Convex Subalgebras and Convex Spectral Topology on Autometrized Algebras

ABSTRACT

In the development of the theory of autometrized algebra, various types of research have been conducted. However, there are some properties like convex subalgebra, prime convex subalgebra, meet closed sets, regular convex subalgebra, and convex spectral topology on autometrized algebras that have not been studied yet. In this paper, we define the notions of convex subalgebras and congruence relations on an autometrized algebra. We demonstrate that the collection of all convex subalgebras of an autometrized algebra forms a lattice and distributive. In particular, we will show that there exists a one-to-one correspondence between the set of all convex subalgebras and the set of all congruences on an np-autometrized algebra. Furthermore, we explore prime convex subalgebras, meet closed subsets, and regular convex subalgebras and obtain some related results. For instance, we show that in a semiregular np-autometrized l-algebra, the intersection of a chain of prime convex subalgebra is a prime convex subalgebra. We also prove that any convex subalgebra in an autometrized algebra is the intersection of regular convex subalgebras. Lastly, we introduce the convex spectral topology of proper prime convex subalgebras in an autometrized l-algebra and discuss some fundamental facts. We also prove that a convex spectrum is compact in an np-autometrized l-algebra A if and only if A is generated by some element. Specifically, we demonstrate that the convex spectrum is a T₁ - space and Hausdorff space.

KEYWORDS:

• autometrized algebra
• convex subalgebra
• congruence relation
• np-autometrized algebra
• prime convex subalgebra
• convex spectrum
• convex spectral topology

1. Introduction

Swamy (1964) introduced the concept of autometrized algebra to formulate a comprehensive theory that includes the existing autometrized algebras such as Boolean algebras (Blumenthal (1952) and Ellis (1951)), Brouwerian algebras (Nordhaus & Lapidus, 1954), Newman algebras (Roy, 1960), autometrized lattices (Nordhaus & Lapidus, 1954), and commutative lattice ordered groups or l-groups (Narasimha Swamy, 1964). Swamy and Rao (1977), Rachŭnek (1987); Rachŭnek (1989); Rachŭnek (1990); Rachŭnek (1998), Hansen (1994), Kovář (2000), and Chajda and Rachunek (2001) further developed the theory of autometrized algebra.

Moreover, the notion of representable autometrized algebras was explored by Subba Rao and Yedlapalli (2018), as well as by Rao et al. (2019); Rao et al. (2021); Rao et al. (2022). Tilahun et al. (2023b); Tilahun et al. (2023a) established the theory of strong ideals and monoid autometrized algebras. They also studied the relationships among normal autometrized semialgebras, normal autometrized l-algebras, and representable autometrized algebras.

The studies mentioned above did not investigate the properties of convex subalgebras, prime convex subalgebras, and regular convex subalgebras. Moreover, the convex spectral topology on autometrized algebras remained unexplored. This fact strongly motivates us to conduct further research in these areas.

The purpose of this paper is to introduce the concept of convex subalgebra in an autometrized algebra. This includes discussing prime convex subalgebras and regular convex subalgebras. We will also be constructing a topology known as the convex spectral topology of proper prime convex subalgebras in an autometrized l-algebra. This might be seen as a continuation of the research carried out by Tilahun et al. (2023b) and an expansion of the spectral topology introduced by Rachŭnek (1998).

This paper will be organized as follows. Section 2 recalls some definitions and terms. In Section 3, we introduce the concepts of convex subalgebra, prime convex subalgebra, and regular convex subalgebra. Section 4 presents the convex spectral topology of proper prime convex subalgebras in an autometrized l-algebra and discusses some fundamental facts. In Section 5, we discuss major results. Finally, Section 6 concludes the paper.

2. Preliminaries

This section covers some basic concepts, definitions, and terms needed in other sections.

Definition 2.1.

Swamy (1964) A system A = $(A,+,0,\text{break}\le ,\ast )$ is called an autometrized algebra if

(i)	=	$(A,+,0)$ is a commutative monoid.
(ii)	=	$(A,\le )$ is a partial ordered set, and ≤ is translation invariant, that is, $\mathrm{\forall }a,b,c\in A;a\le b\Rightarrow a+c\le b+c$.
(iii)	=	$\ast :A\times A\to A$ is autometric on A, that is, $\ast$ satisfies metric operation axioms: $(M_1)$ = $\mathrm{\forall }a,b\in A;a\ast b\ge 0$ and, $a\ast b=0\iff a=b$, $(M_2)$ = $\mathrm{\forall }a,b\in A;a\ast b=b\ast a$, $(M_3)$ = $\mathrm{\forall }a,b,c\in A$; $a\ast c\le a\ast b+b\ast c$.

The following definitions of autometrized algebra (Definition (2.2)—Definition (2.5)) was suggested by Swamy and Rao (1977).

Definition 2.2.

An autometrized algebra A = $(A,+,0,\text{break}\le ,\ast )$ is called normal if and only if

(i)	=	$a\le a\ast 0\mathrm{\forall }a\in A$.
(ii)	=	$(a+c)\ast (b+d)\le (a\ast b)+(c\ast d)\mathrm{\forall }a,b,c,d\in A$.
(iii)	=	$(a\ast c)\ast (b\ast d)\le (a\ast b)+(c\ast d)\mathrm{\forall }a,b,c,d\in A$.
(iv)	=	For any a and b in A, $a\le b\Rightarrow \mathrm{\exists }x\ge 0$ such that $a+x=b$.

Definition 2.3.

Let $A=(A,+,0,\le ,\ast )$ be a system. Then A is said to be a lattice ordered autometrized algebra (or) autometrized l-algebra if

(i)	=	$(A,+,0)$ is a commutative semigroup with 0.
(ii)	=	$(A,\le )$ is a lattice, and ≤ is translation invariant, that is, $\mathrm{\forall }a,b,c\in A$; $\begin{array}{ll}a+(b\vee c)=& (a+b)\vee (a+c)\\ a+(b\wedge c)=& (a+b)\wedge (a+c)\end{array}$
(iii)	=	$\ast :A\times A\to A$ is autometric on A, that is, $\ast$ satisfies metric operation axioms: M1, M2 and M3.

Definition 2.4.

Let A be an autometrized algebra. Then A is said to be semiregular if for any $a\in A$, $a\ge 0\Rightarrow a\ast 0=a$.

Definition 2.5.

Let A is a normal autometrized algebra. An equivalence relation Θ on A is called a congruence relation if and only if

(i)	=	$(a,b),(c,d)\in \mathrm{\Theta }\Rightarrow (a+c,b+d)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d\in A$,
(ii)	=	$(a,b),(c,d)\in \mathrm{\Theta }\Rightarrow (a\ast c,b\ast d)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d\in A$,
(iii)	=	$(a,b)\in \mathrm{\Theta }\text{and}x\ast y\le a\ast b\Rightarrow (x,y)\in \mathrm{\Theta }\mathrm{\forall }a,b,x,y\in A$.

Definition 2.6.

Adhikari and Adhikari (2022) A topological space $(X,\tau )$ is said to be a T₁ - space if x and y are two distinct points of X, then there exist two open sets, one containing x but not y, and the other containing y but not x.

Definition 2.7.

Adhikari and Adhikari (2022) A topological space $(X,\tau )$ is said to be a Hausdorff space or T₂ - space if any two distinct points of X have disjoint neighborhoods. The topology of such a space is said to be a Hausdorff topology.

3. Convex subalgebras of autometrized algebras

This section introduces the concepts of convex subalgebras and congruence relations in an autometrized algebra. In particular, we also develop one-to-one correspondence between the set of all convex subalgebras and the set of all congruences in an np-autometrized algebra. Also, we discuss prime convex subalgebra and regular convex subalgebra and prove different properties.

Definition 3.1.

Let A be an autometrized algebra. Let $B\subseteq A$. Then B is said to be a subalgebra of A if;

(i)	=	$(B,+,0)$ is a commutative monoid.
(ii)	=	$(B,\le )$ is a subposet and $a\le b\Rightarrow a+c\le b+c$ for any $a,b,c\in B$.
(iii)	=	$\ast \mid B:B\times B\to B$.

Example 3.2.

Let $A=\{0,a,b,c\}$ with $0\le a\le b\le c$. Define $\ast$ and + by the following tables.

$\begin{array}{ll}\begin{array}{lllll}\ast & 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& 0& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{b}& 0& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& 0\end{array}& \begin{array}{lllll}+& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}\end{array}\end{array}$

It is clear to show that A is an autometrized algebra. Let $B=\{0,a\}$. We know that any subset of a poset is a poset. Therefore, B is a poset. We see that $0+a=a+0=a$ and $0\ast a=a$. That implies that B is a commutative monoid and is closed under $\ast$. Hence, B is a subalgebra of A.

Definition 3.3.

Let A be an autometrized l-algebra. Let $S\subseteq A$. Then S is said to be a subalgebra of A if S is itself an autometrized l-algebra.

Example 3.4.

It is clear that in Example (3.2), A is an autometrized l-algebra. And we know that $B=\{0,a\}$ is an autometrized algebra. Since $0\wedge a=0$ and $0\vee a=a$; implies B is a lattice. Hence, B is an autometrized l-algebra.

Definition 3.5.

Let A be an autometrized algebra. Let $S\subseteq A$. Then S is said to be convex if $x,y\in S$ and $x\le z\le y\Rightarrow z\in S$.

Example 3.6.

We know that in Example (3.2); A is an autometrized algebra. Let $B=\{0,a\}$ and $S=\{0,a,b\}$. B and S are subalgebras of A. Since $0\le a\le b\Rightarrow a\in S$; shows that S is a convex subalgebra. Clearly, B is a convex subalgebra. Therefore, B and S are convex subalgebras.

Definition 3.7.

Let A be an autometrized l-algebra. Let S be a subalgebra of A. Then S is said to be a convex subalgebra of A if S is a convex set in A.

Example 3.8.

Let $A=\{0,a,b,c\}$ with $0\le a,b\le c$ and elements $a,b$ are incomparable. Define $\ast$ and + by the following tables.

$\begin{array}{ll}\begin{array}{lllll}\ast & 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& 0& \mathrm{c}& \mathrm{b}\\ \mathrm{b}& \mathrm{b}& \mathrm{c}& 0& \mathrm{a}\\ \mathrm{c}& \mathrm{c}& \mathrm{b}& \mathrm{a}& 0\end{array}& \begin{array}{lllll}+& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& \mathrm{a}& \mathrm{c}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{c}& \mathrm{b}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}\end{array}\end{array}$

Clearly, A is an autometrized l-algebra. Let us consider the subalgebras $C_1=\{0,a\}$, $C_2=\{0,b\}$ and $C_3=\{0,c\}$. It is easy to show that $C_1,C_2$ are convex subalgebras. We notice that each subalgebra of A may not be a convex subalgebra of A. We know that $0\le a\le c$ and $0\le b\le c$. But $a,b\notin C_3$. Thus, C₃ is not a convex subalgebra of A.

Let A be an autometrized algebra. If A satisfies $a\le b\iff a\ast 0\le b\ast 0\mathrm{\forall }a,b\in A$, then A is said to be a positive autometrized algebra, denoted as p-autometrized algebra. Furthermore, if A is normal and positive, then A is called np-autometrized algebra. Now consider the following examples.

Example 3.9.

We know that in the previous Example (3.8): A is an autometrized algebra. Also, we can easily check that A is a normal autometrized algebra. Clearly, $0\le a\le c\iff 0\le a\ast 0\le c\ast 0$ and $0\le b\le c\iff 0\le b\ast 0\le c\ast 0$. Therefore, A is p-autometrized algebra. Thus, A is an np-autometrized algebra.

Example 3.10.

Let $A=\{0,a,b,c,d,e\}$ with $0\le a\le b\le c\le d\le e$. Define + and $\ast$ by the following tables.

$\begin{array}{ll}\begin{array}{lllllll}+& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}& \mathrm{e}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}& \mathrm{e}\\ \mathrm{a}& \mathrm{a}& \mathrm{b}& \mathrm{b}& \mathrm{e}& \mathrm{e}& \mathrm{e}\\ \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{e}& \mathrm{e}& \mathrm{e}\\ \mathrm{c}& \mathrm{c}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}\\ \mathrm{d}& \mathrm{d}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}\\ \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}& \mathrm{e}\end{array}& \begin{array}{lllllll}\ast & 0& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}& \mathrm{e}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}& \mathrm{d}& \mathrm{e}\\ \mathrm{a}& \mathrm{a}& 0& \mathrm{a}& \mathrm{c}& \mathrm{c}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{a}& 0& \mathrm{c}& \mathrm{c}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& 0& \mathrm{a}& \mathrm{a}\\ \mathrm{d}& \mathrm{d}& \mathrm{c}& \mathrm{c}& \mathrm{a}& 0& \mathrm{a}\\ \mathrm{e}& \mathrm{e}& \mathrm{c}& \mathrm{c}& \mathrm{a}& \mathrm{a}& 0\end{array}\end{array}$

Clearly, A is an autometrized algebra. We know that $a\le c$. But there is no x such that $a+x=c$. Therefore, A is not a normal autometrized algebra. We can easily show that $0\le a\le b\le c\le d\le e\iff 0\le a\ast 0\le b\ast 0\le c\ast 0\le d\ast 0\le e\ast 0$. Hence, A is a p-autometrized algebra but not an np-autometrized algebra.

