CONNECTEDNESS IN INTUITIONISTIC FUZZY TOPOLOGICAL ...

Commun. Korean Math. Soc. 20 (2005), No. 1, pp. 117–134
CONNECTEDNESS IN INTUITIONISTIC
FUZZY TOPOLOGICAL SPACES
Yong Chan Kim and S. E. Abbas
Abstract. We introduce the notion of (r,s)-connected sets in intuitionistic
fuzzy topological spaces and investigate some properties
of them. In particular, we show that every (r,s)-component in
an intuitionistic fuzzy topological space is (r,s)-component in the
stratification of it.
1. Introduction and preliminaries
Chang [4] introduced the notion of a fuzzy topology. Later, Lowen
[16] redefined it which is now known as a stratified fuzzy topology. Pu
and Liu [19] studied fuzzy connectedness in fuzzy topological spaces. It
has been developed in many directions [6, 17, 26]. Šostak [23] introduced
the notion of a smooth topology as an extension of Chang and Lowen’s
fuzzy topology and developed the theory of smooth topological spaces in
[24,25]. After that, several authors [5, 9-13, 18, 20] have reintroduced the
same definition and studied smooth topological spaces being unaware of
Šostak’s work. They referred to the fuzzy topology in the sense of Chang
and Lowen as the topology on fuzzy subsets.
On the other hand, Atanassov [1] introduced the idea of intuitionistic
fuzzy sets. Recently, much work has been done with these concepts
[1, 2, 3]. Çoker [7,8] introduced the notion of intuitionistic fuzzy topological
spaces in a Chang’sense using intuitionistic fuzzy sets. Samanta
and Mondal [21, 22] introduced the notion of an intuitionistic gradation
of openness as an extension of a smooth topology in a Šostak’sense.
In this paper, we introduce the notion of (r,s)-connected fuzzy sets
in intuitionistic fuzzy topological spaces and investigate some properties
Received February 16, 2004.
2000 Mathematics Subject Classification: 54A40, 54D05, 03E72.
Key words and phrases: intuitionistic (stratified) fuzzy topological spaces, (r,s)-
separated ((r,s)-connected) fuzzy sets, (r,s)-components.

118 Yong Chan Kim and S. E. Abbas
of them. We show that every (r,s)-component in an intuitionistic fuzzy
topological space is (r,s)-component in the stratification of it.
In this paper, let X be a nonempty set, I = [0, 1], I 0 = (0, 1] and
I 1 = [0, 1). For α ∈ I, α(x) = α for all x ∈ X. The family of all fuzzy
sets on X denoted by I X .
Definition 1.1. ([22]) An intuitionistic gradation of openness (IGO,
for short) on X is an ordered pair (T , T ∗ ) of mappings from I X to I
such that
(IGO1) T (λ) + T ∗ (λ) ≤ 1, for all λ ∈ I X ,
(IGO2) T (0) = T (1) = 1 and T ∗ (0) = T ∗ (1) = 0,
(IGO3) T (λ 1 ∧ λ 2 ) ≥ T (λ 1 ) ∧ T (λ 2 ) and T ∗ (λ 1 ∧ λ 2 ) ≤ T ∗ (λ 1 ) ∨
T ∗ (λ 2 ), for each λ i ∈ I X , i = 1, 2,
(IGO4) T ( ∨ i∈∆ λ i) ≥ ∧ i∈∆ T (λ i) and
T ∗ ( ∨ i∈∆ λ i) ≤ ∨ i∈∆ T ∗ (λ i ) for each λ i ∈ I X , i ∈ ∆.
The triplet (X, T , T ∗ ) is called an intuitionistic fuzzy topological space
(ifts, for short). T and T ∗ may be interpreted as gradation of openness
and gradation of nonopenness, respectively.
An ifts (X, T , T ∗ ) is called stratified if
(IS) T (α) = 1 and T ∗ (α) = 0 for each α ∈ I.
Let (U, U ∗ ) and (T , T ∗ ) be IGO’s on X. We say (U, U ∗ ) is finer than
(T , T ∗ ) ((T , T ∗ ) is coarser than (U, U ∗ )) if T (λ) ≤ U(λ) and T ∗ (λ) ≥
U ∗ (λ) for all λ ∈ I X .
Definition 1.2. ([14]) Let (X, T , T ∗ ) be an ifts. A function C :
I X × I 0 × I 1 → I X is called an intuitionistic closure operator if for
λ, µ ∈ I X , r ∈ I 0 and s ∈ I 1 with r + s ≤ 1, it satisfies the following
conditions:
(C1) C(0, r, s) = 0.
