$u\tau$-Convergence in locally solid vector lattices

TL;DR

This paper introduces and studies $u\tau$-convergence in locally solid vector lattices, generalizing existing unbounded convergence concepts, and explores the properties and topological aspects of this new convergence mode.

Contribution

The paper defines $u\tau$-convergence, introduces the $u\tau$-topology, and investigates its metrizability and completeness, extending unbounded convergence theories.

Findings

$u\tau$-convergence generalizes unbounded norm and absolute weak convergence.

The $u\tau$-topology's metrizability and completeness are analyzed.

Properties of $u\tau$-convergence are established in locally solid vector lattices.

Abstract

Let $x_\alpha$ be a net in a locally solid vector lattice $(X,\tau)$; we say that $x_\alpha$ is unbounded $\tau$-convergent to a vector $x\in X$ if $\lvert x_\alpha-x \rvert\wedge w \xrightarrow{\tau} 0$ for all $w\in X_+$. In this paper, we study general properties of unbounded $\tau$-convergence (shortly, $u\tau$-convergence). $u\tau$-Convergence generalizes unbounded norm convergence and unbounded absolute weak convergence in normed lattices that have been investigated recently. Besides, we introduce $u\tau$-topology and study briefly metrizabililty and completeness of this topology.

Equations

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

$$\Vert x{\Vert }_u=\inf \{r>0:|x|\le ru\}\text{leavevmode}\text{leavevmode}\text{leavevmode}\text{leavevmode}(x\in X).$$

$$U_{V,w}:=\{x\in X:|x|\wedge w\in V\}.$$

$${\mathcal{N}}_{u\tau }:=\{U_{V,w}:V\in {\mathcal{N}}_{\tau },0\ne w\in X_{+}\}$$

$$U_{V,w}=\{x\in X:|x|\wedge w\in V\}.$$

$$(\alpha y)(t)\wedge w(t)\le x(t)\wedge w(t).$$

$$V_{\epsilon ,w,j}=X\cap {\hat{V}}_{\epsilon ,w,j}\subseteq X\cap [-w,w{]}_{\hat{X}}=[-w,w]\subseteq X.$$

$${\hat{\rho }}_j(x_{{\alpha }_{\epsilon }}-\hat{x})<\epsilon .$$

$${\hat{\rho }}_j(x_{{\alpha }_{\epsilon }}-ne\wedge x_{{\alpha }_{\epsilon }})={\rho }_j(x_{{\alpha }_{\epsilon }}-ne\wedge x_{{\alpha }_{\epsilon }})<\epsilon \text{leavevmode}\text{leavevmode}\text{leavevmode}\text{leavevmode}(\forall n⩾n_{\epsilon }).$$

$${\hat{\rho }}_j(\hat{x}-\hat{x}\wedge ne)\le \epsilon +\epsilon +{\hat{\rho }}_j(x_{{\alpha }_{\epsilon }}-\hat{x})\le 3\epsilon .$$

$$\displaystyle d(s_{n},s_{m})=d(s_{n}-s_{m},0)=d\Big{(}\sum_{k=m+1}^{n}ky_{\beta_{k}},0\Big{)}$$

$$\le \sum _{k=m+1}^n\frac{1}{2^k}\to 0,\text{ as }n,m\to \infty .$$

$$k|x_{\alpha }|⩽k{y}_{{\beta }_k}⩽u(\forall \alpha ⩾{\alpha }_{{\beta }_k})$$

$$x_{n,H}:=\sum _{\gamma \in H}x\wedge n{e}_{\gamma }.$$

$$x_{n,H}\le x\text{for all}(n,H)\in \mathbb{N}\times \mathcal{F}(\Gamma ).$$

$$x_{n,H}=\sum _{\gamma \in H}x\wedge n{e}_{\gamma }\le n\sum _{\gamma \in H}{e}_{\gamma },$$

$$x_{\epsilon }\le n_{\epsilon }\sum _{\gamma \in H_{\epsilon }}{e}_{\gamma }.$$

$$w:=x\wedge \sum _{\gamma \in H_{\epsilon }}{n}_{\epsilon }{e}_{\gamma }.$$

$$0\le y_{\gamma }\le n_{\epsilon }{e}_{\gamma }$$

$$w=\sum _{\gamma \in H_{\epsilon }}{y}_{\gamma }.$$

$$y_{\gamma }\le x\text{leavevmode}\text{leavevmode}\text{leavevmode}\text{leavevmode}(\forall \gamma \in H_{\epsilon }).$$

$$w=\sum _{\gamma \in H_{\epsilon }}{y}_{\gamma }\le \sum _{\gamma \in H_{\epsilon }}x\wedge n_{\epsilon }{e}_{\gamma }=x_{n_{\epsilon },H_{\epsilon }}.$$

$$0\le x_{\epsilon }\le w.$$

$$|x_{\alpha }|\wedge u$$

$$\le |x_{\alpha }|\wedge (u-u_{n,H})+|x_{\alpha }|\wedge u_{n,H}$$

$$\le (u-u_{n,H})+|x_{\alpha }|\wedge \sum _{\gamma \in H}u\wedge n{e}_{\gamma }$$

$$\le (u-u_{n,H})+|x_{\alpha }|\wedge \sum _{\gamma \in H}n{e}_{\gamma }$$

$$\displaystyle\leq(u-u_{n,H})+n\big{(}\lvert x_{\alpha}\rvert\wedge\sum\limits_{\gamma\in H}e_{\gamma}\big{)}$$

$$=(u-u_{n,H})+n\sum _{\gamma \in H}|x_{\alpha }|\wedge e_{\gamma }.$$

$${\rho }_j(u-u_{n_{\epsilon },H_{\epsilon }})<\epsilon .$$

$$\rho_{j}\big{(}n_{\varepsilon}(e_{\gamma}\wedge\lvert x_{\alpha}\rvert)\big{)}<\frac{\varepsilon}{\lvert H_{\varepsilon}\rvert}\ \ \ \ (\forall\alpha\geq\alpha_{\varepsilon,H_{\varepsilon}},\ \forall\gamma\in H_{\varepsilon}).$$

$${\rho }_j(|x_{\alpha }|\wedge u)$$

$$\displaystyle\leq\varepsilon+\sum\limits_{\gamma\in H_{\varepsilon}}\rho_{j}\big{(}n_{\varepsilon}(e_{\gamma}\wedge\lvert x_{\alpha}\rvert)\big{)}<\varepsilon+\sum\limits_{\gamma\in H_{\varepsilon}}\frac{\varepsilon}{\lvert H_{\varepsilon}\rvert}=2\varepsilon,$$

$$\mathcal{R}:=\{{\rho }_j:{\rho }_j((x_n{)}_{n\in \mathbb{N}}):=|x_j|\}$$

Full text

1. Introduction and preliminaries

The subject of “unbounded convergence” has attracted many researchers [25, 23, 11, 13, 9, 8, 27, 15, 5, 17, 16, 12, 22]. It is well-investigated in vector lattices and normed lattices [11, 14, 13, 27]. In the present paper, we study unbounded convergence in locally solid vector lattices. Results in this article extend previous works [8, 13, 15, 27].

For a net $x_{\alpha}$ in a vector lattice $X$, we write $x_{\alpha}\xrightarrow{\mathit{o}}x$, if $x_{\alpha}$ converges to $x$ in order. This means that 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 $x_{\alpha}$ is unbounded order convergent to a vector $x\in X$ if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\mathit{o}}0$ for every $u\in X_{+}$. We write $x_{\alpha}\xrightarrow{\mathit{uo}}x$ and say that $x_{\alpha}$ uo-converges to $x$. Clearly, order convergence implies $uo$-convergence and they coincide for order bounded nets. For a measure space $(\Omega,\Sigma,\mu)$ and for a sequence $f_{n}$ in $L_{p}(\mu)$ ($0\leq p\leq\infty$), $f_{n}\xrightarrow{\mathit{uo}}0$ iff $f_{n}\to 0$ almost everywhere (cf. [13, Rem. 3.4]). It is well known that almost everywhere convergence is not topological in general [18]. Therefore, the $uo$-convergence might not be topological. Quite recently, it has been shown that order convergence is never topological in infinite dimensional vector lattices [7].