Lemma 3.11.

In an autometrized algebra A, the intersection of any collection of convex subalgebras in A is again a convex subalgebra.

Proof.

The proof is obvious.

Let A be an autometrized algebra. Let $X\subseteq A$. Then the convex subalgebra generated by X is denoted by $\mathcal{C}(X)$, defined as $\mathcal{C}(X)$ = intersection of all convex subalgebras of A containing X. Also, the set of all convex subalgebras of A is denoted by $\mathcal{C}(A)$.

Remark 3.12.

Let $\{C_i{\}}_{i\in I}$ be a collection of convex subalgebra of A. Then ${\cup }_{i\in I}{C}_i$ may not be a convex subalgebra of A. See the following example.

Example 3.13.

Consider the subalgebras $C_1=\{0,a\}$ and $C_2=\{0,b\}$ in Example (3.8); clearly $C_1,C_2$ are convex subalgebras of A. That is; $C_1,C_2\in \mathcal{C}(A)$. Since $a+b=c$ and $c\notin C_1\cup C_2$; and hence $C_1\cup C_2=\{0,a,b\}$ is not a convex subalgebra of A.

Clearly, $\mathcal{C}(A)$ is a lattice because if $C_1,C_2\in \mathcal{C}(A)$, then $C_1\vee C_2\in \mathcal{C}(A)$ (that is, the intersection of all convex subalgebras containing $C_1\cup C_2$) is the least upper bound of C₁ and C₂. Also $C_1\wedge C_2\in \mathcal{C}(A)$ (that is, the intersection of the convex subalgebras C₁ and C₂) is the greatest lower bound of C₁ and C₂. Since we can replace the set $\{C_1,C_2\}$ by an arbitrary family of convex subalgebras, the lattice ($\mathcal{C}(A),\vee ,\wedge )$ is a complete lattice.

Theorem 3.14.

Let A be an np-autometrized algebra. Let $X\subseteq A$. Then

$\mathcal{C}(X)=\{x\in A|x\ast 0\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X\}$.

Proof.

Let $D=\{x\in A|x\ast 0\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X\}$.

To claim that D is the convex subalgebra of A generated by X. That is, $D=\mathcal{C}(X)$.

To show that D is a convex subalgebra of A.

(i)	=	To show that $(D,+,0)$ is a commutative monoid. Since $0=0\ast 0\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k$; implies that $0\in D$. Let $x,y\in D$. That implies $x\ast 0\le m_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)$ and, $y\ast 0\le n_1(b_1\ast 0)+\dots +n_l(b_l\ast 0)$ for $m_1,\dots ,m_k,n_1,\dots ,n_l$ are positive integers and $a_1,\cdots ,a_k$, $b_1,\cdots ,b_l\in X$. By the property of normal, we get: $\begin{array}{ll}(x+y)\ast 0& \le (x\ast 0)+(y\ast 0)\\ & \le m_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)\\ & +n_1(b_1\ast 0)+\dots +n_l(b_l\ast 0)\end{array}$ for $m_1,\dots ,m_k,n_1,\dots ,n_l$ are positive integers and $a_1,\cdots ,a_k$, $b_1,\cdots ,b_l\in X$. Therefore $x+y\in D$.
(ii)	=	Clearly, $(D,\le )$ is a subposet and $a\le b\Rightarrow a+c\le b+c$ for any $a,b,c\in D$.
(iii)	=	Let $x,y\in D$. That implies $x\ast 0\le m_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)$ and, $y\ast 0\le n_1(b_1\ast 0)+\dots +n_l(b_l\ast 0)$ for $m_1,\dots ,m_k,n_1,\dots ,n_l$ are positive integers and $a_1,\cdots ,a_k$, $b_1,\cdots ,b_l\in X$. By the property of normal, we get: $\begin{array}{ll}(x\ast y)\ast 0& \le (x\ast 0)+(y\ast 0)\\ & \le m_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)\\ & +n_1(b_1\ast 0)+\dots +n_l(b_l\ast 0)\end{array}$ for $m_1,\dots ,m_k,n_1,\dots ,n_l$ are positive integers and $a_1,\cdots ,a_k$, $b_1,\cdots ,b_l\in X$. Therefore $x\ast y\in D$. Hence, D is a subalgebra.
(iv)	=	Let $a,b\in D$ and $x\in A$. Suppose $a\le x\le b$. Therefore by the given condition, $a\ast 0\le x\ast 0\le b\ast 0$. Since $b\ast 0\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X$; implies that $x\in D$. Hence D is convex.

Now, to show that $X\subseteq D$. Let $x\in X$. We know that $x\ast 0\le 1(x\ast 0)$. Hence $x\in D$. Therefore $X\subseteq D$. Thus, D is a convex subalgebra of A containing X.

Let E be any convex subalgebra of A containing X. To show that $D\subseteq E$. Let $g\in D$. Therefore,

$g\ast 0\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X$. This implies that $0\ast 0\le g\ast 0\le [m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)]\ast 0$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X$. By the given condition, $0\le g\le m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_1,\cdots ,a_k\in X$.

Since $X\subseteq E$; $a_1,\cdots ,a_k\in E$. Which implies that $a_1\ast 0,\cdots ,a_k\ast 0\in E$. Hence, $m_1(a_1\ast 0)+\cdots +m_k(a_k\ast 0)\in E$. Since E is convex; $g\in E$. Hence, $D\subseteq E$. Therefore, D is the smallest convex subalgebra containing X. Hence $D=\mathcal{C}(X)$.

Example 3.15.

We know that in the previous Example (3.10): A is an autometrized algebra. Let $X=\{a,b\}$. Then $\mathcal{C}(X)=\{x\in A|x\ast 0\le m_1(a\ast 0)+m_2(b\ast 0)$ for some positives $m_1,m_2$ and $a,b\in X\}$. Clearly, $0\le m_1(a\ast 0)+m_2(b\ast 0)$, $a\ast 0\le m_1(a\ast 0)+m_2(b\ast 0)$ and $b\ast 0\le m_1(a\ast 0)+m_2(b\ast 0)$. Therefore, $\mathcal{C}(X)=\{0,a,b\}$ is the convex subalgebra of A generated by X.

Theorem 3.16.

Let A be an np-autometrized l-algebra. Let $X,Y\subseteq A$. Then

(i)	=	$\mathcal{C}(X)\vee \mathcal{C}(Y)=\mathcal{C}(X\cup Y)=\mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid \text{break}x\in X,y\in Y\})$.
(ii)	=	$\mathcal{C}(X)\wedge \mathcal{C}(Y)=\mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid x\in X,\text{break}y\in Y\})$.

Proof.

(i)	=	We know that: $X,Y\subseteq X\cup Y$. Therefore, $\mathcal{C}(X),\mathcal{C}(Y)\subseteq \mathcal{C}(X\cup Y)$. Hence, $\mathcal{C}(X\cup Y)$ is an upper bound of $\mathcal{C}(X),\mathcal{C}(Y)\in \mathcal{C}(A)$. Let $T\in \mathcal{C}(A)$ be an upper bound of $\{\mathcal{C}(X),\mathcal{C}(Y)\}$. This implies that $\mathcal{C}(X),\mathcal{C}(Y)\subseteq \text{break}T$. Since $X\subseteq \mathcal{C}(X)\subseteq T$ and $Y\subseteq \mathcal{C}(Y)\subseteq T$; implies $X\cup Y\subseteq T$. Therefore, T is a convex subalgebra containing $X\cup Y$. Hence $\mathcal{C}(X\cup Y)\subseteq T$. Therefore, $\mathcal{C}(X\cup Y)$ is the least upper bound of $\{\mathcal{C}(X),\mathcal{C}(Y)\}$. (1) $\begin{array}{l}\mathcal{C}(X)\vee \mathcal{C}(Y)=\mathcal{C}(X\cup Y).\end{array}$(1) Let $a\in X$. Therefore, $a\ast 0\le (a\ast 0)\vee (b\ast 0)$ for any $b\in Y$. This implies that $0\ast 0\le \text{break}a\ast 0\le [(a\ast 0)\vee (b\ast 0)]\ast 0$. By the given condition, $0\le a\le (a\ast 0)\vee (b\ast 0)$. Since $(a\ast 0)\vee (b\ast 0)\in \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})$; implies that $a\in \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})$. Hence, $X\subseteq \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})$. Similarly; $Y\subseteq \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})$. Therefore, $X\cup Y\subseteq \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})$. Hence, (2) $\begin{array}{l}\mathcal{C}(X\cup Y)\subseteq \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\}).\end{array}$(2) Let $x\in X,y\in Y$. So $x,y\in X\cup Y\subseteq \mathcal{C}(X\cup Y)$. Therefore, $x\ast 0,y\ast 0\in \mathcal{C}(X\cup Y)$ and $(x\ast 0)\vee (y\ast 0)\in \mathcal{C}(X\cup Y)$. Thus, $\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\}\subseteq \mathcal{C}(X\cup Y)$. (3) $\begin{array}{l}\mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})\subseteq \mathcal{C}(X\cup Y).\end{array}$(3) By Equations (1)(1) $\begin{array}{l}\mathcal{C}(X)\vee \mathcal{C}(Y)=\mathcal{C}(X\cup Y).\end{array}$(1) , (2(2) $\begin{array}{l}\mathcal{C}(X\cup Y)\subseteq \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\}).\end{array}$(2) ) and (3(3) $\begin{array}{l}\mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid x\in X,y\in Y\})\subseteq \mathcal{C}(X\cup Y).\end{array}$(3) ); we have: $\begin{array}{ll}\mathcal{C}(X)\vee \mathcal{C}(Y)=\mathcal{C}(X\cup Y)=& \mathcal{C}(\{(x\ast 0)\vee (y\ast 0)\mid \\ & x\in X,y\in Y\}).\end{array}$
(ii)	=	Let $x\in X,y\in Y$. Therefore, $0\le (x\ast 0)\wedge (y\ast 0)\le x\ast 0,y\ast 0$. Since $x\ast 0\in \mathcal{C}(X),y\ast 0\in \mathcal{C}(Y)$ and $\mathcal{C}(X),\mathcal{C}(Y)$ are convex; implies that $(x\ast 0)\wedge (y\ast 0)\in \mathcal{C}(X),\mathcal{C}(Y)$. Therefore, $(x\ast 0)\wedge (y\ast 0)\in \mathcal{C}(X)\cap \mathcal{C}(Y)$. We have, (4) $\begin{array}{ll}& \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)|x\in X,y\in Y\})\\ & \subseteq \mathcal{C}(X)\cap \mathcal{C}(Y)=\mathcal{C}(X)\wedge \mathcal{C}(Y).\end{array}$(4) Let $g\in \mathcal{C}(X)\cap \mathcal{C}(Y)$. Therefore, $g\in \mathcal{C}(X)$ and $g\in \mathcal{C}(Y)$. Therefore, $g\ast 0\le l_1(x_1\ast 0)+\cdots +l_n(x_n\ast 0)$ and $g\ast 0\le k_1(y_1\ast 0)+\cdots +k_m(y_m\ast 0)$ for some positives $l_1,\cdots ,l_n$, $k_1,\cdots ,k_m$ and $x_1,\cdots ,x_n\in X$; $y_1,\cdots ,y_m\in Y$. That implies that: $\begin{array}{ll}& g\ast 0\le [l_1(x_1\ast 0)+\cdots +l_n(x_n\ast 0)]\wedge [k_1(y_1\ast 0)\\ & +\cdots +k_m(y_m\ast 0)].,\le l_1{k}_1[(x_1\ast 0)\wedge (y_1\ast 0)]\\ & +\cdots +l_n{k}_1[(x_n\ast 0)\wedge (y_1\ast 0)]+\cdots +l_1{k}_m\\ & [(x_1\ast 0)\wedge (y_m\ast 0)]+\cdots +l_n{k}_m[(x_n\ast 0)\wedge \\ & (y_m\ast 0)].\in \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid x\in X,y\in Y\}).\end{array}$ Clearly, by the given condition $g\in \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid x\in X,y\in Y\})$. Hence, (5) $\begin{array}{ll}\mathcal{C}(X)\cap \mathcal{C}(Y)\subseteq & \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid \\ & x\in X,y\in Y\}).\end{array}$(5) By Equations (4)(4) $\begin{array}{ll}& \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)|x\in X,y\in Y\})\\ & \subseteq \mathcal{C}(X)\cap \mathcal{C}(Y)=\mathcal{C}(X)\wedge \mathcal{C}(Y).\end{array}$(4) and (5(5) $\begin{array}{ll}\mathcal{C}(X)\cap \mathcal{C}(Y)\subseteq & \mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid \\ & x\in X,y\in Y\}).\end{array}$(5) ); we have: $\mathcal{C}(X)\wedge \mathcal{C}(Y)=\mathcal{C}(\{(x\ast 0)\wedge (y\ast 0)\mid x\in X,y\in Y\}.$

Remark 3.17.