(C2) λ ≤ C(λ, r, s).
(C3) C(λ, r, s) ∨ C(µ, r, s) = C(λ ∨ µ, r, s).
(C4) C(λ, r, s) ≤ C(λ, r 1 , s 1 ) if r ≤ r 1 , s ≥ s 1 with r 1 + s 1 ≤ 1.
(C5) C(C(λ, r, s), r, s) = C(λ, r, s).
Theorem 1.3. ([14]) Let C be an intuitionistic closure operator on
X. Define the functions T C , T ∗ C : IX → I by
T C (λ) = ∨ {r ∈ I 0 | C(1 − λ, r, s) = 1 − λ},
T ∗ C (λ) = ∧ {s ∈ I 1 | C(1 − λ, r, s) = 1 − λ}.
Then, (T C , TC ∗ ) is an IGO on X.

Connectedness in intuitionistic fuzzy topological spaces 119
Theorem 1.4. ([14]) Let (X, T , T ∗ ) be an ifts. Then for each r ∈ I 0 ,
s ∈ I 1 , λ ∈ I X we define an operator C T ,T ∗ : I X × I 0 × I 1 → I X as
follows
C T ,T ∗(λ, r, s) = ∧ {µ ∈ I X | λ ≤ µ, T (1 − µ) ≥ r, T ∗ (1 − µ) ≤ s}.
Then (1) C T ,T ∗, is an intuitionistic closure operator.
(2) T CT ,T ∗ = T and T ∗ C T ,T ∗ = T ∗ .
Proof. (1) It is proved in [14].
(2) Let T (µ) = r and T ∗ (µ) = s. Then C T ,T ∗(1 − µ, r, s) = 1 − µ.
Hence T CT ,T ∗ ≥ T and T ∗ C T ,T ∗ ≤ T ∗ .
Suppose T CT ,T ∗ ≰ T . Then there exists µ with C T ,T ∗(1 − µ, r, s) =
1 − µ such that
T CT ,T ∗ (µ) ≥ r > T (µ).
But, by the definition of C T ,T ∗, T (µ) ≥ r. It is a contradiction. Other
case is similarly proved.
□
Definition 1.5. ([22]) Let (X, T , T ∗ ) and (Y, U, U ∗ ) be ifts’s and
f : X → Y a function. Then, f is called intuitionistic continuous if
U(λ) ≤ T (f −1 (λ)) and U ∗ (λ) ≥ T ∗ (f −1 (λ)) for all λ ∈ I Y .
Theorem 1.6. Let (X, T , T ∗ ) and (Y, U, U ∗ ) be ifts’s and f : (X, T ,
T ∗ ) → (Y, U, U ∗ ) a function. Then the following statements are
equivalent, for each λ ∈ I X , µ ∈ I Y , r ∈ I 0 , s ∈ I 1 .
(1) f is intuitionistic continuous.
(2) f(C T ,T ∗(λ, r, s)) ≤ C U,U ∗(f(λ), r, s).
(3) C T ,T ∗(f −1 (µ), r, s) ≤ f −1 (C U,U ∗(µ, r, s)).
Proof. (1)⇒ (2) Let f be intuitionistic continuous.
Then T (1 −

120 Yong Chan Kim and S. E. Abbas
f −1 (µ)) ≥ U(1 − µ) and T ∗ (1 − f −1 (µ)) ≤ U ∗ (1 − µ). Hence
C U,U ∗(f(λ), r, s)
= ∧ {µ ∈ I Y | f(λ) ≤ µ, U(1 − µ) ≥ r, U ∗ (1 − µ) ≤ s}
≥ ∧ {µ ∈ I Y | λ ≤ f −1 (µ), T (1 − f −1 (µ)) ≥ r,
T ∗ (1 − f −1 (µ)) ≤ s}
≥ ∧ {f(f −1 (µ)) ∈ I Y | λ ≤ f −1 (µ), T (1 − f −1 (µ)) ≥ r,
T ∗ (1 − f −1 (µ)) ≤ s}
≥ f( ∧ {f −1 (µ) ∈ I Y | λ ≤ f −1 (µ), T (1 − f −1 (µ)) ≥ r,
T ∗ (1 − f −1 (µ)) ≤ s})
≥ f(C T ,T ∗(λ, r, s)).
(2)⇒ (3) For all µ ∈ I Y , put λ = f −1 (µ). By (2),
f(C T ,T ∗(f −1 (µ), r, s)) ≤ C U,U ∗(f(f −1 (µ)), r, s) ≤ C U,U ∗(µ, r, s).
