^{(2)}_{2}\in B^{\textnormal{d}}$. $\square$

5 Atoms and discrete elements in pre-Riesz spaces

In the definition of an atom in a vector lattice $X$ the inequality $|x|\leqslant|a|$, where $a,x\in X$, is equivalent to $\left\{a,-a\right\}^{u}\subseteq\left\{x,-x\right\}^{u}$. For pervasive pre-Riesz spaces we make the following observation.

Proposition 26**.**

*Let $X$ be a pervasive pre-Riesz space and $a\in X\backslash\left\{0\right\}$. Assume that for every $x\in X$ with $\left\{a,-a\right\}^{u}\subseteq\left\{x,-x\right\}^{u}$ it follows $x=\lambda a$ for some $\lambda\in\mathbb{R}$. Then we have $a>0$ or $a<0$. *

Proof*.*

Let $(Y,i)$ be a vector lattice cover of $X$. Since $a\neq 0$, we have $|i(a)|>0$. As $X$ is pervasive, there exists an element $x\in X$ such that $0<i(x)\leqslant|i(a)|$. Due to $x>0$ it follows $\left\{a,-a\right\}^{u}\subseteq\left\{x\right\}^{u}=\left\{x,-x\right\}^{u}$. By assumption there is $\lambda\in\mathbb{R}$ with $x=\lambda a$ in $X$. Due to $x>0$ the case $\lambda>0$ leads to $a>0$ and the case $\lambda<0$ leads to $a<0$. $\square$

In view of Proposition 26 we define atoms in the following way. The definition444We stress that in [2, Definition 1.42] the authors introduce this concept using a different term. They call an element $a\in X_{+}$ for which $0\leqslant x\leqslant a$ implies that $x=\lambda a$ for some real $\lambda\geqslant 0$ an extremal vector or a discrete vector of $X_{+}$. However, a similar concept is well-known in the less general setting of vector lattices. Indeed, [20, § 3] gives a definition of an atom in a vector lattice which differs from our notion by the fact that the considered element need not be positive. However, by [24, III.13.1 a)] for every atom $a\in Y$ it follows $a>0$ or $a<0$. Moreover, if $a<0$ is an atom, so is $-a$. These circumstances maybe clarify that, based on [2, Definition 1.42], we define atoms as positive elements. Proposition 26 justifies our choice in Archimedean pervasive pre-Riesz spaces. is based on [2, Definition 1.42].

Definition 27**.**

*Let $X$ be an ordered vector space with the cone $X_{+}$. An element $a\in X_{+}\backslash\left\{0\right\}$ is said to be an atom if $0\leqslant x\leqslant a$ implies that $x=\lambda a$ for some real $\lambda\geqslant 0$. For the set of all atoms in $X$ we write $\mathcal{A}_{X}:=\left\{a\in X_{+}\hskip 2.84544pt\middle|\hskip 2.84544pta\textit{ is an atom}\right\}$. *

The following characterization of atoms is from [2, Lemma 1.43].

Proposition 28**.**

Let $X$ be an ordered vector space. For a non-zero vector $a\in X_{+}$ the following statements are equivalent.

The vector $a$ is an atom in $X$. 2. 2.

If $x_{1},x_{2}\in X_{+}$ satisfy $a=x_{1}+x_{2}$, then the vectors $x_{1}$ and $x_{2}$ are linearly dependent. 3. 3.

The half-ray $\left\{\lambda a\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\mathbb{R}_{\geqslant 0}\right\}$ is a face of $X_{+}$.

We already defined discrete elements in vector lattices. Based on this definition and by courtesy of the disjointness notion in ordered vector spaces given in (1) we can generalize the term discrete element to ordered vector spaces.

Definition 29**.**

*Let $X$ be an ordered vector space. An element $a\in X$, $a>0$, is called a discrete element, if for every $x,z\in X_{+}$ with $x,z\leqslant a$ and $x\perp z$ it follows $x=0$ or $z=0$. *

In vector lattices, discrete elements and atoms are related notions. In a (not necessarily Archimedean) vector lattice $Y$ every atom is a discrete element by [16, Lemma 26.2 (i)], but discrete elements need not be atoms, as one can see in [16, Lemma 26.2 (ii)]. However, if $Y$ is an Archimedean vector lattice, then by [24, III.13.1 b)] and by [16, Lemma 26.2 (ii)] the notions of an atom and of a discrete element are equivalent. Due to this fact, in the vector lattice theory these two concepts are used interchangeably555For instance, in [16, Definition 26.1] the terms atom and discrete element are used in a reversed way with respect to our terminology.. We establish in Theorem 32 below that in Archimedean pervasive pre-Riesz spaces the notions of an atom and of a discrete element coincide as well. However, we see in Proposition 30 and in Example 31 below that in an Archimedean (not necessarily pervasive) pre-Riesz space every atom is discrete, but discrete elements need not be atoms.

Proposition 30**.**

*Let $X$ be a pre-Riesz space and $a\in\mathcal{A}_{X}$. Then $a$ is discrete. *

Proof*.*

Let $a\in\mathcal{A}_{X}$ and let elements $x,z\in X_{+}$ satisfy the inequalities $x,z\leqslant a$ and $x\perp z$. Since $a$ is an atom, there are positive real numbers $\lambda$ and $\mu$ such that $x=\lambda a$ and $z=\mu a$. Let $(Y,i)$ be a vector lattice cover of $X$. By Proposition 8 the relation $x\perp z$ implies that $i(x)\perp i(z)$, i.e. we have $\lambda i(a)\perp\mu i(a)$ with $i(a)\in Y_{+}\backslash\left\{0\right\}$. This yields $\lambda=0$ or $\mu=0$, i.e. $x=0$ or $z=0$. Thus $a$ is a discrete element in $X$. $\square$

There are pre-Riesz spaces that contain no non-trivial disjoint elements. Consider, e.g. $\mathbb{R}^{3}$ endowed with the ice-cream cone, see [7, Proposition 16] and [14, Examples 4.5 and 4.6]. Clearly, in such a pre-Riesz space every element $x>0$ is a discrete element, whereas the atoms are precisely the non-zero elements on the boundary of the cone. We give an example of a finite-dimensional space which contains non-trivial disjoint elements in the cone and where a discrete element need not be an atom.

