Finite Dimensional Intuitionistic Fuzzy Normed Linear Space

ISSN 1998-6262; Copyright ICSRS Publication, 2009c www.i-csrs.org

Finite Dimensional Intuitionistic Fuzzy Normed Linear Space

T. K. Samanta¹ and Iqbal H. Jebril²

1 Department of Mathematics, Uluberia College, Uluberia, Howrah - 711315, West; Bengal, India.

e-mail: mumpu−[email protected]

2 Department of Mathematics, King Faisal University, Saudi Arabia.

e-mail: [email protected]

Abstract

Following the definition of intuitionistic fuzzy n-norm [3], we have introduced the definition of intuitionistic fuzzy norm (in short IFN) over a linear space and there after a few results on intuitionistic fuzzy normed linear space and finite dimensional intuitionistic fuzzy normed linear space. Lastly, we have introduced the definitions of intuitionistic fuzzy continuity and sequentially intuitionistic fuzzy continuity and proved that they are equivalent.

Keywords: Fuzzy set, Membership function, Non - membership function, Intuitionistic fuzzy set, Fuzzy Norm, Intuitionistic fuzzy norm.

1 Introduction

The authors T. Bag and S. K. Samanta [5] introduced the definition of fuzzy norm over a linear space following the definition S. C. Cheng and J.

N. Moordeson [4] and they have studied finite dimensional fuzzy normed linear spaces. Also the definition of intuitionistic fuzzy n-normed linear space was introduced in the paper [3] and established a sufficient condition for an intuitionistic fuzzy n-normed linear space to be complete. In this paper, following the definition of intuitionistic fuzzy n-norm [3], the definition of intuitionistic fuzzy norm (in short IFN) is defined over a linear space. There after a sufficient condition is given for an intuitionistic fuzzy normed linear space to be complete and also it is proved that a finite dimensional intuitionistic

fuzzy norm linear space is complete. In such spaces, it is established that a necessary and sufficient condition for a subset to be compact. Thereafter following the definition of fuzzy continuous mapping [6], the definition of intuitionistic fuzzy continuity, strongly intuitionistic fuzzy continuity and sequentially intuitionistic fuzzy continuity are defined and proved that the concept of intuitionistic fuzzy continuity and sequentially intuitionistic fuzzy continuity are equivalent. There after it is shown that intuitionistic fuzzy continuous image of a compact set is again a compact set.

Definition 1 [3] A binary operation ∗ : [ 0, 1 ] × [ 0, 1 ] −→ [ 0, 1 ] is continuous t - norm if ∗ satisfies the following conditions :

(i) ∗ is commutative and associative , (ii) ∗ is continuous ,

(iii) a ∗ 1 = a ∀ a ε [ 0, 1 ] ,

(iv) a ∗ b ≤ c ∗ d whenever a ≤ c , b ≤ d and

a , b , c , d ε [ 0, 1 ].

Definition 2 [3] A binary operation : [ 0, 1 ] × [ 0, 1 ] −→ [ 0, 1 ] is continuous t - co - norm if satisfies the following conditions : (i) is commutative and associative ,

(ii) is continuous ,

(iii) a 0 = a ∀ a ε [ 0, 1 ] ,

(iv) a b ≤ c d whenever a ≤ c , b ≤ d and

a , b , c , d ε [ 0, 1 ].

Remark 1 [3] (a) For any r₁ , r₂ ε ( 0, 1 ) with r₁ > r₂ , there exist r3 , r4 ε ( 0, 1 ) such that r1 ∗ r3 > r2 and r1 > r4 r2 . (b) For any r₅ ε ( 0 , 1 ) , there exist r₆ , r₇ ε ( 0 , 1 ) such that r6 ∗ r6 ≥ r5 and r7 r7 ≤ r5 .

Definition 3 [3] Let E be any set. An intuitionistic fuzzy set A of E is an object of the form A = { (x , µ_A(x) , ν_A(x) ) : x ε E } , where the functions µ_A : E −→ [ 0 , 1 ] and ν_A : E −→ [ 0 , 1 ] denotes the degree of membership and the non - membership of the element x ε E respectively and for every x ε E , 0 ≤ µ_A(x) + ν_A(x) ≤ 1 .

Definition 4 [3] If A and B are two intuitionistic fuzzy sets of a non - empty set E , then A ⊆ B if and only if for all x ε E ,

µ_A(x) ≤ µ_B(x) and ν_A(x) ≥ ν_B(x) ;

A = B if and only if for all x ε E ,

µA(x) = µB(x) and νA(x) = νB(x) ; A = {(x , ν_A(x), µ_A(x) ) : x ε E };

A∩B = {(x , min (µA(x), µB(x) ), max (νA(x), νB(x) ) ) : x ε E };

A∪B = {(x , max (µ_A(x), µ_B(x) ), min (ν_A(x), ν_B(x) ) ) : x ε E }.

