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Some notes on Tpartial order
Journal of Inequalities and Applications volume 2013, Article number: 219 (2013)
Abstract
In this study, by means of the Tpartial order defined in Karaçal and Kesicioğlu (Kybernetika 47:300314, 2011), an equivalence relation on the class of tnorms on ([0,1],\le ,0,1) is defined. Then, it is showed that the equivalence class of the weakest tnorm {T}_{D} on [0,1] contains a tnorm which is different from {T}_{D}. Finally, defining the sets of some incomparable elements with any x\in (0,1) according to {\u2aaf}_{T}, these sets are studied.
MSC:03E72, 03B52.
1 Introduction
Triangular norms were originally studied in the framework of probabilistic metric spaces [1–4] aiming at an extension of the triangle inequality and following some ideas of Menger [5]. Later on, they turned out to be interpretations of the conjunction in manyvalued logics [6–8], in particular in fuzzy logics, where the unit interval serves as a set of truth values.
In [9], a natural order for semigroups was defined. Similarly, in [10], a partial order defined by means of tnorms on a bounded lattice was introduced. For any elements x, y of a bounded lattice
where T is a tnorm. This order {\u2aaf}_{T} is called a tpartial order of T. Moreover, some connections between the natural order and the tpartial order {\u2aaf}_{T} were studied.
In [10], it was investigated that {\u2aaf}_{T} implies the natural order but its converse needs not be true. It was showed that a partially ordered set is not a lattice with respect to {\u2aaf}_{T}. Some sets which are lattices with respect to {\u2aaf}_{T} under some special conditions were determined. For more details on tnorms on bounded lattices, we refer to [11–17].
In the present paper, we introduce an equivalence on the class of tnorms on ([0,1],\le ,0,1) based on the equality of the sets of all incomparable elements with respect to {\u2aaf}_{T}. The paper is organized as follows. We shortly recall some basic notions in Section 2. In Section 3, we define an equivalence on the class of tnorms on ([0,1],\le ,0,1), and we determine the equivalence class of the weakest tnorm {T}_{D} on [0,1]. Thus, we obtain that the equivalence class of the weakest tnorm {T}_{D} contains a tnorm which is different from a tnorm {T}_{D}. We obtain that for arbitrary m\in {K}_{T}, there exists an element {y}_{m}\in {K}_{T} such that T(m,\cdot ):[0,1]\to [0,m] or T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}] is not continuous.
2 Basic definitions and properties
Definition 2.1 [18]
A triangular norm (tnorm for short) is a binary operation T on the unit interval [0,1], i.e., a function T:[0,1]\times [0,1]\to [0,1], such that for all x,y,z\in [0,1] the following four axioms are satisfied:
(T1) T(x,y)=T(y,x) (commutativity);
(T2) T(x,T(y,z))=T(T(x,y),z) (associativity);
(T3) T(x,y)\le T(x,z) whenever y\le z (monotonicity);
(T4) T(x,1)=x (boundary condition).
Example 2.1 [18]
The following are the four basic tnorms {T}_{M}, {T}_{P}, {T}_{L}, {T}_{D} given by, respectively,
Also, tnorms on a bounded lattice (L,\le ,0,1) are defined in a similar way, and then extremal tnorms {T}_{W} as well as {T}_{\wedge} on L are defined as {T}_{D} and {T}_{M} on [0,1].
Remark 2.1 [18]

(i)
Directly from Definition 2.1, we can deduce that, for all x\in [0,1], each tnorm T satisfies the following additional boundary conditions:
\begin{array}{c}T(0,x)=0,\hfill \\ T(1,x)=x.\hfill \end{array}Therefore, all tnorms coincide on the boundary of the unit square [0,1]\times [0,1].

