Multiplicative order convergence in $f$-algebras

TL;DR

This paper introduces and studies the concept of multiplicative order convergence in $f$-algebras, defining related notions such as $mo$-convergence, $mo$-Cauchy, and $mo$-completeness, and explores their fundamental properties.

Contribution

It presents new definitions and foundational properties of multiplicative order convergence in $f$-algebras, expanding the theoretical framework of order convergence.

Findings

Defined $mo$-convergence, $mo$-Cauchy, and $mo$-complete in $f$-algebras

Established basic properties of these notions

Explored the structure of $mo$-KB-spaces

Abstract

A net $(x_\alpha)$ in an $f$-algebra $E$ is said to be multiplicative order convergent to $x\in E$ if $\x_\alpha-x\u\oc 0$ for all $u\in E_+$. In this paper, we introduce the notions $mo$-convergence, $mo$-Cauchy, $mo$-complete, $mo$-continuous and $mo$-KB-space. Moreover, we study the basic properties of these notions.

Equations

$$\left|({\pi }_{\alpha })f-(\pi )f\right|\mu =\left|\mu {\lambda }_{{\pi }_{\alpha }}f-\mu {\lambda }_{\pi }f\right|=\left|{\lambda }_{\mu }{\lambda }_{{\pi }_{\alpha }}f-{\lambda }_{\mu }{\lambda }_{\pi }f\right|=\left|{\lambda }_{\mu }\right|\left|{\lambda }_{{\pi }_{\alpha }}f-{\lambda }_{\pi }f\right|\stackrel{\mathrm{o}}{\to }0$$

$$\left|x_{\alpha }{y}_{\beta }-xy\right|u$$

$$\begin{array}{rl}\left|x_{\alpha }\vee y_{\beta }-x\vee y\right|u& \le \left|x_{\alpha }\vee y_{\beta }-x_{\alpha }\vee y\right|u+\left|x_{\alpha }\vee y-x\vee y\right|u\\ & \le \left|y_{\beta }-y\right|u+\left|x_{\alpha }-x\right|u\le w_{\lambda }+z_{\gamma }\end{array}$$

$$\left|x_n\right|u=\left|x_n\right|u\wedge \left|x_{n_0}\right|u=\left|x_n\right|u\wedge w\stackrel{\mathrm{o}}{\to }0$$

$$|x_{\alpha }|(u+w)=|x_{\alpha }|u+|x_{\alpha }|w=|x_{\alpha }|u\stackrel{\mathrm{o}}{\to }0$$

$$\left|x_n\right|u\le \left|x_n\right|(u-u\wedge ne)+\left|x_n\right|(u\wedge ne)\le \frac{1}{n}{u}^2\left|x_n\right|+n\left|x_n\right|(u\wedge e)$$

$$0\le |y_{\alpha }|u\le |y_{\alpha }|v\stackrel{\mathrm{o}}{\to }0,$$

$$\left|y_{\alpha }\right|u=\left|y_{\alpha }\right|(u_1+u_2)=\left|y_{\alpha }\right|u_1\stackrel{\mathrm{o}}{\to }0$$

$$\left|y_{\alpha }\right|u\le \left|y_{\alpha }\right|\left|u-y\right|+\left|y_{\alpha }\right|\left|y\right|\le \left|y_{\alpha }\right|\left|x\right|+\left|y_{\alpha }\right|\left|y\right|.$$

$$|x_{\alpha }-x_{\beta }|\le z_{\gamma }$$

Full text

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1. Introductory facts

In spite of the nature of the classical theory of Riesz algebra and $f$-algebra, as far as we know, the concept of convergence in $f$-algebras related to multiplication has not been done before. However, there are some close studies under the name unbounded convergence in some kinds of vector lattices; see for example [2, 3, 4, 5]. Our aim is to introduce the concept of $mo$-convergence by using the multiplication in $f$-algebras.

First of all, let us remember some notations and terminologies used in this paper. Let $E$ be a real vector space. Then $E$ is called ordered vector space if it has an order relation $\leq$ (i.e, $\leq$ is reflexive, antisymmetric and transitive) that is compatible with the algebraic structure of $E$ that means $y\leq x$ implies $y+z\leq x+z$ for all $z\in E$ and $\lambda y\leq\lambda x$ for each $\lambda\geq 0$. An ordered vector $E$ is said to be vector lattice (or, Riesz space) if, for each pair of vectors $x,y\in E$, the supremum $x\vee y=\sup\{x,y\}$ and the infimum $x\wedge y=\inf\{x,y\}$ both exist in E. Moreover, $x^{+}:=x\vee 0$, $x^{-}:=(-x)\vee 0$, and $\lvert x\rvert:=x\vee(-x)$ are called the positive part, the negative part, and the absolute value of $x\in E$, respectively. Also, two vector $x$, $y$ in a vector lattice is said to be disjoint whenever $\lvert x\rvert\wedge\lvert y\rvert=0$. A vector lattice $E$ is called order complete if $0\leq x_{\alpha}\uparrow\leq x$ implies the existence of $\sup{x_{\alpha}}$ in $E$. A subset $A$ of a vector lattice is called solid whenever $\left|x\right|\leq\left|y\right|$ and $y\in A$ imply $x\in A$. A solid vector subspace is referred to as an order ideal. An order closed ideal is referred to as a band. A sublattice $Y$ of a vector lattice is majorizing $E$ if, for every $x\in E$, there exists $y\in Y$ with $x\leq y$. A partially ordered set $I$ is called directed if, for each $a_{1},a_{2}\in I$, there is another $a\in I$ such that $a\geq a_{1}$ and $a\geq a_{2}$ (or, $a\leq a_{1}$ and $a\leq a_{2}$). A function from a directed set $I$ into a set $E$ is called a net in $E$. A net $(x_{\alpha})_{\alpha\in A}$ in a vector lattice $X$ is called order convergent (or shortly, $o$-convergent) to $x\in X$, if there exists another net $(y_{\beta})_{\beta\in B}$ satisfying $y_{\beta}\downarrow 0$, and for any $\beta\in B$ there exists $\alpha_{\beta}\in A$ such that $|x_{\alpha}-x|\leq y_{\beta}$ for all $\alpha\geq\alpha_{\beta}$. In this case, we write $x_{\alpha}\xrightarrow{o}x$; for more details see for example [1, 6, 7].

