Intuitionistic Type-2 Fuzzy Normed Linear Space and Some of Its Basic Properties

Abstract

An intuitionistic fuzzy set is a more generalised tool than a fuzzy set for handling unpredictability as, in an intuitionistic fuzzy set, there is scope for considering a grade of non-membership, independent of the grade of membership, only satisfying the condition that their sum is less or equal to 1. The motivation of this paper is to introduce the notion of intuitionistic type-2 fuzzy normed linear space (IT2FNLS). Here, to each vector x, we assign two fuzzy real number valued grades, one for its norm and the other for the negation of its norm. A theorem of the decomposition of the intuitionistic type-2 fuzzy norm into a family of pairs of Felbin-type fuzzy norms is established. Also, we deal with Cauchyness and convergence of sequences in the IT2FNLS. Later on, in the finite-dimensional IT2FNLS, the completeness property and compactness property are explored. Finally, we define two types of intuitionistic type-2 fuzzy continuity and examine the relations between them.

1. Introduction

Since L. A. Zadeh introduced the notion of a fuzzy set in 1965 [1,2], its applications have covered a wide spectrum of fields of mathematics from fuzzy logic to fuzzy topology, fuzzy functional analysis, fuzzy differential equations, fuzzy optimisation theory and dynamical systems, etc.

Normed linear space is the important pillar of functional analysis, a major branch of modern mathematics. C. Felbin [3] introduced the concept of the fuzzy norm whose metric analogue is of the Kaleva–Seikkala [4] type and defined fuzzy normed linear space. In 1994, Cheng and Mordeson [5] defined a fuzzy norm whose metric analogue is of the Kramosil and Michalek [6] type by giving a grade to a norm of an element by comparing the norm to a real number. In 2003, Bag and Samanta [7] modified the definition of the fuzzy norm given by Cheng and Mordeson [5] and obtained a decomposition theorem from it. On the other hand, Zadeh [2], following the legacy of his own, introduced the definition of a type-n fuzzy set in 1975. In [8], for the first time, the notion of type-2 fuzzy normed linear space (type-2 FNLS) was introduced by Chiney, Biswas and Samanta, and a decomposition theorem was also proved in this setting.

The notion of an intuitionistic fuzzy set (IFS) was introduced by Atanassov [9,10,11,12] as a generalisation of Zadeh’s fuzzy set [1]. There are situations where IFS theory is more appropriate, as dealt with by [13]. IFS theory has successfully been applied in knowledge engineering, medical diagnosis, decision making, career determination, etc. [14,15,16]. With the advancement of time, several researchers have extended various mathematical aspects such as groups, rings, topological spaces, metric spaces, topological groups, topological vector spaces, etc., in an IFS [17,18,19,20,21,22,23,24]. The definition of intuitionistic fuzzy n-normed linear space was introduced by S. Vijayabalaji, N. Thillaigovindan and Y. Bae Jun [25] in 2007. In 2009, T. K. Samanta and Iqbal H. Jebril [26] introduced the definition of an intuitionistic fuzzy norm over a linear space. Recently, research works have been done on intuitionistic fuzzy normed linear spaces [27,28,29].

The main objective of this paper is to give an idea of intuitionistic type-2 fuzzy normed linear space (IT2FNLS) for the first time. Here, we fuzzify the norm of a vector with an intuitionistic version of the type-2 fuzzy norm. We decompose an intuitionistic type-2 fuzzy norm into a family of pairs of Felbin-type fuzzy norms. Basic properties such as the convergent sequence, Cauchy sequence and closed and boundedness of the set are also studied. The finite-dimensional IT2FNLS is shown to be complete, and, in this space, the compactness of a subset can be deduced from the closed and boundedness. We define the continuity of functions as two types, namely, intuitionistic type-2 fuzzy continuity and sequentially intuitionistic type-2 fuzzy continuity. Later on, we discover that every intuitionistic type-2 fuzzy continuous function is sequentially intuitionistic type-2 fuzzy continuous, but its converse is not true in general, which is justified by a counterexample.

2. Preliminaries

Definition 1

([4]). A fuzzy real number is a fuzzy set on $\mathbb{R}$, i.e., a mapping $\eta :\mathbb{R}\to I(=[0,1\left]\right)$ associating each real number to its grade of membership $\eta \left(t\right)$.

Definition 2

([4]). A fuzzy real number η is convex if $\eta \left(t\right)\ge \eta \left(s\right)\wedge \eta \left(r\right)$=min $\left(\eta \right(s),\eta (r\left)\right)$ where $s\le t\le r$.

Definition 3

([4]). If there exists a $t_0\in \mathbb{R}$ such that $\eta \left(t_0\right)=1$, then η is called a normal fuzzy real number.

Definition 4

([4]). The α-level set of a fuzzy real number η, $0<\alpha \le 1$, denoted by ${\left[\eta \right]}_{\alpha }$, is defined as ${\left[\eta \right]}_{\alpha }=\{t:\eta \left(t\right)\ge \alpha \}$.

Proposition 1

([4]). A fuzzy real number η is convex if and only if each of its α-level sets ${\left[\eta \right]}_{\alpha }$, $0<\alpha \le 1$, is a convex set in $\mathbb{R}$.

Definition 5

([4]). A fuzzy real number η is called upper semi-continuous if, for all $t\in \mathbb{R}$ and $ϵ>0$ with $\eta \left(t\right)=a$, there is $c>0$ such that $|s-t|<c=c\left(t\right)\Rightarrow \eta \left(t\right)<a+ϵ$, i.e., ${\eta }^{-1}\left([0,a+ϵ)\right)$ for all $a\in I$, and $ϵ>0$ is open in the usual topology of $\mathbb{R}$.

Note 1.

It can be easily seen that the α-level sets of an upper semi-continuous convex normal fuzzy real number for each α, $0<\alpha \le 1$, is a closed interval $[a^{\alpha },b^{\alpha }]$ where $a^{\alpha }=-\infty$ and $b^{\alpha }=+\infty$ are also admissible. Let us denote the set of all upper semi-continuous normal convex fuzzy real numbers by $\mathbb{R}\left(I\right)$. Since each $r\in \mathbb{R}$ can be considered a fuzzy real number $\overline{r}$,

$$\overline{r}\left(t\right)=\left\{\begin{array}{c}1 if t=r\hfill \\ 0 if t\ne r\hfill \end{array}\right.$$

$\mathbb{R}$ can be embedded in $\mathbb{R}\left(I\right)$.

Definition 6

([4]). A fuzzy real number η is called non-negative if $\eta \left(t\right)=0$ for all $t<0$. The set of all non-negative fuzzy real numbers of $\mathbb{R}\left(I\right)$ is denoted by ${\mathbb{R}}^{*}\left(I\right)$.

Note 2.

If we take the set $\{\eta \in {\mathbb{R}}^{*}\left(I\right):\eta =\overline{0}$ or $\eta \succ \overline{0}\}$, then we denote this set by ${\mathbb{R}}^{+}\left(I\right)$.

Note 3.

Arithmetic operations on fuzzy real numbers, the definition of the partial ordering of fuzzy real numbers and the definition of the convergence of the sequence of fuzzy real numbers are taken from [4].

Definition 7

([4]). Define a partial ordering ’⪯’ in ${\mathbb{R}}^{+}\left(I\right)$ by $\eta ⪯\delta$ if and only if $a_1^{\alpha }\le a_2^{\alpha }$ and $b_1^{\alpha }\le b_2^{\alpha }$ for all $\alpha \in (0,1]$, where ${\left[\eta \right]}_{\alpha }=[a_1^{\alpha },b_1^{\alpha }]$ and ${\left[\delta \right]}_{\alpha }=[a_2^{\alpha },b_2^{\alpha }]$. We write $\eta ⪯\delta$ as $\delta ⪰\eta$ when desired. The strict inequality in ${\mathbb{R}}^{+}\left(I\right)$ is defined by $\eta \prec \delta$ if and only if $a_1^{\alpha }<a_2^{\alpha }$ and $b_1^{\alpha }<b_2^{\alpha }$ for each $\alpha \in (0,1]$.

Definition 8

([2]). A fuzzy set is of type n, n = 2, 3, …, if its membership function ranges over fuzzy sets of type n-1. The membership function of a fuzzy set of type 1 ranges over the interval $[0,1]$.

Definition 9

([3]). Let X be a vector space over $\mathbb{R}$.

Let $||.||:X\to {\mathbb{R}}^{*}\left(I\right)$.

Write

${[||x||]}_{\alpha }={[|||x|||}_1^{\alpha }{,|||x|||}_2^{\alpha }]$ for $x\in X,0<\alpha \le 1$ and suppose, for all $x\in X,x\ne \underline{0}$, that there exists ${\alpha }_0\in (0,1]$ independent of x such that for all $\alpha \le {\alpha }_0$,

(A) 

${|||x|||}_2^{\alpha }<\infty$;

(B) 

inf${|||x|||}_1^{\alpha }>0.$

Then, $(X,||.||)$ is called a fuzzy normed linear space, and $||.||$ is a fuzzy norm if

(i) 

$||x||=\overline{0}$ if and only if $x=\underline{0}$;

(ii) 

$||rx||=|r|||x||,x\in X,r\in \mathbb{R}$;

(iii) 

For all $x,y\in X$, $||x+y||⪯||x||\oplus ||y||$.

Definition 10

([30]). Let X be a linear space over $\mathbb{R}$. Let $||||:X\to {\mathcal{F}}^{+}(={\mathbb{R}}^{*}\left(I\right))$ be a mapping satisfying the following:

(i) 

$||x||=\overline{0}$ iff $x=\underline{0}$;

(ii) 

$||rx||=|r|||x||,x\in X,r\in \mathbb{R}$;

(iii) 

For all $x,y\in X$, $||x+y||⪯||x||\oplus ||y||$

and

($A^{\prime }$): 

$x\ne \underline{0}\Rightarrow ||x||\left(t\right)=0,\forall t\le 0$

Then, $(X,||||)$ is called a fuzzy normed linear space, and $||||$ is called a fuzzy norm on X.

Remark 1

([30]). (i) Condition $\left(A^{\prime }\right)$ in Definition 10 is equivalent to the condition $\left(A^{″}\right)$: for all $x(\ne \underline{0})\in X$, ${||x||}_{\alpha }^1>0$, $\forall \alpha \in (0,1]$, where ${[||x||}_{\alpha }{]=[||x||}_{\alpha }^1{,||x||}_{\alpha }^2]$ and (ii) ${||||}_{\alpha }^i:i=1,2$ are crisp norms on X.

