intuitionistic fuzzy n-normed algebra and continuous product · a systematic development of fuzzy...

Intuitionistic fuzzy n-normed algebra andcontinuous product

Nabanita KonwarNorth Eastern Regional Institute of Science and Tec., India

andPradip Debnath

North Eastern Regional Institute of Science and Tec., IndiaReceived : March 2017. Accepted : April 2017

Proyecciones Journal of MathematicsVol. 37, No 1, pp. 63-83, March 2018.Universidad Catolica del NorteAntofagasta - Chile

Abstract

In this paper we extend the notion of intuitionistic fuzzy n-normedlinear space (IFnNLS) to define an intuitionistic fuzzy n-normed al-gebra (IFnNA). We give a necessary and sufficient condition for anIFnNA to be with continuous product. Further, the concept of multi-plicatively continuous product has been introduced and related resultshave been established. Illustrative examples have been provided in sup-port of our results.

Keywords : Intuitionistic fuzzy n-normed linear space; intuitionis-tic fuzzy continuous mapping; intuitionistic fuzzy continuous linearoperator.

AMS Subject Classification : 55M20, 54H25, 47H09.

64 N. Konwar and P. Debnath

1. Introduction

The fuzzy set theory was introduced by Zadeh [30] in 1965, in order toexplain the situations where data are imprecise or vague. On the otherhand in 1986, a generalized form of fuzzy set namely, intuitionistic fuzzyset was introduced by Atanassaov [1] which deals with both the degree ofmembership (belongingness) and non-membership (non-belongingness) ofan elements within a set.

Situations where the crisp norm is unable to measure the length of avector accurately, the notion of fuzzy norm happens to be useful. The ideaof a fuzzy norm on a linear space was introduced by Katsaras [17] in 1984.In 1992, Felbin [14] introduced an alternative idea of a fuzzy norm whoseassociated metric is of Kaleva and Seikkala [16] type. In 1994, anothernotion of fuzzy norm on a linear space was given by Cheng-Moderson [6]whose associated metric is that of Kramosil-Michalek [20] type. Againin 2003, following Cheng and Mordeson, Bag and Samanta [2] introducedanother concept of fuzzy normed linear space. In this way, there has beena systematic development of fuzzy normed linear spaces (FNLSs) and oneof the important development over FNLS is the notion of intuitionisticfuzzy normed linear space (IFNLS) [25]. Vijayabalaji and Narayanan [28]extended n-normed linear space to fuzzy n-normed linear space while theconcept of IFnNLS was introduced by Vijayabalaji et al [29]. Some morerecent work in similar context may be found in [5, 7, 8, 10, 11, 13, 18, 19,22, 24, 27].

The notion of intuitionistic fuzzy Banach algebra was introduced byDinda et al. [12]. More work in this direction have been carried out recentlyin [4, 21].

In the current paper we introduce the notion of IFnNA and investigatethe concept of continuous product.

2. Preliminaries

First we recall some basic and preliminary definitions and examples whichare related to intuitionistic fuzzy n-normed linear space.

Definition 2.1. [15] Let n ∈ N and X be a real linear space of dimensiond ≥ n (dmay be infinite). A real valued function k.k onX ×X × · · · ×X| {z }

n

=

Xn is called an n-norm on X if it satisfies the following properties:

Intuitionistic fuzzy n-normed algebra and continuous product 65

(i) kx1, x2, . . . , xnk = 0 if and only if x1, x2, . . . , xn are linearly depen-dent,

(ii) kx1, x2, . . . , xnk is invariant under any permutation,

(iii) kx1, x2, . . . , αxnk = |α|kx1, x2, . . . , xnk for any α ∈ R,

(iv) kx1, x2, . . . , xn−1, y+zk ≤ kx1, x2, . . . , xn−1, yk+kx1, x2, . . . , xn−1, zk,and the pair (X, k.k) is called an n-normed linear space.

Definition 2.2. [24] An IFnNLS is the five-tuple (X,µ, ν, ∗, ◦), where Xis a linear space over a field F , ∗ is a continuous t-norm, ◦ is a continuoust-conorm, µ, ν are fuzzy sets on Xn× (0,∞), µ denotes the degree of mem-bership and ν denotes the degree of non-membership of (x1, x2, . . . , xn, t) ∈Xn × (0, 1) satisfying the following conditions for every(x1, x2, . . . , xn) ∈ Xn and s, t > 0:

(i) µ(x1, x2, . . . , xn, t) + ν(x1, x2, . . . , xn, t) ≤ 1,

(ii) µ(x1, x2, . . . , xn, t) > 0,

(iii) µ(x1, x2, . . . , xn, t) = 1 if and only if x1, x2, . . . , xn are linearly depen-dent,

