The concept of order compactness extends the notion of compactness to spaces without a defined topology. This allows for the generalization of key results related to compactness in Riesz spaces, enriching the theory and potentially offering new tools for analysis in these spaces.

If Γ is a subset of the partially ordered set Ω, we say that Γ is up-directed if and only if every finite subset of Γ has an upper bound in Γ. “Directed” will be used to denote “up-directed”. A net in a Riesz space E is an arbitrary function from a non-empty directed set Γ to the space E. Nets will be denoted by ${\left\{x_{\alpha }\right\}}_{\alpha \in \text{Γ}}$, where $x_{\alpha }$ is the point of E assigned to the element $\alpha$ of the directed set Γ. Several authors have given various meanings to the statement “Net ${\left\{x_{\alpha }\right\}}_{\alpha \in \text{Γ}}$ order converges to the element $x$”. In the literature on Riesz spaces theory, nets order convergence is defined in one of three ways.

Definition 1.1 ( [1] ) A net ${\left\{x_{\alpha }\right\}}_{\alpha \in \Gamma }$ is order convergent to $x$, if there exists a net ${\left\{y_{\beta }\right\}}_{\beta \in \Delta }$ such that:

1) $y_{\beta }\downarrow 0$, and

2) $\left|x_{\alpha }-x\right|\le y_{\beta }$ for all $\alpha \in \text{Γ}$.

Definition 1.2 ( [2] ) A net ${\left\{x_{\alpha }\right\}}_{\alpha \in \Gamma }$ is order convergent to $x$, written $x_{\alpha }\stackrel{o}{\to }x$, if there exists a net ${\left\{y_{\beta }\right\}}_{\beta \in \Delta }$ such that:

1) $y_{\beta }\downarrow 0$, and

2) for each $\beta \in \text{Δ}$ there exists some ${\alpha }_0\in \text{Γ}$ satisfying $\left|x_{\alpha }-x\right|\le y_{\beta }$ for all $\alpha \ge {\alpha }_0$.

It should be noticed that definition 1.1 does not satisfy our understanding of the word “convergence”. A converging net must remain converging even if we attach additional terms at the “beginning of the net. On the other hand, we hope that the definition adopted will not conflict with that used when the trellis is provided with a topology: If E is any topological (Riesz) space with a family $T$ of “open” sets, the convergence of a net ${\left\{x_{\alpha }\right\}}_{\alpha \in \text{Γ}}$ of points of E is defined by the rule: $x_{\alpha }\stackrel{T}{\to }x$ means that for every open set U containing $x$, ${\beta }_U$ exists such that $x_{\alpha }\in U$ for all $\alpha \ge {\beta }_U$.

We now give the Aliprantis-Border definition of order convergence of a net.

Definition 1.3 ( [2] , p. 322) A net $\left\{x_{\alpha }\right\}$ in a Riesz space E is order convergent to some $x\in E$, written $x_{\alpha }\stackrel{o}{\to }x$, if there is a net $\left\{y_{\alpha }\right\}$ (with the same directed set) satisfying:

1) $y_{\alpha }\stackrel{o}{\to }0$ and

2) $\left|x_{\alpha }-x\right|\le y_{\alpha }$ for each $\alpha$

In order to ensure compatibility with the results when we introduce a topology in the Riesz space, we will adopt this last definition.

Finally, we will need the following definition.

Definition 1.4 ( [2] , p. 31) A net ${\left\{y_{\lambda }\right\}}_{\lambda \in \Lambda }$ is a subnet of a net ${\left\{x_{\alpha }\right\}}_{\alpha \in \Gamma }$ if there is a function $\phi :\Lambda \to \Gamma$ satisfying

1) $y_{\lambda }=x_{{\phi }_{\lambda }}$ for each $\lambda \in \text{Λ}$, where ${\phi }_{\lambda }$ stands for $\phi \left(\lambda \right)$; and

2) for each ${\alpha }_0\in \text{Γ}$ there exists some ${\lambda }_0\in \text{Λ}$ such that $\lambda \ge {\lambda }_0$ implies ${\phi }_{\lambda }\ge {\alpha }_0$.

