Complete decomposable MS-algebras

Abstract

According to the characterization of decomposable MS-algebras in terms of triples (M,D,φ), where M is a de Morgan algebra, D is a distributive lattice with 1 and φ is a (0,1)-homomorphism of M into F(D), the filter lattice of D, we characterize complete decomposable MS-algebras in terms of complete decomposable MS-triples. Also, we describe the complete homomorphisms of complete decomposable MS-algebras by means of complete decomposable MS-triples.

Introduction

Morgan Stone algebras (or simply MS-algebras) are introduced and characterized by T.S. Blyth and J.C. Varlet [1] as a generalization of both de Morgan algebras and Stone algebras. In [2], T.S. Blyth and J.C. Varlet described the lattice Λ(MS) of subclasses of the class MS of all MS-algebras. A. Badawy, D. Guffova, and M. Haviar [3] introduced and characterized decomposable MS-algebras by means of decomposable MS-triples. Moreover, they constructed a one-to-one correspondence between decomposable MS-algebras and decomposable MS-triples. A. Badawy and R. El-Fawal [4] studied many properties of decomposable MS-algebras in terms of decomposable MS-triples as homomorphisms and subalgebras. Also, they formulated and solved some fill in problems concerning homomorphisms and subalgebras of decomposable MS-algebras. A. Badawy [5] introduced the notion of d_L-filters of principal MS-algebras. Recently, A. Badawy [6] studied the relationship between de Morgan filters and congruences of decomposable MS-algebras. Also, many properties of ideals of MS-algebras are given in [7] and [8].

Several authors studied complete p-algebras, like C.C. Chain and G. Gr$\ddot {a}$tzer [9] for Stone algebras, S. El-Assar, and M. Atallah [10] for distributive p-algebras and P. Mederly [11] for modular p-algebras.

In this paper, we introduce complete decomposable MS-algebras and complete decomposable MS-triples. We show that a decomposable MS-algebra L constructed from the decomposable MS-triple (M,D,φ) is complete if and only if the triple (M,D,φ) is complete. Also, a description of complete homomorphisms of decomposable MS-algebras is given in terms of complete decomposable MS-triples.

Preliminaries

In this section, we present definitions and main results which are needed through this paper. We refer the reader to [1–4, 12–15] for more details.

A de Morgan algebra is an algebra (L;∨,∧,⁻,0,1) of type (2,2,1,0,0) where (L;∨,∧,0,1) is a bounded distributive lattice and the unary operation of involution ⁻ satisfies

$\overline {\overline {x}}=x,\overline {(x\vee y)}=\overline {x}\wedge \overline {y},\overline {(x\wedge y)}=\overline {x}\vee \overline {y}.$

An MS-algebra is an algebra (L;∨,∧, ^∘,0,1) of type (2,2,1,0,0) where (L;∨,∧,0,1) is a bounded distributive lattice and the unary operation ^∘ satisfies

x≤x^∘∘,(x∧y)^∘=x^∘∨y^∘,1^∘=0.

The following Theorem gives the basic properties of MS-algebras.

Theorem 1

([1,12]). For any two elements a,b of an MS-algebra L, we have (1) 0^∘=1, (2) a≤b⇒b^∘≤a^∘, (3) a^∘∘∘=a^∘, (4) (a∨b)^∘=a^∘∧b^∘, (5) (a∨b)^∘∘=a^∘∘∨b^∘∘, (6) (a∧b)^∘∘=a^∘∘∧b^∘∘.

Lemma 1

([1,3]). Let L be an MS-algebra. Then (1) L^∘∘={x∈L:x=x^∘∘} is a de Morgan subalgebra of L, (2) D(L)={x∈L:x^∘=0} is a filter (filter of dense elements) of L.

For any lattice L, let F(L) denotes the set of all filters of L. It is known that, (F(L);∧,∨) is a distributive lattice if and only if L is a distributive lattice, where the operation ∧ and ∨ are given by

F∧G=F∩G and F∨G={x∈L:x≥f∧g,f∈F,g∈G},respectively for everyF,G∈F(L).

Also, [a)={x∈L:x≥a} is a principal filter of L generated by a.

Definition 1

[9]. Let $\phantom {\dot {i}\!}L=(L;\vee,\wedge,0_{L},1_{L})$ and $\phantom {\dot {i}\!}L_{1}=(L_{1};\vee,\wedge,0_{L_{1}},1_{L_{1}})$ be bounded lattices. The map h:L→L₁ is called (0,1)-lattice homomorphism if (1) $\phantom {\dot {i}\!}0_{L}h=0_{L_{1}}$ and $\phantom {\dot {i}\!}1_{L}h=1_{L_{1}}$, (2) h preserves joins, that is, (x∨y)h=xh∨yh for every x,y∈L, (3) h preserves meets, that is, (x∧y)h=xh∧yh for every x,y∈L.

Definition 2

[14] A (0,1)-lattice homomorphism h:L→L₁ of an MS-algebra L into an MS-algebra L₁ is called a homomorphism if x^∘h=xh^∘ for all x∈L. If L and L₁ are de Morgan algebras, then h is called a de Morgan homomorphism.