Definition 5 Let ∗ be a continuous t - norm , be a continuous t - co - norm and V be a linear space over the field F ( = R or C ) . An intuitionistic fuzzy norm or in short IFNon V is an object of the form A = { ( (x , t), N(x , t), M(x , t) ) : (x , t) ε V × R⁺ } , where N , M are fuzzy sets on V × R⁺ , N denotes the degree of membership and M denotes the degree of non - membership (x , t) ε V × R⁺ satisfying the following conditions :

(i) N(x , t) + M(x , t) ≤ 1 ∀ (x , t) ε V × R⁺; (ii) N(x , t) > 0 ;

(iii) N(x , t) = 1 if and only if x = 0 ; (iv) N(c x , t) = N(x , _|^t_c|) c 6= 0, c ε F ; (v) N(x , s) ∗ N(y , t) ≤ N(x + y , s + t) ;

(vi) N(x , ·) is non - decreasing function of R⁺ and

tlim→ ∞ N (x , t) = 1;

(vii) M(x , t) > 0 ;

(viii) M(x , t) = 0 if and only if x = 0 ; (ix) M(c x , t) = M(x , _|^t_c|) c 6= 0, c ε F; (x) M(x , s) M(y , t) ≥ M(x + y , s + t) ;

(xi) M(x , ·) is non - increasing function of R⁺ and

tlim→ ∞ M (x , t) = 0.

Example 1 Let (V = R , k · k) be a normed linear space where kxk = |x| ∀ x ε R . Define a ∗ b = min{ a , b } and a b = max{ a , b } for all a , b ε [ 0 , 1 ] . Also define N(x , t) = _t+k^t_|x| and M(x , t) = _t+^k|_k^x_|^|_x| where k > 0 . We now consider A = {( (x , t), N(x , t), M(x , t) ) : (x , t) ε V × R⁺} . Here A is an IFN on V .

Proof: Obviously follows from the calculation of the example 3.2 [ 3 ] . Definition 6 If A is an IFN on V ( a linear space over the field F ( = R or C ) ) then (V , A) is called an intuitionistic fuzzy normed linear space or in short IFNLS.

Definition 7 [3] A sequence {x_n }_n in an IFNLS (V , A) is said to converge to x ε V if given r > 0 , t > 0 , 0 < r < 1 there exists an integer n0 ε N such that N(xn − x , t) > 1 − r and M(x_n − x , t) < r for all n ≥ n₀ .

Theorem 1 In an IFNLS (V , A) , a sequence {x_n }_n converges to x ε V if and only if lim

n → ∞ N (x_n − x , t) = 1 and

nlim→ ∞ M (x_n − x , t) = 0 .

Proof: The proof directly follows from the proof of the theorem 3.4 [3] . Theorem 2 If a sequence {xn }_n in an IFNLS (V , A) is convergent , it’s limit is unique .

Proof: Let lim

n → ∞ x_n = x and lim

n → ∞ x_n = y . Also let s , t ε R⁺ . Now ,

nlim→ ∞ x_n = x ⇒

_lim

n→ ∞ N(xn−x , t) = 1

nlim→ ∞ M(xn−x , t) = 0

nlim→ ∞ xn = y ⇒

_lim

n→ ∞N(xn−y , t) = 1

nlim→ ∞M(xn−y , t) = 0

N (x − y , s + t) = N (x − xn + xn − y , s + t)

≥ N (x − x_n, s) ∗ N (x_n − y , t)

= N (x_n − x , s) ∗ N (x_n − y , t) Taking limit , we have

N(x−y , s+t) ≥ lim

n→ ∞ N (x_n − x , s)∗ lim

n→ ∞ N (x_n − y , t) = 1

=⇒ N(x − y , s + t) = 1 =⇒ x − y = 0 =⇒ x = y This completes the proof .

Theorem 3 If lim

n→ ∞ xn = x and lim

n→ ∞ yn = y then

nlim→ ∞ x_n + y_n = x + y in an IFNLS (V , A) .

Proof: Let s , t ε R⁺ . Now ,

nlim→ ∞ xn = x ⇒

_lim

n→ ∞ N(xn−x , t) = 1

nlim→ ∞ M(xn−x , t) = 0

nlim→ ∞ y_n = y ⇒

_lim

n→ ∞ N(yn−y , t) = 1

nlim→ ∞ M(yn−y , t) = 0

Now,

N ( (x_n + y_n) − (x + y) , s + t) = N ( (x_n − x) + (y_n − y), s + t)

≥ N (x_n − x , s) ∗ N (y_n − y , t) Taking limit, we have

nlim→ ∞ N ( (x_n + y_n) − (x + y) , s + t)

≥ lim

n→ ∞N (x_n − x , s) ∗ lim

n→ ∞ N (y_n − y , t)

= 1 ∗ 1 = 1

⇒ lim

n→ ∞ N ( (x_n + y_n) − (x + y) , s + t) = 1 Again,

M ( (x_n + y_n) − (x + y) , s + t) = M ( (x_n − x) + (y_n − y), s + t)

≤ M (x_n − x , s) M (y_n − y , t) Taking limit, we have

nlim→ ∞ M ( (x_n + y_n) − (x + y) , s + t)

≤ lim

n→ ∞M (x_n − x , s) lim

n→ ∞ M (y_n − y , t)

= 0 0 = 0

⇒ lim

n→ ∞ M ( (xn + yn) − (x + y) , s + t) = 0 Thus, we see that lim

n→ ∞ xn + yn = x + y . Theorem 4 If lim

n → ∞ xn = x and c(6= 0 ) ε F then lim

n→ ∞ c xn = c x in an IFNLS (V , A) .

Proof: Obvious.

Theorem 5 In an IFNLS (V , A), every subsequence of a convergent sequence converges to the limit of the sequence .

Proof: Obvious.