(iv) µ(x1, x2, . . . , xn, t) is invariant under any permutation of x1, x2, . . . , xn,

(v) µ(x1, x2, . . . , cxn, t) = µ(x1, x2, . . . , xn,t|c|) if c 6= 0, c ∈ F ,

(vi) µ(x1, x2, . . . , xn, s)∗µ(x1, x2, . . . , x0n, t) ≤ µ(x1, x2, . . . , xn+x

0n, s+ t),

(vii) µ(x1, x2, . . . , xn, t) : (0,∞)→ [0, 1] is continuous in t,

(viii) limt→∞ µ(x1, x2, . . . , xn, t) = 1 and limt→0 µ(x1, x2, . . . , xn, t) = 0,

(ix) ν(x1, x2, . . . , xn, t) < 1,

(x) ν(x1, x2, . . . , xn, t) = 0 if and only if x1, x2, . . . , xn are linearly depen-dent,

(xi) ν(x1, x2, . . . , xn, t) is invariant under any permutation of x1, x2, . . . , xn,

(xii) ν(x1, x2, . . . , cxn, t) = ν(x1, x2, . . . , xn,t|c|) if c 6= 0, c ∈ F ,

(xiii) ν(x1, x2, . . . , xn, s) ◦ ν(x1, x2, . . . , x0n, t) ≥ ν(x1, x2, . . . , xn+x

0n, s+ t),


(xiv) ν(x1, x2, . . . , xn, t) : (0,∞)→ [0, 1] is continuous in t,

(xv) limt→∞ ν(x1, x2, . . . , xn, t) = 0 and limt→0 ν(x1, x2, . . . , xn, t) = 1.

a ∧ b = min{a, b}, a · b = ab(usual multiplication in [0,1]) and a ∗L b =max{a+ b− 1, 0} (the Lukasiewicz t-norm).Definition 2.3. [19] Suppose ∗, ∗0 are t-norms. Then ∗0 dominates ∗ if(x1 ∗

0x2) ∗ (y1 ∗

0y2) ≤ (x1 ∗ y1) ∗

0(x2 ∗ y2) for all x1, x2, y1, y2 ∈ [0, 1] and

it is denoted by ∗0 >> ∗.Definition 2.4. [22] Let (X,µ, ν, ∗, ◦) be an IFnNLS. We say that a se-quence x = {xk} in X is convergent to l ∈ X with respect to the intuition-istic fuzzy n-norm (µ, ν)n if, for every > 0, t > 0 and y1, y2, . . . , yn−1 ∈X, there exists k0 ∈ N such that µ(y1, y2, . . . , yn−1, xk − L, t) > 1 −and ν(y1, y2, . . . , yn−1, xk − l, t) < for all k ≥ k0. It is denoted by

(µ, ν)n − limx = l or xk(µ,ν)n→ l as k →∞.

Definition 2.5. [22] Let (X,µ, ν, ∗, ◦) be an IFnNLS. Then the sequencex = {xk} in X is called a Cauchy sequence with respect to the intuitionisticfuzzy n-norm (µ, ν)n if, for every > 0, t > 0 and y1, y2, . . . , yn−1 ∈ X,there exists k0 ∈ N such that µ(y1, y2, . . . , yn−1, xk − xm, t) > 1 − andν(y1, y2, . . . , yn−1, xk − xm, t) < for all k,m ≥ k0.

Definition 2.6. [3] Let (X,µ, ν, ∗, ◦) be an IFnNLS. Then (X,µ, ν, ∗, ◦) issaid to be complete if any Cauchy sequence in X is convergent to a pointin X. A complete IFnNLS is called a intuitionistic fuzzy Banach space.

Definition 2.7. [8] Let (X,µ, ν, ∗, ◦) be an IFnNLS. For t > 0, we de-fine an open ball B(x, r, t) with centre at x ∈ X, radius 0 < r < 1 andy1, y2, . . . , yn−1 ∈ X as

B(x, r, t) = {y ∈ X : µ(y1, y2, . . . , yn−1, x− y, t) >r and ν(y1, y2, . . . , yn−1, x− y, t) < 1− r}

Then, Tµ,ν = {T ⊂ X : x ∈ T if and only if, there exists t > 0, r ∈(0, 1) such that B(x, r, t) ⊆ T}is a topology on X.

Moreover, if the t-norm ∗ and t-conorm ◦ satisfies the condition supx∈(0,1)x∗x = 1 and infx∈(0,1)x ◦ x = 0, then (X,Tµ,ν) is Hausdorff.

The following result is a straightforward conclusion from [24].

Theorem 2.2. Let (X,µ, ν, ∗, ◦) be an IFnNLS. Then (X,Tµ,ν) is a metriz-able topological vector space.