A vector lattice $E$ under an associative multiplication is said to be a Riesz algebra whenever the multiplication makes $E$ an algebra (with the usual properties), and in addition, it satisfies the following property: $x,y\in E_{+}$ implies $xy\in E_{+}$. A Riesz algebra $E$ is called commutative if $xy=yx$ for all $x,y\in E$. A Riesz algebra $E$ is called $f$-algebra if $E$ has additionally property that $x\wedge y=0$ implies $(xz)\wedge y=(zx)\wedge y=0$ for all $z\in E_{+}$; see for example [1]. A vector lattice $E$ is called Archimedean whenever $\frac{1}{n}x\downarrow 0$ holds in $E$ for each $x\in E_{+}$. Every Archimedean $f$-algebra is commutative; see Theorem 140.10 [7]. Assume $E$ is an Archimedean $f$-algebra with a multiplicative unit vector $e$. Then, by applying Theorem 142.1(v) [7], in view of $e=ee=e^{2}\geq 0$, it can be seen that $e$ is a positive vector. In this article, unless otherwise, all vector lattices are assumed to be real and Archimedean, and so $f$-algebras are commutative.

Recall that a net $(x_{\alpha})$ in a vector lattice $E$ is unbounded order convergent (or shortly, $uo$-convergent) to $x\in E$ if $|x_{\alpha}-x|\wedge u\xrightarrow{o}0$ for every $u\in E_{+}$. In this case, we write $x_{\alpha}\xrightarrow{uo}x$; see for example [5] and [2, 3, 4]. Motivated from this definition, we give the following notion.

Definition 1.1**.**

Let $E$ be an $f$-algebra. A net $(x_{\alpha})$ in $E$ is said to be multiplicative order convergent to $x\in E$ (shortly, $(x_{\alpha})$ $mo$-converges to $x$) if $\left|x_{\alpha}-x\right|u\operatorname{\xrightarrow[]{o}}0$ for all $u\in E_{+}$. Abbreviated as $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$.

It is clear that $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ in an $f$-algebra $E$ implies $x_{\alpha}y\operatorname{\xrightarrow[]{mo}}xy$ for all $y\in E$ because of $\left|xy\right|=\left|x\right|\left|y\right|$ for all $x,y\in E$. We shall keep in mind the following useful lemma, obtained from the property of $xy\in E_{+}$ for every $x,y\in E_{+}$.

Lemma 1.2**.**

If $y\leq x$ is provided in an $f$-algebra $E$ then $uy\leq ux$ for all $u\in E_{+}$.

Recall that multiplication by a positive element in $f$-algebras is a vector lattice homomorphism, i.e., $u(x\wedge y)=(ux)\wedge(uy)$ and $u(x\vee y)=(ux)\vee(uy)$ for every positive element $u$; see for example Theorem 142.1(i) [7]. We will denote an $f$-algebra $E$ as infinite distributive $f$-algebra whenever the following condition holds: if $\inf(A)$ exists for any subset $A$ of $E_{+}$ then the infimum of the subset $uA$ exists and $\inf(uA)=u\inf(A)$ for each positive vector $u\in E_{+}$. For a net $(x_{\alpha})\downarrow 0$ in an infinite distributive $f$-algebra, the net $(ux_{\alpha})$ is also decreasing to zero for all positive vector $u$.

Remark 1.3**.**

The order convergence implies the $mo$-convergence in infinite distributive $f$-algebras. The converse holds true in $f$-algebras with multiplication unit. Indeed, assume a net $(x_{\alpha})_{\alpha\in A}$ order converges to $x$ in an infinite distributive $f$-algebra $E$. Then there exists another net $(y_{\beta})_{\beta\in B}$ satisfying $y_{\beta}\downarrow 0$, and, for any $\beta\in B$, there exists $\alpha_{\beta}\in A$ such that $|x_{\alpha}-x|\leq y_{\beta}$. Hence, we have $|x_{\alpha}-x|u\leq y_{\beta}u$ for all $\alpha\geq\alpha_{\beta}$ and for each $u\in E_{+}$. Since $y_{\beta}\downarrow$, we have $uy_{\beta}\downarrow$ for each $u\in E_{+}$ by Lemma 1.2, and $\inf(uy_{\beta})=u\inf(y_{\beta})=0$ because of $\inf(y_{\beta})=0$. Therefore, $\left|x_{\alpha}-x\right|u\operatorname{\xrightarrow[]{o}}0$ for each $u\in E_{+}$. That means $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$.

