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Complete decomposable MS-algebras
Journal of the Egyptian Mathematical Society volume 27, Article number: 23 (2019)
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→L_{1} 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→L_{1} of an MS-algebra L into an MS-algebra L_{1} is called a homomorphism if x^{∘}h=xh^{∘} for all x∈L. If L and L_{1} 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
is a decomposable MS-algebra, if we define
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→L_{1} of a complete lattice L into a complete lattice L_{1} 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→L_{1} of a complete MS-algebra L into a complete MS-algebra L_{1} 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 z_{1},z_{2}∈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
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).
Then it is sufficient to prove the following equality.
Let \(t\in \bar {b}\varphi \vee [s)\). Then
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
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
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 (M_{1},D_{1},φ_{1}) be decomposable MS-triples. A triple homomorphism of the triple (M,D,φ) into (M_{1},D_{1},φ_{1}) is a pair (f,g), where f is a homomorphism of M into M_{1},g is a homomorphism of D into D_{1} preserving 1 such that for every a∈M,
Lemma 5
[4] Let (f,g) be a triple homomorphism of a decomposable MS-triple (M,D,φ) into a decomposable MS-triple (M_{1},D_{1},φ_{1}). Let a,b∈M and x,y,t∈D. Then (i) aφ∩[y)=[t) implies afφ_{1}∩[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 L_{1} be decomposable MS-algebras, (M,D,φ) and (M_{1},D_{1},φ_{1}) be the associated decomposable MS-triples, respectively. Let h be a homomorphism of L into L_{1} 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 L_{1} with h_{M}=f,h_{D}=g by the following rule:
where x=x^{∘∘}∧d for some d∈D(L).
If L and L_{1} are represented as in the construction Theorem then (3) reads
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 L_{1}=(M_{1},D_{1},φ_{1}) be decomposable MS-algebras, we will write h=(f,g) to indicate that (f,g):(M,D,φ)→(M_{1},D_{1},φ_{1}) is the triple homomorphism of decomposable MS-triples corresponding to the homomorphism h of L into L_{1}.
Lemma 6
Let h=(f,g) be a homomorphism of a decomposable MS-algebra L onto a decomposable MS-algebra L_{1}. Then for each a∈L^{∘∘}, we have
aφg=afφ_{1}.
Proof
We have, aφg⊆afφ_{1} by (2). It remains to show that afφ_{1}⊆aφg. Let y∈afφ_{1}. Then
y∈[(af)^{∘})∩D(L_{1})=[(ah)^{∘})∩D(L_{1}) implies y∈[(ah)^{∘}) and y∈D(L_{1}).
Then y≥(ah)^{∘}=a^{∘}h. Since h is onto, then g:D(L)→D(L_{1}) 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(L_{1})=([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 (M_{1},D_{1},φ_{1}) is called complete if the following conditions are satisfied (i) f is a complete homomorphism of M and M_{1}, (ii) g is a complete homomorphism of D and D_{1}, (iii) (maxM_{E})f= maxM_{1Eg} for each ϕ≠E⊆D.
Remark 1
First, we observe that the map g:D→D_{1} 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 L_{1}=(M_{1},D_{1},φ_{1}) be complete decomposable MS-algebras and let h=(f,g) be a homomorphism of L onto L_{1}. Then h is complete if and only if (f,g) is complete.
Proof
The decomposable MS-triples (M,D,φ) and (M_{1},D_{1},φ_{1}) are associated with L and L_{1}, respectively. Let h=(f,g) be a complete homomorphism of L onto L_{1}. Then f is a de Morgan homomorphism of M onto M_{1} and g is a lattice homomorphism of D onto D_{1} preserving 1. We have to verify that f and g are complete. Let ϕ≠N⊆M. Then
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 L_{1}. 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 L_{1} is complete and Hh⊂L_{1} 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 z_{1}= sup{x_{γ}g: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=z_{1}. We have cfφ_{1}=cφg by Lemma 6 and \(\bar {c}\varphi g\vee [tg)=(\bar {c}\varphi \vee [t))g\) by Lemma 5(1). Then
which implies z_{1}=zg. Therefore, \((\sup _{L}H)h=\sup _{L_{1}}Hh\) and h is complete. □
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Badawy, AM., Gaber, A. Complete decomposable MS-algebras. J Egypt Math Soc 27, 23 (2019). https://doi.org/10.1186/s42787-019-0027-8
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DOI: https://doi.org/10.1186/s42787-019-0027-8
Keywords
- MS-algebras
- Complete lattice
- Complete decomposable MS-algebras
- Complete decomposable MS-triples
- Triple homomorphisms
- Complete homomorphisms
AMS Mathematics Subject Classification (2010)
- Primary 06D30
- Secondary 06D15.