Proposition 2

([3]). Let $\{x_1,x_2,\dots ,x_n\}$ be a linearly independent set of vectors in a fuzzy normed linear space $(X,||.||)$ (of any dimension). Then, there is an $\eta \succ \overline{0}(\eta \in {\mathbb{R}}^{*}\left(I\right))$ with $su{p}_{\alpha \in (0,1]}{b}^{\alpha }<\infty$ where ${\left[\eta \right]}_{\alpha }=[a^{\alpha },b^{\alpha }]$ and such that, for every choice of scalars $a_1,a_2,\dots ,a_n$, we have $||a_1{x}_1+\dots +a_n{x}_n||⪰(|a_1|+\dots +|a_n|)\eta$.

Definition 11

([25]). Let E be any set. An intuitionistic fuzzy set A of E is an object of the form A = {$(x,{\mu }_A\left(x\right),{\nu }_A\left(x\right)):x\in E$}, where the functions ${\mu }_A:E\to [0,1]$ and ${\nu }_A:E\to [0,1]$ denote the degree of membership and the non-membership of the element $x\in E$, respectively, and, for every $x\in E$, $0\le {\mu }_A\left(x\right)+{\nu }_A\left(x\right)\le 1$.

Theorem 1

([31]). Suppose that $\{u_t:t\in \mathsf{\Omega }\}\subset E^1$ is bounded. Then, its supremum and infimum must exist and are determined by two pairs of usual functions of λ on $[0,1]$

$$(u_{s,\mathsf{\Omega }}^{-}\left(\lambda \right),u_{s,\mathsf{\Omega }}^{+}\left(\lambda \right)),(u_{I,\mathsf{\Omega }}^{-}\left(\lambda \right),u_{I,\mathsf{\Omega }}^{+}\left(\lambda \right)),$$

where

$$u_{s,\mathsf{\Omega }}^{-}\left(\lambda \right)=\left\{\begin{array}{c}{u}_s^{-}\left(\lambda \right)for \lambda \in (0,1]\hfill \\ u_s^{-}(0+0)for \lambda =0\hfill \end{array}\right.$$

$$u_{s,\mathsf{\Omega }}^{+}\left(\lambda \right)=\left\{\begin{array}{c}{u}_s^{+}\left(\lambda \right)for \lambda \in [0,1]\setminus \left\{{\lambda }_m^s\right\}\hfill \\ u_s^{+}({\lambda }_m^s-0)for \lambda ={\lambda }_m^s(m=1,2,\dots )\hfill \end{array}\right.$$

$$u_{I,\mathsf{\Omega }}^{-}\left(\lambda \right)=\left\{\begin{array}{c}{u}_I^{-}\left(\lambda \right)for \lambda \in [0,1]\setminus \left\{{\lambda }_m^{\prime }\right\}\hfill \\ u_I^{-}({\lambda }_m^{\prime }-0)for \lambda ={\lambda }_m^{\prime }(m=1,2,\dots )\hfill \end{array}\right.$$

$$u_{I,\mathsf{\Omega }}^{+}\left(\lambda \right)=\left\{\begin{array}{c}{u}_I^{+}\left(\lambda \right)for \lambda \in (0,1]\hfill \\ u_I^{+}(0+0)for \lambda =0\hfill \end{array}\right.$$

$$u_s^{-}\left(\lambda \right)=su{p}_{t\in \mathsf{\Omega }}{\left(u_t\right)}_{\lambda }^{-},u_s^{+}\left(\lambda \right)=su{p}_{t\in \mathsf{\Omega }}{\left(u_t\right)}_{\lambda }^{+}$$

$$u_I^{-}\left(\lambda \right)=in{f}_{t\in \mathsf{\Omega }}{\left(u_t\right)}_{\lambda }^{-},u_I^{+}\left(\lambda \right)=in{f}_{t\in \mathsf{\Omega }}{\left(u_t\right)}_{\lambda }^{+}$$

and $\left\{{\lambda }_m^s\right\}$ and $\left\{{\lambda }_m^{\prime }\right\}$ are all noncontinuous points of $u_s^{+}\left(\lambda \right)$ and $u_t^{-}\left(\lambda \right)$ on $[0,1]$, respectively.

Note 4.

$\mathcal{I}$ denotes the set of all fuzzy sets from $[0,1]$ to $[0,1]$, i.e, $\mathcal{I}=\{A|A:[0,1]\to [0,1]\}$, and let ${\mathcal{I}}_0$ denotes the set ${\mathcal{I}}_0=\{\zeta \in \mathcal{I}:\overline{0}\prec \zeta \prec \overline{1}\}$.

Definition 12

([8]). Let X be a linear space over $\mathbb{R}$. A fuzzy subset $\mathcal{N}:X\times {\mathbb{R}}^{+}\left(I\right)\to \mathcal{I}$ is called a type-2 fuzzy norm on X if, for all $x,u\in X,c\in \mathbb{R}$ and $\eta \in {\mathbb{R}}^{+}\left(I\right)$,

• (N1) When $\eta =\overline{0},\mathcal{N}(x,\eta )=\overline{0}$;
• (N2) ($\forall \eta \succ \overline{0},\mathcal{N}(x,\eta )=\overline{1})$ if $x=\underline{0}$;
• (N3) $(\forall \eta \in {\mathbb{R}}^{+}\left(I\right),\eta \succ \overline{0},\mathcal{N}(cx,\eta )=\mathcal{N}(x,\frac{1}{|c|}\eta )$ if $c\ne 0$;
• (N4) $\forall {\eta }_1,{\eta }_2\in {\mathbb{R}}^{+}\left(I\right)$ and $x,u\in X$
  $\mathcal{N}(x+u,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x,{\eta }_1)$,$\mathcal{N}(u,{\eta }_2)\}$;
• (N5) $\mathcal{N}(x,.)$ is a non-decreasing function of ${\mathbb{R}}^{+}\left(I\right)$, which means, if ${\eta }_1,{\eta }_2\in {\mathbb{R}}^{+}\left(I\right)$ with ${\eta }_2⪰{\eta }_1$, then $\mathcal{N}(x,{\eta }_2)⪰\mathcal{N}(x,{\eta }_1)$, and $li{m}_{\eta \to \overline{\infty }}\mathcal{N}(x,\eta )=\overline{1}$.

The pair $(X,\mathcal{N})$ is called the type-2 fuzzy normed linear space (type-2 FNLS).

Theorem 2

([8]). Let $(X,\mathcal{N})$ be a type-2 FNLS. Assume further that

• (N6) $\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )\succ \overline{0}\}=\overline{0}$ implies $x=\underline{0}$.

Define ${||x||}^{\zeta }=\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta \},\zeta \in {\mathcal{I}}_0$.

Then $\{||.|{|}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is a Felbin-type fuzzy norm on X, and $(X,||.|{|}^{\zeta })$ is a Felbin-type fuzzy normed linear space. Also, $\{||x|{|}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is a family of Felbin-type fuzzy norms on X such that ${\zeta }_2\succ {\zeta }_1$$({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}^{{\zeta }_2}⪰{||x||}^{{\zeta }_1}.$

Theorem 3

([8]). Let $\{||x|{|}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ be a family of Felbin-type fuzzy norms on a linear space X such that ${\zeta }_2\succ {\zeta }_1$$({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}^{{\zeta }_2}⪰{||x||}^{{\zeta }_1}$. Now we define a function ${\mathcal{N}}^{\prime }:X\times {\mathbb{R}}^{+}\left(I\right)\to \mathcal{I}$ as

$$\begin{array}{ccc}\hfill {\mathcal{N}}^{\prime }(x,\eta )& =& \vee {\{\zeta :||x||}^{\zeta }⪯\eta \},when(x,\eta )\ne (\underline{0},\overline{0})\hfill \\ & =& \overline{0},when(x,\eta )=(\underline{0},\overline{0}){or\{\zeta :||x||}^{\zeta }⪯\eta \}=\varphi \hfill \end{array}$$

Then ${\mathcal{N}}^{\prime }$ is a type-2 fuzzy norm on X.

Lemma 1

([8]). Let $(X,\mathcal{N})$ be a type-2 FNLS satisfying (N6) and $\{e_1,e_2,\dots ,e_n\}$ be a finite set of linearly independent elements of X. Then, for each $\zeta \in {\mathcal{I}}_0$, there exists a ${\lambda }^{\zeta }\succ \overline{0}$ with $su{p}_{\alpha \in (0,1]}{b}^{\alpha }<\infty$, where ${\left[{\lambda }^{\zeta }\right]}_{\alpha }=[a^{\alpha },b^{\alpha }]$, such that, for every choice of scalars $a_1,a_2,\dots ,a_n$, we have

$||a_1{e}_1+a_2{e}_2+\dots +a_n{e}_n{||}^{\zeta }⪰(|a_1|+|a_2|+\dots +|a_n|){\lambda }^{\zeta }$.

Lemma 2

([8]). If $\left\{u_n\right\}$ is a sequence of positive real numbers and $\eta \succ \overline{0}$ such that ${\left[\eta \right]}_{\alpha }=[a^{\alpha },b^{\alpha }]$, then ${\left[u_n\eta \right]}_{\alpha }=[u_n{a}^{\alpha },u_n{b}^{\alpha }]$. Also, if $u_n\to \infty$ as $n\to \infty$, then $u_n\eta \to \overline{\infty }$ as $n\to \infty$.

3. Finite-Dimensional Intuitionistic Type-2 Fuzzy Normed Linear Spaces

3.1. Intuitionistic Type-2 Fuzzy Normed Linear Spaces

In this section, we introduce the notion of an intuitionistic type-2 fuzzy normed linear space with an example. In addition, we show that the decomposition of an intuitionistic type-2 fuzzy norm gives us a family of pairs of Felbin-type fuzzy norms.

Definition 13.