Definition 8 A sequence {xn }_n in an IFNLS (V , A) is said to be a Cauchy sequence if lim

n→ ∞ N (x_n+p − x_n, t) = 1 and

nlim→ ∞ M (x_n+p − x_n, t) = 0 , p = 1 , 2 , 3 , · · · , t > 0 .

Theorem 6 In an IFNLS (V , A) , every convergent sequence is a Cauchy sequence.

Proof: Let {xn}_n be a convergent sequence in the IFNLS (V , A) with

nlim→ ∞ x_n = x . Let s , t ε R⁺ and p = 1 , 2, 3, · · · , we have N (x_n+p − x_n , s + t) = N (x_n+p − x + x − x_n, s + t)

≥ N (x_n+p − x , s) ∗ N (x − x_n, t)

= N (x_n+p − x , s) ∗ N (x_n − x , t) Taking limit , we have

nlim→ ∞N (x_n+p − x_n , s + t)

≥ lim

n→ ∞N (x_n+p − x , s) ∗ lim

n→ ∞N (x_n − x , t)

= 1 ∗ 1 = 1

⇒ lim

n→ ∞ N (xn+p − xn , s + t) = 1 ∀ s , t ε R⁺ and p = 1, 2, 3, · · ·

Again ,

M (x_n+p − x_n, s + t) = M (x_n+p − x + x − x_n, s + t)

≤ M (xn+p − x , s) M (x − xn, t)

= M (x_n+p − x , s) M (x_n − x , t) Taking limit , we have

nlim→ ∞M (x_n+p − x_n , s + t)

≤ lim

n→ ∞M (xn+p − x , s) lim

n→ ∞M (xn − x , t)

= 0 0 = 0

⇒ lim

n→ ∞ M (x_n+p − x_n , s + t) = 0 ∀ s , t ε R⁺ and p = 1, 2, 3, · · ·

Thus, {x_n }_n is a Cauchy sequence in the IFNLS (V , A) .

Note 1 The converse of the above theorem is not necessarily true . It is verified by the following example .

Example 2 Let (V , k · k) be a normed linear space and define a ∗ b = min{a , b} and a b = max{a , b} for all a , b ε [ 0, 1 ] . For all

t > 0, define N(x , t) = _t+k^t_kxk and M(x , t) = _t+^kk_k^x_k^k_xk where k > 0 . It is easy to see that A = {( (x , t), N(x , t), M(x , t) ) : (x , t) ε V × R⁺ } is an IFN on V . We now show that

(a) {x_n}_n is a Cauchy sequence in (V , k · k) if and only if {x_n }_n is a Cauchy sequence in the IFNLS (V , A) .

(b) {xn }_n is a convergent sequence in (V , k · k) if and only if {x_n}_n is a convergent sequence in the IFNLS (V , A) .

Proof: (a) Let {x_n}_n be a Cauchy sequence in (V , k · k) and t > 0 .

⇐⇒ lim

n→ ∞ kx_n+p − x_nk = 0 f or p = 1, 2, · · ·

⇐⇒ lim

n→ ∞

t

t + k kx_n+p − x_nk = 1 and lim

n→ ∞

k kx_n+p − x_nk

t + k kx_n+p − x_nk = 0

⇐⇒ lim

n→ ∞ N (x_n+p − x_n, t) = 1and lim

n→ ∞ M (x_n+p − x_n, t) = 0

⇐⇒ {x_n}_n is a Cauchy sequence in (V , A)

(b) Let {x_n}_n be a convergent sequence in (V , k · k) and t > 0 .

⇐⇒ lim

n→ ∞ kxn − xk = 0

⇐⇒ lim

n→ ∞

t

t + kkxn−xk = 1 and lim

n→ ∞

kkxn−xk

t + kkxn−xk = 0

⇐⇒ lim

n→ ∞ N (xn − x , t) = 1and lim

n→ ∞ M (xn − x , t) = 0

⇐⇒ {x_n}_n is a convegent sequence in (V , A).

Theorem 7 Let (V , A) be an IFNLS , such that every Cauchy sequence in (V , A) has a convergent subsequence. Then (V , A) is complete .

Proof: Let {x_n}_n be a Cauchy sequence in (V , A) and {x_nk}_k be a subsequence of {x_n}_n that converges to x ε V and t > 0 . Since {x_n}_n is a Cauchy sequence in (V , A) , we have

lim

n , k→ ∞N

x_n − x_k , t 2

= 1 and lim

n , k→ ∞M

x_n − x_k , t 2

= 0 Again since {x_nk}_k converges to x , we have

klim→ ∞N

xn_k − x , t 2

= 1 and lim

k→ ∞M

xn_k − x , t 2

= 0 Now,

N (x_n − x , t) = N (x_n − x_nk + x_nk − x , t)

≥ N x_n − x_nk , ₂^t

∗ N x_nk − x , ₂^t

=⇒ lim

n→ ∞ N (x_n − x , t) = 1 Again, we see that

M (x_n − x , t) = M (x_n − x_nk + x_nk − x , t)

≤ M x_n − x_nk , ₂^t

M x_nk − x , ₂^t

=⇒ lim

n→ ∞ M (x_n − x , t) = 0

Thus, {x_n}_n converges to x in (V , A) and hence (V , A) is complete.

