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Some notes on T-partial order

Abstract

In this study, by means of the T-partial order defined in Karaçal and Kesicioğlu (Kybernetika 47:300-314, 2011), an equivalence relation on the class of t-norms on ([0,1],,0,1) is defined. Then, it is showed that the equivalence class of the weakest t-norm T D on [0,1] contains a t-norm which is different from T D . Finally, defining the sets of some incomparable elements with any x(0,1) according to T , these sets are studied.

MSC:03E72, 03B52.

1 Introduction

Triangular norms were originally studied in the framework of probabilistic metric spaces [14] 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 many-valued logics [68], 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 t-norms on a bounded lattice was introduced. For any elements x, y of a bounded lattice

x T y:T(,y)=xfor some ,

where T is a t-norm. This order T is called a t-partial order of T. Moreover, some connections between the natural order and the t-partial order T were studied.

In [10], it was investigated that 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 T . Some sets which are lattices with respect to T under some special conditions were determined. For more details on t-norms on bounded lattices, we refer to [1117].

In the present paper, we introduce an equivalence on the class of t-norms on ([0,1],,0,1) based on the equality of the sets of all incomparable elements with respect to 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 t-norms on ([0,1],,0,1), and we determine the equivalence class of the weakest t-norm T D on [0,1]. Thus, we obtain that the equivalence class of the weakest t-norm T D contains a t-norm which is different from a t-norm T D . We obtain that for arbitrary m K T , there exists an element y m K T such that T(m,):[0,1][0,m] or T( y m ,):[0,1][0, y m ] is not continuous.

2 Basic definitions and properties

Definition 2.1 [18]

A triangular norm (t-norm for short) is a binary operation T on the unit interval [0,1], i.e., a function T:[0,1]×[0,1][0,1], such that for all x,y,z[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)T(x,z) whenever yz (monotonicity);

(T4) T(x,1)=x (boundary condition).

Example 2.1 [18]

The following are the four basic t-norms T M , T P , T L , T D given by, respectively,

T M ( x , y ) = min ( x , y ) , T P ( x , y ) = x y , T L ( x , y ) = max ( x + y 1 , 0 ) , T D ( x , y ) = { 0 if  ( x , y ) [ 0 , 1 [ 2 , min ( x , y ) otherwise .

Also, t-norms on a bounded lattice (L,,0,1) are defined in a similar way, and then extremal t-norms T W as well as T on L are defined as T D and T M on [0,1].

Remark 2.1 [18]

  1. (i)

    Directly from Definition 2.1, we can deduce that, for all x[0,1], each t-norm T satisfies the following additional boundary conditions:

    T ( 0 , x ) = 0 , T ( 1 , x ) = x .

    Therefore, all t-norms coincide on the boundary of the unit square [0,1]×[0,1].

  2. (ii)

    The monotonicity of a t-norm 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 )T( x 2 , y 2 )whenever x 1 x 2  and  y 1 y 2 .
    (2.1)

Definition 2.2 [18]

A function F:[0,1]×[0,1][0,1] is called continuous if for all convergent sequences ( x n ) n N , ( y n ) n N [ 0 , 1 ] N ,

F ( lim n x n , lim n y n ) = lim n F( x n , y n ).

Proposition 2.1 [18]

A function F:[0,1]×[0,1][0,1] which is non-decreasing, 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 [0,1], both the vertical section F( x 0 ,):[0,1][0,1] and the horizontal section F(, y 0 ):[0,1][0,1] are continuous functions in one variable.

Proposition 2.2 [18]

A non-decreasing function F:[0,1]×[0,1][0,1] is lower semicontinuous if and only if it is left-continuous in each component, i.e., if for all x 0 , y 0 [0,1] and for all sequences ( x n ) n N , ( y n ) n N [ 0 , 1 ] N , we have

sup { F ( x n , y 0 ) n N } = F ( sup { x n n N } , y 0 ) , sup { F ( x 0 , y n ) n N } = F ( x 0 , sup { y n n N } ) .

By the same token, the upper semicontinuity of a non-decreasing function F:[0,1]×[0,1][0,1] is equivalent to its right-continuity in each component.

Proposition 2.3 [7]

For a non-decreasing function F: I n R (I an interval), the following conditions are equivalent:

  1. (i)

    F is continuous;

  2. (ii)

    F is continuous in each variable, i.e., for any x I n and any i{1,2,,n}, the unary function

    uF( x 1 ,, x i 1 ,u, x i + 1 ,, x n )

    is continuous;

  3. (iii)

    F has the intermediate value property: For any x,y I n , with xy, and any c[F(x),F(y)], there exists z I n , with xzy, such that F(z)=c.

Definition 2.3 [10]

Let L be a bounded lattice, let T be a t-norm on L. The order defined as follows is called a t-order (triangular order) for a t-norm T.

x T y:T(,y)=xfor some L.

Example 2.2 [10]

Let L={0,a,b,c,1} and consider the order ≤ on L as in Figure 1.

Figure 1
figure 1

The order on L .

We choose the t-norm T W . Then ab, but a T W b. We suppose that a T W b. Then there exists an element L such that T W (,b)=a. If =0, then it is obtained that a=0, a contradiction. If =a,b or c, then we have that T W (,b)=0=a, a contradiction. If =1, then it is obtained that T W (1,b)=b=a, which is not possible. So, there does not exist any element L satisfying T W (,b)=a. Thus, a T W b. Then the order T W on L is as in Figure 2.

Figure 2
figure 2

The order T W on L .

Proposition 2.4 [10]

Let L be a bounded lattice, let T be a t-norm on L. Then the binary relation T is a partial order on L.

Definition 2.4 [10]

This partial order T is called a T-partial order on L.