Definition 3

[3] An MS-algebra L is called decomposable MS-algebra if for every x∈L there exists d∈D(L) such that x=x^∘∘∧d.

Definition 4

[3] A decomposable MS-triple is (M,D,φ), where (i) $(M;\vee,\wedge,\bar {},0,1)$ is a de Morgan algebra, (ii) (D;∨,∧,1) is a distributive lattice with 1, (iii) φ is a (0,1)-homomorphism from M into F(D) such that for every element a∈M and for every y∈D there exists an element t∈D with aφ∩[y)=[t).

Theorem 2

[3] (Construction Theorem) Let (M,D,φ) be a decomposable MS-triple. Then

$$L=\left\{(a,\bar{a}\varphi \vee \left[x\right)):a\in M,x\in D\right\} $$

is a decomposable MS-algebra, if we define

$$(a,\bar{a}\varphi \vee \left[x\right))\vee (b,\bar{b}\varphi \vee \left[y\right))=\left(a\vee b,\overline{(a\vee b)}\varphi \vee \left[t\right)\right)\text{for some}t\in D, $$

$$(a,\bar{a}\varphi \vee \left[x\right))\wedge (b,\bar{b}\varphi \vee \left[y\right))=\left(a\wedge b,\overline{(a\wedge b)}\varphi \vee \left[x\wedge y\right)\right), $$

$$\left(a,\bar{a}\varphi \vee \left[x\right)\right)^{\circ }=(\bar{a},a\varphi), $$

$$1_{L}=(1,\left[1\right)), $$

$$0_{L}=(0,D). $$

Conversely, every decomposable MS-algebra L can be associated with the decomposable MS-triple (L^∘∘,D(L),φ(L)), where

aφ(L)=[a^∘)(L),a∈L^∘∘.

The decomposable MS-algebra L constructed in Theorem 2 is called the decomposable MS-algebra associated with the decomposable MS-triple (M,D,φ) and the construction of L described in Theorem 2 is called a decomposable MS-construction.

Corollary 1

[3] Let L be a decomposable MS-algebra associated with the decomposable MS-triple (M,D,φ). Then (1) $L^{\circ \circ }=\left \{(a,\bar {a}\varphi):a\in M\right \}$, (2) D(L)={(1,[x)):x∈D}, (3) D≅D(L) and M≅L^∘∘, (4) The order of L is given as follows: $(a,\bar {a}\varphi \vee [x))\leq (b,\bar {b}\varphi \vee [y))$ iff a≤b and $\bar {a}\varphi \vee [x)\supseteq \bar {b}\varphi \vee [y)$.

Definition 5

[14] A lattice L is called complete if infLH and supLH exist for each ϕ≠H⊆L.

Definition 6

[14] A lattice L is called conditionally complete if every upper bounded subset of L has a supermum in L and every lower bounded subset of L has an infimum in L.

An MS-algebra L is called complete if it is complete as a lattice.

Definition 7

[14] A lattice homomorphism h:L→L₁ of a complete lattice L into a complete lattice L₁ is called complete if

$(\inf _{L}H)h=\inf _{L_{1}}Hh$ and $(\sup _{L}H)h=\sup _{L_{1}}Hh$ for each ϕ≠H⊆L.

A homomorphism h:L→L₁ of a complete MS-algebra L into a complete MS-algebra L₁ is called complete if it is complete as a lattice homomorphism.

Characterization of complete decomposable MS-algebras via triples

In this section, we introduce and characterize complete decomposable MS-triples of complete decomposable MS-algebras.

Let L be a decomposable MS-algebra L. For ϕ≠N⊆L, define N^∘ as follows:

N^∘={n^∘:n∈N}.

Lemma 2

If L is a complete decomposable MS-algebra, then for ϕ≠N⊆L,ϕ≠C⊆L^∘∘ and ϕ≠E⊆D(L), we have (1) (supLN)^∘= infLN^∘, (2) $\sup _{L^{\circ \circ }}C=(\sup _{L}C)^{\circ \circ }=(\inf _{L} C^{\circ })^{\circ }$, (3) $\inf _{L^{\circ \circ }} C=\inf _{L}C$, (4) infD(L)E= infLE and supD(L)E= supLE.