The famous topological properties lacking in the general context of Riesz spaces, we take for definition of order compactness an equivalent for the order of its important topological property, namely the Bolzano-Weierstrass characterization. For a complete account of Riesz spaces, the reader is referred to [3] - [9] .

Definition 2.1 A subset $\mathcal{K}$ of a Riesz space E is:

Note that if $\mathcal{K}$ is order compact then $\mathcal{K}$ is $\sigma$-order compact.

Proposition 2.1 Let E be a Riesz space. The following assertions are fulfilled:

i) Finite unions of order compact subsets are order compact.

ii) Intersections of order compact parts are order compact.

iii) Order closed subsets of order compact subsets are order compact.

Proof. i) Let $\left\{x_{\alpha }\right\}$ be a net in $\mathcal{K}={\displaystyle \underset{1\le i\le p}{\cup }{\mathcal{K}}_i}$, where ${\mathcal{K}}_1,{\mathcal{K}}_2,\cdots ,{\mathcal{K}}_p$ are order compact subsets. It is easy to see that at least one ${\mathcal{K}}_i$ contains an infinity of terms $x_{\alpha }$. The order compactness of ${\mathcal{K}}_i$ ensures the existence of a convergent subnet of $\left\{x_{\alpha }\right\}$ in ${\mathcal{K}}_i\subset \mathcal{K}$, and the proof of assertion (i) is complete.

ii) is trivial.

iii) derives easily from the fact that a subset $\mathcal{K}$ of a Riesz space is order closed if $\left\{x_{\alpha }\right\}\subset \mathcal{K}$ and $x_{\alpha }\stackrel{o}{\to }x$ imply $x\in \mathcal{K}$. ◼

With the same terms as for the whole space, the definition of the order completion for a part can be formulated as follows.

Definition 2.2 A subset $\mathcal{K}$ of a Riesz space $E$ is order complete, or Dedekind complete if both of the following conditions are fulfilled:

1) all its non-empty parts that are order bounded from above have a supremum in $\mathcal{K}$.

2) Every nonempty subset of $\mathcal{K}$ that is bounded from below has an infimum in $\mathcal{K}$.

Theorem 2.1 Let $\mathcal{K}$ be an order compact subset of a Dedekind complete Riesz space E such that $x\vee y\in \mathcal{K}$ and $x\wedge y\in \mathcal{K}$ for all $x,y\in \mathcal{K}$. Then $\mathcal{K}$ is order complete.

Proof. Let A be an order bounded from the above subset of $\mathcal{K}$. The order ≥ on the set $S\subset \mathcal{K}$ of suprema of finite subsets of A is a direction: for each pair $x,y\in S$, we have $x\le x\vee y$, $y\le x\vee y$, and $x\vee y\in S$. Furthermore, S has the same upper bounds as A. Let $a$ be upper bound of A in E. Then $a$ is upper bound of S in E. So, there exists ${\left\{x_{\alpha }\right\}}_{\alpha \in L}$, $L\subset S$ such that $x_{\alpha }\uparrow a$. Since $\mathcal{K}$ is order compact, it follows that there exists some subnet of $\left\{x_{\alpha }\right\}$ convergent in $\mathcal{K}$. Thus $a\in \mathcal{K}$.

Now if A is an order bounded from below subset of $\mathcal{K}$. Condition (2) of the above definition is derived from the same method applied to the set $I\subset \mathcal{K}$ of infimum of finite subsets of A. The order ≤ on I is a direction: for each pair $x,y\in I$, we have $x\ge x\wedge y$, $y\ge x\wedge y$, and $x\wedge y\in I\subset \mathcal{K}$. ◼

We need to manipulate functions that are defined only on parts of a Riesz space, a fact that allows the following definition.

Definition 2.3 A function $f:X\to F$ from a subset X of a Riesz space E into a Riesz space F is order continuous on X if every net $\left\{x_{\alpha }\right\}\subset X$ satisfies: $x_{\alpha }\stackrel{0}{\to }x$ in E (with $x\in X$) implies $f\left(x_{\alpha }\right)\stackrel{0}{\to }f\left(x\right)$ in F.

A subset $\mathcal{S}$ of a Riesz space is order closed if $\left\{x_{\alpha }\right\}\subset \mathcal{S}$ and $x_{\alpha }\stackrel{o}{\to }x$ imply $x\in \mathcal{S}$.

