For negative soft photon, the soft factors are given by its complex conjugate.

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# semi unbounded order convergent in ordered vector spaces

Masoumeh Ebrahimzadeh  Islamic Azad University, Department of Mathematics,
Sarab Branch, Sarab, Iran<EMAIL_ADDRESS>and  Kazem Haghnejad
Azar∗ Department of Mathematics and Applications, Faculty of Sciences,
University of Mohaghegh Ardabili, Ardabil, Iran<EMAIL_ADDRESS>
###### Abstract.

Let $X$ be an ordered vector space. The net $\\{x_{\alpha}\\}\subseteq X$ is
semi unbounded order convergent to $x$ (in symbol
$x_{\alpha}\xrightarrow{suo}x$), if there is a net $\\{y_{\beta}\\}$, possibly
over a different index set, such that $y_{\beta}\downarrow 0$ and for every
$\beta$ there exists $\alpha_{0}$ such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. In vector lattice
$E$, semi unbounded order convergence is equivalent with unbounded order
convergence. We study some properties of this convergence and some of its
relationships with others known order convergence.

###### Key words and phrases:

ordered vector space, pre-Riesz space, unbounded order convergent, semi
unbounded order convergent.
∗Corresponding author

###### 2010 Mathematics Subject Classification:

Primary 47B65; Secondary 46B40, 46B42.

##  1\. Introduction

First we will ours motivate to write this article. By studying of [6], we
became interested in working on ordered vector spaces. In this article, we
have discussed convergence in ordered vector spaces that similar to the
unbounded order convergent that has been done in many articles, including [3]
and [4].

Here we bring some definitions of need. Let $X$ be a real vector space and $K$
be a cone in $X$, that is, $K$ is a wedge ($x,y\in K$ and $\lambda,\mu\geq 0$
imply $\lambda x+\mu y\in K$) and $K\cap(-K)=\\{0\\}$. In $X$ a partial order
is defined by $x\leq y$ whenever $y-x\in K$. The space $(X,K)$ (or, loosely
$X$) is then called a (partially) ordered vector space. A subspace $M\subseteq
X$ is majorizing in $X$ if for every $x\in X$ there exists some $m\in M$ with
$x\leq m$ (or, equivalently, if for each $x\in X$ there exists some $m\in M$
with $m\leq x$). A subspace $M\subseteq X$ is called directed if for every
$x,y\in M$ there is an element $z\in M$ such that $x\leq z$ and $y\leq z$. An
ordered vector space $X$ is directed if and only if $X_{+}$ is generating in
$X$, that is, $X=X_{+}-X_{+}$. An ordered vector space $X$ is called
Archimedean if for every $x,y\in X$ with $nx\leq y$ for every $n\in\mathbb{N}$
one has $x\leq 0$. The ordered vector space $X$ has the Riesz decomposition
property (RDP) if for every $x_{1},x_{2},z\in K$ with $z\leq x_{1}+x_{2}$
there exist $z_{1},z_{2}\in K$ such that $z=z_{1}+z_{2}$ with $z_{1}\leq
x_{1}$ and $z_{2}\leq x_{2}$. We call a linear subspace $M$ of an ordered
vector space $X$ order dense in $X$ if for every $x\in X$ we have