Thus C T ,T ∗(f −1 (µ), r, s) ≤ f −1 (C U,U ∗(µ, r, s)).
(3)⇒ (1) It follows from C U,U ∗(µ, r, s) = µ implies C T ,T ∗(f −1 (µ), r, s)
= f −1 (µ). □
2. Connectedness in intuitionistic fuzzy topological spaces
Definition 2.1. Let (X, T , T ∗ ) be an ifts. For λ, µ, ρ ∈ I X , λ and
µ are called (r,s)-separated if for r ∈ I 0 and s ∈ I 1 ,
C T ,T ∗(λ, r, s) ∧ µ = C T ,T ∗(µ, r, s) ∧ λ = 0.
A fuzzy set ρ is called (r,s)-connected if there not exist (r,s)-separated
λ, µ ∈ I X − {0} such that ρ = λ ∨ µ. A fuzzy set ρ is called connected
if it is (r,s)-connected for all r ∈ I 0 and s ∈ I 1 . A triplet (X, T , T ∗ ) is
called (r,s)-connected if 1 is (r,s)-connected.
Remark 2.2. Let λ and µ be (r,s)-separated. For each ρ ∈ I X and
r 1 ≤ r, s 1 ≥ s, since C T ,T ∗(ρ, r 1 , s 1 ) ≤ C T ,T ∗(ρ, r, s) , λ and µ are
(r 1 , s 1 )-separated. Furthermore, from this fact, if ρ is (r 1 , s 1 )-connected
for r 1 ≤ r, s 1 ≥ s, ρ is (r,s)-connected.

Connectedness in intuitionistic fuzzy topological spaces 121
Example 2.3. Let X = {x, y} be a set. We define an IGO (T , T ∗ )
on X as follows: for each λ ∈ I X ,
⎧
1 if λ = 0 or 1 ,
⎪⎨ 1
3
if λ = χ {x} ,
T (λ) =
1
2
if λ = χ {y} ,
⎪⎩
0 otherwise,
⎧
0 if λ = 0 or 1 ,
⎪⎨ 2
T ∗ 3
if λ = χ {x} ,
(λ) =
1
2
if λ = χ {y} ,
⎪⎩
1 otherwise.
We can obtain
⎧
0 if λ = 0, r ∈ I 0 , s ∈ I 1 ,
⎪⎨ χ {x} if 0 ≠ λ ≤ χ {x} , r ≤ 1 2
C T ,T ∗(λ, r, s) =
, s ≥ 1 2 ,
χ {y} if 0 ≠ λ ≤ χ {y} , r ≤
⎪⎩
1 3 , s ≥ 2 3 ,
1 otherwise.
If r ≤ 1 3 , s ≥ 2 3 , then (χ {x} = C T ,T ∗(χ {x} , r, s)) ∧ χ {y} = 0 and χ {x} ∧
(χ {y} = C T ,T ∗(χ {y} , r, s)) = 0. Thus, 1 X = χ {x} ∨ χ {y} is not (r,s)-
connected for r ≤ 1 3 and s ≥ 2 3 . If r > 1 3 and s < 2 3 , (X, T , T ∗ ) is
(r,s)-connected.
Theorem 2.4. Let (X, T , T ∗ ) be an ifts. The following statements
are equivalent.
(1) (X, T , T ∗ ) is (r,s)-connected.
(2) If λ ∨ µ = 1 and λ ∧ µ = 0 for (T (λ) ≥ r, T ∗ (λ) ≤ s) and
(T (µ) ≥ r, T ∗ (µ) ≤ s), then λ = 0 or µ = 0.
(3) If λ ∨ µ = 1 and λ ∧ µ = 0 for (T (1 − λ) ≥ r, T ∗ (1 − λ) ≤ s) and
(T (1 − µ) ≥ r, T ∗ (1 − µ) ≤ s), then λ = 0 or µ = 0.
Proof. (1) ⇒ (2) Suppose that there exist λ, µ ∈ I X − {0} such
that for (T (λ) ≥ r, T ∗ (λ) ≤ s) and (T (µ) ≥ r, T ∗ (µ) ≤ s), λ ∨ µ =
1, λ ∧ µ = 0. It implies
(1 − λ) ∧ (1 − µ) = 0, (1 − λ) ∨ (1 − µ) = 1.
Since C T ,T ∗(1 − λ, r, s) = 1 − λ and C T ,T ∗(1 − µ, r, s) = 1 − µ from
Theorem 1.4, 1 − λ and 1 − µ are (r,s)-separated. Suppose λ = 1. Then
µ = λ ∧ µ = 0. It is a contradiction. Thus, 1 − λ ∈ I X − {0}. Similarly,
1 − µ ∈ I X − {0}. Furthermore, (1 − λ) ∨ (1 − µ) = 1. Hence 1 is not
(r,s)-connected.