Let E and F be Riesz spaces. The Cartesian product $E\times F$ is a Riesz space under the usual ordering where $a_1=\left(x_1,y_1\right)\ge a_2=\left(x_2,y_2\right)$ whenever $x_1\ge x_2$ and $y_1\ge y_2$. The infimum and supremum of two vectors $a_1$ and $a_2$ are given by $a_1\vee a_2=\left(x_1\vee x_2,y_1\vee y_2\right)$ and $a_1\wedge a_2=\left(x_1\wedge x_2,y_1\wedge y_2\right)$. Consider a function $f:\mathcal{S}\to \mathcal{K}$ between two subsets of E and F, respectively.

Remember that the graph of $f$ is $G_f=\left\{\left(x,f\left(x\right)\right):x\in \mathcal{S}\right\}$. If $f$ is order continuous then the graph of $f$ is order closed. Indeed, let $\left\{\left(x_{\alpha },f\left(x_{\alpha }\right)\right)\right\}\subset G_f$ such that $\left(x_{\alpha },f\left(x_{\alpha }\right)\right)\stackrel{o}{\to }\left(x,y\right)$, which in terms of the lattice product, reads as $x_{\alpha }\stackrel{o}{\to }x$ and $f\left(x_{\alpha }\right)\stackrel{o}{\to }y$. It follows from the order continuity of $f$ and the uniqueness of the order limit, that $f\left(x_{\alpha }\right)\stackrel{o}{\to }f\left(x\right)=y$. Thus $\left(x,y\right)\in G_f$ and $G_f$ is order closed.

Proposition 2.2 If a subset $\mathcal{K}$ of a Riesz space E is order compact then it is order closed.

Proof. Let $\left\{x_{\alpha }\right\}$ a net in $\mathcal{K}$ such that $x_{\alpha }\stackrel{o}{\to }x\in E$. From the order compactness of $\mathcal{K}$ we infer that there exists a subnet $\left\{x_{\phi \left(\alpha \right)}\right\}$ of $\left\{x_{\alpha }\right\}$ such that $\left\{x_{\phi \left(\alpha \right)}\right\}\stackrel{o}{\to }a\in \mathcal{K}$. The uniqueness of the order limit of $\left\{x_{\phi \left(\alpha \right)}\right\}$ implies that

$x=a\in \mathcal{K}$. ◼

Proposition 2.3 Let ${\mathcal{K}}_1,{\mathcal{K}}_3,\cdots ,{\mathcal{K}}_n$ be order compact subsets of Riesz spaces $E_1,E_2,\cdots ,E_n$, respectively. Then ${\mathcal{K}}_1\times {\mathcal{K}}_2\times \cdots \times {\mathcal{K}}_n$ is order compact in the Riesz space $E_1\times E_2\times \cdots \times E_n$

Proof. Let ${\left\{\left(x_{\alpha 1},x_{\alpha 2},\cdots ,x_{\alpha n}\right)\right\}}_{\alpha }$ a net in ${\mathcal{K}}_1\times {\mathcal{K}}_2\times \cdots \times {\mathcal{K}}_n$. The order compactness of ${\mathcal{K}}_1$ implies that there exists a subnet ${\left\{x_{{\phi }_1\left(\alpha \right)1}\right\}}_{{\phi }_1\left(\alpha \right)}$ which is order convergent to $a_1$ in ${\mathcal{K}}_1$. The net ${\left\{x_{{\phi }_1\left(\alpha \right)2}\right\}}_{{\phi }_1\left(\alpha \right)}$ contains a subnet ${\left\{x_{{\phi }_2\circ {\phi }_1\left(\alpha \right)2}\right\}}_{{\phi }_2\circ {\phi }_1\left(\alpha \right)}$ which is order convergent to $a_2$ in ${\mathcal{K}}_2$. A finite induction gives that ${\left\{\left(x_{\alpha 1},x_{\alpha 2},\cdots ,x_{\alpha n}\right)\right\}}_{\alpha }$ admits a subnet