$x=\inf\\{z\in M:x\leq z\\},$

that is, the greatest lower bound of set $\\{z\in M:x\leq z\\}$ exists in $X$
and equals to $x$, see page $360$ of [1]. Clearly, if $M$ is order dense in
$X$, then $M$ is majorizing in $X$. Denote for a subset $M$ of $X$, the set of
all upper bounds (resp, down bounds) by $M^{u}=\\{x\in X:x\geq m\ for\ all\
m\in M\\}$ (resp, $M^{l}=\\{x\in X:x\leq m\ for\ all\ m\in M\\}$). It is clear
that for every subset $A,B$ of $X$, $(A^{l}+B^{l})\subseteq(A+B)^{l}$.
Moreover, if $X$ has Riesz decomposition property (in short RNP) and
$A,B\subseteq K$, then $(A+B)^{l}\subseteq A^{l}+B^{l}$.
The elements $x,y\in X$ are called disjoint, in symbols $x\perp y$, if
$\\{\pm(x+y)\\}^{u}=\\{\pm(x-y)\\}^{u}$. The disjoint complement of a subset
$M\subseteq X$ is $M^{d}=\\{x\in X\mid\forall y\in M:x\perp y\\}$. A sequence
$\\{x_{n}\\}\subseteq X$ is said to be disjoint, if for every $n\neq m$,
$x_{n}\perp x_{m}$. A linear subspace $B$ of an ordered vector space $X$ is
called a band in $X$ if $B=B^{dd}$. A subset $M$ of an ordered vector space
$X$ is called solid if for every $x\in X$ and $y\in M$ the relation $\\{\pm
y\\}^{u}\subseteq\\{\pm x\\}^{u}$ implies $x\in M$. A solid subspace $M$ of
$X$ is called an ideal.
Recall that a linear map $i:X\rightarrow Y$ is said to be bipositive if for
every $x\in X$ one has $i(x)\geq 0$ if and only if $x\geq 0$. A partially
ordered vector space $(X,K)$ is called pre-Riesz space if for every $x,y,z\in
X$ the inclusion $\\{x+y,x+z\\}^{u}\subseteq\\{y,z\\}^{u}$ implies $x\in K$.
Clearly, each vector lattice is a pre-Riesz space, since the inclusion in
definition of pre-Riesz space reduces to inequality $(x+y)\vee(x+z)\geq y\vee
z$, so $x+(y\vee z)\geq y\vee z$, which implies $x\geq 0$. By Theorem 4.3 of
[2], partially ordered vector space $X$ is a pre-Riesz space if and only if
there exist a vector lattice $Y$ and a bipositive linear map $i:X\rightarrow
Y$ such that $i(X)$ is order dense in $Y$. The pair $(Y,i)$ (or, loosely $Y$)
is then called a vector lattice cover of $X$. The theory of pre-Riesz spaces
and their vector lattice covers is due to van Haandel, see [8]. A net
$\\{x_{\alpha}\\}\subseteq X$ is said to be decreasing (in symbols,
$x_{\alpha}\downarrow$), whenever $\alpha\geq\beta$ implies $x_{\alpha}\leq
x_{\beta}$. For $x\in X$ the notation $x_{\alpha}\downarrow x$ means that
$x_{\alpha}\downarrow$ and $\inf_{\alpha}\\{x_{\alpha}\\}=x$ both hold. The
meanings of $x_{\alpha}\uparrow x$ are analogous. We say that a net
$\\{x_{\alpha}\\}\subseteq X$ $(o)$-converges (resp, $\tilde{o}$-converges) to
$x\in X$ (in symbols, $x_{\alpha}\xrightarrow{o}x$, resp,
$x_{\alpha}\xrightarrow{\tilde{o}}x$), if there is a net
$\\{y_{\alpha}\\}\subseteq X$ (resp, $\\{y_{\beta}\\}$ possibly over a
different index set) such that $y_{\alpha}\downarrow 0$ (resp,
$y_{\beta}\downarrow 0$) and for all $\alpha$ (resp, for every $\beta$ there
exists $\alpha_{0}$ that for all $\alpha\geq\alpha_{0}$ ) one has
$\pm(x_{\alpha}-x)\leq y_{\alpha}$, (resp, $\pm(x_{\alpha}-x)\leq y_{\beta}$).
For two elements $y,z\in K$ with $y\leq z$ denote the according order interval
by $[y,z]=\\{x\in X:y\leq x\leq z\\}$. A set $M\subset X$ is called order
bounded if there are $y,z\in X$ such that $M\subseteq[y,z]$. Let $X$ and $Z$
be ordered vector spaces.

Recall that the net $\\{x_{\alpha}\\}$ in vector lattice $E$ is said to be
unbounded order convergent to $x$ (or, $uo$-convergent for short) to $x$ if
there is a net $\\{y_{\beta}\\}$, possibly over a different index set, such
that $y_{\beta}\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$
such that $\mid x_{\alpha}-x\mid\wedge y\leq y_{\beta}$, whenever
$\alpha\geq\alpha_{0}$ and for all $0\leq y\in E$. In this case, we write
$x_{\alpha}\xrightarrow{uo}x$.

##  2\. semi unbounded order convergence on ordered vector spaces

######  Definition 2.1.

Let $X$ be an ordered vector space. The net $\\{x_{\alpha}\\}\subseteq X$ is
semi unbounded order convergent (or, $semi$-$uo$-convergent for short) to $x$
if there is a net $\\{y_{\beta}\\}$, possibly over a different index set, such
that $y_{\beta}\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$
such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. In this case, we
write $x_{\alpha}\xrightarrow{suo}x$.

######  Remark 2.2.

${\tilde{o}}$-convergence implies ${suo}$-convergence, but the convergence, in
general, not holds. Let $\\{x_{\alpha}\\}\subseteq X$ and
$x_{\alpha}\xrightarrow{\tilde{o}}x$. Therefore, there exists a net
$\\{y_{\beta}\\}\subseteq X$ and $y_{\beta}\downarrow 0$ and for all $\beta$,
there is an $\alpha_{0}$ that for each $\alpha$, $\pm(x_{\alpha}-x)\leq
y_{\beta}$. Let $z\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$ for all $0\leq
y\in X$. Since for $\alpha\geq\alpha_{0}$, $\pm(x_{\alpha}-x)\leq y_{\beta}$,
hence $z\leq y_{\beta}$. So
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$.
Therefore, $x_{\alpha}\xrightarrow{suo}x$.