For a net $x_{\alpha}$ in a normed lattice $(X,\lVert\cdot\rVert)$, 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 $x\in X$ (or $x_{\alpha}$ un-converges to $x$) if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\lVert\cdot\rVert}0$ for every $u\in X_{+}$. We write $x_{\alpha}\xrightarrow{\mathit{un}}x$. 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 $x_{\alpha}$ is unbounded absolute weak convergent to $x\in X$ (or $x_{\alpha}$ $uaw$-converges to $x$) if $\lvert x_{\alpha}-x\rvert\wedge u\xrightarrow{\mathit{w}}0$ for all $u\in X_{+}$, where “$\mathit{w}$” refers the weak convergence. We write $x_{\alpha}\xrightarrow{\mathit{uaw}}x$. Absolute weak convergence implies $uaw$-convergence. The notions of $\mathit{uaw}$-convergence and $uaw$-topology were introduced in [27].

If $X$ is a vector lattice, and $\tau$ is a linear topology on $X$ that has a base at zero consisting of solid sets, then the pair $(X,\tau)$ is called a locally solid vector lattice. It should be noted that all topologies considered throughout this article are assumed to be Hausdorff. It follows from [2, Thm. 2.28] that a linear topology $\tau$ on a vector lattice $X$ is locally solid iff it is generated by a family $\{\rho_{j}\}_{j\in J}$ of Riesz pseudonorms. Moreover, if a family of Riesz pseudonorms generates a locally solid topology $\tau$ on a vector lattice $X$, then $x_{\alpha}\xrightarrow{\tau}x$ in $X$ iff $\rho_{j}(x_{\alpha}-x)\xrightarrow[\alpha]{}0$ in $\mathbb{R}$ for each $j\in J$. Since $X$ is Hausdorff, then the family $\{\rho_{j}\}_{j\in J}$ of Riesz pseudonorms is separating; i.e., if $\rho_{j}(x)=0$ for all $j\in J$, then $x=0$. In this article, unless otherwise, the pair $(X,\tau)$ refers to as a locally solid vector lattice.

A subset $A$ in a topological vector space $(X,\tau)$ is called topologically bounded (or simply $\tau$-bounded) if, for every $\tau$-neighborhood $V$ of zero, there exists some $\lambda>0$ such that $A\subseteq\lambda V$. If $\rho$ is a Riesz pseudonorm on a vector lattice $X$ and $x\in X$, then $\frac{1}{n}\rho(x)\leq\rho(\frac{1}{n}x)$ for all $n\in\mathbb{N}$. Indeed, if $n\in\mathbb{N}$ then $\rho(x)=\rho(n\frac{1}{n}x)\leq n\rho(\frac{1}{n}x)$. The following standard fact is included for the sake of completeness.

Proposition 1**.**

Let $(X,\tau)$ be a locally solid vector lattice with a family of a Riesz pseudonorms $\{\rho_{j}\}_{j\in J}$ that generates the topology $\tau$. If a subset $A$ of $X$ is $\tau$-bounded then $\rho_{j}(A)$ is bounded in $\mathbb{R}$ for any $j\in J$.

Proof.

Let $A\subseteq X$ be $\tau$-bounded and $j\in J$. Put $V:=\{x\in X:\rho_{j}(x)<1\}$. Clearly, $V$ is a neighborhood of zero in $X$. Since $A$ is $\tau$-bounded, there is $\lambda>0$ satisfying $A\subseteq\lambda V$. Thus $\rho_{j}(\frac{1}{\lambda}a)\leq 1$ for all $a\in A$. There exists $n\in\mathbb{N}$ with $n>\lambda$. Now, $\frac{1}{n}\rho_{j}(a)\leq\rho_{j}(\frac{1}{n}a)\leq\rho_{j}(\frac{1}{\lambda}a)\leq 1$ for all $a\in A$. Hence, $\sup_{a\in A}\rho_{j}(a)\leq n<\infty$. ∎

Next, we discuss the converse of the proposition above.

Let $\{\rho_{j}\}_{j\in J}$ be a family of Riesz pseudonorms for a locally solid vector lattice $(X,\tau)$. For $j\in J$, let $\tilde{\rho}_{j}:=\frac{\rho_{j}}{1+\rho_{j}}$. Then $\tilde{\rho}_{j}$ is a Riesz pseudonorm on $X$. Moreover, the family $(\tilde{\rho}_{j})_{j\in J}$ generates the topology $\tau$ on $X$. Clearly, $\tilde{\rho}_{j}(A)\leq 1$ for any subset $A$ of $X$, but still we might have a subset that is not $\tau$-bounded.

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 $x_{\alpha}\xrightarrow{\mathit{o}}0$ implies $x_{\alpha}\xrightarrow{\tau}0$; and $(X,\tau)$ is said to have the $\sigma$-Lebesgue property if $x_{n}\downarrow 0$ in $X$ implies $x_{n}\xrightarrow{\tau}0$. Finally, $(X,\tau)$ is said to have the Levi property if $0\leq x_{\alpha}\uparrow$ and the net $x_{\alpha}$ is $\tau$-bounded, then $x_{\alpha}$ has the supremum in $X$; and $(X,\tau)$ is said to have the $\sigma$-Levi property if $0\leq x_{n}\uparrow$ and $x_{n}$ is $\tau$-bounded, then $x_{n}$ has supremum in $X$, see [2, Def. 3.16].

Let $X$ be a vector lattice, and take $0\neq u\in X_{+}$. Then a net $x_{\alpha}$ in $X$ is said to be u-uniformly convergent to a vector $x\in X$ if, for each $\varepsilon>0$, there exists some $\alpha_{\varepsilon}$ such that $\lvert x_{\alpha}-x\rvert\leq\varepsilon u$ holds for all $\alpha\geqslant\alpha_{\varepsilon}$; and $x_{\alpha}$ is said to be u-uniformly Cauchy if, for each $\varepsilon>0$, there exists some $\alpha_{\varepsilon}$ such that, for all $\alpha,\alpha^{\prime}\geqslant\alpha_{\varepsilon}$, we have $\lvert x_{\alpha}-x_{\alpha^{\prime}}\rvert\leq\varepsilon u$. A vector lattice $X$ is said to be u-uniformly complete if every $u$-uniformly Cauchy sequence in $X$ is $u$-uniformly convergent; and $X$ is said to be uniformly complete if $X$ is $u$-uniformly complete for each $0\neq u\in X_{+}$.

Let $X$ be a vector lattice. An element $0\neq e\in X_{+}$ is called a strong unit if $I_{e}=X$ (equivalently, for every $x\geqslant 0$, there exists $n\in\mathbb{N}$ such that $x\leqslant ne$), and $0\neq e\in X_{+}$ is called a weak unit if $B_{e}=X$ (equivalently, $x\wedge ne\uparrow x$ for every $x\in X_{+}$). Here $B_{e}$ denotes the band generated by $e$. 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 $X$ [20, Def. II.6.1]. It is known that

[TABLE]

Recall that a Banach lattice $X$ is called an $AM$-space if $\lVert x\vee y\rVert=\max\{\lVert x\rVert,\lVert y\rVert\}$ for all $x,y\in X$ with $x\wedge y=0$.

Let $(X,\tau)$ be a sequentially complete locally solid vector lattice. Then it follows from the proof of [4, Cor. 2.59] that it is uniformly complete. So, for each $0\neq u\in X_{+}$, let $I_{u}$ be the ideal generated by $u$ and $\lVert\cdot\rVert_{u}$ be the norm on $I_{u}$ given by

[TABLE]

Then, by [4, Thm. 2.58], the pair $(I_{u},\lVert.\rVert_{u})$ is a Banach lattice. Now Theorem 3.4 in [1] implies that $(I_{u},\lVert\cdot\rVert_{u})$ is an $AM$-space with a strong unit $u$, and then, by [1, Thm. 3.6], it is lattice isometric (uniquely, up to a homeomorphism) to $C(K)$ for some compact Hausdorff space $K$ in such a way, that the strong unit $u$ is identified with the constant function $\mathbbold{1}$ on $K$.

For unexplained terminologies and notions we refer to [2, 3].

2. Unbounded $\tau$-convergence

Suppose $(X,\tau)$ is a locally solid vector lattice. Let $x_{\alpha}$ be a net in $X$. We say that $x_{\alpha}$ is unbounded $\tau$-convergent to $x\in X$ if, for any $w\in X_{+}$, we have $\lvert x_{\alpha}-x\rvert\wedge w\xrightarrow{\tau}0$. In this case, we write $x_{\alpha}\xrightarrow{\mathit{u}\tau}x$ and say that $x_{\alpha}$ $u\tau$-converges to $x$. Obviously, if $x_{\alpha}\xrightarrow{\tau}x$ then $x_{\alpha}\xrightarrow{\mathit{u}\tau}x$. The converse holds if the net $x_{\alpha}$ is order bounded. Note also that $u\tau$-convergence respects linear and lattice operations. It is clear that $u\tau$-convergence is a generalization of $un$-convergence [8, 15] and, of $uaw$-convergence [27].