Let A be a semiregular np-autometrized l-algebra. Let $X,Y\subseteq A$. Then

$\begin{array}{ll}(i)\mathcal{C}(X)\vee \mathcal{C}(Y)=& \mathcal{C}(X\cup Y).\\ =& \mathcal{C}(\{x\vee y\mid x\in X,y\in Y\})\\ & forx,y\ge 0.\\ (ii)\mathcal{C}(X)\wedge \mathcal{C}(Y)=& \mathcal{C}(\{x\wedge y\mid x\in X,y\in Y\})\\ & forx,y\ge 0.& & & & & & & & & & & & & & \end{array}$

And,

$\begin{array}{llllllllllllllll}(i)\mathcal{C}(x)\vee \mathcal{C}(y)& =\mathcal{C}(x\vee y)forx,y\ge 0.& & & & & & & & & & & & & & \\ (ii)\mathcal{C}(x)\wedge \mathcal{C}(y)& =\mathcal{C}(x\wedge y)forx,y\ge 0.& & & & & & & & & & & & & & \end{array}$

Theorem 3.18.

Let A be an np-autometrized l-algebra. Then $\mathcal{C}(A)$ infinitely meet distributive lattice. Hence $\mathcal{C}(A)$ is a distributive lattice.

Proof.

To show that $\mathcal{C}(A)$ is infinitely meet distributive. Let $C\in \mathcal{C}(A)$ and $\{C_i{\}}_{i\in I}\subseteq \mathcal{C}(A)$.

To show that $C\wedge (\underset{i\in I}{\bigvee }{C}_i)=\underset{i\in I}{\bigvee }(C\wedge C_i)$. Clearly $\underset{i\in I}{\bigvee }(C\wedge C_i)\subseteq C\wedge (\underset{i\in I}{\bigvee }{C}_i)$.

Let $g\in C\wedge (\underset{i\in I}{\bigvee }{C}_i)=C\cap \mathcal{C}(\bigcup _{i\in I}{C}_i)$. Therefore, $g\in C$ and $g\in \mathcal{C}(\bigcup _{i\in I}{C}_i)$. Then consider $g\ast 0\le k_1(x_1\ast 0)+\cdots +k_n(x_n\ast 0)$ for some positives $k_1,\cdots ,k_n$ and for $x_j\in C_{i_j}$. Since A is l-algebra;

(6) $\begin{array}{ll}g\ast 0=& (g\ast 0)\wedge [k_1(x_1\ast 0)+\cdots +k_n(x_n\ast 0)].,\\ & \le k_1[(g\ast 0)\wedge (x_1\ast 0)]+\cdots \\ & +k_n[(g\ast 0)\wedge (x_n\ast 0)].\end{array}$(6)

Since $x_j\in C_{i_j}$; implies that $x_j\ast 0\in C_{i_j}$. Since $g\in C$; implies that $g\ast 0\in C$. Therefore,

(7) $\begin{array}{l}(g\ast 0)\wedge (x_j\ast 0)\in C\wedge C_{i_j}.\end{array}$(7)

By Equations (6)(6) $\begin{array}{ll}g\ast 0=& (g\ast 0)\wedge [k_1(x_1\ast 0)+\cdots +k_n(x_n\ast 0)].,\\ & \le k_1[(g\ast 0)\wedge (x_1\ast 0)]+\cdots \\ & +k_n[(g\ast 0)\wedge (x_n\ast 0)].\end{array}$(6) and (7(7) $\begin{array}{l}(g\ast 0)\wedge (x_j\ast 0)\in C\wedge C_{i_j}.\end{array}$(7) ), we have,

(8) $\begin{array}{ll}0\ast 0& \le g\ast 0\le \{k_1[(g\ast 0)\wedge (x_1\ast 0)]\\ & +\cdots +k_n[(g\ast 0)\wedge (x_n\ast 0)]\}\ast 0.\end{array}$(8)

Because $k_1[(g\ast 0)\wedge (x_1\ast 0)]+\cdots +k_n[(g\ast 0)\wedge (x_n\ast 0)]\in \mathcal{C}(\bigcup _{i\in I}(C\wedge C_i))$; by the given condition,

$\begin{array}{l}g\in \mathcal{C}(\bigcup _{i\in I}(C\wedge C_i))=\underset{i\in I}{\bigvee }(C\wedge C_i).\end{array}$

Therefore, $C\wedge (\underset{i\in I}{\bigvee }{C}_i)\subseteq \underset{i\in I}{\bigvee }(C\wedge C_i)$. Hence, $C\wedge (\underset{i\in I}{\bigvee }{C}_i)=\underset{i\in I}{\bigvee }(C\wedge C_i)$. Therefore, $\mathcal{C}(A)$ is infinitely meet distributive. Hence $\mathcal{C}(A)$ is distributive.

Example 3.19.

Let $A=\{0,a,b,x,c\}$ with $0\le a,b,x\le c$ and elements $a,b,x$ are incomparable. Define + and $\ast$ by the following tables.

$\begin{array}{ll}\begin{array}{llllll}+& 0& \mathrm{a}& \mathrm{b}& \mathrm{x}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{x}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& \mathrm{a}& \mathrm{c}& \mathrm{c}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{c}& \mathrm{b}& \mathrm{c}& \mathrm{c}\\ \mathrm{x}& \mathrm{x}& \mathrm{c}& \mathrm{c}& \mathrm{x}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}\end{array}& \begin{array}{llllll}\ast & 0& \mathrm{a}& \mathrm{b}& \mathrm{x}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{x}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& 0& \mathrm{c}& \mathrm{c}& \mathrm{b}\\ \mathrm{b}& \mathrm{b}& \mathrm{c}& 0& \mathrm{c}& \mathrm{a}\\ \mathrm{x}& \mathrm{x}& \mathrm{c}& \mathrm{c}& 0& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{b}& \mathrm{a}& \mathrm{c}& 0\end{array}\end{array}$

Clearly, A is a p-autometrized algebra. We know that $(0+x)\ast (x+a)=x\ast c=c$ and $a\ast 0=a$. This implies that $(0+x)\ast (x+a)=x\ast c=c\le a\ast 0=a$. This is a contradiction. Therefore, A is not a normal autometrized algebra. Hence, A is not an np-autometrized algebra. Now consider the convex subalgebras $C_1=\{0,a\}$, $C_2=\{0,b\}$ and $C_3=\{0,x\}$. Therefore, $C_1\cap (C_2\vee C_3)=\{0,a\}\cap \mathcal{C}(\{0,b,x\})=\{0,a\}\cap A=\{0,a\}$. On the other hand, $(C_1\cap C_2)\vee (C_1\cap C_3)=\{0\}\vee \{0\}=\{0\}$. Therefore, $C_1\cap (C_2\vee C_3)\ne (C_1\cap C_2)\vee (C_1\cap C_3)$. Hence A is not distributive.

Definition 3.20.

Let A be an autometrized algebra. An equivalence relation Θ on A is called a congruence relation if and only if

(i)	=	$(a,b),(c,d)\in \mathrm{\Theta }\Rightarrow (a+c,b+d)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d\in A$,
(ii)	=	$(a,b),(c,d)\in \mathrm{\Theta }\Rightarrow (a\ast c,b\ast d)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d\in A$,
(iii)	=	$(a,b)\in \mathrm{\Theta }$ and $x\ast y\le a\ast b\Rightarrow (x,y)\in \mathrm{\Theta }\mathrm{\forall }a,b,x,y\in A$, or
$(iii{)}^{\prime }$	=	$(a,b),(c,d)\in \mathrm{\Theta }$ and $a\ast b\le x\ast y\le c\ast d\Rightarrow (x,y)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d,x,y\in A$.

Remark 3.21.

we have $(iii)\iff (iii{)}^{\prime }$. Suppose $(a,b)\in \mathrm{\Theta }$ and $x\ast y\le a\ast b\Rightarrow (x,y)\in \mathrm{\Theta }\mathrm{\forall }a,b,x,y\in A$. Let $(a,b),(c,d)\in \mathrm{\Theta }$. Suppose $a\ast b\le x\ast y\le \text{break}c\ast d\mathrm{\forall }a,b,c,d,x,y\in A$. To show that $(x,y)\in \mathrm{\Theta }$. Since $x\ast y\le c\ast d$; by the given condition $(x,y)\in \mathrm{\Theta }$.

Conversely, suppose that $(a,b),(c,d)\in \mathrm{\Theta }$ and $a\ast b\le x\ast y\le c\ast d\Rightarrow (x,y)\in \mathrm{\Theta }\mathrm{\forall }a,b,c,d,x,y\in A$. Let $(a,b)\in \mathrm{\Theta }$. Suppose $x\ast y\le a\ast b\mathrm{\forall }a,b,x,y\in A$. Clearly, $(a,a)\in \mathrm{\Theta }$. As a result, $a\ast a=0\le x\ast y$. Therefore, $a\ast a=0\le x\ast y\le a\ast b\mathrm{\forall }a,b,x,y\in A$. Hence by the given condition, $(x,y)\in \mathrm{\Theta }$.

Example 3.22.

Let $A=\{0,a,b,c\}$ with $0\le a,b\le c$ and elements $a,b$ are incomparable. Define $\ast$ and + by the following tables.

$\begin{array}{ll}\begin{array}{lllll}\ast & 0& a& b& c\\ 0& 0& a& b& c\\ a& a& 0& c& b\\ b& b& c& 0& a\\ c& c& b& a& 0\end{array}& \begin{array}{lllll}+& 0& a& b& c\\ 0& 0& a& b& c\\ a& a& a& c& c\\ b& b& c& b& c\\ c& c& c& c& c\end{array}\end{array}$

So, A becomes an autometrized algebra. Let us consider the equivalent relations $\mathrm{△}=\{(0,0),(a,a),(b,b),(c,c)\}$, ${\mathrm{\Theta }}_1=\{(0,0),(a,a),(b,b),(c,c),(0,a),(a,0),(b,c),$ $(c,b)\}$, $$\begin{gathered} {\mathrm{\Theta }}_2=\{(0,0),(a,a),(b,b),(c,c),(a,c),(c,a), \\ (0,b),(b,0)\} \end{gathered}$$ and $\mathrm{\nabla }=A\times A$ on A. Then $\mathrm{△}$, Θ₁, Θ₂ and $\mathrm{\nabla }$ are all the congruence relations on A.

Definition 3.23.

Let A be an autometrized algebra. Let Θ be a congruence relation on A. Then the subset $S_{\mathrm{\Theta }}$ which is defined by $S_{\mathrm{\Theta }}=\{x\in A|(x,0)\in \mathrm{\Theta }\}=\mathrm{\Theta }(0)$ is called the subset induced by Θ.

Definition 3.24.

Let A be an autometrized algebra. Let S be a convex subalgebra of A. The relation ${\mathrm{\Theta }}_S$ on A which is defined by ${\mathrm{\Theta }}_S=\{(x,y)\in A\times A\mid x\ast y\in S\}$ is called the relation induced by S.

Theorem 3.25.

Let A be a p-autometrized algebra. Let $\mathrm{\Theta }\in Con(A)$. Then $S_{\mathrm{\Theta }}$ is a convex subalgebra of A.

Proof.

To show that $S_{\mathrm{\Theta }}\in \mathcal{C}(A)$. To show that $S_{\mathrm{\Theta }}$ is a convex subalgebra of A.

(i)	=	To show that $S_{\mathrm{\Theta }}$ is subalgebra of A. To show that $(S_{\mathrm{\Theta }},+,0)$ is a commutative monoid. Since $(0,0)\in \mathrm{\Theta }$; implies that $0\in S_{\mathrm{\Theta }}$. Let $a,b\in S_{\mathrm{\Theta }}$. Then $(a,0),(b,0)\in \mathrm{\Theta }$. Since Θ is a congruence relation; $(a+b,0+0)\in \mathrm{\Theta }$. Which implies that $(a+b,0)\in \mathrm{\Theta }$. Therefore, $a+b\in S_{\mathrm{\Theta }}$.
(ii)	=	Clearly, $(S_{\mathrm{\Theta }},\le )$ is a subposet and $a\le b\Rightarrow a+c\le b+c$ for any $a,b,c\in S_{\mathrm{\Theta }}$.
(iii)	=	Let $a,b\in S_{\mathrm{\Theta }}$. Then, $(a,0),(b,0)\in \mathrm{\Theta }$. Since Θ is a congruence relation: $(a\ast b,0\ast 0)\in \mathrm{\Theta }$. Then, $(a\ast b,0)\in \mathrm{\Theta }$. Therefore, $a\ast b\in S_{\mathrm{\Theta }}$. Hence, $S_{\mathrm{\Theta }}$ is a subalgebra.
(iv)	=	Let $a,b\in S_{\mathrm{\Theta }}$ and $x\in A$. Suppose $a\le x\le b$. By the given condition, $a\ast 0\le x\ast 0\le b\ast 0$. Then $(a,0),(b,0)\in \mathrm{\Theta }$. Since Θ is a congruence relation; which implies that $(x,0)\in \mathrm{\Theta }$ and thus $x\in S_{\mathrm{\Theta }}$. Therefore, $S_{\mathrm{\Theta }}$ is convex. Hence, $S_{\mathrm{\Theta }}$ is a convex subalgebra of A.