Theorem 8 Let (V , A) be an IFNLS , we further assume that , (xii) ^a_{a ∗} ^a_a ⁼₌ ^a_a } ∀ a ε [ 0 , 1 ]

(xiii) N(x , t) > 0 ∀ t > 0 =⇒ x = 0 (xiv) M(x , t) > 0 ∀ t > 0 =⇒ x = 0

Define kxk¹_α = ∧ {t : N(x , t) ≥ α} and kxk²_α = ∨ {t : M(x , t) ≤ α} , α ε ( 0, 1 ) . Then both { kxk¹_α : α ε ( 0, 1 )} and { k xk²_α : α ε ( 0, 1 )} are ascending family of norms on V . We call these norms as α - norm on V corresponding to the IFN A on V .

Proof: Let α ε ( 0, 1 ) . To prove kxk¹_α is a norm on V , we will prove the followings :

( 1 ) kxk¹_α ≥ 0 ∀ x ε V ; ( 2 ) kxk¹_α = 0 ⇐⇒ x = 0 ; ( 3 ) kc xk¹_α = |c| kxk¹_α ;

( 4 ) kx + y k¹_α ≤ kxk¹_α + kyk¹_α .

The proof of ( 1 ) , ( 2 ) and ( 3 ) directly follows from the proof of the theorem 2.1 [5] . So, we now prove ( 4 ) .

kxk¹_α + kyk¹_α = ∧ {s : N(x , s) ≥ α} + ∧ {t : N(y , t) ≥ α} = ∧ {s + t : N(x , s) ≥ α , N(y , t) ≥ α} =

∧ {s + t : N(x , s) ∗ N(y , t) ≥ α ∗ α} ≥ ∧ {s + t : N(x + y , s + t) ≥ α} = kx + yk¹_α, which proves ( 4 ) . Let 0 < α1 < α2 < 1 . kxk¹_α

1 = ∧ {t : N(x , t) ≥ α1} and kxk¹_α

2 = ∧ {t : N(x , t) ≥ α₂} . Since α₁ < α₂ , {t : N(x , t) ≥ α₂} ⊂ {t : N(x , t) ≥ α₁} =⇒ ∧{t : N(x , t) ≥ α₂} ≥ ∧{t : N(x , t) ≥ α₁} =⇒ kxk¹_α

2 ≥

kxk¹_α

1 . Thus, we see that { kxk¹_α : α ε ( 0 , 1 ) } is an ascending

family of norms on V .

Now we shall prove that { kxk²_α : α ε ( 0 , 1 ) } is also an ascending family of norms on V . Let α ε ( 0 , 1 ) and x , y ε V . It is obvious that kxk²_α ≥ 0 . Let kxk²_α = 0 . Now, kxk²_α = 0 =⇒ ∨{t : M(x , t) ≤ α} = 0 =⇒ M(x , t) >

α > 0 ∀ t > 0 =⇒ x = 0 . Conversely, we assume that x = 0 =⇒ M(x , t) = 0 ∀ t > 0 =⇒ ∨{t : M(x , t) ≤ α} = 0 =⇒ kxk²_α = 0 .

It is easy to see that kc xk²_α = |c| kxk²_α ∀ c ε F.

kxk²_α + ky k²_α = ∨ {s : M(x , s) ≤ α} + ∨ {t : M(y , t) ≤ α} = ∨ {s + t : M(x , s) ≤ α , M(y , t) ≤ α} = ∨ {s + t : M(x , s) M(y , t) ≤ α α} ≥

∨ {s + t : M(x + y , s + t) ≤ α} = kx + yk²_α , that is kx + yk²_α ≤ kxk²_α + kyk²_α ∀ x , y ε V .

Let 0 < α₁ < α₂ < 1 . Therefore, kxk²_α

1 = ∨ {t : M(x , t) ≤ α₁} and kxk²_α

2 = ∨ {t : M(x , t) ≤ α₂} . Since α₁ < α₂ , we have

{t : M(x , t) ≤ α₁} ⊂ {t : M(x , t) ≤ α₂}

=⇒ ∨{t : M(x , t) ≤ α₁} ≤ ∨{t : M(x , t) ≤ α₂}

=⇒ kxk²_α

1 ≤ kxk²_α

2. Thus we see that { kxk²_α : α ε ( 0, 1 )} is an ascending family of norms on V .

Lemma 1 [5] Let (V , A) be an IFNLS satisfying the condition (Xiii) and {x₁ , x₂ , · · · , x_n} be a finite set of linearly independent vectors of V . Then for each α ε ( 0, 1 ) there exists a constant C_α > 0 such that for any scalars α₁ , α₂ , · · · , α_n ,

kα₁x₁ + α₂x₂ + · · · + α_nx_nk¹_α ≥ C_α

n

X

i= 1

|α_i | where k · k¹_α is defined in the previous theorem.

Theorem 9 Every finite dimensional IFNLS satisfying the conditions (Xii) and (Xiii) is complete .