On the other hand, the standard basis of $c_{0}$, $\\{e_{n}\\}_{n=1}^{\infty}$
is ${suo}$-convergence, but not ${\tilde{o}}$-convergence. Note that for order
bounded nets, $semi$-$uo$-convergence is equivalent to
$\tilde{o}$-convergence.

######  Proposition 2.3.

If $E$ is a vector lattice, then the net $\\{x_{\alpha}\\}\subseteq E$ is
$uo$-convergent to $x$ iff is $semi$-$uo$-convergent to $x$.

###### Proof.

Let $\\{x_{\alpha}\\}\subseteq E$ and $x_{\alpha}\xrightarrow{uo}x$. There
exists a net $\\{y_{\beta}\\}\subseteq E$ that $y_{\beta}\downarrow 0$ and for
each $\beta$ there exists an $\alpha_{0}$ that for each $\alpha\geq\alpha_{0}$
we have $|x_{\alpha}-x|\wedge y\leq y_{\beta}$ for all $y\in E^{+}$. Let
$z\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$. It is clear that
$z\leq|x_{\alpha}-x|\wedge y$ and therefore for each $\alpha\geq\alpha_{0}$,
$z\leq y_{\beta}$. Therefore,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$. It
means that $x_{\alpha}\xrightarrow{suo}x$.

For conversely, let $x_{\alpha}\xrightarrow{suo}x$. It is clear that
$(|x_{\alpha}-x|\wedge y)\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$ for all
$y\in E^{+}$. Therefore, for each $\beta$ there exists $\alpha_{0}$ such that
for each $\alpha\geq\alpha_{0}$, $(|x_{\alpha}-x|\wedge
y)\in\\{y_{\beta}\\}^{l}$. Hence $|x_{\alpha}-x|\wedge y\leq y_{\beta}$ for
all $y\in E^{+}$ whenever $\alpha\geq\alpha_{0}$. It means that
$x_{\alpha}\xrightarrow{uo}x$. ∎

######  Lemma 2.4.

Let $X$ be an ordered vector space and $\\{x_{\alpha}\\}\subseteq X$, then

  1. (1)

$x_{\alpha}\xrightarrow{suo}x$ iff $(x_{\alpha}-x)\xrightarrow{suo}0$.

  2. (2)

if for each $\alpha$, $x_{\alpha}\leq y$ and $x_{\alpha}\xrightarrow{suo}x$,
then $x\leq y$.

  3. (3)

if $0\leq x_{\alpha}\xrightarrow{suo}x$, then $0\leq x$.

Moreover, if $X$ has the $RDP$ property, then

  4. (4)

if $x_{\alpha}\xrightarrow{suo}x$ and $y_{\alpha}\xrightarrow{suo}y$, then
$\lambda x_{\alpha}+\mu y_{\alpha}\xrightarrow{suo}\lambda x+\mu y$ for each
scalar $\lambda,\mu\in\mathbb{R}$.

  5. (5)

if $x_{\alpha}\xrightarrow{suo}x$, $z_{\alpha}\xrightarrow{suo}z$ and
$x_{\alpha}\leq z_{\alpha}$ for all $\alpha$, then $x\leq z$.

  6. (6)

if $x_{\alpha}\xrightarrow{suo}x$ and $x_{\alpha}\xrightarrow{suo}y$, then
$x=y$.

  7. (7)

if $x_{\alpha}\xrightarrow{suo}0$, then for each subnet $\\{z_{\gamma}\\}$ of
$\\{x_{\alpha}\\}$, $z_{\gamma}\xrightarrow{suo}0$.

###### Proof.

  1. (1)

By Definition 2.1, it is established.

  2. (2)

By assumption, it is clear that $\\{x_{\alpha}\\}$ is order bounded.
Therefore, $x_{\alpha}\xrightarrow{\tilde{o}}x$. So by Lemma 2.1 of [6], the
proof is clear.

  3. (3)

Let $0\leq x_{\alpha}\xrightarrow{suo}x$, therefore there exists net
$\\{y_{\beta}\\}\subseteq X$ such that $y_{\beta}\downarrow 0$ and for every
$\beta$ there exists $\alpha_{0}$ such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$ whenever
$\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. We know that
$\\{x_{\alpha}-x\\}\subseteq\\{\pm(x_{\alpha}-x)\\}$ for all $\alpha$.
Therefore, it is clear that for all $\alpha$,
$\\{\pm(x_{\alpha}-x)\\}^{u}\subseteq\\{x_{\alpha}-x\\}^{u}$. So we have
$\\{\\{x_{\alpha}-x\\}^{u},y\\}^{l}\subseteq\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$
for all $\alpha$. Therefore, for $\alpha\geq\alpha_{0}$,
$\\{\\{x_{\alpha}-x\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$ for all
$0\leq y\in X$. It is clear that $-x\in\\{\\{x_{\alpha}-x\\}^{u},y\\}^{l}$.
Hence $-x\leq y_{\beta}$. So $x\geq 0$.