Let $\mathcal{N}_{\tau}$ be a neighborhood base at zero consisting of solid sets for $(X,\tau)$. For each $0\neq w\in X_{+}$ and $V\in\mathcal{N}_{\tau}$, let

[TABLE]

It can be easily shown that the collection

[TABLE]

forms a neighborhood base at zero for a locally solid topology; we call it $u\tau$-topology, where $u$ refers to as unbounded. Moreover, $x_{\alpha}\xrightarrow{\mathit{{u\tau}}}0$ iff $x_{\alpha}\rightarrow 0$ with respect to $u\tau$-topology. Indeed, suppose $x_{\alpha}\xrightarrow{\mathit{{u\tau}}}0$. Given a neighborhood $U_{V,w}\in\mathcal{N}_{u\tau}$. Then there are $0\neq w\in X_{+}$ and $V\in\mathcal{N}_{\tau}$ such that

[TABLE]

Now, $x_{\alpha}\xrightarrow{\mathit{{u\tau}}}0$ implies $\lvert x_{\alpha}\rvert\wedge w\xrightarrow{\tau}0$. So, there is $\alpha_{0}$ such that, for all $\alpha\geq\alpha_{0}$, we have $\lvert x_{\alpha}\rvert\wedge w\in V$. That is $x_{\alpha}\in U_{V,w}$ for all $\alpha\geq\alpha_{0}$. Thus, $x_{\alpha}\rightarrow 0$ in the $u\tau$-topology.

Conversely, assume $x_{\alpha}\to 0$ in the $u\tau$-topology. Given $0\neq w\in X_{+}$ and $V\in\mathcal{N_{\tau}}$. Then, $U_{V,w}$ is a zero neighborhood in the $u\tau$-topology. So, there is $\alpha^{\prime}$ such that $x_{\alpha}\in U_{V,w}$ for all $\alpha\geq\alpha^{\prime}$. That is, $\lvert x_{\alpha}\rvert\wedge w\in V$ for all $\alpha\geq\alpha^{\prime}$. Thus, $\lvert x_{\alpha}\rvert\wedge w\xrightarrow{\tau}0$ or $x_{\alpha}\xrightarrow{\mathit{{u\tau}}}0$. The locally solid $u\tau$-topology will be referred to as unbounded $\tau$-topology.

The neighborhood base at zero for the $u\tau$-topology on $X$ has an equivalent representation in terms of a family $(\rho_{j})_{j\in J}$ of Riesz pseudonorms that generates the topology $\tau$. For $\varepsilon>0$, $j\in J$, and $0\neq w\in X_{+}$, let $V_{\varepsilon,w,j}\coloneqq\{x\in X:\rho_{j}(\lvert x\rvert\wedge w)<\varepsilon\}$. Clearly, the collection $\{V_{\varepsilon,w,j}:\varepsilon>0,0\neq w\in X_{+},j\in J\}$ generates the $u\tau$-topology.

It is known that the topology of any linear topological space can be derived from a unique translation-invariant uniformity, i.e., any linear topological space is uniformisable (cf. [21, Thm. 1.4]). It follows from [10, Thm. 8.1.20] that any linear topological space is completely regular. In particular, the unbounded $\tau$-convergence is completely regular.

Since $x_{\alpha}\xrightarrow{\tau}0$ implies $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$, then the $\tau$-topology in general is finer than $u\tau$-topology. The next result should be compared with [15, Lm. 2.1].

Lemma 1**.**

Let $(X,\tau)$ be a sequentially complete locally solid vector lattice, where $\tau$ is generated by a family $(\rho_{j})_{j\in J}$ of Riesz pseudonorms. Let $\varepsilon>0$, $j\in J$, and $0\neq w\in X_{+}$. Then either $V_{\varepsilon,w,j}$ is contained in $[-w,w]$, or it contains a non-trivial ideal.

Proof.

Suppose that $V_{\varepsilon,w,j}$ is not contained in $[-w,w]$. Then there exists $x\in V_{\varepsilon,w,j}$ such that $x\not\in[-w,w]$. Replacing $x$ with $\lvert x\rvert$, we may assume $x>0$. Since $x\not\in[-w,w]$, then $y=(x-w)^{+}>0$. Now, letting $z=x\vee w$, we have that the ideal $I_{z}$ generated by $z$, is lattice and norm isomorphic to $C(K)$ for some compact and Hausdorff space $K$, where $z$ corresponds to the constant function $\mathbbm{1}$. Also $x$, $y$, and $w$ in $I_{z}$ correspond to $x(t)$, $y(t)$, and $w(t)$ in $C(K)$ respectively.

Our aim is to show that for all $\alpha\geq 0$ and $t\in K$, we have

[TABLE]

For this, note that $y(t)=(x-w)^{+}(t)=(x-w)(t)\vee 0$.

Let $t\in K$ be arbitrary.

• •

Case (1): If $(x-w)(t)>0$, then $x(t)\wedge w(t)=w(t)\geq(\alpha y)(t)\wedge w(t)$ for all $\alpha\geq 0$, as desired.

• •

Case (2): If $(x-w)(t)<0$, then $(\alpha y)(t)\wedge w(t)\leq(\alpha y)(t)=\alpha(x-w)(t)\vee 0=0\leq x(t)\wedge w(t)$, as desired.

Hence, for all $\alpha\geq 0$ and $t\in K$, we have $(\alpha w)(t)\wedge w(t)\leq x(t)\wedge w(t)$ and so $(\alpha y)\wedge w\leq x\wedge w$ for all $\alpha\geq 0$ . Note, that $\alpha y,w,x\in X_{+}$. Thus $\rho_{j}(\left|\alpha y\right|\wedge w)\leq\rho_{j}(\left|x\right|\wedge w)<\varepsilon$, so $\alpha y\in V_{\varepsilon,w,j}$ and, since $V_{\varepsilon,w,j}$ is solid, then $I_{z}\subseteq V_{\varepsilon,w,j}$. ∎

Note that the sequential completeness in Lemma 1 can be removed, as we see in the following corollary.

Theorem 1**.**

Let $(X,\tau)$ be a locally solid vector lattice, where $\tau$ is generated by a family $(\rho_{j})_{j\in J}$ of Riesz pseudonorms. Let $\varepsilon>0$, $j\in J$, and $0\neq w\in X_{+}$. Then either $V_{\varepsilon,w,j}$ is contained in $[-w,w]$ or $V_{\varepsilon,w,j}$ contains a non-trivial ideal.

Proof.

Given $\varepsilon>0$, $j\in J$, and $0\neq w\in X_{+}$. Let $(\hat{X},\hat{\tau})$ be the topological completion of $(X,\tau)$. In particular, $(\hat{X},\hat{\tau})$ is sequentially complete. Let ${\hat{V}}_{\varepsilon,w,j}=\{\hat{x}\in\hat{X}\colon{\hat{\rho}}_{j}(\lvert\hat{x}\rvert\wedge w)<\varepsilon\}$. Then $V_{\varepsilon,w,j}=X\cap{\hat{V}}_{\varepsilon,w,j}$. By Lemma 1, either ${\hat{V}}_{\varepsilon,w,j}$ is a subset of $[-w,w]_{\hat{X}}$ in $\hat{X}$ or ${\hat{V}}_{\varepsilon,w,j}$ contains a non-trivial ideal of $\hat{X}$. If ${\hat{V}}_{\varepsilon,w,j}\subseteq[-w,w]_{\hat{X}}$, then

[TABLE]

If ${\hat{V}}_{\varepsilon,w,j}$ contains a non-trivial ideal, then ${\hat{V}}_{\varepsilon,w,j}\nsubseteq[-w,w]_{\hat{X}}$. So, there is $\hat{x}\in\hat{V}_{\varepsilon,w,j}$ with $\hat{x}\notin[-w,w]_{\hat{X}}$. Since $[-w,w]_{\hat{X}}$ is $\hat{\tau}$-closed, then there is a solid neighborhood ${N}_{\hat{x}}$ of $\hat{x}$ in $\hat{X}$ such that ${N}_{\hat{x}}\cap[-w,w]_{\hat{X}}=\emptyset$. Hence, ${N}_{\hat{x}}\cap{\hat{V}}_{\varepsilon,w,j}\cap[-w,w]_{\hat{X}}=\emptyset$, and $N_{\hat{x}}\cap{\hat{V}}_{\varepsilon,w,j}$ is open in $\hat{X}$ with $\hat{x}\in{N}_{\hat{x}}\cap{\hat{V}}_{\varepsilon,w,j}$. By $\tau$-density of $X$ in $\hat{X}$, we may take $x\in X\cap{N}_{\hat{x}}\cap{\hat{V}}_{\varepsilon,w,j}$. Since $\lvert x\rvert\in X\cap{N}_{\hat{x}}\cap{\hat{V}}_{\varepsilon,w,j}$, we may also assume that $x\in X_{+}$.