Example 3.26.

In Example (3.22) A is a p-autometrized algebra. Consider the congruence relation $$\begin{gathered} {\mathrm{\Theta }}_1=\{(0,0),(a,a),(b,b),(c,c),(0,a),(a,0),(b,c), \\ (c,b)\} \end{gathered}$$. Then $S_{{\mathrm{\Theta }}_1}=\{0,a\}$ is a convex subalgebra of A.

Theorem 3.27.

Let A be a normal autometrized algebra. Let S be a convex subalgebra of A. Then ${\mathrm{\Theta }}_S$ is a congruence relation on A.

Proof.

To show that ${\mathrm{\Theta }}_S\in Con(A)$.

First, to show that ${\mathrm{\Theta }}_S$ is an equivalence relation on A.

(i)	=	Reflexive: Let $a\in A$. Since $0\in S$; $a\ast a=0\in S$. Therefore, $(a,a)\in {\mathrm{\Theta }}_S\mathrm{\forall }a\in A$.
(ii)	=	Symmetric: Suppose $(a,b)\in {\mathrm{\Theta }}_S$. Then $a\ast b\in S$. Since $a\ast b=b\ast a\in S$. Hence, $(b,a)\in {\mathrm{\Theta }}_S$.
(iii)	=	Transitivity: Suppose $(a,b),(b,c)\in {\mathrm{\Theta }}_S$. Therefore, $a\ast b,b\ast c\in S$ and $(a\ast b)+(b\ast c)\in S$. By metric property; $a\ast c\le (a\ast b)+(b\ast c)$. Since S is convex subalgebra and $0\in S$; $0\le a\ast c\le (a\ast b)+(b\ast c)$. Therefore, $a\ast c\in S$. Which implies that $(a,c)\in {\mathrm{\Theta }}_S$. Hence, ${\mathrm{\Theta }}_S$ is an equivalence relation. To show that ${\mathrm{\Theta }}_S$ is congruence on A.
(a)	=	Let $(a,b),(c,d)\in {\mathrm{\Theta }}_S$. Then, $a\ast b\in S$ and $c\ast d\in S$. Which implies that $(a\ast b)+(c\ast d)\in S$. Since A is normal; $(a+c)\ast (b+d)\le (a\ast b)+(c\ast d)$. This implies that: $0\le (a+c)\ast (b+d)\le (a\ast b)+(c\ast d)$. Since S is convex; $(a+c)\ast (b+d)\in S$. Hence, $(a+c,b+d)\in {\mathrm{\Theta }}_S$.
(b)	=	Let $(a,b),(c,d)\in {\mathrm{\Theta }}_S$. Then, $a\ast b\in S$ and $c\ast d\in S$. Which implies that $(a\ast b)+(c\ast d)\in S$. Since A is normal; $(a\ast c)\ast (b\ast d)\le (a\ast b)+(c\ast d)$. This implies that: $0\le (a\ast c)\ast (b\ast d)\le (a\ast b)+(c\ast d)$. Since S is convex; $(a\ast c)\ast (b\ast d)\in S$. Hence, $(a\ast c,b\ast d)\in {\mathrm{\Theta }}_S$.
(c)	=	Let $(a,b),(c,d)\in {\mathrm{\Theta }}_S$. Suppose that $a\ast b\le x\ast y\le c\ast d$. Which implies that $a\ast b,c\ast d\in S$. Therefore, $x\ast y\in S$. Hence, $(x,y)\in {\mathrm{\Theta }}_S$. Thus, ${\mathrm{\Theta }}_S\in Con(A)$. Therefore, ${\mathrm{\Theta }}_S$ is a congruence relation on A.

Example 3.28.

In Example (3.22), A is a normal autometrized algebra. Consider the convex subalgebra $S=\{0,b\}$. Then ${\mathrm{\Theta }}_{1_S}=\{(0,0),(a,a),\text{break}(b,b),(c,c),(a,c),(c,a),(0,b),(b,0)\}$ is a congruence relation.

Theorem 3.29.

Let A be an np-autometrized algebra. Let S be a convex subalgebra of A. Then $S_{{\mathrm{\Theta }}_s}=S$.

Proof.

$\begin{array}{ll}{S}_{{\mathrm{\Theta }}_s}=& {\mathrm{\Theta }}_s(0)\\ =& \{x\in A|(x,0)\in {\mathrm{\Theta }}_S\}.\\ =& \{x\in A|x\ast 0\in S\}.\end{array}$

Since $0\ast 0\le x\ast 0\le (x\ast 0)\ast 0$; hence by the given condition $0\le x\le x\ast 0$.

$\begin{array}{ll}{S}_{{\mathrm{\Theta }}_s}=& \{x\in A|x\in S\}.\\ =& S.\end{array}$

Theorem 3.30.

Let A be a normal autometrized algebra. Let Θ be a congruence relation on A. Then ${\mathrm{\Theta }}_{S_{\mathrm{\Theta }}}=\mathrm{\Theta }$.

Proof.

$\begin{array}{ll}{\mathrm{\Theta }}_{S_{\mathrm{\Theta }}}=& \{(a,b)|a\ast b\in S_{\mathrm{\Theta }}\}.\\ =& \{(a,b)|(a\ast b,0)\in \mathrm{\Theta }\}.\end{array}$

Since $(a\ast b,0),(0,0)\in \mathrm{\Theta }$ and $0\ast 0\le a\ast b\le (a\ast b)\ast 0$. By the definition of congruence, $(a,b)\in \mathrm{\Theta }$. Therefore, ${\mathrm{\Theta }}_{S_{\mathrm{\Theta }}}=\{(a,b)|(a,b)\in \mathrm{\Theta }\}=\mathrm{\Theta }$.

Theorem 3.31.

Correspondence theorem Let A be an np-autometrized algebra. There is a bijection between the set of all convex subalgebras of A and the set of all congruence relations on A.

Proof.

Define $\psi :\mathcal{C}(A)\to Con(A)$ by $\psi (s)={\mathrm{\Theta }}_s$.

Let $s_1,s_2\in \mathcal{C}(A)$. Suppose $s_1=s_2$. Which implies that ${\mathrm{\Theta }}_{s_1}={\mathrm{\Theta }}_{s_2}$. Then $\psi (s_1)=\psi (s_2)$. Therefore, ψ is well-defined.

Now to show that ψ is onto.

Let $\mathrm{\Theta }\in Con(A)$. Then, there exists $S_{\mathrm{\Theta }}\in \mathcal{C}(A)$ such that:

$\begin{array}{ll}\psi (S_{\mathrm{\Theta }})& ={\mathrm{\Theta }}_{S_{\mathrm{\Theta }}}.[Bytheorem(3.30)]\end{array}$

Therefore, $\psi (S_{\mathrm{\Theta }})=\{(a,b)|(a,b)\in \mathrm{\Theta }\}=\mathrm{\Theta }$. Thus, ψ is onto.

Finally, let us show that ψ is one-to-one.

Suppose $\psi (s_1)=\psi (s_2)$. This implies that ${\mathrm{\Theta }}_{s_1}={\mathrm{\Theta }}_{s_2}$. Thus, $S_{{\mathrm{\Theta }}_{s_1}}=S_{{\mathrm{\Theta }}_{s_2}}$. Consider,

$\begin{array}{ll}{S}_{{\mathrm{\Theta }}_s}& ={\mathrm{\Theta }}_s(0)=S.[Bytheorem(3.29)]\end{array}$

Whence, $s_1=s_2$. Hence ψ is one-to-one and onto. Then ψ is one-to-one correspondence. Therefore $\mathcal{C}(A)$ and Con(A) are in one-to-one correspondence.

Example 3.32.

In Example (3.22) A is an np-autometrized algebra. Let us consider the convex subalgebras $S_1=\{0\}$, $S_2=\{0,a\}$, $S_3=\{0,b\}$ and $S_4=A$. We know that $\mathrm{△}$, Θ₁, Θ₂ and $\mathrm{\nabla }$ are congruence relations on A. Then $\psi (S_1)=\mathrm{\Delta }$, $\psi (S_2)={\mathrm{\Theta }}_1$, $\psi (S_3)={\mathrm{\Theta }}_2$ and $\psi (S_4)=\mathrm{\nabla }$. Therefore $\mathcal{C}(A)$ and Con(A) are in one-to-one correspondence.

3.1. Prime convex subalgebra

Definition 3.33.

Let A be an autometrized algebra. Let $C_1\in \mathcal{C}(A)$. Then C₁ is said to be prime convex subalgebra of A if for any $C_2,C_3\in \mathcal{C}(A)$;

$\begin{array}{l}{C}_2\cap C_3=C_1\Rightarrow C_2=C_{1}or{C}_3=C_1.\end{array}$

Example 3.34.

In Example (3.22), A is an autometrized algebra. We know that $C_1=\{0\}$, $C_2=\{0,a\}$, $C_3=\{0,b\}$ and $C_4=A$ are convex subalgebras. Clearly, $C_2\cap C_3=C_1$. But $C_1\ne C_2$ and $C_1\ne C_3$. Therefore, C₁ is not prime convex subalgebra. On the other hand, $C_2\cap C_4=C_2$ and $C_3\cap C_4=C_3$. Hence, C₂ and C₃ are prime convex subalgebras.

Theorem 3.35.

Let A be an autometrized algebra. Let $H\in \mathcal{C}(A)$. Let $x\in A$. Let $x\notin H$. Then there exists a prime convex subalgebra in A containing H and not containing x.

Proof.

Let $H\in \mathcal{C}(A)$. Let $x\in A$. Let $x\notin H$. Let

$T=\{C\in \mathcal{C}(A)\mid H\subseteq C,x\notin C\}.$

Since $x\notin H$ and $H\subseteq H$; implies that $H\in T$ and thus $T\ne \mathrm{\varnothing }$. Then $(T,\subseteq )$ is a non-empty poset. Let us consider an arbitrary chain system $\{C_{\alpha }{\}}_{\alpha \in \mathrm{\Gamma }}$ of elements in T and let $K={\cup }_{\alpha \in \mathrm{\Gamma }}{C}_{\alpha }$. Clearly, $K\in T$ is an upper bound of $\{C_{\alpha }{\}}_{\alpha ϵ\mathrm{\Gamma }}$.

That implies that every chain in T has an upper bound in T. By Zorn’s lemma, T has a maximum element say L. That is, $L\in T$ is the maximum in T. Thus $L\in \mathcal{C}(A)$ containing H and $x\notin L$.

Now to show that L is prime convex subalgebra. Let $M,N\in \mathcal{C}(A)$ and $M\cap N=L$. To show that $M\subseteq L$ or $N\subseteq L$. Suppose $M⊈L$ and $N⊈L$. Therefore $L⫋M$ and $L⫋N$. If $x\notin M$ and $x\notin N$, then $M,N\in T$. This is a contradiction. Because L is maximal. Therefore $x\in M$ and $x\in N$; then $x\in M\cap N=L$. So $x\in L$. This is a contradiction. Therefore L is prime. Hence L is prime convex subalgebra containing H and $x\notin L$.

Theorem 3.36.

Let A be an autometrized algebra. Let $x(\ne 0)\in A$. Then there exists a prime convex subalgebra in A not containing the element x.

Proof.

Since $\{0\}\in \mathcal{C}(A)$ and take $H=\{0\}$ in the above theorem (3.35); then there exists a prime convex subalgebra in A containing H and not containing x.

Theorem 3.37.

Let A be a semiregular np-autometrized l-algebra. Let $C\in \mathcal{C}(A)$. Then the following are equivalent.

(i)	=	C is a prime convex subalgebra.
(ii)	=	$\mathrm{\forall }{C}_1,C_2\in \mathcal{C}(A);C_1\cap C_2\subseteq C\Rightarrow C_1\subseteq Cor{C}_2\subseteq C$.
(iii)	=	$\mathrm{\forall }x,y\in A;0\le x\wedge y\in C\Rightarrow x\in Cory\in C$.

Proof.