For the converse, assume $E$ is an $f$-algebra with multiplication unit $e$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ in $E$. That is, $\left|x_{\alpha}-x\right|u\operatorname{\xrightarrow[]{o}}0$ for all $u\in E_{+}$. Since $e\in E_{+}$, in particular, choose $u=e$, and so we have $\left|x_{\alpha}-x\right|=\left|x_{\alpha}-x\right|e\operatorname{\xrightarrow[]{o}}0$, or $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$ in $E$.

By considering Example 141.5 [7], we give the following example.

Example 1.4**.**

Let $[a,b]$ be a closed interval in $\mathbb{R}$ and let $E$ be vector lattice of all reel continuous functions on $[a,b]$ such that the graph of functions consists of a finite number of line segments. In view of Theorem 141.1 [7], every positive orthomorphism $\pi$ in $E$ is trivial orthomorphism, i.e., there is a reel number $\lambda$ such that $\pi(f)=\lambda f$ for all $f\in E$. Therefore, a net of positive orthomorphism $(\pi_{\alpha})$ is order convergent to $\pi$ iff it is $mo$-convergent to $\pi$ whenever the multiplication is the natural multiplicative, i.e., $\pi_{1}\pi_{2}(f)=\pi_{1}(\pi_{2}f)$ for all $\pi_{1},\pi_{2}\in Orth(E)$ and all $f\in E$. Indeed, $Orth(E)$ is Archimedean $f$-algebra with the identity operator as a unit element; see Theorem 140.4 [7]. So, by applying Remark 1.3, the $mo$-convergence implies the order convergence of the net $(\pi_{\alpha})$.

Conversely, assume the net of positive orthomorphisms $\pi_{\alpha}\operatorname{\xrightarrow[]{o}}\pi$ in $Orth(E)$. Then we have $\pi_{\alpha}(f)\operatorname{\xrightarrow[]{o}}\pi(f)$ for all $f\in E$; see Theorem VIII.2.3 [6]. For fixed $0\leq\mu\in Orth(E)$, there is a reel number $\lambda_{\mu}$ such that $\mu(f)=\lambda_{\mu}f$ for all $f\in E$. Since $\left|\pi_{\alpha}(f)-\pi(f)\right|=\left|\lambda_{\pi_{\alpha}}f-\lambda_{\pi}f\right|\operatorname{\xrightarrow[]{o}}0$, we have

[TABLE]

for all $f\in E$. Since $\mu$ is arbitrary, we get $\pi_{\alpha}\operatorname{\xrightarrow[]{mo}}\pi$.

2. Main results

We begin the section with the next list of properties of the $mo$-convergence which follows directly from Lemma 1.2, and the inequalities $\left|x-y\right|\leq\left|x-x_{\alpha}\right|+\left|x_{\alpha}-y\right|$ and $\left|\left|x_{\alpha}\right|-\left|x\right|\right|\leq\left|x_{\alpha}-x\right|$.

Lemma 2.1**.**

Let $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\alpha}\operatorname{\xrightarrow[]{mo}}y$ in an $f$-algebra $E$. Then the following holds:

1. (i)

$x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$* iff $(x_{\alpha}-x)\operatorname{\xrightarrow[]{mo}}0$;* 2. (ii)

if $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ then $y_{\beta}\operatorname{\xrightarrow[]{mo}}x$ for each subnet $(y_{\beta})$ of $(x_{\alpha})$; 3. (iii)

suppose $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$ then $ax_{\alpha}+by_{\beta}\operatorname{\xrightarrow[]{mo}}ax+by$ for any $a,b\in\mathbb{R}$; 4. (iv)

if $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}y$ then $x=y$; 5. (v)

if $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ then $\lvert x_{\alpha}\rvert\operatorname{\xrightarrow[]{mo}}\lvert x\rvert$.

Recall that an order complete vector lattice $E^{\delta}$ is said to be an order completion of the vector lattice $E$ whenever $E$ is Riesz isomorphic to a majorizing order dense vector lattice subspace of $E^{\delta}$. Every Archimedean Riesz space has a unique order completion; see Theorem 2.24 [1].

Proposition 2.2**.**

Let $(x_{\alpha})$ be a net in an $f$-algebra $E$. Then $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$ iff $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in the order completion $E^{\delta}$ of $E$.

Proof 2.3**.**

Assume $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$. Then $\left|x_{\alpha}\right|u\operatorname{\xrightarrow[]{o}}0$ in $E$ for all $u\in E_{+}$, and so $\left|x_{\alpha}\right|u\operatorname{\xrightarrow[]{o}}0$ in $E^{\delta}$ for all $u\in E_{+}$; see Corollary 2.9 [5]. Now, let’s fix $v\in E^{\delta}_{+}$. Then there exists $x_{v}\in E_{+}$ such that $v\leq x_{v}$ because $E$ majorizes $E^{\delta}$. Then we have $\left|x_{\alpha}\right|v\leq\left|x_{\alpha}\right|x_{v}$. From $\left|x_{\alpha}\right|x_{v}\operatorname{\xrightarrow[]{o}}0$ in $E^{\delta}$ it follows that $\left|x_{\alpha}\right|v\operatorname{\xrightarrow[]{o}}0$ in $E^{\delta}$, that is, $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in the order completion $E^{\delta}$ because $v\in E^{\delta}_{+}$ is arbitrary.