${\left\{\left(x_{{\phi }_n\circ {\phi }_{n-1}\circ \cdots \circ {\phi }_1\left(\alpha \right)1},x_{{\phi }_n\circ {\phi }_{n-1}\circ \cdots \circ {\phi }_1\left(\alpha \right)2},\cdots ,x_{{\phi }_n\circ {\phi }_{n-1}\circ \cdots \circ {\phi }_1\left(\alpha \right)n}\right)\right\}}_{{\phi }_n\circ {\phi }_{n-1}\circ \cdots \circ {\phi }_1\left(\alpha \right)}$

which is order convergent to $\left(a_1,a_2,\cdots ,a_n\right)$ in ${\mathcal{K}}_1\times {\mathcal{K}}_2\times \cdots \times {\mathcal{K}}_n$. ◼

Proposition 2.4 Let $\mathcal{K}$ be an order compact in an order complete Riesz space E such that $x\vee y\in \mathcal{K}$ and $x\wedge y\in \mathcal{K}$ for all $x,y\in \mathcal{K}$. Then $\mathcal{K}$ is order bounded.

Proof. It follows from Proposition 1.3 and Theorem 1.4 that $\mathcal{K}\times \mathcal{K}$ is order compact and the order continuous mappings $\left(x,y\right)\to x\vee y$, $\left(x,y\right)\to x\wedge y$ from $\mathcal{K}\times \mathcal{K}$ to $\mathcal{K}$ achieve their maximum and minimum respectively. Which means that $\mathcal{K}$ is order bounded. ◼

We will show that continuous order mappings preserve order compactness.

Theorem 2.2 Let $f:\mathcal{K}\to F$ be an order continuous function from an order compact $\mathcal{K}$ of Riesz E into a Riesz space F. Then $f\left(\mathcal{K}\right)$ is an order compact subset of F.

Proof. Let $\left\{y_{\alpha }\right\}$ be a net in $f\left(\mathcal{K}\right)$. There is $\left\{x_{\alpha }\right\}$ in $\mathcal{K}$ such that $f\left(x_{\alpha }\right)=y_{\alpha }$ for all $\alpha$. Since $\mathcal{K}$ is order compact, there exists a subnet $\left\{x_{\phi \left(\alpha \right)}\right\}$ which converges in order to $x\in \mathcal{K}$. It follows from the order continuity of $f$ that $\left\{y_{\phi \left(\alpha \right)}\right\}=\left\{f\left(x_{\phi \left(\alpha \right)}\right)\right\}\stackrel{o}{\to }y=f\left(x\right)\in f\left(\mathcal{K}\right)$. Thus $f\left(\mathcal{K}\right)$ is an order compact in F. ◼

A similar result is obtained for functions with values in a Riesz space when a stability condition is required (which is obviously verified in the case of the real line)

Theorem 2.3 Let $\mathcal{K}$ be an order compact subset of a Riesz space E. Let $f:\mathcal{K}\to F$ be an order continuous function defined on $\mathcal{K}$ into an order complete Riesz space F such that $x\vee y\in f\left(\mathcal{K}\right)$ and $x\wedge y\in f\left(\mathcal{K}\right)$ for all $x,y\in f\left(\mathcal{K}\right)$. Then $f$ achieves its supremum and infimum values.

Proof. According to Theorem 1.1 and Theorem 1.2, $f\left(\mathcal{K}\right)$ is order compact and order complete. So, to conclude the proof, it suffices to apply the definition 1.2. ◼

Theorem 2.4 Let $\mathcal{K}$ be an order compact subset of an order complete Riesz space E. Denote by $f_{\mathcal{K}}$ the mapping $x\to \inf \left\{\left|x-a\right|:a\in \mathcal{K}\right\}$ from E to $E_{+}$. Then $f_{\mathcal{K}}$ is well defined and satisfies:

(i) $\forall x\in E$, $x\in \mathcal{K}\iff f_{\mathcal{K}}\left(x\right)=0$

(ii) $\forall x,y\in E$, $\left|f_{\mathcal{K}}\left(x\right)-f_{\mathcal{K}}\left(y\right)\right|\le \left|x-y\right|$.