$(i)\Rightarrow (ii)$	=	Suppose C is a prime convex subalgebra. Let $C_1,C_2\in \mathcal{C}(A)$. Suppose $C_1\cap C_2\subseteq C$. Since $\mathcal{C}(A)$ is distributive by a theorem (3.18); $C=C\vee (C_1\wedge C_2)=(C\vee C_1)\wedge (C\vee C_2)$. Since C is prime convex subalgebra; either $C=C\vee C_1$ or $C=C\vee C_2$. Therefore $C_1\subseteq C$ or $C_2\subseteq C$.
$(ii)\Rightarrow (iii)$	=	Suppose $\mathrm{\forall }{C}_1,C_2\in \mathcal{C}(A);C_1\cap C_2\subseteq C\Rightarrow C_1\subseteq C$ or $C_2\subseteq C$. Let $x,y\in A$ and $x\wedge y\ge 0$. Suppose $x\wedge y\in C$. Then $\mathcal{C}(x\wedge y)\subseteq C$. Since $0\le x\wedge y\le x,y$; by remark (3.17) $\mathcal{C}(x)\cap \mathcal{C}(y)\subseteq C$. By the given condition; either $\mathcal{C}(x)\subseteq C$ or $\mathcal{C}(y)\subseteq C$. Therefore either $x\in C$ or $y\in C$.
$(iii)\Rightarrow (i)$	=	Suppose $\mathrm{\forall }x,y\in A;0\le x\wedge y\in C\Rightarrow x\in C$ or $y\in C$. To claim that C is a prime convex subalgebra. Let $C_1,C_2\in \mathcal{C}(A)$. Suppose $C=C_1\cap C_2$. So $C\subseteq C_1$ and $C\subseteq C_2$. To show that $C=C_1$ or $C=C_2$. Assume that $C\ne C_1$ and $C\ne C_2$. Clearly $C⫋C_1$ and $C⫋C_2$. Then $\mathrm{\exists }x\in C_1$ such that $x\notin C$ and $\mathrm{\exists }y\in C_2$ such that $y\notin C$. Since C is convex subalgebra; $0\in C$. Therefore x ≠ 0 and y ≠ 0. We know $x\ast 0\ge 0$ and $y\ast 0\ge 0$. If $x\ast 0=0$ and $y\ast 0=0$, then this implies x = 0 and y = 0. This is a contradiction. Therefore $x\ast 0\ne 0$ and $y\ast 0\ne 0$. So $x\ast 0>0$ and $y\ast 0>0$. This implies $0\le (x\ast 0)\wedge (y\ast 0)\le x\ast 0,y\ast 0$. Since $x\in C_1$ implies $x\ast 0\in C_1$ and $y\in C_2$ implies $y\ast 0\in C_2$. And $x,y\notin C$ implies $x\ast 0,y\ast 0\notin C$. Suppose not; that is $x\ast 0,y\ast 0\in C$. Since A is normal; $0\ast 0\le x\ast 0\le (x\ast 0)\ast 0$. By the given condition, $0\le x\le x\ast 0$. Therefore $x\in C$. This is a contradiction. Clearly $(x\ast 0)\wedge (y\ast 0)\in C_1\cap C_2=C$. By the given condition $x\ast 0\in C$ or $y\ast 0\in C$. This is a contradiction. Therefore $C=C_1$ or $C=C_2$.

Corollary 3.38.

Let A be a semiregular np-autometrized l-algebra. Let C be a prime convex subalgebra of A. Then for any $x,y\in A$;

$\begin{array}{l}x\wedge y=0\Rightarrow eitherx\in Cory\in C.\end{array}$

Proof.

Let $x,y\in A$. Suppose $x\wedge y=0$. Since $0\in C$; $x\wedge y\in C$. By theorem (3.37) (iii); either $x\in C$ or $y\in C$.

Theorem 3.39.

Let A be a semiregular np-autometrized l-algebra. Let $\{C_i{\}}_{i\in \mathrm{\Gamma }}$ be a chain of prime convex subalgebra. Then $C={\cap }_{i\in \mathrm{\Gamma }}{C}_i$ is a prime convex subalgebra in A.

Proof.

To claim that C is a prime convex subalgebra. Let $x,y\in A$ and $0\le x\wedge y$. Suppose $x\wedge y\in C$. To show that either $x\in C$ or $y\in C$. Suppose that $x\notin C$ and $y\notin C$. Then $\mathrm{\exists }i,j\in \mathrm{\Gamma }$ such that $x\notin C_i$ and $y\notin C_j$.

Suppose $C_i\subseteq C_j$. Then $x\notin C_i$ and $y\notin C_i$. By the assumption $x\wedge y\in C_i$. Since C_i is prime; either $x\in C_i$ or $y\in C_i$. This is a contradiction. Because $x\notin C_i$ and $y\notin C_i$. Our assumption is false. Therefore P is a prime convex subalgebra.

Corollary 3.40.

Let A be a semiregular np-autometrized l-algebra. Then every prime convex subalgebra of A contains a minimal prime convex subalgebra.

Proof.

Let C be a prime convex subalgebra of A.

If C is minimal, then stop the process. Suppose C is not minimal. There exists a prime convex subalgebra C₁ such that $C_1⫋C$.

If C₁ is minimal, then stop the process. Suppose C₁ is not minimal. There exists a prime convex subalgebra C₂ such that $C_2⫋C_1$.

Continuing like this we get a chain of prime convex subalgebras: $\dots ⫋C_2⫋C_1⫋C_0=C$.

By a theorem (3.39); $Q=\bigcap _{i=0}^{\mathrm{\infty }}{C}_i$ is a prime convex subalgebra. To show that Q is minimal. Assume that Q is not minimal. There exists a prime convex subalgebra $Q_1⫋Q$. So $Q_1⫋\bigcap _{i=0}^{\mathrm{\infty }}{C}_i$. Therefore $Q_1\subseteq C_i\text{break}\mathrm{\forall }i=0,\dots ,\mathrm{\infty }$.

If Q₁ = C_i for some i, then $Q\subseteq C_i=Q_1$. This is a contradiction. Because $Q_1⫋Q$.

If $Q_1\ne C_i\mathrm{\forall }i$, then $Q_1⫋C_i\mathrm{\forall }i$. Therefore

$\begin{array}{ll}Q=& Q_1\cap \bigcap _{i=0}^{\mathrm{\infty }}{C}_i.\\ =& Q_1\cap Q=Q_1.\end{array}$

This is a contradiction. Hence Q is minimal.

Theorem 3.41.

Let A be a semiregular np-autometrized l-algebra. Let $P\in \mathcal{C}(A)$. Let $\mathcal{C}=\{C\in \mathcal{C}(A)\mid P\subseteq C\}$. If $\mathcal{C}$ is a chain, then P is a prime convex subalgebra in A.

Proof.

Let P be a convex subalgebra of A satisfying the condition of the assumption. To claim that P is a prime convex subalgebra in A. Assume that P is not prime. Then $\mathrm{\exists }x,y\in A$ with $0\le x\wedge y\in P$ such that $x\notin P$ and $y\notin P$. Clearly $0\le x\wedge y\le x,y$. Since $x,y\notin P$ and $0\in P$; implies that $0<x,y$. Let

$\begin{array}{ll}L& =\{a\in A|(a\ast 0)\wedge y\in P\}\\ M& =\{b\in A|(b\ast 0)\wedge x\in P\}.\end{array}$

Let $a\in P$, then $a\ast 0\in P$. Therefore $0\le (a\ast 0)\wedge y\le a\ast 0$. Hence, $(a\ast 0)\wedge y\in P$ which implies $a\in L$. Therefore $P\subseteq L$. Similarly if $b\in P$, then $b\ast 0\in P$. Therefore $0\le (b\ast 0)\wedge x\le b\ast 0$. Hence, $(b\ast 0)\wedge x\in P$; and $b\in M$. Therefore $P\subseteq M$.

Now, to show that $L,M\in \mathcal{C}(A)$. Let $a,b\in L$. Therefore $(a\ast 0)\wedge y$ and $(b\ast 0)\wedge y\in P$. Clearly, $[(a\ast 0)\wedge y]+[(b\ast 0)\wedge y]\in P$. Since A is normal, we get:

$\begin{array}{ll}0\le [(a+b)\ast 0]\wedge y& \le [(a\ast 0)+(b\ast 0)]\wedge y.\\ & \le [(a\ast 0)\wedge y]+[(b\ast 0)\wedge y].\end{array}$

Therefore $[(a+b)\ast 0]\wedge y\in P$. Hence $a+b\in L$.

Also, let $a,b\in L$. Therefore $(a\ast 0)\wedge y$ and $(b\ast 0)\wedge y\in P$. Clearly, $[(a\ast 0)\wedge y]+[(b\ast 0)\wedge y]\in P$. Since A is normal, we get:

$\begin{array}{ll}0\le [(a\ast b)\ast 0]\wedge y& \le [(a\ast 0)+(b\ast 0)]\wedge y.\\ & \le [(a\ast 0)\wedge y]+[(b\ast 0)\wedge y].\end{array}$

Therefore $[(a\ast b)\ast 0]\wedge y\in P$. Hence $a\ast b\in L$.

Further, let $a,b\in L$, and $c\in A$. This implies that $(a\ast 0)\wedge y\in P$ and $(b\ast 0)\wedge y\in P$. Suppose $a\le c\le b$. By the given condition $a\ast 0\le c\ast 0\le b\ast 0$. Therefore, we get :

$\begin{array}{l}(a\ast 0)\wedge y\le (c\ast 0)\wedge y\le (b\ast 0)\wedge y.\end{array}$

Therefore, $(c\ast 0)\wedge y\in P$. Therefore $c\in L$. Hence $L\in \mathcal{C}(A)$. Similarly, $M\in \mathcal{C}(A)$. Therefore $L,M\in \mathcal{C}$.

Lastly, let’s show that $L⊈M$ and $M⊈L$. We know that $x\wedge y\in P,x\notin P$ and $y\notin P$. Since A is semiregular, $(x\ast 0)\wedge y=x\wedge y\in P$. Then $x\in L$. In addition, $(x\ast 0)\wedge x\notin P$; because A is semiregular $x=x\ast 0=x\wedge x$. Therefore $x\notin M$ and so $x\in L\mathrm{\setminus }M$.

Similarly A is semiregular; $(y\ast 0)\wedge x=x\wedge y\in P$. Then $y\in M$. In addition, $(y\ast 0)\wedge y\notin P$; because A is semiregular $y=y\ast 0=y\wedge y$. Therefore $y\notin L$ and so $y\in M\mathrm{\setminus }L$.

Hence $P\subseteq L,P\subseteq M;L⊈M,M⊈L$. That implies that M and L are incomparable. Whence $\mathcal{C}$ is not a chain. This is a contradiction. Since $\mathcal{C}$ is a chain. Therefore P is a prime convex subalgebra in A.

Definition 3.42.

Let A be an autometrized algebra. Let $S\subseteq A$. Then S is said to be a meet closed subset of A if $x,y\in S\Rightarrow x\wedge y\in S$.

Remark 3.43.

Let A be an autometrized l-algebra. Any subalgebra S of A is a meet closed subset of A.

Theorem 3.44.

Let A be a semiregular np-autometrized l-algebra. Let C be a convex subalgebra of A. Then C is prime if and only if $A\setminus C$ is meet closed subset of A.

Proof.

Suppose that C is a prime convex subalgebra. To show that $A\setminus C$ is meet closed subset. Let $x,y\in A\setminus C$. Therefore, $x,y\in A$ and $x,y\notin C$.

If $0\le x\wedge y\in C$, then either $x\in C$ or $y\in C$. This is a contradiction. Therefore, $x\wedge y\notin C$. Since $x\wedge y\in A$; $x\wedge y\in A\setminus C$. Hence $A\setminus C$ is meet closed subset.

Conversely, suppose that $A\setminus C$ is meet closed. To show that C is prime.

Let $x,y\in A$ and $x\wedge y=0$. Clearly, $x,y\ge 0$. To show that either $x\in C$ or $y\in C$. Assume that $x\notin C$ and $y\notin C$. Therefore, $x,y\in A\setminus C$. Since $A\setminus C$ is meet closed; $x\wedge y\in A\setminus C$. Therefore, $x\wedge y\notin C$. As a result, $0\notin C$. This is a contradiction. Therefore, either $x\in C$ or $y\in C$. Thus, C is prime.

Theorem 3.45.

Let A be a semiregular np-autometrized l-algebra. Then S is a meet closed subset of A and C is a convex subalgebra of A which is maximal with respect to $C\cap S=\mathrm{\varnothing }\Rightarrow C$ is prime.

Proof.

To show that C is prime. Let $x,y\in A$. Suppose that $x\wedge y=0$. Clearly, $x,y\ge 0$.

Let $C_1=\mathcal{C}(C,x)$ = the convex subalgebra generated by $C\cup \{x\}$ = $C\vee \mathcal{C}(x)$.

Let $C_2=\mathcal{C}(C,y)$ = the convex subalgebra generated by $C\cup \{y\}$ = $C\vee \mathcal{C}(y)$.

Therefore,

$\begin{array}{ll}{C}_1\cap C_2=& \mathcal{C}(C,x)\cap \mathcal{C}(C,y).\\ =& [C\vee \mathcal{C}(x)]\cap [C\vee \mathcal{C}(y)].\\ =& C\vee [\mathcal{C}(x)\cap \mathcal{C}(y)].\\ & [bytheorem(3.18);distributve]\\ =& C\vee \mathcal{C}((x\ast 0)\wedge (y\ast 0)).\\ =& C\vee \mathcal{C}(x\wedge y).\\ =& C\vee \mathcal{C}(0).\\ =& C.\end{array}$

Clearly, $C\subseteq C_1,C_2$.

