$um$-Topology in multi-normed vector lattices

TL;DR

This paper introduces the unbounded m-topology in multi-normed vector lattices, characterizes its metrizability and completeness, and explores conditions for compactness and properties like Lebesgue and Levi.

Contribution

It develops the theory of $um$-convergence and $um$-topology in MNVLs, providing new characterizations of metrizability, completeness, and compactness related to Lebesgue and Levi properties.

Findings

$um$-topology is metrizable iff $X$ has a countable topological orthogonal system.

Characterization of MNVLs with Lebesgue's and Levi's properties via $um$-completeness.

$um$-compactness of bounded, closed sets occurs iff the space is atomic with Lebesgue's and Levi's properties.

Abstract

Let $\mathcal{M}=\{m_\lambda\}_{\lambda\in\Lambda}$ be a separating family of lattice seminorms on a vector lattice $X$, then $(X,\mathcal{M})$ is called a multi-normed vector lattice (or MNVL). We write $x_\alpha \xrightarrow{\mathrm{m}} x$ if $m_\lambda(x_\alpha-x)\to 0$ for all $\lambda\in\Lambda$. A net $x_\alpha$ in an MNVL $X=(X,\mathcal{M})$ is said to be unbounded $m$-convergent (or $um$-convergent) to $x$ if $\lvert x_\alpha-x \rvert\wedge u \xrightarrow{\mathrm{m}} 0$ for all $u\in X_+$. $um$-Convergence generalizes $un$-convergence \cite{DOT,KMT} and $uaw$-convergence \cite{Zab}, and specializes $up$-convergence \cite{AEEM1} and $u\tau$-convergence \cite{DEM2}. $um$-Convergence is always topological, whose corresponding topology is called unbounded $m$-topology (or $um$-topology). We show that, for an $m$-complete metrizable MNVL $(X,\mathcal{M})$, the $um$-topology is metrizable iff $X$ has a countable topological orthogonal system. In terms of $um$-completeness, we present a characterization of MNVLs possessing both Lebesgue's and Levi's properties. Then, we characterize MNVLs possessing simultaneously the $\sigma$-Lebesgue and $\sigma$-Levi properties in terms of sequential $um$-completeness. Finally, we prove that any $m$-bounded and $um$-closed set is $um$-compact iff the space is atomic and has Lebesgue's and Levi's properties.

Equations

$$\text{strong unit}\Rightarrow \text{quasi-interior point}\Rightarrow \text{weak unit}.$$

$$V_{\epsilon ,u,\lambda }=\{x\in X:m_{\lambda }(|x|\wedge u)<\epsilon \}(\epsilon >0,0\ne u\in X_{+},\lambda \in \Lambda )$$

$$f_n=\{\begin{array}{ll}n& \text{if }x\in [0,\frac{1}{n}],\\ n^2(1-n)x+n^2& \text{if }x\in [\frac{1}{n},\frac{1}{n-1}],\\ 0& \text{if }x\in [\frac{1}{n-1},1].\end{array}$$

$$f_n\wedge \text{mathbbold}1=\{\begin{array}{ll}1& \text{if }x\in [0,\frac{n+1}{n^2}],\\ n^2(1-n)x+n^2& \text{if }x\in [\frac{n+1}{n^2},\frac{1}{n-1}],\\ 0& \text{if }x\in [\frac{1}{n-1},1].\end{array}$$

$${\rho }_{\mathcal{M}}(x,y)=\sum _{k=1}^{\infty }\frac{m_k(x-y)}{2^k(m_k(x-y)+1)}(x,y\in X).$$

$$d(x,y)=\sum _{k,n=1}^{\infty }\frac{1}{2^{k+n}}\cdot \frac{m_k(|x-y|\wedge e_n)}{1+m_k(|x-y|\wedge e_n)},$$

$$V_{\epsilon ,u_n,k}=\{x\in X:m_k(|x|\wedge u_n)<\epsilon \}.$$

$$e:=\sum _{n=1}^{\infty }\frac{u_n}{2^n{c}_n}.$$

$$\displaystyle\sum_{n=1}^{\infty}m_{k}\bigg{(}\dfrac{u_{n}}{2^{n}c_{n}}\bigg{)}=\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}m_{k}\bigg{(}\frac{u_{n}}{c_{n}}\bigg{)}$$

$$x_{\alpha }\wedge u_n\le 2^n{c}_n{x}_{\alpha }\wedge 2^n{c}_{n}e=2^n{c}_n(x_{\alpha }\wedge e)\stackrel{\mathrm{m}}{\to }0(\alpha \to \infty ),$$

$$m_{\lambda }(|x|)$$

$$m_{\lambda }(|x-a_{\alpha }|\wedge |x|)+\underset{a\in A}{\sup }{m}_{\lambda }(a)=m_{\lambda }(|x-a_{\alpha }|\wedge |x|)+M_{\lambda }.$$

$$|x_{\alpha }\wedge ke-x_{\beta }\wedge ke|\le |x_{\alpha }-x_{\beta }|\wedge ke,$$

$$m_{\lambda }(y_k)$$

$$\le m_{\lambda }(y_k-x_{\alpha }\wedge ke)+m_{\lambda }(x_{\alpha })$$

$$\le m_{\lambda }(y_k-x_{\alpha }\wedge ke)+\underset{\alpha }{\sup }{m}_{\lambda }(x_{\alpha }).$$

$$|x_{\alpha }\wedge ke-x_{\beta }\wedge ke|\wedge e$$

$$m_{\lambda }(|x_{\alpha }\wedge ke-x_{\beta }\wedge ke|\wedge e)\le m_{\lambda }(|x_{\alpha }-x_{\beta }|\wedge e).$$

$$m_{\lambda }(|x_{\alpha }\wedge ke-y_k|\wedge e)\le \underset{\beta }{\lim }{m}_{\lambda }(|x_{\alpha }-x_{\beta }|\wedge e).$$

$$m_{\lambda }(|x_{\alpha }-y|\wedge e)\le \underset{\beta }{\lim }{m}_{\lambda }(|x_{\alpha }-x_{\beta }|\wedge e).$$

$$\underset{\alpha }{\lim }{m}_{\lambda }(|x_{\alpha }-y|\wedge e)\le \underset{\alpha ,\beta }{\lim }{m}_{\lambda }(|x_{\alpha }-x_{\beta }|\wedge e)=0.$$

$$P_{\delta }{x}_{\alpha }\stackrel{\mathrm{um}}{\to }{z}_{\delta }\ge 0\text{leavevmode}\text{inleavevmodenobreak }{B}_{\delta }(\alpha \to \infty ).$$

$$s_m\wedge w-s_n\wedge w$$

$$=\sum _{k=n+1}^{m}w\wedge x_k=w\wedge \sum _{k=n+1}^m{x}_k=w\wedge |s_m-s_n|.$$

$$\sum_{n=1}^{\infty}m_{k}\bigg{(}\frac{x_{n}}{2^{n}}\bigg{)}=\sum_{n=1}^{\infty}\frac{1}{2^{n}}m_{k}(x_{n})\leq c_{k}\sum_{n=1}^{\infty}\frac{1}{2^{n}}<\infty,$$