Proof

(1). Let x= supLN. Then x≥n for all n∈N implies x^∘≤n^∘. Hence x^∘ is a lower bound of N^∘. Let y be a lower bound of N^∘. Then y≤n^∘ for all n∈N implies y^∘≥n^∘∘≥n. So, y^∘ is an upper bound of N. Thus x≤y^∘ as x= supLN. This gives x^∘≥y^∘∘≥y. Therefore x^∘= infLN^∘=(supLN)^∘. (2) Let supLC=x. Then x^∘∘=(supLC)^∘∘. We have to show that $x^{\circ \circ }=\sup _{L^{\circ \circ }}C$. Since supLC=x, then x≥c for all c∈C. so, x^∘∘≥c^∘∘=c for all c∈C. Therefore x^∘∘ is an upper bound of C. Let y be another upper bound of C in L^∘∘. Then y≥c for all c∈C. Thus y^∘∘≥c^∘∘=c. Hence y^∘∘ is an upper bound of C. So y^∘∘≥x as x= supLC. It follows that y=y^∘∘≥x^∘∘. Hence x^∘∘ is the least upper bound of C. Since x^∘∘∈L^∘∘, then $x^{\circ \circ }=\sup _{L^{\circ \circ }}C$. By (1) we have (supLC)^∘∘=(infLC^∘)^∘. (3) Let x= infLC. Then x≤c for all c∈C. Then x^∘∘≤c^∘∘=c. Hence x^∘∘ is a lower bound of C. Thus x≥x^∘∘ as x= infLC. But x≤x^∘∘. Then x^∘∘=x and x∈L^∘∘. Thus $\inf _{L^{\circ \circ }}C=x$. (4) Let x= infLE and y= infD(L)E. Then x≤e and y≤e for all e∈E imply that x=y. Now we prove supD(L)E= supLE. Let y= supLE. Then y≥e for all e∈E. It follows that y^∘≤e^∘=0. Then y∈D(L) implies y= supD(L)E. □

Let (M,D,φ) be a decomposable MS-triple. For any ∅≠E⊆D, consider the set M_E as follows:

$M_{E}=\left \{a\in M : \bar {a}\varphi \vee [z)\supset E \ \text {for some} \ z\in D\right \}$.

Lemma 3

Let (M,D,φ) be a decomposable MS-triple. For any ∅≠E⊆D, we have (1) M_E is an ideal of M, (2) [E)=∪{[t):t∈E}, (2) M_E=M_[E).

Proof

(1). Let a,b∈M_E. Then $\bar {a}\varphi \vee [z_{1})\supset E$ and $\bar {b}\varphi \vee [z_{2})\supset E$ for some z₁,z₂∈D. Hence $E\subset (\bar {a}\varphi \vee [z_{1})) \cap (\bar {b}\varphi \vee [z_{2}))=\overline {(a\vee b)}\varphi \vee [t)$ for some t∈D (see Theorem 2). It follows that a∨b∈M_E. Now, let a∈M_E and c∈M. Then, ∃z∈D such that $\bar {a}\varphi \vee [z)\supset E$. Since a∧c≤a, then $\overline {a\wedge c}\geq \bar {a}$. This gives $\overline {(a\wedge c)}\varphi \supseteq \bar {a}\varphi$. It follows that $\overline {(a\wedge c)}\varphi \vee [z)\supseteq \bar {a}\varphi \vee [z)\supset E$. Then a∧c∈M_E. Consequently, M_E is an ideal of M. (2) Obvious. (3) Clearly, M_[E)⊆M_E. Let a∈M_E. Then, ∃z∈D such that $\bar {a}\varphi \vee [z)\supset E$. Since $\bar {a}\varphi \vee [z)$ is a filter of D and [E) is the smallest filter of D containing E, then $\bar {a}\varphi \vee [z)\supset [E)$. Hence, a∈M_[E) and M_E⊆M_[E). Therefore, M_E=M_[E). □

Definition 8

A complete decomposable MS-triple is a decomposable MS-triple (M,D,φ) satisfying the following conditions: (i) M is complete, (ii) D is conditionally complete, (iii) For each ∅≠E⊆D, the set M_E has the greatest element in M.

Theorem 3

Let L be a complete decomposable MS-algebra constructed from the decomposable MS-triple (M,D,φ). Then, the triple (M,D,φ) is complete.

Proof

Since L is associated with the decomposable MS-triple (M,D,φ), then by Theorem 2, we have

$L=\left \{(a,\bar {a}\varphi \vee [x)): a\in M,x\in D\right \}$.

Corollary 1(1)-(3), gives

$L^{\circ \circ }=\left \{(a,\bar {a}\varphi): a\in M\right \}\cong M$ and D(L)={(1,[x)):x∈D}≅D.

We have to prove that a decomposable MS-triple (M,D,φ) is complete. So we proceed to prove (i)–(iii) of Definition 8. For (i), let ∅≠C⊆M. Consider a subset $\acute {C}=\{(c,\bar {c}\varphi):c\in C\}$ of L^∘∘ corresponding to C. Since L is complete, then $\inf _{L}\acute {C}=(a,\bar {a}\varphi \vee [x))$ for some $(a,\bar {a}\varphi \vee [x))\in L$. Thus, $(a,\bar {a}\varphi \vee [x))\leq (c,c\varphi)$ for all c∈C. Then a≤c for all c∈C implies that a is a lower bound of C. We verify that a is the greatest lower bound of C in M. Let b be a lower bound of C. Then b≤c for all c∈C. This gives $\bar {b}\varphi \supseteq \bar {c}\varphi$. Therefore, $(b,\bar {b}\varphi)\leq (c,\bar {c}\varphi)$ for all c∈C and (b,bφ) is a lower bound of $\acute {C}$. Then $(a,\bar {a}\varphi \vee [x))\geq (b,b\varphi)$ as $\inf _{L}C=(a,\bar {a}\varphi \vee [x))$. Consequently, a≥b and a= infMC. Since a= infMC and M is bounded above by 1, then, M is complete.Now we prove (ii). Let ϕ≠E⊆D. Consider $\acute {E}\subseteq D(L)$ corresponding to E. Then

$\acute {E}=\left \{(1,[e)): e\in D\right \}$.