Example 31**.**

In an Archimedean pre-Riesz space with an order unit a discrete element need not be an atom.

Let $X:=\mathbb{R}^{3}$ be the three-dimensional Euclidean space. Consider in $X$ the four vectors

[TABLE]

and let the order on $X$ be induced by the four-ray-cone $K_{4}:=\textnormal{pos}\left\{v_{1},v_{2},v_{3},v_{4}\right\}$, i.e. by the positive-linear hull of the vectors $v_{1},\ldots,v_{4}$. Clearly, $(X,K_{4})$ has an order unit. In [14, Example 4.8] it is shown that $(X,K_{4})$ can be order densely embedded into $\mathbb{R}^{4}$ with the standard cone $\mathbb{R}^{4}_{+}$. Indeed, consider the four functionals $f_{i}\in X^{\prime}$ on $X$, where $X^{\prime}$ is identified with $\mathbb{R}^{3}$:

[TABLE]

Then the map $i\colon X\rightarrow\mathbb{R}^{4}$, given by $i:x\mapsto(f_{1}(x),\ldots,f_{4}(x))^{T}$, is a bipositive embedding of the pre-Riesz space $(X,K_{4})$ into $(\mathbb{R}^{4},\mathbb{R}^{4}_{+})$. Notice that by [7, Proposition 13] the space $(\mathbb{R}^{4},\mathbb{R}^{4}_{+})$ is the Riesz completion of $(X,K_{4})$. The pre-Riesz space $(X,K_{4})$ is not pervasive, see [15, Example 4.4.18].

By Proposition 28 an element $a\in X_{+}$ is an atom if and only if the half-ray $\left\{\lambda a\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\mathbb{R}_{\geqslant 0}\right\}$ is a face of the cone $K_{4}$. Hence, $\mathcal{A}_{X}=\left\{\lambda v_{k}\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\mathbb{R}_{>0},k\in\left\{1,\ldots,4\right\}\right\}$. Let $d:=v_{1}+v_{2}$. Clearly, $d\notin\mathcal{A}_{X}$. However, $d$ is a discrete element. Indeed, by [12, Example 4.6], we have $\left\{v_{1}\right\}^{\textnormal{d}}=\mathop{\textnormal{span}}\left\{v_{3}\right\}$, $\left\{v_{2}\right\}^{\textnormal{d}}=\mathop{\textnormal{span}}\left\{v_{4}\right\}$, $\left\{v_{3}\right\}^{\textnormal{d}}=\mathop{\textnormal{span}}\left\{v_{1}\right\}$, $\left\{v_{4}\right\}^{\textnormal{d}}=\mathop{\textnormal{span}}\left\{v_{2}\right\}$ and $\left\{x\right\}^{\textnormal{d}}=\left\{0\right\}$ for every $x\in K_{4}\backslash\left(\mathop{\textnormal{span}}\left\{v_{1}\right\}\cup\mathop{\textnormal{span}}\left\{v_{2}\right\}\cup\mathop{\textnormal{span}}\left\{v_{3}\right\}\cup\mathop{\textnormal{span}}\left\{v_{4}\right\}\right)$. Since $[0,d]$ is the convex hull of the set $\left\{0,v_{1},v_{2},d\right\}$, it follows that there are no disjoint elements $x,z\in X_{+}\backslash\left\{0\right\}$ such that $x,z\leqslant d$. Therefore $d$ is discrete.

The next result yields for a pervasive pre-Riesz space $X$ relationships between atoms in $X$ and in its vector lattice cover.

Theorem 32**.**

Let $X$ be an Archimedean pervasive pre-Riesz space and $(Y,i)$ a vector lattice cover of $X$. Let $a\in X$.

We have $a\in\mathcal{A}_{X}$ if and only if $i(a)\in\mathcal{A}_{Y}$. 2. 2.

The element $a$ is discrete in $X$ if and only if $i(a)$ is discrete in $Y$. 3. 3.

We have $a\in\mathcal{A}_{X}$ if and only if $a$ is a discrete element. 4. 4.

If $a\in\mathcal{A}_{X}$, then we have $i(\mathcal{I}_{a})=\mathcal{I}_{i(a)}$. 5. 5.

If $\tilde{a}\in\mathcal{A}_{Y}$, then $\tilde{a}\in i(X)$ and $i^{-1}(\tilde{a})\in\mathcal{A}_{X}$.