$$|P_{\gamma }{x}_{\alpha }|\wedge b=P_{\gamma }|x_{\alpha }|\wedge b\le |x_{\alpha }|\wedge b\stackrel{\mathrm{m}}{\to }0,$$

$$x_{\alpha }\wedge u=\sum _{\gamma \in F}{P}_{\gamma }(x_{\alpha }\wedge u)+\sum _{\gamma \in \Gamma \setminus F}{P}_{\gamma }(x_{\alpha }\wedge u).$$

$$\sum _{\gamma \in F}{P}_{\gamma }(x_{\alpha }\wedge u)=\sum _{\gamma \in F}{P}_{\gamma }{x}_{\alpha }\wedge P_{\gamma }u\stackrel{\mathrm{m}}{\to }0.$$

$$\sum _{\gamma \in \Gamma \setminus F}{P}_{\gamma }(x_{\alpha }\wedge u)\le \frac{1}{n}\sum _{\gamma \in F}{P}_{\gamma }u+\sum _{\gamma \in \Gamma \setminus F}{P}_{\gamma }u,$$

$$y_{\delta }=\frac{1}{n}\sum _{\gamma \in F}{P}_{\gamma }u+\sum _{\gamma \in \Gamma \setminus F}{P}_{\gamma }u.$$

$$\frac{1}{n_1}\sum _{\gamma \in F_1}{P}_{\gamma }u\ge \frac{1}{n_2}\sum _{\gamma \in F_1}{P}_{\gamma }u.$$

$$\frac{1}{n_2}\sum _{\gamma \in F_2}{P}_{\gamma }u=\frac{1}{n_2}\sum _{\gamma \in F_1}{P}_{\gamma }u+\frac{1}{n_2}\sum _{\gamma \in F_2\setminus F_1}{P}_{\gamma }u.$$

Full text

1. Introduction and preliminaries

Unbounded convergences have attracted many researchers (see for instance [13, 9, 10, 8, 7, 25, 15, 3, 19, 17, 16, 11, 12, 21, 6]. Unbounded convergences are well-investigated in vector and normed lattices (cf. [7, 10, 15, 22, 24]). In the present paper, we also extend several previous results from [7, 10, 15, 22, 24, 25] to multi-normed setting. This work is a continuation of [6], in which unbounded topological convergence was studied in locally solid vector lattices.

For a net $x_{\alpha}$ in a vector lattice $X$, we write $x_{\alpha}\xrightarrow{\mathrm{o}}x$ if $x_{\alpha}$ converges to $x$ in order. That is, there is a net $y_{\beta}$, possibly over a different index set, such that $y_{\beta}\downarrow 0$ and, for every $\beta$, there exists $\alpha_{\beta}$ satisfying $\lvert x_{\alpha}-x\rvert\leqslant y_{\beta}$ whenever $\alpha\geqslant\alpha_{\beta}$. A net $x_{\alpha}$ in a vector lattice $X$ is unbounded order convergent (uo-convergent) to $x\in X$ if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\mathrm{o}}0$ for every $u\in X_{+}$. We write $x_{\alpha}\xrightarrow{\mathrm{uo}}x$ in this case. Clearly, order convergence implies $uo$-convergence and they coincide for order bounded nets. For a measure space $(\Omega,\Sigma,\mu)$ and a sequence $f_{n}$ in $L_{p}(\mu)$ ($0\leq p\leq\infty$), $f_{n}\xrightarrow{\mathrm{uo}}0$ iff $f_{n}\to 0$ almost everywhere [10, Rem.3.4]. It is known that almost everywhere convergence is not topological. Therefore, $uo$-convergence might not be topological in general. It was also shown recently that order convergence is never topological in infinite dimensional vector lattices [5].

Let $(X,\lVert\cdot\rVert)$ be a normed lattice. For a net $x_{\alpha}$ in $X$, we write $x_{\alpha}\xrightarrow{\lVert\cdot\rVert}x$ if $x_{\alpha}$ converges to $x$ in norm. We say that $x_{\alpha}$ unbounded norm converges to $x$ ($x_{\alpha}$ un-converges to $x$ or $x_{\alpha}\xrightarrow{\mathrm{un}}x$) if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\lVert\cdot\rVert}0$ for every $u\in X_{+}$. Clearly, norm convergence implies $un$-convergence. The $un$-convergence is topological, and the corresponding topology (which is known as un-topology) was investigated in [15]. A net $x_{\alpha}$ uaw-converges to $x$ if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\mathit{w}}0$ for all $u\in X_{+}$, where “$\mathit{w}$” stands for the weak convergence. Absolute weak convergence implies $uaw$-convergence. $uaw$-Convergence and $uaw$-topology were introduced and investigated in [25].

All topologies considered throughout this article are assumed to be Hausdorff. If a linear topology $\tau$ on a vector lattice $X$ has a base at zero consisting of solid sets, then the pair $(X,\tau)$ is called a locally solid vector lattice. Furthermore, if $\tau$ has base at zero consisting of convex-solid sets, then $(X,\tau)$ is called a locally convex-solid vector lattice. It is known that a linear topology $\tau$ on $X$ is locally convex-solid iff there exists a family $\mathcal{M}=\{m_{\lambda}\}_{\lambda\in\Lambda}$ of lattice seminorms that generates $\tau$ (cf. [1, Thm.2.25]). Moreover, for such $\mathcal{M}$, $x_{\alpha}\xrightarrow{\tau}x$ iff $m_{\lambda}(x_{\alpha}-x)\xrightarrow[\alpha]{}0$ in $\mathbb{R}$ for each $m_{\lambda}\in\mathcal{M}$. Since $\tau$ is Hausdorff then the family $\mathcal{M}$ is separating.

A subset $A$ in a topological vector space $(X,\tau)$ is called $\tau$-bounded if, for every $\tau$-neighborhood $V$ of zero, there exists $\lambda>0$ such that $A\subseteq\lambda V$. In the case when the topology $\tau$ is generated by a family $\{m_{\lambda}\}_{\lambda\in\Lambda}$ of seminorms, a subset $A$ of $X$ is $\tau$-bounded iff $\sup_{a\in A}m_{\lambda}(a)<\infty$ for all $\lambda\in\Lambda$.

Recall that a locally solid vector lattice $(X,\tau)$ is said to have the Lebesgue property if $x_{\alpha}\downarrow 0$ in $X$ implies $x_{\alpha}\xrightarrow{\tau}0$ or, equivalently, if $x_{\alpha}\xrightarrow{\mathrm{o}}0$ implies $x_{\alpha}\xrightarrow{\tau}0$; $(X,\tau)$ is said to have the $\sigma$-Lebesgue property if $x_{n}\downarrow 0$ in $X$ implies $x_{n}\xrightarrow{\tau}0$; and $\left(X,\tau\right)$ is said to have the pre-Lebesgue property if $0\leq x_{n}\uparrow\ \leq x$ implies only that $x_{n}$ is $\tau$-Cauchy. Finally, $\left(X,\tau\right)$ is said to have the Levi property if, when $0\leq x_{\alpha}\uparrow$ and $x_{\alpha}$ is $\tau$-bounded, then $x_{\alpha}\uparrow x$ for some $x\in X$; $(X,\tau)$ is said to have the $\sigma$-Levi property if $x_{n}$ has supremum in $X$ provided by $0\leq x_{n}\uparrow$ and by the $\tau$-boundedness of $x_{n}$, see [1, Def. 3.16].