  4. (4)

Let $\lambda>0$ and $x_{\alpha}\xrightarrow{suo}x$. We have $\\{\\{\pm(\lambda
x_{\alpha}-\lambda
x)\\}^{u},y\\}^{l}=\lambda(\\{\\{\pm(x_{\alpha}-x)\\}^{u},y^{\prime}\\}^{l})\subseteq\\{\lambda
y_{\beta}\\}^{l}$. Note that if $\lambda<0$, then $\lambda(\pm
x_{\alpha})=-\lambda(\pm x_{\alpha})$.

We have
$\\{\\{\pm(x_{\alpha}-x)\\}^{u}+\\{\pm(y_{\alpha}-y)\\}^{u}\\}\subseteq\\{\pm(x_{\alpha}-x+y_{\alpha}-y)\\}^{u}$.
Therefore,
$\\{\\{\pm(x_{\alpha}-x+y_{\alpha}-y)\\}^{u},y\\}^{l}\subseteq\\{\\{\\{\pm(x_{\alpha}-x)\\}^{u}+\\{\pm(y_{\alpha}-y)\\}^{u}\\},y\\}^{l}$
for all $0\leq y\in X$. Since $X$ has the $RDP$ property,
$\\{\\{\pm(x_{\alpha}-x+y_{\alpha}-y)\\}^{u},y\\}^{l}\subseteq\\{\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}+\\{\\{\pm(y_{\alpha}-y)\\}^{u}\\},y\\}^{l}$.
Hence $x_{\alpha}+y_{\alpha}\xrightarrow{suo}x+y$.

  5. (5)

For all $y\in X^{+}$ we have
$\\{\\{x-z\\}^{u},y\\}=\\{\\{x-x_{\alpha}+x_{\alpha}-z\\}^{u},y\\}$. It is
clear that
$\\{\\{x-x_{\alpha}+z_{\alpha}-z\\}^{u},y\\}\subseteq\\{\\{x-x_{\alpha}+x_{\alpha}-z\\}^{u},y\\}$
and therefore
$\\{\\{x-x_{\alpha}+x_{\alpha}-z\\}^{u},y\\}^{l}\subseteq\\{\\{x-x_{\alpha}+z_{\alpha}-z\\}^{u},y\\}^{l}$.
Since $X$ has the $RDP$ property, the rest of the proof is clear.

  6. (6)

Similar to the proof of 5, since for each $\alpha$, we have $x_{\alpha}\leq
x_{\alpha}$, therefore $x\leq y$ and $y\leq x$. Therefore, $x=y$.

  7. (7)

We have $\\{z_{\gamma}\\}\subseteq\\{x_{\alpha}\\}$. Therefore,
$\\{x_{\alpha}\\}^{u}\subseteq\\{z_{\gamma}\\}^{u}$ and so
$\\{\\{z_{\gamma}\\}^{u},y\\}^{l}\subseteq\\{\\{x_{\alpha}\\}^{u},y\\}^{l}$
for each $0\leq y\in X$. Hence the proof is complete.

∎

######  Proposition 2.5.

Let $B$ be a projection band of ordered vector space $X$ and $P_{B}$ the
corresponding band projection. If $\\{x_{\alpha}\\}\subseteq X$ is
$semi$-$uo$-null in $X$, then $\\{P_{B}(x_{\alpha})\\}$ is $semi$-$uo$-null in
$B$.

###### Proof.

Let $x_{\alpha}\xrightarrow{suo}0$ in $X$, therefore there is a net
$\\{y_{\beta}\\}$, possibly over a different index set, such that
$y_{\beta}\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$ such
that $\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. We know that $\pm
P_{B}(x_{\alpha})\leq\pm(x_{\alpha})$ for all $\alpha$. Therefore,
$\\{\pm(x_{\alpha})\\}^{u}\subseteq\\{\pm(P_{B}(x_{\alpha}))\\}^{u}$. So we
have
$\\{\\{\pm(P_{B}(x_{\alpha}))\\}^{u},P_{B}(y)\\}^{l}\subseteq\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$.
Note that $P_{B}(y_{\beta})\downarrow 0$ in $B$ and $P_{B}(y_{\beta})\leq
y_{\beta}$ for all $\beta$. Hence $P_{B}(y_{\beta})\in\\{y_{\beta}\\}^{l}$. It
means that for each $z\in X$ that
$z\in\\{\\{\pm(P_{B}(x_{\alpha}))\\}^{u},P_{B}(y)\\}^{l}$, we have
$z\leq(y_{\beta})$. It is obvious that
$P_{B}(z)\in\\{\\{\pm(P_{B}(x_{\alpha}))\\}^{u},P_{B}(y)\\}^{l}$,
$P_{B}(z)\leq z$ and therefore $P_{B}(z)\in\\{P_{B}(y_{\beta})\\}^{l}$.
Therefore,
$\\{\pm(P_{B}(x_{\alpha}))\\}^{u},P_{B}(y)\\}^{l}\subseteq\\{P_{B}(y_{\beta})\\}^{l}$
and so $P_{B}(x_{\alpha})\xrightarrow{suo}0$. ∎

######  Definition 2.6.

A net $\\{x_{\alpha}\\}$ in ordered vector space $X$ is said to be
$suo$-Cauchy, if $\\{x_{\alpha}-x_{\beta}\\}_{(\alpha,\beta)}$ $semi$-$uo$\-
converges to $0$ in $X$.