An intuitionistic type-2 fuzzy norm, or, in short, an IT2FN, on X (where X is a linear space over $\mathbb{R}$) is an object of the form $A=\{((x,\eta ),\mathcal{N}(x,\eta ),\mathcal{M}(x,\eta )):(x,\eta )\in X\times {\mathbb{R}}^{+}\left(I\right)\}$ where $\mathcal{N}$, $\mathcal{M}$ are functions from $X\times {\mathbb{R}}^{+}\left(I\right)$ to $\mathcal{I}$ satisfying the following conditions:

• (i) When $\eta =\overline{0}$, $\mathcal{N}(x,\eta )=\overline{0}$;
• (ii) ($\forall \eta \succ \overline{0},\mathcal{N}(x,\eta )=\overline{1}$) if and only if $x=\underline{0}$;
• (iii) $\mathcal{N}(cx,\eta )=\mathcal{N}(x,\frac{1}{|c|}\eta )$, if $c\ne 0$;
• (iv) $\mathcal{N}(x+y,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x,{\eta }_1),\mathcal{N}(y,{\eta }_2)\}$;
• (v) $\mathcal{N}(x,.)$ is a non-decreasing function of ${\mathbb{R}}^{+}\left(I\right)$, which means, if ${\eta }_1,{\eta }_2\in {\mathbb{R}}^{+}\left(I\right)$ with ${\eta }_2⪰{\eta }_1$, then $\mathcal{N}(x,{\eta }_2)⪰\mathcal{N}(x,{\eta }_1)$ and $li{m}_{\eta \to \overline{\infty }}\mathcal{N}(x,\eta )=\overline{1}$;
• (vi) When $\eta =\overline{0}$, $\mathcal{M}(x,\eta )=\overline{1}$;
• (vii) ($\forall \eta \succ \overline{0},\mathcal{M}(x,\eta )=\overline{0}$) if and only if $x=\underline{0}$;
• (viii) $\mathcal{M}(cx,\eta )=\mathcal{M}(x,\frac{1}{|c|}\eta )$, if $c\ne 0$;
• (ix) $\mathcal{M}(x+y,{\eta }_1\oplus {\eta }_2)⪯max\{\mathcal{M}(x,{\eta }_1),\mathcal{M}(y,{\eta }_2)\}$;
• (x) $\mathcal{M}(x,.)$ is a non-increasing function of ${\mathbb{R}}^{+}\left(I\right)$, which means, if ${\eta }_1,{\eta }_2\in {\mathbb{R}}^{+}\left(I\right)$ with ${\eta }_2⪰{\eta }_1$, then $\mathcal{M}(x,{\eta }_2)⪯\mathcal{M}(x,{\eta }_1)$ and $li{m}_{\eta \to \overline{\infty }}\mathcal{M}(x,\eta )=\overline{0}$.

Definition 14.

If A is an IT2FN on X (a linear space over the field $\mathbb{R}$), then $(X,A)$ is called an intuitionistic type-2 fuzzy normed linear space or, in short, an IT2FNLS.

The following is an example of an intuitionistic type-2 fuzzy normed linear space.

Example 1.

Let $(X=\mathbb{R},|.|)$ be the usual normed linear space. For any fuzzy real number $\eta \succ \overline{0}$, define ${\eta }_0$ as follows:

$${\eta }_0=\overline{\frac{a+b}{2}}$$

where $[a,b]$ denotes the closure of support of the fuzzy real number η. Let $k>0$ be any fixed real number. Define

$$\mathcal{N}(x,\eta )=\left\{\begin{array}{cc}{\eta }_0\oslash ({\eta }_0\oplus \overline{k|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{0},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

$$\mathcal{M}(x,\eta )=\left\{\begin{array}{cc}\overline{k|x|}\oslash ({\eta }_0\oplus \overline{k|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{1},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

Then $(X,A)$ is an intuitionistic type-2 fuzzy normed linear space.

Solution 1.

(i) When $\eta =\overline{0}$, we have, from the definition, $\mathcal{N}(x,\eta )=\overline{0}$.

(ii) $\forall \eta \in {\mathbb{R}}^{+}\left(I\right)$ with $\eta \succ \overline{0}$, $\mathcal{N}(x,\eta )=\overline{1}$.

$\iff {\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})=\overline{1}$

$\iff \frac{\frac{a+b}{2}}{\frac{a+b}{2}+||x||}=1$

$\iff ||x||=0$

$\iff x=\underline{0}$.

(iii) $\forall \eta \in {\mathbb{R}}^{+}\left(I\right)$ with $\eta \succ \overline{0}$ and $c(\ne 0)\in \mathbb{R}$, we obtain $\mathcal{N}(cx,\eta )={\eta }_0\oslash ({\eta }_0\oplus \overline{||cx||})$.

Again, $\mathcal{N}(x,\frac{1}{c}\eta )=\left(\frac{1}{c}{\eta }_0\right)\oslash (\frac{1}{c}{\eta }_0\oplus \overline{||x||})=\left(\overline{\frac{a+b}{2c}}\right)\oslash (\overline{\frac{a+b}{2c}}\oplus \overline{||x||})=\left(\overline{\frac{a+b}{2}}\right)\oslash (\overline{\frac{a+b}{2}}\oplus \overline{||cx||})={\eta }_0\oslash ({\eta }_0\oplus \overline{||cx||})$.

(iv) We have to show that $\forall \eta ,\xi \in {\mathbb{R}}^{+}\left(I\right)$ and $\forall x,y\in X$,

$\mathcal{N}(x+y,\eta \oplus \xi )⪰$$min\left\{\mathcal{N}\right(x,\eta ),\mathcal{N}(y,\xi \left)\right\}$.

If

(a) 

$\eta \oplus \xi =\overline{0},$

(b) 

$\eta \oplus \xi \succ \overline{0};\eta \succ \overline{0},\xi =\overline{0};\eta =\overline{0},\xi \succ \overline{0}$, then, in these cases, the relation is obvious.

If

(c) 

$\eta \oplus \xi \succ \overline{0};\eta \succ \overline{0},\xi \succ \overline{0}$, then

$\mathcal{N}(x+y,\eta \oplus \xi )=({\eta }_0\oplus {\xi }_0)\oslash ({\eta }_0\oplus {\xi }_0\oplus \overline{||x+y||})⪰({\eta }_0\oplus {\xi }_0)\oslash ({\eta }_0\oplus {\xi }_0\oplus \overline{||x||}\oplus \overline{||y||})$.

Now {${\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}⪰\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||y||})\}$

$\{{\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}\ominus \{{\xi }_0\oslash ({\xi }_0\oplus \overline{||y||})\}⪰\overline{0}$

$\Rightarrow ({\eta }_0\odot \overline{||y||})\ominus ({\xi }_0\odot \overline{||x||})⪰\overline{0}$....................$\left(i\right)$.

So $\{({\eta }_0\oplus {\xi }_0)\oslash ({\eta }_0\oplus {\xi }_0\oplus \overline{||x||}\oplus \overline{||y||})\}\ominus \{{\xi }_0\oslash ({\xi }_0\oplus \overline{||y||})\}=\{({\eta }_0\odot \overline{||y||})\ominus ({\xi }_0\odot \overline{||x||})\}\oslash \{({\eta }_0\oplus {\xi }_0\oplus \overline{||x||}\oplus \overline{||y||})\odot ({\xi }_0\oplus \overline{||y||})\}⪰\overline{0}$ by (i)

$\Rightarrow \{({\eta }_0\oplus {\xi }_0)\oslash ({\eta }_0\oplus {\xi }_0\oplus \overline{||x||}\oplus \overline{||y||})\}⪰\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||y||})\}$.

Similarly, if $\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||y||})\}⪰\{{\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}$, we have

$\{({\eta }_0\oplus {\xi }_0)\oslash ({\eta }_0\oplus {\xi }_0\oplus \overline{||x||}\oplus \overline{||y||})\}⪰\{{\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}$.

Thus, $\mathcal{N}(x+y,\eta \oplus \xi )⪰$$min\left\{\mathcal{N}\right(x,\eta ),\mathcal{N}(y,\xi \left)\right\}$.

(v) We consider the case $\eta \prec \xi$.

If $\eta \prec \xi$ and $\eta =\overline{0}$, then $\xi \succ \overline{0}$,

and $\mathcal{N}(x,\eta )=\overline{0}$, $\mathcal{N}(x,\xi )=\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||x||})\}⪰\overline{0}$

So $\mathcal{N}(x,\xi )⪰\mathcal{N}(x,\eta )$.

If $\eta \prec \xi$ and $\eta \succ \overline{0}$, then $\xi \succ \overline{0}$ and hence

$\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||x||})\}\ominus \{{\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}=\{({\xi }_0\odot \overline{||x||})\ominus ({\eta }_0\odot \overline{||x||})\}\oslash \{({\xi }_0\oplus \overline{||x||})\odot ({\eta }_0\oplus \overline{||x||}\}⪰\overline{0}$ [since ${\xi }_0\succ {\eta }_0$]

So $\{{\xi }_0\oslash ({\xi }_0\oplus \overline{||x||})\}⪰\{{\eta }_0\oslash ({\eta }_0\oplus \overline{||x||})\}$

$\Rightarrow \mathcal{N}(x,\xi )⪰\mathcal{N}(x,\eta )$, for all $x\in X$.

Thus, $\mathcal{N}(x,.)$ is a non-decreasing function of ${\mathbb{R}}^{+}\left(I\right)$.

When $\eta \to \overline{\infty }$, then $b\to \infty$, and hence ${\eta }_0\to \overline{\infty }$. So, $li{m}_{\eta \to \overline{\infty }}\mathcal{N}(x,\eta )=\overline{1}$.

Similarly, $\mathcal{M}$ satisfies all conditions of $\left(vi\right)-\left(x\right)$ of Definition 13.

Theorem 4.

Let $(X,A)$ be an IT2FNLS. We assume further that

(xi) 

$\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )\succ \overline{0}\}=\overline{0}$ implies $x=\underline{0}$;

(xii) 

$\wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )\prec \overline{1}\}=\overline{0}$ implies $x=\underline{0}.$

Define ${||x||}_{\mathcal{N}}^{\zeta }=\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta \},\zeta \in {\mathcal{I}}_0$

and ${||x||}_{\mathcal{M}}^{\zeta }=\wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯\zeta \},\zeta \in {\mathcal{I}}_0$.

Then (A) $\{||.|{|}_{\mathcal{N}}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is a family of Felbin-type fuzzy norms on X such that ${\zeta }_2⪰{\zeta }_1$$({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}_{\mathcal{N}}^{{\zeta }_2}⪰{||x||}_{\mathcal{N}}^{{\zeta }_1}$

and (B) $\{||.|{|}_{\mathcal{M}}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is a family of Felbin-type fuzzy norms on X such that ${\zeta }_1⪯{\zeta }_2$$({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}_{\mathcal{M}}^{{\zeta }_1}⪰{||x||}_{\mathcal{M}}^{{\zeta }_2}.$

Proof. 

(A) If ${||x||}_{\mathcal{N}}^{\zeta }=\overline{0}$, then $\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta \succ \overline{0}\}=\overline{0}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )\succ \overline{0}\}=\overline{0}$

$\Rightarrow x=\underline{0}$ by (xi).

Conversely, let $x=\underline{0}$

$\Rightarrow \mathcal{N}(x,\eta )=\overline{1}$ for all $\eta \succ \overline{0}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta \}=\overline{0}$.

