In MV-algebras, filters and ideals are dual notions, but this is not necessarily the case in general residuated lattices (see [
4,
5,
7,
8]). The authors in [
5,
8] have studied ideals and established a link with filters in residuated lattices by using different operators. In this section, we start by introducing the notion of ideal in the framework of triangle algebras. To establish the link between filters and ideals, we first prove that the operators used by [
5,
8] in residuated lattices are equal. Then, we prove that applying this operator to an ideal does not yield a filter in triangle algebras. Hence, we set up a new operator to link ideals to filters in the case of triangle algebras. We end by investigating the link between ideals and congruences, and some characterizations of the ideal generated by a given subset.
3.1. Definition and Characterizations of the Notion of Ideal
Since we want an ideal of a triangle algebra to be an ideal of the underlying residuated lattice, and taking into account the approximation operators and , we propose the following definition.
Definition 1. An ideal of is any non-empty subset I of A satisfying, for all ,
- (I1)
and ;
- (I2)
and ;
- (IT)
.
We denote by the set of all ideals of .
The following lemma will be helpful in the sequel.
Lemma 1. The following properties hold in any triangle algebra , for all .
- 1.
and ;
- 2.
and ;
- 3.
and (triangle inequalities).
Proof. Since (T1), we have , thus ((T′3) and ).
Since , we have by ( and ).
Hence, we have
.
By replacing x and y in the above by and , respectively, and using and , we have .
Since and , we have . Therefore, .
□
The following propositions are some characterizations of the notion of ideal in the case of triangle algebras. Some proofs are omitted, since in any residuated lattice we have the following equivalences [
19]:
((I1) + (I2)) ⇔ ((I1) + (I3)) ⇔ ((I′1) + (I4)) ⇔ ((I′1) + (I5)).
Proposition 1. A non-empty subset I of A is an ideal of if for all
- (I1)
and ;
- (I3)
and ;
- (IT)
.
Proposition 2. A non-empty subset I of A is an ideal of if for all
- (I′1)
;
- (I4)
and ;
- (IT)
.
Proposition 3. A non-empty subset I of A is an ideal of if for all
- (I′1)
;
- (I5)
and ;
- (IT)
.
The following result will be useful in some proofs.
Proposition 4. Let I be an ideal of and , then the following conditions are equivalent:
- (N1)
;
- (N2)
;
- (N3)
;
- (N4)
;
- (N5)
.
Proof. From (I6), we have , i.e, .
From (I1), (IT) and (T′1), we have , i.e, .
From (I6), we have , i.e, .
Since , we have . Hence, , which implies .
Since , we have .
From (a) and (I1), we have , i.e, .
From (b) and (I1), we have , i.e, .
Thus, . □
3.2. Pseudo-Duality
In the case of residuated lattices, in [
5,
8], the authors studied how to obtain a filter from an ideal, and vice versa, an ideal from a filter.
Let us first recall the procedure followed in [
5]. Given
F a filter and
I an ideal of
L, we obtain that
is an ideal and
is a filter of
. One can also prove that
,
and
.
In [
8], for
F and
I a filter and an ideal of
L, respectively, it was shown that
is a filter and
is an ideal of
L. This induces the following properties:
, and .
When a filter F satisfies , it is said to be an N-involutive filter.
In the following, we prove that for the pseudo-duality in residuated lattices the operators proposed by [
5,
8] are equal.
First, to simplify the proofs, we will use the following characterization of .
Proposition 5. Let F be a filter of a residuated lattice . Then, L; .
Proof. Let , then there exists such that . Since , we have . Hence, L; .
Let such that there exists , , then . Hence, .
□
Proposition 6. Let F be a filter and I an ideal of a residuated lattice , then
- (i)
;
- (ii)
.
Proof. - (i)
Let , then . Putting , we have , .
Reciprocally, let , then there exists such that , which implies . Since , we have , by (I6), thus , by (I1), i.e., .
- (ii)
Let , then . Since , we have , with .