######  Proposition 2.7.

Let $X$ be an ordered vector space with the property $RDP$. Then

  1. (1)

if an $semi$-$uo$-Cauchy net $\\{x_{\alpha}\\}$ has an $semi$-$uo$-convergent
subnet whose $semi$-$uo$-limit is $x$, then $x_{\alpha}\xrightarrow{suo}x$.

  2. (2)

each $semi$-$uo$-convergent net is a $semi$-$uo$-Cauchy net.

###### Proof.

  1. (1)

Let $\\{x_{\alpha}\\}$ be a $semi$-$uo$-Cauchy net in $X$ and
$\\{z_{\gamma}\\}$ be a subnet of $\\{x_{\alpha}\\}$ such that
$z_{\gamma}\xrightarrow{suo}x$. We have
$x_{\alpha}-x=x_{\alpha}-z_{\gamma}+z_{\gamma}-x$ and
$-(x_{\alpha}-x)=-(x_{\alpha}-z_{\gamma}+z_{\gamma}-x)$. We have
$\\{\\{\pm(x_{\alpha}-z_{\gamma})\\}^{u}+\\{\pm(z_{\gamma}-x)\\}^{u}\\}\subseteq\\{\pm(x_{\alpha}-z_{\gamma}+z_{\gamma}-x)\\}^{u}$.
Therefore,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}=\\{\\{\pm(x_{\alpha}-z_{\gamma}+z_{\gamma}-x)\\}^{u},y\\}^{l}\subseteq\\{\\{\\{\pm(x_{\alpha}-z_{\gamma})\\}^{u}+\\{\pm(z_{\gamma}-x)\\}^{u}\\},y\\}^{l}$
for all $0\leq y\in X$. Since $X$ has the $RDP$ property,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{\\{\\{\pm(x_{\alpha}-z_{\gamma})\\}^{u},y\\}^{l}+\\{\\{\pm(z_{\gamma}-x)\\}^{u}\\},y\\}^{l}$.
Because by assumption, $x_{\alpha}-z_{\gamma}\xrightarrow{suo}0$ and
$z_{\gamma}\xrightarrow{suo}x$, the proof is complete.

  2. (2)

Without loss generality, by Lemma 2.4 we assume that $\\{x_{\alpha}\\}$ be a
$semi$-$uo$-convergent to $0$ in $X$. Then there is a net $\\{y_{\sigma}\\}$
such that $y_{\sigma}\downarrow 0$ and for every $\sigma$ there exists
$\alpha_{0}$ such that
$\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}\subseteq\\{y_{\sigma}\\}^{l}$, whenever
$\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. Obviously that

$\\{\pm(x_{\alpha})\\}^{u}+\\{\pm(x_{\beta})\\}^{u}\subseteq\frac{1}{2}\\{\pm(x_{\alpha}-x_{\beta})\\}^{u}.$

Let $y>0$. By property $RDP$, we have

$\\{\frac{1}{2}\\{\pm(x_{\alpha}-x_{\beta})\\}^{u},y\\}^{l}\subseteq\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}+\\{\\{\pm(x_{\beta})\\}^{u},y\\}^{l}.$

It follows that
$\\{\frac{1}{2}\\{\pm(x_{\alpha}-x_{\beta})\\}^{u},y\\}^{l}\subseteq
2\\{y_{\sigma}\\}^{l}$, and so proof follows.

∎

######  Theorem 2.8.

Let $X$ be an order dense subspace of an ordered vector space $Y$ and
$\\{x_{\alpha}\\}\subseteq X$. $x_{\alpha}\xrightarrow{suo}x$ in $X$ iff
$x_{\alpha}\xrightarrow{suo}x$ in $Y$.

###### Proof.

Let $x_{\alpha}\xrightarrow{suo}x$ in $X$ and
$z\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$ for all $0\leq y\in Y$.
Therefore, $z\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$ for all $0\leq y\in
X$. By assumption there is a net $\\{y_{\beta}\\}\subseteq X$ that
$y_{\beta}\downarrow 0$ in $X$ and $z\in\\{y_{\beta}\\}^{l}$. By Proposition
5.1 of [2], $y_{\beta}\downarrow 0$ in $Y$. Therefore, there is a net
$\\{y_{\beta}\\}\subseteq Y$ that $y_{\beta}\downarrow 0$ and for each $\beta$
there is an $\alpha_{0}$ that for each $\alpha\geq\alpha_{0}$,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$ for all
$0\leq y\in Y$. So $x_{\alpha}\xrightarrow{suo}x$ in $Y$.