Proof. (i) Pick any $x$ in E, in view of ( [2] (a) Corollary 8.7, p. 318) we can conclude that the mapping $g_x$: $a\to \left|a-x\right|$ from $\mathcal{K}$ to $E_{+}$ satisfies

$\left|g_x\left(a\right)-g_x\left(b\right)\right|=\left|\left|x-a\right|-\left|x-b\right|\right|\le \left|a-b\right|$

for all $a,b\in \mathcal{K}$.

Now take ${\left\{a_{\alpha }\right\}}_{\alpha }$ a net in $\mathcal{K}$ with $a_{\alpha }\stackrel{o}{\to }a$. and $\left|a_{\alpha }-a\right|\le u_{\alpha }\downarrow 0$. Then $\left|g_x\left(a_{\alpha }\right)-g_x\left(a\right)\right|\le \left|a_{\alpha }-a\right|\le u_{\alpha }\downarrow 0$ and $g_x$ is order continuous. It follows from Theorem.1.4 that $g_x$ achieves its infimums values in $\mathcal{K}$. Then the function $f_{\mathcal{K}}$ is well defined and the proof of (i) is obviously obtained.

(ii) Let $x,y\in E$, for every $a\in \mathcal{K}$ we have we have:

$\begin{array}{c}\left|x-a\right|\le \left|x-a-y+a\right|+\left|y-a\right|\\ \le \left|x-y\right|+\left|y-a\right|\end{array}$

Taking the infimum on a, we get

$f_{\mathcal{K}}\left(x\right)\le \left|x-y\right|+f_{\mathcal{K}}\left(y\right)$ (1)

Interchanging $x$ and $y$ in the identity (1), we obtain (ii) ◼

One of the theorems that is applied in analysis is that of the fixed point. Here is a version of the order.

Theorem 2.5 Let $\mathcal{K}$ be a nonempty order compact subset of an Archimedean Riesz space E and let $f$ be a function from $\mathcal{K}$ to $\mathcal{K}$. If there is some $c\in \left]0,1\right[$ such that $\left|f\left(x\right)-f\left(y\right)\right|\le c\left|x-y\right|$, for all $x,y\in \mathcal{K}$, then $f$ has a unique fixed point.

Proof. Pick any point $x_0\in \mathcal{K}$, and then define a sequence $\left\{x_n\right\}$ inductively by the formula $x_{n+1}=f\left(x_n\right)$, $n=0,1,\cdots$. For $n\ge 1$ we have

$\left|x_{n+1}-x_n\right|=\left|f\left(x_n\right)-f\left(x_{n-1}\right)\right|\le c\left|x_n-x_{n-1}\right|$

and by induction, we see that $\left|x_{n+1}-x_n\right|\le c^n\left|x_1-x_0\right|$. Hence, for $m>n$ the triangle inequality yields

$\begin{array}{c}\left|x_m-x_n\right|\le \underset{k=n+1}{\overset{n}{{\displaystyle \sum }}}\left|x_k-x_{k-1}\right|\\ \le \underset{k=n+1}{\overset{n}{{\displaystyle \sum }}}{c}^{k-1}\left|x_1-x_0\right|\\ \le \left[\underset{k=n+1}{\overset{+\infty }{{\displaystyle \sum }}}{c}^{k-1}\right]\left|x_1-x_0\right|\\ \le \frac{c^n}{1-c}\left|x_1-x_0\right|.\end{array}$ (2)

Now, since $\mathcal{K}$ is order compact, there exists a subsequence $\left\{x_{n_k}\right\}$ which converges in order to some $a\in \mathcal{K}$. According to (2), for every $n_k>n$ we have $\left|x_{n_k}-x_n\right|\le \frac{c^n}{1-c}\left|x_1-x_0\right|$. By letting $n_k\to \infty$, it follows from ( [2] Lemma 8.15) that $\left|a-x_n\right|\le \frac{c^n}{1-c}\left|x_1-x_0\right|$. From this, the hypothesis on $c$, and the fact that $E$ is Archimedean, we conclude that $\left\{x_n\right\}\stackrel{o}{\to }a$ and consequently its subsequence $\left\{x_{n+1}\right\}\stackrel{o}{\to }a$. On the other hand, the order continuity of $f$ implies that $\left\{x_{n+1}\right\}=\left\{f\left(x_n\right)\right\}\stackrel{o}{\to }f\left(a\right)$. However, a sequence can have at most one order limit, then $a=f\left(a\right)$. That is, $a$ is also a fixed point of $f$.