Proof*.*

1: Let $a\in\mathcal{A}_{X}$. Let $y\in Y$ be such that $0<y\leqslant i(a)$. Since $X$ is pervasive, by Proposition 7 we have for the set $S:=\left\{z\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0<z\leqslant y\right\}$ the relation $y=\sup S$. For every $z\in S$ it holds that $0<z\leqslant y\leqslant i(a)$, i.e. for $x:=i^{-1}(z)$ we have $0<x\leqslant a$. Since $a\in X$ is an atom, there exists a real $\lambda>0$ such that $x=\lambda a$. Therefore we have $S=\left\{\lambda i(a)\hskip 2.84544pt\middle|\hskip 2.84544pt0<\lambda i(a)\leqslant y\right\}$. Due to $Y$ being Archimedean, the set $\Lambda:=\left\{\lambda\in\mathbb{R}\hskip 2.84544pt\middle|\hskip 2.84544pt0<\lambda i(a)\leqslant y\right\}$ is bounded above in $\mathbb{R}$ and therefore its supremum $\sup\Lambda\geqslant 0$ exists in $\mathbb{R}$. For every $\lambda\in\Lambda$ we have $\lambda i(a)\leqslant y$, thus due to Proposition 4 taking the supremum leads to $(\sup\Lambda)i(a)\leqslant y=\sup S$. On the other hand, for every $\lambda i(a)\in S$ we have $\lambda i(a)\leqslant(\sup\Lambda)i(a)$ and therefore $\sup S\leqslant(\sup\Lambda)i(a)$. It follows $y=\sup S=(\sup\Lambda)i(a)$, i.e. $i(a)$ is an atom in $Y$.

Conversely, let $a\in X$ be such that $i(a)\in\mathcal{A}_{Y}$. Let $x\in X$ with $0<x\leqslant a$. As $i(a)$ is an atom, $0<i(x)\leqslant i(a)$ implies that there is $\lambda\in\mathbb{R}$ with $i(x)=\lambda i(a)$. It follows $x=\lambda a$, i.e. $a$ is an atom in $X$.

2: Let $a\in X$ be a discrete element in $X$. We show that $i(a)$ is a discrete element in $Y$. Let $y_{1},y_{2}\in Y$ be such that $0\leqslant y_{1},y_{2}\leqslant i(a)$ and $y_{1}\perp y_{2}$. As $X$ is pervasive, by Proposition 7 for the sets

[TABLE]

we have the relations $y_{n}=\sup S_{n}$. Due to $y_{1}\perp y_{2}$ we obtain for every $x_{1}\in X$ with $i(x_{1})\in S_{1}$ and every $x_{2}\in X$ with $i(x_{2})\in S_{2}$ the relations $x_{1}\perp x_{2}$ and $0\leqslant x_{1},x_{2}\leqslant a$. Assume $y_{1}\neq 0$, then there is $x\in X_{+}\backslash\left\{0\right\}$ with $i(x)\in S_{1}$. As $a$ is discrete and $i(x)\neq 0$, it follows for every $x_{2}\in X$ with $i(x_{2})\in S_{2}$ that $x_{2}=0$, i.e. $S_{2}=\left\{0\right\}$. Due to $y_{2}=\sup S_{2}$ we conclude $y_{2}=0$. Therefore $i(a)$ is discrete.

Conversely, let $i(a)$ be discrete. Let $x_{1},x_{2}\in X$ be such that $0\leqslant x_{1},x_{2}\leqslant a$ and $x_{1}\perp x_{2}$. For $y_{1}:=i(x_{1})$ and $y_{2}:=i(x_{2})$ we have $0\leqslant y_{1},y_{2}\leqslant i(a)$ and $y_{1}\perp y_{2}$ by Proposition 8. As $i(a)$ is discrete, it follows $y_{1}=0$ or $y_{2}=0$, i.e. $x_{1}=0$ or $x_{2}=0$. That is, $a$ is discrete.

3: If $a\in\mathcal{A}_{X}$, then by Proposition 30 the element $a$ is discrete. Conversely, let $a\in X$ be a discrete element. By 2 the element $i(a)$ is discrete in $Y$ as well. By Proposition 12 every discrete element in an Archimedean vector lattice is an atom, i.e $i(a)\in\mathcal{A}_{Y}$. By 1 it follows $a\in\mathcal{A}_{X}$.

4: Let $a\in\mathcal{A}_{X}$. The inclusion $i(\mathcal{I}_{a})\subseteq\mathcal{I}_{i(a)}$ is immediate. By 1 we have $i(a)\in\mathcal{A}_{Y}$. For every $y\in\mathcal{I}_{i(a)}$ it follows that there exists $\lambda\in\mathbb{R}$ such that $y=\lambda i(a)\in i(\mathcal{I}_{a})$. Thus we have $i(\mathcal{I}_{a})\supseteq\mathcal{I}_{i(a)}$. We conclude $i(\mathcal{I}_{a})=\mathcal{I}_{i(a)}$.

5: Let $\tilde{a}\in\mathcal{A}_{Y}$. Since $X$ is pervasive, there exists $x\in X$ such that $0<i(x)\leqslant\tilde{a}$. As $\tilde{a}$ is an atom in $Y$, there exists $\lambda\in\mathbb{R}$ such that $\tilde{a}=\lambda i(x)=i(\lambda x)\in\mathcal{A}_{Y}$. By 1 it follows $a:=i^{-1}(\tilde{a})=\lambda x\in\mathcal{A}_{X}$. $\square$

Notice that in part 1 of the above result pervasiveness of $X$ is not needed for the implication “$\Leftarrow$”. This leads to the following observation.

Proposition 33**.**

*Let $X$ be an Archimedean pre-Riesz space and $Y$ a vector lattice cover of $X$. Then we have $\mathcal{A}_{Y}\cap i(X)\subseteq i(\mathcal{A}_{X})$. *

Remark**.**

We can reformulate 5 of Theorem 32 as $\mathcal{A}_{Y}\subseteq i(\mathcal{A}_{X})$. 2. 2.

If $X$ is not Archimedean, then the equivalence in Theorem 32 3 is not true, in general. As already mentioned, without the condition of Archimedeanity even in a vector lattice a discrete element need not be an atom.

6 Finite-dimensional pervasive pre-Riesz spaces

With the help of atoms, we now investigate finite-dimensional Archimedean pervasive pre-Riesz spaces. In Theorem 38 below we establish that all such spaces are, in fact, vector lattices. We start with two technical statements.