Proof: Let (V , A) be a finite dimensional IFNLS satisfying the conditions (Xii) and (Xiii) . Also, let dim V = k and {e1 , e2 , · · · , ek} be a basis of V . Consider {x_n}_n as an arbitrary Cauchy sequence in (V , A) . Let x_n = β₁⁽ⁿ⁾ e₁ + β⁽₂ⁿ⁾ e₂ + · · · + β_k⁽ⁿ⁾ e_k where β₁⁽ⁿ⁾ , β₂⁽ⁿ⁾ , · · · , β_k⁽ⁿ⁾ are suitable scalars. Then by the same calculation of the theorem 2.4 [5], there exist β₁ , β₂ , · · · , β_k ε F such that the sequence {β_i⁽ⁿ⁾}_n converges to β_i for i = 1 , 2 , · · · , k. Clearly x =

k

P

i= 1

β_ie_i ε V . Now, for all t > 0 ,

N(x_n − x , t) = N(

k

P

i= 1

β_i⁽ⁿ⁾e_i −

k

P

i= 1

β_ie_i , t)

= N(

k

P

i= 1

(β_i⁽ⁿ⁾ − β_i)e_i , t)

≥ N( (β₁⁽ⁿ⁾ − β1) e1 , _k^t ) ∗ · · · ∗ N( (β_k⁽ⁿ⁾ − β_k)e_k, _k^t )

= N(e₁ , ^t

k|β₁⁽ⁿ⁾−β1| ) ∗ · · · ∗ N(e_k , ^t

k|β⁽_kⁿ⁾−βk| ) Since lim

n→ ∞

t k

˛

˛

˛β_i⁽ⁿ⁾−βi

˛

˛

˛

= ∞, we see that lim

n→ ∞ N(e_i , ^t

k|β_i⁽ⁿ⁾−βi| ) = 1

=⇒ lim

n→ ∞ N(xn − x , t) ≥ 1 ∗ · · · ∗ 1 = 1 ∀ t > 0

=⇒ lim

n→ ∞ N(x_n − x , t) = 1 ∀ t > 0 Again, for all t > 0 ,

M(xn − x , t) = M(

k

P

i= 1

β_i⁽ⁿ⁾ei −

k

P

i= 1

βiei , t)

= M(

k

P

i= 1

(β_i⁽ⁿ⁾ − β_i)e_i , t)

≤ M( (β₁⁽ⁿ⁾ − β₁) e₁ , _k^t ) · · · M( (β_k⁽ⁿ⁾ − β_k)e_k, _k^t )

= M(e1 , ^t

k|β₁⁽ⁿ⁾−β1| ) · · · M(ek , ^t

k|β_k⁽ⁿ⁾−β_k| ) Since lim

n→ ∞

t k

˛

˛

˛β_i⁽ⁿ⁾−βi

˛

˛

˛

= ∞, we see that lim

n→ ∞ M(e_i , ^t

k|β_i⁽ⁿ⁾−βi| ) = 0

=⇒ lim

n→ ∞ M(x_n − x , t) ≤ 1 · · · 1 = 1 ∀ t > 0

=⇒ lim

n→ ∞ M(x_n − x , t) = 0 ∀ t > 0 .

Thus, we see that {x_n}_n is an arbitrary Cauchy sequence that converges to x ε V , hence the IFNLS (V , A) is complete .

Definition 9 Let (V , A) be an IFNLS. A subset P of V is said to be closed if for any sequence {x_n}_n in P converges to x ε P , that is,

nlim→ ∞ N(x_n − x , t) = 1, and lim

n→ ∞ M(x_n − x , t) = 0 =⇒ x ε P.

Definition 10 Let (V , A) be an IFNLS. A subset Q of V is said to be the closure of P ( ⊂ V ) if for any x ε Q , there exists a sequence {xn}n in P such that

nlim→ ∞ N(x_n − x , t) = 1, and lim

n→ ∞ M(x_n − x , t) = 0 ∀ t ε R⁺. We denote the set Q by P .

Definition 11 A subset P of an IFNLS is said to be bounded if and only if there exist t > 0 and 0 < r < 1 such that

N(x , t) > 1 − r and M(x , t) < r ∀ x ε P.

Definition 12 Let (V , A) be an IFNLS. A subset P of of V is said to be compact if any sequence {x_n}_n in P has a subsequence converging to an element of P .

Theorem 10 Let (V , A) be an IFNLS satisfying the condition (Xii) . Every Cauchy sequence in (V , A) is bounded .

Proof: Let {x_n}_n be a Cauchy sequence in the IFNLS (V , A) . Then we have

nlim→ ∞N(xn+p−xn, t) = 1

nlim→ ∞M(xn+p−xn, t) = 0

∀ t > 0 , p = 1 , 2 , · · ·. Choose a fixed r₀ with 0 < r₀ < 1 . Now we see that

nlim→ ∞ N(x_n − x_n+p , t) = 1 > r₀ ∀ t > 0 , p = 1, 2, · · ·

=⇒ F or t⁰ > 0 ∃ n0 = n0(t⁰) such that N(xn − xn+p , t⁰) >

r₀ ∀ n ≥ n₀ , p = 1 , 2 , · · · Since, lim

t→ ∞ N(x , t) = 1 , we have for each x_i , ∃t_i > 0 such that N(x_i , t) > r₀ ∀ t > t_i , i = 1 , 2, · · ·

Let t₀ = t⁰ + max{t₁ , t₂ , · · · , t_n0} . Then , N(x_n, t₀) ≥ N(x_n , t⁰ + t_n0)

= N(xn − xn0 + xn0 , t⁰ + tn0)

≥ N(x_n − x_n0 , t⁰) ∗ N(x_n0 , t_n0)

