Open Access

Some notes on T-partial order

Journal of Inequalities and Applications20132013:219

https://doi.org/10.1186/1029-242X-2013-219

Received: 4 December 2012

Accepted: 18 April 2013

Published: 30 April 2013

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.

Keywords

triangular norm T-partial order

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 ) = x for 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 y z (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
    u F ( 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 x y , and any c [ F ( x ) , F ( y ) ] , there exists z I n , with x z y , 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 ) = x for 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

The order on L .

We choose the t-norm T W . Then a b , 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

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 x y , 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 x y , 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. □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

Authors’ Affiliations

(1)
Department of Mathematics, Karadeniz Technical University

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© Karaçal and Aşıcı; licensee Springer 2013

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