2. Multi-Normed Vector Lattices

Let $(X,\tau)$ be a locally convex-solid vector lattice with an upward directed family $\mathcal{M}=\{m_{\lambda}\}_{\lambda\in\Lambda}$ of lattice seminorms generating $\tau$. Throughout this article, the pair $(X,\mathcal{M})$ will be referred to as a multi-normed vector lattice $($MNVL$)$. Also, $\tau$-convergence, $\tau$-Cauchy, $\tau$-complete, etc. will be denoted by $m$-convergence, $m$-Cauchy, $m$-complete, etc.

Let $X$ be a vector space, $E$ be a vector lattice, and $p:X\to E_{+}$ be a vector norm (i.e.  $\,p(x)=0\Leftrightarrow x=0$, $p(\lambda x)=|\lambda|p(x)$ for all $\lambda\in\mathbb{R}$, $x\in X$, and $p(x+y)\leq p(x)+p(y)$ for all $x,y\in X$), then $(X,p,E)$ is called a lattice-normed space, abbreviated as LNS, see [18]. If $X$ is a vector lattice, and the vector norm $p$ is monotone (i.e. $\lvert x\rvert\leq\lvert y\rvert\Rightarrow p(x)\leq p(y)$), then the triple $(X,p,E)$ is called a lattice-normed vector lattice, abbreviated as LNVL (cf. [3, 4]).

Given an LNS $(X,p,E)$. Recall that a net $x_{\alpha}$ in $X$ is said to be p-convergent to $x$ (see [3]) if $p(x_{\alpha}-x)\xrightarrow{\mathrm{o}}0$ in $E$. In this case, we write $x_{\alpha}\xrightarrow{\mathrm{p}}x$. A subset $A$ of $X$ is called p-bounded if there exists $e\in E$ such that $p(a)\leq e$ for all $a\in A$.

Proposition 1**.**

Every MNVL induces an LNVL. Moreover, for arbitrary nets, $p$-convergence in the induced LNVL implies $m$-convergence, and they coincide in the case of $p$-bounded nets.

Proof.

Let $(X,\mathcal{M})$ be an MNVL, then there is a separating family $\{m_{\lambda}\}_{\lambda\in\Lambda}$ of lattice seminorms on $X$. Let $E=\mathbb{R}^{\Lambda}$ be the vector lattice of all real-valued functions on $\Lambda$, and define $p:x\mapsto p_{x}$ from $X$ into $E_{+}$ such that $p_{x}[\lambda]\coloneqq m_{\lambda}(x)$.

It is clear that $p$ is a monotone vector norm on $X$. Therefore $(X,p,E)$ is an LNVL. Let $x_{\alpha}$ be a net in $X$. If $x_{\alpha}\xrightarrow{\mathrm{p}}0$, then ${p_{x}}_{\alpha}\xrightarrow{\mathrm{o}}0$ in $\mathbb{R}^{\Lambda}$, and so ${p_{x}}_{\alpha}[\lambda]\to 0$ or $m_{\lambda}(x_{\alpha})\to 0$ for all $\lambda\in\Lambda$. Hence $x_{\alpha}\xrightarrow{\mathrm{m}}0$.

Finally, assume a net $x_{\alpha}$ to be $p$-bounded. If $x_{\alpha}\xrightarrow{\mathrm{m}}0$, then $m_{\lambda}(x_{\alpha})\to 0$ or ${p_{x}}_{\alpha}[\lambda]\to 0$ for each $\lambda\in\Lambda$. Since $x_{\alpha}$ is $p$-bounded, then ${p_{x}}_{\alpha}\xrightarrow{\mathrm{o}}0$ in $\mathbb{R}^{\Lambda}$. That is $x_{\alpha}\xrightarrow{\mathrm{p}}0$. ∎

Let $X$ be a vector lattice. An element $0\neq e\in X_{+}$ is called a strong unit if the ideal $I_{e}$ generated by $e$ is $X$ or, equivalently, for every $x\geqslant 0$, there exists $n\in\mathbb{N}$ such that $x\leqslant ne$; a weak unit if the band $B_{e}$ generated by $e$ is $X$ or, equivalently, $x\wedge ne\uparrow x$ for every $x\in X_{+}$. If $(X,\tau)$ is a topological vector lattice, then $0\neq e\in X_{+}$ is called a quasi-interior point if the principal ideal $I_{e}$ is $\tau$-dense in $X$ (see Definition 6.1 in [20]). It is known that

[TABLE]

The following proposition characterizes quasi-interior points, and should be compared with [2, Thm.4.85].

Proposition 2**.**

Let $(X,\mathcal{M})$ be an MNVL, then the following statements are equivalent$:$

1. (1)

$e\in X_{+}$* is a quasi-interior point;* 2. (2)

for all $x\in X_{+}$, $x-x\wedge ne\xrightarrow{\mathrm{m}}0$ as $n\rightarrow\infty$; 3. (3)

$e$* is strictly positive on $X^{*}$, i.e., $0<f\in X^{*}$ implies $f(e)>0$, where $X^{*}$ denotes the topological dual of $X$.*

Proof.

(1)$\Rightarrow$(2) Suppose that $e$ is a quasi-interior point of $X$, then ${\overline{I_{e}}}^{m}=X$. Let $x\in X_{+}$. Then $x\in{\overline{I_{e}}}^{m}$, so there exists a net $x_{\alpha}$ in $I_{e}$ that $m$-converges to $x$. But $x_{\alpha}\xrightarrow{\mathrm{m}}x$ implies $\lvert x_{\alpha}\rvert\xrightarrow{\mathrm{m}}\lvert x\rvert=x$. Moreover, $x_{\alpha}\wedge x\xrightarrow{\mathrm{m}}x\wedge x=x$, and $x_{\alpha}\wedge x\leq x_{\alpha}$ implies that $x_{\alpha}\wedge x\in I$, because $I_{e}$ is an ideal. So we can assume also that $x_{\alpha}\leq x$. Hence, for any $x\in X_{+}$, there is a net $0\leq x_{\alpha}\in I_{e}$ and $x_{\alpha}\leq x$. Then $0\leq x_{\alpha}\wedge ne\leq x\wedge ne\leq x$ for all $n\in\mathbb{N}$. Now, take $\lambda\in\Lambda$, and let $\varepsilon>0$, then there is $\alpha_{\varepsilon}$ such that $m_{\lambda}(x-x_{\alpha_{\varepsilon}})<\varepsilon$. But $0\leq x_{\alpha_{\varepsilon}}\in I_{e}$, so $0\leq x_{\alpha_{\varepsilon}}\leq k_{\varepsilon}e$ for some $k_{\varepsilon}\in\mathbb{N}$. Since $0\leq x_{\alpha_{\varepsilon}}=x_{\alpha_{\varepsilon}}\wedge k_{\varepsilon}e\leq x\wedge k_{\varepsilon}e\leq x$, then $m_{\lambda}(x-x\wedge ne)\leq m_{\lambda}(x-x\wedge k_{\varepsilon}e)\leq m_{\lambda}(x-x_{\alpha}\wedge k_{\varepsilon}e)=m_{\lambda}(x-x_{\alpha_{\varepsilon}})<\varepsilon$ for all $n\geq k_{\varepsilon}$. Hence $m_{\lambda}(x-x\wedge ne)\to 0$ as $n\to\infty$. Since $\lambda\in\Lambda$ was chosen arbitrary, we get $x-x\wedge ne\xrightarrow{\mathrm{m}}0$.