3. Main Results

In this section we discuss our main results.

First we define an intuitionistic fuzzy n-normed algebra (IFnNA) andthen explain some examples and propositions on the same.

Definition 3.1. Suppose (X,µ1, ν1, ∗1, ◦1) and (Y, µ2, ν2, ∗2, ◦2) are twoIFnNLSs. A mapping T : X −→ Y is said to be intuitionistic fuzzycontinuous (if-continuous) at x0 ∈ X if, for every > 0, α ∈ (0, 1),y1, y2, . . . , yn−1 ∈ X and T (y1), T (y2), . . . , T (yn−1) ∈ Y there exist δ =δ( , α, x0) > 0, β = β( , α, x0) ∈ (0, 1), such that for all x ∈ X,

µ1(y1, y2, . . . , yn−1, x−x0, δ) > β and ν1(y1, y2, . . . , yn−1, x−x0, δ) < 1−β.

implies,

µ2(T (y1), T (y2), . . . , T (yn−1), T (x)− T (x0), ) > α and

ν2(T (y1), T (y2), . . . , T (yn−1), T (x)− T (x0), ) < 1− α.

If T is if-continuous at each point of X, then T is called if-continuouson X.

Definition 3.2. The seven-tuple (X,µ, ν, ∗, ◦, ∗1, ◦1) is said to be an IFnNAif

(i) ∗ and ∗1 are continuous t-norms and ◦ and ◦1 are continuous t-conorms;

(ii) X is an algebra;

(iii) (X,µ, ν, ∗, ◦) is an IFnNLS;

(iv) µ(x1, x2, . . . , xn−1, xy, ts) ≥ µ(x1, x2, . . . , xn−1, x, t)∗1µ(x1, x2, . . . , xn−1, y, s)and ν(x1, x2, . . . , xn−1, xy, ts) ≤ ν(x1, x2, . . . , xn−1, x, t)◦1ν(x1, x2, . . . , xn−1, y, s),for all x1, x2, . . . , xn−1 ∈ X; x, y ∈ X and t, s ≥ 0.

If (X,µ, ν, ∗, ◦) is an intuitionistic fuzzy Banach space (IFBS), then (X,µ, ν, ∗, ◦, ∗1, ◦1)will be called intuitionistic fuzzy Banach algebra (IFBA).


Example 3.3. Suppose (X, k · k) is an n-normed algebra, ∗ and ∗1 arecontinuous t-norms, ◦ and ◦1 are continuous t-conorms and µ, ν : X ×[0,∞) −→ [0, 1] defined by-

µ(x1, x2, . . . , xn−1, x, t) =

(0, t ≤ kxk1, t > kxk

and

ν(x1, x2, . . . , xn−1, x, t) =

(1, t ≤ kxk0, t > kxk

Then (X,µ, ν, ∗, ◦, ∗1, ◦1) is an IFnNA.

Proof. It is a routine verification to prove that (X,µ, ν, ∗, ◦) is anIFnNLS.

Next we have to show that (X,µ, ν, ∗, ◦, ∗1, ◦1) is an IFnNA.

(i) ∗ and ∗1 are continuous t-norms and ◦ and ◦1 are continuous t-conorms;

(ii) As (X, k · k) is n-normed algebra, therefore X is an algebra as well;

(iii) (X,µ, ν, ∗, ◦) is an IFnNLS.

Finally, we have to show that

µ(x1, x2, . . . , xn−1, xy, ts) ≥ µ(x1, x2, . . . , xn−1, x, t)∗1µ(x1, x2, . . . , xn−1, y, s),

and

ν(x1, x2, . . . , xn−1, xy, ts) ≤ ν(x1, x2, . . . , xn−1, x, t)◦1ν(x1, x2, . . . , xn−1, y, s).

Let x1, x2, . . . , xn−1 ∈ X; x, y ∈ X and t, s ∈ [0,∞). If kxyk ≥ ts, thent ≤ kxk or s ≤ kyk (if t > kxk and s > kyk, then ts > kxk.kyk ≥ kxyk,contradiction.)

If t ≤ kxk, then µ(x1, x2, . . . , xn−1, x, t) = 0 and ν(x1, x2, . . . , xn−1, x, t) =1.

If s ≤ kyk, then µ(x1, x2, . . . , xn−1, y, s) = 0 and ν(x1, x2, . . . , xn−1, y, s) =1.

Thus,

µ(x1, x2, . . . , xn−1, x, t) ∗1 µ(x1, x2, . . . , xn−1, y, s) = 0,


andν(x1, x2, . . . , xn−1, x, t) ◦1 ν(x1, x2, . . . , xn−1, y, s) = 1.