(ii)
The monotonicity of a tnorm T in its second component described by (T3) is, together with the commutativity (T1), equivalent to the monotonicity in both components, i.e., to
T({x}_{1},{y}_{1})\le T({x}_{2},{y}_{2})\phantom{\rule{1em}{0ex}}\text{whenever}\phantom{\rule{1em}{0ex}}{x}_{1}\le {x}_{2}\text{and}{y}_{1}\le {y}_{2}.(2.1)
Definition 2.2 [18]
A function F:[0,1]\times [0,1]\to [0,1] is called continuous if for all convergent sequences {({x}_{n})}_{n\in \mathbb{N}},{({y}_{n})}_{n\in \mathbb{N}}\in {[0,1]}^{\mathbb{N}},
Proposition 2.1 [18]
A function F:[0,1]\times [0,1]\to [0,1] which is nondecreasing, i.e., which satisfies (2.1), is continuous if and only if it is continuous in each component, i.e., if for all {x}_{0},{y}_{0}\in [0,1], both the vertical section F({x}_{0},\cdot ):[0,1]\to [0,1] and the horizontal section F(\cdot ,{y}_{0}):[0,1]\to [0,1] are continuous functions in one variable.
Proposition 2.2 [18]
A nondecreasing function F:[0,1]\times [0,1]\to [0,1] is lower semicontinuous if and only if it is leftcontinuous in each component, i.e., if for all {x}_{0},{y}_{0}\in [0,1] and for all sequences {({x}_{n})}_{n\in \mathbb{N}},{({y}_{n})}_{n\in \mathbb{N}}\in {[0,1]}^{\mathbb{N}}, we have
By the same token, the upper semicontinuity of a nondecreasing function F:[0,1]\times [0,1]\to [0,1] is equivalent to its rightcontinuity in each component.
Proposition 2.3 [7]
For a nondecreasing function F:{\mathbb{I}}^{n}\to \mathbb{R} (\mathbb{I} an interval), the following conditions are equivalent:

(i)
F is continuous;

(ii)
F is continuous in each variable, i.e., for any x\in {\mathbb{I}}^{n} and any i\in \{1,2,\dots ,n\}, the unary function
u\to F({x}_{1},\dots ,{x}_{i1},u,{x}_{i+1},\dots ,{x}_{n})is continuous;

(iii)
F has the intermediate value property: For any x,y\in {\mathbb{I}}^{n}, with x\le y, and any c\in [F(x),F(y)], there exists z\in {\mathbb{I}}^{n}, with x\le z\le y, such that F(z)=c.
Definition 2.3 [10]
Let L be a bounded lattice, let T be a tnorm on L. The order defined as follows is called a torder (triangular order) for a tnorm T.
Example 2.2 [10]
Let L=\{0,a,b,c,1\} and consider the order ≤ on L as in Figure 1.
We choose the tnorm {T}_{W}. Then a\le b, but a{\u22e0}_{{T}_{W}}b. We suppose that a{\u2aaf}_{{T}_{W}}b. Then there exists an element \ell \in L such that {T}_{W}(\ell ,b)=a. If \ell =0, then it is obtained that a=0, a contradiction. If \ell =a,b or c, then we have that {T}_{W}(\ell ,b)=0=a, a contradiction. If \ell =1, then it is obtained that {T}_{W}(1,b)=b=a, which is not possible. So, there does not exist any element \ell \in L satisfying {T}_{W}(\ell ,b)=a. Thus, a{\u22e0}_{{T}_{W}}b. Then the order {\u2aaf}_{{T}_{W}} on L is as in Figure 2.
Proposition 2.4 [10]
Let L be a bounded lattice, let T be a tnorm on L. Then the binary relation {\preccurlyeq}_{T} is a partial order on L.
Definition 2.4 [10]
This partial order {\preccurlyeq}_{T} is called a Tpartial order on L.
Definition 2.5 Let T be a tnorm on [0,1] and let {K}_{T} be defined by
We will use the notation {K}_{T} to denote the set of all incomparable elements with respect to {\u2aaf}_{T}.