Let $y:=(x-w)^{+}$, then $y>0$ and $y\in X_{+}$. By the same argument in Lemma 1, we get $(\alpha y)\wedge w\leq x\wedge w$ for all $\alpha\in\mathbb{R}_{+}$. Since $x\in{\hat{V}}_{\varepsilon,w,j}$, then $\alpha y\in{\hat{V}}_{\varepsilon,w,j}$ for all $\alpha\in\mathbb{R}_{+}$. But $\alpha y\in X_{+}$ for all $\alpha\in\mathbb{R}_{+}$ and, since $V_{\varepsilon,w,j}=X\cap{\hat{V}}_{\varepsilon,w,j}$, we get $\alpha y\in V_{\varepsilon,w,j}$ for all $\alpha\in\mathbb{R}_{+}$. Since $V_{\varepsilon,w,j}$ is solid, we conclude that the principal ideal $I_{y}$ taken in $X$ is a subset of $V_{\varepsilon,w,j}$. ∎

Lemma 2**.**

Let $(X,\tau)$ be a locally solid vector lattice, where $\tau$ is generated by a family $(\rho_{j})_{j\in J}$ of Riesz pseudonorms. If $V_{\varepsilon,w,j}$ is contained in $[-w,w]$, then $w$ is a strong unit.

Proof.

Suppose $V_{\varepsilon,w,j}\subseteq[-w,w]$. Since $V_{\varepsilon,w,j}$ is absorbing, for any $x\in X_{+}$, there exist $\alpha>0$ such that $\alpha x\in V_{\varepsilon,w,j}$ , and so $\alpha x\in[-w,w]$, or $x\leq\frac{1}{\alpha}w$. Thus $w$ is a strong unit, as desired. ∎

Proposition 2**.**

Let $e\in X_{+}$. Then $e$ is a quasi-interior point in $(X,\tau)$ iff $e$ is a quasi-interior point in the topological completion $(\hat{X},\hat{\tau})$.

Proof.

The backward implication is trivial.

For the forward implication let $\hat{x}\in\hat{X}_{+}$. Our aim is to show that $\hat{x}-\hat{x}\wedge ne\xrightarrow{\tau}0$ in $\hat{X}$ as $n\to\infty$. By [2, Thm. 2.40], $\hat{X}_{+}={\overline{X}^{\hat{\tau}}_{+}}$. So, there is a net $x_{\alpha}$ in $X_{+}$ such that $x_{\alpha}\xrightarrow{\mathit{\hat{\tau}}}\hat{x}$ in $\hat{X}$. Let $j\in J$ and $\varepsilon>0$. Since $\hat{\rho}_{j}(x_{\alpha}-\hat{x})\to 0$, then there is $\alpha_{\varepsilon}$ satisfying

[TABLE]

Since $e$ is a quasi-interior point in $X$ and $x_{\alpha_{\varepsilon}}\in X_{+}$, then $x_{\alpha_{\varepsilon}}-x_{\alpha_{\varepsilon}}\wedge ne\xrightarrow{\tau}0$ in $X$ as $n\to\infty$. Thus, there is $n_{\varepsilon}\in\mathbb{N}$ such that

[TABLE]

Now, $0\leq\hat{x}-\hat{x}\wedge ne=\hat{x}-x_{\alpha_{\varepsilon}}+x_{\alpha_{\varepsilon}}-ne\wedge x_{\alpha_{\varepsilon}}+ne\wedge x_{\alpha_{\varepsilon}}-\hat{x}\wedge ne$. So $\hat{\rho}_{j}(\hat{x}-\hat{x}\wedge ne)\leq\hat{\rho}_{j}(\hat{x}-x_{\alpha_{\varepsilon}})+\hat{\rho}_{j}(x_{\alpha_{\varepsilon}}-ne\wedge x_{\alpha_{\varepsilon}})+\hat{\rho}_{j}(ne\wedge x_{\alpha_{\varepsilon}}-\hat{x}\wedge ne)$. For $n\geqslant n_{\varepsilon}$, we have, by (2.1), (2.2), and [3, Thm. 1.9(2)], that

[TABLE]

Therefore, $e$ is a quasi-interior point in $\hat{X}$. ∎

The technique used in the proof of [15, Thm. 3.1] can be used in the following theorem as well, and so we omit its proof.

Theorem 2**.**

Let $(X,\tau)$ be a sequentially complete locally solid vector lattice, where $\tau$ is generated by a family $(\rho_{j})_{j\in J}$ of Riesz pseudonorms. Let $e\in X_{+}$. The following are equivalent:

1. (1)

$e$* is a quasi-interior point;* 2. (2)

for every net $x_{\alpha}$ in $X_{+}$, if $x_{\alpha}\wedge e\xrightarrow{\tau}0$ then $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$; 3. (3)

for every sequence $x_{n}$ in $X_{+}$, if $x_{n}\wedge e\xrightarrow{\tau}0$ then $x_{n}\xrightarrow{\mathit{u}\tau}0$.

3. Unbounded $\tau$-convergence in sublattices

Let $Y$ be a sublattice of a locally solid vector lattice $(X,\tau)$. If $y_{\alpha}$ is a net in $Y$ such that $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$, then clearly, $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $Y$. The converse does not hold in general. For example, the sequence $e_{n}$ of standard unit vectors is $un$-null in $c_{0}$, but not in $\ell_{\infty}$. In this section, we study when the $u\tau$-convergence passes from a sublattice to the whole space.

Recall that a sublattice $Y$ of a vector lattice $X$ is majorizing if, for every $x\in X_{+}$, there exists $y\in Y_{+}$ with $x\leqslant y$. The following theorem extends [15, Thm. 4.3] to locally solid vector lattices.

Theorem 3**.**

Let $(X,\tau)$ be a locally solid vector lattice and $Y$ be a sublattice of $X$. If $y_{\alpha}$ is a net in $Y$ and $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $Y$, then $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$ in each of the following cases$:$

1. (1)

$Y$* is majorizing in $X$;*** 2. (2)

$Y$* is $\tau$-dense in $X$;*** 3. (3)

$Y$* is a projection band in $X$.*

Proof.

1. (1)

Trivial. 2. (2)

Let $u\in X_{+}$. Fix $\varepsilon>0$ and take $j\in J$. Since $Y$ is $\tau$-dense in $X$, then there is $v\in Y_{+}$ such that $\rho_{j}(u-v)<\varepsilon$. But $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $Y$ and so, in particular, $\rho_{j}(\lvert y_{\alpha}\rvert\wedge v)\to 0$. So there is $\alpha_{0}$ such that $\rho_{j}(\lvert y_{\alpha}\rvert\wedge v)<\varepsilon$ for all $\alpha\geqslant\alpha_{0}$. It follows from $u\leq v+\lvert u-v\rvert$, that $\lvert y_{\alpha}\rvert\wedge u\leq\lvert y_{\alpha}\rvert\wedge v+\lvert u-v\rvert$, and so $\rho_{j}(\lvert y_{\alpha}\rvert\wedge u)<\rho_{j}(\lvert y_{\alpha}\rvert\wedge v)+\rho_{j}(u-v)<2\varepsilon$. Thus, $\rho_{j}(\lvert y_{\alpha}\rvert\wedge u)\to 0$ in $\mathbb{R}$. Since $j\in J$ was chosen arbitrary, we conclude that $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$. 3. (3)

Let $u\in X_{+}$. Then $u=v+w$, where $v\in Y_{+}$ and $w\in Y_{+}^{d}$. Now $\lvert y_{\alpha}\rvert\wedge u=\lvert y_{\alpha}\rvert\wedge v+\lvert y_{\alpha}\rvert\wedge w=\lvert y_{\alpha}\rvert\wedge v$, since $y_{\alpha}\in Y$. Then $\lvert y_{\alpha}\rvert\wedge u=\lvert y_{\alpha}\rvert\wedge v\xrightarrow{\tau}0$ in $X$.

∎

Corollary 1**.**

If $(X,\tau)$ is a locally solid vector lattice and $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$, then $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in the Dedekind completion $X^{\delta}$ of $X$.

