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 ME 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) ME is an ideal of M, (2) [E)=∪{[t):t∈E}, (2) ME=M[E).
Proof
(1). Let a,b∈ME. 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∈ME. Now, let a∈ME 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∈ME. Consequently, ME is an ideal of M. (2) Obvious. (3) Clearly, M[E)⊆ME. Let a∈ME. 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 ME⊆M[E). Therefore, ME=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 ME 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 ME. 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∈ME. Now, let c∈ME. 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 ME 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= maxMF. 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{xγ:γ∈Γ}. We prove that ∩γ∈Γ[xγ)=[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∈[xγ) for all γ∈Γ. It follows that z≥xγ for all γ∈Γ. This means that z is an upper bound of the set {xγ:γ∈Γ}. Then s≤z as s= supD{xγ: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∈MF. 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∈ME. 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 ME is an ideal of M (see Lemma 1(1)), then M is complete and ME is a principal ideal of M. Therefore, ME 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 \(\bar {a}\varphi \vee [z)\). □