Unbounded Order Convergence and Universal Completions

TL;DR

This paper characterizes vector lattices where unbounded order convergence becomes order bounded, providing a solution to a previously posed problem and advancing the understanding of convergence properties in vector lattices.

Contribution

It offers a new characterization of vector lattices with unbounded order convergence that is eventually order bounded, solving an open problem in the field.

Findings

Characterization of vector lattices with unbounded order convergence

Solution to Problem 23 in Azema's work

Advancement in understanding convergence in vector lattices

Abstract

We characterize vector lattices in which unbounded order convergence is eventually order bounded. Among other things, the characterization provides a solution to \cite[Probl.23]{Az}.

Equations

$$x_{(n,m)}(k):=\{\begin{array}{ll}0& \text{if }k\ne n\wedge m,\\ n\vee m& \text{if }k=n\wedge m.\end{array}$$

$$y\ge x_{(n,m)}(\forall (n,m)\ge (n_0,m_0)).$$

$$y(m_0)\ge x_{(n,m_0)}(m_0)=n\vee m_0=n(\forall n\ge n_0\vee m_0)$$

$$\underset{(n,m)\to \infty }{\lim }p{r}_{\{e_k\}}(x_{(n,m)})=\underset{(n,m)\to \infty }{\lim }(n\vee m)p{r}_{\{e_k\}}(e_{n\wedge m})=0(\forall k\in \mathbb{N}),$$

$$y\ge x_{(n,m)}(\forall (n,m)\ge (n_0,m_0)).$$

$$y\ge x_{(n,m_0)}=(n\vee m_0)\cdot e_{n\wedge m_0}=(n\vee m_0)\cdot e_{m_0}=n\cdot e_{m_0}>0(\forall n\ge n_0\vee m_0)$$

Full text

March 19, 2024

1. Preliminaries

Throughout the paper, $X$ stands for a vector lattice and all vector lattices are assumed to be real and Archimedean. We refer to [2, 7, 5] for unexplained terminology and standard facts on vector lattice theory.

We recall a few standard definitions and results related to vector lattices. $X$ is said to be Dedekind ($\sigma$-Dedekind) complete if every order bounded (countable) subset of $X$ has a supremum. A Dedekind complete ($\sigma$-Dedekind complete) $X$ is said to be universally ($\sigma$-universally) complete if every pairwise disjoint (countable) subset of $X_{+}$ has a supremum. Every universally complete vector lattices has a weak unit. It is well known that $X$ possesses a unique up to lattice isomorphism Dedekind (universal) completion, which will be denoted by $X^{\delta}$ (by $X^{u}$). Dealing with the completions, we always suppose that $X\subseteq X^{\delta}\subseteq X^{u}$, whereas $X^{\delta}$ sits as an ideal in $X^{u}$.

A sublattice $Y$ of $X$ is called regular if $y_{\alpha}\downarrow 0$ in $Y$ implies $y_{\alpha}\downarrow 0$ in $X$. $Y$ is said to be order dense if for every $0\neq x\in X_{+}$ there exists $0\neq y\in Y_{+}$ such that $y\leq x$. It is well known that ideals and order dense sublattices are regular. Furthermore, $X$ is atomic iff it is lattice isomorphic to an order dense sublattice of ${\mathbb{R}^{A}}$ (cf. [2, Thm.1.78]).

A net $(x_{\alpha})_{\alpha\in A}$ in $X$ $o$-converges to $x$ if there exists a net $(y_{\gamma})_{\gamma\in\Gamma}$ in $X$ satisfying $y_{\gamma}\downarrow 0$ and for each $\gamma\in\Gamma$ there is $\alpha_{\gamma}\in A$ with $|x_{\alpha}-x|\leq y_{\gamma}$ for all $\alpha\geq\alpha_{\gamma}$. In this case we write $x_{\alpha}\xrightarrow{o}x$. This definition is used e.g. in [7, 5]. In some of the literature (cf. [2]) a slightly different definition of the order convergence is used, namely a net $(x_{\alpha})_{\alpha\in A}$ in $X$ is said to be $o$-convergent to $x$ if there exists a net $(z_{\alpha})_{\alpha\in A}$ in $X$ such that $z_{\alpha}\downarrow 0$ and $|x_{\alpha}-x|\leq z_{\alpha}$ for all $\alpha$. Notice that both notions coincide in the case of order bounded nets in a Dedekind complete vector lattice (cf. [5, Rem.2.2] ). We refer to [1] for further discussion of definitions of $o$-convergence. It should be noted that $o$-convergence in $X$ is never topological unless $\dim(X)<\infty$ [4, Thm.1] (cf. also [6, Thm.2]).