$\Rightarrow {||x||}_{\mathcal{N}}^{\zeta }=\overline{0}$

(ii) If $c\ne 0$, then

$$\begin{array}{ccc}\hfill {||cx||}_{\mathcal{N}}^{\zeta }& =& \wedge \{\eta \succ \overline{0}:\mathcal{N}(cx,\eta )⪰\zeta \}\hfill \\ & =& \wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\frac{1}{|c|}\eta )⪰\zeta \}\hfill \\ & =& \wedge \{|c|\delta \succ \overline{0}:\mathcal{N}(x,\delta )⪰\zeta \}\hfill \\ & =& |c|[\wedge \{\delta \succ \overline{0}:\mathcal{N}(x,\delta )⪰\zeta \}]\hfill \\ & =& {|c|||x||}_{\mathcal{N}}^{\zeta }\hfill \end{array}$$

(iii)

$$\begin{array}{ccc}\hfill {||x||}_{\mathcal{N}}^{\zeta }\oplus {||y||}_{\mathcal{N}}^{\zeta }& =& \wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta :\zeta \in {\mathcal{I}}_0\}\oplus \wedge \{\delta \succ \overline{0}:\mathcal{N}(y,\delta )⪰\zeta :\zeta \in {\mathcal{I}}_0\}\hfill \\ & =& \wedge \{\eta \oplus \delta \succ \overline{0}:\mathcal{N}(x,\eta ),\mathcal{N}(y,\delta )⪰\zeta \}\hfill \\ & ⪰& \wedge \{\mu \succ \overline{0}:\mathcal{N}(x+y,\mu )⪰\zeta \}[Since,\mathcal{N}(x,\eta ),\mathcal{N}(y,\delta )⪰\zeta \Rightarrow \mathcal{N}(x+y,\eta \oplus \delta )⪰\zeta ]\hfill \end{array}$$

Therefore, $||x+{y||}_{\mathcal{N}}^{\zeta }⪯{||x||}_{\mathcal{N}}^{\zeta }\oplus {||y||}_{\mathcal{N}}^{\zeta }$.

Now take $\overline{0}\prec {\zeta }_1⪯{\zeta }_2$ and ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$.

Since ${\zeta }_1⪯{\zeta }_2$, we have $\{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰{\zeta }_2\}\subset \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰{\zeta }_1\}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰{\zeta }_2\}⪰\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰{\zeta }_1\}$

$\Rightarrow {||x||}_{\mathcal{N}}^{{\zeta }_2}⪰{||x||}_{\mathcal{N}}^{{\zeta }_1}$.

Thus, $\{||.|{|}_{\mathcal{N}}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is a family of Felbin-type fuzzy norms on X such that ${\zeta }_2⪰{\zeta }_1$$({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}_{\mathcal{N}}^{{\zeta }_2}⪰{||x||}_{\mathcal{N}}^{{\zeta }_1}$.

(B) Now we will prove that $\{||x|{|}_{\mathcal{M}}^{\zeta }:\zeta \in {\mathcal{I}}_0\}$ is also a family of Felbin-type fuzzy norms on X such that ${\zeta }_2⪰{\zeta }_1({\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0)\Rightarrow {||x||}_{\mathcal{M}}^{{\zeta }_2}⪰{||x||}_{\mathcal{M}}^{{\zeta }_1}.$

Let $\zeta \in {\mathcal{I}}_0$ and ${||x||}_{\mathcal{M}}^{\zeta }=\overline{0}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯\zeta \}=\overline{0}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )\prec \overline{1}\}=\overline{0}$

$\Rightarrow x=\underline{0}$ by (xii).

Conversely, we assume that $x=\overline{0}$

$\Rightarrow \mathcal{M}(x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯\zeta \}=\overline{0}$

$\Rightarrow {||x||}_{\mathcal{M}}^{\zeta }=\overline{0}$.

By definition, it is quite obvious that ${||cx||}_{\mathcal{M}}^{\zeta }={|c|||x||}_{\mathcal{M}}^{\zeta }$ $\forall c\in \mathbb{R}$.

$$\begin{array}{ccc}\hfill {||x||}_{\mathcal{M}}^{\zeta }\oplus {||y||}_{\mathcal{M}}^{\zeta }& =& \wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯\zeta \}\oplus \wedge \{{\eta }^{\prime }\succ \overline{0}:\mathcal{M}(y,{\eta }^{\prime })⪯\zeta \}\hfill \\ & ⪰& \wedge \{\eta \oplus {\eta }^{\prime }\succ \overline{0}:\mathcal{M}(x,\eta )⪯\zeta ,\mathcal{M}(y,{\eta }^{\prime })⪯\zeta \}\hfill \\ & ⪰& \wedge \{\eta \oplus {\eta }^{\prime }\succ \overline{0}:\mathcal{M}(x+y,\eta \oplus {\eta }^{\prime })⪯\zeta \}\hfill \\ & ⪰& \wedge \{{\eta }_1\succ \overline{0}:\mathcal{M}(x+y,{\eta }_1)⪯\zeta \}\hfill \\ & =& ||x+{y||}_{\mathcal{M}}^{\zeta }.\hfill \end{array}$$

Let ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$ and ${\zeta }_1⪯{\zeta }_2$.

Now $\{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯{\zeta }_1\}\subset \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯{\zeta }_2\}$

$\Rightarrow \wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯{\zeta }_1\}⪰\wedge \{\eta \succ \overline{0}:\mathcal{M}(x,\eta )⪯{\zeta }_2\}$

$\Rightarrow {||x||}_{\mathcal{M}}^{{\zeta }_1}⪰{||x||}_{\mathcal{M}}^{{\zeta }_2}$. □

Lemma 3.

Let $(X,A)$ be an IT2FNLS satisfying the condition (xi) and $\{e_1,e_2,\dots ,e_n\}$ be a finite set of linearly independent elements of X. Then, for each $\zeta \in {\mathcal{I}}_0$, there exists a ${\lambda }^{\zeta }\succ \overline{0}$ with $su{p}_{\alpha \in (0,1]}{b}^{\alpha }<\infty$, where ${\left[{\lambda }^{\zeta }\right]}_{\alpha }=[a^{\alpha },b^{\alpha }]$ such that, for every choice of scalars $a_1,a_2,\dots ,a_n$, we have

$||a_1{e}_1+a_2{e}_2+\dots +a_n{e}_n{||}^{\zeta }⪰(|a_1|+|a_2|+\dots +|a_n|){\lambda }^{\zeta }$.

Proof. 

From Theorem 4, it follows that, if $(X,A)$ is an intuitionistic type-2 fuzzy normed linear space satisfying (xi), then ${||x||}^{\zeta }=\wedge \{\eta \succ \overline{0}:\mathcal{N}(x,\eta )⪰\zeta \}$ is a Felbin-type fuzzy norm for each $\zeta \in {\mathcal{I}}_0$. Therefore, by Proposition 2, for each $\zeta \in {\mathcal{I}}_0$, there exists a ${\lambda }^{\zeta }$ such that $||a_1{e}_1+a_2{e}_2+\dots +a_n{e}_n{||}^{\zeta }⪰(|a_1|+|a_2|+\dots +|a_n|){\lambda }^{\zeta }$. □

3.2. Convergence in Intuitionistic Type-2 Fuzzy Normed Linear Space

The idea of the convergence of sequences and some of the basic results related to convergence are studied in this subsection.

Definition 15.

In an IT2FNLS $(X,A)$, a sequence $\left\{x_n\right\}$ is said to be convergent to $x\in X$ if $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$ and is denoted by $li{m}_{n\to \infty }{x}_n=x$.

Theorem 5.

If a sequence $\left\{x_n\right\}$ in an IT2FNLS $(X,A)$ is convergent, its limit is unique.

Proof. 

Let $lim{x}_n=x$ and $lim{x}_n=y$.

Then, $li{m}_{n\to \infty }\mathcal{N}(x_n-x,{\eta }_1)=li{m}_{n\to \infty }\mathcal{N}(x_n-y,{\eta }_2)=\overline{1},\forall {\eta }_1,{\eta }_2\succ \overline{0}$.

Now $\mathcal{N}(x-y,{\eta }_1\oplus {\eta }_2)=\mathcal{N}(x-x_n+x_n-y,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x-x_n,{\eta }_1),\mathcal{N}(x_n-y,{\eta }_2)\}$

i.e, $\mathcal{N}(x-y,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x_n-x,{\eta }_1),\mathcal{N}(x_n-y,{\eta }_2)\}$.

Now

$li{m}_{n\to \infty }\mathcal{N}(x_n-x,{\eta }_1)=\overline{1},\forall {\eta }_1\succ \overline{0}$,

$li{m}_{n\to \infty }\mathcal{N}(x_n-y,{\eta }_2)=\overline{1},\forall {\eta }_2\succ \overline{0}$.

Thus, $li{m}_{n\to \infty }\mathcal{N}(x-y,{\eta }_1\oplus {\eta }_2)=\overline{1},\forall {\eta }_1,{\eta }_2\succ \overline{0}$

$\Rightarrow x-y=0$

$\Rightarrow x=y$. □

Theorem 6.

If $li{m}_{n\to \infty }{x}_n=x$ and $li{m}_{n\to \infty }{y}_n=y$, then $li{m}_{n\to \infty }(x_n+y_n)=x+y$ in an IT2FNLS $(X,A).$

Proof. 

Since $li{m}_{n\to \infty }{x}_n=x$ and $li{m}_{n\to \infty }{y}_n=y$,

$li{m}_{n\to \infty }\mathcal{N}(x_n-x,\frac{\eta }{2})=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\frac{\eta }{2})=\overline{0}$ $\forall \eta \succ \overline{0}$,

$li{m}_{n\to \infty }\mathcal{N}(y_n-y,\frac{\eta }{2})=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(y_n-y,\frac{\eta }{2})=\overline{0}$ $\forall \eta \succ \overline{0}$.

$$\begin{array}{ccc}\hfill \mathcal{N}(x_n+y_n-x-y,\eta )& =& \mathcal{N}(x_n-x+y_n-y,\eta )\hfill \\ & ⪰& min\{\mathcal{N}(x_n-x,\frac{\eta }{2}),\mathcal{N}(y_n-y,\frac{\eta }{2})\}\hfill \end{array}$$

Now, proceeding to the limit as $n\to \infty$, we obtain $li{m}_{n\to \infty }\mathcal{N}(x_n+y_n-x-y,\eta )=\overline{1}$ $\forall \eta \succ \overline{0}$..................(i).