Proposition 34**.**

Let $X$ be an Archimedean pre-Riesz space and $a_{1},a_{2}\in\mathcal{A}_{X}$.

If $a_{1}\perp a_{2}$, then $a_{1}$ and $a_{2}$ are linearly independent. 2. 2.

Let $X$ be additionally pervasive. If $a_{1}$ and $a_{2}$ are linearly independent, then $a_{1}\perp a_{2}$.

Proof*.*

1: Let $a_{1},a_{2}\in\mathcal{A}_{X}$ with $a_{1}\perp a_{2}$. Then in a vector lattice cover $(Y,i)$ of $X$ we have by Proposition 8 that $i(a_{1})\perp i(a_{2})$. That is, $i(a_{1})\wedge i(a_{2})=0$. By Proposition 6 2 the infimum $a_{1}\wedge a_{2}$ exists in $X$ and equals [math]. this implies that $a_{1}$ and $a_{2}$ are linearly independent. Indeed, let on the contrary, $a_{1}=\lambda a_{2}$ for some $\lambda\in\mathbb{R}_{>0}$. Then $0=a_{1}\wedge a_{2}=a_{1}\wedge(\lambda a_{1})=\min\left\{1,\lambda\right\}a_{1}>0$, a contradiction.

2: Let $a_{1},a_{2}\in\mathcal{A}_{X}$. We show that if $a_{1}\not\perp a_{2}$, then $a_{1},a_{2}$ are linearly dependent.

Let $a_{1}\not\perp a_{2}$, i.e. $i(a_{1})\wedge i(a_{2})>0$. Since $X$ is pervasive, there exists an element $a\in X$ with $0<i(a)\leqslant i(a_{1})\wedge i(a_{2})$, i.e. $0<a\leqslant a_{1}$ and $0<a\leqslant a_{2}$. Since $a_{1},a_{2}$ are atoms, there exist $\lambda_{1},\lambda_{2}\in\mathbb{R}_{>0}$ with $\lambda_{1}a_{1}=a=\lambda_{2}a_{2}$, i.e. $a_{1}$ and $a_{2}$ are linearly dependent. $\square$

In Proposition 34 the condition of $X$ being pervasive can not be omitted. Indeed, in the Archimedean non-pervasive pre-Riesz space in Example 31 the linearly independent vectors $v_{1}$ and $v_{2}$ are atoms, but they are not disjoint.

The idea for the next result originated during a discussion with H. van Imhoff.

Proposition 35**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space and the atoms $a_{1},\ldots,a_{n}\in\mathcal{A}_{X}$ pairwise linearly independent. Then the set $\left\{a_{1},\ldots,a_{n}\right\}$ of atoms is linearly independent. *

Proof*.*

Assume that the set $\left\{a_{1},\ldots,a_{n}\right\}\subseteq\mathcal{A}_{X}$ of pairwise linearly independent atoms is linearly dependent, that is, there exist $\lambda_{k}\in\mathbb{R}$ for $k\in\left\{2,\ldots,n\right\}$ such that

[TABLE]

Due to Proposition 34 the atoms $a_{1},\ldots,a_{n}$ are pairwise disjoint. Thus in a vector lattice cover $(Y,i)$ of $X$ for $k,j\in\left\{1,\ldots,n\right\}$ with $k\neq j$ we have $i(a_{k})\perp\lambda_{j}i(a_{j})$. Therefore we can apply Lemma 31 and obtain the following contradiction:

[TABLE]

$\square$

Observe that the conclusion of Proposition 35 is not true if the pre-Riesz space is not pervasive, see [2, p. 38].

To characterize finite-dimensional pervasive pre-Riesz spaces in Theorem 38 below, we first recall some basics. Let $X$ be an ordered vector space. A convex set $B\subseteq X_{+}$ is called a base of the cone $X_{+}$ if every $x\in X_{+}\backslash\left\{0\right\}$ has a unique representation $x=\lambda b$, where $\lambda\in\mathbb{R}_{>0}$ and $b\in B$. The following result can be found in [2, Theorem 1.48].

Proposition 36**.**

*Let $X$ be an ordered vector space such that the cone $X_{+}$ has a base $B$. Let $b\in B$. Then $b\in\mathcal{A}_{X}$ if and only if $b$ is an extreme point of $B$. *

From [25, Theorems IV.1.1, VII.1.1] we obtain the following.

Proposition 37**.**

*Let $X$ be a finite-dimensional ordered Banach space. If the cone $X_{+}$ is (norm) closed, then $X_{+}$ has a norm bounded base. *

Theorem 38**.**

*Let $X$ be an $n$-dimensional Archimedean pre-Riesz space. Then $X$ is pervasive if and only if $X$ is a vector lattice. *

Proof*.*

Clearly, if $X$ is a vector lattice, then it is pervasive in its Riesz completion $X^{\varrho}=X$. Conversely, let $X$ be an $n$-dimensional Archimedean pervasive pre-Riesz space. Endowed with the Euclidean norm, $X$ is an ordered Banach space. Since $X$ is Archimedean, the cone $X_{+}$ is closed. By Proposition 37 the cone $X_{+}$ has a norm bounded base $B$. Let $\mathcal{A}:=\mathcal{A}_{X}\cap B$. We first establish $X_{+}=\textnormal{pos}\mathcal{A}$. The base $B$ is a convex set and, as $X$ is finite-dimensional, also compact. By Proposition 36 a vector $a\in B$ is an extreme point of $B$ if and only if $a$ is an atom. By Minkowski’s theorem666Minkowski’s theorem states that in a finite-dimensional normed vector space every element of a compact convex set $S\subseteq X$ is a convex combination of extreme points of $S$, see e.g. [19, p. 1]. every $b\in B$ is a convex combination of atoms in $\mathcal{A}$. From $X_{+}=\bigcup_{\lambda\in\mathbb{R}_{\geqslant 0}}\lambda B$, it follows $X_{+}=\textnormal{pos}\mathcal{A}$. Observe that $\mathcal{A}$ contains pairwise linearly independent atoms. Indeed, if $a,b\in\mathcal{A}$ with $a\neq b$ such that $a=\lambda b$, then $a$ has two representations with respect to the basis, which is a contradiction.