Corollary 2**.**

If $(X,\tau)$ is a locally solid vector lattice and $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$, then $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in the topological completion $\hat{X}$ of $X$.

The next result generalizes Corollary 4.6 in [15] and Proposition 16 in [27].

Theorem 4**.**

Let $(X,\tau)$ be a topologically complete locally solid vector lattice that possesses the Lebesgue property, and $Y$ be a sublattice of $X$. If $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $Y$, then $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$.

Proof.

Suppose $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $Y$. By Theorem 3(1), $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in the ideal $I(Y)$ generated by $Y$ in $X$. By Theorem 3(2), $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in the closure $\overline{\{I(Y)\}}^{\tau}$ of $I(Y)$. It follows from [2, Thm. 3.7] that $\overline{\{I(Y)\}}^{\tau}$ is a band in $X$. Now, [2, Thm. 3.24] assures that $X$ is Dedekind complete, and so $\overline{\{I(Y)\}}^{\tau}$ is a projection band in $X$. Then $y_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X$, in view of Theorem 3(3). ∎

Suppose that $(X,\tau)$ is a locally solid vector lattice possessing the Lebesgue property. Then, in view of [2, Thms. 3.23 and 3.26], its topological completion $(\hat{X},\hat{\tau})$ possesses the Lebesgue property as well. Hence, by [2, Thm. 3.24], $\hat{X}$ is Dedekind complete. Since $X\subseteq\hat{X}$, there holds $X^{\delta}\subseteq(\hat{X})^{\delta}=\hat{X}$. So, $X\subseteq X^{\delta}\subseteq\hat{X}$. Now, Theorem 4 assures that, given a net $z_{\alpha}$ in $X^{\delta}$, if $z_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $X^{\delta}$ then $z_{\alpha}\xrightarrow{\mathit{u}\tau}0$ in $\hat{X}$.

4. unbounded relatively uniformly convergence

In this section we discuss unbounded relatively uniformly convergence. Recall that a net $x_{\alpha}$ in a vector lattice $X$ is said to be relatively uniformly convergent to $x\in X$ if, there is $u\in X_{+}$ such that for any $n\in\mathbb{N}$, there exists $\alpha_{n}$ satisfying $\lvert x_{\alpha}-x\rvert\leq\frac{1}{n}u$ for $\alpha\geqslant\alpha_{n}$. In this case we write $x_{\alpha}\xrightarrow{\mathit{ru}}x$ and the vector $u\in X_{+}$ is called regulator, see [24, Def. III.11.1].

If $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$ in a locally solid vector lattice $(X,\tau)$, then $x_{\alpha}\xrightarrow{\mathit{{\tau}}}0$. Indeed, let $V$ be a solid neighborhood at zero. Since $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$, then there is $u\in X_{+}$ such that, for a given $\varepsilon>0$, there is $\alpha_{\varepsilon}$ satisfying $\lvert x_{\alpha}\rvert\leq\varepsilon u$ for all $\alpha\geq\alpha_{\varepsilon}$. Since $V$ is absorbing, there is $c\geq 1$ such that $\frac{1}{c}u\in V$. There is some $\alpha_{0}$ such that $\lvert x_{\alpha}\rvert\leq\frac{1}{c}u$ for all $\alpha\geq\alpha_{0}$. Since $V$ is solid and $\lvert x_{\alpha}\rvert\leq\frac{1}{c}u$ for all $\alpha\geq\alpha_{0}$, then $x_{\alpha}\in V$ for all $\alpha\geq\alpha_{0}$. That is $x_{\alpha}\xrightarrow{\mathit{{\tau}}}0$.

The following result might be considered as an $ru$-version of Theorem 1 in [7].

Theorem 5**.**

Let $X$ be a vector lattice. Then the following conditions are equivalent.

$(1)$* There exists a linear topology $\tau$ on $X$ such that, for any net $x_{\alpha}$ in $X$: $x_{\alpha}\xrightarrow{\mathit{ru}}0$ iff $x_{\alpha}\xrightarrow{\mathit{{\tau}}}0$.*

$(2)$* There exists a norm $\|\cdot\|$ on $X$ such that, for any net $x_{\alpha}$ in $X$: $x_{\alpha}\xrightarrow{\mathit{ru}}0$ iff $\|x_{\alpha}\|\to 0$.*

$(3)$* $X$ has a strong order unit.*

Proof.

$(1)\Rightarrow(3)$ It follows from [7, Lem. 1].

$(3)\Rightarrow(2)$ Let $e\in X$ be a strong order unit. Then $x_{\alpha}\xrightarrow{\mathit{ru}}0$ iff $\|x_{\alpha}\|_{e}\to 0$, where $\|x\|_{e}:=\inf\{r:|x|\leqslant re\}$.

$(2)\Rightarrow(1)$ It is trivial. ∎

Let $X$ be a vector lattice. A net $x_{\alpha}$ in $X$ is said to be unbounded relatively uniformly convergent to $x\in X$ if $\lvert x_{\alpha}-x\rvert\wedge w\xrightarrow{\mathit{{ru}}}0$ for all $w\in X_{+}$. In this case, we write $x_{\alpha}\xrightarrow{\mathit{{uru}}}x$. Clearly, if $x_{\alpha}\xrightarrow{\mathit{{uru}}}0$ in a locally solid vector lattice $(X,\tau)$, then $x_{\alpha}\xrightarrow{\mathit{{u\tau}}}0$.

In general, $uru$-convergence is also not topological. Indeed, consider the vector lattice $L_{1}[0,1]$. It satisfies the diagonal property for order convergence by [19, Thm. 71.8]. Now, by combining Theorems 16.3, 16.9, and 68.8 in [19] we get that for any sequence $f_{n}$ in $L_{1}[0,1]$ $f_{n}\xrightarrow{\mathit{o}}0$ iff $f_{n}\xrightarrow{\mathit{{ru}}}0$. In particular, $f_{n}\xrightarrow{\mathit{{uo}}}0$ iff $f_{n}\xrightarrow{\mathit{{uru}}}0$. But the $uo$-convergence in $L_{1}[0,1]$ is equivalent to $a.e$.-convergence which is not topological, see [18].

However, in some vector lattices the $uru$-convergence could be topological. For example, if $X$ is a vector lattice with a strong unit $e$, It follows from Theorem 5, that $ru$-convergence is equivalent to the norm convergence $\lVert\cdot\lVert_{e}$, where $\lVert x\lVert_{e}\coloneqq\inf\{\lambda>0:\lvert x\rvert\leq\lambda e\}$, $x\in X$. Thus $uru$-convergence in $X$ is topological.

Consider vector lattice $c_{00}$ of eventually zero sequences. It is well known that in $c_{00}$: $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$ iff $x_{\alpha}\xrightarrow{\mathit{o}}0$. For the sake of completeness we include a proof of this fact. Clearly, $x_{\alpha}\xrightarrow{\mathit{{ru}}}0\Rightarrow x_{\alpha}\xrightarrow{\mathit{o}}0$. For the converse, suppose $x_{\alpha}\xrightarrow{\mathit{o}}0$ in $c_{00}$. Then there is a net $y_{\beta}\downarrow 0$ in $c_{00}$ such that, for any $\beta$, there is $\alpha_{\beta}$ satisfying $\lvert x_{\alpha}\rvert\leq y_{\beta}$ for all $\alpha\geq\alpha_{\beta}$. Let $e_{n}$ denote the sequence of standard unit vectors in $c_{00}$. Fix $\beta_{0}$. Then $y_{\beta_{0}}=c_{1}^{\beta_{0}}e_{k_{1}}+\dots+c_{n}^{\beta_{0}}e_{k_{n}},\ c_{i}^{\beta_{0}}\in\mathbb{R},i=1,\dots,n$. Since $y_{\beta}$ is decreasing, then $y_{\beta}\leq y_{\beta_{0}}$ for all $\beta\geq\beta_{0}$. So, $y_{\beta}=c_{1}^{\beta}e_{k_{1}}+\dots+c_{n}^{\beta}e_{k_{n}}$ for all $\beta\geq\beta_{0},c_{i}^{\beta}\in\mathbb{R},i=1,\dots,n$. Since $y_{\beta}\downarrow 0$ then $\lim_{\beta}c_{i}^{\beta}=0$ for all $i=1,\dots,n$. Let $u=e_{k_{1}}+\dots+e_{k_{n}}$. Given $\varepsilon>0$. Then, there is $\beta_{\varepsilon}\geq\beta_{0}$ such that $c_{i}^{\beta}<\varepsilon$ for all $\beta\geq\beta_{\varepsilon}$ for $i=1,\dots,n$. Consider $y_{\beta_{\varepsilon}}$ then there is $\alpha_{\varepsilon}$ such that $\lvert x_{\alpha}\rvert\leq y_{\beta_{\varepsilon}}$ for all $\alpha\geq\beta_{\varepsilon}$. But $y_{\beta_{\varepsilon}}=c_{1}^{\beta_{\varepsilon}}e_{k_{1}}+\dots+c_{n}^{\beta_{\varepsilon}}e_{k_{n}}\leq\varepsilon u$. So, $\lvert x_{\alpha}\rvert\leq\varepsilon u$ for all $\alpha\geq\alpha_{\varepsilon}$. That is $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$. Thus, the $uru$-convergence in $c_{00}$ coincides with the $uo$-convergence which is pointwise convergence and, therefore, is topological.