Therefore,


and


Next, we consider kxyk < ts, then we have t > kxk or s > kyk.Then, if t > kxk, we have µ(x1, x2, . . . , xn−1, x, t) = 1 and ν(x1, x2, . . . , xn−1, x, t) =

0,and if s > kyk, we have µ(x1, x2, . . . , xn−1, y, s) = 1 and ν(x1, x2, . . . , xn−1, y, s) =0.

Thus,

µ(x1, x2, . . . , xn−1, x, t) ∗1 µ(x1, x2, . . . , xn−1, y, s) = 1

andν(x1, x2, . . . , xn−1, x, t) ◦1 ν(x1, x2, . . . , xn−1, y, s) = 0.

Therefore,

µ(x1, x2, . . . , xn−1, xy, ts) = 1 and ν(x1, x2, . . . , xn−1, xy, ts) = 0.

Thus,


and


Therefore the inequality holds.Hence (X,µ, ν, ∗, ◦, ∗1, ◦1) is an IFnNA.2

Example 3.4. Suppose (X, k · k) is a n-normed algebra and µ, ν : X ×[0,∞) −→ [0, 1] be defined by-

µ(x1, x2, . . . , xn−1, xn, t) =

(t

t+kx1,x2,...,xnk , t > 0

0, t = 0,


and

ν(x1, x2, . . . , xn−1, x, t) =

( kx1,x2,...,xnkt+kx1,x2,...,xnk , t > 0

1, t = 0.

Then (X,µ, ν,∧,∨, ·, ·) is an IFnNA.

The next theorem gives a necessary and sufficient condition for anIFnNA to be with continuous product.

Theorem 3.5. Let (X,µ, ν, ∗, ◦, ∗1, ◦1) be an IFnNA. Then X is with con-tinuous product if and only if, for all α ∈ (0, 1) there exist β = β(α) ∈ (0, 1)and M =M(α) > 0 such that for all x, y ∈ X; s, t > 0 and x1, x2, . . . , xn ∈X:

µ(x1, x2, . . . , xn−1, x, s) > β and ν(x1, x2, . . . , xn−1, x, s) < 1− β,

and

µ(x1, x2, . . . , xn−1, y, t) > β and ν(x1, x2, . . . , xn−1, y, t) < 1− β.

imply

µ(x1, x2, . . . , xn−1, xy,Mst) > α and ν(x1, x2, . . . , xn−1, xy,Mst) < 1− α.

Proof. Suppose α ∈ (0, 1) and V = {u ∈ X : µ(x1, x2, . . . , xn−1, u, 1) >α and ν(x1, x2, . . . , xn−1, u, 1) < 1− α} be an open neighborhood of zero.

Define a function X×X −→ X such that (x, y) −→ x.y, where (x, y) ∈X ×X and x.y ∈ X.

As this function is continuous at (0, 0), there exist 1 = 1(α) > 0,

2 = 2(α) > 0, γ1 = γ1(α) ∈ (0, 1), γ2 = γ2(α) ∈ (0, 1) such that for allu1, u2 ∈ X:

µ(x1, x2, . . . , xn−1, u1, 1) > γ1 and ν(x1, x2, . . . , xn−1, u1, 1) < 1− γ1,

and

µ(x1, x2, . . . , xn−1, u2, 2) > γ2 and ν(x1, x2, . . . , xn−1, u2, 2) < 1− γ2.

imply


µ(x1, x2, . . . , xn−1, u1.u2, 1) > α and ν(x1, x2, . . . , xn−1, u1.u2, 1) < 1− α.

Let β = max{γ1, γ2} ∈ (0, 1), M = 11. 2

> 0.Let x, y ∈ X, s, t > 0 and x1, x2, . . . , xn−1 ∈ X such that



Then

µ(x1, x2, . . . , xn−1,x

s, 1) > β ≥ γ1 and ν(x1, x2, . . . , xn−1,

x

s, 1) < 1−β ≤ 1−γ1,

µ(x1, x2, . . . , xn−1,y

t, 1) > β ≥ γ2 and ν(x1, x2, . . . , xn−1,

y

t, 1) < 1−β ≤ 1−γ2.

Let u1 = 1.xs , u2 =

2.ys . Then we have,

µ(x1, x2, . . . , xn−1, u1, 1) = µ(x1, x2, . . . , xn−1,u1

1, 1)

= µ(x1, x2, . . . , xn−1,x

s, 1) > γ1,

and

ν(x1, x2, . . . , xn−1, u1, 1)

= ν(x1, x2, . . . , xn−1,u1

1, 1) = ν(x1, x2, . . . , xn−1,

x

s, 1) < 1− γ1.