A net $x_{\alpha}$ is $X$ is unbounded order convergent (shortly, $uo$-convergent) to $x\in X$ if $|x_{\alpha}-x|\wedge y\xrightarrow{o}0$ for every $y\in X_{+}$. In this case we, write $x_{\alpha}\xrightarrow{uo}x$. Following Nakano [9], $uo$-convergence is considered as a natural generalization of convergence almost everywhere (see [5, 8, 3, 10] and references therein). Clearly, $o$-convergence agrees with eventually order bounded $uo$-convergence. By [5, Thm.3.2], $uo$-convergence passes freely between $X$, $X^{\delta}$, and $X^{u}$. It was shown in [5, Cor.3.5] that if $X$ has a weak unit $u$, then for a net $x_{\alpha}$ in $X$, $x_{\alpha}\xrightarrow{uo}x\Leftrightarrow|x_{\alpha}-x|\wedge u\xrightarrow{o}0$. It was also proved in [5, Cor.3.12] that any $uo$-null sequence in $X$ is $o$-null in $X^{u}$. Accordingly to [8, Ex.2.6], it is no longer true for nets in $\ell_{\infty}$. Theorem 1 shows that all $uo$-null nets in $X$ are $o$-null in $X^{u}$ only in the trivial case $\dim(X)<\infty$. For further purposes, we include the following modification of [8, Ex.2.6].

Example 1**.**

A net $(x_{(n,m)})_{(n,m)\in\mathbb{N}^{2}}$ in $X=c_{00}$ is defined by

[TABLE]

Shortly, $x_{(n,m)}=(n\vee m)\cdot\mathbb{I}_{\{n\wedge m\}}$. Since $\lim\limits_{(n,m)\to\infty}x_{(n,m)}(k)=0$ for every $k\in\mathbb{N}$, then $x_{(n,m)}\xrightarrow{uo}0$ $($e.g., by [4, Prop.1.]$)$. Suppose $x_{(n,m)}$ is eventually order bounded by some $y\in\mathbb{R}^{\mathbb{N}}$. Then there exists $(n_{0},m_{0})\in\mathbb{N}^{2}$ such that

[TABLE]

Since $n\wedge m_{0}=m_{0}$ and $(n,m_{0})\geq(n_{0},m_{0})$ for $n\geq n_{0}\vee m_{0}$, then

[TABLE]

which is impossible. Therefore, the $uo$-null net $x_{(n,m)}$ is not eventually order bounded in $\mathbb{R}^{\mathbb{N}}$ and hence is not $o$-convergent in $X^{u}=(c_{00})^{u}=\mathbb{R}^{\mathbb{N}}$.

Although the $uo$-convergence is not topological in most of important cases (e.g., in $L_{1}[0,1]$ and in $C[0,1]$), it is topological in all atomic vector lattices [4, Thm.2].

A net $x_{\alpha}$ in $X$ is called $o$-Cauchy ($uo$-Cauchy) if the double net $(x_{\alpha}-x_{\beta})_{(\alpha,\beta)}$ $o$-converges ($uo$-converges) to [math]. It was noticed in [8, Lm.2.1] with a reference to [5, Prop.5.7] that every order bounded positive increasing net in $X$ is $o$-Cauchy. A net $x_{\alpha}$ in a Dedekind complete vector lattice with a weak unit $u$ is $uo$-Cauchy iff $\inf_{\alpha}\sup_{\beta,\gamma\geq\alpha}|x_{\beta}-x_{\gamma}|\wedge u=0$ [8, Lm.2.7]. It is well known that the completeness with respect to the $o$-convergence is equivalent to the Dedekind completeness. By [5, Cor.3.12], a sequence $x_{n}$ in X is $uo$-Cauchy in $X$ iff it is $o$-convergent in $X^{u}$. In the same paper, authors proved that a sequence in a $\sigma$-universally complete vector lattice is $uo$-Cauchy iff it is $o$-convergent [5, Thm.3.10]. Theorem 1 shows that there is no net-version of [5, Thm.3.10] unless $\dim(X)<\infty$.

In [10, Thm.3.9] (see, also [3, Thm.28]), it was shown that a vector lattice is $\sigma$-universally complete iff it is sequentially $uo$-complete. It was also proved in [3, Thm.17] that the $uo$-completeness is equivalent to the universal completeness. Therefore, there is no need in considering $uo$-completions (sequential $uo$-completions) of vector lattices.