To show that either $x\in C$ or $y\in C$. Assume that $x\notin C$ and $y\notin C$. So $C⫋C_1$ and $C⫋C_2$. If $C_1\cap S=\mathrm{\varnothing }$, then C is maximal and $C_1\subseteq C$. This is a contradiction. Therefore, $C_1\cap S\ne \mathrm{\varnothing }$ and $C_2\cap S\ne \mathrm{\varnothing }$. Choose $s\in C_1\cap S\ne \mathrm{\varnothing }$ and $t\in C_2\cap S\ne \mathrm{\varnothing }$. Therefore, $s\in C_1,t\in C_2$ and $s,t\in S$. Clearly, $s\ast 0\in C_1$ and $t\ast 0\in C_2$. Consider, $0\le (s\ast 0)\wedge (t\ast 0)\le s\ast 0,t\ast 0$. Since $C_1\cap C_2$ is convex; $(s\ast 0)\wedge (t\ast 0)\in C_1\cap C_2$. We know that $s\wedge t\le (s\ast 0)\wedge (t\ast 0)$. By the given condition; $0\ast 0\le (s\wedge t)\ast 0\le [(s\ast 0)\wedge (t\ast 0)]\ast 0$. This implies that $0\le s\wedge t\le (s\ast 0)\wedge (t\ast 0)$. Therefore, $s\wedge t\in C$. Since S is meet closed, $s\wedge t\in S$. Therefore, $s\wedge t\in C\cap S$. This is a contradiction. Since $C\cap S\ne \mathrm{\varnothing }$. Therefore, either $x\in C$ or $y\in C$. Hence C is prime.

Theorem 3.46.

Let A be a semiregular np-autometrized l-algebra. If C is minimal prime, then $A\setminus C$ is meet closed which is maximal with respect to not containing 0.

Proof.

Suppose C is a minimal prime convex subalgebra of A. By theorem (3.44), we have $A\setminus C$ is meet closed. Since $0\in C$; $0\notin A\setminus C$.

To show that $A\setminus C$ is maximal meet closed with respect to not containing 0.

Let M be a meet closed subset of A such that

(9) $\begin{array}{l}A\setminus C\subseteq Mand0\notin M.\end{array}$(9)

Let Q be a convex subalgebra of A that is maximal with respect to $Q\cap M=\mathrm{\varnothing }$. By theorem (3.45), we have Q is prime. Since $Q\cap M=\mathrm{\varnothing }$; implies that $M\subseteq A\setminus Q$. As a result $A\setminus C\subseteq A\setminus Q$. Therefore $Q\subseteq C$. But C is minimal prime; we have Q = C. Therefore, $C\cap M=Q\cap M=\mathrm{\varnothing }$. As a result,

(10) $\begin{array}{l}M\subseteq A\setminus C.\end{array}$(10)

By Equations (9)(9) $\begin{array}{l}A\setminus C\subseteq Mand0\notin M.\end{array}$(9) and (10(10) $\begin{array}{l}M\subseteq A\setminus C.\end{array}$(10) ); $M=A\setminus C$. Hence $A\setminus C$ is maximal meet closed with respect to not containing 0.

3.2. Regular convex subalgebra

Definition 3.47.

Let A be an autometrized algebra. Let $R\in \mathcal{C}(A)$. Then R is called a regular convex subalgebra in A if $R=\bigcap _{\alpha \in \mathrm{\Gamma }}{C}_{\alpha }$, where $C_{\alpha }\in \mathcal{C}(A)$ for each $\alpha \in \mathrm{\Gamma }$ implies $R=C_{\beta }$ for some $\beta \in \mathrm{\Gamma }$.

Example 3.48.

Let $A=\{0,a,b,c\}$ with $0\le a\le b\le c$. Define $\ast$ and + by the following tables.

$\begin{array}{ll}\begin{array}{lllll}\ast & 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& 0& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{b}& 0& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& 0\end{array}& \begin{array}{lllll}+& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ 0& 0& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{a}& \mathrm{a}& \mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{b}& \mathrm{c}\\ \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}& \mathrm{c}\end{array}\end{array}$

It is clear to show that A is an autometrized algebra. We know that $C_1=\{0\}$, $C_2=\{0,a\}$, $C_3=\{0,a,b\}$ and $C_4=A$ are convex subalgebras. Also, $C_1=\bigcap _{i=1}^4{C}_i$, $C_2=C_2\cap C_3\cap C_4$ and $C_3=C_3\cap C_4$. Hence, C₁, C₂, and C₃ are regular convex subalgebras in A.

Remark 3.49.

Every regular convex subalgebra is a prime convex subalgebra. The converse is also true.

Remark 3.50.

Let A be an autometrized algebra. Let R be a regular convex subalgebra in A and R ≠ A. Denote $R^{\ast }$ the intersection of all convex subalgebras in A strictly containing R. Clearly, $R⫋R^{\ast }$ and $R^{\ast }$ is a unique cover of R in the lattice $\mathcal{C}(A)$.

Definition 3.51.

Let A be an autometrized algebra. Let $x(\ne 0)\in A$. Let $R\in \mathcal{C}(A)$ be a maximal convex subalgebra in A not containing “x”. Then R is called a value of the element “x” in A. The set of all values of “x” will be denoted by V(x).

Example 3.52.

In Example (3.22), A is an autometrized algebra. We know that $C_2=\{0,a\}$ and $C_3=\{0,b\}$ are convex subalgebras. Also, C₂ and C₃ are maximal convex subalgebras in A not containing “c”. Then C₂ and C₃ are values of the element “c” in A. Hence, $V(c)=\{C_2,C_3\}$.

Theorem 3.53.

Let A be an autometrized algebra. Let $R\in \mathcal{C}(A)$. Then R is regular if and only if there exists $x\in A$ such that $R\in V(x)$.

Proof.

Suppose R is regular. We know that by remark (3.50) $R^{\ast }$ is a unique cover of R in the lattice $\mathcal{C}(A)$. That is, $R⫋R^{\ast }$. Then $\mathrm{\exists }x\in R^{\ast }$ such that $x\notin R$. As a result, R is a convex subalgebra not containing x. Now, to show that R is a maximal convex subalgebra with respect to not containing x. Let $J\in \mathcal{C}(A)$ and $x\notin J$. Suppose $R\subseteq J$. So to show that R = J. Assume that R ≠ J. Clearly $R⫋J$ and $R^{\ast }\subseteq J$. Since $x\in R^{\ast }$ implies $x\in J$. This is a contradiction. Hence R = J. This implies R is maximal convex subalgebra not containing x. Then R is a value of x. Thus $R\in V(x)$.

Conversely, suppose $\mathrm{\exists }x\in A$ such that $R\in V(x)$. To show R is regular. Suppose that $R=\bigcap _{\alpha \in \mathrm{\Gamma }}{C}_{\alpha }$, where $\{C_{\alpha }{\}}_{\alpha \in \mathrm{\Gamma }}\subseteq \mathcal{C}(A)$. Since $x\notin R$; $\mathrm{\exists }\beta \in \mathrm{\Gamma }$ such that $x\notin C_{\beta }$. So C_β is a convex subalgebra not containing x and $R\subseteq C_{\beta }$. Since R is maximal with respect to not containing x, then $R=C_{\beta }$. Hence R is regular.

Theorem 3.54.

Let A be an autometrized algebra. Let $R\in \mathcal{C}(A)$. Let $x\in A\setminus R$. Then there exists $J\in V(x)$ such that $R\subseteq J$.

Proof.

The proof is obvious.

Theorem 3.55.

Let A be an autometrized algebra. Then any convex subalgebra in A is the intersection of regular convex subalgebras in A.

Proof.

Let $R\in \mathcal{C}(A)$ and $R⫋A$. Let $x\in A\setminus R$. By theorem (3.54); $\mathrm{\exists }{C}_x\in V(x)$ such that $R\subseteq C_x$. Hence $R\subseteq \bigcap _{x\in A\setminus R}{C}_x$. Conversely, let $y\in \bigcap _{x\in A\setminus R}{C}_x$. Then $y\in C_x\mathrm{\forall }x\in A\setminus R$. If $y\in A\setminus R$, then $y\in C_y$. So $y\notin A\setminus R$ and implies that $y\in R$. As result $\bigcap _{x\in A\setminus R}{C}_x\subseteq R$. Hence $R=\bigcap _{x\in A\setminus R}{C}_x$. Since $C_x\in V(x)$; by a theorem (3.53) C_x is a regular convex subalgebra. Therefore R is the intersection of regular convex subalgebras.

4. Convex spectral topologies of autometrized L-Algebras

In this section, we introduce the convex spectral topology of proper prime convex subalgebras in an np-autometrized l-algebra and discuss some fundamental facts. We also show that the convex spectrum is a Hausdorff space.

Definition 4.1.

Let A be an autometrized algebra. Let CSpec(A) = the set of all proper prime convex subalgebra in A. For any $N\subseteq A$, let

$\begin{array}{ll}O(N)=& \{P\in CSpec(A)\mid N⊈P\}.\\ K(N)=& \{P\in CSpec(A)\mid N\subseteq P\}.\end{array}$

Put $N=\{x\}$; then

$\begin{array}{l}O(x)=\{P\in CSpec(A)\mid x\notin P\}.\\ K(x)=\{P\in CSpec(A)\mid x\in P\}.\end{array}$

Example 4.2.

In Example (3.48), A is an autometrized algebra. We know that $C_1=\{0\}$, $C_2=\{0,a\}$, $C_3=\{0,a,b\}$ are proper prime convex subalgebras. Therefore, $CSpec(A)=\{C_1,C_2,C_3\}$. Let $N=\{a,b\}$. Therefore,

$\begin{array}{ll}O(N)=& \{P\in CSpec(A)\mid N⊈P\}.\\ =& \{C_1,C_2\}.\\ K(N)=& \{P\in CSpec(A)\mid N\subseteq P\}.\\ =& \{C_3\}.\end{array}$

Also, it is clear that $\mathcal{C}(N)=C_3=\{0,a,b\}$. Therefore,

$\begin{array}{ll}O(\mathcal{C}(N))=O(\mathcal{C}(N))=& \{P\in CSpec(A)\mid \mathcal{C}(N)⊈P\}.\\ =& \{C_1,C_2\}.\\ K(\mathcal{C}(N))=K(\mathcal{C}(N))=& \{P\in CSpec(A)\mid \mathcal{C}(N)\subseteq P\}.\\ =& \{C_3\}.\end{array}$

As a result, $O(\mathcal{C}(N))=O(N)$ and $K(\mathcal{C}(N))=K(N)$.

From this example, we conclude the following.

Remark 4.3.

Let A be an autometrized algebra. Let CSpec(A) = the set of all proper prime convex subalgebra in A. For any $N\subseteq A$, let $O(\mathcal{C}(N))=O(N)$ and $K(\mathcal{C}(N))=K(N)$.

Proof.

Let $P\in O(\mathcal{C}(N))$. Then $\mathcal{C}(N)⊈P$. If $N\subseteq P$, then $\mathcal{C}(N)\subseteq P$. This is a contradiction. Therefore $N⊈P$. Then $P\in O(N)$. Hence $O(\mathcal{C}(N))\subseteq O(N)$. Conversely, $P\in O(N)$. Which implies that $N⊈P$; since $N\subseteq \mathcal{C}(N)$ implies $\mathcal{C}(N)⊈P$. Thus, $P\in O(\mathcal{C}(N))$ and $O(N)\subseteq O(\mathcal{C}(N))$. Hence $O(N)=O(\mathcal{C}(N))$.

Similarly, let $P\in K(\mathcal{C}(N))$. Then $\mathcal{C}(N)\subseteq P$; and implies that $N\subseteq P$. So $P\in K(N)$. Hence $K(\mathcal{C}(N))\subseteq K(N)$. Conversely, $P\in K(N)$. That implies that $N\subseteq P$; and implies $\mathcal{C}(N)\subseteq P$. Then $P\in K(\mathcal{C}(N))$. Therefore $K(N)\subseteq K(\mathcal{C}(N))$. Hence $K(N)=K(\mathcal{C}(N))$.

Hence we will consider only

$\begin{array}{l}\{O(N)\mid N\subseteq A\}=\{O(C)\mid C\in \mathcal{C}(A)\}.\\ \{K(N)\mid N\subseteq A\}=\{K(C)\mid C\in \mathcal{C}(A)\}.\end{array}$

For any $x\in A$;

$\begin{array}{l}O(x)=O(\mathcal{C}(x)).\\ K(x)=K(\mathcal{C}(x)).\end{array}$

Lemma 4.4.