3 The equivalence of any two tnorms
Let L be a lattice and let T be any tnorm on L. In [10], a partial order for a tnorm T on L was defined. In this section, we define an equivalence relation with the help of the sets of all incomparable elements with respect to {\u2aaf}_{T}. The above introduced Tpartial order allows us to introduce the next equivalence relation on the class of all tnorms on ([0,1],\le ,0,1).
Definition 3.1 Let ([0,1],\le ,0,1) be the unit interval. Define a relation ∼ on the class of all tnorms on ([0,1],\le ,0,1) by {T}_{1}\sim {T}_{2} if and only if the set of all incomparable elements with respect to the {T}_{1}partial order is equal to the set of all incomparable elements with respect to the {T}_{2}partial order, that is,
Proposition 3.1 The relation ∼ given in Definition 3.1 is an equivalence relation.
Proof Let {T}_{1}, {T}_{2} and {T}_{3} be tnorms on ([0,1],\le ,0,1). Since {K}_{{T}_{1}}={K}_{{T}_{1}}, it is obtained that {T}_{1}\sim {T}_{1}. Thus, the reflexivity is satisfied.
Let {T}_{1}\sim {T}_{2}. Then we have that {K}_{{T}_{1}}={K}_{{T}_{2}}, and since {K}_{{T}_{2}}={K}_{{T}_{1}}, it is obtained that {T}_{2}\sim {T}_{1}. Thus, the symmetry is satisfied. Let {T}_{1}\sim {T}_{2} and {T}_{2}\sim {T}_{3}. Then we have {K}_{{T}_{1}}={K}_{{T}_{2}} and {K}_{{T}_{2}}={K}_{{T}_{3}}. Since {K}_{{T}_{1}}={K}_{{T}_{3}}, it is obtained that {T}_{1}\sim {T}_{3}. This means that the relation ∼ satisfies the transitivity. So, we have that ∼ is an equivalence relation. □
Definition 3.2 For a given tnorm T on ([0,1],\le ,0,1), we denote by \overline{T} the ∼ equivalence class linked to T, i.e.,
Proposition 3.2 shows that the equivalence class of the tnorm {T}_{D} contains a tnorm which is different from {T}_{D}.
Proposition 3.2 Let the tnorm {T}_{D} be on [0,1]. Then \overline{{T}_{D}}\ne \{{T}_{D}\}.
We give a contrary example as follows for the proof of Proposition 3.2.
Example 3.1 Consider the tnorm T:[0,1]\times [0,1]\to [0,1] defined by
Then {K}_{T}={K}_{{T}_{D}}. Firstly, let us show that {K}_{T}=(0,1). Let x\in (0,1) and y=\frac{2x}{3}. Then y<x, but y{\u22e0}_{T}x. Suppose that y{\preccurlyeq}_{T}x. Then, for some ℓ, T(\ell ,x)=\frac{2x}{3}. Since x\ne y, it is not possible \ell =1. Then \frac{2x}{3}=T(\ell ,x)=\frac{\ell x}{2}, whence it is obtained that \ell =\frac{4}{3}, a contradiction. Since for any x\in (0,1) there exists an element y=\frac{2x}{3} such that \frac{2x}{3}<x but \frac{2x}{3}{\u22e0}_{T}x, x\in {K}_{T}. Conversely, for any tnorm T, it is clear that {K}_{T}\subseteq (0,1). So, it is obtained that {K}_{T}=(0,1).
Now, we will show that {K}_{{T}_{D}}=(0,1). Let x\in (0,1). For any y\in (0,1) with x<y, it is obvious that x{\u22e0}_{{T}_{D}}y. Otherwise, it would be {T}_{D}(\ell ,y)=x for some ℓ. Since x\ne y, \ell \ne 1. Thus, it must be x=0 for \ell ,y\in (0,1), a contradiction. Since there is an element y with x<y such that x{\u22e0}_{{T}_{D}}y, x\in {K}_{{T}_{D}}. This shows that {K}_{{T}_{D}}=(0,1). So, it is obtained that {K}_{T}={K}_{{T}_{D}}.