Consequently, let $x_{\alpha}\xrightarrow{suo}x$ in $Y$. Therefore, there
exists net $\\{y_{\beta}\\}\subseteq Y$ such that $y_{\beta}\downarrow 0$ and
for every $\beta$ there exists $\alpha_{0}$ such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in Y$. Since $X$ is order
dense in $Y$, therefore $X$ is majorizing in $Y$. Hence for each $\beta$,
there is a $z\in X$ such that $y_{\beta}\leq z$. We define $z_{\beta}=\\{z\in
X:y_{\beta}\leq z\\}$. $z_{\beta}\downarrow 0$ in $X$. It is obvious that
$\\{y_{\beta}\\}^{l}\subseteq\\{z_{\beta}\\}^{l}$. Therefore, for each $\beta$
there is an $\alpha_{0}$ that for each $\alpha\geq\alpha_{0}$,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{z_{\beta}\\}^{l}$ for all
$0\leq y\in X$. ∎

######  Corollary 2.9.

Let $X$ be a pre-Riesz space with vector lattice cover $(Y,i)$ and
$\\{x_{\alpha}\\}\subseteq X$ is $semi$-$uo$-null in $X$. By Theorem 2.8,
$x_{\alpha}\xrightarrow{uo}0$ in $Y$. If $Y$ has a weak unit $u$ and $u\in X$,
then by Lemma 3.2 of [4], $x_{\alpha}\xrightarrow{suo}0$ in $X$ iff there is a
net $\\{y_{\beta}\\}$, possibly over a different index set, such that
$y_{\beta}\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$ such
that $\\{\\{\pm(x_{\alpha}-x)\\}^{u},u\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$.

######  Theorem 2.10.

Let $X$ be a pre-Riesz space with vector lattice cover $(Y,i)$. Then the
following assertions are true.

  1. (1)

$\\{x_{\alpha}\\}\subseteq X$ is $semi$-$uo$-null iff $\\{i(x_{\alpha})\\}$ is
$semi$-$uo$-null in $Y$.

  2. (2)

If the sequence $\\{x_{n}\\}\subseteq X$ is disjoint, then
$x_{n}\xrightarrow{suo}0$ in $X$.

Moreover, if $X$ is normed space and

  3. (3)

$Y,Y^{\prime}$ have order continuous norm and $\\{x_{\alpha}\\}\subseteq X$ is
$semi$-$uo$-null, then $\\{x_{\alpha}\\}$ is $w$-null.

  4. (4)

$\\{x_{\alpha}\\}\subseteq X$ is order bounded, $x_{\alpha}\xrightarrow{suo}0$
and $Y$ is a Banach lattice with order continuous norm, then
$\\{x_{\alpha}\\}$ is norm-null.

###### Proof.

  1. (1)

Let $\\{x_{\alpha}\\}\subseteq X$ is $semi$-$uo$-null in $X$. Therefore, there
is a net $\\{y_{\beta}\\}$, possibly over a different index set, such that
$y_{\beta}\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$ such
that $\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. By Lemma 1 of [5],
$i(y_{\beta})\downarrow 0$ in $i(X)$. It is obvious that $\\{i(x_{\alpha})\\}$
is $semi$-$uo$-null in $i(X)$. Since $i(X)$ is order dense in $Y$, then by
Proposition 2.8, $\\{i(x_{\alpha})\\}$ is $semi$-$uo$-null in $Y$.

Conversely, let $\\{x_{\alpha}\\}\subseteq X$ and
$x_{\alpha}\xrightarrow{suo}0$ in $Y$. Hence $x_{\alpha}\xrightarrow{uo}0$ in
$Y$ and therefore it is $semi$-$uo$-null in $i(X)$. Therefore, there is a net
$\\{i(y_{\beta})\\}$, possibly over a different index set, such that
$i(y_{\beta})\downarrow 0$ and for every $\beta$ there exists $\alpha_{0}$
such that

$\\{\\{\pm i(x_{\alpha})\\}^{u},i(y)\\}^{l}\subseteq\\{i(y_{\beta})\\}^{l},$

whenever $\alpha\geq\alpha_{0}$ and for all $0\leq i(y)\in i(X)$. We have
$i(\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l})=\\{\\{\pm
i(x_{\alpha})\\}^{u},i(y)\\}^{l}\subseteq\\{i(y_{\beta})\\}^{l}=i(\\{y_{\beta}\\}^{l}).$
Since $i$ is a bipositive operator, therefore
$\\{\\{\pm(x_{\alpha})\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$ and so
$x_{\alpha}\xrightarrow{suo}0$ in $X$.