And,

µ(x1, x2, . . . , xn−1, u2, 2)

= µ(x1, x2, . . . , xn−1,u2

2, 1) = µ(x1, x2, . . . , xn−1,

y

t, 1) > γ2,

and

ν(x1, x2, . . . , xn−1, u2, 2) = ν(x1, x2, . . . , xn−1,u2

2, 1) = ν(x1, x2, . . . , xn−1,

y

t, 1) < 1−γ2.


Hence, µ(x1, x2, . . . , xn−1, u1.u2, 1) > α⇒ µ(x1, x2, . . . , xn−1, 1xs . 2y

t , 1) >α⇒ µ(x1, x2, . . . , xn−1, xy,

st1. 2

> α⇒ µ(x1, x2, . . . , xn−1, xy,Mst) > α.

And

ν(x1, x2, . . . , xn−1, u1.u2, 1) < 1 − α ⇒ ν(x1, x2, . . . , xn−1, 1xs . 2y

t , 1) <1− α⇒ ν(x1, x2, . . . , xn−1, xy,

st1. 2) < 1− α

⇒ ν(x1, x2, . . . , xn−1, xy,Mst) < 1− α. i.e.


Converse Part:

First we will prove that for each y0 ∈ X the mapping X −→ X suchthat x −→ xy0, where x ∈ X and xy0 ∈ X is continuous.

Let > 0, α ∈ (0, 1). Thus there exist β = β(α) ∈ (0, 1), M =M(α) >0 such that


and


This implies,


As, limt−→∞µ(x1, x2, . . . , xn−1, y0, t) = 1, there exists t0 > 0 such that

µ(x1, x2, . . . , xn−1, y0, t0) > β,

and as limt−→∞ν(x1, x2, . . . , xn−1, y0, t) = 0, there exists t0 > 0 suchthat

ν(x1, x2, . . . , xn−1, y0, t0) < 1− β.

Let δ = δ(α, ) = t0.Mand β(α, ) = β.

Let x ∈ X such that µ(x1, x2, . . . , xn−1, x, δ) > β and ν(x1, x2, . . . , xn−1, x, δ) <1− β.


As µ(x1, x2, . . . , xn−1, y0, t0) > β and ν(x1, x2, . . . , xn−1, y0, t0) < 1− βwe obtain that

µ(x1, x2, . . . , xn−1, xy0,Mt0δ) > α and ν(x1, x2, . . . , xn−1, xy0,Mt0δ) < 1−α.

This implies that

µ(x1, x2, . . . , xn−1, xy0, ) > α and ν(x1, x2, . . . , xn−1, xy0, ) < 1− α.

Similarly, we can establish that, for each x0 ∈ X, the mapping y −→x0y, where y ∈ X and x0y ∈ X is continuous.

Next, we will prove that (X,µ, ν, ∗, ◦, ∗1, ◦1) is with continuous product.Let xn −→ x0, yn −→ y0. Thus xny0 −→ x0y0 and x0yn −→ x0y0.Hence,

limn−→∞µ(x1, x2, . . . , xn−1, xny0 − x0y0, s) = 1 and

limn−→∞ν(x1, x2, . . . , xn−1, xny0 − x0y0, s) = 0,

andlimn−→∞µ(x1, x2, . . . , xn−1, x0yn − x0y0, t) = 1 and

limn−→∞ν(x1, x2, . . . , xn−1, x0yn − x0y0, t) = 0,

for all s, t > 0.Therefore,µ(x1, x2, . . . , xn−1, xnyn − x0y0, t) = µ(x1, x2, . . . , xn−1, (xn − x0)(yn −

y0) + (xn − x0)y0 + x0(yn − y0), t)≥ µ(x1, x2, . . . , xn−1, (xn−x0)(yn−y0), t3)∗µ(x1, x2, . . . , xn−1, (xn−x0)y0,

t3)

∗ µ(x1, x2, . . . , xn−1, x0(yn − y0),t3)

≥ (µ(x1, x2, . . . , xn−1, xn − x0,√t√3) ∗1 µ(x1, x2, . . . , xn−1, yn − y0,

√t√3))

∗ µ(x1, x2, . . . , xn−1, (xn − x0)y0,t3) ∗ µ(x1, x2, . . . , xn−1, x0(yn − y0),

t3)

−→ 1.And

ν(x1, x2, . . . , xn−1, xnyn − x0y0, t)= ν(x1, x2, . . . , xn−1, (xn − x0)(yn − y0) + (xn − x0)y0 + x0(yn − y0), t)≤ ν(x1, x2, . . . , xn−1, (xn−x0)(yn−y0), t3)◦ν(x1, x2, . . . , xn−1, (xn−x0)y0,

t3)◦

ν(x1, x2, . . . , xn−1, x0(yn − y0),t3)

≤ (ν(x1, x2, . . . , xn−1, xn − x0,√t√3) ◦1 ν(x1, x2, . . . , xn−1, yn − y0,

√t√3)) ◦

ν(x1, x2, . . . , xn−1, (xn − x0)y0,t3) ◦ ν(x1, x2, . . . , xn−1, x0(yn − y0),

t3)

−→ 0.Hence, xnyn −→ x0y0.Hence X is with continuous product. 2


Lemma 3.6. Any continuous t-norm ∗ and t-conorm ◦ satisfy:for all γ ∈ (0, 1) there exist α, β ∈ (0, 1) such that α ∗ β = γ and

(1− α) ◦ β = 1− γ.