(2) ⇒ (3) By the De Morgan’s law, it is easily proved.

122 Yong Chan Kim and S. E. Abbas
(3) ⇒ (1) If (X, T , T ∗ ) is not (r,s)-connected, then there exist (r,s)-
separated λ, µ ∈ I X −{0} such that λ∨µ = 1. Since λ∧µ ≤ C T ,T ∗(λ, r, s)∧
µ = 0, we have λ ∧ µ = 0. So, (1 − λ) ∧ (1 − µ) = 0 implies 1 − µ ≤ λ.
Furthermore, C T ,T ∗(λ, r, s)∧µ = 0 implies C T ,T ∗(λ, r, s) ≤ 1−µ. Hence
C T ,T ∗(λ, r, s) ≤ λ. By Definition 1.2(C2), we have C T ,T ∗(λ, r, s) = λ.
From Theorem 1.4, we have T (1 − λ) ≥ r and T ∗ (1 − λ) ≤ s. Similarly,
we have T (1 − µ) ≥ r and T ∗ (1 − µ) ≤ s. It does not satisfy the
condition of (3).
□
Lemma 2.5. Let (X, T , T ∗ ) be an ifts and λ, µ, ρ ∈ I X . If µ and ρ
are (r,s)-separated, λ ∧ µ and λ ∧ ρ are (r,s)-separated.
Proof. Let µ and ρ be (r,s)-separated. So,
C T ,T ∗(λ ∧ µ, r, s) ∧ (λ ∧ ρ) ≤ C T ,T ∗(µ, r, s) ∧ ρ = 0.
Similarly, (λ ∧ µ) ∧ C T ,T ∗(λ ∧ ρ, r, s) = 0. Hence λ ∧ µ and λ ∧ ρ are
(r,s)-separated.
□
Theorem 2.6. Let (X, T , T ∗ ) be an ifts and λ ∈ I X . The following
statements are equivalent.
(1) λ is (r,s)-connected.
(2) If µ and ρ are (r,s)-separated such that λ ≤ µ ∨ ρ, then λ ∧ µ = 0
or λ ∧ ρ = 0.
(3) If µ and ρ are (r,s)-separated such that λ ≤ µ ∨ ρ, then λ ≤ µ or
λ ≤ ρ.
Proof. (1) ⇒ (2) Let µ and ρ be (r,s)-separated such that λ ≤ µ∨ρ.
By Lemma 2.5, λ ∧ µ and λ ∧ ρ are (r,s)-separated. Since λ is (r,s)-
connected and λ = λ ∧ (µ ∨ ρ) = (λ ∧ µ) ∨ (λ ∧ ρ), then λ ∧ µ = 0 or
λ ∧ ρ = 0.
(2) ⇒ (3) It easily proved from the following statements. If λ∧µ = 0,
then
λ = λ ∧ (µ ∨ ρ) = (λ ∧ µ) ∨ (λ ∧ ρ) = 0 ∨ (λ ∧ ρ) = λ ∧ ρ.
Hence λ ≤ ρ. If λ ∧ ρ = 0, similarly, λ ≤ µ.
(3) ⇒ (1) Let µ and ρ be (r,s)-separated such that λ = µ ∨ ρ. By (3),
λ ≤ µ or λ ≤ ρ. If λ ≤ µ and µ and ρ are (r,s)-separated, then
ρ = ρ ∧ λ ≤ ρ ∧ µ ≤ ρ ∧ C T (µ, r, s) = 0.
Hence ρ = 0. If λ ≤ ρ, similarly µ = 0.
□

Connectedness in intuitionistic fuzzy topological spaces 123
Theorem 2.7. Let (X, T , T ∗ ) be an ifts and λ, µ ∈ I X .
(1) If λ is (r, s)-connected and λ ≤ µ ≤ C T ,T ∗(λ, r, s), then µ is
(r, s)-connected.
(2) If λ and µ are (r, s)-connected fuzzy sets which are not (r, s)-
separated, then λ ∨ µ is (r, s)-connected.
Proof. (1) Let ν and ρ be (r,s)-separated such that µ = ν ∨ ρ. Put
ν 1 = λ ∧ ν and ρ 1 = λ ∧ ρ. Then ν 1 and ρ 1 are (r,s)-separated such that
λ = ν 1 ∨ ρ 1 . Since λ is (r,s)-connected, ν 1 = 0 or ρ 1 = 0. If ν 1 = 0, then
λ = ρ 1 = λ ∧ ρ ⇒ λ ≤ ρ. It implies
µ ≤ C T ,T ∗(λ, r, s) ≤ C T ,T ∗(ρ, r, s).