Conversely, assume $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E^{\delta}$. Then, for all $u\in E^{\delta}_{+}$, we have $\left|x_{\alpha}\right|u\operatorname{\xrightarrow[]{o}}0$ in $E^{\delta}$. In particular, for all $x\in E_{+}$, $\left|x_{\alpha}\right|x\operatorname{\xrightarrow[]{o}}0$ in $E^{\delta}$. By Corollary 2.9 [5], we get $\left|x_{\alpha}\right|x\operatorname{\xrightarrow[]{o}}0$ in $E$ for all $x\in E_{+}$. Hence $x_{\alpha}\operatorname{\xrightarrow[]{mo}}$ in $E$.

The multiplication in $f$-algebra is $mo$-continuous in the following sense.

Theorem 2.4**.**

Let $E$ be an infinite distributive $f$-algebra, and $(x_{\alpha})_{\alpha\in A}$ and $(y_{\beta})_{\beta\in B}$ be two nets in $E$. If $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$ for some $x,y\in E$ and each positive element of $E$ can be written as a multiplication of two positive elements then $x_{\alpha}y_{\beta}\operatorname{\xrightarrow[]{mo}}xy$.

Proof 2.5**.**

Assume $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$. Then $\left|x_{\alpha}-x\right|u\operatorname{\xrightarrow[]{o}}0$ and $\left|y_{\beta}-y\right|u\operatorname{\xrightarrow[]{o}}0$ for every $u\in E_{+}$. Let’s fix $u\in E_{+}$. So, there exist another two nets $(z_{\gamma})_{\gamma\in\Gamma}\downarrow 0$ and $(z_{\xi})_{\xi\in\Xi}\downarrow 0$ in $E$ such that, for all $(\gamma,\xi)\in\Gamma\times\Xi$ there are $\alpha_{\gamma}\in A$ and $\beta_{\xi}\in B$ with $\left|x_{\alpha}-x\right|u\leq z_{\gamma}$ and $\left|y_{\beta}-y\right|u\leq z_{\xi}$ for all $\alpha\geq\alpha_{\gamma}$ and $\beta\geq\beta_{\xi}$.

Next, we show the $mo$-convergence of $(x_{\alpha}y_{\beta})$ to $xy$. By considering the equality $\left|xy\right|=\left|x\right|\left|y\right|$ and Lemma 1.2, we have

[TABLE]

The second and the third terms in the last inequality both order converge to zero as $\beta\to\infty$ and $\alpha\to\infty$ respectively because of $\left|x\right|u,\left|y\right|u\in E_{+}$, $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$.

Now, let’s show the convergence of the first term of last inequality. There are two positive elements $u_{1},u_{2}\in E_{+}$ such that $u=u_{1}u_{2}$ because the positive element of $E$ can be written as a multiplication of two positive elements. So, we get $\left|x_{\alpha}-x\right|\left|y_{\beta}-y\right|u=(\left|x_{\alpha}-x\right|u_{1})(\left|y_{\beta}-y\right|u_{2})$. Since $(z_{\gamma})_{\gamma\in\Gamma}\downarrow 0$ and $(z_{\xi})_{\xi\in\Xi}\downarrow 0$, the multiplication $(z_{\gamma}z_{\xi})\downarrow 0$. Indeed, we firstly show that the multiplication is decreasing. For indexes $(\gamma_{1},\xi_{1})(\gamma_{2},\xi_{2})\in\Gamma\times\Xi$, we have $z_{\gamma_{2}}\leq z_{\gamma_{1}}$ and $z_{\xi_{2}}\leq z_{\xi_{1}}$ because both of them are decreasing. Since the nets are positive, it follows from $z_{\xi_{2}}\leq z_{\xi_{1}}$ that $z_{\gamma_{2}}z_{\xi_{2}}\leq z_{\gamma_{2}}z_{\xi_{1}}\leq z_{\gamma_{1}}z_{\xi_{1}}$. As a result $(z_{\gamma}z_{\xi})_{(\gamma,\xi)\in\Gamma\times\Xi}\downarrow$. Now, we show that the infimum of multiplication is zero. For a fixed index $\gamma_{0}$, we have $z_{\gamma}z_{\xi}\leq z_{\gamma_{0}}z_{\xi}$ for $\gamma\geq\gamma_{0}$ because $(z_{\gamma})$ is decreasing. Thus, we get $\inf(z_{\gamma}z_{\xi})=0$ because of $\inf(z_{\gamma_{0}}z_{\xi})=z_{\gamma_{0}}\inf(z_{\xi})=0$. Therefore, we see $(\left|x_{\alpha}-x\right|u_{1})(\left|y_{\beta}-y\right|u_{2})\operatorname{\xrightarrow[]{o}}0$. Hence, we get $x_{\alpha}y_{\beta}\operatorname{\xrightarrow[]{mo}}xy$.

The lattice operations in an $f$-algebra are $mo$-continuous in the following sense.

Proposition 2.6**.**

Let $(x_{\alpha})_{\alpha\in A}$ and $(y_{\beta})_{\beta\in B}$ be two nets in an $f$-algebra $E$. If $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$ then $(x_{\alpha}\vee y_{\beta})_{(\alpha,\beta)\in A\times B}\operatorname{\xrightarrow[]{mo}}x\vee y$. In particular, $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ implies $x_{\alpha}^{+}\operatorname{\xrightarrow[]{mo}}x^{+}$.