Theorem 3.7. Let (X,µ, ν, ∗, ◦, ∗1, ◦1) be any IFnNA. Then X is withcontinuous product.

Proof. Let α ∈ (0, 1). Then there exists > 0 such that α + ∈ (0, 1)and (1− α)− ∈ (0, 1).

As ∗1 is a continuous t-norm and ◦1 is a continuous t-conorm by lemma 3.6,we obtain that there exist βα, γα ∈ (0, 1) such that

α+ = βα ∗1 γα and (1− α)− = (1− βα) ◦1 γα

We suppose that βα ≥ γα and (1 − βα) ≤ γα (the case βα ≤ γγ issimilar).

We choose M =M(α) = 1.Let x, y ∈ X, s, t > 0 and x1, x2, . . . , xn−1 ∈ X such that

µ(x1, x2, . . . , xn−1, x, s) > βα and ν(x1, x2, . . . , xn−1, x, s) < 1− βα,

and

µ(x1, x2, . . . , xn−1, y, t) > βα and ν(x1, x2, . . . , xn−1, y, t) < 1− βα.

Thenµ(x1, x2, . . . , xn−1, xy,Mst)

≥ µ(x1, x2, . . . , xn−1, x, s) ∗1 µ(x1, x2, . . . , xn−1, y, t)≥ βα ∗1 βα≥ βα ∗1 γα= α+> α.

Andν(x1, x2, . . . , xn−1, xy,Mst)

≤ µ(x1, x2, . . . , xn−1, x, s) ◦1 µ(x1, x2, . . . , xn−1, y, t)< (1− βα) ◦1 (1− βα)< (1− βα) ◦1 γα= (1− α)−< 1− α.

Hence X is with continuous product.2


Definition 3.8. Let (X,µ, ν, ∗, ◦, ∗1, ◦1) be the IFnNA. Then X is said tobe with multiplicatively continuous product if for all α ∈ (0, 1); x, y ∈ X;s, t > 0 and x1, x2, . . . , xn−1 ∈ X:

µ(x1, x2, . . . , xn−1, x, s) > α and ν(x1, x2, . . . , xn−1, x, s) < 1− α,

and

µ(x1, x2, . . . , xn−1, y, t) > α and ν(x1, x2, . . . , xn−1, y, t) < 1− α.

imply,

µ(x1, x2, . . . , xn−1, xy, st) ≥ α and ν(x1, x2, . . . , xn−1, xy, st) ≤ 1− α.

Example 3.9. (IFnNA with multiplicatively continuous product) Let(X,µ, ν, ∗, ◦, ∗1, ◦1) be the IFnNA from the example 3.3 is with multiplica-tively continuous product.

Proof. Let α ∈ (0, 1), for all x, y ∈ X, for all x1, x2, . . . , xn−1 ∈ X, ands, t > 0 such that


and


Then,

µ(x1, x2, . . . , xn−1, x, s) = 1 and ν(x1, x2, . . . , xn−1, x, s) = 0,

and

µ(x1, x2, . . . , xn−1, y, t) = 1 and ν(x1, x2, . . . , xn−1, y, t) = 0.

Thus we have,kxk < s, kyk < t.Therefore, kxyk ≤ kxk.kyk < st.Hence,

µ(x1, x2, . . . , xn−1, xy, st) = 1 > α and ν(x1, x2, . . . , xn−1, xy, st) = 0 < 1−α.

Therefore X is with multiplicatively continuous product. 2


Example 3.10. (IFnNA which is not with multiplicatively continuous prod-uct) We consider the IFnNA from example 3.4 where X = R and the normon X is the absolute value k · k. Then (R, µ, ν, ∗, ◦, ∗1, ◦1) is not withmultiplicatively continuous product.

Proof. Suppose, α = 17 , 1− α = 6

7 , x =72s, y =

72 t and s, t > 0.

Then we have,

µ(x1, x2, . . . , xn−1, x, s) =s

s+ kxk =s

s+ 72s=2

9>1

7.

i.e.