The proofs of the implications (2)$\Rightarrow$(3), and (3)$\Rightarrow$(1) are similar to the proofs of the corresponding implications of Theorem 4.85 in [2].

∎

3. $um$-Topology

In this section we introduce the $um$-topology in a analogous manner to the $un$-topology [15] and $uaw$-topology [25]. First we define the $um$-convergence.

Definition 1**.**

Let $(X,\mathcal{M})$ be an MNVL, then a net $x_{\alpha}$ is said to be unbounded m-convergent to $x$, if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\mathrm{m}}0$ for all $u\in X_{+}$. In this case, we say $x_{\alpha}$ $um$-converges to $x$ and write $x_{\alpha}\xrightarrow{\mathrm{um}}x$.

Clearly, that $um$-convergence is a generalization of $un$-convergence. The following result generalizes [15, Cor.4.5].

Proposition 3**.**

If $(X,\mathcal{M})$ is an MNVL possessing the Lebesgue and Levi properties, and $x_{\alpha}\xrightarrow{\mathrm{um}}0$ in $X$, then $x_{\alpha}\xrightarrow{\mathrm{um}}0$ in $X^{**}$.

Proof.

It follows from Theorem 6.63 of [1] that $(X,\mathcal{M})$ is $m$-complete and $X$ is a band in $X^{**}$. Now, [1, Thm.2.22] shows that $X^{**}$ is Dedekind complete, and so $X$ is a projection band in $X^{**}$. The conclusion follows now from [6, Thm.3(3)]. ∎

In a similar way as in [7, Section 7], one can show that $\mathcal{N}_{0}$, the collection of all sets of the form

[TABLE]

forms a neighborhood base at zero for some Hausdorff locally solid topology $\tau$ such that, for any net $x_{\alpha}$ in $X$: $x_{\alpha}\xrightarrow{\mathrm{um}}0$ iff $x_{\alpha}\xrightarrow{\tau}0$. Thus, the $um$-convergence is topological, and we will refer to its topology as the um-topology.

Clearly, if $x_{\alpha}\xrightarrow{\mathrm{m}}0$, then $x_{\alpha}\xrightarrow{\mathrm{um}}0$, and so the $m$-topology, in general, is finer than $um$-topology. On the contrary to Theorem 2.3 in [15], the following example provides an MNVL which has a strong unit, yet the $m$-topology and $um$-topology do not agree.

Example 1**.**

Let $X=C[0,1]$. Let $\mathcal{A}:=\{[a,b]\subseteq[0,1]:a<b\}$. For $[a,b]\in\mathcal{A}$ and $f\in X$, let $m_{[a,b]}(f):=\frac{1}{b-a}\int_{a}^{b}\lvert f(t)\rvert dt$. Then $\mathcal{M}=\{m_{[a,b]}:[a,b]\in\mathcal{A}\}$ is a separating family of lattice seminorms on $X$. Thus, $(X,\mathcal{M})$ is an MNVL. For each $2\leq n\in\mathbb{N}$, let

[TABLE]

So we have

[TABLE]

Now, let $0<b\leq 1$, then there is $n_{0}\in\mathbb{N}$ such that $\frac{1}{n_{0}-1}<b$. So, for $n\geq n_{0}$, we have $\frac{1}{n-1}<b$, and so we get $m_{[0,b]}(f_{n})=\frac{1}{b}(1+\frac{1}{n-1})\to\frac{1}{b}\neq 0$ as $n\to\infty$. Thus, $f_{n}\not\xrightarrow{\mathrm{m}}0$. On the other hand, if $[a,b]\in\mathcal{A}$ then there is $n_{0}\in\mathbb{N}$ such that $\frac{1}{n_{0}-1}<b$ so, for $n\geq(n_{0}-1)$, we have $m_{[a,b]}(f_{n}\wedge\mathbbold{1})=\frac{1}{b-a}(\frac{n+1}{n^{2}}+\frac{1}{2n^{2}(n-1)})\to 0$ as $n\to\infty$. Since $\mathbbold{1}$ is a strong unit in $X$ then, by [6, Cor.5], $f_{n}\xrightarrow{\mathrm{um}}0$.

4. Metrizabililty of $um$-topology

The main result in this section is Proposition 4, which shows that the $um$-topology is metrizable iff the space has a countable topological orthogonal system.

It is well known (cf. [1, Thm.2.1]) that a topological vector space is metrizable iff it has a countable neighborhood base at zero. Furthermore, an MNVL $(X,\mathcal{M})$ is metrizable iff the $m$-topology is generated by a countable family of lattice seminorms, see [23, Theorem VII.8.2].

Notice that, in an MNVL $(X,\mathcal{M})$ with countable $\mathcal{M}=\{m_{k}\}_{k\in\mathbb{N}}$, an equivalent translation-invariant metric $\rho_{\mathcal{M}}$ can be constructed by the formula

[TABLE]

Since the function $t\to\frac{t}{t+1}$ is increasing on $[0,\infty)$, $|x|\leqslant|y|$ in $X$ implies that $\rho_{\mathcal{M}}(x,0)\leqslant\rho_{\mathcal{M}}(y,0)$.

Recall that a collection $\{e_{\gamma}\}_{\gamma\in\Gamma}$ of positive vectors in a vector lattice $X$ is called an orthogonal system if $e_{\gamma}\wedge e_{\gamma^{\prime}}=0$ for all $\gamma\neq\gamma^{\prime}$. If, moreover, $x\wedge e_{\gamma}=0$ for all $\gamma\in\Gamma$ implies $x=0$, then $\{e_{\gamma}\}_{\gamma\in\Gamma}$ is called a maximal orthogonal system. It follows from the Zorn’s lemma that every vector lattice containing at least one non-zero element has a maximal orthogonal system. Next, we recall the following notion.

Definition 2**.**

[6, Def.1]** Let $(X,\tau)$ be a topological vector lattice. An orthogonal system $Q=\{e_{\gamma}\}_{\gamma\in\Gamma}$ of non-zero elements in $X_{+}$ is said to be a topological orthogonal system, if the ideal $I_{Q}$ generated by $Q$ is $\tau$-dense in $X$.