Next we consider three cases for the cardinality of $\mathcal{A}$. If $|\mathcal{A}|<n$, then $X_{+}$ is polyhedral. As $X_{+}=\textnormal{pos}\mathcal{A}$, the cone is not generating and the space is not even pre-Riesz, a contradiction. Let $|\mathcal{A}|>n$ and pick $m>n$ atoms $a_{1},\ldots,a_{m}\in\mathcal{A}$. Then due to $a_{1},\ldots,a_{m}$ being pairwise linearly independent, by Proposition 35 the set of atoms $\left\{a_{1},\ldots,a_{m}\right\}$ is linearly independent. But the number $m$ of linearly independent elements is greater than the dimension $n$ of the space, a contradiction. Finally, let $|\mathcal{A}|=n$, i.e. $X_{+}$ has $n$ extreme rays. Then $(X,X_{+})$ is order isomorphic to $(\mathbb{R}^{n},\mathbb{R}^{n}_{+})$. By Proposition 5 it follows that $X$ is a vector lattice. $\square$

The combination of Proposition 5 and Theorem 38 leads to the following.

Corollary 39**.**

Let $(X,K)$ be an $n$-dimensional Archimedean pre-Riesz space. Each of the properties 1 to 3 of Proposition 5 is equivalent to

1. (iv)

$(X,K)$* is pervasive.*

7 Atoms and principal bands in pre-Riesz spaces

The following properties of atoms were partially shown for vector lattices in [24, Proposition III.13], see also Proposition 13. We generalize them for pervasive pre-Riesz spaces in the following technical result.

Lemma 40**.**

Let $X$ be an Archimedean pervasive pre-Riesz space. Let $a\in\mathcal{A}_{X}$ and $x\in X_{+}$. Then we have the following.

There exists $\lambda\in\mathbb{R}_{\geqslant 0}$ such that $x-\lambda a\geqslant 0$ and $(x-\lambda a)\perp a$. 2. 2.

The real number $\lambda\in\mathbb{R}$ in 1 is unique. Moreover,

1. (a)

$\lambda a\leqslant x$* and for every $\varepsilon>0$ we have $(\lambda+\varepsilon)a\not\leqslant x$, i.e. $\lambda$ is the greatest number $\mu\in\mathbb{R}$ which satisfies the inequality $\mu a\leqslant x$,* 2. (b)

$\lambda=0$* holds if and only if $x\perp a$.*

Proof*.*

Let $(Y,i)$ be a vector lattice cover of $X$. We consider four different cases and first show 1 and 22(a) in each of them.

Case 1: Let first $0<a\leqslant x$, i.e. $0<i(a)\leqslant i(x)$.

1: Since $a$ is an atom, by Theorem 32 1 it follows that $i(a)$ is an atom in $Y$. By Proposition 13 there exists $\lambda\in\mathbb{R}_{>0}$ such that $i(x)-\lambda\hskip 2.84544pti(a)\geqslant 0$ and $(i(x)-\lambda\hskip 2.84544pti(a))\perp i(a)$. It follows $x-\lambda a\geqslant 0$, and by Proposition 8 we have $(x-\lambda a)\perp a$ in $X$, which establishes 1.

2: We prove the statement 2(a). Assume that there are two numbers $\lambda,\mu\in\mathbb{R}_{\geqslant 0}$ with $\mu\neq\lambda$ such that $x-\lambda a\geqslant 0$, $(x-\lambda a)\perp a$ and $x-\mu a\geqslant 0$, $(x-\mu a)\perp a$. Let without loss of generality $\mu>\lambda$. Since $(x-\lambda a)\perp a$, it follows

[TABLE]

or, equivalently, $0=[(i(x)-\mu i(a))+(\mu-\lambda)i(a)]\wedge i(a)$. Due to $(x-\mu a)\perp(\mu-\lambda)a$ we have $(i(x)-\mu i(a))\perp(\mu-\lambda)i(a)$, from which (using Lemma 31 in the second step) it follows

[TABLE]

a contradiction. Thus the number $\lambda\in\mathbb{R}$ with the properties as in 1 is uniquely defined.

Moreover, for every $\varepsilon>0$ we have $(\lambda+\varepsilon)a\not\leqslant x$. Indeed, let $\mu:=\lambda+\varepsilon$ and assume $\mu a\leqslant x$. Then due to $0\leqslant x-\mu a<x-\lambda a$ and $(x-\lambda a)\perp a$ it follows $(x-\mu a)\perp a$, i.e. $\mu\in\mathbb{R}_{>0}$ with $\mu\neq\lambda$ is another real number which satisfies the conditions in 1, a contradiction to the uniqueness of $\lambda$.

Case 2: Let $0\leqslant x<a$.

1: Due to $a$ being an atom there is $\lambda\in\mathbb{R}_{\geqslant 0}$ with $x=\lambda a$. Thus $x-\lambda a=0$ and therefore $(x-\lambda a)\wedge a=0$.