Let z be a lower bound of E. Since L is complete, then $\inf _{L}\acute {E}$ exists. Let $\inf _{L}\acute {E}=(a,\bar {a}\varphi \vee [x))$. Since z≤e for all e∈E as z is a lower bound of E. Then, [z)⊇[e) and (1,[z))≤(1,[e)). Thus, (1,z) is a lower bound of $\acute {E}$. Then, $(a,\bar {a}\varphi \vee [x))\geq (1,[z))$ because of $\inf _{L}\acute {E}=(a,\bar {a}\varphi \vee [x))$. This implies that a≥1 and $\bar {a}\varphi \vee [x)\subseteq [z)$. Consequently, a=1 and $\bar {a}\varphi \vee [x)=0\varphi \vee [x)=[x)$. Thus [x)⊆[z) implies x≥z. This shows that x is the greatest lower bound of E in D and x= infDE. Using a similar way, we can show that, if E has an upper bound, then supDE exists. Therefore, D is a conditionally complete lattice as required.

Now we prove (iii). Let ∅≠E⊆D. Consider $\acute {E}\subseteq D(L)$ corresponding to E. Then

$$\begin{array}{@{}rcl@{}} \acute{E}=\left\{(1,[x)): x\in E\right\}. \end{array} $$

Since L is complete, then $\inf _{L} \acute {E}$ exists. Let $(b,\bar {b}\varphi \vee [z))=\inf _{L} \acute {E}$. We show that b is the largest element of M_E. Since $(b,\bar {b}\varphi \vee [z))=\inf _{L} \acute {E}$, then $(b,\bar {b}\varphi \vee [z))\leq (1,[x)), \ \forall x\in E$. This gives b≤1 and $\bar {b}\varphi \vee [z)\supseteq [x), \ \forall x\in E$. Therefore, $\bar {b}\varphi \vee [z)\supseteq \cup _{x\in E}[x)=[E)\supset E$. Thus, b∈M_E. Now, let c∈M_E. Then $\bar {c}\varphi \vee [y)\supset E$ for some y∈D. It follows that $\bar {c}\varphi \vee [y)\supseteq [E)\supseteq [x)$ for all x∈E. Hence, $(1,[x))\leq (c,\bar {c}\varphi \vee [y))$ for all x∈E. Thus, $(c,\bar {c}\varphi \vee [y))$ is a lower bound of $\acute {E}$ and therefore $(c,\bar {c}\varphi \vee [y))\leq (b,\bar {b}\varphi \vee [z))$. Then, c≤b. This deduce that b is the largest element of M_E in M. Therefore, (M,D,φ) is a complete decomposable MS-triple. □

The converse of the above theorem is given in the following.

Theorem 4

Let L be a decomposable MS-algebra constructed from the complete decomposable MS-triple (M,D,φ). Then L is complete.

Proof

Let (M,D,φ) be a complete decomposable MS-triple. Then –(iii) of Definition 8 hold. Let ∅≠N⊆L, where L is constructed as in construction Theorem from the decomposable MS-triple (M,D,φ) as follows:

$L=\left \{(a,\bar {a}\varphi \vee [x)):a\in M,x\in D\right \}$.

Since L is bounded, it is enough to show the existence of infLN. Denote a= infMN^∘∘ and $F=\cup \left \{[t): (c,\bar {c}\varphi \vee [t))\in N \text {for some} \ c\in M\right \}$ (∪ means the union in F(D)). Let b= maxM_F. Now, we prove that there exists an element z∈D such that $\bar {b}\varphi \vee [z)\supset F$ and if $\bar {b}\varphi \vee [y)\supset F$ for some y∈D then $\bar {b}\varphi \vee [y) \supseteq \bar {b}\varphi \vee [z)$. For this purpose, consider the following set:

$\left \{x_{\gamma }:\gamma \in \Gamma \text {for all}x_{\gamma }\text {with}\bar {b}\varphi \vee [x_{\gamma })\supset F\right \}$.

Thus, we have to find a z∈D with $\bar {b}\varphi \vee [y)\supset F$ and $\bar {b}\varphi \vee [y)\supseteq \bar {b}\varphi \vee [z)$ for all γ∈Γ. The set $\left \{x_{\gamma }:\gamma \in \Gamma \text {for all} x_{\gamma }\text {with}\bar {b}\varphi \vee [x_{\gamma })\supset F\right \}$ is bounded from above. Then, by (ii), there exists s= supD{x_γ:γ∈Γ}. We prove that ∩_γ∈Γ[x_γ)=[s).