Also,

$$\begin{array}{ccc}\hfill \mathcal{M}(x_n+y_n-x-y,\eta )& =& \mathcal{M}(x_n-x+y_n-y,\eta )\hfill \\ & ⪯& max\{\mathcal{M}(x_n-x,\frac{\eta }{2}),\mathcal{M}(y_n-y,\frac{\eta }{2})\}\hfill \end{array}$$

Now, proceeding to the limit as $n\to \infty$, we obtain $li{m}_{n\to \infty }\mathcal{M}(x_n+y_n-x-y,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$...................(ii).

Combining (i) and (ii), we obtain $li{m}_{n\to \infty }(x_n+y_n)=x+y$. □

Theorem 7.

If $li{m}_{n\to \infty }{x}_n=x$ and $c(\ne 0)\in \mathbb{R}$, then $li{m}_{n\to \infty }c{x}_n=cx$ in an IT2FNLS $(X,A)$.

Proof. 

Since $li{m}_{n\to \infty }{x}_n=x$,

we have $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.

Now $\mathcal{N}(c{x}_n-cx,\eta )=\mathcal{N}(x_n-x,\frac{1}{|c|}\eta )$

and so $li{m}_{n\to \infty }\mathcal{N}(c{x}_n-cx,\eta )=li{m}_{n\to \infty }\mathcal{N}(x_n-x,\frac{1}{|c|}\eta )=\overline{1}$ $\forall \eta \succ \overline{0}$.

Proceeding similarly, we obtain $li{m}_{n\to \infty }\mathcal{M}(c{x}_n-cx,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.

Thus, we have $li{m}_{n\to \infty }c{x}_n=cx$. □

Theorem 8.

In an IT2FNLS $(X,A)$, every subsequence of a convergent sequence is convergent and converges to the same limit.

Proof. 

Let $\left\{x_n\right\}$ be a convergent sequence in $(X,A)$ with $li{m}_{n\to \infty }{x}_n=x$ and $\left\{x_{n_r}\right\}$ be a subsequence of $\left\{x_n\right\}$.

Then, $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.................(i)

Now, as $r\to \infty$, then $n_r\to \infty$, and, from (i), we easily obtain that

$li{m}_{n\to \infty }\mathcal{N}(x_{n_r}-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_{n_r}-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.

Thus, we see that the sequence $\left\{x_{n_r}\right\}$ is convergent, and $li{m}_{r\to \infty }{x}_{n_r}=x$. □

Definition 16.

A sequence $\left\{x_n\right\}$ in an IT2FNLS $(X,A)$ is said to be a Cauchy sequence if $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_{n+p}-x_n,\eta )=\overline{0}$, $p=1,2,3,\dots \dots$ and $\forall \eta \succ \overline{0}$.

Theorem 9.

In an IT2FNLS $(X,A)$, every convergent sequence is a Cauchy sequence.

Proof. 

Suppose $\left\{x_n\right\}$ is convergent and $lim{x}_n=x$.

Then $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.

Now $\mathcal{N}(x_{n+p}-x_n,{\eta }_1\oplus {\eta }_2)=\mathcal{N}(x_{n+p}-x+x-x_n,{\eta }_1\oplus {\eta }_2)$.

We also have $\mathcal{N}(x_{n+p}-x+x-x_n,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x_{n+p}-x,{\eta }_1),\mathcal{N}(x-x_n,{\eta }_2)\}$

i.e, $\mathcal{N}(x_{n+p}-x_n,{\eta }_1\oplus {\eta }_2)⪰min\{\mathcal{N}(x_{n+p}-x,{\eta }_1),\mathcal{N}(x_n-x,{\eta }_2)\}$.

Now, $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x,{\eta }_1)=li{m}_{n\to \infty }\mathcal{N}(x_n-x,{\eta }_2)=\overline{1},\forall {\eta }_1,{\eta }_2\succ \overline{0}$, $p=1,2,3,\dots$

So, $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,{\eta }_1\oplus {\eta }_2)=\overline{1},\forall {\eta }_1,{\eta }_2\succ \overline{0}$, $p=1,2,3,\dots$

Proceeding similarly, we can prove that $li{m}_{n\to \infty }\mathcal{M}(x_{n+p}-x_n,{\eta }_1\oplus {\eta }_2)=\overline{0},\forall {\eta }_1,{\eta }_2\succ \overline{0}$, $p=1,2,3,\dots$

Hence, $\left\{x_n\right\}$ is a Cauchy sequence in $(X,A)$. □

3.3. Completeness and Finite Dimensionality in Intuitionistic Type-2 Fuzzy Normed Linear Space

Basic properties related to the completeness, boundedness, compactness and finite dimensionality in the IT2FNLS are studied in this subsection.

Definition 17.

Let $(X,A)$ be an IT2FNLS. If every Cauchy sequence in $(X,A)$ is convergent, then we call $(X,A)$ to be complete.

Theorem 10.

Let $(X,A)$ be an IT2FNLS, such that every Cauchy sequence in $(X,A)$ has a convergent subsequence. Then, $(X,A)$ is complete.

Proof. 

Let $\left\{x_n\right\}$ be a Cauchy sequence in $(X,A)$ and $\left\{x_{n_r}\right\}$ be a convergent subsequence of $\left\{x_n\right\}$ with $li{m}_{r\to \infty }{x}_{n_r}=x$.

Then $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,\frac{\eta }{2})=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_{n+p}-x_n,\frac{\eta }{2})=\overline{0}$, $p=1,2,3,\dots \dots$ and $\forall \eta \succ \overline{0}$.

Also, $li{m}_{r\to \infty }\mathcal{N}(x_{n_r}-x,\frac{\eta }{2})=\overline{1}$ and $li{m}_{r\to \infty }\mathcal{M}(x_{n_r}-x,\frac{\eta }{2})=\overline{0}$ $\forall \eta \succ \overline{0}$.

Now

$$\begin{array}{ccc}\hfill \mathcal{N}(x_n-x,\eta )& =& \mathcal{N}(x_n-x_{n_r}+x_{n_r}-x,\eta )\hfill \\ & ⪰& min\{\mathcal{N}(x_n-x_{n_r},\frac{\eta }{2}),\mathcal{N}(x_{n_r}-x,\frac{\eta }{2})\}\hfill \end{array}$$

Then, proceeding to the limit as $n\to \infty$, we obtain $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ $\forall \eta \succ \overline{0}$...................(i).

Similarly, going by the previous approach, we can prove that $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$......................(ii).

Combining (i) and (ii), we have the sequence $\left\{x_n\right\}$, which is convergent, and $li{m}_{n\to \infty }{x}_n=x$.

Thus, we see that every Cauchy sequence in $(X,A)$ is convergent, and so $(X,A)$ is complete. □

Theorem 11.

Every finite- dimensional IT2FNLS satisfying the conditions (xi) and (xii) is complete.

Proof. 

Let $(X,A)$ be an IT2FNLS and dim $X=k$.

Let $\{e_1,e_2,\dots ,e_k\}$ be a basis of X and $\left\{x_n\right\}$ be a Cauchy sequence in X.

Let $x_n={\beta }_1^n{e}_1+{\beta }_2^n{e}_2+\dots +{\beta }_k^n{e}_k$, ${\beta }_i^n\in \mathbb{R}$, $i=1,2,\dots k$, $n\in \mathbb{N}$.

Now we have $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,\eta )=\overline{1},\forall \eta \succ \overline{0}$, $p=1,2,3,\dots$

$\Rightarrow li{m}_{n\to \infty }\mathcal{N}({\sum }_{i=1}^k{\beta }_i^{n+p}{e}_i-{\sum }_{i=1}^k{\beta }_i^n{e}_i,\eta )=\overline{1}$

$\Rightarrow li{m}_{n\to \infty }\mathcal{N}({\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i^n)e_i,\eta )=\overline{1},\forall \eta \succ \overline{0}$, $p=1,2,3,\dots$

Choose $\zeta \in {\mathcal{I}}_0$ and $\eta \succ \overline{0}$.

Then there exists a positive integer $n_0(\eta ,\zeta )$ such that

$\mathcal{N}({\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i)e_i,\eta )\succ \zeta ,\forall n\ge n_0(\eta ,\zeta )$, $p=1,2,3,\dots$

Now $||{\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i)e_i{||}^{\zeta }=\wedge \{\xi \succ \overline{0}:\mathcal{N}({\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i)e_i,\xi )⪰\zeta \}$.

Therefore, $||{\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i)e_i{||}^{\zeta }⪯\eta ,\forall n\ge n_0(\eta ,\zeta )$.

Since $\eta \succ \overline{0}$ is arbitary,

$||{\sum }_{i=1}^k({\beta }_i^{n+p}-{\beta }_i^n)e_i{||}^{\zeta }\to \overline{0}$ as $n\to \infty$ for each $\zeta \in {\mathcal{I}}_0$, $p=1,2,3,\dots$

$\Rightarrow ({\sum }_{i=1}^k|{\beta }_i^{n+p}-{\beta }_i^n|){\lambda }^{\zeta }\to \overline{0}$ as $n\to \infty$ where ${\lambda }^{\zeta }\succ \overline{0}$, by the Lemma 3,

$\Rightarrow {\sum }_{i=1}^k|{\beta }_i^{n+p}-{\beta }_i^n|\to 0$ as $n\to \infty$, since ${\lambda }^{\zeta }\succ \overline{0}$ for all $\zeta \in {\mathcal{I}}_0$

$\Rightarrow {\left\{{\beta }_i^n\right\}}_n$ is a Cauchy sequence in $\mathbb{R}$ for each $i=1,2,\dots ,k$.

Since $\mathbb{R}$ is complete, ${\left\{{\beta }_i^n\right\}}_n$ converges for each $i=1,2,\dots k$.

Let $li{m}_{n\to \infty }{\beta }_i^n={\beta }_i$ for $i=1,2,\dots k$ and $x={\sum }_{i=1}^k{\beta }_i{e}_i$.

Then, $x\in X$.

Now, for all $\eta \succ \overline{0}$,

$$\begin{array}{ccc}\hfill \mathcal{N}(x_n-x,\eta )& =& \mathcal{N}(\sum _{i=1}^k{\beta }_i^n{e}_i-\sum _{i=1}^k{\beta }_i{e}_i,\eta )\hfill \\ & =& \mathcal{N}(\sum _{i=1}^k({\beta }_i^n-{\beta }_i)e_i,\eta )\hfill \\ & ⪰& mi{n}_i\mathcal{N}(e_i,\frac{1}{k|{\beta }_i^n-{\beta }_i|}\eta ).....................(i)\hfill \end{array}$$

Let $u_n=\frac{1}{k|{\beta }_i^n-{\beta }_i|}$. Then $u_n\to \infty$ as $n\to \infty$ [Since $|{\beta }_i^n-{\beta }_i|\to 0+$ as $n\to \infty$ for $i=1,2,\dots k$].