  2. (2)

Since $X$ is an order dense subspace of $Y$, by Proposition 5.9 of [2],
$\\{x_{n}\\}$ is disjoint in $Y$. By Corollary 3.6 of [3],
$x_{n}\xrightarrow{uo}0$ in $Y$. Therefore, $x_{n}\xrightarrow{suo}0$ in
$i(X)$. By 1, $x_{n}\xrightarrow{suo}0$ in $X$.

  3. (3)

Let $\\{x_{\alpha}\\}\subseteq X$ be $semi$-$uo$-null. Then by Proposition
2.8, $x_{\alpha}\xrightarrow{uo}0$ in $Y$. Since $Y$ and $Y^{\prime}$ have
order continuous norm, then by Theorem 5 of [9], $x_{\alpha}\xrightarrow{w}$
in $Y$. By Theorem 3.6 of [7], $x_{\alpha}\xrightarrow{w}0$ in $X$.

  4. (4)

We have $x_{\alpha}\xrightarrow{uo}0$ in $Y$. Since $Y$ is a Banach lattice
with order continuous norm and the net $\\{x_{\alpha}\\}$ is almost order
bounded in $Y$, therefore by Proposition 3.7 of [4], $\\{x_{\alpha}\\}$ is
norm-null in $Y$ and so is norm-null in $X$.

∎

######  Definition 2.11.

A subset $J$ of ordered vector space $X$ is said to be $semi$-$uo$-closed, if
$\\{x_{\alpha}\\}\subseteq J$ with $x_{\alpha}\xrightarrow{suo}x$ implies
$x\in J$.

######  Remark 2.12.

Let $J\subseteq X$ be a $semi$-$uo$-closed set and $\\{x_{\alpha}\\}\subseteq
J$ that $x_{\alpha}\xrightarrow{\tilde{o}}x$. It is obvious that
$x_{\alpha}\xrightarrow{suo}x$. Since $J$ is $semi$-$uo$-closed, therefore
$x\in J$. So $J$ is $\tilde{o}$-closed.

######  Theorem 2.13.

The following assertions are true.

  1. (1)

Let $X$ be an order dense subspace of an ordered vector space $Y$. If
$J\subseteq Y$ is $semi$-$uo$-closed in $Y$, then $J\cap X$ is
$semi$-$uo$-closed in $X$.

  2. (2)

Let $X$ be a pre-Riesz space with vector lattice cover $(Y,i)$. $J\subseteq X$
is $semi$-$uo$-closed iff $i(J)$ is $semi$-$uo$-closed in $i(X)$.

###### Proof.

  1. (1)

Let $\\{x_{\alpha}\\}\subseteq J\cap X$ and $x_{\alpha}\xrightarrow{suo}x$.
There exists a net $\\{y_{\beta}\\}\subseteq X$ that $y_{\beta}\downarrow 0$
and for each $\beta$ there exists $\alpha_{0}$ that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. Since $X$ is order
dense in $Y$, then for each $z\in Y$,
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},z\\}^{l}\subseteq\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$
and $y_{\beta}\downarrow 0$ in $Y$. Because $\\{x_{\alpha}\\}\subseteq J$ and
$J$ is $semi$-$uo$-closed in $Y$, therefore $x\in J$.

  2. (2)

Let $\\{x_{\alpha}\\}\subseteq J$ and $x_{\alpha}\xrightarrow{suo}x$. It is
obvious that $i(x_{\alpha})\xrightarrow{suo}i(x)$. Since $i(J)$ is
$semi$-$uo$-closed, therefore $i(x)\in i(J)$. So $x\in J$.

Conversely, let $\\{i(x_{\alpha})\\}\subseteq i(J)$ and
$i(x_{\alpha})\xrightarrow{suo}i(x)$. There exists a net
$\\{i(y_{\beta})\\}\subseteq i(X)$ that $i(y_{\beta})\downarrow 0$ and for
each $\beta$ there exists $\alpha_{0}$ that
$\\{\\{\pm(i(x_{\alpha}-x)\\}^{u},i(y)\\}^{l}\subseteq\\{i(y_{\beta})\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq i(y)\in i(X)$. By Lemma 1
of [5], $y_{\beta}\downarrow 0$ in $X$. Therefore,
$x_{\alpha}\xrightarrow{suo}x$. Since $\\{x_{\alpha}\\}\subseteq J$ and $J$ is
$semi$-$uo$-closed, hence $x\in J$ and therefore $i(x)\in i(J)$.

∎

######  Theorem 2.14.

Let $J$ be a $semi$-$uo$-closed subspace in Archimedean ordered vector space
$X$ and $\\{x_{\alpha}\\}\subseteq J$. $x_{\alpha}\xrightarrow{suo}x$ in $J$
iff $x_{\alpha}\xrightarrow{suo}x$ in $X$.

###### Proof.

Let $\\{x_{\alpha}\\}\subseteq J$ and $x_{\alpha}\xrightarrow{suo}x$ in $X$.
Since $J$ is $semi$-$uo$-closed it is obvious that
$x_{\alpha}\xrightarrow{suo}x$ in $J$.