$$\begin{array}{@{}rcl@{}} y\in\cap_{\gamma\in\Gamma}[x_{\gamma})&\Leftrightarrow& y\in[x_{\gamma}),~ \ \forall\gamma\in\Gamma\\&\Leftrightarrow& y\geq x_{\gamma},~ \ \forall\gamma\in\Gamma\\ &\Leftrightarrow& y \ \text{is an upper bound of} \ \{x_{\gamma}: \gamma\in\Gamma\}\\ &\Leftrightarrow& y\geq s\ \text{as}\ s=\sup_{D}\{x_{\gamma} : \gamma\in\Gamma\}\\&\Leftrightarrow& y\in[s). \end{array} $$

Then it is sufficient to prove the following equality.

$$\begin{array}{@{}rcl@{}} \cap_{\gamma\in\Gamma}(\bar{b}\varphi\vee[x_{\gamma}))=\bar{b}\varphi\vee\cap_{\gamma\in\Gamma}[x_{\gamma})=\bar{b}\varphi\vee[s). \end{array} $$

(1)

Let $t\in \bar {b}\varphi \vee [s)$. Then

$$\begin{array}{@{}rcl@{}} t\in \bar{b}\varphi\vee[s)&\Rightarrow& t\geq t_{1}\wedge s\ \text{where}\ t_{1}\in \bar{b}\varphi\\ &\Rightarrow& t\geq t_{1}\wedge (s\vee x_{\gamma})\ \text{as}\ s\geq x_{\gamma}\text{for all}\ \gamma\in \Gamma\\ &\Rightarrow&t\geq(t_{1}\wedge s)\vee (t_{1}\wedge x_{\gamma})\\ &\Rightarrow&t\geq t_{1}\wedge x_{\gamma}\\ &\Rightarrow& t\in \bar{b}\varphi\vee[x_{\gamma})\ \text{for all}\ \gamma\in \Gamma. \end{array} $$

Then $\bar {b}\varphi \vee \cap _{\gamma \in \Gamma }[x_{\gamma })\subseteq \bar {b}\varphi \vee [x_{\gamma })$ implies $\bar {b}\varphi \vee \cap _{\gamma \in \Gamma }[x_{\gamma })\subseteq \cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))$. Conversely, let $y\in \cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))$. Then $y\in \bar {b}\varphi \vee [x_{\gamma })$ for all γ∈Γ. Hence y≥t∧z for $t\in \bar {b}\varphi$ and z∈[x_γ) for all γ∈Γ. It follows that z≥x_γ for all γ∈Γ. This means that z is an upper bound of the set {x_γ:γ∈Γ}. Then s≤z as s= supD{x_γ:γ∈Γ}. Now

$$\begin{array}{@{}rcl@{}} y&\geq& t\wedge z\\ &=&t\wedge(s\vee z)\ \text{as}\ s\leq z\\ &=&(t\wedge s)\vee(t\wedge z)\ \text{by distributivity of}\ D\\ &\geq&t\wedge s\in \bar{b}\varphi\vee[s). \end{array} $$

Then $y\in \bar {b}\varphi \vee [s)$. Therefore, $\cap _{\gamma \in \Gamma }(\bar {b}\varphi \vee [x_{\gamma }))\subseteq \bar {b}\varphi \vee [s)$.

We prove the existence of infLN. First, we claim that

$i=\left (a\wedge b,\overline {(a\wedge b)}\varphi \vee [z)\right)=\inf _{L} N~(\text {we put then} z=s)$.

First, we show that i is a lower bound of N. Let $(f,\bar {f}\varphi \vee [y))\in N.$ Since a= infMN^∘∘, we get a≤f. So, a∧b≤a≤f. Then a∧b≤f implies that $\overline {a\wedge b}\geq \bar {f}$. Consequently, $\overline {(a\wedge b)}\varphi =\bar {a}\varphi \vee \bar {b}\varphi \supseteq \bar {f}\varphi$. Moreover, $[y)\subseteq F\subseteq \bar {b}\varphi \vee [z)$ as y∈F. Then

$$\begin{array}{@{}rcl@{}} \overline{(a\wedge b)}\varphi\vee[z)&=&(\bar{a}\vee \bar{b})\varphi\vee[z)\\ &=&(\bar{a}\varphi\vee \bar{b}\varphi)\vee(\bar{b}\varphi\vee[z))\\ &\supseteq&\bar{f}\varphi\vee[y). \end{array} $$