Let A be an np-autometrized l-algebra. Then

(i)	=	$O(0)=\mathrm{\varnothing },O(A)=CSpec(A)$.
(ii)	=	$O(C_1\cap C_2)=O(C_1)\cap O(C_2);C_1,C_2\in \mathcal{C}(A)$.
(iii)	=	For any $\{C_{\alpha }{\}}_{\alpha \in \mathrm{\Delta }}\subseteq \mathcal{C}(A)$; we have $\begin{array}{l}O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })=\bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha }).\end{array}$

Proof.

(i)	=	Since every convex subalgebra contains 0; $O(0)=\{P\in CSpec(A)\mid 0\notin P\}=\mathrm{\varnothing }$. Since each $P\in CSpec(A)$ is proper; $O(A)=\{P\in CSpec(A)\mid A⊈P\}=CSpec(A)$.
(ii)	=	Let $C_1,C_2\in \mathcal{C}(A)$. Let $P\in O(C_1\cap C_2)$. Then, $C_1\cap C_2⊈P$. If either $C_1\subseteq P$ or $C_2\subseteq P$; then $C_1\cap C_2\subseteq C_1,C_2\subseteq P$. This is a contradiction; since $C_1⊈P$ and $C_2⊈P$. Which implies $P\in O(C_1)$ and $P\in O(C_2)$; then $P\in O(C_1)\cap O(C_2)$. Hence (11) $\begin{array}{l}O(C_1\cap C_2)\subseteq O(C_1)\cap O(C_2).\end{array}$(11) Let $P\in O(C_1)\cap O(C_2)$. So $P\in O(C_1)$ and $P\in O(C_2)$. Which implies that $C_1⊈P$ and $C_2⊈P$. Since P is prime convex subalgebra; $C_1\cap C_2⊈P$. Then $P\in O(C_1\cap C_2)$. (12) $\begin{array}{l}O(C_1)\cap O(C_2)\subseteq O(C_1\cap C_2).\end{array}$(12) By Equations (11)(11) $\begin{array}{l}O(C_1\cap C_2)\subseteq O(C_1)\cap O(C_2).\end{array}$(11) and (12(12) $\begin{array}{l}O(C_1)\cap O(C_2)\subseteq O(C_1\cap C_2).\end{array}$(12) ); $O(C_1\cap C_2)=O(C_1)\text{break}\cap O(C_2)$.
(iii)	=	Let $\{C_{\alpha }{\}}_{\alpha \in \mathrm{\Delta }}\subseteq \mathcal{C}(A)$. Let $P\in O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })$. Then $\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }⊈P$. If $C_{\alpha }\subseteq P\mathrm{\forall }\alpha \in \mathrm{\Delta }$, then $\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }\subseteq P$. This is a contradiction. Then $\mathrm{\exists }\beta \in \mathrm{\Delta }$ such that $C_{\beta }⊈P$. Therefore $P\in O(C_{\beta })$. Thus $P\in \bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha })$. (13) $\begin{array}{l}O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })\subseteq \bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha }).\end{array}$(13) Let $P\in \bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha })$. Then $P\in O(C_{\beta })$ for some $\beta \in \mathrm{\Delta }$. Thus, $C_{\beta }⊈P$. Since $C_{\beta }\subseteq \underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }$; then $\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }⊈P$. Hence $P\in O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })$. (14) $\begin{array}{l}\bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha })\subseteq O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }).\end{array}$(14) By Equations (13)(13) $\begin{array}{l}O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })\subseteq \bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha }).\end{array}$(13) and (14(14) $\begin{array}{l}\bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha })\subseteq O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }).\end{array}$(14) ); (15) $\begin{array}{l}O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })=\bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha }).\end{array}$(15)

Lemma 4.5.

Let A be an np-autometrized l-algebra. Then

(i)	=	$O((x\ast 0)\vee (y\ast 0))=O(x)\cup O(y)\mathrm{\forall }x,y\in A$.
(ii)	=	$O((x\ast 0)\wedge (y\ast 0))=O(x)\cap O(y)\mathrm{\forall }x,y\in A$.

Proof.

(i)	=	By theorem (3.16); we have for any $x,y\in A$ $\begin{array}{ll}\mathcal{C}(x)\vee \mathcal{C}(y)=& \mathcal{C}((x\ast 0)\vee (y\ast 0)).\\ O(\mathcal{C}(x)\vee \mathcal{C}(y))=& O(\mathcal{C}((x\ast 0)\vee (y\ast 0))).\\ O(\mathcal{C}(x))\cup O(\mathcal{C}(y))=& O(\mathcal{C}((x\ast 0)\vee (y\ast 0))).\\ & [Bytheorem(4.4)]\end{array}$ Therefore, $O(x)\cup O(y)=O((x\ast 0)\vee (y\ast 0))$.
(ii)	=	By theorem (3.16); we have for any $x,y\in A$ $\begin{array}{ll}\mathcal{C}(x)\wedge \mathcal{C}(y)=& \mathcal{C}((x\ast 0)\wedge (y\ast 0)).\\ O(\mathcal{C}(x)\wedge \mathcal{C}(y))=& O(\mathcal{C}((x\ast 0)\wedge (y\ast 0))).\\ O(\mathcal{C}(x))\cap O(\mathcal{C}(y))=& O(\mathcal{C}((x\ast 0)\wedge (y\ast 0))).\\ & [Bytheorem(4.4)]\end{array}$ Therefore, $O(x)\cap O(y)=O((x\ast 0)\wedge (y\ast 0))$.

Corollary 4.6.

Let A be an np-autometrized l-algebra. Let $\tau =\{O(C)\mid C\in \mathcal{C}(A)\}$. Then τ is a topology of CSpec(A).

Proof.

(i)	=	We know that $\{0\},A\in \mathcal{C}(A)$. By lemma (4.4); $\mathrm{\varnothing }=O(0)\in \tau$ and $CSpec(A)=O(A)\in \tau$.
(ii)	=	Let $O(C_1),O(C_2)\in \tau$. Where $C_1,C_2\in \mathcal{C}(A)$. Then $C_1\cap C_2\in \mathcal{C}(A)$. By lemma (4.4); $O(C_1)\cap O(C_2)=O(C_1\cap C_2)\in \tau$. Hence τ is closed under finite intersections.
(iii)	=	Let $\{O(C_{\alpha }){\}}_{\alpha \in \mathrm{\Delta }}\subseteq \tau$. Where $\{C_{\alpha }{\}}_{\alpha \in \mathrm{\Delta }}\subseteq \mathcal{C}(A)$. So $\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }\in \mathcal{C}(A)$. By lemma (4.4);

$\begin{array}{l}\bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha })=O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })\in \tau .\end{array}$

That means an arbitrary union of elements of τ is again an element of τ. Hence τ is a topology on CSpec(A).

Definition 4.7.

Let A be an np-autometrized l-algebra. Then, the topology τ is called Convex Spectral Topology on CSpec(A). The topological space $(CSpec(A),\tau )$ is called the Convex Spectrum of A.

Example 4.8.

In Example (4.2), A is an np-autometrized l-algebra and we have $CSpec(A)=\{C_1,C_2,C_3\}$, also $O(C_1)=\mathrm{\varnothing }$, $O(C_2)=\{C_1\}$, $O(C_3)=\{C_1,C_2\}$ and $O(A)=CSpec(A)=\{C_1,C_2,C_3\}$ . Then, $$\begin{gathered} \tau =\{\mathrm{\varnothing },\{C_1\}, \\ \{C_1,C_2\},\{C_1,C_2,C_3\}\} \end{gathered}$$ is a convex spectral topology. Hence, $(CSpec(A),\tau )$ is a convex spectrum of A.

Theorem 4.9.

Let A be an np-autometrized l-algebra. Let $\mathcal{B}=\{O(x)\mid x\in A\}$. Then $\mathcal{B}$ is a basis for τ.

Proof.

Let $O(C)\in \tau$ where $C\in \mathcal{C}(A)$. We can write $C=\underset{x\in C}{\bigvee }\mathcal{C}(x)$. Then, $O(C)=O(\underset{x\in C}{\bigvee }\mathcal{C}(x))$. By lemma (4.4), $O(C)=\bigcup _{x\in C}O(\mathcal{C}(x))$. Hence $O(C)=\bigcup _{x\in C}O(x)$.

Therefore, O(C) = union of elements of $\mathcal{B}$. Hence $\mathcal{B}$ is basis for τ.

Proposition 4.10.

Let A be an np-autometrized l-algebra. Define a map $\theta :\mathcal{C}(A)\to \tau$ by $\theta (C)=O(C)$. Then θ is a lattice isomorphism of $\mathcal{C}(A)$ onto τ.

Proof.

We know that $\mathcal{C}(A),\tau$ are lattices. Clearly, θ is on to map. Consider,

$\begin{array}{ll}\theta (C_1\wedge C_2)& =\theta (C_1\cap C_2)=O(C_1\cap C_2).\end{array}$

By lemma (4.4);

$\begin{array}{ll}\theta (C_1\wedge C_2)& =O(C_1)\cap O(C_2)=\theta (C_1)\cap \theta (C_2).\end{array}$

Also, consider,

$\begin{array}{ll}\theta (C_1\vee C_2)& =O(C_1\vee C_2).\end{array}$

By lemma (4.4);

$\begin{array}{ll}\theta (C_1\vee C_2)& =O(C_1)\cup O(C_2)=\theta (C_1)\cup \theta (C_2).\end{array}$

Hence θ is a homomorphism.

Suppose $\theta (C_1)=\theta (C_2)$. Then $O(C_1)=O(C_2)$. By theorem (3.55), every convex subalgebra is the intersection of regular convex subalgebra(hence prime convex subalgebra) containing it. Therefore we can write:

$\begin{array}{ll}{C}_1& =\bigcap \{P\in CSpec(A)\mid C_1\subseteq P\}.\\ & =\bigcap \{P\in CSpec(A)\mid P\in K(C_1)\}.\end{array}$

Similarly,

$\begin{array}{ll}{C}_2& =\bigcap \{Q\in CSpec(A)\mid Q\in K(C_2)\}.\end{array}$

Since $O(C_1)=O(C_2)$, we get $K(C_1)=K(C_2)$. Thus $C_1=C_2$ and θ is one-to-one. Therefore $\mathcal{C}(A)\cong \tau$.

Theorem 4.11.

Let A be an np-autometrized l-algebra.

(i)	=	O(x) is compact for every $x\in A$.
(ii)	=	If B is open compact set of CSpec(A), then $B=O(x)$ for some $x\in A$.

Proof.

(i)	=	Let $x\in A$. To claim that O(x) is compact. Suppose, $\begin{array}{ll}O(x)\subseteq & \bigcup _{\alpha \in \mathrm{\Delta }}O(C_{\alpha }).\\ =& O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }).\end{array}$ Then, $O(\mathcal{C}(x))\subseteq O(\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })$. By theorem (4.9); $\theta (\mathcal{C}(x))\subseteq \theta (\underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha })$. Since θ is order preserving, $\mathcal{C}(x)\subseteq \underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }$. Which implies, $x\in \underset{\alpha \in \mathrm{\Delta }}{\bigvee }{C}_{\alpha }$. Therefore $x\ast 0\le m_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)$ for some positives $m_1,\cdots ,m_k$ and $a_i\in C_{{\alpha }_i}$ for $i=1,2,\dots ,k$. So, $\begin{array}{ll}{m}_1(a_1\ast 0)+\dots +m_k(a_k\ast 0)& \in \underset{i=1}{\overset{k}{\bigvee }}{C}_{{\alpha }_i}.\end{array}$ By the given condition, $x\in \underset{i=1}{\overset{k}{\bigvee }}{C}_{{\alpha }_i}$. Therefore $\begin{array}{ll}O(x)& \subseteq O(\underset{i=1}{\overset{k}{\bigvee }}{C}_{{\alpha }_i})=\bigcup _{i=1}^{k}O(C_{{\alpha }_i}).\end{array}$ Hence O(x) is compact.
(ii)	=	Suppose B is compact open set in CSpec(A). Therefore $B\in \tau$. So $B=\bigcup _{i=1}^{\mathrm{\infty }}O(x_i)$= union of basis element. Since B is compact; $B=\bigcup _{i=1}^{n}O(x_i)$. Hence $B=O(\underset{i=1}{\overset{n}{\bigvee }}{x}_i)=O(y)$ where $y=\underset{i=1}{\overset{n}{\bigvee }}\text{break}{x}_i\in A$.

Corollary 4.12.

Let A be an np-autometrized l-algebra. CSpec(A) is compact if and only if $A=\mathcal{C}(x)$ for some $x\in A$.

Proof.

Suppose CSpec(A) is compact. That is, CSpec(A) is compact in $(CSpec(A),\tau )$. By lemma (4.4); we know that: $O(A)=CSpec(A)\in \tau$. Therefore, $O(A)=CSpec(A)$ is open and compact. By above theorem (4.11);

$\begin{array}{ll}O(A)& =O(x)\text{for some}x\in A.\\ & =O(\mathcal{C}(x)).\end{array}$

Thus, $\theta (A)=\theta (\mathcal{C}(x))$. Since θ is an isomorphism. Hence $A=\mathcal{C}(x)$.