Definition 2.5 Let T be a t-norm on [0,1] and let K T be defined by

K T = { x [ 0 , 1 ]  for some  y [ 0 , 1 ] , [ x y  implies  x T y ]  or  [ y x  implies  y T x ] } .

We will use the notation K T to denote the set of all incomparable elements with respect to T .

3 The equivalence of any two t-norms

Let L be a lattice and let T be any t-norm on L. In [10], a partial order for a t-norm 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 T . The above introduced T-partial order allows us to introduce the next equivalence relation on the class of all t-norms on ([0,1],,0,1).

Definition 3.1 Let ([0,1],,0,1) be the unit interval. Define a relation on the class of all t-norms on ([0,1],,0,1) by T 1 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,

T 1 T 2 : K T 1 = K T 2 .

Proposition 3.1 The relation given in Definition  3.1 is an equivalence relation.

Proof Let T 1 , T 2 and T 3 be t-norms on ([0,1],,0,1). Since K T 1 = K T 1 , it is obtained that T 1 T 1 . Thus, the reflexivity is satisfied.

Let T 1 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 T 1 . Thus, the symmetry is satisfied. Let T 1 T 2 and T 2 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 T 3 . This means that the relation satisfies the transitivity. So, we have that is an equivalence relation. □

Definition 3.2 For a given t-norm T on ([0,1],,0,1), we denote by T ¯ the equivalence class linked to T, i.e.,

T ¯ = { T T  is a  t -norm on  [ 0 , 1 ]  and  T T } .

Proposition 3.2 shows that the equivalence class of the t-norm T D contains a t-norm which is different from T D .

Proposition 3.2 Let the t-norm T D be on [0,1]. Then T D ¯ { T D }.

We give a contrary example as follows for the proof of Proposition 3.2.

Example 3.1 Consider the t-norm T:[0,1]×[0,1][0,1] defined by

T(x,y)={ x y 2 if  ( x , y ) [ 0 , 1 ) 2 , min ( x , y ) otherwise .

Then K T = K T D . Firstly, let us show that K T =(0,1). Let x(0,1) and y= 2 x 3 . Then y<x, but y T x. Suppose that y T x. Then, for some , T(,x)= 2 x 3 . Since xy, it is not possible =1. Then 2 x 3 =T(,x)= x 2 , whence it is obtained that = 4 3 , a contradiction. Since for any x(0,1) there exists an element y= 2 x 3 such that 2 x 3 <x but 2 x 3 T x, x K T . Conversely, for any t-norm T, it is clear that K T (0,1). So, it is obtained that K T =(0,1).

Now, we will show that K T D =(0,1). Let x(0,1). For any y(0,1) with x<y, it is obvious that x T D y. Otherwise, it would be T D (,y)=x for some . Since xy, 1. Thus, it must be x=0 for ,y(0,1), a contradiction. Since there is an element y with x<y such that x T D y, x 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 t-norm on [0,1] and let K x , K x be defined by

K x = { y [ 0 , 1 ] y x  and  y T x } , K x = { y [ 0 , 1 ] x y  and  x T y } for any  x ( 0 , 1 ) .

Lemma 3.1 Let T be a t-norm on [0,1] and x K T be arbitrarily chosen. If T is continuous at (x,y) for all y[0,1], then K x =.

Proof Let T be a t-norm on [0,1] and let x K T be arbitrarily chosen. Suppose that K x . Then there exists an element y 0 [0,1] such that y 0 x, but y 0 T x. Since the t-norm T(x,):[0,1][0,x] is continuous, there exists an element z[0,1] such that T(x,z)= y 0 for y 0 [0,x] by Proposition 2.3. So, it is obtained that y 0 T x, a contradiction. Therefore we have that K x =. □

Lemma 3.2 Let T be a t-norm on [0,1] and the function T( x 0 ,):[0,1][0, x 0 ] be continuous. Then, for all y[0,1] with y x 0 , we have that y T x 0 .

Proof Let T be a t-norm on [0,1] and the function T( x 0 ,):[0,1][0, x 0 ] be continuous. Suppose that there exists an element y 0 [0,1] such that y 0 x 0 and y 0 T x 0 . Since T( x 0 ,):[0,1][0, x 0 ] is continuous, there exists an element t[0,1] such that T( x 0 ,t)= y 0 for y 0 [0, x 0 ] by Proposition 2.3. Thus, it is obtained that y 0 T x 0 , a contradiction. Therefore, for all y[0,1] with y x 0 , we have that y T x 0 . □

Theorem 3.1 Let T be a t-norm on [0,1] and K T . Then, for arbitrary m K T , there exists an element y m K T such that T(m,):[0,1][0,m] or T( y m ,):[0,1][0, y m ] is not continuous.

Proof Let T be a t-norm on [0,1] and K T . Suppose that T(x,):[0,1][0,x] is continuous for x K T . Choose m K T arbitrarily. Then there exists an element y m K T such that m< y m but m T y m , or y m <m but y m T m. Let m< y m but m T y m . Since T( y m ,):[0,1][0, y m ] is continuous, then it is obtained that m T y m by Lemma 3.2, a contradiction. Let y m <m but y m T m. Since T(m,):[0,1][0,m] is continuous, then it is obtained that y m T m by Lemma 3.2, a contradiction. Therefore, for arbitrary m K T , there exists an element y m K T such that T(m,):[0,1][0,m] or T( y m ,):[0,1][0, y m ] is not continuous. □

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Acknowledgements

Dedicated to Professor Hari M Srivastava.

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The present study was proposed by FK. All authors read and approved the final manuscript.

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Karaçal, F., Aşıcı, E. Some notes on T-partial order. J Inequal Appl 2013, 219 (2013). https://doi.org/10.1186/1029-242X-2013-219

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Keywords

  • triangular norm
  • T-partial order