Reciprocally, let , then there exists such that , which implies . Since , we have , i.e., .
□
Let us now, in the case of triangle algebras, consider the pseudo-duality between filters and ideals. We first show that the image of an ideal by the above operator in the case of triangle algebras is not always a filter.
Example 2. Let us consider the triangle algebra of Example 1.
is an ideal of , but is not a filter of . Indeed, , but , so is not true.
Hence, in the following, we introduce a new operator for the pseudo-duality in the case of triangle algebras.
Proposition 7. Let I be an ideal of . Then is a filter of .
Proposition 8. We have to show that is a filter of .
- For (F1),
let such that and , we have to show that .
It holds that implies , thus . Since , there exists such that , so by transitivity, , thus .
- For (F2),
let , we have to show that .
Since , there exists such that .
Since , there exists such that . Then, by , , thus , i.e., .
Since and I is an ideal, and we have by [4] , we obtain It follows that , thus .
- For (FT),
let . We have to show that . Since , there exists such that . Since , , we have , thus
The following is a characterization of the above operator.
Proposition 9. Let I be an ideal of , then .
Proof. Let such that . Since , we have , for .
Let . Hence, there exists such that , which implies . Since , by (I6), , so , by (I1).
□
According to the relation between and , we introduce the following operator for filters of triangle algebras.
Proposition 10. Let F be a filter of , then is an ideal of .
Proof. - For (I1),
let such that and . We have that and imply . Hence, , because F is a filter. Thus .
- For (I3),
let . Then, and . Hence, .
Therefore, . From the triangle inequality , we have . Hence, . Thus, .
- For (IT),
it holds because .
□
In the following, we prove that the above operator is equal to the one defined, in the case of residuated lattices, on filters, and some characterizations are given.
Proposition 11. Let F be a filter of , then .
Proof. Let , we have to find such that .
Since , we have . Because , we have , with .
Let . We have to find such that . Since , there exists such that . Hence, . Because , we have , which implies , by (F1). Since and , we have . Thus, , with .
Let us prove that . Let . Then, there exists such that . Therefore, . Hence, . Because , we have , by (FT). Since , we have , and then , by (F1). Thus, .
Next, we have to prove that .
Let , then , which implies by (FT) . Since , we have . Thus .
Let , then , i.e., , by . Since , we have , by . Thus .
□
Proposition 12. Let I be an ideal and F a filter of . Then,
- (1)
;
- (2)
;
- (3)
;
- (4)
.
Proof. - (1)
- (2)
- (3)
Let . Then, , and since , we have , which implies . Thus .
- (4)
Since is an ideal, then , by Proposition 12 (1).
□
We have in the following example an illustration of the previous Propositions; it is also highlighted that in Proposition 12 (3), we do not always have an equality.
Example 3. Let
. We define below,
Tables of ⇒ and ∗⇒ | 0 | v | a | b | 1 | | ∗ | 0 | v | a | b | 1 |
0 | 1 | 1 | 1 | 1 | 1 | | 0 | 0 | 0 | 0 | 0 | 0 |
v | 0 | 1 | 1 | 1 | 1 | | v | 0 | v | v | v | v |
a | 0 | b | 1 | b | 1 | | a | 0 | v | a | v | 1 |
b | 0 | a | a | 1 | 1 | | b | 0 | v | v | b | b |
1 | 0 | v | a | b | 1 | | 1 | 0 | v | a | b | 1 |
, , . Then, is a triangle algebra [23] whose lattice is given by Figure 1. Let , , and then:
- (i)
are filters and I is an ideal;
- (ii)
, , , , and . We have , but and , so .
Guided by the above observations, for pseudo-duality in triangle algebras, we will use the operator
T, i.e., for any ideal
I and filter
F of
,
3.3. Ideals and Congruences
Let us now investigate the link between ideals and congruences. In particular, we will explain how to build a congruence from an ideal and vice versa.
Definition 2. Acongruence relationon is an equivalence relation which is compatible with all operations on .