A series $\sum_{i=1}^{\infty}x_{i}$ in a multi-normed space $(X,\mathcal{M})$ is called absolutely $m$-convergent if $\sum_{i=1}^{\infty}m_{\lambda}(x_{i})<\infty$ for all $\lambda\in\Lambda$; and the series is $m$-convergent, if the sequence $s_{n}\coloneqq\sum_{i=1}^{n}x_{i}$ of partial sums is $m$-convergent. The following lemma can be proven by combining a diagonal argument with the proof of [14, Prop. 3 in Section 3.3] and therefore we omit its proof.

Lemma 1**.**

A metrizable multi-normed space $(X,\mathcal{M})$ is $m$-complete iff every absolutely $m$-convergent series in $X$ is $m$-convergent.

The following result extends [15, Thm.3.2].

Proposition 4**.**

Let $(X,\mathcal{M})$ be a metrizable $m$-complete MNVL. Then the following conditions are equivalent$:$

$(i)$* $X$ has a countable topological orthogonal system;*

$(ii)$* the $um$-topology is metrizable;*

$(iii)$* $X$ has a quasi interior point.*

Proof.

Since $(X,\mathcal{M})$ is metrizable, we may suppose that $\mathcal{M}=\{m_{k}\}_{k\in\mathbb{N}}$ is countable and directed.

$(i)\Rightarrow(ii)$ It follows directly from [6, Prop.5]. Notice also that a metric $d_{um}$ of the $um$-topology can be constructed by the following formula:

[TABLE]

where $\{e_{n}\}_{n\in\mathbb{N}}$ is a countable topological orthogonal system for $X$.

$(ii)\Rightarrow(iii)$ Assume that the $um$-topology is generated by a metric $d_{um}$ on $X$. For each $n\in\mathbb{N}$, let $B_{um}(0,\frac{1}{n})=\{x\in X:d_{um}(x,0)<\frac{1}{n}\}$. Since the $um$-topology is metrizable, then, for each $n\in\mathbb{N}$, there are $k_{n}\in\mathbb{N},0<u_{n}\in X_{+}$, and $\varepsilon_{n}>0$ such that $V_{\varepsilon_{n},u_{n},k_{n}}\subseteq B_{um}(0,\frac{1}{n})$, where

[TABLE]

Notice that $\{V_{\varepsilon,u_{n},k}\}_{\varepsilon>0,n,k\in\mathbb{N}}$ is a base at zero of the $um$-topology on $X$.

Let $B_{m}(0,1)=\{x\in X:d_{m}(x,0)<1\}$, where $d_{m}$ is the metric generating the $m$-topology. There is a zero neighborhood $V$ in the $m$-topology such that $V\subseteq B_{m}(0,1)$. Since $V$ is absorbing, then, for every $n\in\mathbb{N}$, there is $c_{n}\geq 1$ such that $\frac{1}{c_{n}}u_{n}\in V$. Thus $\frac{1}{c_{n}}u_{n}\in V\subseteq B_{m}(0,1)$ for each $n\in\mathbb{N}$. Hence, the sequence $\frac{1}{c_{n}}u_{n}$ is $d_{m}$-bounded and so it is bounded with respect to the multi-norm $\mathcal{M}=\{m_{k}\}_{k\in\mathbb{N}}$. Let

[TABLE]

Fix $k\in\mathbb{N}$. Since the sequence $\frac{u_{n}}{c_{n}}$ is bounded with respect to $\mathcal{M}$, there exists $r_{k}\in\mathbb{R}_{+}$ such that $m_{k}(\frac{u_{n}}{c_{n}})\leq r_{k}<\infty$ for all $n\in\mathbb{N}$. Hence,

[TABLE]

Thus, the series $\sum_{n=1}^{\infty}\dfrac{u_{n}}{2^{n}c_{n}}$ is absolutely $m$-convergent. Since $X$ is $m$-complete, Lemma 1 assures that the series $\sum_{n=1}^{\infty}\dfrac{u_{n}}{2^{n}c_{n}}$ is $m$-convergent to some $e\in X$.

Now, we use Theorem 2 in [6] to show that $e$ is a quasi-interior point in $X$. Let $x_{\alpha}$ be a net in $X_{+}$ such that $x_{\alpha}\wedge e\xrightarrow{\mathrm{m}}0$. Our aim is to show that $x_{\alpha}\xrightarrow{\mathrm{um}}0$. Since

[TABLE]

then $x_{\alpha}\wedge u_{n}\xrightarrow{\mathrm{m}}0$ for all $n\in\mathbb{N}$. In particular, $m_{k_{n}}(x_{\alpha}\wedge u_{n})\to 0$. Thus, there exists $\alpha_{n}$ such that $m_{k_{n}}(x_{\alpha}\wedge u_{n})<\varepsilon_{n}$ for all $\alpha\geq\alpha_{n}$. That is $x_{\alpha}\in V_{\varepsilon_{n},u_{n},k_{n}}$ for all $\alpha\geq\alpha_{n}$, which implies $x_{\alpha}\in B_{um}(0,\frac{1}{n})$. Therefore, $x_{\alpha}\xrightarrow{\mathrm{d_{um}}}0$ and so $x_{\alpha}\xrightarrow{\mathrm{um}}0$. Hence, $e$ is a quasi interior point.

$(iii)\Rightarrow(i)$ It is trivial. ∎

Similar to [15, Prop.3.3], we have the following result.

Proposition 5**.**

Let $(X,\mathcal{M})$ be an $m$-complete metrizable MNVL. The $um$-topology is stronger than a metric topology iff $X$ has a weak unit.

Proof.

The sufficiency follows from [6, Prop.6].

For the necessity, suppose that the $um$-topology is stronger than the topology generated by a metric $d$. Let $e$ be as in (4.3) above. Assume $x\wedge e=0$. Since $e\geq\frac{u_{n}}{2^{n}c_{n}}$ for all $n\in\mathbb{N}$, we get $x\wedge\frac{u_{n}}{2^{n}c_{n}}=0$, and hence $x\wedge u_{n}=0$ for all $n$. Then $x\in V_{\varepsilon_{n},u_{n},k_{n}}$ for all $n$, and $x\in B(0,\frac{1}{n})=\{x\in X:d(x,0)<\frac{1}{n}\}$ for each $n\in\mathbb{N}$. So $x=0$, which means that $e$ is a weak unit. ∎

5. $um$-Completeness

A subset $A$ of an MNVL $(X,\mathcal{M})$ is said to be $($sequentially$)$ $um$-complete if, it is (sequentially) complete in the $um$-topology. In this section, we characterize $um$-complete subsets of $X$ in terms of the Lebesgue and Levi properties. We begin with the following technical lemma.

Lemma 2**.**

Let $(X,\mathcal{M})$ be an MNVL, and $A\subseteq X$ be $m$-bounded, then $\overline{A}^{um}$ is $m$-bounded.