Hence ν = ν ∧ µ ≤ ν ∧ C T ,T ∗(ρ, r, s) = 0. If ρ 1 = 0, similarly, ρ = 0.
Thus, µ is (r,s)-connected.
(2) Let ν and ρ be (r,s)-separated such that λ ∨ µ = ν ∨ ρ. Since
λ is (r,s)-connected, by Theorem 2.6(3), λ ≤ ν or λ ≤ ρ. Say λ ≤ ν.
Suppose µ ≤ ρ. Since (λ∨µ)∧ν = λ and (λ∨µ)∧ρ = µ, by Lemma 2.5,
λ and µ are (r,s)-separated. It is a contradiction. Hence µ ≤ ν. Thus
λ ∨ µ ≤ ν, by Theorem 2.6(3), λ ∨ µ is (r,s)-connected.
□
Theorem 2.8. Let (X, T , T ∗ ) be an ifts. Let A = {λ i | i ∈ Γ} be
a family of (r, s)-connected fuzzy sets in (X, T , T ∗ ) such that no two
members of A are (r, s)-separated. Then ∨ i∈Γ λ i is (r, s)-connected.
Proof. Put λ = ∨ i∈Γ λ i. Let µ and ρ be (r,s)-separated such that
λ = µ ∨ ρ. Since any two member λ i , λ j ∈ A are not (r,s)-separated,
by Theorem 2.7 (2), λ i ∨ λ j is (r,s)-connected. From Theorem 2.6(3),
λ i ∨ λ j ≤ µ or λ i ∨ λ j ≤ ρ, say λ i ∨ λ j ≤ µ. It implies λ ≤ µ. Hence λ
is (r,s)-connected.
□
The following corollary is obvious from Theorem 2.8.
Corollary 2.9. Let (X, T , T ∗ ) be an ifts. Let {λ i | i ∈ Γ} be a
family of (r, s)-connected fuzzy sets in (X, T , T ∗ ). If ∧ i∈Γ λ i ≠ 0, then
∨
i∈Γ λ i is (r, s)-connected.
Theorem 2.10. Let (X, T 1 , T1 ∗ ) and (Y, T 2 , T2 ∗ ) be ifts’s. If f :
(X, T 1 , T1 ∗ ) → (Y, T 2 , T2 ∗ ) is intuitionistic continuous and λ is (r, s)-
connected, then f(λ) is (r, s)-connected.

124 Yong Chan Kim and S. E. Abbas
Proof. Let µ and ρ be (r,s)-separated such that f(λ) = µ ∨ ρ. We
have
λ ≤ f −1 (f(λ)) = f −1 (µ ∨ ρ) = f −1 (µ) ∨ f −1 (ρ).
Since f is intuitionistic continuous, by Theorem 1.6(3),
Hence
C T1 ,T ∗
(f −1 (µ), r, s) ≤ f −1 (C
1 T2 ,T2 ∗ (µ, r, s)).
C T1 ,T ∗
1 (f −1 (µ), r, s) ∧ f −1 (ρ) ≤ f −1 (C T2 ,T ∗
2 (µ, r, s)) ∧ f −1 (ρ)
= f −1 (C T2 ,T2 ∗ (µ, r, s) ∧ ρ)
= f −1 (0) = 0.
Similarly, we have f −1 (µ) ∧ C T1 ,T ∗
1 (f −1 (ρ), r, s) = 0. So, f −1 (µ) and
f −1 (ρ) are (r,s)-separated. Since λ is (r,s)-connected, by Theorem 2.6(3),
λ ≤ f −1 (µ) or λ ≤ f −1 (ρ), say λ ≤ f −1 (µ). Then f(λ) ≤ f(f −1 (µ)) ≤
µ. Therefore, f(λ) is (r,s)-connected. □
Definition 2.11. Let (X, T , T ∗ ) be an ifts. A fuzzy set λ is a (r,s)-
component in (X, T , T ∗ ) if λ is a maximal (r,s)-connected fuzzy set in
(X, T , T ∗ ), i.e. if µ ≥ λ and µ is (r,s)-connected, then µ = λ.
Theorem 2.12. Let (X, T , T ∗ ) be an ifts.
(1) If λ is (r, s)-component, then C T ,T ∗(λ, r, s) = λ.
(2) If λ 1 and λ 2 are (r, s)-components in (X, T , T ∗ ) such that λ 1 ∧λ 2 =
0, then λ 1 and λ 2 are (r, s)-separated.