Then $(a\wedge b,\overline {(a\wedge b)}\varphi \vee [z))\leq (f,\bar {f}\vee [y))$ for all $(f,\bar {f}\vee [y))\in N$. Therefore, i is a lower bound of N. It remains to show that i is the greatest lower bound of N. Let $(c,\bar {c}\varphi \vee [x))$ be a lower bound of N. Then, $(c,\bar {c}\varphi \vee [x))\leq (f,\bar {f}\varphi \vee [y)), \ \forall (f,\bar {f}\varphi \vee [y))\in N$. So, c≤f, ∀f∈N^∘∘. Then c is a lower bound of N^∘∘. Thus c≤a as $a=\inf \limits _{M} N^{\circ \circ }$. On the other hand, $\bar {c}\varphi \vee [x)\supseteq \bar {f}\varphi \vee [y), \ \forall (f,\bar {f}\varphi \vee [y))\in N$. So, $\bar {c}\varphi \vee [x)\supseteq [y), \ \forall y\in F$. Therefore, $\bar {c}\varphi \vee [x)\supseteq F$. Hence, $\bar {c}\varphi \vee [x)\supseteq \bar {b}\varphi \vee [z)$ by using equality (1). Then $\bar {c}\varphi \vee [x)\supseteq F$ implies that c∈M_F. So, c≤b as $b=\max \limits _{M} M_{F}\in M$. Now, we have c≤a and c≤b. Then c≤a∧b. Moreover, we have $\bar {c}\varphi \supseteq \bar {a}\varphi$ because of c≤a. Also, $\bar {c}\varphi \vee [x)\supseteq \bar {b}\varphi \vee [z)$. So, $\bar {c}\varphi \vee [x)\supseteq \bar {a}\varphi \vee \bar {b}\varphi \vee [z)=\overline {(a\wedge b)}\varphi \vee [z)$. Therefore, $(c,\bar {c}\varphi \vee [x))\leq i$. Then i= infLN and L is complete. □

Corollary 2

If M and D are complete, then so is L.

Proof

. We need only to prove that the condition (iii) of Definition 8 holds. Let E⊆D and t= infDE. Then, [t)=[ infDE)⊇E. So, $(1,\bar {1}\varphi \vee [t))=(1,[t))\in L$. Therefore, 1∈M_E. Hence, by the above Theorem, L is complete. □

Corollary 3

If M is finite and D is conditionally complete, then L is complete.

Proof

Since M is finite and M_E is an ideal of M (see Lemma 1(1)), then M is complete and M_E is a principal ideal of M. Therefore, M_E contains the greatest element in M. So, the conditions (i)– (iii) of Definition 8 are satisfied and consequently, L is complete. □

Combining Theorems 3 and 4, we get the following theorem.

Theorem 5

Let L be a decomposable MS-algebra constructed from the decomposable MS-triple (M,D,φ). Then L is complete if and only if (M,D,φ) is complete.

Let L be a complete decomposable MS-algebra. In the proof of Theorem 4 arbitrary meets in L are described. In the following Lemma, we describe joins in L.

Lemma 4

Let L be a complete decomposable MS-algebra constructed from the decomposable MS-triple (M,D,φ). Let ϕ≠N⊆L and a= supMN^∘∘. Then there exists an element z∈D such that $[z)=\bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi$ and $\sup N=(a,\bar {a}\varphi \vee [z))$.

Proof

Let ϕ≠N⊆L and $\sup _{L} N=(b,\bar {b}\varphi \vee [z))$. We can assume that z∈aφ. We prove that b=a= supMN^∘∘. Using Lemma 2(2), we get

$\sup _{M}N^{\circ \circ }=(\sup _{L}N)^{\circ \circ }=(b,\bar {b}\varphi \vee [z))^{\circ \circ }=(b,\bar {b}\varphi)$.

But $a=(a,\bar {a}\varphi)=\sup _{M}N^{\circ \circ }$. Then b=a. Hence, $\bar {a}\varphi \vee [z)$ is the greatest filter of the form $\bar {a}\varphi \vee [x), x\in D$ with

$\bar {a}\varphi \vee [z))\subset \bar {c}\varphi \vee [t)$ for each $(c,\bar {c}\varphi \vee [t))\in N$.

The last condition is equivalent to

$[z)\subset \bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi$.

Let $\bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in N\right \}\cap a\varphi =R$. If [z)≠R, then there is y∈R,y≧̸z. It follows that y∧z<z and y∧z∈R. Then [z)⊂[y∧z) implies $\bar {a}\varphi \vee [z)\subset \bar {a}\varphi \vee [y\wedge z)$. Since y∧z∈R then $[y\wedge z)\subset \bar {c}\varphi \vee [t)$ for all $(c,\bar {c}\varphi \vee [t))\in N$. Since a≥c (as a= supMN^∘∘) then $\bar {a}\leq \bar {c}$. It follows that $\bar {a}\varphi \leq \bar {c}\varphi$. Therefore, $\bar {a}\varphi \vee [y\wedge z)\subset \bar {c}\varphi \vee [t)$ for all $(c,\bar {c}\varphi \vee [t))\in N$. Consequently,

$\bar {a}\varphi \vee [z)\subset \bar {a}\varphi \vee [y\wedge z)\subset \bar {c}\varphi \vee [t)$ for all $(c,\bar {c}\varphi \vee [t))\in N$,

which contradicts the maximality of $\bar {a}\varphi \vee [z)$. □

Complete homomorphisms via complete triple homomorphisms

In this section, we introduce complete triple homomorphisms of complete decomposable MS-algebras. Then, we characterize complete homomorphisms of complete decomposable MS-algebras in terms of complete triple homomorphisms. For this purpose, we recall from [4], the notion of triple homomorphism of decomposable MS-triples and related properties which will be used in rest of the paper.