2. Main result

We begin with the following generalization of Example 1. Given a nonempty subset $A\subset X$, $pr_{A}$ stands for the band projection in $X^{u}$ onto the band in $X^{u}$ generated by $A$.

Example 2**.**

In any infinite-dimensional Archimedean vector lattice $X$ there exists a $uo$-null net which is not eventually order bounded in $X^{u}$.

As $\dim(X)=\infty$, there is a sequence $e_{n}$ of pairwise disjoint positive nonzero elements of $X$. Let  $\mathbb{N}^{2}$ be the coordinatewise directed set of pairs of naturals. A net in $X$ is defined via $x_{(n,m)}=(n\vee m)\cdot e_{n\wedge m}$. Since $\{x_{(n,m)}:(n,m)\in\mathbb{N}^{2}\}\subseteq B_{\{e_{k}:k\in\mathbb{N}\}}$ and

[TABLE]

then $x_{(n,m)}\xrightarrow{uo}0$ as $(n,m)\to\infty$ $($e.g., it can be seen by use of [5, Cor.3.5.] for a weak unit $u$ in $X^{u}$ s.t. $u\wedge e_{k}=e_{k}$ for all k$$). If $x_{(n,m)}$ is eventually order bounded by some $y\in X^{u}$, then for some $(n_{0},m_{0})\in\mathbb{N}^{2}$ we have

[TABLE]

Since $n\wedge m_{0}=m_{0}$ and $(n,m_{0})\geq(n_{0},m_{0})$ for $n\geq n_{0}\vee m_{0}$, then

[TABLE]

which is impossible. Therefore, the net  $x_{(n,m)}$ is not eventually order bounded in $X^{u}$.

Theorem 1**.**

*Let $X$ be an Archimedean vector lattice. TFAE:

$(1)$ $\dim(X)<\infty$;

$(2)$ every $uo$-Cauchy net in $X$ is eventually order bounded in $X^{u}$;

$(3)$ every $uo$-Cauchy net in $X$ is $o$-convergent in $X^{u}$;

$(4)$ every $uo$-null net in $X$ is $o$-null in $X^{u}$;

$(5)$ every $uo$-null net in $X$ is eventually order bounded in $X^{u}$;

$(6)$ every $uo$-convergent net in $X$ is eventually order bounded in $X^{u}$;

$(7)$ every $uo$-convergent net in $X$ is eventually order bounded in $X$.*

Proof.

$(1)\Rightarrow(2)$, $(4)\Rightarrow(5)\Leftrightarrow(6)$, and $(7)\Rightarrow(6)$ are trivial.

$(2)\Rightarrow(3)$: Suppose $x_{\alpha}$ is $uo$-Cauchy in $X$. Then $x_{\alpha}$ is $uo$-Cauchy in $X^{u}$ by [5, Thm.3.2], because $X$ is regular in $X^{u}$. It follows from [3, Thm.17] that $x_{\alpha}\xrightarrow{uo}y$ for some $y\in X^{u}$. Since $x_{\alpha}$ is eventually order bounded in $X^{u}$ by the assumption, then $x_{\alpha}\xrightarrow{o}y$.

$(3)\Rightarrow(4)$ follows since every $uo$-null net is $uo$-Cauchy, $o$-convergent implies $uo$-convergent, and the $uo$-limit of any $uo$-convergent net is unique.

$(5)\Rightarrow(1)$ is Example 2.

$(6)\Rightarrow(7)$ follows from the equivalence $(6)\Leftrightarrow(1)$ because $(1)\Rightarrow(7)$ is obvious. ∎

The equivalence $(1)\Leftrightarrow(7)$ of Theorem 1 justifies use of term “unbounded order convergence” for the $uo$-convergence because the $uo$-convergence for nets in $X$ is automatically eventually order bounded only if $X$ is finite-dimensional.

The following question “suppose that $X$ is an arbitrary Dedekind complete but not universally complete vector lattice. Is there a $uo$-Cauchy net in $X$ that fails to be $o$-convergent in $X^{u}$?” was asked in [3, Prob.23]. Since $X\neq X^{u}$ implies $\dim(X)=\infty$, the equivalence $(1)\Leftrightarrow(3)$ of Theorem 1 gives a positive answer to this question for an arbitrary non-universally complete Archimedean vector lattice $X$.