(3) Each fuzzy point x t is connected.
(4) Every (r, s)-component is a crisp set.
Proof. (1) Since λ is (r,s)-connected and λ ≤ C T ,T ∗(λ, r, s), by Theorem
2.7(1), C T ,T ∗(λ, r, s) is (r,s)-connected. Since λ is (r,s)-component,
C T ,T ∗(λ, r, s) = λ.
(2) By (1), it is trivial.
(3) Let λ and µ be (r,s)-separated such that x t = λ ∨ µ. Then x t = λ
or x t = µ. If x t = λ, then
µ = µ ∧ (λ ∨ µ) = µ ∧ x t = µ ∧ λ ≤ µ ∧ C T ,T ∗(λ, r, s) = 0.
Similarly, if x t = µ, then λ = 0. Hence x t is connected.
(4) Let λ be a (r,s)-component with x ∈ supp(λ) = {x ∈ X | λ(x) >
0} and µ a (r,s)-component containing x 1 . Since λ ∧ µ ≥ x λ(x) ∧ x 1 =
x λ(x) ≠ 0, by Corollary 2.9, λ ∨ µ is (r,s)-connected. Thus λ = µ = λ ∨ µ
is (r,s)-component. So, x ∈ supp(λ) implies λ(x) = 1, that is, λ is a
crisp set.
□

Connectedness in intuitionistic fuzzy topological spaces 125
3. Stratification of intuitionistic fuzzy topological spaces
Theorem 3.1. Let (X, T , T ∗ ) be an ifts. Define the functions T st , T ∗
st :
I X → I as follows: for each λ ∈ I X ,
T st (λ) = ∨ { ∧
j∈J
T (λ j ) | λ = ∨ j∈J
(λ j ∧ α j ) }
where the first ∨ ∨
is taken over all families {λ j | j ∈ J} with λ =
j∈J (λ j ∧ α j ),
Tst(λ) ∗ = ∧ { ∨
(λ j ∧ α j ) }
j∈J
T ∗ (λ j ) | λ = ∨ j∈J
where the first ∧ is taken over all families {λ j | j ∈ J} with λ =
∨j∈J (λ j ∧α j ). Then (T st , T ∗
st) is the coarsest stratified IGO on X which
is finer than (T , T ∗ ).
Proof. First, we will show that (T st , T ∗
st) is a stratified IGO on X.
(IGO1) Suppose there exists λ ∈ I X such that
T st (λ) + T ∗
st(λ) > 1.
There exist r ∈ I and a family {λ j | j ∈ J} with λ = ∨ j∈J (λ j ∧ α j )
such that
T st (λ) ≥ ∧ T (λ j ) > r > 1 − Tst(λ).
∗
j∈J
Since T (λ j ) > r for each j ∈ J, there exist r j and s j such that
Hence
T (λ j ) ≥ r j > r, T ∗ (λ j ) ≤ s j ≤ 1 − r j .
Tst(λ) ∗ ≤ ∨ T ∗ (λ j ) ≤ ∨ s j ≤ ∨ − r j ) ≤ 1 − r.
j∈J
j∈J j∈J(1
It is contradiction.
(IGO2) and (IS). For each α ∈ I, there exists a family {1} with
α = α ∧ 1, we have T st (α) ≥ T (1) = 1 and Tst(α) ∗ ≤ T ∗ (1) = 0. Hence
T st (α) = 1 and Tst(α) ∗ = 0.

126 Yong Chan Kim and S. E. Abbas
(IGO3) Suppose there exist µ, ν ∈ I X and r ∈ I 0 , s ∈ I 1 with
T st (µ ∧ ν) < r < T st (µ) ∧ T st (ν) and T ∗
st(µ ∧ ν) > s > T ∗
st(µ) ∨ T ∗
st(ν).
Since (T st (µ) > r and T st (ν) > r) and (Tst(µ) ∗ < s and Tst(ν) ∗ < s), by
the definition of (T st , Tst), ∗ there exist two families {µ j | j ∈ J} with
µ = ∨ j∈J (µ j ∧ α j ) and {ν k | k ∈ K} with ν = ∨ k∈K (ν k ∧ α k ) such that
T st (µ) ≥ ∧ j∈J
T (µ j ) > r and T st (ν) ≥ ∧
k∈K
T (ν k ) > r
Tst(µ) ∗ ≤ ∨ T ∗ (µ j ) < s and Tst(ν) ∗ ≤ ∨
T ∗ (ν k ) < s.