Proof.

Given $\lambda\in\Lambda$, then $M_{\lambda}=\sup_{a\in A}m_{\lambda}(a)<\infty$. Let $x\in\overline{A}^{um}$, then there is a net $a_{\alpha}$ in $A$ such that $a_{\alpha}\xrightarrow{\mathrm{um}}x$. So $m_{\lambda}(\lvert a_{\alpha}-x\rvert\wedge u)\rightarrow 0$ for any $u\in X_{+}$. In particular,

[TABLE]

Letting $\alpha\rightarrow\infty$, we get $m_{\lambda}(x)=m_{\lambda}(|x|)\leq M_{\lambda}<\infty$ for all $x\in\overline{A}^{um}$. ∎

Theorem 1**.**

Let $(X,\mathcal{M})$ be an MNVL and let $A$ be an $m$-bounded and $um$-closed subset in $X$. If $X$ has the Lebesgue and Levi properties, then $A$ is $um$-complete.

Proof.

Suppose that $x_{\alpha}$ is $um$-Cauchy in $A$, then, without lost of generality, we may assume that $x_{\alpha}$ consists of positive elements.

Case (1): If $X$ has a weak unit $e$, then $e$ is a quasi-interior point, by the Lebesgue property of $X$ and Proposition 2. Note that, for each $k\in\mathbb{N}$,

[TABLE]

hence the net $(x_{\alpha}\wedge ke)_{\alpha}$ is $m$-Cauchy in $X$. Now, [1, Thm.6.63] assures that $X$ is $m$-complete, and so the net $(x_{\alpha}\wedge ke)_{\alpha}$ is $m$-convergent to some $y_{k}\in X$. Given $\lambda\in\Lambda$. Then

[TABLE]

Taking limit over $\alpha$, we get $m_{\lambda}(y_{k})\leq\sup_{\alpha}m_{\lambda}(x_{\alpha})<\infty$. Hence the sequence $y_{k}$ is $m$-bounded in $X$. Note also that $y_{k}$ is increasing in $X$, but $X$ has the Lebesgue and Levi properties, so, by [1, Thm.6.63], $y_{k}$ $m$-converges to some $y\in X$.

It remains to show that $y$ is the $um$-limit of $x_{\alpha}$. Given $\lambda\in\Lambda$. Note that, by Birkhoff’s inequality,

[TABLE]

Thus

[TABLE]

Taking limit over $\beta$, we get

[TABLE]

Now taking limit over $k$, we have

[TABLE]

Finally, as $x_{\alpha}$ is $um$-Cauchy, taking limit over $\alpha$, yields

[TABLE]

Thus, $x_{\alpha}\xrightarrow{\mathrm{um}}y$ and, since $A$ is $um$-closed, $y\in A$.

Case (2): If $X$ has no weak unit. Let $\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a maximal orthogonal system in $X$. Let $\Delta$ be the collection of all finite subsets of $\Gamma$. For each $\delta\in\Delta,\leavevmode\nobreak\ \delta=\{\gamma_{1},\gamma_{2},\dots,\gamma_{n}\}$, consider the band $B_{\delta}$ generated by $\{e_{\gamma_{1}},e_{\gamma_{1}},\dots,e_{\gamma_{n}}\}$. It follows from [1, Thm.3.24] that $B_{\delta}$ is a projection band. Then $B_{\delta}$ is an $m$-complete MNVL in its own right. Moreover, the $m$-topology restricted to $B_{\delta}$ possesses the Lebesgue and Levi properties. Note that $B_{\delta}$ has a weak unit, namely $e_{\gamma_{1}}+e_{\gamma_{2}}+\dots+e_{\gamma_{n}}$. Let $P_{\delta}$ be the band projection corresponding to $B_{\delta}$.

For $\delta\in\Delta$, since $x_{\alpha}$ is $um$-Cauchy in $X$ and $P_{\delta}$ is a band projection, then $P_{\delta}x_{\alpha}$ is $um$-Cauchy in $B_{\delta}$. Lemma 2 assures that $\overline{P_{\delta}(A)}^{um}$ is $m$-bounded in $B_{\delta}$. Thus, by Case (1), there is $z_{\delta}\in B_{\delta}$ such that

[TABLE]

Since $B_{\delta}$ is a projection band, then $P_{\delta}x_{\alpha}\xrightarrow{\mathrm{um}}z_{\delta}\geq 0\leavevmode\nobreak\ \text{in\leavevmode\nobreak\ }X\leavevmode\nobreak\ \leavevmode\nobreak\ \leavevmode\nobreak\ (\text{over\leavevmode\nobreak\ }\alpha)$. It is easy to see that $0\leq z_{\delta}\uparrow$, and $z_{\delta}$ is $m$-bounded. Since $X$ has the Lebesgue and Levi properties, it follows from [1, Thm.6.63], that there is $z\in X_{+}$ such that $z_{\delta}\xrightarrow{\mathrm{m}}z$, and so $z_{\delta}\uparrow z$. It remains to show that $x_{\alpha}\xrightarrow{\mathrm{um}}z$. The argument is similar to the proof of [13, Thm.4.7], and we leave it as an exercise. Since $A$ is $um$-closed, then $z\in A$ and so $A$ is $um$-complete. ∎

The following lemma and its proof are analogous to Lemma 1.2 in [15].

Lemma 3**.**

Let $(X,\mathcal{M})$ be an MNVL. If $x_{\alpha}$ is an increasing net in $X$ and $x_{\alpha}\xrightarrow{\mathrm{um}}x$, then $x_{\alpha}\uparrow x$ and $x_{\alpha}\xrightarrow{\mathrm{m}}x$.

Lemma 4**.**

Let $(X,\mathcal{M})$ be an MNVL possessing the pre-Lebesgue property. Let $x_{n}$ be a positive disjoint sequence which is not $m$-null. Put $s_{n}\coloneqq\sum_{k=1}^{n}x_{k}$. Then the sequence $s_{n}$ is $um$-Cauchy, which is not $um$-convergent.

Proof.

The sequence $s_{n}$ is monotone increasing and, since $x_{n}$ is not $m$-null, $s_{n}$ is not $m$-convergent. Hence, by Lemma 3, the sequence $s_{n}$ is not $um$-convergent. To show that $s_{n}$ is $um$-Cauchy, fix any $\varepsilon>0$ and take $0\neq w\in X_{+}$. Since $x_{n}$ is a positive disjoint sequence, we have $s_{n}\wedge w=\sum_{k=1}^{n}w\wedge x_{k}$. The sequence $s_{n}\wedge w$ is increasing and order bounded by $w$, hence it is $m$-Cauchy, by [1, Thm.3.22]. Let $\lambda\in\Lambda$. We can find $n_{\varepsilon_{\lambda}}$ such that $m_{\lambda}(s_{m}\wedge w-s_{n}\wedge w)<\varepsilon$ for all $m\geq n\geq n_{\varepsilon_{\lambda}}$. Observe that

[TABLE]

It follows $m_{\lambda}(\lvert s_{m}-s_{n}\rvert\wedge w)<\varepsilon$ for all $m\geq n\geq n_{\varepsilon_{\lambda}}$. But $\lambda\in\Lambda$ was chosen arbitrary. Hence $s_{n}$ is $um$-Cauchy. ∎

Next theorem generalizes Theorem 6.4 in [15].