µ(x1, x2, . . . , xn−1, x, s) > α,

and

ν(x1, x2, . . . , xn−1, x, s) =kxk

s+ kxk =72s

s+ 72s=7

9<6

7.

i.e.

ν(x1, x2, . . . , xn−1, x, s) < 1− α

.But,

µ(x1, x2, . . . , xn−1, xy, st) =st

st+ kxyk =st

st+ 72 ·

72st

=4

53<1

7= α,

and

ν(x1, x2, . . . , xn−1, xy, st) =kxyk

st+ kxyk =494 st

st+ 494 st

=49

53>1

7= 1− α.

i.e.

µ(x1, x2, . . . , xn−1, xy, st) < α and ν(x1, x2, . . . , xn−1, xy, st) > 1− α.

Thus (R, µ, ν, ∗, ◦, ∗1, ◦1) is not with multiplicatively continuous prod-uct. 2

Theorem 3.11. Let (X,µ, ν, ∗, ◦, ∗1, ◦1) be the IFnNA such that α∗1α ≥ αand (1− α) ◦1 (1− α) ≤ 1− α for all α ∈ (0, 1). Thus (X,µ, ν, ∗, ◦, ∗1, ◦1)is with multiplicatively continuous product.


Proof. Let α ∈ (0, 1), for all x, y ∈ X, for all x1, x2, . . . , xn−1 ∈ X, ands, t > 0 such that


and


Then,

µ(x1, x2, . . . , xn−1, xy, st) ≥ µ(x1, x2, . . . , xn−1, x, s)∗1µ(x1, x2, . . . , xn−1, y, t)> α ∗1 α≥ α.

and

ν(x1, x2, . . . , xn−1, xy,Mst) ≤ µ(x1, x2, . . . , xn−1, x, s)◦1µ(x1, x2, . . . , xn−1, y, t)< (1− α) ◦1 (1− α)≤ 1− α.

Thus X is with multiplicatively continuous product. 2

Remark 3.12. Theorem 3.11 is sufficient but not necessary.

Theorem 3.13. Let (X,µ1, ν1, ∗, ◦, ∗1, ◦1) and (Y, µ2, ν2, ∗, ◦, ∗1, ◦1) be twoIFnNAs. If t-norm ∗0 denotes both ∗ and ∗1 and t-conorm ◦0 denotes both◦ and ◦1, then ((X × Y ), µ, ν, ∗0 , ◦0 , ∗1, ◦1) is a IFnNA where,

µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), t)

= µ1(x1, x2, . . . , xn−1, x, t) ∗0µ2(y1, y2, . . . , yn−1, y, t),

and

ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), t)

= ν1(x1, x2, . . . , xn−1, x, t) ◦0ν2(y1, y2, . . . , yn−1, y, t),

for all (x1, y1), (x2, y2), . . . , (xn−1, yn−1) ∈ X × Y , x1, x2, . . . , xn−1 ∈ X,y1, y2, . . . , yn−1 ∈ Y , (x, y) ∈ X × Y and s, t > 0.


Proof. We have to prove thatµ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

≥ µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s)∗1 µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t),

andν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

≤ ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s)◦1 ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t),

for all (x1, y1), (x2, y2), . . . , (xn−1, yn−1) ∈ X×Y , x, x0 ∈ X, y, y

0,∈ Y , and

s, t > 0.Now,µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

= µ1(x1, x2, . . . , xn−1, xx0, st) ∗0 µ2(y1, y2, . . . , yn−1, yy

0, st)

≥ (µ1(x1, x2, . . . , xn−1, x, s) ∗1 µ1(x1, x2, . . . , xn−1, x0, t)) ∗0

(µ2(y1, y2, . . . , yn−1, y, s) ∗1 µ2(y1, y2, . . . , yn−1, y0, t))

≥ [µ1(x1, x2, . . . , xn−1, x, s) ∗0µ2(y1, y2, . . . , yn−1, y, s)] ∗1

[µ1(x1, x2, . . . , xn−1, x0, t) ∗0 µ2(y1, y2, . . . , yn−1, y

0, t)]

= µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) ∗1µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t).

Therefore,µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

≥ µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s)∗1 µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t).

Andν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

= ν1(x1, x2, . . . , xn−1, xx0, st) ◦0 ν2(y1, y2, . . . , yn−1, yy

0, st)

≤ (ν1(x1, x2, . . . , xn−1, x, s) ◦1 ν1(x1, x2, . . . , xn−1, x0, t)) ◦0

(ν2(y1, y2, . . . , yn−1, y, s) ◦1 ν2(y1, y2, . . . , yn−1, y0, t))

≤ [ν1(x1, x2, . . . , xn−1, x, s) ◦0ν2(y1, y2, . . . , yn−1, y, s)] ◦1

[ν1(x1, x2, . . . , xn−1, x0, t) ◦0 ν2(y1, y2, . . . , yn−1, y

0, t)]

= ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) ◦1ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t).