2: Let $\varepsilon>0$. Then we have $(\lambda+\varepsilon)a>x$, that is $(\lambda+\varepsilon)a\not\leqslant x$, i.e. there does not exist a real number greater than $\lambda$ satisfying 1. Moreover, if $0<\mu<\lambda$ for some $\mu\in\mathbb{R}$, then we have $0<\mu a\leqslant x$. But since $x-\mu a=(\lambda-\mu)a>0$, it follows $(x-\mu a)\not\perp a$, i.e. $\lambda$ is the only real number which satisfies the properties in 1.

Case 3: Let $a$ and $x$ be not comparable and let there exist $\gamma>0$ such that $0<\gamma a\leqslant x$. Then we can apply Case 1 to show 1 and 2.

Case 4: Finally, consider the case where $a$ and $x$ are not comparable, but there does not exist $\gamma>0$ such that $0<\gamma a\leqslant x$.

1: Let $\lambda=0$. We show $(x-\lambda a)\perp a$, i.e. $x\perp a$, by contradiction. Assume that $x\not\perp a$. Then it follows $i(x)\wedge i(a)\neq 0$. Due to $X$ being pervasive there exists $\tilde{a}\in X$ such that $0<i(\tilde{a})\leqslant i(x)\wedge i(a)\leqslant i(a)$. Since $a$ is an atom, by Theorem 32 1 the element $i(a)$ is also an atom and therefore there exists $\gamma>0$ with $0<i(\tilde{a})=\gamma\hskip 2.84544pti(a)$, i.e. $0<\tilde{a}=\gamma a$. Moreover, we have $0<i(\tilde{a})\leqslant i(x)$, i.e. $0<\gamma a\leqslant x$, a contradiction.

2: We have $x-\lambda a=x\geqslant 0$. Moreover, for $\gamma>0$ by assumption we obtain $0<(\lambda+\gamma)a=\gamma a\not\leqslant x$. This leads to the uniqueness of $\lambda$.

2(b) Notice that $\lambda=0$ holds only in Case 4. Due to the uniqueness of the number $\lambda$ in each of the four cases, Case 4 is the only case in which we have $x\perp a$. That is, $\lambda=0$ holds if and only if $x\perp a$. $\square$

By Lemma 40 we can decompose every $x\in X$ as $x=\lambda a+(x-\lambda a)$, where the two summands are disjoint. In particular, one summand belongs to the band $\mathcal{B}_{a}$ generated by $a$. The next example depicts such a situation in a specific pre-Riesz space. Similar to the vector lattice case, we call an ordered vector space $X$ atomic, if for every $x\in X$ with $x>0$ there is an atom $a\in\mathcal{A}_{X}$ such that $0<a\leqslant x$.

Example 41**.**

An Archimedean atomic pervasive pre-Riesz space $X$ which is not a vector lattice.

For the space $C^{1}[0,1]$ of continuously differentiable functions on $[0,1]$ and characteristic functions $\mathds{1}_{\{x\}}$ of some singleton $\{x\}\subseteq[0,1]$ we consider the space

[TABLE]

with pointwise order. Clearly, $X$ is atomic with $\mathcal{A}_{X}=\left\{\lambda\mathds{1}_{\{x\}}\hskip 2.84544pt\middle|\hskip 2.84544ptx\in[0,1],\lambda\in\mathbb{R}_{>0}\right\}$. Moreover, $X$ is Archimedean and directed, therefore $X$ is pre-Riesz. However, $X$ is not a vector lattice, as the two differentiable linear functions $x_{1}(t):=t$ and $x_{2}(t)=1-t$ have no infimum in $X$. In analogy to $C[0,1]$ being a vector lattice cover of $C^{1}[0,1]$, a vector lattice cover of $X$ is given by

[TABLE]

It is easy to see that $X$ is pervasive, since for every $y\in Y_{+}$, $y\neq 0$, we can find an atom $a\in X$ with $0<i(a)\leqslant y$.

For every $a\in\mathcal{A}_{X}$ we have for the principal ideal $\mathcal{I}_{a}$ generated by $a$ and for the principal band $\mathcal{B}_{a}$ generated by $a$ the equality $\mathcal{I}_{a}=\mathcal{B}_{a}$ in $X$. Moreover, the band $\mathcal{B}_{a}$ is one-dimensional.

Theorem 42**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space and $a\in\mathcal{A}_{X}$. Then the principal ideal and the principal band generated by $a$ coincide, i.e. $\mathcal{I}_{a}=\mathcal{B}_{a}$. *

Proof*.*

Let $a\in\mathcal{A}_{X}$. It follows that the ideal $\mathcal{I}_{a}$ is one-dimensional, i.e. $\mathcal{I}_{a}=\left\{\lambda a\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\mathbb{R}\right\}$. Consider the band $\mathcal{B}_{a}=\left\{a\right\}^{\textnormal{dd}}$. Assume that there exists an element $y\in\mathcal{B}_{a}\backslash\mathcal{I}_{a}$. That is, $y\in\mathcal{B}_{a}$ and for every $\lambda\in\mathbb{R}$ we have $y\neq\lambda a$. We intend to apply Lemma 40, for which we need an appropriate positive element. Since $X$ is pervasive, by Proposition 7 it follows

[TABLE]