Proof 2.7**.**

Assume $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $y_{\beta}\operatorname{\xrightarrow[]{mo}}y$. Then there exist two nets $(z_{\gamma})_{\gamma\in\Gamma}$ and $(w_{\lambda})_{\lambda\in\Lambda}$ in $E$ satisfying $z_{\gamma}\downarrow 0$ and $w_{\lambda}\downarrow 0$, and for all $(\gamma,\lambda)\in\Gamma\times\Lambda$ there are $\alpha_{\gamma}\in A$ and $\beta_{\lambda}\in B$ such that $\left|x_{\alpha}-x\right|u\leq z_{\gamma}$ and $\left|y_{\beta}-y\right|u\leq w_{\lambda}$ for all $\alpha\geq\alpha_{\gamma}$ and $\beta\geq\beta_{\lambda}$ and for every $u\in E_{+}$. It follows from the inequality $|a\vee b-a\vee c|\leq|b-c|$ in vector lattices that

[TABLE]

for all $\alpha\geq\alpha_{\gamma}$ and $\beta\geq\beta_{\lambda}$ and for every $u\in E_{+}$. Since $(w_{\lambda}+z_{\gamma})\downarrow 0$, $\left|x_{\alpha}\vee y_{\beta}-x\vee y\right|u$ order converges to [math] for all $u\in E_{+}$. That is, $(x_{\alpha}\vee y_{\beta})_{(\alpha,\beta)\in A\times B}\operatorname{\xrightarrow[]{mo}}x\vee y$.

Lemma 2.8**.**

Let $(x_{\alpha})$ be a net in an $f$-algebra $E$. Then

1. (i)

$0\leq x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$* implies $x\in E_{+}$.* 2. (ii)

if $(x_{\alpha})$ is monotone and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ then implies $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$.

Proof 2.9**.**

$(i)$* Assume $0\leq x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$. Then we have $x_{\alpha}=x_{\alpha}^{+}\operatorname{\xrightarrow[]{mo}}x^{+}=0$ by Proposition 2.6. Hence, we get $x\in E_{+}$.*

$(ii)$* We show that $x_{\alpha}\uparrow$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ implies $x_{\alpha}\uparrow x$. Fix an index $\alpha$. Then we have $x_{\beta}-x_{\alpha}\in X_{+}$ for $\beta\geq\alpha$. By $(i)$, $x_{\beta}-x_{\alpha}\operatorname{\xrightarrow[]{mo}}x-x_{\alpha}\in X_{+}$. Therefore, $x\geq x_{\alpha}$ for any $\alpha$. Since $\alpha$ is arbitrary, then $x$ is an upper bound of $(x_{\alpha})$. Assume $y$ is another upper bound of $(x_{\alpha})$, i.e., $y\geq x_{\alpha}$ for all $\alpha$. So, $y-x_{\alpha}\operatorname{\xrightarrow[]{mo}}y-x\in X_{+}$, or $y\geq x$, and so $x_{\alpha}\uparrow x$.*

The following simple observation is useful in its own right.

Proposition 2.10**.**

Every disjoint decreasing sequence in an $f$-algebra $mo$-converges to zero.

Proof 2.11**.**

Suppose $(x_{n})$ is a disjoint and decreasing sequence in an $f$-algebra $E$. So, $\left|x_{n}\right|u$ is also a disjoint sequence in $E$ for all $u\in E_{+}$; see Theorem 142.1(iii) [7]. Fix $u\in E_{+}$, by Corollary 3.6 [5], we have $\left|x_{n}\right|u\operatorname{\xrightarrow[]{uo}}0$ in $E$. So, $\left|x_{n}\right|u\wedge w\operatorname{\xrightarrow[]{o}}0$ in $E$ for all $w\in E_{+}$. Thus, in particular for fixed $n_{0}$, taking $w$ as $\left|x_{n_{0}}\right|u$. Then, for all $n\geq n_{0}$, we get

[TABLE]

because of $\left|x_{n}\right|u\leq\left|x_{n_{0}}\right|u$. Therefore, $x_{n}\operatorname{\xrightarrow[]{mo}}0$ in $E$.

For the next two facts, observe the following fact. Let $E$ be a vector lattice, $I$ be an order ideal of $E$ and $(x_{\alpha})$ be a net in $I$. If $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$ in $I$ then $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$ in $E$. Conversely, if $(x_{\alpha})$ is order bounded in $I$ and $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$ in $E$ then $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$ in $I$.

Proposition 2.12**.**

Let $E$ be an $f$-algebra, $B$ be a projection band of $E$ and $P_{B}$ be the corresponding band projection. If $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ in $E$ then $P_{B}(x_{\alpha})\operatorname{\xrightarrow[]{mo}}P_{B}(x)$ in both $E$ and $B$.

Proof 2.13**.**

It is known that $P_{B}$ is a lattice homomorphism and $0\leq P_{B}\leq I$. It follows from $\lvert P_{B}(x_{\alpha})-P_{B}(x)\rvert=P_{B}\lvert x_{\alpha}-x\rvert\leq\lvert x_{\alpha}-x\rvert$ that $\lvert P_{B}(x_{\alpha})-P_{B}(x)\rvert u\leq\lvert x_{\alpha}-x\rvert u$ for all $u\in E_{+}$. Then it follows easily that $P_{B}(x_{\alpha})\operatorname{\xrightarrow[]{mo}}P_{B}(x)$ in both X and B.