Lemma 2. Let be a triangle algebra, θ be an equivalence of and [4] .
If for all , and , then, for all , [4] ( and .
Proof. Let assume , for all . Let , such that and ; we have to show that .
Since , then, , and . Also, since , then and . Furthermore , which implies . Likely, , which implies □
The following proposition shows that any ideal of a triangle algebra induces a congruence relation.
Proposition 13. Let I be an ideal of . Then, the binary relation defined by and , for all , is a congruence on .
Proof. is an equivalence relation.
Reflexivity: Let , we have
Symmetry: is symmetric by definition.
Transitivity: Let such that and , we have to show that .
Since , we have and ;
since , we have and ;
since , then by (T3) and , by residuation . Furthermore, , then by transitivity . Hence, . Since , we have
Since , we have ; then, . Furthermore, and , which implies , by Proposition 3.
Likewise, we can show that .
is compatible with the operations of
Let , such that , that is and .
We have to show that .
We have , then , since is increasing. Consequently, . Hence, , since . Likewise, we show that . Thus .
We have to show that .
We have by , then . Consequently . Hence, , since . Likewise, we show that . Thus .
We have to show that .
We have by , then . Hence, , then , since . Likewise, we show that . Thus .
We have to show that .
We have by and residuation. Hence, , then Therefore , since . Likewise, we show that . Thus .
We have to show that .
We have by and residuation. Hence, , then It follows , since . Likewise, we show that . Thus .
We have to show that .
We have by (T7). Then , hence . We have , since , . Likewise, we show that . Thus .
We have to show that .
We have by . Therefore, by . Hence, , and .Likewise, we show that . Thus .
□
For any ideal I of , the -class of any is denoted by and the set of all -class is denoted by .
The following proposition establishes a one-to-one correspondence between the set of ideals and the set of congruences of .
Proposition 14. Let I be an ideal of , then .
Proof. Let , then , by Proposition 4.
Moreover, . Hence, , and consequently .
Since I is an ideal, it holds that .
Let , then , which implies , by Proposition 4.
□
The following result proves that any congruence relation on a triangle algebra induces an ideal.
Proposition 15. For any congruence ∼ of , is an ideal.
Proof. - For (I1),
let such that and . Since , then , which implies , because ∼ is a congruence. Therefore, , because . Hence, .
- For (I3),
let . Since , then . Consequently, . Similarly, . Hence, , therefore , i.e., . Thus, .
- For (IT),
let . Since , then thus , which implies .
Hence, is an ideal. □
Given a congruence relation ∼ on , it is not clear whether holds. This remains an open question.
As a consequence of Proposition 13, the following result establishes that the quotient of a triangle algebra by a congruence induced by an ideal is a triangle algebra.
Proposition 16. Let I be an ideal of , then is a triangle algebra, where for all , and .
Proof. This is a direct consequence of the fact that is a congruence. □
Definition 3 ([
24]).
A
homomorphism
between and two triangle algebras is any mapping , that satisfies for all a, , , , , , , ; the
kernel
of f, denoted by , is define by . The following establish that the kernel of any homomorphism is an ideal.
Proposition 17. Let be two triangle algebras and a homomorphism, then is an ideal of .
Proof. Let , such that and .
Since , we have . Hence, . Because f is a homomorphism, we have . Then .
Since , we have . Therefore, . Hence, .
Let . Then and . Furthermore, because f is a homomorphism, we have . Thus, .
Let . We have . Therefore, .
Thus, is an ideal of . □
The following proposition establishes that any ideal is the kernel of a homomorphism of triangle algebras.
Proposition 18. Let I be an ideal of a triangle algebra , then there exists a homomorphism f of triangle algebras such that .