So $u_n\eta \to \overline{\infty }$ as $n\to \infty$, by Lemma 2.

Hence, we have $li{m}_{n\to \infty }\mathcal{N}(e_i,\frac{1}{k|{\beta }_i^n-{\beta }_i|}\eta )=li{m}_{n\to \infty }\mathcal{N}(e_i,u_n\eta )=\overline{1}$ for $i=1,2,$…k ....................(ii).

From (i) and (ii), we obtain $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1},\forall \eta \succ \overline{0}$.

Again, for all $\eta \succ \overline{0}$,

$$\begin{array}{ccc}\hfill \mathcal{M}(x_n-x,\eta )& =& \mathcal{M}(\sum _{i=1}^k{\beta }_i^n{e}_i-\sum _{i=1}^k{\beta }_i{e}_i,\eta )\hfill \\ & =& \mathcal{M}(\sum _{i=1}^k({\beta }_i^n-{\beta }_i)e_i,\eta )\hfill \\ & ⪯& ma{x}_i\mathcal{M}(e_i,\frac{1}{k|{\beta }_i^n-{\beta }_i|}\eta )......................(iii)\hfill \end{array}$$

Now $li{m}_{n\to \infty }\mathcal{M}(e_i,\frac{1}{k|{\beta }_i^n-{\beta }_i|}\eta )=li{m}_{n\to \infty }\mathcal{M}(e_i,u_n\eta )=\overline{0}$ for $i=1,2,$…k ....................(iv).

From (iii) and (iv), we obtain $li{m}_{n\to \infty }\mathcal{M}(x_n-x,\eta )=\overline{0},\forall \eta \succ \overline{0}$.

Hence, lim$x_n=x$, i.e, the sequence $\left\{x_n\right\}$ is convergent.

Thus, we see that the IT2FNLS $(X,A)$ is complete. □

Definition 18.

A subset U in an IT2FNLS $(X,A)$ is said to be bounded if, for any ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$ with ${\zeta }_1\oplus {\zeta }_2⪯\overline{1}$, $\exists \eta \succ \overline{0}$ such that $\mathcal{N}(x,\eta )\succ {\zeta }_1$ and $\mathcal{M}(x,\eta )\prec {\zeta }_2,\forall x\in U$.

Theorem 12.

Let $(X,A)$ be an IT2FNLS. Then, every Cauchy sequence in $(X,A)$ is bounded.

Proof. 

Let us consider a Cauchy sequence $\left\{x_n\right\}$ in an IT2FNLS $(X,A)$.

Then $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,\eta )=\overline{1}$ and $\mathcal{M}(x_{n+P}-x_n,\eta )=\overline{0},\forall \eta \succ \overline{0}$ and $p=1,2,3,\dots$

Let ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$ with ${\zeta }_1\oplus {\zeta }_2⪯\overline{1}$.

Then we have $li{m}_{n\to \infty }\mathcal{N}(x_{n+p}-x_n,\eta )=li{m}_{n\to \infty }\mathcal{N}(x_n-x_{n+p},\eta )=\overline{1}\succ {\zeta }_1,\forall \eta \succ \overline{0}$ and $p=1,2,3,..$.

For ${\eta }^{\prime }\succ \overline{0},\exists n_0=n_0({\eta }^{\prime },{\zeta }_1)$ such that $\mathcal{N}(x_n-x_{n+p},{\eta }^{\prime })\succ {\zeta }_1,\forall n\ge n_0$, $p=1,2,3,\dots$

Since $li{m}_{\eta \to \overline{\infty }}\mathcal{N}(x,\eta )=\overline{1},\exists {\eta }_i\succ \overline{0}$ such that $\mathcal{N}(x_i,{\eta }_i)\succ {\zeta }_1$, for all $i=1,2,3,\dots n_0$.

Let ${\eta }_0={\eta }^{\prime }\oplus {\eta }_1\oplus {\eta }_2\oplus \dots \oplus {\eta }_{n_0}$.

Then $\mathcal{N}(x_n,{\eta }_0)⪰\mathcal{N}(x_n,{\eta }^{\prime }\oplus {\eta }_{n_0})=\mathcal{N}(x_n-x_{n_0}+x_{n_0},{\eta }^{\prime }\oplus {\eta }_{n_0})⪰min\{\mathcal{N}(x_n-x_{n_0},{\eta }^{\prime }),\mathcal{N}(x_{n_0},{\eta }_{n_0})\}.$

Therefore, $\mathcal{N}(x_n,{\eta }_0)⪰{\zeta }_1,\forall n>n_0$.

Also, $\mathcal{N}(x_n,{\eta }_0)⪰\mathcal{N}(x_n,{\eta }_n)⪰{\zeta }_1,\forall n=1,2,\dots n_0$.

Hence, $\mathcal{N}(x_n,{\eta }_0)⪰{\zeta }_1,\forall n$..........................(i)

Also, we have $li{m}_{n\to \infty }\mathcal{M}(x_{n+p}-x_n,\eta )=li{m}_{n\to \infty }\mathcal{M}(x_n-x_{n+p},\eta )=\overline{0}\prec {\zeta }_2,\forall \eta \succ \overline{0}$ and $p=1,2,3,..$.

For ${\eta }_0^{\prime }\succ \overline{0},\exists n_0^{\prime }=n_0^{\prime }({\eta }_0^{\prime },{\zeta }_2)$ such that $\mathcal{M}(x_n-x_{n+p},{\eta }_0^{\prime })\prec {\zeta }_2,\forall n\ge n_0^{\prime }$, $p=1,2,3,\dots$

Since $li{m}_{\eta \to \overline{\infty }}\mathcal{M}(x,\eta )=\overline{0},\exists {\eta }_i^{\prime }\succ \overline{0}$ such that $\mathcal{M}(x_i,{\eta }_i^{\prime })\prec {\zeta }_2$, for all $i=1,2,3,\dots n_0^{\prime }$.

Let ${\eta }_1={\eta }_0^{\prime }\oplus {\eta }_1^{\prime }\oplus {\eta }_2^{\prime }\oplus \dots \oplus {\eta }_{n_0^{\prime }}^{\prime }$.

Then $\mathcal{M}(x_n,{\eta }_1)⪯\mathcal{M}(x_n,{\eta }_0^{\prime }\oplus {\eta }_{n_0^{\prime }}^{\prime })=\mathcal{M}(x_n-x_{n_0^{\prime }}+x_{n_0^{\prime }},{\eta }_0^{\prime }\oplus {\eta }_{n_0^{\prime }}^{\prime })⪯max\{\mathcal{M}(x_n-x_{n_0^{\prime }},{\eta }_0^{\prime }),\mathcal{M}(x_{n_0^{\prime }},{\eta }_{n_0^{\prime }}^{\prime })\}.$

Therefore, $\mathcal{M}(x_n,{\eta }_1)⪯{\zeta }_2,\forall n>n_0^{\prime }$.

Also, $\mathcal{M}(x_n,{\eta }_1)⪯\mathcal{M}(x_n,{\eta }_n^{\prime })⪯{\zeta }_2,\forall n=1,2,\dots n_0^{\prime }$.

Hence, $\mathcal{M}(x_n,{\eta }_1)⪯{\zeta }_2,\forall n$...........................(ii)

Let ${\eta }_2={\eta }_0\oplus {\eta }_1$.

Then, from (i) and (ii), we obtain $\mathcal{N}(x_n,{\eta }_2)⪰\mathcal{N}(x_n,{\eta }_0)⪰{\zeta }_1$ and $\mathcal{M}(x_n,{\eta }_2)⪯\mathcal{M}(x_n,{\eta }_1)⪯{\zeta }_2$ $\forall n$, i.e, $\mathcal{N}(x_n,{\eta }_2)⪰{\zeta }_1$ and $\mathcal{M}(x_n,{\eta }_2)⪯{\zeta }_2$ $\forall n$ with ${\zeta }_1\oplus {\zeta }_2⪯\overline{1}$.

This implies that $\left\{x_n\right\}$ is bounded in $(X,A)$. □

Remark 2.

The converse of the above theorem is not true in general, which follows from the following example.

Example 2.

Let $(X=\mathbb{R},|.|)$ be the usual real normed linear space and $||.||$ be the usual Felbin-type fuzzy norm on $\mathbb{R}$, i.e,

$$||x||\left(t\right)=\left\{\begin{array}{c}1,ift=|x|\hfill \\ 0,otherwise\hfill \end{array}\right..$$

Define

$$\mathcal{N}(x,\eta )=\left\{\begin{array}{c}\overline{1}if||x||\prec \eta \hfill \\ \overline{0}otherwise\hfill \end{array}\right.$$

$$\mathcal{M}(x,\eta )=\left\{\begin{array}{c}\overline{0}if||x||\prec \eta \hfill \\ \overline{1}otherwise\hfill \end{array}\right..$$

Then $(X,A)$ is an intuitionistic type-2 fuzzy normed linear space. Take the sequence $\left\{x_n\right\}$ where $x_n={(-1)}^{n+1},\forall n\in \mathbb{N}$. Then, we see that the sequence $\left\{x_n\right\}$ is bounded but not Cauchy.

In fact, for the above sequence, we have $||x_n||=\overline{1}$ $\forall n\in \mathbb{N}$.

Now choose any ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$ such that ${\zeta }_1\oplus {\zeta }_2⪯\overline{1}$.

Then, if we take any ${\eta }_0\succ \overline{1}$, then $\mathcal{N}(x_n,{\eta }_0)=\overline{1}\succ {\zeta }_1$ and also $\mathcal{M}(x_n,{\eta }_0)=\overline{0}\prec {\zeta }_2$.

Thus, the sequence $\left\{x_n\right\}$ is bounded.

Also, for the given sequence, we see that $||x_{n+p}-x_n||=\overline{0}$ or $\overline{2}$ for any $n\in \mathbb{N}$, and $p=1,2,3,\dots$

If $\eta \succ \overline{2}$, then $\mathcal{N}(x_{n+1}-x_n,\eta )=\overline{1}$ $\forall n\in \mathbb{N}$, and, if $\eta \prec \overline{1}$, then $\mathcal{N}(x_{n+1}-x_n,\eta )=\overline{0}$ $\forall n\in \mathbb{N}$.

Hence, the sequence $\left\{x_n\right\}$ is not Cauchy.

Definition 19.