Theorem 2**.**

Let $(X,\mathcal{M})$ be an $m$-complete MNVL with the pre-Lebesgue property. Then $X$ has the Lebesgue and Levi properties iff every $m$-bounded $um$-closed subset of $X$ is $um$-complete.

Proof.

The necessity follows directly from Theorem 1.

For the sufficiency, first notice that, in an $m$-complete MNVL, the pre-Lebesgue and Lebesgue properties coincide [1, Thm.3.24].

If $X$ does not have the Levi property then, by [1, Thm.6.63], there is a disjoint sequence $x_{n}\in X_{+}$, which is not $m$-null, such that its sequence of partial sums $s_{n}=\sum_{j=1}^{n}x_{j}$ is $m$-bounded. Let $A=\overline{\{s_{n}:n\in\mathbb{N}\}}^{um}$. By Lemma 2, we have that $A$ is $m$-bounded. By Lemma 4, the sequence $s_{n}$ is $um$-Cauchy in $X$ and so in $A$, in contrary with that the sequence $s_{n+1}-s_{n}=x_{n+1}$ is not $m$-null. ∎

Theorem 3**.**

Let $(X,\mathcal{M})$ be an $m$-complete metrizable MNVL, and let $A$ be an $m$-bounded sequentially $um$-closed subset of $X$. If $X$ has the $\sigma$-Lebesgue and $\sigma$-Levi properties then $A$ is sequentially $um$-complete. Moreover, the converse holds if, in addition, $X$ is Dedekind complete.

Proof.

Suppose $\mathcal{M}=\{m_{k}\}_{k\in\mathbb{N}}$. Let $0\leq x_{n}$ be a $um$-Cauchy sequence in $A$. Let $e\coloneqq\sum_{n=1}^{\infty}\frac{x_{n}}{2^{n}}$. For $k\in\mathbb{N}$,

[TABLE]

where $m_{k}(a)\leq c_{k}<\infty$ for all $a\in A$. Since $\sum_{n=1}^{\infty}\frac{x_{n}}{2^{n}}$ is absolutely $m$-convergent, then, by Lemma 1, $\sum_{n=1}^{\infty}\frac{x_{n}}{2^{n}}$ is $m$-convergent in $X$. Note that, $x_{n}\leq 2^{n}e$, so $x_{n}\in B_{e}$ for all $n\in\mathbb{N}$. Since $X$ has the Levi property, then $X$ is $\sigma$-order complete (see [1, Definition 3.16]). Thus $B_{e}$ is a projection band. Also $e$ is a weak unit in $B_{e}$. Then, by the same argument as in Theorem 1, we get that there is $x\in B_{e}$ such that $x_{n}\xrightarrow{\mathrm{um}}x$ in $B_{e}$ and so $x_{n}\xrightarrow{\mathrm{um}}x$ in $X$. Since $A$ is sequentially $um$-closed, we get $x\in A$. Thus $A$ is sequentially $um$-complete.

The converse follows from Proposition 8 in [6]. ∎

6. $um$-Compact sets

A subset $A$ of an MNVL $(X,\mathcal{M})$ is said to be $($sequentially$)$ $um$-compact if, it is (sequentially) compact in the $um$-topology. In this section, we characterize $um$-compact subsets of $X$ in terms of the Lebesgue and Levi properties. We begin with the following result which shows that $um$-compactness can be “localized” under certain conditions.

Theorem 4**.**

Let $(X,\mathcal{M})$ be an MNVL possessing the Lebesgue property. Let $\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a maximal orthogonal system. For each $\gamma\in\Gamma$, let $B_{\gamma}$ be the band generated by $e_{\gamma}$, and $P_{\gamma}$ be the corresponding band projection onto $B_{\gamma}$. Then $x_{\alpha}\xrightarrow{\mathrm{um}}0\text{\leavevmode\nobreak\ in \leavevmode\nobreak\ }X$ iff $P_{\gamma}x_{\alpha}\xrightarrow{\mathrm{um}}0$ in $B_{\gamma}$ for all $\gamma\in\Gamma$.

Proof.

For the forward implication, we assume that $x_{\alpha}\xrightarrow{\mathrm{um}}0$ in $X$. Let $b\in(B_{\gamma})_{+}$. Then

[TABLE]

that implies $P_{\gamma}x_{\alpha}\xrightarrow{\mathrm{um}}0$ in $B_{\gamma}$.

For the backward implication, without lost of generality, we may assume that $x_{\alpha}\geq 0$ for all $\alpha$. Let $u\in X_{+}$. Our aim is to show that $x_{\alpha}\wedge u\xrightarrow{\mathrm{m}}0$. It is known that $x_{\alpha}\wedge u=\sum_{\gamma\in\Gamma}P_{\gamma}(x_{\alpha}\wedge u)$. Let $F$ be a finite subset of $\Gamma$. Then

[TABLE]

Note

[TABLE]

We have to control the second term in (6.1).

[TABLE]

where $n\in\mathbb{N}$. Let $\mathscr{F}(\Gamma)$ be the collection of all finite subsets of $\Gamma$. Let $\Delta=\mathscr{F}(\Gamma)\times\mathbb{N}$. For each $\delta=(F,n)$, put

[TABLE]

We show that $y_{\delta}$ is decreasing. Let $\delta_{1}\leq\delta_{2}$ then $\delta_{1}=(F_{1},n_{1}),\delta_{2}=(F_{2},n_{2})$. Then $\delta_{1}\leq\delta_{2}$ iff $F_{1}\subseteq F_{2}$ and $n_{1}\leq n_{2}$. But $n_{1}\leq n_{2}$ iff $\frac{1}{n_{1}}\geq\frac{1}{n_{2}}$. So,

[TABLE]

Note also

[TABLE]

Since $F_{1}\subseteq F_{2}$, then $\Gamma\setminus F_{1}\supseteq\Gamma\setminus F_{2}$ and hence, $\sum_{\gamma\in\Gamma\setminus F_{1}}P_{\gamma}u\geq\sum_{\gamma\in\Gamma\setminus F_{2}}P_{\gamma}u$. Note, that

[TABLE]

Now,

[TABLE]

Combining (6.6) and (6.7), we get

[TABLE]

Adding (6.4) and (6.8), we get

[TABLE]

It follows from (6.5), that

[TABLE]

that is $y_{\delta_{1}}\geq y_{\delta_{2}}$. Next, we show $y_{\delta}\downarrow 0$. Assume $0\leq x\leq y_{\delta}$ for all $\delta\in\Delta$. Let $\gamma_{0}\in\Gamma$ be arbitrary and fix it. Let

[TABLE]

We apply $P_{\gamma_{0}}$ for the expression above, so $0\leq P_{\gamma_{0}}x\leq\frac{1}{n}P_{\gamma_{0}}u$ for all $n\in\mathbb{N}$, and so $P_{\gamma_{0}}x=0$. Since $\gamma_{0}\in\Gamma$ was chosen arbitrary, we get $P_{\gamma_{0}}x=0$ for all $\gamma\in\Gamma$. Hence, $x=0$ and so $y_{\delta}\downarrow 0$. Since $(X,\mathcal{M})$ has the Lebesgue property, we get $y_{\delta}\xrightarrow{\mathrm{m}}0$. Therefore, by (6.3),

[TABLE]

Hence (6.1), (6.2), and (6.9) imply $x_{\alpha}\wedge u\xrightarrow{\mathrm{m}}0$. ∎

The following result and its proof are similar to Theorem 7.1 in [15]. Therefore we omit its proof.