Conversely, let $\\{x_{\alpha}\\}\subseteq J$ and
$x_{\alpha}\xrightarrow{suo}x$ in $J$. Therefore, there is a net
$\\{y_{\beta}\\}\subseteq J$, possibly over a different index set, such that
$y_{\beta}\downarrow 0$ in $J$ and for every $\beta$ there exists $\alpha_{0}$
such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in B$. It is obvious that
$y_{\beta}\downarrow$ in $X$. Since $X$ is Archimedean, that there exists a
$z$ that $y_{\beta}\downarrow z$ in $X$. It is clear that
$y_{\beta}\xrightarrow{\tilde{o}}z$ in $X$ and so
$y_{\beta}\xrightarrow{suo}z$ in $X$. Therefore, $y_{\beta}\xrightarrow{suo}z$
in $J$. Since by Lemma 2.4, $semi$-$uo$-limits are unique, therefore $z=0$.
Hence $y_{\beta}\downarrow 0$ in $X$. Let
$k\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}$ for all $0\leq y\in X$. It is
clear that $k\in\\{\\{\pm(x_{\alpha}-x)\\}^{u},y^{\prime}\\}^{l}$ for all
$0\leq y^{\prime}\in J$. So
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$ for all
$0\leq y\in X$. Therefore, $x_{\alpha}\xrightarrow{suo}x$ in $X$. ∎

######  Proposition 2.15.

A solid subset $J$ of an ordered vector space $X$ is $semi$-$uo$-closed iff
$\\{x_{\alpha}\\}\subseteq J$and $0\leq x_{\alpha}\uparrow x$ imply $x\in J$.

###### Proof.

Let $\\{x_{\alpha}\\}\subseteq J$ and $0\leq x_{\alpha}\uparrow x$. It is
clear that $x_{\alpha}\xrightarrow{o}x$ and therefore
$x_{\alpha}\xrightarrow{suo}x$. Because $J$ is $semi$-$uo$-closed imply $x\in
J$.

Conversely, let a net $\\{x_{\alpha}\\}\subseteq J$ and
$x_{\alpha}\xrightarrow{suo}x$. Therefore, there is a net
$\\{y_{\beta}\\}\subseteq X$, possibly over a different index set, such that
$y_{\beta}\downarrow 0$ in $X$ and for every $\beta$ there exists $\alpha_{0}$
such that
$\\{\\{\pm(x_{\alpha}-x)\\}^{u},y\\}^{l}\subseteq\\{y_{\beta}\\}^{l}$,
whenever $\alpha\geq\alpha_{0}$ and for all $0\leq y\in X$. It is clear that
$\\{\pm x_{\alpha}\\}^{u}\subseteq\\{\pm(x-y_{\beta})\\}^{u}$. Because $J$ is
solid, we have $\\{x-y_{\beta}\\}\subseteq J$. It is obvious that $0\leq
x-y_{\beta}\uparrow x$. It follows that $x\in J$. Hence $J$ is
$semi$-$uo$-closed. ∎

######  Definition 2.16.

Let $B$ be an ideal in ordered vector space $X$. $B$ is said to be a
$semi$-$uo$-band in $X$ if $B$ is $semi$-$uo$-closed in $X$.

######  Corollary 2.17.

  1. (1)

Let $X$ be an order dense subspace of directed ordered vector space $Y$. If
$J\subseteq Y$ is $semi$-$uo$-band in $Y$, then by Theorem 2.13, $J\cap X$ is
$semi$-$uo$-closed in $X$ and by Proposition 5.3 of [2], $J\cap X$ is an ideal
in $X$. Therefore, $J\cap X$ is a $semi$-$uo$-band in $X$.

Let $X$ be a pre-Riesz space with vector lattice cover $(Y,i)$.

  2. (2)

It is clear that $J\subseteq X$ is an ideal in $X$ iff $i(J)$ is an ideal in
$i(X)$ and by Theorem 2.13, $J\subseteq X$ is $semi$-$uo$-closed in $X$ iff
$i(J)$ is $semi$-$uo$-closed in $i(X)$. So, $J\subseteq X$ is a
$semi$-$uo$-band in $X$ iff $i(J)$ is $semi$-$uo$-band in $i(X)$.

######  Proposition 2.18.

Let $X$ be a vector lattice. If an ideal $B\subseteq X$ is a $semi$-$uo$-band
in $X$, then it is a band in $X$.

###### Proof.

Let an ideal $B$ be a $semi$-$uo$-band in $X$ and $\\{x_{\alpha}\\}\subseteq
X$ be a net that $x_{\alpha}\xrightarrow{o}x$. Therefore,
$x_{\alpha}\xrightarrow{uo}x$ in $X$. Since $X$ is a vector lattice, then by
Proposition 2.3, $x_{\alpha}\xrightarrow{suo}x$ in $X$. By assumption, $x\in
B$. So $B$ is a band in $X$. ∎

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