Let $S:=\left\{z\in i(X)\hskip 2.84544pt\middle|\hskip 2.84544pt0<z\leqslant|i(y)|\right\}$. First we show by contradiction that there exists a positive element $v\in S$ such that for $x:=i^{-1}(v)$ and for every $\lambda\in\mathbb{R}$ we have $x\neq\lambda a$. Assume, on the contrary, that for every $z\in S$ there is $\lambda_{z}\in\mathbb{R}_{\geqslant 0}$ such that $z=\lambda_{z}i(a)$. Since $Y$ is Archimedean, by Proposition 4 for $\lambda:=\sup\left\{\lambda_{z}\hskip 2.84544pt\middle|\hskip 2.84544ptz\in S\right\}\in\mathbb{R}$ we have $|i(y)|=\lambda i(a)$. This contradicts $y\notin\mathcal{I}_{a}$. Thus there exists $x\in X$ such that $i(x)\in S$, and for every $\lambda\in\mathbb{R}$ we have $x\neq\lambda a$. Moreover, due to $0<i(x)\leqslant|i(y)|$ we have $\left\{x,-x\right\}^{u}\supseteq\left\{y,-y\right\}^{u}$. Due to $\mathcal{B}_{a}$ being an ideal, it follows $x\in\mathcal{B}_{a}$. We conclude that there is a positive element $x\in X$, $x\neq 0$, with $x\in\mathcal{B}_{a}$, such that for every $\lambda\in\mathbb{R}$ we have $x\neq\lambda a$.

Second, by Lemma 40 there exists $\lambda\in\mathbb{R}_{\geqslant 0}$ such that $(x-\lambda a)\perp a$. Due to $x,\lambda a\in\mathcal{B}_{a}$ it follows $w:=x-\lambda a\in\mathcal{B}_{a}$. Thus $w\in\mathcal{B}_{a}=\left\{a\right\}^{\textnormal{dd}}$, and, due to $w\perp a$, we get $w\in\left\{a\right\}^{\textnormal{d}}$. We have $\left\{a\right\}^{\textnormal{dd}}\cap\left\{a\right\}^{\textnormal{d}}=\left\{0\right\}$, which implies $w=0$. This leads to $x=\lambda a$, a contradiction.

We established for every $y\in\mathcal{B}_{a}$ the relation $y=\lambda a$ for some $\lambda\in\mathbb{R}$. That is, $\mathcal{B}_{a}=\mathcal{I}_{a}$. $\square$

Theorem 43**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space and let $a\in\mathcal{A}_{X}$. Then we have $X=\mathcal{B}_{a}\oplus\mathcal{B}_{a}^{\textnormal{d}}$. *

Proof*.*

Throughout the proof we use Theorem 42. We show that every element $x\in X$ has a unique disjoint decomposition.

Case 1: Let $x\in X_{+}$.

Case 1.1: Let $0<a\leqslant x$. By Lemma 40 1 there exists a real $\lambda>0$ such that $x-\lambda a\geqslant 0$ and $(x-\lambda a)\perp a$. It follows $(x-\lambda a)\perp\lambda a$. That is, $x$ has a disjoint decomposition into positive elements

[TABLE]

By Lemma 40 2 this decomposition is unique.

Case 1.2: Let $x\in X_{+}$ be arbitrary. If $x\perp a$, then $x\in\left\{a\right\}^{\textnormal{d}}$ and thus $x$ has the unique disjoint decomposition $x=0+x$, where $0\in\left\{a\right\}^{\textnormal{dd}}$ and $x\in\left\{a\right\}^{\textnormal{d}}$. If $x\not\perp a$, then $i(x)\wedge i(a)\neq 0$. Thus we have $0<i(x)\wedge i(a)\leqslant i(a)$. Since $a$ is an atom in $X$, by Theorem 32 1 the element $i(a)$ is an atom in $Y$. Hence there exists a real $\lambda>0$ such that $i(x)\wedge i(a)=\lambda i(a)$. By Proposition 6, the infimum $x\wedge a$ exists in $X$, and we have $x\wedge a=\lambda a$ in $X$. The element $\tilde{a}:=\lambda a$ is an atom in $X$ and we obtain $0<x\wedge a=\tilde{a}\leqslant x$. By Case 1.1 there exists a unique disjoint decomposition into positive elements as in (4), i.e. there exists $\mu\in\mathbb{R}$ such that $x=\mu\tilde{a}+(x-\mu\tilde{a})$. Since for some real $\gamma>0$ we have $\mu\tilde{a}=\gamma a$, the element $x$ has a unique disjoint decomposition $x=\gamma a+(x-\gamma a)$.

Case 2: Let $x\in X$. Due to Case 1 we conclude that, since $x$ can be represented as a difference of two positive elements, it has a disjoint decomposition

[TABLE]

We have to show that this decomposition is unique.

First we show that the decomposition of positive elements is compatible with addition. Indeed, for $y,z\in X_{+}$ the positive element $y+z$ has a unique decomposition $y+z=(y+z)_{a}+(y+z)_{d}$ with $(y+z)_{a}\in\mathcal{B}_{a}$ and $(y+z)_{d}\in\mathcal{B}_{a}^{\textnormal{d}}$. Moreover, both $y$ and $z$ have a decomposition $y=y_{a}+y_{d}$ and $z=z_{a}+z_{d}$, where $y_{a},z_{a}\in\mathcal{B}_{a}$ and $y_{d},z_{d}\in\mathcal{B}_{a}^{\textnormal{d}}$. Therefore the element $y+z$ has a second decomposition $y+z=y_{a}+z_{a}+y_{d}+z_{d}$. Since the decompositions for positive elements are unique, it follows that the parts of $y+z$ which belong to $\mathcal{B}_{a}$ coincide, i.e. $(y+z)_{a}=y_{a}+z_{a}$, and analogously $(y+z)_{d}=y_{d}+z_{d}$. We conclude that the decomposition of positive elements is compatible with addition, i.e. for $y,z\in X_{+}$ we have

[TABLE]

Finally, we establish the uniqueness of the disjoint decomposition in (5). Let $x\in X$. Since $X$ is directed, $x$ can be written as a difference of two positive elements.