Proof. Let us consider
is a homomorphism between the triangle algebras
and
. Let
, then
, i.e.,
. Hence, by Proposition 4,
. □
3.4. Ideals Generated by a Subset
Before we dive into the case of ideals, let us first highlight that we have identified an inaccuracy in the characterization of the filter generated by a subset of a triangle algebra as proposed in [
16]. Here, we provide an example to counter the arguments in the proof given in ([
16], Proposition 4.2). Subsequently, we provide some accurate characterizations of the filter generated by a subset of a triangle algebra. Furthermore, we describe the ideal induced by a subset of a triangle algebra.
Recall that for any subset X of a triangle algebra , the filter generated by X, which we will denote , is the smallest filter of containing X.
In ([
16], Proposition 4.2) they mentioned that for any subset
S of a triangle algebra
,
, which we claim is not accurate, as we can see in Example 4 below.
Example 4. Let us consider the triangle algebra of Example 1. In the following two cases, we have .
- 1.
For , S is a filter of the residuated lattice , but S is not a filter of the triangle algebra (see Example 2). Using the fact that is the smallest filter of containing S, we have ; however, using the characterization above given in [16], we have . Hence, . - 2.
For . Since , we have , . Therefore, , . Hence, . Thus, .
To mend this problem, we introduce the following characterization of the filter induced by a subset of a triangle algebra.
Proposition 19. Let S be a non-empty subset of A. Then, Proof. Let . We will show that .
First, we show that B is a filter.
- For (F1),
let such that and . Therefore, there exists , such that . Since , then . Thus, .
- For (F2),
let . Then, there exists , such that and there exists , such that . By , we have . Thus .
- For (FT),
let . Therefore, there exists , such that . Hence, by (T3), we have which implies, by and : . Thus, .
Next, we show that .
Let . Since , we have .
Let F be a filter of containing S. Let , then there exists , , such that . Let , then , thus , so by (FT) . Because this is true for all , we have, by (F2): . Since , we have, by (F1): .
Concluding, . □
Below is another characterization of the filter induced by a subset of a triangle algebra.
Proposition 20. Let S be a non-empty subset of A.
Then .
Proof. Let , then (FT), , so .
Let , then there exists , , such that . Since (by (F1)), it holds that .
□
We recall that, for any subset S of a triangle algebra , the ideal generated by S is the smallest ideal of containing S, it is denoted by . Next, we introduce a characterization of the ideal induced by a non-empty subset of a triangle algebra.
Proposition 21. Let S be a non-empty subset of a triangle algebra . We have .
Proof. Let .
We first show that I is an ideal.
- For (I1),
this is obvious.
- For (I3),
let . Therefore, , with , and , with . Hence, . Thus .
- For (IT),
let
. Then, we have
, with
. Hence,
Thus, .
Let . Since , (T′1), we have . Thus, .
Let J be an ideal of containing S. Let , we have , with . Let , then . Because J is an ideal of , (IT), for all . Hence, by (I3), and by (I1), . Thus .
In conclusion, . □
Any ideal generated by will be denoted by and will be called principal ideal.
Let us now explore the algebraic structure of , the set of ideals of . One can observe that, if I and J are two ideals of , then is an ideal of .
In what follows, we establish that the set of ideals of is a Heyting algebra.
Proposition 22. is a complete lattice, with, , for all.
Proof. Let ;
Idempotence: ;
Symmetry: ;
Associativity: . Moreover,
Since , and , we have , thus , i.e., . Moreover, , thus , consequently, , i.e., .
We show likewise , therefore, .
Absorption laws: ; .
Bounds: Let a be non-empty family of elements of . For this family, the infimum is given by and the supremum .
□
Proposition 23. is a Heyting algebra, where, for all.
Proof. Let us show first that is a residuated lattice.
- For RL1,
is a bounded lattice.
- For RL2,
is a commutative ordered monoid.
- For RL3,
let . We will prove that .
- ∗
Assume that and let .
Let , then and . Hence, . Therefore, , i.e., . Thus .
- ∗
Assume that .
Let , then and . Then , since . Hence, . Thus .
Since is a residuated lattice and the product and the conjunction coincide, it holds that is a Heyting algebra.
□