Proposition 3**.**

Let $X$ be Lebesgue and complete metrizable locally solid vector lattice. then $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$ iff $x_{\alpha}\xrightarrow{\mathit{o}}0$.

Proof.

The necessity is obvious. For the sufficiency assume that $x_{\alpha}\xrightarrow{\mathit{o}}0$. Then there exists $y_{\beta}\downarrow 0$ such that for any $\beta$ there is $\alpha_{\beta}$ with $|x_{\alpha}|\leqslant y_{\beta}$ as $\alpha\geqslant\alpha_{\beta}$. Since $d(y_{\beta},0)\to 0$, there exists an increasing sequence $(\beta_{k})_{k}$ of indeces with $d(ky_{\beta_{k}},0)\leqslant\frac{1}{2^{k}}$. Let $s_{n}=\sum_{k=1}^{n}ky_{\beta_{k}}$. We show the sequence $s_{n}$ is Cauchy. For $n>m$,

[TABLE]

Since $X$ is complete, then the sequence $s_{n}$ converges to some $u\in X_{+}$. That is, $u:=\sum\limits_{k=1}^{\infty}ky_{\beta_{k}}$. Then

[TABLE]

which means that $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$. ∎

Let $X=\mathbb{R}^{\Omega}$ be the vector lattice of all real-valued functions on a set $\Omega$.

Proposition 4**.**

In the vector lattice $X=\mathbb{R}^{\Omega}$, the following conditions are equivalent$:$

$(1)$* for any net $f_{\alpha}$ in $X$: $f_{\alpha}\xrightarrow{\mathit{o}}0$ iff $f_{\alpha}\xrightarrow{\mathit{{ru}}}0$;*

$(2)$* $\Omega$ is countable.*

Proof.

$(1)\Rightarrow(2)$ Suppose $f_{\alpha}\xrightarrow{\mathit{o}}0\Leftrightarrow f_{\alpha}\xrightarrow{\mathit{{ru}}}0$ for any sequence $f_{\alpha}$ in $X=\mathbb{R}^{\Omega}$. Our aim is to show that $\Omega$ is countable. Assume, in contrary, that $\Omega$ is uncountable. Let $\mathcal{F}(\Omega)$ be the collection of all finite subsets of $\Omega$. For each $\alpha\in\mathcal{F}(\Omega)$, put $f_{\alpha}=\mathcal{X}_{\alpha}$. Clearly, $f_{\alpha}\uparrow\mathbbold{1}$, where $\mathbbold{1}$ denotes the constant function one on $\Omega$. Then $\mathbbold{1}-f_{\alpha}\downarrow 0$ or $\mathbbold{1}-f_{\alpha}\xrightarrow{\mathit{o}}0$ in $\mathbb{R}^{\Omega}$. So, there is $0\leq g\in\mathbb{R}^{\Omega}$ such that, for any $\varepsilon>0$, there exists $\alpha_{\varepsilon}$ satisfying $\mathbbold{1}-f_{\alpha}\leq\varepsilon g$ for all $\alpha\geqslant\alpha_{\varepsilon}$. Let $n\in\mathbb{N}$. Then there is a finite set $\alpha_{n}\subseteq\Omega$ such that $\mathbbold{1}-f_{\alpha_{n}}\leq\frac{1}{n}g$. Consequently, $g(x)\geqslant n$ for all $x\in\Omega\setminus\alpha_{n}$. Let $S=\cup_{n=1}^{\infty}\alpha_{n}$. Then $S$ is countable and $\Omega\setminus S\neq\emptyset$. Moreover, for each $x\in\Omega\setminus S$, we have $g(x)\geqslant n$ for all $n\in\mathbb{N}$, which is impossible.

$(2)\Rightarrow(1)$ Suppose that $\Omega$ is countable. So, we may assume that $X=s$, the space of all sequences. Since, from $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$ always follows that $x_{\alpha}\xrightarrow{\mathit{o}}0$, it is enough to show that if $x_{\alpha}\xrightarrow{\mathit{o}}0$ then $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$. To see this, let $(x_{\alpha}^{n})_{n}=x_{\alpha}\xrightarrow{\mathit{o}}0$. Then, the net $x_{\alpha}$ is eventually bounded, say $|x_{\alpha}|\leqslant u=(u_{n})_{n}\in s$. Take $w:=(nu_{n})_{n}\in s$. We show that $x_{\alpha}\xrightarrow{\mathit{{ru}}}0$ with the regulator $w$. Let $k\in\mathbb{N}$. Since $x_{\alpha}\xrightarrow{\mathit{o}}0$, then for each $n\in\mathbb{N}$, $x_{\alpha}^{n}\to 0$ in $\mathbb{R}$. Hence, there is $\alpha_{k}$ such that $k\lvert x_{\alpha}^{1}\rvert<u_{1}$, $k\lvert x_{\alpha}^{2}\rvert<u_{2}$, $\cdots$, $k\lvert x_{\alpha}^{k-1}\rvert<u_{k-1}$ for all $\alpha\geqslant\alpha_{k}$. Note that for $n\geqslant k$, $k\lvert x_{\alpha}^{n}\rvert<u_{n}$. Therefore, $k\lvert x_{\alpha}\rvert<w$ for all $\alpha\geqslant\alpha_{k}$. ∎

It follows from Proposition 4 that, for countable $\Omega$, the $uru$-convergence in $\mathbb{R}^{\Omega}$ coincides with the $uo$-convergence (which is pointwise) and therefore is topological. We do not know, whether or not the countability of $\Omega$ is necessary for the property that $uru$-convergence is topological in $\mathbb{R}^{\Omega}$.

5. Topological orthogonal systems and metrizabililty

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 Zorn’s Lemma that every vector lattice containing at least one non-zero element has a maximal orthogonal system. Motivated by Definition III.5.1 in [20], we introduce the following notion.

Definition 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 $X$.

Lemma 3**.**

If $Q=\{e_{\gamma}\}_{\gamma\in\Gamma}$ is a topological orthogonal system in a topological vector lattice $(X,\tau)$, then $Q$ is a maximal orthogonal system in $X$.

Proof.

Assume $x\wedge e_{\gamma}=0$ for all $\gamma\in\Gamma$. By the assumption, there is a net $x_{\alpha}$ in the ideal $I_{Q}$ such that $x_{\alpha}\xrightarrow{\tau}x$. Without lost of generality, we may assume $0\leq x_{\alpha}\leq x$ for all $\alpha$. Since $x_{\alpha}\in I_{Q}$, then there are $0<\mu_{\alpha}\in\mathbb{R}$ and $\gamma_{1},\leavevmode\nobreak\ \gamma_{2},\dots,\leavevmode\nobreak\ \gamma_{n}$, such that $0\leq x_{\alpha}\leq\mu_{\alpha}(e_{\gamma_{1}}+e_{\gamma_{2}}+\dots+e_{\gamma_{n}})$. So $0\leq x_{\alpha}=x_{\alpha}\wedge x\leq\mu_{\alpha}(e_{\gamma_{1}}+e_{\gamma_{2}}+\dots+e_{\gamma_{n}})\wedge x$ $=\mu_{\alpha}e_{\gamma_{1}}\wedge x+\dots+\mu_{\alpha}e_{\gamma_{n}}\wedge x$ $=0$. Hence $x_{\alpha}=0$ for all $\alpha$, and so $x=0$. ∎

We recall the following construction from [20, p.169]. Let $X$ be a vector lattice and $Q=\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a maximal orthogonal system of $X$. Let ${\mathscr{F}}(\Gamma)$ denote the collection of all finite subsets of $\Gamma$ ordered by inclusion. For each $(n,H)\in\mathbb{N}\times\mathscr{F}(\Gamma)$ and $x\in X_{+}$, define

[TABLE]

Clearly $\{x_{n,H}:(n,H)\in\mathbb{N}\times\mathscr{F}(\Gamma)\}$ is directed upward, and

[TABLE]

Moreover, Proposition II.1.9 in [20] implies $x_{n,H}\uparrow x$.