Theorem 2.14**.**

Let $E$ be an $f$-algebra and $I$ be an order ideal and sub-$f$-algebra of $E$. For an order bounded net $(x_{\alpha})$ in $I$, $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $I$ iff $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$.

Proof 2.15**.**

Suppose $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$. Then for any $u\in I_{+}$, we have $\lvert x_{\alpha}\rvert u\operatorname{\xrightarrow[]{o}}0$ in $E$. So, the preceding remark implies $\lvert x_{\alpha}\rvert u\operatorname{\xrightarrow[]{o}}0$ in $I$ because $\lvert x_{\alpha}\rvert u$ is order bounded in $I$ . Therefore, we get $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $I$.

Conversely, assume that $(x_{\alpha})$ $mo$-converges to zero in $I$. For any $u\in I_{+}$, we have $\lvert x_{\alpha}\rvert u\operatorname{\xrightarrow[]{o}}0$ in $I$, and so in $E$. Then, by applying Theorem 142.1(iv) [7], we have $x_{\alpha}w=0$ for all $w\in I^{d}=\{x\in E:x\perp y\ for\ all\ y\in I\}$ and for each $\alpha$ because $(x_{\alpha})$ in $I$. For any $u\in I_{+}$ and each $0\leq w\in I^{d}$, it follows that

[TABLE]

in $E$. So that, for each $z\in(I\oplus I^{d})_{+}$, we get $\lvert x_{\alpha}\rvert z\operatorname{\xrightarrow[]{o}}0$ in $E$. It is known that $I\oplus I^{d}$ is order dense in $E$; see Theorem 1.36 [1]. Fix $v\in E_{+}$. Then there exists some $u\in(I\oplus I^{d})$ such that $v\leq u$. Thus, we have $\lvert x_{\alpha}\rvert v\leq\lvert x_{\alpha}\rvert u\operatorname{\xrightarrow[]{o}}0$ in $E$. Therefore, $\lvert x_{\alpha}\rvert v\operatorname{\xrightarrow[]{o}}0$, and so $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$.

The following proposition extends Theorem 3.8 [2] to the general setting.

Theorem 2.16**.**

Let $E$ be an infinite distributive $f$-algebra with a unit $e$ and $(x_{n})\downarrow$ be a sequence in $E$. Then $x_{n}\operatorname{\xrightarrow[]{mo}}0$ iff $\left|x_{n}\right|(u\wedge e)\operatorname{\xrightarrow[]{o}}0$ for all $u\in E_{+}$.

Proof 2.17**.**

For the forward implication, assume $x_{n}\operatorname{\xrightarrow[]{mo}}0$. Hence, $\left|x\right|u\operatorname{\xrightarrow[]{o}}0$ for all $u\in E_{+}$, and so $\left|x_{n}\right|(u\wedge e)\leq\left|x_{n}\right|u\operatorname{\xrightarrow[]{o}}0$ because of $e\in E_{+}$. Therefore, $\left|x_{n}\right|(u\wedge e)\operatorname{\xrightarrow[]{o}}0$.

For the reverse implication, fix $u\in E_{+}$. By applying Theorem 2.57 [1] and Theorem 142.1(i) [7], note that

[TABLE]

Since $(x_{n})\downarrow$ and $E$ is Archimedean, we have $\frac{1}{n}u^{2}\left|x_{n}\right|\downarrow 0$. Furthermore, it follows from $\left|x_{n}\right|(u\wedge e)\operatorname{\xrightarrow[]{o}}0$ for each $u\in E_{+}$ that there exists another sequence $(y_{m})_{m\in B}$ satisfying $y_{m}\downarrow 0$, and for any $m\in B$, there exists $n_{m}$ such that $\left|x_{n}\right|(u\wedge e)\leq\frac{1}{n}y_{m}$, or $n\left|x_{n}\right|(u\wedge e)\leq y_{m}$ for all $n\geq n_{m}$. Hence, we get $n\left|x_{n}\right|(u\wedge e)\operatorname{\xrightarrow[]{o}}0$. Therefore, we have $\left|x_{n}\right|u\operatorname{\xrightarrow[]{o}}0$, and so $x_{n}\operatorname{\xrightarrow[]{mo}}0$.

The $mo$-convergence passes obviously to any sub-$f$-algebra $Y$ of $E$, i.e., for any net $(y_{\alpha})$ in $Y$, $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$ implies $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $Y$. For the converse, we give the following theorem.

Theorem 2.18**.**

Let $Y$ be a sub-$f$-algebra of an $f$-algebra $E$ and $(y_{\alpha})$ be a net in $Y$. If $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $Y$ then it $mo$-converges to zero in $E$ for each of the following cases;

1. (i)

$Y$* is majorizing in $E$;* 2. (ii)

$Y$* is a projection band in $E$;* 3. (iii)

if, for each $u\in E$, there are element $x,y\in Y$ such that $\left|u-y\right|\leq|x|$.

Proof 2.19**.**

Assume $(y_{\alpha})$ is a net in $Y$ and $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $Y$. Let’s fix $u\in E_{+}$.