Now if $b=f\left(b\right)$, then clearly $\left|a-b\right|=\left|f\left(a\right)-f\left(b\right)\right|\le c\left|a-b\right|$ and $\left(c-1\right)\left|a-b\right|\in -E_{+}\cap E_{+}=\left\{0\right\}$, so $b=a$. Hence, $a$ is the only fixed point of $f$.

◼

Recall that a linear topology $\tau$ on a Riesz space E is locally solid (and $\left(E,\tau \right)$ is called a locally solid Riesz space) if $\tau$ has a base at zero consisting of solid neighborhoods. A function $f:E\to F$ between two topological vector spaces is uniformly continuous if for each neighborhood W of zero in F there is a neighborhood V of zero in E such that $x-y\in V$ implies $f\left(x\right)-f\left(y\right)\in W$. It is well known that $\left(E,\tau \right)$ is a locally solid Riesz space if and only if the lattice operations are uniformly continuous with respect to $\tau$. A locally solid linear topology $\tau$ on a Riesz space E which is moreover locally convex is called locally convex-solid topology. In other words, $\tau$ has a base $ℬ$ at zero consisting of neighborhoods that are simultaneously closed, solid, and convex.

Let $\left(E,\tau \right)$ be a locally convex-solid Riesz space. The gauge $p_V$ of $V\in ℬ$ $p_V\left(x\right)=\text{inf}\left\{\lambda >0:x\in \lambda V\right\}$ is a lattice seminorm. With the notation above, we have:

Definition 3.1 A locally convex-solid topology $\tau$ is:

Theorem 3.1 Let $\mathcal{K}$ be an order compact subset of a locally convex-solid Riesz space $\left(E,\tau \right)$. If $\tau$ is order continuous then $\mathcal{K}$ is compact (in the topology sense).

Proof. According to ( [2] Theorem 2.31 p. 39), it is enough to show that every net of $\mathcal{K}$ has a $\tau$-convergent subnet. To this end, consider a net ${\left\{x_{\alpha }\right\}}_{\alpha }$ in $\mathcal{K}$. It follows from the order compactness of $\mathcal{K}$, that there is a subnet ${\left\{x_{\phi \left(\alpha \right)}\right\}}_{\phi \left(\alpha \right)}\stackrel{o}{\to }a\in \mathcal{K}$. So, there is a net ${\left\{u_{\phi \left(\alpha \right)}\right\}}_{\phi \left(\alpha \right)}$ in $E_{+}$ satisfying: $u_{\phi \left(\alpha \right)}\downarrow 0$ and $\left|x_{\phi \left(\alpha \right)}-a\right|\le u_{\phi \left(\alpha \right)}$. Since the locally convex-solid topology $\tau$ is order continuous, we get $p_V\left(x_{\phi \left(\alpha \right)}-a\right)\le p_V\left(u_{\phi \left(\alpha \right)}\right)$ and $p_V\left(u_{\phi \left(\alpha \right)}\right)\downarrow 0$ for all $V\in ℬ$. This shows that ${\left\{x_{\phi \left(\alpha \right)}\right\}}_{\phi \left(\alpha \right)}\stackrel{\tau }{\to }a$. This proves that $\mathcal{K}$ is topologically compact. ◼

An important special case of topological Riesz spaces is that of Banach lattices (which means that the Banach norm $\Vert \Vert$ is compatible with order: i.e., $\left|x\right|\le \left|y\right|$ in E implies $\Vert x\Vert \le \Vert y\Vert$). The norm is order continuous if $x_{\alpha }\downarrow 0$ implies $\Vert x_{\alpha }\Vert \downarrow 0$.

A simple consequence of Theorem 3.1 is the following.

Theorem 3.2 Let E be a Banach lattice with order continuous norm. Every order compact of E is compact (with respect to this norm).

In the Banach lattice literature, disjoint sequences are important. The following result combines them with order compactness.