> r₀ ∗ r₀ = r₀ ∀ n > n₀ Thus , we have

N(x_n , t₀) > r₀ ∀ n > n₀

Also , N(x_n , t₀) ≥ N(x_n , t_n) > r₀ f orall n = 1, 2, · · · , n0

So, we have

N(x_n , t₀) > r₀ ∀ n = 1 , 2, · · · ( 1 ) Now, lim

n→ ∞ M(xn − xn+p , t) = 0 < ( 1 − r0) ∀ t > 0 , p = 1, 2, · · ·

=⇒ F or t⁰ > 0 ∃ n⁰₀ = n⁰₀(t⁰) such that M(x_n − x_n+p , t⁰) <

( 1 − r0) ∀ n ≥ n⁰₀ , p = 1 , 2 , · · ·

Since, lim

t→ ∞ M(x , t) = 0 , we have for each x_i , ∃t⁰_i > 0 such that M(x_i , t) < ( 1 − r₀) ∀ t > t⁰_i , i = 1, 2, · · ·

Let t⁰₀ = t⁰ + max{t⁰₁ , t⁰₂ , · · · , t⁰_n0} . Then , M(x_n, t⁰₀) ≤ M(x_n, t⁰ + t⁰_n0)

= M(x_n − x_n⁰

0 + x_n⁰

0 , t⁰ + t⁰_n

0)

≤ M(x_n − x_n⁰

0 , t⁰) M(x_n⁰

0 , t⁰_n0)

< ( 1 − r₀) ( 1 − r₀) = ( 1 − r₀) ∀ n > n⁰₀ Thus , we have

M(x_n , t⁰₀) < ( 1 − r₀) ∀ n > n⁰₀

Also , M(x_n , t⁰₀) ≤ M(x_n , t⁰_n) < ( 1 − r₀) f orall n = 1, 2, · · · , n⁰₀

So, we have

M(x_n , t⁰₀) < ( 1 − r₀) ∀ n = 1 , 2, · · · ( 2 ) Let t⁰⁰₀ = max{t₀ , t⁰₀} . Hence from ( 1 ) and ( 2 ) we see that

N(^{xn, t00}0) ^{> r0}

M(^xn, t⁰⁰₀)^{< ( 1−r}0)

∀ n = 1 , 2 , · · · This implies that {x_n}_n is bounded in (V , A) .

Theorem 11 In a finite dimensional IFNLS (V , A) satisfying the conditions (Xii), (Xiii) and (Xiv) , a subset P of V is compact if and only if P is closed and bounded in (V , A).

Proof: =⇒ part : Proof of this part directly follows from the proof of the theorem 2.5 [5].

⇐= part : In this part, we suppose that P is closed and bounded in the finite dimensional IFNLS (V , A) . To show P is compact, consider {x_n}_n, an arbitrary sequence in P. Since V is finite dimensional, let dim V = n and {e₁ , e₂ , · · · , e_n} be a basis of V . So, for each x_k,

∃ β₁^k , β^k₂ , · · · , β_n^k ε F such that

x_k = β₁^ke₁ + β₂^k e₂ + · · · + β_n^k e_n , k = 1 , 2, · · ·

Since P is bounded, {x_k}_k is also bounded. So, ∃ t₀ > 0 and r₀ where 0 < r₀ < 1 such that

N(xk, t0) > 1−r0=α0

M(xk, t0)< r0

o ∀ k · · · ( 1 )

Let kxk_α = ∧ {t : N(x , t) ≥ α}, α ε ( 0, 1 ) . So,we have kxkα0 ≤ t0 · · · ( 2 ) (By( 1 ) )

Since {e₁ , e₂ , · · · , e_n} is linearly independent, by Lemma ( 1 ) , ∃ a constant c > 0 such that ∀ k = 1, 2, · · · ,

kxkkα0 = k

n

X

i= 1

β_i^kei kα0 > c

n

X

i= 1

|β_i^k| · · · ( 3 ) From ( 2 ) and ( 3 ) we have

n

X

i= 1

|β_i^k| ≤ t₀

c f or k = 1, 2, · · ·

=⇒ F oreach i,

|β_i^k| ≤

n

X

i= 1

|β_i^k| ≤ t₀

c f or k = 1 , 2, · · ·

=⇒ {β_i^k}_k is a bounded sequence for each i = 1, 2, · · · , n

=⇒ {β_i^k}_k has a convergent subsequence say {β_i^kl}_l .

=⇒ {β₁^kl}_l , {β₂^kl}_l , · · · , {β_n^kl}_l are all convergent . Let x_kl = β₁^kl e₁ + β₂^kl e₂ + · · · + β_n^kl e_n and β₁ = lim

n→ ∞ β₁^kl , β₂ =

nlim→ ∞ β₂^kl , · · · , β_n = lim

n→ ∞ β_n^kl and x = β₁e₁ + β₂e₂ + · · · + β_ne_n. Now ∀ t > 0 , we have

N(x_kl − x , t) = N(

n

P

i= 1

β_i^kle_i −

n

P

i= 1

β_ie_i , t)

= N(

n

P

i= 1

(β_i^kl − β_i)e_i , t)

≥ N( (β₁^kl −β₁)e₁ , _n^t )∗ · · · ∗ N( (β_n^kl −β_n)e_n, _n^t )