$(i)$* Since $Y$ is majorizing in $E$, there exists $v\in Y_{+}$ such that $u\leq v$. It follows from*

[TABLE]

that $|y_{\alpha}|u\operatorname{\xrightarrow[]{o}}0$ in $E$. That is, $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$.

$(ii)$* Since $Y$ is a projection band in $E$, we have $Y=Y^{\bot\bot}$ and $E=Y\oplus Y^{\bot}$. Hence $u=u_{1}+u_{2}$ with $u_{1}\in Y_{+}$ and $u_{2}\in Y^{\bot}_{+}$. Thus, we have $y_{\alpha}\wedge u_{2}=0$ because $(y_{\alpha})$ in $Y$ and $u_{2}\in Y^{\bot}$. Hence, by applying Theorem 142.1(iii) [7], we see $y_{\alpha}u=0$ for all index $\alpha$. It follows from*

[TABLE]

tha $\left|y_{\alpha}\right|u\operatorname{\xrightarrow[]{o}}0$ in $E$. Therefore, $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$.

$(iii)$* For the given $u\in E_{+}$, there exists elements $x,y\in Y$ with $\left|u-y\right|\leq|x|$. Then*

[TABLE]

By $mo$-convergence of $(y_{\alpha})$ in $Y$, we have $\left|y_{\alpha}\right|\left|x\right|\operatorname{\xrightarrow[]{o}}0$ and $\left|y_{\alpha}\right|\left|y\right|\operatorname{\xrightarrow[]{o}}0$, and so $\left|y_{\alpha}\right|u\operatorname{\xrightarrow[]{o}}0$. That means $y_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$ because $u$ is arbitrary in $E_{+}$.

We continue with some basic notions in $f$-algebra, which are motivated by their analogies from vector lattice theory.

Definition 2.20**.**

Let $(x_{\alpha})_{\alpha\in A}$ be a net in $f$-algebra $E$. Then

1. (i)

$(x_{\alpha})$* is said to be $mo$-Cauchy if the net $(x_{\alpha}-x_{\alpha^{\prime}})_{(\alpha,\alpha^{\prime})\in A\times A}$ $mo$-converges to [math],* 2. (ii)

$E$* is called $mo$-complete if every $mo$-Cauchy net in $E$ is $mo$-convergent,* 3. (iii)

$E$* is called $mo$-continuous if $x_{\alpha}\operatorname{\xrightarrow[]{o}}0$ implies $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$,* 4. (iv)

$E$* is called a $mo$-KB-space if every order bounded increasing net in $E_{+}$ is $mo$-convergent.*

Remark 2.21**.**

An $f$-algebra $E$ is $mo$-continuous iff $x_{\alpha}\downarrow 0$ in $E$ implies $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$. Indeed, the implication is obvious. For the converse, consider a net $x_{\alpha}\operatorname{\xrightarrow[]{o}}0$. Then there exists a net $z_{\beta}\downarrow 0$ in $X$ such that, for any $\beta$ there exists $\alpha_{\beta}$ so that $|x_{\alpha}|\leq z_{\beta}$ for all $\alpha\geq\alpha_{\beta}$. Hence, by $mo$-continuity of $E$, we have $z_{\beta}\operatorname{\xrightarrow[]{mo}}0$, and so $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$.

Proposition 2.22**.**

Let $(x_{\alpha})$ be a net in an $f$-algebra $E$. If $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $(x_{\alpha})$ is an $o$-Cauchy net then $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$. Moreover, if $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $(x_{\alpha})$ is $uo$-Cauchy then $x_{\alpha}\operatorname{\xrightarrow[]{uo}}x$.

Proof 2.23**.**

Assume $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ and $(x_{\alpha})$ is an order Cauchy net in $E$. Then $x_{\alpha}-x_{\beta}\operatorname{\xrightarrow[]{o}}0$ as $\alpha,\beta\to\infty$. Thus, there exists another net $z_{\gamma}\downarrow 0$ in $E$ such that, for every $\gamma$, there exists $\alpha_{\gamma}$ satisfying

[TABLE]

for all $\alpha,\beta\geq\alpha_{\gamma}$. By taking $f$-limit over $\beta$ the above inequality and applying Proposition 2.6, i.e., $\left|x_{\alpha}-x_{\beta}\right|\operatorname{\xrightarrow[]{mo}}\left|x_{\alpha}-x\right|$, we get $|x_{\alpha}-x|\leq z_{\gamma}$ for all $\alpha\geq\alpha_{\gamma}$. That means $x_{\alpha}\operatorname{\xrightarrow[]{o}}x$. The similar argument can be applied for the $uo$-convergence case, and so the proof is omitted.

In the case of $mo$-complete in $f$-algebras, we have conditions for $mo$-continuity.