Theorem 5**.**

Let $(X,\mathcal{M})$ be an MNVL possessing the Lebesgue and Levi properties. Let $\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a maximal orthogonal system. Let $A$ be a $um$-closed $m$-bounded subset of $X$. Then $A$ is $um$-compact iff $P_{\gamma}(A)$ is $um$-compact in $B_{\gamma}$ for each $\gamma\in\Gamma$, where $B_{\gamma}$ is the band generated by $e_{\gamma}$ and $P_{\gamma}$ is the band projection corresponding to $B_{\gamma}$.

Theorem 6**.**

Let $(X,\mathcal{M})$ be an $MNVL$. The following are equivalent$:$

1. (1)

Any $m$-bounded and $um$-closed subset $A$ of $X$ is $um$-compact. 2. (2)

$X$* is an atomic vector lattice and $(X,\mathcal{M})$ has the Lebesgue and Levi properties.*

Proof.

(1) $\Rightarrow$ (2). Let $[a,b]$ be an order interval in $X$. For $x\in[a,b]$, we have $a\leq x\leq b$ and so $0\leq x-a\leq b-a$. Consider the order interval $[0,b-a]\subseteq X_{+}$. Clearly, $[0,b-a]$ is $m$-bounded and $um$-closed in $X$. By (1), the order interval $[0,b-a]$ is $um$-compact. Let $x_{\alpha}$ be a net in $[0,b-a]$. Then there is a subset $x_{\alpha_{\beta}}$ such that $x_{\alpha_{\beta}}\xrightarrow{\mathrm{um}}x$ in $[0,b-a]$. That is $\lvert x_{\alpha_{\beta}}-x\rvert\wedge u\xrightarrow{\mathrm{m}}0$ for all $u\in[0,b-a]$. Hence, $\lvert x_{\alpha_{\beta}}-x\rvert=\lvert x_{\alpha_{\beta}}-x\rvert\wedge(b-a)\xrightarrow{\mathrm{m}}0$. So, $x_{\alpha_{\beta}}\xrightarrow{\mathrm{m}}x$ in $[0,b-a]$. Thus, $[0,b-a]$ is $m$-compact. Consider the following shift operator $T_{a}:X\rightarrow X$ given by $T_{a}(x)\coloneqq x+a$. Clearly, $T_{a}$ is continuous, and so $T_{a}([0,b-a])=[a,b]$ is $m$-compact.

Since any order interval in $X$ is $m$-compact, then it follows from [1, Cor.6.57] that $X$ is atomic and has the Lebesgue property. It remains to show that $X$ has the Levi property. Suppose $0\leq x_{\alpha}\uparrow$ and is $m$-bounded. Let $A=\overline{\{x_{\alpha}\}}^{um}$. Then $A$ is $um$-closed and, by Lemma 2, $A$ is an $m$-bounded subset of $X$. Thus, $A$ is $um$-compact and so, there are a subnet $x_{\alpha_{\beta}}$ and $x\in A$ such that $x_{\alpha_{\beta}}\xrightarrow{\mathrm{um}}x$. Hence, by Lemma 3, $x_{\alpha_{\beta}}\uparrow x$, and so $x_{\alpha}\uparrow x$. Hence, $X$ has the Levi property.

(2) $\Rightarrow$(1). Let $A$ be an $m$-bounded and $um$-closed subset of $X$. We show that $A$ is $um$-compact. Since $X$ is atomic, there is a maximal orthogonal system $\{e_{\gamma}\}_{\gamma\in\Gamma}$ of atoms. For each $\gamma\in\Gamma$, let $P_{\gamma}$ be the band projection corresponding to $e_{\gamma}$. Clearly, $P_{\gamma}(A)$ is $m$-bounded. Now, by the same argument as in the proof of Theorem 7.1 in [15], we get that $P_{\gamma}(A)$ is $um$-closed in $\prod_{\gamma\in\Gamma}B_{\gamma}$, and so it is $um$-closed in $B_{\gamma}$. But $um$-closedness implies $m$-closedness. So $P_{\gamma}(A)$ is $m$-bounded and $m$-closed in $B_{\gamma}$ for all $\gamma\in\Gamma$. Since each $e_{\gamma}$ is an atom in $X$, then $B_{\gamma}=\operatorname{span}\{e_{\gamma}\}$ is a one-dimensional subspace. It follows from the Heine-Borel theorem that $P_{\gamma}(A)$ is $m$-compact in $B_{\gamma}$, and so it is $um$-compact in $B_{\gamma}$ for all $\gamma\in\Gamma$. Therefore, Theorem 5 implies that $A$ is $um$-compact in $X$. ∎

Proposition 6**.**

Let $A$ be a subset of an $m$-complete metrizable MNVL $(X,\mathcal{M})$.

1. (1)

If $X$ has a countable topological orthogonal system, then $A$ is sequentially $um$-compact iff $A$ is $um$-compact. 2. (2)

Suppose that $A$ is $m$-bounded, and $X$ has the Lebesgue property. If $A$ is $um$-compact, then $A$ is sequentially $um$-compact.

Proof.

(1). It follows immediately from Proposition 4.

(2). Let $x_{n}$ be a sequence in $A$. Find $e\in X_{+}$ such that $x_{n}$ is contained in $B_{e}$ (e.g., take $e=\sum_{n=1}^{\infty}\frac{\lvert x_{n}\rvert}{2^{n}}$). Since $A$ is $um$-compact, then $A\cap B_{e}$ is $um$-compact in $B_{e}$. Now, since $X$ is $m$-complete and has the Lebesgue property, then $B_{e}$ is also $m$-complete and has the Lebesgue property. Moreover, $e$ is a quasi-interior point of $B_{e}$. Hence, by Proposition 4, the $um$-topology on $B_{e}$ is metrizable, consequently, $A\cap B_{e}$ is sequentially $um$-compact in $B_{e}$. It follows that there is a subsequence $x_{n_{k}}$ that $um$-converges in $B_{e}$ to some $x\in A\cap B_{e}$. Since $B_{e}$ is a projection band, then [6, Thm.3(3)] implies $x_{n_{k}}\xrightarrow{\mathrm{um}}x$ in $X$. Thus, $A$ is sequentially $um$-compact. ∎