= N(e₁ , ^t

n|β₁^{k l}−β1| ) ∗ · · · ∗ N(e_n, ^t

n|β^{k l}n −βn| ) Since lim

l→ ∞

t n

˛

˛

˛β_i^{k l}−βi

˛

˛

˛

= ∞, we see that lim

l→ ∞ N(e_i , ^t

n|β_i^{k l} −βi| ) = 1

=⇒ lim

l→ ∞ N(x_kl − x , t) ≥ 1 ∗ · · · ∗ 1 = 1 ∀ t > 0

=⇒ lim

l→ ∞ N(x_kl − x , t) = 1 ∀ t > 0 · · · ( 4 ) Again, for all t > 0 ,

M(x_kl − x , t) = M(

n

P

i= 1

β_i^kle_i −

n

P

i= 1

β_ie_i , t)

= M(

n

P

i= 1

(β_i^kl − β_i)e_i , t)

≤ M( (β₁^kl − β₁) e₁ , _n^t ) · · · M( (β_n^kl − β_n)e_n, _n^t )

= M(e₁ , ^t

n|β₁^{k l}−β1| ) · · · M(e_n, ^t

n|β^{k l}n −βn| ) Since lim

l→ ∞

t n

˛

˛

˛β_i^{k l}−βi

˛

˛

˛

= ∞, we see that lim

l→ ∞ M(e_i , ^t

n|β_i^{k l}−βi| ) = 0

=⇒ lim

l→ ∞ M(x_kl − x , t) ≤ 0 · · · 0 = 0 ∀ t > 0

=⇒ lim

l→ ∞ M(x_kl − x , t) = 0 ∀ t > 0 · · · ( 5 )

Thus, from ( 4 ) and ( 5 ) we see that lim

l→ ∞ x_kl = x =⇒ x ε A [ Since A is closed ] .

=⇒ A is compact.

Definition 13 Let (U , A) and (V , B) be two IFNLS over the same field F . A mapping f from (U , A) to (V , B) is said to be intuitionistic fuzzy continuous ( or in short IFC ) at x₀ ε U, if for any given ε > 0 , α ε ( 0, 1 ) , ∃ δ = δ(α , ε) > 0, β = β(α , ε) ε ( 0, 1 ) such that for all x ε U,

N_U(x − x₀ , δ) > β =⇒ N_V (f(x) − f(x₀) , ε) > α and

M_U(x − x₀ , δ) < 1 − β =⇒ M_V (f(x) − f(x₀) , ε) <

1 − α.

If f is continuous at each point of U , f is said to be IFC on U .

Definition 14 A mapping f from (U , A) to (V , B) is said to be strongly intuitionistic fuzzy continuous ( or in short strongly IFC ) at x₀ ε U, if for any given ε > 0 , ∃ δ = δ(α , ε) > 0 such that for all x ε U,

N_V (f(x) − f(x₀), ε) ≥ N_U(x − x₀ , δ) and M_V (f(x) − f(x₀), ε) < M_U(x − x₀ , δ) .

f is said to be strongly IFC on U if f is strongly IFC at each point of U .

Definition 15 A mapping f from (U , A) to (V , B) is said to be sequentially intuitionistic fuzzy continuous ( or in short sequentially IFC ) at x₀ ε U , if for any sequence {x_n}_n , x_n ε U ∀ n , with xn −→ x0 in (U , A) implies f(xn) −→ f(x0) in (V , B) , that is ,

nlim→ ∞N_U(x_n − x₀ , t) = 1 and lim

n→ ∞ M_U(x_n − x₀ , t) = 0

=⇒ lim

n→ ∞ N_V (f(x_n) − f(x₀) , t) = 1 and lim

n→ ∞ M_V (f(x_n) − f(x₀), t) = 0

If f is sequentially IFC at each point of U then f is said to be sequentially IFC on U .

Theorem 12 Let f be a mapping from (U , A) to (V , B). If f strongly IFC then it is sequentially IFC but not conversely .

Proof: Let f : (U , A) −→ (V , B) be strongly IFC on U and x₀ ε U. Then for each ε > 0 , ∃ δ = δ(x₀ , ε) > 0 such that for all x ε U ,

N_V (f(x) − f(x₀), ε) ≥ N_U(x − x₀ , δ) and

M_V (f(x) − f(x₀), ε) < M_U(x − x₀ , δ)

Let {x_n}_n be a sequence in U such that x_n −→ x₀ , that is , for all t > 0 ,

nlim→ ∞ N_U(x_n −x₀ , t) = 1 and lim

n→ ∞ M_U(x_n− x₀ , t) = 0

Thus, we see that

N_V (f(x_n) − f(x₀), ε) ≥ N_U(x_n − x₀ , δ) and M_V (f(x_n) − f(x₀), ε) < M_U(x_n − x₀ , δ)

which implies that

nlim→ ∞ N_V (f(x_n) − f(x₀) , ε) = 1 and lim

n→ ∞ M_V (f(x_n) − f(x₀), ε) = 0

that is , f(x_n) −→ f(x₀) in (V , B) .

To show that the sequentially IFC of f does not imply strongly IFC of f on U , consider the following example .