Definition 9

[4] Let (M,D,φ) and (M₁,D₁,φ₁) be decomposable MS-triples. A triple homomorphism of the triple (M,D,φ) into (M₁,D₁,φ₁) is a pair (f,g), where f is a homomorphism of M into M₁,g is a homomorphism of D into D₁ preserving 1 such that for every a∈M,

$$ a\varphi g\subseteq af\varphi_{1} $$

(2)

Lemma 5

[4] Let (f,g) be a triple homomorphism of a decomposable MS-triple (M,D,φ) into a decomposable MS-triple (M₁,D₁,φ₁). Let a,b∈M and x,y,t∈D. Then (i) aφ∩[y)=[t) implies afφ₁∩[yg)=[tg), (ii) $\left (\bar {a}f\varphi _{1}\vee \lbrack xg\right))\cap (\bar {b}f\varphi _{1}\vee \lbrack yg))=\overline {(a\vee b)}f\varphi _{1}\vee \lbrack tg).$

Theorem 6

[4] Let L and L₁ be decomposable MS-algebras, (M,D,φ) and (M₁,D₁,φ₁) be the associated decomposable MS-triples, respectively. Let h be a homomorphism of L into L₁ and h_M,h_D the restrictions of h to M and D, respectively. Then (h_M,h_D) is a triple homomorphism of the decomposable MS-triples. Conversely, every triple homomorphism (f,g) of the decomposable MS-triples uniquely determines a homomorphism h of L into L₁ with h_M=f,h_D=g by the following rule:

$$ xh=x^{\circ \circ }f\wedge dg,\text{for all}\ x\in L, $$

(3)

where x=x^∘∘∧d for some d∈D(L).

If L and L₁ are represented as in the construction Theorem then (3) reads

$$ (a,\bar{a}\varphi\vee[x))h=(af,\overline{(af)}\varphi\vee[xg))\ \text{for all}\ (a,\bar{a}\varphi\vee[x))\in L. $$

(4)

In the following, we will write L=(M,D,φ) to indicate that (M,D,φ) is the decomposable MS-triple associated with L, that is, L^∘∘=M,D(L)=D, and φ(L)=φ. Let L=(M,D,φ) and L₁=(M₁,D₁,φ₁) be decomposable MS-algebras, we will write h=(f,g) to indicate that (f,g):(M,D,φ)→(M₁,D₁,φ₁) is the triple homomorphism of decomposable MS-triples corresponding to the homomorphism h of L into L₁.

Lemma 6

Let h=(f,g) be a homomorphism of a decomposable MS-algebra L onto a decomposable MS-algebra L₁. Then for each a∈L^∘∘, we have

aφg=afφ₁.

Proof

We have, aφg⊆afφ₁ by (2). It remains to show that afφ₁⊆aφg. Let y∈afφ₁. Then

y∈[(af)^∘)∩D(L₁)=[(ah)^∘)∩D(L₁) implies y∈[(ah)^∘) and y∈D(L₁).

Then y≥(ah)^∘=a^∘h. Since h is onto, then g:D(L)→D(L₁) is also onto. Hence, there exists x∈D(L) such that xh=y. Evidently, a^∘∨x∈[a^∘)∩D(L) and

(a^∘∨x)h=a^∘h∨xh=xh as xh=y≥a^∘h.

Therefore, y∈[a^∘h)∩D(L₁)=([a^∘)h∩Dg)=([a^∘)∩D)g=aφg. □

Now, we introduce the concept of complete triple homomorphism.

Definition 10

A triple homomorphism (f,g) of a decomposable MS-triple (M,D,φ) into a decomposable MS-triple (M₁,D₁,φ₁) is called complete if the following conditions are satisfied (i) f is a complete homomorphism of M and M₁, (ii) g is a complete homomorphism of D and D₁, (iii) (maxM_E)f= maxM_1Eg for each ϕ≠E⊆D.

Remark 1

First, we observe that the map g:D→D₁ is a complete means that $(\sup _{D}E)g=\sup _{D_{1}}Eg$ for any E⊆D and if infDE and $\inf _{D_{1}}Mg$ exist then $(\inf _{D}E)g=\inf _{D_{1}}Eg$.

Theorem 7

Let L=(M,D,φ) and L₁=(M₁,D₁,φ₁) be complete decomposable MS-algebras and let h=(f,g) be a homomorphism of L onto L₁. Then h is complete if and only if (f,g) is complete.

Proof

The decomposable MS-triples (M,D,φ) and (M₁,D₁,φ₁) are associated with L and L₁, respectively. Let h=(f,g) be a complete homomorphism of L onto L₁. Then f is a de Morgan homomorphism of M onto M₁ and g is a lattice homomorphism of D onto D₁ preserving 1. We have to verify that f and g are complete. Let ϕ≠N⊆M. Then

$$\begin{array}{@{}rcl@{}} \left(\inf_{M}N\right)f=\left(\inf_{L}N\right)f=\left(\inf_{L}N\right)h=\inf_{L_{1}}Nh=\inf_{L_{1}}Nf=\inf_{M_{1}}Nf\ \text{by Lemma 2(3)},\\ \left(\sup_{M}N\right)f=\left(\sup_{L}N\right)^{\circ\circ}f=\left(\left(\sup_{L}N\right)h\right)^{\circ\circ}=\left(\sup_{L_{1}}Nh\right)^{\circ\circ}=\sup_{M_{1}}Nf\ \text{by Lemma 2(2)}. \end{array} $$