Theorem 6**.**

Let $Q=\{e_{\gamma}\}_{\gamma\in\Gamma}$ be an orthogonal system of a locally solid vector lattice $(X,\tau)$, then $Q$ is a topological orthogonal system iff we have $x_{n,H}\xrightarrow{\tau}x$ over $(n,H)\in\mathbb{N}\times\mathscr{F}(\Gamma)$ for each $x\in X_{+}$.

Proof.

For the backward implication take $x\in X_{+}$. Since

[TABLE]

then $x_{n,H}\in I_{Q}$ for each $(n,H)\in\mathbb{N}\times\mathscr{F}(\Gamma)$. Also, we have, by assumption, $x_{n,H}\xrightarrow{\tau}x$. Thus, $x\in\overline{I}^{\tau}_{Q}$, i.e., $Q$ is a topological orthogonal system of $X$.

For the forward implication, note that $Q$ is a maximal orthogonal system, by Lemma 3. Let $x\in X_{+}$, and $j\in J$. Given $\varepsilon>0$. Let $V_{\varepsilon,x,j}\coloneqq\{z\in X:\rho_{j}(z-x)<\varepsilon\}$. Then $V_{\varepsilon,x,j}$ is a neighborhood of $x$ in the $\tau$-topology. Since $I_{Q}$ is dense in $X$ with respect to the $\tau$-topology, there is $x_{\varepsilon}\in I_{Q}$ with $0\leq x_{\varepsilon}\leq x$ such that $\rho_{j}(x_{\varepsilon}-x)<\varepsilon$. Now, $x_{\varepsilon}\in I_{Q}$ implies that there are $H_{\varepsilon}\in\mathscr{F}(\Gamma)$ and $n_{\varepsilon}\in\mathbb{N}$ such that

[TABLE]

Let

[TABLE]

It follows from $0\leq w\leq\sum\limits_{\gamma\in H_{\varepsilon}}n_{\varepsilon}e_{\gamma}$ and the Riesz decomposition property, that, for each $\gamma\in H_{\varepsilon}$, there exists $y_{\gamma}$ with

[TABLE]

such that

[TABLE]

From (5.3) and (5.5), we have

[TABLE]

Also, (5.4) and (5.6) imply that $y_{\gamma}\leq n_{\varepsilon}e_{\gamma}\wedge x$. Now

[TABLE]

But, from (5.2) and (5.3), we get

[TABLE]

Thus, it follows from (5.7), (5.8), and (5.1), that $0\leq x_{\varepsilon}\leq x_{n_{\varepsilon},H_{\varepsilon}}\leq x$. Hence, $0\leq x-x_{n_{\varepsilon},H_{\varepsilon}}\leq x-x_{\varepsilon}$ and so $\rho_{j}(x-x_{n,H})\leq\rho_{j}(x-x_{n_{\varepsilon},H_{\varepsilon}})\leq\rho_{j}(x-x_{\varepsilon})$ for each $(n,H)\geq(n_{\varepsilon},H_{\varepsilon})$. Therefore $x_{n,H}\xrightarrow{\tau}x$. ∎

The following corollary can be proven easily.

Corollary 3**.**

Let $(X,\tau)$ be a locally solid vector lattice. The following statements are equivalent:

1. (1)

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

for each $x\in X_{+},\leavevmode\nobreak\ x-x\wedge ne\xrightarrow{\mathit{{\tau}}}0$ as $n\rightarrow\infty$.

Corollary 4**.**

Let $(X,\tau)$ be a locally solid vector lattice possessing the $\sigma$-Lebesgue property. Then every weak unit in $X$ is a quasi-interior point.

Proof.

Let $x\in X^{+}$, and let $e$ be a weak unit. Then $x\wedge ne\uparrow x$. So, by the $\sigma$-Lebesgue property, we get $x-x\wedge ne\xrightarrow{\mathit{{\tau}}}0$ as $n\rightarrow\infty$. ∎

Theorem 7**.**

Let $(X,\tau)$ be a locally solid vector lattice, and $Q=\{e_{\gamma}\}_{\gamma\in\Gamma}$ be a topological orthogonal system of $(X,\tau)$. Then $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ iff $\lvert x_{\alpha}\rvert\wedge e_{\gamma}\xrightarrow{\tau}0$ for every ${\gamma\in\Gamma}$.

Proof.

The forward implication is trivial. For the backward implication, assume $\lvert x_{\alpha}\rvert\wedge e_{\gamma}\xrightarrow{\tau}0$ for every ${\gamma\in\Gamma}$. Let $u\in X_{+}$, $j\in J$. Fix $\varepsilon>0$. We have

[TABLE]

Now, Theorem 6 assures that $u_{n,H}\xrightarrow{\tau}u$, and so, there exists $(n_{\varepsilon},H_{\varepsilon})\in\mathbb{N}\times\mathscr{F}(\Gamma)$ such that

[TABLE]

Thus, $\lvert x_{\alpha}\rvert\wedge u\leq u-u_{n_{\varepsilon},H_{\varepsilon}}+\sum\limits_{\gamma\in H_{\varepsilon}}n_{\varepsilon}(e_{\gamma}\wedge\lvert x_{\alpha}\rvert)$. But, by the assumption, $e_{\gamma}\wedge\lvert x_{\alpha}\rvert\xrightarrow{\tau}0$ for all $\gamma\in\Gamma$, and so $n_{\varepsilon}(e_{\gamma}\wedge\lvert x_{\alpha}\rvert)\xrightarrow{\tau}0$. Hence, there is $\alpha_{\varepsilon,H_{\varepsilon}}$ such that

[TABLE]

Here $\lvert H_{\varepsilon}\rvert$ denotes the cardinality of $H_{\varepsilon}$. For $\alpha\geq\alpha_{\varepsilon,H_{\varepsilon}}$, we have

[TABLE]

where the second inequality follows from (5.9) and the third one from (5.10). Therefore, $\rho_{j}(\lvert x_{\alpha}\rvert\wedge u)\to 0$, and so $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$. ∎

The following corollary is immediate.

Corollary 5**.**

Let $(X,\tau)$ be a locally solid vector lattice, and $e\in X_{+}$ be a quasi-interior point. Then $x_{\alpha}\xrightarrow{\mathit{u}\tau}0$ iff $\lvert x_{\alpha}\rvert\wedge e\xrightarrow{\tau}0$.

Recall that a topological vector space is metrizable iff it has a countable neighborhood base at zero, [2, Thm. 2.1]. In particular, a locally solid vector lattice $(X,\tau)$ is metrizable iff its topology $\tau$ is generated by a countable family $(\rho_{k})_{k\in\mathbb{N}}$ of Riesz pseudonorms. The following result gives a sufficient condition for the metrizabililty of $u\tau$-topology.

Proposition 5**.**

Let $(X,\tau)$ be a complete metrizable locally solid vector lattice. If $X$ has a countable topological orthogonal system, then the $u\tau$-topology is metrizable.

Proof.

First note that, since $(X,\tau)$ is metrizable, $\tau$ is generated by a countable family $(\rho_{k})_{k\in\mathbb{N}}$ of Riesz pseudonorms.

Now suppose $(e_{n})_{n\in\mathbb{N}}$ to be a topological orthogonal system. For each $n\in\mathbb{N}$, put $d_{n}(x,y)\coloneqq\sum\limits_{k=1}^{\infty}\frac{1}{2^{k}}\frac{\rho_{k}(\lvert x-y\rvert\wedge e_{n})}{1+\rho_{k}(\lvert x-y\rvert\wedge e_{n})}$. Note that each $d_{n}$ is a semimetric, and $d_{n}(x,y)\leq 1$ for all $x,y\in X$. If $d_{n}(x,y)=0$, then $\rho_{k}(\lvert x-y\rvert\wedge e_{n})=0$ for all $k\in\mathbb{N}$, so $(\lvert x-y\rvert\wedge e_{n})=0$. For $x,y\in X$, let $d(x,y)\coloneqq\sum\limits_{n=1}^{\infty}\frac{1}{2^{n}}d_{n}(x,y)$. Clearly, $d(x,y)$ is nonnegative and satisfies the triangle inequality, and $d(x,y)=d(y,x)$ for all $x,y\in X$. Now $d(x,y)=0$ iff $d_{n}(x,y)=0$ for all $n\in\mathbb{N}$ iff $\rho_{k}(\lvert x-y\rvert\wedge e_{n})=0$ for all $k\in\mathbb{N}$ iff $(\lvert x-y\rvert\wedge e_{n})=0$ for all $n\in\mathbb{N}$ iff $\lvert x-y\rvert=0$ iff $x=y$. Thus $(X,d)$ is a metric space. Finally, it is easy to see from Theorem 7 that $d$ generates the $u\tau$-topology. ∎

Recall that a topological space $X$ is called submetrizable if its topology is finer that some metric topology on $X$.