Example 3 Let (X =R , k · k) be a normed linear space where kxk =

|x| ∀ x ε R. Define a ∗ b = min{a , b} and a b = max{a , b} for all a , b ε [ 0, 1 ]. Also, define

N₁ , M₁ , N₂ , M₂ : X × R⁺ −→ [ 0, 1 ] by

N₁(x , t) = t

t + |x| , M₁(x , t) = |x| t + |x|

N₂(x , t) = t

t + k|x| , M₁(x , t) = k|x|

t + k|x| k > 0

Let A = {( (x , t), N₁ , M₁) : (x , t) ε X × R⁺}and B = {( (x , t), N2 , M2) : (x , t) ε X × R⁺}

It is easy to see that (X , A) and (X , B) are IFNLS . Let us now define, f(x) = _{1 +}^x4_x2 ∀ x ε X. Let x₀ ε X and {x_n}_n be a sequence in X such that x_n −→ x₀ in (X , A) , that is , for all t > 0 ,

nlim→ ∞ N1(xn − x0 , t) = 1 and lim

n→ ∞M1(xn − x0 , t) = 0 that is

nlim→ ∞

t

t+|xn−x0| = 1 and lim

n→ ∞

|xn−x0|

t+|xn−x0| = 0

=⇒ lim

n→ ∞ |x_n − x₀| = 0 Now , for all t > 0

N₂(f(x_n)−f(x₀), t) = _t+k|f(x ^t

n)−f(x0)|

= ^t

t+k| ^x4n

1 +x2 n

− ^x

4 0 1 +x2

0

|

= _{t( 1+x2} ^{t( 1+x2n) ( 1+x20)}

n)( 1+x²₀)+k|x⁴_n( 1 +x²₀)−x⁴₀( 1 +x²_n)|

= _{t( 1+x2} ^{t( 1+x2n)( 1+x20)}

n)( 1+x²₀)+k|(x²_n−x²₀) (x²_n+x²₀)+x²_nx²₀(x²_n−x²₀)|

=⇒ lim

n→ ∞ N2(f(xn) − f(x0), t) = 1.

=⇒ M₂(f(x_n)−f(x₀), t) = _t(1+x2 ^{k|(x2n−x20)(x2n+x20)+x2nx20(x2n−x20)|}

n)(1+x²₀)+k|(x²_n−x²₀) (x²_n+x²₀) +x²_nx²₀(x²_n−x²₀)|

=⇒ lim

n→ ∞ M₂(f(x_n) − f(x₀), t) = 0.

Thus, f is sequentially continuous on X . From the calculation of the example [6] , it follows that f is not strongly IFC .

Theorem 13 Let f be a mapping from the IFNLS (U , A) to (V , B). Then f is IFC on U if and only if it is sequentially IFC on U .

Proof: =⇒ part : Suppose f is IFC at x0 ε U and {xn}n is a sequence in U such that x_n −→ x₀ in (U , A) . Let ε > 0 and α ε ( 0 , 1 ) . Since f is IFC at x₀ , ∃ δ = δ(ε , α) > 0 and

∃ β = β(ε , α) ε ( 0, 1 ) such that for all x ε U,

N_U(x − x₀ , δ) > β =⇒ N_V (f(x) − f(x₀), ε) > α M_U(x − x₀ , δ) < 1−β =⇒ M_V (f(x)−f(x₀), ε)<1−α. Since xn −→ x0 in (U , A) , there exists a positive integer n0

such that for all n ≥ n₀

N_U(x_n − x₀ , δ) > β and M_U(x_n − x₀ , δ) < 1 − β

=⇒ NV (f(xn) − f(x0), ε) > α and MV (f(xn) − f(x0), ε) <

1 − α

=⇒ f(x_n) −→ f(x₀) in (V , B) , that is , f is sequentially IFC at x0 .

⇐= part : Let f be sequentially IFC at x₀ ε U. If possible, we suppose that f is not IFC at x₀.

=⇒ ∃ ε > 0 and α ε ( 0, 1 ) such that for any δ > 0 and β ε ( 0, 1 ),

∃ y ( depending on δ , β ) such that

N_U(x₀ − y , δ) > β but N_V (f(x₀) − f(y) , ε) ≤ α and

M_U(x₀−y, δ)<1−β but M_V (f(x₀)−f(y), ε) ≥ 1−α.

Thus for β = 1 − _{n+ 1}¹ , δ = _{n+ 1}¹ , n = 1, 2, · · · , ∃ y_n such that NU(x0 −yn, _{n+ 1}¹ ) > 1− _{n+ 1}¹ but NV (f(x0)−f(yn), ε) ≤ α, M_U(x₀ − y_n, _{n+ 1}¹ ) < _{n+ 1}¹ but M_V (f(x₀) − f(y), ε) ≥ 1 − α. Taking δ > 0 , ∃ n₀ such that _{n+ 1}¹ < δ ∀ n ≥ n₀ .

N_U(x₀ − y_n , δ) ≥ N_U(x₀ − y_n , _{n+ 1}¹ ) > 1 − _{n+ 1}¹ ∀ n ≥ n₀ , M_U(x₀ − y_n , δ) ≤ M_U(x₀ − y_n, _{n+ 1}¹ ) < _{n+ 1}¹ ∀ n ≥ n₀ .

=⇒ lim

n→ ∞ N_U(x₀ − y_n, δ) = 1 and lim

n→ ∞ M_U(x₀ − y_n , δ) = 0 But , NV (f(x0) − f(yn) , ε) ≤ α =⇒ lim

n→ ∞NV (f(x0) − f(y_n), ε) 6= 1.

Thus, {f(y_n)}_n does not converge to f(x₀) where as y_n −→ x₀ in