j∈J
k∈K
Since I is completely distributive lattice, we have
µ ∧ ν = ( ∨
(µ j ∧ α j ) ) ∧ ( ∨
(ν k ∧ α k ) )
j∈J
k∈K
= ∨ j,k(µ j ∧ ν k ) ∧ (α j ∧ α k )
= ∨ j,k(µ j ∧ ν k ) ∧ α jk . (α jk = α j ∧ α k )
Moreover, since T (µ j ∧ ν k ) ≥ T (µ j ) ∧ T (ν k ) and T ∗ (µ j ∧ ν k ) ≤ T ∗ (µ j )
∨T ∗ (ν k ), we have
T st (µ ∧ ν) ≥ ∧ j,k
T (µ j ∧ ν k )
≥ ∧ (
T (µj ) ∧ T (ν k ) )
j,k
= ( ∧
T (µ j ) ) ∧ ( ∧
T (ν k ) ) > r,
j∈J
k∈K
T ∗
st(µ ∧ ν) ≤ ∨ j,k
T ∗ (µ j ∧ ν k )
≤ ∨ (
T ∗ (µ j ) ∨ T ∗ (ν k ) )
j,k
= ( ∨
T ∗ (µ j ) ) ∨ ( ∨
T ∗ (ν k ) ) < s.
j∈J
k∈K

Connectedness in intuitionistic fuzzy topological spaces 127
It is a contradiction. Hence, for all µ, ν ∈ I X ,
T st (µ ∧ ν) ≥ T st (µ) ∧ T st (ν) and T ∗
st(µ ∧ ν) ≤ T ∗
st(µ) ∨ T ∗
st(ν).
(IGO4) Suppose there exists a family {µ i ∈ I X | i ∈ Γ} and r ∈ I 0 ,
s ∈ I 1 with
T st ( ∨ i∈Γ
µ j ) < r < ∧ i∈Γ
T st (µ i ), Tst( ∨ ∗ µ j ) > s > ∨ Tst(µ ∗
i ).
i∈Γ
i∈Γ
Since T st (µ i ) > r and T ∗
st(µ i ) < s for each i ∈ Γ, there exists a family
{µ ij | j ∈ J i } with µ i = ∨ j∈J i
(µ ij ∧ α j ) such that
T st (µ i ) ≥ ∧
T (µ ij ) > r and Tst(µ ∗
i ) ≤ ∨
j∈J i
Since ∨ i∈Γ µ i = ∨ i∈Γ
T st ( ∨ i∈Γ
j∈J i
T ∗ (µ ij ) < s.
( ∨
j∈J i
(µ ij ∧ α j ) ) = ∨ i,j (µ ij ∧ α j ), we have
µ i ) ≥ ∧ i,j
T (µ ij ) = ∧ i∈Γ
( ∧
j∈J i
T (µ ij ) ) ≥ r,
Tst( ∨ ∗ µ i ) ≤ ∨ T ∗ (µ ij ) = ∨ ( ∨
T ∗ (µ ij ) ) ≤ s.
i∈Γ i,j
i∈Γ j∈J i
It is a contradiction. Hence, for any {µ i } i∈Γ ⊂ I X ,
T st ( ∨ i∈Γ
µ i ) ≥ ∧ i∈Γ
T st (µ i ) and Tst( ∨ ∗ µ i ) ≤ ∨ Tst(µ ∗
i ).
i∈Γ i∈Γ
Second, for each λ ∈ I X , there exists a family {1} with λ = 1 ∧ λ
such that T st (λ) ≥ T (λ) and Tst(λ) ∗ ≤ T ∗ (λ). Hence (T st , Tst) ∗ is finer
than (T , T ∗ ). Finally, if a stratified IGO (U, U ∗ ) is finer than (T , T ∗ ),
we will show that T st (λ) ≤ U(λ) and Tst(λ) ∗ ≥ U ∗ (λ) for all λ ∈ I X .
Suppose there exist µ ∈ I X and r ∈ I 0 , s ∈ I 1 such that
T st (µ) > r > U(µ) and T ∗
st(µ) < s < U ∗ (µ).
Since T st (µ) > r and T ∗
st(µ) < s, there exists a family {µ j | j ∈ J} with
µ = ∨ j∈J (µ j ∧ α j ) such that
T st (µ) ≥ ∧ j∈J
T (µ j ) > r and T ∗
st(µ) ≤ ∨ j∈J
T ∗ (µ j ) < s.