Conversely, suppose $A=\mathcal{C}(x)$ for some $x\in A$. Then $O(A)=O(\mathcal{C}(x))$. Which implies $CSpec(A)=O(x)$. By theorem (4.11), we have O(x) is compact. Hence CSpec(A) is compact.

Definition 4.13.

Let A be an autometrized l-algebra. For any $M\subseteq CSpec(A)$; define

$\begin{array}{l}{C}_M=\bigcap \{P\mid P\in M\}.\end{array}$

And if $M\subseteq N\subseteq CSpec(A)$. Then $C_N\subseteq C_M$.

Proposition 4.14.

Let A be an np-autometrized l-algebra.

(i)	=	The closed sets in CSpec(A) are exactly all K(C) where $C\in \mathcal{C}(A)$.
(ii)	=	$M\subseteq CSpec(A)\Rightarrow \bar{M}=K(C_M)$.

Proof.

(i)	=	For any $C\in \mathcal{C}(A)$, we have $K(C)=CSpec(A)\text{break}\setminus O(C)$ = closed set; since O(C) is open set. Let $C\subseteq CSpec(A)$ be a closed set. Then $CSpec(A)\setminus C$ is open set; and $CSpec(A)\setminus C=O(C)$ for some $C\in \mathcal{C}(A)$. Hence $C=CSpec(A)\setminus O(C)=K(C)$.
(ii)	=	Let $M\subseteq CSpec(A)$. We know that $K(C_M)=\{P\in CSpec(A)\mid C_M\subseteq P\}$. Let $P\in M$. By the definition of $C_M;C_M\subseteq P$. Then $P\in K(C_M)$. Hence, (16) $\begin{array}{l}M\subseteq K(C_M).\end{array}$(16) By (i); $K(C_M)$ is closed set; implies $K(C_M)$ is a closed set containing M. But $\bar{M}$ = the smallest closed set containing M; thus $\bar{M}\subseteq K(C_M)$. Therefore, (17) $\begin{array}{l}{C}_{K(C_M)}\subseteq C_{\bar{M}}.\end{array}$(17) Now, let us show that $C_{K(C_M)}=C_M$. By Equation (16)(16) $\begin{array}{l}M\subseteq K(C_M).\end{array}$(16) ; we have $M\subseteq K(C_M)$. So $C_{K(C_M)}\subseteq C_M$. Consider $\begin{array}{ll}{C}_{K(C_M)}=& \bigcap \{Q\mid Q\in K(C_M)\}.\\ =& \bigcap \{Q\mid C_M\subseteq Q\}.\\ =& \bigcap \{Q\mid \bigcap \{P\mid P\in M\}\subseteq Q\}.\end{array}$ Since Q is prime convex subalgebra, $\begin{array}{ll}& =\bigcap \{Q\mid P\subseteq Q\text{for some}P\in M\}.\end{array}$ Let $z\in C_M$. Then $z\in P\mathrm{\forall }P\in M$. If $P\subseteq Q$ for some $P\in M$, then $z\in Q$. Then, $\begin{array}{ll}z& \in \bigcap \{Q\mid P\subseteq Q\text{for some}P\in M\}.\end{array}$ So $z\in C_{K(C_M)}$. Therefore $C_M\subseteq C_{K(C_M)}$. Hence, (18) $\begin{array}{ll}{C}_{K(C_M)}& =C_M.\end{array}$(18) By Equations (17)(17) $\begin{array}{l}{C}_{K(C_M)}\subseteq C_{\bar{M}}.\end{array}$(17) and (18(18) $\begin{array}{ll}{C}_{K(C_M)}& =C_M.\end{array}$(18) ); we can write $C_M\subseteq C_{\bar{M}}$. We know that $M\subseteq \bar{M}$; $C_{\bar{M}}\subseteq C_M$. (19) $\begin{array}{ll}{C}_M& =C_{\bar{M}}.\end{array}$(19) Hence, (20) $\begin{array}{ll}K(C_M)& =K(C_{\bar{M}}).\end{array}$(20) Now to show that $K(C_{\bar{M}})=\bar{M}$. By Equation (16)(16) $\begin{array}{l}M\subseteq K(C_M).\end{array}$(16) we have; $\bar{M}\subseteq K(C_{\bar{M}})$. Then $M\subseteq \bar{M}\subseteq K(C_{\bar{M}})$. Which implies that $K(C_{\bar{M}})$ is a closed set containing M. Let K(C) be a closed set containing M where $C\in \mathcal{C}(A)$. Therefore $M\subseteq K(C)$; and $\bar{M}\subseteq K(C)$. Whence $C_{K(C)}\subseteq C_{\bar{M}}$; and implies that $K(C_{\bar{M}})\subseteq K(C_{K(C)})$. Since $\begin{array}{ll}{C}_{K(C)}& =\bigcap \{P\mid P\in K(C)\}.\end{array}$ Then, $\begin{array}{ll}K(C_{K(C)})=& \{Q\mid C_{K(C)}\subseteq Q\}.\\ =& \{Q\mid \bigcap \{P\mid P\in K(C)\}\subseteq Q\}.\\ =& \{Q\mid P\subseteq Q\text{for some}P\in K(C)\}.\\ =& \{Q\mid P\subseteq Q\text{for, some }P\text{such, that}\\ & C\subseteq P\}.\\ =& \{Q\mid C\subseteq Q\}.\\ =& K(C).\end{array}$ As a result $K(C_{\bar{M}})\subseteq K(C)$. Thus $K(C_{\bar{M}})$ is the smallest closed set containing M; and (21) $\begin{array}{lllll}\bar{M}=K(C_{\bar{M}}).& & & & \end{array}$(21) By Equations (20)(20) $\begin{array}{ll}K(C_M)& =K(C_{\bar{M}}).\end{array}$(20) and (21(21) $\begin{array}{lllll}\bar{M}=K(C_{\bar{M}}).& & & & \end{array}$(21) ); we have: $\bar{M}=K(C_M)$.

Corollary 4.15.

Let A be an np-autometrized l-algebra. Let $M\subseteq CSpec(A)$. Then M is dense if and only if $\bigcap \{P\mid P\in M\}=\{0\}$.

Proof.

Suppose $M\subseteq CSpec(A)$. Let M be dense. To show that $\bigcap \{P\mid P\in M\}=\{0\}$. Since M is dense; $\bar{M}=CSpec(A)$. By proposition (4.14); $\bar{M}=K(C_M)$ where $C_M=\bigcap \{P\mid P\in M\}$; and then $CSpec(A)=K(C_M)$. Also $CSpec(A)=K(\{0\})=\{P\in CSpec(A)\mid \{0\}\subseteq P\}$. Which implies that $K(C_M)=K(\{0\})$. We get $O(C_M)=O(\{0\})$. Since θ is an isomorphism; $C_M=\{0\}$. Hence $\bigcap \{P\mid P\in M\}=\{0\}$.

Conversely, suppose $\bigcap \{P\mid P\in M\}=\{0\}$. Let $M\subseteq CSpec(A)$. To show that M is dense. That is $\bar{M}=CSpec(A)$. Since $\bigcap \{P\mid P\in M\}=\{0\}$; we get $C_M=\{0\}$. By proposition (4.14); $M\subseteq CSpec(A)\Rightarrow \bar{M}=K(C_M)$. Therefore $\bar{M}=K(\{0\})$. Hence $\bar{M}=CSpec(A)$.

Theorem 4.16.

Let A be an np-autometrized l-algebra. Then $(CSpec(A),\tau )$ is T₁ - space.

Proof.

Let C₁ and C₂ be two distinct prime convex subalgebra of A. Clearly, there exist two open sets:

$\begin{array}{l}O(C_1)=\{C\in CSpec(A)\mid C_1⊈C\},\\ O(C_2)=\{C\in CSpec(A)\mid C_2⊈C\},\end{array}$

such that $C_1\in O(C_2)$ but $C_2\notin O(C_2)$ and $C_2\in O(C_1)$ but $C_1\notin O(C_1)$. Hence $(CSpec(A),\tau )$ is T₁ - space.

Theorem 4.17.

Let A be an np-autometrized l-algebra with for any $C_i,C_j\in CSpec(A)$ with $C_i\cap C_j=\{0\}$ for any i ≠ j. Then $(CSpec(A),\tau )$ is a Hausdorff space.

Proof.

Let C₁ and C₂ be two distinct prime convex subalgebra of A. Clearly, there exist two open sets:

$\begin{array}{l}O(C_1)=\{C\in CSpec(A)\mid C_1⊈C\},\\ O(C_2)=\{C\in CSpec(A)\mid C_2⊈C\},\end{array}$

such that $C_1\in O(C_2)$ but $C_2\notin O(C_2)$ and $C_2\in O(C_1)$ but $C_1\notin O(C_1)$.

Now to show that $O(C_1)\cap O(C_2)=\mathrm{\varnothing }$. Clearly, $O(C_1)\cap O(C_2)=O(C_1\cap C_2)$. It is clear that $C_1\cap C_2=\{0\}$. Therefore, $O(C_1)\cap O(C_2)=\mathrm{\varnothing }$. Hence $(CSpec(A),\tau )$ is a Hausdorff space.

Remark 4.18.

In general, suppose there exist i and j such that $C_i,C_j\in CSpec(A)$ with $C_i\cap C_j\ne \{0\}$ for any i ≠ j, then $(CSpec(A),\tau )$ is not necessary to be Hausdorff space. In Example (4.8), we know that $(CSpec(A),\tau )$ is a convex spectrum of A. And we have, $O(C_1)=\mathrm{\varnothing }$, $O(C_2)=\{C_1\}$, $O(C_3)=\{C_1,C_2\}$ and $O(A)=CSpec(A)=\{C_1,C_2,C_3\}$. But $O(C_2)\cap O(C_3)=\{C_1\}\cap \{C_1,C_2\}=\{C_1\}\ne \mathrm{\varnothing }$. Thus $(CSpec(A),\tau )$ is not a Hausdorff space.

5. Discussion

Our study introduces the concept of convex subalgebras and demonstrates that the set of all convex subalgebras of an autometrized algebra forms a distributive lattice. This finding is significant because it can help in constructing a distributive lattice. Additionally, we explore the concept of congruence relations in an autometrized algebra, which extends the congruence relations in normal autometrized algebra as introduced by Swamy and Rao (1977). We establish a one-to-one correspondence between the set of all convex subalgebras and the set of all congruences in an np-autometrized algebra. Our findings provide valuable insights into algebraic structures and their properties.

Furthermore, in this study, we explored the concepts of prime convex subalgebra and regular convex subalgebra, which have not been previously studied in previous articles. Our findings helped us understand the relationships between prime convex subalgebra, meet closed set, and regular convex subalgebra. Specifically, we discussed the similarities and differences between prime convex subalgebra and regular convex subalgebra, and how they relate to meet closed sets. Overall, this study sheds light on important aspects of algebraic structures that were previously unexplored.

Also, we discovered that a set of proper prime convex subalgebras construct a topology called convex spectral topology. This new topology is an extension of the spectral topology introduced by Rachŭnek (1998). Additionally, we have established several results in convex spectral topology. Notably, our findings demonstrate that the convex spectrum is both a T₁ space and a Hausdorff space.

6. Conclusion

This paper presented the notions of convex subalgebras and congruence relations in an autometrized algebra. We also demonstrated that the collection of all convex subalgebras of an autometrized algebra forms a lattice and distributive. Additionally, we established a one-to-one correspondence between the set of all convex subalgebras and the set of all congruences on an np-autometrized algebra. Furthermore, we explored the concepts of prime convex subalgebra and regular convex subalgebra. We also introduced the idea of meet closed subsets. We showed that in a semiregular np-autometrized l-algebra, the intersection of a chain of prime convex subalgebra is a prime convex subalgebra. We also proved that any convex subalgebra in an autometrized algebra is the intersection of regular convex subalgebras.

Lastly, we introduced the convex spectral topology of proper prime convex subalgebras in an autometrized l-algebra and discussed some fundamental facts. We also proved that a convex spectrum is compact in an np-automatized l-algebra A if and only if A is generated by some element. Specifically, we demonstrated that the convex spectrum is a T₁ - space and Hausdorff space. In the future, we may investigate the notions of quotient convex subalgebra and get more results in the convex spectral topology of autometrized l-algebras.

Acknowledgements

The authors are very grateful to the anonymous reviewers for their helpful comments and suggestions that helped to improve the paper’s quality.

Disclosure statement

No potential conflict of interest was reported by the author(s).

Supplementary material

Supplemental data for this article can be accessed online at https://doi.org/10.1080/27684830.2023.2283261

Additional information

Funding

The authors received no direct funding from public or non-public organizations for this research work.

Notes on contributors

Gebrie Yeshiwas Tilahun

Gebrie Yeshiwas Tilahun is a Ph.D. scholar at the Department of Mathematics, College of Natural Sciences, Arba Minch University, Ethiopia. His research interest includes the development of an autometrized algebra.