Theorem 3.3 Let $\mathcal{K}$ be an order compact in a Banach lattice with order continuous norm such that $x\vee y\in \mathcal{K}$ and $x\wedge y\in \mathcal{K}$ for all $x,y\in \mathcal{K}$. If $0_E\notin \mathcal{K}$ then there is now disjoint sequence in $\mathcal{K}$.

Proof. If ${\left\{x_n\right\}}_n$ is a disjoint sequence in $\mathcal{K}$. From ( [2] (3) Theorem 9.22, p. 356) follows that E is order complete. Looking at Proposition1.4 and ( [2] (7) Theorem 9.22), we end up with ${\left\{x_n\right\}}_n\stackrel{\Vert \Vert }{\to }{0}_E$ and $0_E\in \mathcal{K}$. ◼

The following gives the converse of Theorem 3.1 in the case of a Banach lattice with the order continuous norm. In this situation, the two notions of compactness are identical.

Theorem 3.4 Let $\mathcal{K}$ be a compact subset of a Banach lattice $\left(E,\Vert \Vert \right)$ with order continuous norm. Then $\mathcal{K}$ is order compact.

Proof. Suppose that $\mathcal{K}$ is a compact of $\left(E,\Vert \Vert \right)$ and consider a net ${\left\{x_{\alpha }\right\}}_{\alpha }$ in $\mathcal{K}$. Then there is a subnet ${\left\{x_{\phi \left(\alpha \right)}\right\}}_{\phi \left(\alpha \right)}$ which satisfies $\left(x_{\phi \left(\alpha \right)}-a\right)\stackrel{\Vert \Vert }{\to }0$. Therefore, the fact that Banach lattices are local-solid-convex Hausdorff topological Riesz spaces ( [2] Theorem 8.41, p. 334) implies that $\left(\left|x_{\phi \left(\alpha \right)}-a\right|\right)\stackrel{\Vert \Vert }{\to }0$. We can assume without loss of generality that $\Vert \left(\left|x_{\phi \left(\alpha \right)}-a\right|\right)\Vert \downarrow 0$. According to ( [2] (2) Theorem 8.43), we can conclude that $\left|x_{\phi \left(\alpha \right)}-a\right|\downarrow 0$. Thus $x_{\phi \left(\alpha \right)}\stackrel{o}{\to }a$, as required. ◼

If the norm of a Banach lattice E is not order continuous then the two compactnesses are different in E. Indeed, there exists a net $\left\{x_{\alpha }\right\}\subset E$ satisfying $x_{\alpha }\downarrow 0$ but, $\Vert x_{\alpha }\Vert \nrightarrow 0$. Obviously, $\mathcal{K}=\left\{x_{\alpha }\right\}\cup \left\{0\right\}$ is order compact without being norm compact.

Theorem 3.5 Let $T:E\to F$ be a lattice homomorphism from a Banach lattice with order continuous norm into an Archimedean Riesz space. If $\mathcal{K}$ is a compact in E then $T\left(\mathcal{K}\right)$ is order compact in F.

Proof. Let $\mathcal{K}$ be a compact of E. It follows from Theorem 3.4 that $\mathcal{K}$ is order compact. Then the desired conclusion follows from Theorem 2.2. ◼

In functional analysis, the space $C\left(X\right)$ (X being a topological space) has nice properties and allows to achieve important results. In this section, we will establish order version of the Banach-Stone Theorem. ${\mathcal{C}}_n\left(\mathcal{K}\right)$ means the Riesz space, under pointwise algebraic operations and order, of order continuous functions from $\mathcal{K}$ into $ℝ$. Let $\mathcal{K}$ and $ℒ$ be order compacts in the Riesz spaces E and F, respectively. Now, let $\varphi :{\mathcal{C}}_n\left(\mathcal{K}\right)\to {\mathcal{C}}_n\left(ℒ\right)$ be a Riesz isomorphism satisfying the condition

(C): $0\notin f\left(\mathcal{K}\right)$ if and only if $0\notin \varphi \left(f\right)\left(ℒ\right)$ for each $f\in {\mathcal{C}}_n\left(\mathcal{K}\right)$.

We will show that $\mathcal{K}$ is Lattice homeomorphic to $ℒ$. The proof will be based upon the following order version of the Urysohn’s lemma.

The author thanks the referees for their helpful recommendations.