Therefore,ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

≤ ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s)◦1 ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t).

Therefore X × Y is an IFnNA.2

Theorem 3.14. Let ∗1 be a t-norm and ◦1 be a t-conorm satisfying α∗1α ≥


α and (1−α)◦1(1−α) ≤ (1−α) for all α ∈ (0, 1) and let (X,µ1, ν1, ∗, ◦, ∗1, ◦1)and (Y, µ2, ν2, ∗, ◦, ∗1, ◦1) be two IFnNAs with multiplicatively continuousproduct. If a t-norm ∗0 denotes both ∗ and ∗1 and t-conorm ◦

0denotes both

◦ and ◦1, then ((X×Y ), µ, ν, ∗0 , ◦0 , ∗1, ◦1) is a IFnNA with multiplicativelycontinuous product.

Proof. Let α ∈ (0, 1), (x1, y1), (x2, y2), . . . , (xn−1, yn−1) ∈ X×Y ,(x, y) ∈X × Y ,(x

0, y

0) ∈ X × Y and s, t > 0 such that

µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) > α;

ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) < 1− α.

and

µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x0, y

0), t) > α;

ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x0, y

0), t) < 1− α.

Then we have successively:µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

= µ1(x1, x2, . . . , xn−1, xx0, st) ∗0 µ2(y1, y2, . . . , yn−1, yy

0, st)

≥ (µ1(x1, x2, . . . , xn−1, x, s) ∗1 µ1(x1, x2, . . . , xn−1, x0, t)) ∗0

(µ2(y1, y2, . . . , yn−1, y, s) ∗1 µ2(y1, y2, . . . , yn−1, y0, t))

≥ [µ1(x1, x2, . . . , xn−1, x, s) ∗0µ2(y1, y2, . . . , yn−1, y, s)] ∗1

[µ1(x1, x2, . . . , xn−1, x0, t) ∗0 µ2(y1, y2, . . . , yn−1, y

0, t)]

= µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) ∗1µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x

0, y

0), t)

> α ∗1 α≥ α.

i.e.

µ((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx0, yy

0), st) ≥ α.

Andν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx

0, yy

0), st)

= ν1(x1, x2, . . . , xn−1, xx0, st) ◦0 ν2(y1, y2, . . . , yn−1, yy

0, st)

≤ (ν1(x1, x2, . . . , xn−1, x, s) ◦1 ν1(x1, x2, . . . , xn−1, x0, t)) ◦0

(ν2(y1, y2, . . . , yn−1, y, s) ◦1 ν2(y1, y2, . . . , yn−1, y0, t))

≤ [ν1(x1, x2, . . . , xn−1, x, s) ◦0ν2(y1, y2, . . . , yn−1, y, s)] ◦1

[ν1(x1, x2, . . . , xn−1, x0, t) ◦0 ν2(y1, y2, . . . , yn−1, y

0, t)]

= ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x, y), s) ◦1


ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (x0, y

0), t)

< (1− α) ◦1 (1− α)≤ (1− α).i.e.

ν((x1, y1), (x2, y2), . . . , (xn−1, yn−1), (xx0, yy

0), st) ≤ (1− α).

Therefore X×Y is an IFnNA with multiplicatively continuous product.2

Example 3.15. Let (X,µ, ν, ∗, ◦, ∗1, ◦1) be an IFnNAwith multiplicativelycontinuous product and let S ⊂ X be a linear closed algebra of X. Then(S, µ, ν, ∗, ◦, ∗1, ◦1) is an IFnNA with multiplicatively continuous product.

4. Conclusion

In this paper we have introduced the notion of IFnNA and given a char-acterization for an IFnNA to be with continuous product. Further, thenotion of multiplicatively continuous product has also been introduced andstudied. The spectral properties of an IFnNLS will be an interesting topicfor future research in this regard which may be applied in the study ofdynamical systems and particle physics.

Competing interestsThe authors declare that they have no competing interests.

Authors’ contributionsAll authors contributed equally and significantly in writing this article.

All authors read and approved the final manuscript.


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Nabanita KonwarDepartment of MathematicsNorth Eastern Regional Institute of Science and TechnologyNirjuli, Arunachal Pradesh - 791109,Indiae-mail : [email protected]

and

Pradip DebnathDepartment of MathematicsNorth Eastern Regional Institute of Science and TechnologyNirjuli, Arunachal Pradesh - 791109,Indiae-mail : [email protected]

intuitionistic fuzzy n-normed algebra and continuous product · a systematic development of fuzzy...

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