Proposition 6**.**

Let $(X,\tau)$ be a metrizable locally solid vector lattice. If $X$ has a weak unit, then the $u\tau$-topology is submetrizable.

Proof.

Note that, since $(X,\tau)$ is metrizable, then $\tau$ is generated by a countable family $(\rho_{k})_{k\in\mathbb{N}}$ of Riesz pseudonorms.

Suppose that $e\in X_{+}$ is a weak unit. Put $d(x,y)\coloneqq\sum\limits_{k=1}^{\infty}\frac{1}{2^{k}}\frac{\rho_{k}(\lvert x-y\rvert\wedge e)}{1+\rho_{k}(\lvert x-y\rvert\wedge e)}$. Note that $d(x,y)=0$ iff $\rho_{k}(\lvert x-y\rvert\wedge e)=0$ for all $k\in\mathbb{N}$ iff $\lvert x-y\rvert\wedge e=0$ and, since $e$ is a weak unit, $x=y$. It can easily be shown that $d$ satisfies the triangle inequality. Assume $x_{\alpha}\xrightarrow{\mathit{u}\tau}x$. Then, for all $u\in X_{+}$, $\rho_{k}(\lvert x-y\rvert\wedge u)\to 0$ for all $k\in\mathbb{N}$. In particular, $\rho_{k}(\lvert x-y\rvert\wedge e)\to 0$ for all $k\in\mathbb{N}$. Then in a similar argument to [24, p.200], it can be shown that $x_{\alpha}\xrightarrow{\mathit{d}}x$. Therefore, the $u\tau$-topology is finer than the metric topology generated by $d$, and hence $u\tau$-topology is submetrizable. ∎

We do not know whether the converse of propositions 5, and 6 is true or not.

6. Unbounded $\tau$-Completeness

A subset $A$ of a locally solid vector lattice $(X,\tau)$ is said to be $($sequentially$)$ $u\tau$-complete if, it is (sequentially) complete in the $u\tau$-topology. In this section, we relate sequential $u\tau$-completeness of subsets of $X$ with the Lebesgue and Levi properties. First, we remind the following theorem.

Theorem 8**.**

[26, Thm. 1]** If $(X,\tau)$ is a locally solid vector lattice, then the following statements are equivalent$:$

1. (1)

$(X,\tau)$* has the Lebesgue and Levi properties**;*** 2. (2)

$X$* is $\tau$-complete, and $c_{0}$ is not lattice embeddable in $(X,\tau)$.*

Recall that two locally solid vector lattices $(X_{1},\tau_{1})$ and $(X_{2},\tau_{2})$ are said to be isomorphic, if there exists a lattice isomorphism from $X_{1}$ onto $X_{2}$ that is also a homeomorphism; in other words, if there exists a mapping from $X_{1}$ onto $X_{2}$ that preserves the algebraic, the lattice, and the topological structures. A locally solid vector lattice $(X_{1},\tau_{1})$ is said to be lattice embeddable into another locally solid vector lattice $(X_{2},\tau_{2})$ if there exists a sublattice $Y_{2}$ of $X_{2}$ such that $(X_{1},\tau_{1})$ and $(Y_{2},\tau_{2})$ are isomorphic.

Note that $(X,\tau)$ can have the Lebesgue and Levi properties and simultaneously contains $c_{0}$ as a sublattice, but not as a lattice embeddable copy. The following example illustrates this.

Example 1**.**

Let $s$ denote the vector lattice of all sequences in $\mathbb{R}$ with coordinatewise ordering. Clearly, $c_{0}$ is a sublattice of $s$. Define the following separating family of Riesz pseudonorms

[TABLE]

for each $j\in\mathbb{N}$ and $(x_{n})_{n}\in s$. Then $\mathcal{R}$ generates a locally solid topology $\tau$ on $s$. It can be easily shown that $(s,\tau)$ has the Lebesgue and Levi properties. Although $c_{0}$ is a sublattice of $s$, but $(c_{0},\lVert\cdot\rVert_{\infty})$ is not lattice embeddable in $(s,\tau)$. To see this, consider the sequence $e_{n}$ of the standard unit vectors in $c_{0}$. Then the sequence $e_{n}$ is not norm null in $(c_{0},\lVert\cdot\rVert_{\infty})$, whereas $e_{n}\xrightarrow{\tau}0$ in $(s,\tau)$.

Proposition 7**.**

Let $(X,\tau)$ be a complete locally solid vector lattice. If every $\tau$-bounded subset of $X$ is sequentially $u\tau$-complete, then $X$ has the Lebesgue and Levi properties.

Proof.

Suppose $X$ does not possess the Lebesgue or Levi properties. Then, by Theorem 8, $c_{0}$ is lattice embeddable in $(X,\tau)$. Let $s_{n}=\sum_{k=1}^{n}e_{k}$, where $e_{k}$’s denote the standard unit vectors in $c_{0}$. Clearly, the sequence $s_{n}$ is norm-bounded in $c_{0}$ and so it is $\tau$-bounded in $(X,\tau)$. Note that $\lVert e_{k}\rVert_{\infty}=1\nrightarrow 0$, and so $e_{k}$ is not $\tau$-null. It follows from [15, Lm. 6.1] that $s_{n}$ is $un$-Cauchy in $c_{0}$, but is not $un$-convergent in $c_{0}$. That is $s_{n}$ is $u\tau$-Cauchy which is not $u\tau$-convergent, a contradiction. ∎

Using the proof of the previous result and [26, Thm. 1′], one can easily prove the following result.

Proposition 8**.**

Let $X$ be a Dedekind complete vector lattice equipped with a sequentially complete topology $\tau$. If every $\tau$-bounded subset of $X$ is sequentially $u\tau$-complete, then $X$ has the $\sigma$-Lebesgue and $\sigma$-Levi properties.

Clearly, every finite dimensional locally solid vector lattice $(X,\tau)$ is $u\tau$-complete. On the contrary of [15, Prop. 6.2], we provide an example of a $\tau$-complete locally solid vector lattice $(X,\tau)$ possessing the Lebesgue property such that it is $u\tau$-complete and $\dim X=\infty$.

Example 2**.**

*Let $X=s$ and $\mathcal{R}=(\rho_{j})_{j\in\mathbb{N}}$ such that $\rho_{j}((x_{n}))\coloneqq\lvert x_{j}\rvert$, where $(x_{n})_{n\in\mathbb{N}}\in s$. It is easy to see that $(X,\mathcal{R})$ is $\tau$-complete and has the Lebesgue property. Now, we show that $(X,\mathcal{R})$ is $u\tau$-complete. Suppose $x^{\alpha}$ is $u\tau$-Cauchy net. Then, for each $u\in X_{+}$, we have $|x^{\alpha}-x^{\beta}|\wedge u\xrightarrow{\tau}0$. Now, $u=u_{n}$ and, $x^{\alpha}=x_{n}^{\alpha}$. Let $j\in\mathbb{N}$, then $\rho_{j}(|x^{\alpha}-x^{\beta}|\wedge u)\rightarrow 0$ in $\mathbb{R}$ over $\alpha,\beta$ iff $|x_{j}^{\alpha}-x_{j}^{\beta}|\wedge u_{j}\rightarrow 0$ in $\mathbb{R}$ iff $|x_{j}^{\alpha}-x_{j}^{\beta}|\rightarrow 0$ in $\mathbb{R}$ over $\alpha,\beta$.

Thus, $(x_{j}^{\alpha})_{\alpha}$ is Cauchy in $\mathbb{R}$ and so there is $x_{j}\in\mathbb{R}$ such that $x_{j}^{\alpha}\rightarrow x_{j}$ in $\mathbb{R}$ over $\alpha$. Let $x=(x_{j})_{j\in\mathbb{N}}\in s$, then, clearly, $x^{\alpha}\xrightarrow{\mathit{u}\tau}x$.*