Definition 3.3 Let T be a tnorm on [0,1] and let {K}_{x\downarrow}, {K}_{x\uparrow} be defined by
Lemma 3.1 Let T be a tnorm on [0,1] and x\in {K}_{T} be arbitrarily chosen. If T is continuous at (x,y) for all y\in [0,1], then {K}_{x\downarrow}=\mathrm{\varnothing}.
Proof Let T be a tnorm on [0,1] and let x\in {K}_{T} be arbitrarily chosen. Suppose that {K}_{x\downarrow}\ne \mathrm{\varnothing}. Then there exists an element {y}_{0}\in [0,1] such that {y}_{0}\le x, but {y}_{0}{\u22e0}_{T}x. Since the tnorm T(x,\cdot ):[0,1]\to [0,x] is continuous, there exists an element z\in [0,1] such that T(x,z)={y}_{0} for {y}_{0}\in [0,x] by Proposition 2.3. So, it is obtained that {y}_{0}{\u2aaf}_{T}x, a contradiction. Therefore we have that {K}_{x\downarrow}=\mathrm{\varnothing}. □
Lemma 3.2 Let T be a tnorm on [0,1] and the function T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}] be continuous. Then, for all y\in [0,1] with y\le {x}_{0}, we have that y{\u2aaf}_{T}{x}_{0}.
Proof Let T be a tnorm on [0,1] and the function T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}] be continuous. Suppose that there exists an element {y}_{0}\in [0,1] such that {y}_{0}\le {x}_{0} and {y}_{0}{\u22e0}_{T}{x}_{0}. Since T({x}_{0},\cdot ):[0,1]\to [0,{x}_{0}] is continuous, there exists an element t\in [0,1] such that T({x}_{0},t)={y}_{0} for {y}_{0}\in [0,{x}_{0}] by Proposition 2.3. Thus, it is obtained that {y}_{0}{\u2aaf}_{T}{x}_{0}, a contradiction. Therefore, for all y\in [0,1] with y\le {x}_{0}, we have that y{\u2aaf}_{T}{x}_{0}. □
Theorem 3.1 Let T be a tnorm on [0,1] and {K}_{T}\ne \mathrm{\varnothing}. Then, for arbitrary m\in {K}_{T}, there exists an element {y}_{m}\in {K}_{T} such that T(m,\cdot ):[0,1]\to [0,m] or T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}] is not continuous.
Proof Let T be a tnorm on [0,1] and {K}_{T}\ne \mathrm{\varnothing}. Suppose that T(x,\cdot ):[0,1]\to [0,x] is continuous for x\in {K}_{T}. Choose m\in {K}_{T} arbitrarily. Then there exists an element {y}_{m}\in {K}_{T} such that m<{y}_{m} but m{\u22e0}_{T}{y}_{m}, or {y}_{m}<m but {y}_{m}{\u22e0}_{T}m. Let m<{y}_{m} but m{\u22e0}_{T}{y}_{m}. Since T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}] is continuous, then it is obtained that m{\u2aaf}_{T}{y}_{m} by Lemma 3.2, a contradiction. Let {y}_{m}<m but {y}_{m}{\u22e0}_{T}m. Since T(m,\cdot ):[0,1]\to [0,m] is continuous, then it is obtained that {y}_{m}{\u2aaf}_{T}m by Lemma 3.2, a contradiction. Therefore, for arbitrary m\in {K}_{T}, there exists an element {y}_{m}\in {K}_{T} such that T(m,\cdot ):[0,1]\to [0,m] or T({y}_{m},\cdot ):[0,1]\to [0,{y}_{m}] is not continuous. □
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Acknowledgements
Dedicated to Professor Hari M Srivastava.
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Karaçal, F., Aşıcı, E. Some notes on Tpartial order. J Inequal Appl 2013, 219 (2013). https://doi.org/10.1186/1029242X2013219
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DOI: https://doi.org/10.1186/1029242X2013219
Keywords
 triangular norm
 Tpartial order