Case 2.1: Let $x=x^{(1)}-x^{(2)}=x^{(3)}-x^{(4)}$ with $x^{(1)},x^{(2)},x^{(3)},x^{(4)}\geqslant 0$ and assume that $x^{(1)}$ and $x^{(3)}$ are comparable, e.g. $x^{(1)}>x^{(3)}$. Then with $\tilde{x}:=x^{(1)}-x^{(3)}>0$ we have

[TABLE]

This and the definition of $\tilde{x}$ yield

[TABLE]

Moreover, for the positive elements $x^{(1)},x^{(2)},x^{(3)},x^{(4)}$ and $\tilde{x}$ by Case 1 we have unique disjoint decompositions, i.e. $x^{(1)}=x^{(1)}_{a}+x^{(1)}_{d}$ with $x^{(1)}_{a}\in\mathcal{B}_{a}$ and $x^{(1)}_{d}\in\mathcal{B}_{a}^{\textnormal{d}}$, etc. Since the decomposition is compatible with addition, as in (6), from (7) it follows

[TABLE]

i.e. $x_{a}:=x^{(1)}_{a}-x^{(2)}_{a}=x^{(3)}_{a}-x^{(4)}_{a}$ is independent of the choice of representatives. Analogously it follows $x_{d}:=x^{(1)}_{d}-x^{(2)}_{d}=x^{(3)}_{d}-x^{(4)}_{d}$. Thus the element $x=x_{a}+x_{d}$ has a unique disjoint decomposition with $x_{a}\in\mathcal{B}_{a}$ and $x_{d}\in\mathcal{B}_{a}^{\textnormal{d}}$.

Case 2.2: Let there be two decompositions of $x\in X$ as $x=x^{(3)}-x^{(4)}=x^{(5)}-x^{(6)}$, with $x^{(3)}$ and $x^{(5)}$ not comparable. Since $X$ is directed, there exists an element $x^{(1)}\in X$ such that $x^{(1)}\geqslant x^{(3)},x^{(5)},x$. The element $x$ has a decomposition $x=x^{(1)}-(x^{(1)}+x)$ with $x^{(1)}\geqslant 0$ and $x^{(2)}:=x^{(1)}-x\geqslant 0$. Moreover, we have $x^{(1)}>x^{(3)}$. By Case 2.1 the disjoint decomposition of $x$ as $x=x_{a}+x_{d}$ is independent of the choice of representatives with respect to $x=x^{(1)}-x^{(2)}=x^{(3)}-x^{(4)}$. Similarly, due to $x^{(1)}>x^{(5)}$ it is independent of the choice of representatives with respect to $x=x^{(1)}-x^{(2)}=x^{(5)}-x^{(6)}$. Hence the decomposition is independent with respect to $x=x^{(3)}-x^{(4)}=x^{(5)}-x^{(6)}$.

Altogether, we established that for every $x\in X$ there exists a unique (disjoint) decomposition

[TABLE]

that is, $X=\mathcal{B}_{a}\oplus\mathcal{B}_{a}^{\textnormal{d}}$. $\square$

As a consequence of Theorem 141 and Theorem 43 we obtain the following.

Corollary 44**.**

*Let $X$ be an Archimedean pervasive pre-Riesz space and $a\in\mathcal{A}_{X}$. Then there is an order projection $P\colon X\rightarrow X$ with range $\mathcal{B}_{a}$. *

The following example demonstrates that the conclusion of Corollary 44 is not true, in general, if the pre-Riesz space is not pervasive.

Example 45**.**

An Archimedean pre-Riesz space with atoms and only trivial order projections.

We return to Example 31, where we considered the pre-Riesz space $X=\mathbb{R}^{3}$ endowed with the four-ray cone $K_{4}$. Recall that $\mathcal{A}_{X}=\left\{\lambda v_{k}\hskip 2.84544pt\middle|\hskip 2.84544pt\lambda\in\mathbb{R}_{>0}\textnormal{ and }v_{k}\in\left\{v_{1},\ldots,v_{4}\right\}\right\}$ and that by Theorem 141 order projections and band projections on $X$ coincide. Due to Theorem 142 it is sufficient to consider only projections onto directed bands. In [12, Example 4.6] it is shown that $X$ has precisely four non-trivial directed bands. They are given by $B_{1}:=\left\{v_{1}\right\}^{\textnormal{dd}}=\mathop{\textnormal{span}}\left\{v_{1}\right\}$, $B_{2}:=\mathop{\textnormal{span}}\left\{v_{2}\right\}$, $B_{3}:=\mathop{\textnormal{span}}\left\{v_{3}\right\}$ and $B_{4}:=\mathop{\textnormal{span}}\left\{v_{4}\right\}$. We show that the projections onto the bands $B_{1}$ to $B_{4}$ are trivial. First, let $P$ be a projection onto $B_{1}$. For every $k\in\left\{2,3,4\right\}$ we have

[TABLE]

Due to $P(X)\subseteq B_{1}$ and $0\leqslant P\leqslant I$, where $I$ is the identity map on $X$, the equation (8) implies for $k\in\left\{2,3,4\right\}$ that $P(v_{k})\in B_{1}\cap[0,v_{k}]\subseteq B_{1}\cap B_{k}=\left\{0\right\}$, i.e. $P(v_{k})=0$. As the vectors $v_{2},v_{3}$ and $v_{4}$ form a base of $X$, it follows $P=0$. Analogously, the band projection onto any of the bands $B_{2},B_{3}$ and $B_{4}$ is trivial. We conclude that $X$ has only the trivial order projections [math] and $I$.

To sum up, we obtain for Archimedean pervasive pre-Riesz spaces the extension property for projection bands and a theory of bands generated by atoms similar to the vector lattice case. As one can see in Theorems 21, 32 and 42, pervasiveness turns out to be the key condition in these results.