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 $([0,1],\le ,0,1)$ 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 $x\in (0,1)$ according to ${⪯}_T$, these sets are studied.

MSC:03E72, 03B52.

1 Introduction

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

$$x{⪯}_{T}y:\iff T(\ell ,y)=x\text{for some }\ell ,$$

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 [11–17].

In the present paper, we introduce an equivalence on the class of t-norms on $([0,1],\le ,0,1)$ 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 $([0,1],\le ,0,1)$, 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 $m\in K_T$, 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 (t-norm 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 t-norms $T_M$, $T_P$, $T_L$, $T_D$ given by, respectively,

$$\begin{array}{c}{T}_M(x,y)=min(x,y),\hfill \\ T_P(x,y)=x\cdot y,\hfill \\ T_L(x,y)=max(x+y-1,0),\hfill \\ T_D(x,y)=\{\begin{array}{cc}0\hfill & \text{if }(x,y)\in [0,1{[}^2,\hfill \\ min(x,y)\hfill & \text{otherwise}.\hfill \end{array}\hfill \end{array}$$

Also, t-norms on a bounded lattice $(L,\le ,0,1)$ are defined in a similar way, and then extremal t-norms $T_W$ 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 t-norm T satisfies the following additional boundary conditions:

  $$\begin{array}{c}T(0,x)=0,\hfill \\ T(1,x)=x.\hfill \end{array}$$

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

• (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

  $$T(x_1,y_1)\le T(x_2,y_2)\text{whenever}{x}_1\le x_2\text{ and }{y}_1\le y_2.$$

  (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 ${(x_n)}_{n\in \mathbb{N}},{(y_n)}_{n\in \mathbb{N}}\in {[0,1]}^{\mathbb{N}}$,

$$F(\underset{n\to \mathrm{\infty }}{lim}{x}_n,\underset{n\to \mathrm{\infty }}{lim}{y}_n)=\underset{n\to \mathrm{\infty }}{lim}F(x_n,y_n).$$

Proposition 2.1 [18]

A function $F:[0,1]\times [0,1]\to [0,1]$ 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 $x_0,y_0\in [0,1]$, 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 non-decreasing function $F:[0,1]\times [0,1]\to [0,1]$ is lower semicontinuous if and only if it is left-continuous in each component, i.e., if for all $x_0,y_0\in [0,1]$ and for all sequences ${(x_n)}_{n\in \mathbb{N}},{(y_n)}_{n\in \mathbb{N}}\in {[0,1]}^{\mathbb{N}}$, we have

$$\begin{array}{c}sup\{F(x_n,y_0)\mid n\in \mathbb{N}\}=F(sup\{x_n\mid n\in \mathbb{N}\},y_0),\hfill \\ sup\{F(x_0,y_n)\mid n\in \mathbb{N}\}=F(x_0,sup\{y_n\mid n\in \mathbb{N}\}).\hfill \end{array}$$

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

Proposition 2.3 [7]

For a non-decreasing 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

  $$u\to F(x_1,\dots ,x_{i-1},u,x_{i+1},\dots ,x_n)$$

  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 t-norm on L. The order defined as follows is called a t-order (triangular order) for a t-norm T.

$$x{\preccurlyeq }_{T}y:\iff T(\ell ,y)=x\text{for some }\ell \in L.$$

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 $a\le b$, but $a{⋠}_{T_W}b$. We suppose that $a{⪯}_{T_W}b$. 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, $a{⋠}_{T_W}b$. Then the order ${⪯}_{T_W}$ on L is as in Figure 2.

Figure 2

The order ${\mathbf{⪯}}_{{\mathit{T}}_{\mathit{W}}}$ on L .

Proposition 2.4 [10]

Let L be a bounded lattice, let T be a t-norm on L. Then the binary relation ${\preccurlyeq }_T$ is a partial order on L.

Definition 2.4 [10]

This partial order ${\preccurlyeq }_T$ 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

$$K_T=\{x\in [0,1]\mid \text{ for some }y\in [0,1],[x\le y\text{ implies }x{⋠}_{T}y]\text{ or }[y\le x\text{ implies }y{⋠}_{T}x]\}.$$

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 $([0,1],\le ,0,1)$.

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

$$T_1\sim T_2:\iff K_{T_1}=K_{T_2}.$$

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 $([0,1],\le ,0,1)$. 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 t-norm T on $([0,1],\le ,0,1)$, we denote by $\overline{T}$ the ∼ equivalence class linked to T, i.e.,

$$\overline{T}=\{T^{\prime }\mid T^{\prime }\text{ is a }t\text{-norm on }[0,1]\text{ and }T\sim T^{\prime }\}.$$

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 $\overline{T_D}\ne \{T_D\}$.

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

Example 3.1 Consider the t-norm $T:[0,1]\times [0,1]\to [0,1]$ defined by

$$T(x,y)=\{\begin{array}{cc}\frac{xy}{2}\hfill & \text{if }(x,y)\in {[0,1)}^2,\hfill \\ min(x,y)\hfill & \text{otherwise}.\hfill \end{array}$$

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 $y{⋠}_{T}x$. 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 $\frac{2x}{3}{⋠}_{T}x$, $x\in K_T$. Conversely, for any t-norm 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 $x{⋠}_{T_D}y$. 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 $x{⋠}_{T_D}y$, $x\in K_{T_D}$. 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 t-norm on $[0,1]$ and let $K_{x\downarrow }$, $K_{x\uparrow }$ be defined by

$$\begin{array}{c}{K}_{x\downarrow }=\{y\in [0,1]\mid y\le x\text{ and }y{⋠}_{T}x\},\hfill \\ K_{x\uparrow }=\{y^{\prime }\in [0,1]\mid x\le y^{\prime }\text{ and }x{⋠}_T{y}^{\prime }\}\text{for any }x\in (0,1).\hfill \end{array}$$

Lemma 3.1 Let T be a t-norm 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 t-norm 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 $y_0{⋠}_{T}x$. Since the t-norm $T(x,\cdot ):[0,1]\to [0,x]$ 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 $y_0{⪯}_{T}x$, a contradiction. Therefore we have that $K_{x\downarrow }=\mathrm{\varnothing }$. □

Lemma 3.2 Let T be a t-norm 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 $y{⪯}_T{x}_0$.

Proof Let T be a t-norm 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 $y_0{⋠}_T{x}_0$. 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 $y_0{⪯}_T{x}_0$, a contradiction. Therefore, for all $y\in [0,1]$ with $y\le x_0$, we have that $y{⪯}_T{x}_0$. □

Theorem 3.1 Let T be a t-norm 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 t-norm 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 $m{⋠}_T{y}_m$, or $y_m<m$ but $y_m{⋠}_{T}m$. Let $m<y_m$ but $m{⋠}_T{y}_m$. Since $T(y_m,\cdot ):[0,1]\to [0,y_m]$ 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 $T(m,\cdot ):[0,1]\to [0,m]$ is continuous, then it is obtained that $y_m{⪯}_{T}m$ 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. □