Let $(X,A)$ be an IT2FNLS. A subset V of X is said to be closed if any sequence $\left\{x_n\right\}$ in V converges to $x\in V$ that is $li{m}_{n\to \infty }\mathcal{N}(x_n-x,\eta )=\overline{1}$ and $\mathcal{M}(x_n-x,\eta )=\overline{0}$ $\forall \eta \succ \overline{0}\Rightarrow$$x\in V$.

Definition 20.

Let $(X,A)$ be an IT2FNLS. A subset U of X is said to be compact if any sequence $\left\{x_n\right\}$ in U has a subsequence converging to an element of U.

Theorem 13.

Let $(X,A)$ be a finite-dimensional IT2FNLS satisfying the conditions (xi) and (xii) and $U\subset X$. Then U is compact if it is closed and bounded.

Proof. 

Let dim$X=k$ and $\{e_1,e_2,\dots ,e_k\}$ be a basis of X.

We take a sequence $\left\{x_n\right\}$ in U and let $x_n={\beta }_1^n{e}_1+{\beta }_2^n{e}_2+\dots +{\beta }_k^n{e}_k$, where ${\beta }_i^n\in \mathbb{R}$, $i=1,2,\dots k$.

Since U is bounded, $\left\{x_n\right\}$ is bounded. Then, for each ${\zeta }_1,{\zeta }_2\in {\mathcal{I}}_0$ with ${\zeta }_1\oplus {\zeta }_2⪯\overline{1}$, $\exists {\eta }_1\succ \overline{0}$ such that $\mathcal{N}(x_n,{\eta }_1)\succ {\zeta }_1$ and $\mathcal{M}(x_n,{\eta }_1)\prec {\zeta }_2$ for all $n\in \mathbb{N}$......................(i).

Now $||x_n{||}^{{\zeta }_1}=\wedge \{\eta \succ \overline{0}:\mathcal{N}(x_n,\eta )⪰{\zeta }_1\}$.

Then, from (i), we obtain $||x_n{||}^{{\zeta }_1}⪯{\eta }_1$................(ii).

Since $\{e_1,e_2,\dots ,e_k\}$ is a linearly independent set, by the Lemma 3, there exists a ${\lambda }^{{\zeta }_1}\succ \overline{0}$ such that $||x_n{||}^{{\zeta }_1}=||{\sum }_{i=1}^k{\beta }_i^n{e}_i{||}^{{\zeta }_1}⪰({\sum }_{i=1}^k|{\beta }_i^n|){\lambda }^{{\zeta }_1}$ (n = 1,2,…)...............(iii).

From (ii) and (iii), we obtain $({\sum }_{i=1}^k|{\beta }_i^n|){\lambda }^{{\zeta }_1}⪯{\eta }_1$.

Now, if we take ${\left[{\lambda }^{{\zeta }_1}\right]}_{\alpha }=[a^{\alpha },b^{\alpha }]$ and ${\left[{\eta }_1\right]}_{\alpha }=[a_1^{\alpha },b_1^{\alpha }]$ for $0<\alpha \le 1$, then $({\sum }_{i=1}^k|{\beta }_i^n|)a^{\alpha }\le a_1^{\alpha }$ and $({\sum }_{i=1}^k|{\beta }_i^n|)b^{\alpha }\le b_1^{\alpha }$.

Then $({\sum }_{i=1}^k|{\beta }_i^n|)\le \frac{a_1^{\alpha }}{a^{\alpha }}$ for n = 1,2,…, and so, for each $i=1,2,\dots k$, ${\left\{{\beta }_i^n\right\}}_n$ is a bounded sequence of real numbers.

Now ${\left\{{\beta }_1^n\right\}}_n$ is bounded, and so, by the Bolzano–Weierstrass theorem, it has a convergent subsequence. Let ${\beta }_1$ denote the limit of that subsequence and let ${\left\{x_{(1,n)}\right\}}_n$ denote the corresponding subsequence of $\left\{x_n\right\}$. By the same argument, $\left\{x_{(1,n)}\right\}$ has a subsequence $\left\{x_{(2,n)}\right\}$ for which the corresponding subsequence of real ${\beta }_2^n$ converges. Let ${\beta }_2$ denote the limit of that subsequence. Continuing in this way after k steps, we obtain a sequence ${\left\{x_{(k,n)}\right\}}_n=\{x_{(k,1)},x_{(k,2)},\dots \}$ of $\left\{x_n\right\}$ whose elements are of the form $x_{(k,n)}={\sum }_{i=1}^k{\gamma }_i^n{e}_i$ with scalars ${\gamma }_i^n$ satisfying ${\gamma }_i^n\to {\beta }_i$ as $n\to \infty$ for $i=1,2,\dots ,k$.

Let $x={\sum }_{i=1}^k{\beta }_i{e}_i$. Then, $x\in X$.

Now, for $\eta \succ \overline{0}$, we have

$$\begin{array}{ccc}\hfill \mathcal{N}(x_{(k,n)}-x,\eta )& =& \mathcal{N}(\sum _{i=1}^k({\gamma }_i^n-{\beta }_i)e_i,\eta )\hfill \\ & ⪰& mi{n}_i\mathcal{N}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )\hfill \end{array}$$

$\Rightarrow li{m}_{n\to \infty }\mathcal{N}(x_{(k,n)}-x,\eta )⪰mi{n}_{i}li{m}_{n\to \infty }\mathcal{N}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )$.....................(iv).

Let $u_n=\frac{1}{k|{\gamma }_i^n-{\beta }_i|}$.

Then $u_n\to \infty$ as $n\to \infty$, since $|{\gamma }_i^n-{\beta }_i|\to 0+$, when $n\to \infty$.

So $u_n\eta \to \overline{\infty }$ as $n\to \infty$ by Lemma 2.

Then we have $li{m}_{n\to \infty }\mathcal{N}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )=li{m}_{n\to \infty }\mathcal{N}(e_i,u_n\eta )=\overline{1}$...............(v).

Combining (iv) and (v), we obtain $li{m}_{n\to \infty }\mathcal{N}(x_{(k,n)}-x,\eta )=\overline{1},\forall \eta \succ \overline{0}$......................(vi).

Again, for all $\eta \succ \overline{0}$,

$$\begin{array}{ccc}\hfill \mathcal{M}(x_{(k,n)}-x,\eta )& =& \mathcal{M}(\sum _{i=1}^k({\gamma }_i^n-{\beta }_i)e_i,\eta )\hfill \\ & ⪯& ma{x}_i\mathcal{M}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )\hfill \end{array}$$

$\Rightarrow li{m}_{n\to \infty }\mathcal{M}(x_{(k,n)}-x,\eta )⪯ma{x}_{i}li{m}_{n\to \infty }\mathcal{N}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )$.....................(vii)

Let $u_n=\frac{1}{k|{\gamma }_i^n-{\beta }_i|}$.

Then, $u_n\to \infty$ as $n\to \infty$, since $|{\gamma }_i^n-{\beta }_i|\to 0+$, when $n\to \infty$.

So $u_n\eta \to \overline{\infty }$ as $n\to \infty$ by Lemma 2.

Then we have $li{m}_{n\to \infty }\mathcal{M}(e_i,\frac{1}{k|{\gamma }_i^n-{\beta }_i|}\eta )=li{m}_{n\to \infty }\mathcal{M}(e_i,u_n\eta )=\overline{0}$...............(viii).

Combining (vi) and (vii), we obtain $li{m}_{n\to \infty }\mathcal{M}(x_{(k,n)}-x,\eta )=\overline{0},\forall \eta \succ \overline{0}$.................(ix).

Combining (vi) and (xi), we obtain $li{m}_{n\to \infty }\mathcal{N}(x_{(k,n)}-x,\eta )=\overline{1}$ and $li{m}_{n\to \infty }\mathcal{M}(x_{(k,n)}-x,\eta )=\overline{0},\forall \eta \succ \overline{0}$

$\Rightarrow li{m}_{n\to \infty }{x}_{(k,n)}=x$, i.e, ${\left\{x_{(k,n)}\right\}}_n$ is a convergent subsequence of ${\left\{x_n\right\}}_n$ which converges to x. Since U is closed, $x\in U$. Hence, every sequence in U has a subsequence which converges in U. Thus, U is compact. □

4. Intuitionistic Fuzzy Continuous Functions in Intuitionistic Type-2 Fuzzy Normed Linear Space

We know that, in classical normed linear space, continuity of function at a point can be characterised by the convergence of the sequence at that point. But here we discover that, in an IT2FNLS setting, the sequential criterion for continuity holds partially. For this, we introduce two types of continuous functions, namely, intuitionistic type-2 fuzzy continuous and sequentially intuitionistic type-2 fuzzy continuous functions. Here, we show that every intuitionistic type-2 fuzzy continuous function is sequentially intuitionistic type-2 fuzzy continuous, but its converse is not true, which is justified by a counterexample.

Definition 21.

Let $(X,A)$ and $(Y,B)$ be two IT2FNLSs. A mapping $T:X\to Y$ is said to be intuitionistic type-2 fuzzy continuous (or, in short, IT2FC) at $x_0\in X$ if, for each ${\eta }_1\succ \overline{0}$, $\exists {\eta }_2\succ \overline{0}$ such that $\forall x\in X$,

${\mathcal{N}}_B(T\left(x\right)-T\left(x_0\right),{\eta }_1)⪰{\mathcal{N}}_A(x-x_0,{\eta }_2)$ and

${\mathcal{M}}_B(T\left(x\right)-T\left(x_0\right),{\eta }_1)⪯{\mathcal{M}}_A(x-x_0,{\eta }_2)$.

T is said to be IT2FC on X if T is IT2FC at each point of X.

Definition 22.

Let $(X,A)$ and $(Y,B)$ be two IT2FNLSs. A mapping $T:X\to Y$ is said to be sequentially intuitionistic type-2 fuzzy continuous (or, in short, sequentially IT2FC) at $x_0\in X$ if any sequence $\left\{x_n\right\}$ in X with $x_n\to x_0$ implies $T{x}_n\to T{x}_0$, i.e, $li{m}_{n\to \infty }{\mathcal{N}}_A(x_n-x_0,\eta )=\overline{1}$ and $li{m}_{n\to \infty }{\mathcal{M}}_A(x_n-x_0,\eta )=\overline{0}$, $\forall \eta \succ \overline{0}$

$\Rightarrow li{m}_{n\to \infty }{\mathcal{N}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{1}$ and $li{m}_{n\to \infty }{\mathcal{M}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{0}$, $\forall \eta \succ \overline{0}$.

If T is sequentially IT2FC at each point of X, then T is said to be sequentially IT2FC on X.

Theorem 14.