Thus, f is complete. We prove that g is complete. Let ϕ≠E⊆D. Then

$(\sup _{D}E)g=(\sup _{L}E)g=(\sup _{L}N)h=\sup _{L_{1}}Nh=\sup _{D_{1}}Eg\ \text {by Lemma 2(4)}.$

If infDE and $\inf _{D_{1}}Eg$ exist, then

$(inf_{D}E)g=(\inf _{L}E)g=(\inf _{L}N)h=\inf _{L_{1}}Nh=\inf _{D_{1}}Eg\ \text {by Lemma 2(4)}.$

Now, we prove (iii). Let ϕ≠E⊆D. Consider E corresponding the set $\acute {E}$ on D(L), where

$\acute {E}=\left \{(1,[x)):x\in E\right \}\subseteq D(L)$.

By (4), we have

$\acute {E}h=\{(1,[xg)):x\in E\}\subseteq D(L_{1})$.

Since h is complete, then $(\inf _{L}E)h=\inf _{L_{1}}Eh$ for each ϕ≠E⊆L. Hence, (infLE)^∘∘= maxM_E (see the proof of Theorem 3) and similarly $(\inf _{L_{1}}Eh)^{\circ \circ }=\max M_{1Eg}$. Conversely, assume that (i)–(iii) hold and let h=(f,g) be a homomorphism of L onto L₁. We have to show that h is complete. First we prove that for $\phi \not = H\subseteq L, (\inf _{L}H)h=\inf _{L_{1}}Hh$ holds. Consider $E=\bigcup \left \{[t):(c,\bar {c}\varphi \vee [x))\in M\right \}$. Let maxM_E=b and infMH^∘∘=a. Then according to the proof of Theorem 4, we get

$i=\left (a\wedge b,\overline {(a\wedge b)}\varphi \vee [z)\right)=\inf _{L}H$, where $z=\sup _{D}\left \{x_{\gamma }:\bar {b}\varphi \vee [x_{\gamma })\supset E\right \}$. Using (4), we have

$Hh=\left \{(cf,\bar {cf}\varphi \vee [xg)):(c,\bar {c}\varphi \vee [x))\in H\right \}$,

and

$ih=\left ((a\wedge b)f,\overline {(a\wedge b)f}\varphi \vee [zg)\right)=(\inf _{L}H)h$.

Now, $\inf _{L_{1}}(Hf)^{\circ \circ }=(\inf _{M}H^{\circ \circ })f= af$ by (i) and maxM_1Eg=(maxM_E)f=bf by (iii). Since L₁ is complete and Hh⊂L₁ then again according to the proof of Theorem 4, we get

$\inf _{L_{1}}Hh=\left ((a\wedge b)f,\overline {(a\wedge b)f}\varphi \vee [z_{1})\right)=ih$, where z₁= sup{x_γg:γ∈Γ}=(sup{x_γ:γ∈Γ})g=zg as g is an onto homomorphism. Therefore, $\inf _{L} Mh=(\inf _{L_{1}}M)h$.

Now, we prove that $(\sup _{L}H)h=\sup _{L_{1}}Hh$. By Lemma 4, $\sup _{L}(M)=(a,\bar {a}\varphi \vee [z))$, where a= supMH^∘∘ and $[z)=\bigcap \left \{\bar {c}\varphi \vee [t):(c,\bar {c}\varphi \vee [t))\in H\right \}\cap a\varphi$. Then $\sup _{L_{1}}Hh=(a_{1},\bar {a_{1}}\varphi _{1}\vee [z_{1}))$, where $a_{1}=\sup _{M_{1}}(Hh)^{\circ \circ }=\sup _{L_{1}}(Hh)^{\circ \circ }=\sup _{L_{1}}H^{\circ \circ }h=(\sup _{L}M^{\circ \circ })h=(\sup _{M}H^{\circ \circ })h=ah=af$(by using Lemma 2(2) and (i) of Definition 9) and $[z_{1})=\bigcap \left \{\bar {cf}\varphi _{1}\vee [tg):(c,\bar {c}\varphi \vee [t))\in H\right \}\cap a_{1}\varphi _{1}$. We show that zg=z₁. We have cfφ₁=cφg by Lemma 6 and $\bar {c}\varphi g\vee [tg)=(\bar {c}\varphi \vee [t))g$ by Lemma 5(1). Then

$$\begin{array}{@{}rcl@{}} [z_{1})&=&\bigcap\{\bar{c}\varphi\vee[t))g:(c,\bar{c}\varphi\vee[t))\in H\}\cap a\varphi g\\ &=&\left(\bigcap\left\{\bar{c}\varphi\vee[t):(c,\bar{c}\varphi\vee[t))\in H\right\}\cap a\varphi\right)g\\ &=&[zg) \end{array} $$

which implies z₁=zg. Therefore, $(\sup _{L}H)h=\sup _{L_{1}}Hh$ and h is complete. □