Theorem 2.24**.**

For an $mo$-complete $f$-algebra $E$, the following statements are equivalent:

1. (i)

$E$* is $mo$-continuous;* 2. (ii)

if $0\leq x_{\alpha}\uparrow\leq x$ holds in $E$ then $x_{\alpha}$ is a $mo$-Cauchy net; 3. (iii)

$x_{\alpha}\downarrow 0$* implies $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ in $E$.*

Proof 2.25**.**

$(i)\Rightarrow(ii)$* Consider the increasing and bounded net $0\leq x_{\alpha}\uparrow\leq x$ in $E$. Then there exists a net $(y_{\beta})$ in $E$ such that $(y_{\beta}-x_{\alpha})_{\alpha,\beta}\downarrow 0$; see Lemma 12.8 [1]. Thus, by applying Remark 2.21, we have $(y_{\beta}-x_{\alpha})_{\alpha,\beta}\operatorname{\xrightarrow[]{mo}}0$, and so the net $(x_{\alpha})$ is $mo$-Cauchy because of $\left|x_{\alpha}-x_{\alpha^{\prime}}\right|_{\alpha,\alpha^{\prime}\in A}\leq\left|x_{\alpha}-y_{\beta}\right|+\left|y_{\beta}-x_{\alpha^{\prime}}\right|$.*

$(ii)\Rightarrow(iii)$* Suppose that $x_{\alpha}\downarrow 0$ in $E$, and fix arbitrary $\alpha_{0}$. Then we have $x_{\alpha}\leq x_{\alpha_{0}}$ for all $\alpha\geq\alpha_{0}$. Thus we can get $0\leq(x_{\alpha_{0}}-x_{\alpha})_{\alpha\geq\alpha_{0}}\uparrow\leq x_{\alpha_{0}}$. So, it follows from $(ii)$ that the net $(x_{\alpha_{0}}-x_{\alpha})_{\alpha\geq\alpha_{0}}$ is $mo$-Cauchy, i.e., $(x_{\alpha^{{}^{\prime}}}-x_{\alpha})\operatorname{\xrightarrow[]{mo}}0$ as $\alpha_{0}\leq\alpha,\alpha^{{}^{\prime}}\to\infty$. Then there exists $x\in E$ satisfying $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ as $\alpha_{0}\leq\alpha\to\infty$ because $E$ is $mo$-complete. Since $x_{\alpha}\downarrow$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$, it follows from Lemma 2.8 that $x_{\alpha}\downarrow 0$, and so we have $x=0$. Therefore, we get $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$.*

$(iii)\Rightarrow(i)$* It is just the implication of Remark 2.21.*

Corollary 2.26**.**

Let $E$ be an $mo$-continuous and $mo$-complete $f$-algebra. Then $E$ is order complete.

Proof 2.27**.**

Suppose $0\leq x_{\alpha}\uparrow\leq u$ in $E$. We show the existence of supremum of $(x_{\alpha})$. By considering Theorem 2.24 $(ii)$, we see that $(x_{\alpha})$ is an $mo$-Cauchy net. Hence, there is $x\in E$ such that $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ because $E$ is $mo$-complete. It follows from Lemma 2.8 that $x_{\alpha}\uparrow x$ because of $x_{\alpha}\uparrow$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$. Therefore, $E$ is order complete.

Proposition 2.28**.**

Every $mo$-KB-space is $mo$-continuous.

Proof 2.29**.**

Assume $x_{\alpha}\downarrow 0$ in $E$. From Theorem 2.24, it is enough to show $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$. Let’s fix an index $\alpha_{0}$, and define another net $y_{\alpha}:=x_{\alpha_{0}}-x_{\alpha}$ for $\alpha\geq\alpha_{0}$. Then it is clear that $0\leq y_{\alpha}\uparrow\leq x_{\alpha_{0}}$, i.e., $(y_{\alpha})$ is increasing and order bounded net in $E$. Since $E$ is a $mo$-KB-space, there exists $y\in E$ such that $y_{\alpha}\operatorname{\xrightarrow[]{mo}}y$. Thus, by Lemma 2.8, we have $y_{\alpha}\operatorname{\xrightarrow[]{o}}y$. Hence, $y=\sup\limits_{\alpha\geq\alpha_{0}}y_{\alpha}=\sup\limits_{\alpha\geq\alpha_{0}}(x_{\alpha_{0}}-x_{\alpha})=x_{\alpha_{0}}$ because of $x_{\alpha}\downarrow 0$. Therefore, we get $y_{\alpha}=x_{\alpha_{0}}-x_{\alpha}\operatorname{\xrightarrow[]{mo}}x_{\alpha_{0}}$ or $x_{\alpha}\operatorname{\xrightarrow[]{mo}}0$ because of $y_{\alpha}\operatorname{\xrightarrow[]{mo}}y$.

Proposition 2.30**.**

Every $mo$-KB-space is order complete.

Proof 2.31**.**

Suppose $0\leq x_{\alpha}\uparrow\leq z$ is an order bounded and increasing net in an $mo$-KB-space $E$ for some $z\in E_{+}$. Then $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$ for some $x\in E$ because $E$ is $mo$-KB-space. By Lemma 2.8, we have $x_{\alpha}\uparrow x$ because of $x_{\alpha}\uparrow$ and $x_{\alpha}\operatorname{\xrightarrow[]{mo}}x$. So, $E$ is order complete.

Proposition 2.32**.**

Let $Y$ be an sub-$f$-algebra and order closed sublattice of an $mo$-KB-space $E$. Then $Y$ is also a $mo$-KB-space.

Proof 2.33**.**

Let $(y_{\alpha})$ be a net in $Y$ such that $0\leq y_{\alpha}\uparrow\leq y$ for some $y\in Y_{+}$. Since $E$ is a $mo$-KB-space, there exists $x\in E_{+}$ such that $y_{\alpha}\operatorname{\xrightarrow[]{mo}}x$. By Lemma 2.8, we have $y_{\alpha}\uparrow x$, and so $x\in Y$ because $Y$ is order closed. Thus $Y$ is a $mo$-KB-space.