Let $(X,A)$ and $(Y,B)$ be two IT2FNLSs and $T:X\to Y$ be a mapping. If T is IT2FC on X, then it is sequentially IT2FC on X.

Proof. 

Let $x_0$ be an arbitrary point of X and T be IT2FC on X. Then, for an ${\eta }_1\succ \overline{0}$, $\exists {\eta }_2\succ \overline{0}$ such that $\forall x\in X$

${\mathcal{N}}_B(T\left(x\right)-T\left(x_0\right),{\eta }_1)⪰{\mathcal{N}}_A(x-x_0,{\eta }_2)$..........................(i)

and ${\mathcal{M}}_B(T\left(x\right)-T\left(x_0\right),{\eta }_1)⪯{\mathcal{M}}_A(x-x_0,{\eta }_2)$.........................(ii).

Let $\left\{x_n\right\}$ be a sequence in X such that $x_n\to x_0$, that is, for all $\eta \succ \overline{0}$, $li{m}_{n\to \infty }{\mathcal{N}}_A(x_n-x_0,\eta )=\overline{1}$ and $li{m}_{n\to \infty }{\mathcal{M}}_A(x_n-x_0,\eta )=\overline{0}$.

Then, for all $\eta \succ \overline{0}$ from (i) and (ii), we obtain $li{m}_{n\to \infty }{\mathcal{N}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{1}$ and $li{m}_{n\to \infty }{\mathcal{M}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{0}$, that is, $T\left(x_n\right)\to T\left(x_0\right)$ in $(Y,B)$, and so T is sequentially IT2FC on X. □

The following example will show that a sequentially IT2FC T does not guarantee the IT2FC of T.

Example 3.

Let $(X=\mathbb{R},|.|)$ be a normed linear space. For any fuzzy real number $\eta \succ \overline{0}$, define ${\eta }_0$ as follows:

$${\eta }_0=\overline{\frac{a+b}{2}}$$

where $[a,b]$ denotes the closure of support of the fuzzy real number η. Let $k,k_1>0$ be any fixed real number. Define

$${\mathcal{N}}_A(x,\eta )=\left\{\begin{array}{cc}{\eta }_0\oslash ({\eta }_0\oplus \overline{k|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{0},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

$${\mathcal{M}}_A(x,\eta )=\left\{\begin{array}{cc}\overline{k|x|}\oslash ({\eta }_0\oplus \overline{k|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{1},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

and

$${\mathcal{N}}_B(x,\eta )=\left\{\begin{array}{cc}{\eta }_0\oslash ({\eta }_0\oplus \overline{k_1|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{0},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

$${\mathcal{M}}_B(x,\eta )=\left\{\begin{array}{cc}\overline{k_1|x|}\oslash ({\eta }_0\oplus \overline{k_1|x|}),\hfill & if\eta \succ \overline{0}\hfill \\ \overline{1},\hfill & if\eta =\overline{0}\hfill \end{array}\right.$$

Then, $(X=\mathbb{R},A)$ and $(Y=\mathbb{R},B)$ are two IT2FNLSs.

We define a function $T:X\to Y$ by $T\left(x\right)=\frac{x^4}{1+x^2}$. Then, T is sequentially IT2FC but not IT2FC.

In fact, by Example 1, we can see that both $(X,A)$ and $(Y,B)$ are two ITFNLSs.

We choose a sequence $\left\{x_n\right\}$, $x_n\in X$ such that $x_n\to x_0$.

Now $\forall \eta \succ \overline{0}$, we have

$li{m}_{n\to \infty }{\mathcal{N}}_A(x_n-x_0,\eta )=\overline{1}$ and $li{m}_{n\to \infty }{\mathcal{M}}_A(x_n-x_0,\eta )=\overline{0}$

$\Rightarrow li{m}_{n\to \infty }\{(\overline{\frac{a+b}{2}})\oslash (\overline{\frac{a+b}{2}}\oplus \overline{|x_n-x_0|})\}=\overline{1}$ and $li{m}_{n\to \infty }\{(\overline{k|x_n-x_0|})\oslash (\overline{\frac{a+b}{2}}\oplus \overline{k|x_n-x_0|})\}=\overline{0}$

$\Rightarrow li{m}_{n\to \infty }\frac{\frac{a+b}{2}}{\frac{a+b}{2}+|x_n-x_0|}=1$ and $li{m}_{n\to \infty }\frac{k|x_n-x_0|}{\frac{a+b}{2}+k|x_n-x_0|}=0$

$\Rightarrow li{m}_{n\to \infty }|x_n-x_0|=0$........................(i).

Now ${\mathcal{N}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\{(\overline{\frac{a+b}{2}})\oslash (\overline{\frac{a+b}{2}}\oplus \overline{k_1|\frac{x_n^4}{1+x_n^2}-\frac{x_0^4}{1+x_0^2}}|)\}$

Now $li{m}_{n\to \infty }(k_1|\frac{x_n^4}{1+x_n^2}-\frac{x_0^4}{1+x_0^2}|)=0$ by using (i)

$\Rightarrow li{m}_{n\to \infty }{\mathcal{N}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{1}$ $\forall \eta \succ \overline{0}$.

Also, $\forall \eta \succ \overline{0}$

$li{m}_{n\to \infty }{\mathcal{M}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\{(\overline{k_1|\frac{x_n^4}{1+x_n^2}-\frac{x_0^4}{1+x_0^2}|})\oslash (\overline{\frac{a+b}{2}}\oplus \overline{k_1|\frac{x_n^4}{1+x_n^2}-\frac{x_0^4}{1+x_0^2}|})\}$

Now $li{m}_{n\to \infty }(k_1|\frac{x_n^4}{1+x_n^2}-\frac{x_0^4}{1+x_0^2}|)=0$ by using (i)

$\Rightarrow li{m}_{n\to \infty }{\mathcal{M}}_B(T\left(x_n\right)-T\left(x_0\right),\eta )=\overline{0}$ $\forall \eta \succ \overline{0}$.

Thus, $T:(X,A)\to (Y,B)$ is sequentially IT2FC.

Let ${\eta }_2\succ \overline{0}$ be given. Then, for any ${\eta }_1\succ \overline{0}$,

${\mathcal{N}}_B(T\left(x\right)-T\left(x_0\right),{\eta }_2)⪰{\mathcal{N}}_A(x-x_0,{\eta }_1)$ $\forall x\in X$

$\Rightarrow {\mathcal{N}}_B(\frac{x^4}{1+x^2}-\frac{x_0^4}{1+x_0^2},{\eta }_2)⪰{\mathcal{N}}_A(x-x_0,{\eta }_1)$

$\Rightarrow \{(\overline{\frac{a_2+b_2}{2}})\oslash (\overline{\frac{a_2+b_2}{2}}\oplus \overline{k_1|\frac{x^4}{1+x^2}-\frac{x_0^4}{1+x_0^2}}|)\}⪰\{(\overline{\frac{a_1+b_1}{2}})\oslash (\overline{\frac{a_1+b_1}{2}}\oplus \overline{k|x-x_0|})\}$ [where $[a_1,b_1]$, $[a_2,b_2]$ denotes, respectively, the closure of support of the fuzzy real numbers ${\eta }_1$, ${\eta }_2$]

$\Rightarrow \frac{\frac{a_2+b_2}{2}}{\frac{a_2+b_2}{2}+k_1|\frac{x^4}{1+x^2}-\frac{x_0^4}{1+x_0^2}|}\ge \frac{\frac{a_1+b_1}{2}}{\frac{a_1+b_1}{2}+k|x-x_0|}$

$\Rightarrow \frac{r_2}{r_2+k_1|\frac{x^4}{1+x^2}-\frac{x_0^4}{1+x_0^2}|}\ge \frac{r_1}{r_1+k|x-x_0|}$ where $r_1=\frac{a_1+b_1}{2}$ and $r_2=\frac{a_2+b_2}{2}$

$\Rightarrow \frac{r_2|1+x^2||1+x_0^2|}{r_2|1+x^2||1+x_0^2|+k_1|x^4+x^4{x}_0^2-x_0^4-x_0^4{x}^2|}\ge \frac{r_1}{r_1+k|x-x_0|}$

$\Rightarrow \frac{r_2|1+x^2||1+x_0^2|}{r_2|1+x^2||1+x_0^2|+k_1|x-x_0||x+x_0||x^2+x_0^2+x^2{x}_0^2|}\ge \frac{r_1}{r_1+k|x-x_0|}$

$\Rightarrow k_1{r}_1|x-x_0||x+x_0||x^2+x_0^2+x^2{x}_0^2|\le r_{2}k|1+x^2||1+x_0^2||x-x_0|$

$\Rightarrow r_1\le \frac{r_{2}k|1+x^2||1+x_0^2||x-x_0|}{k_1|x-x_0||x+x_0||x^2+x_0^2+x^2{x}_0^2|}$.......................(ii)

If T is IT2FC on X, then (ii) is satisfied $\forall x(\ne x_0)\in X$.

Now $in{f}_{x(\ne x_0)\in X}\frac{r_{2}k|1+x^2||1+x_0^2||x-x_0|}{k_1|x-x_0||x+x_0||x^2+x_0^2+x^2{x}_0^2|}=0$.

Hence, from (ii), we obtain $r_1=0$

$\Rightarrow \frac{a_1+b_1}{2}=0$

$\Rightarrow a_1+b_1=0$

$\Rightarrow a_1=b_1=0$ [Since $a_1,b_1\ge 0$]

$\Rightarrow {\eta }_1=\overline{0}$, which is not possible.

Therefore, T is not IT2FC.

5. Conclusions

In this work we have presented for the first time the concept of intuitionistic type-2 fuzzy normed linear space (IT2FNLS). A decomposition theorem of the intuitionistic type-2 fuzzy norm has been established. The later convergence and Cauchyness of a sequence in IT2FNLS have been examined. Also, associated properties of various classical concepts such as completeness, compactness, boundedness and finite dimensionality have been studied with examples and counterexamples in this newly defined IT2FNLS. Afterwards, two types of continuity are introduced in IT2FNLS, viz. IT2FC and sequentially IT2FC, and final relations between them have been analysed with examples, and we found some dissimilarity with the corresponding results in normed linear spaces.

There is a possibility of further study in the following directions:

• The boundedness of linear operators between two IT2FNLSs can be defined.
• The relation between the continuity and boundedness of linear operators can be studied.
• Four fundamental theorems, viz. the Hahn–Banach theorem, open mapping theorem, closed graph theorem and the uniform boundedness principle theorem, can be extended in IT2FNLS.
• Some geometric properties such as strict convexity and uniform convexity, etc., can be defined in these spaces.
• Some fixed-point theorems in these spaces can also be studied.