128 Yong Chan Kim and S. E. Abbas
On the other hand, since U(µ j ) ≥ T (µ j ) and U ∗ (µ j ) ≤ T ∗ (µ j ) for each
j ∈ J, we have
U(µ) = U( ∨ j∈J
(µ j ∧ α j )) ≥ ∧ j∈J
U(µ j ∧ α j )
≥ ∧ (U(µ j ) ∧ U(α j )) = ∧ U(µ j )
j∈J
j∈J
≥ ∧ j∈J
T (µ j ) > r,
U ∗ (µ) = U ∗ ( ∨ (µ j ∧ α j )) ≤ ∨ U ∗ (µ j ∧ α j )
j∈J
j∈J
≤ ∨ (U ∗ (µ j ) ∨ U ∗ (α j )) = ∨ U ∗ (µ j )
j∈J
j∈J
≤ ∨ j∈J
T ∗ (µ j ) < s.
It is a contradiction.
□
Definition 3.2. In the above theorem (T st , T ∗
st) is called the stratification
of an IGO (T , T ∗ ) on X.
Example 3.3. Let X = {a, b} be a set. Let µ, ρ ∈ I X as follows:
µ(x) = 0.5, µ(y) = 0.5 and ρ(x) = 0.4, ρ(y) = 0.6.
We define IGO on X as follows: for each λ ∈ I X
⎧
⎧
1, if λ = 1, 0
0, if λ = 1, 0
1
3
if λ = µ
⎪⎨ 1
2
if λ = ρ
T (λ) =
3
4
if λ = µ ∨ ρ
2
⎪⎩
3
if λ = µ ∧ ρ
0 otherwise,
2
3
if λ = µ
⎪⎨ 1
T ∗ 2
if λ = ρ
(λ) =
1
4
if λ = µ ∨ ρ
1
⎪⎩
3
if λ = µ ∧ ρ
1 otherwise.
If λ(x) = α for 0.5 < α < 0.6 and λ(y) = 0.6, for each β ≥ 0.6, since
λ = (α ∧ 1) ∨ (β ∧ (µ ∨ ρ))
= (α ∧ 1) ∨ (β ∧ ρ)

Connectedness in intuitionistic fuzzy topological spaces 129
we have T st (λ) = [T (1) ∧ T (µ ∨ ρ)] ∨ [T (1) ∧ T (ρ)] = 3 4 , and T st(λ) ∗ =
[T ∗ (1) ∨ T ∗ (µ ∨ ρ)] ∧ [T ∗ (1) ∨ T ∗ (ρ)] = 1 4 .
If λ(x) = α for 0.5 < α < 0.6 and λ(y) = β for 0.5 < α, β < 0.6 and
α < β, we have T st (λ) = 3 4 and T st(λ) ∗ = 1 4 .
If λ(x) = 0.5 and λ(y) = 0.6, since for α ≥ 0.5 and β ≥ 0.6,
λ = β ∧ (µ ∨ ρ) = (α ∧ µ) ∨ (β ∧ ρ),
we have T st (λ) = 3 4 and T ∗
st(λ) = 1 4 .
If λ(x) = 0.5 and λ(y) = β for 0.5 < β < 0.6, since
λ = β ∧ (µ ∨ ρ) = (β ∧ µ) ∨ (β ∧ ρ),
we have T st (λ) = 3 4 and T ∗
st(λ) = 1 4 .
If λ(x) = α and λ(y) = β for 0.4 < α, β < 0.5 and α < β, since, for
λ 1 ∈ {1, µ, µ ∨ ρ}, λ 2 = {ρ, µ ∧ ρ},
λ = (α ∧ λ 1 ) ∨ (β ∧ λ 2 ),
we have Tst(λ) ∗ = 2 3 and T st(λ) ∗ = 1 3
. By a similar method as the above
cases, we can obtain the following:
T st (λ) =
⎧
1 if λ = α, ∀α ∈ I,
3
⎪⎨ 4
if λ(x) = α and λ(y) = β for 0.5 ≤ α, β ≤ 0.6, α < β,
1
2
if λ(x) = α for 0.4 ≤ α < 0.5, λ(y) = β for 0.5 < β ≤ 0.6,
2
3
if λ(x) = α, λ(y) = β for 0.4 ≤ α, β ≤ 0.5, α < β,
⎪⎩
0 otherwise,
T ∗
st(λ) =
⎧
0 if λ = α, ∀α ∈ I,
1
⎪⎨ 4
if λ(x) = α and λ(y) = β for 0.5 ≤ α, β ≤ 0.6, α < β,
1
2
if λ(x) = α for 0.4 ≤ α < 0.5, λ(y) = β for 0.5 < β ≤ 0.6,
1
3
if λ(x) = α, λ(y) = β for 0.4 ≤ α, β ≤ 0.5, α < β,
⎪⎩
1 otherwise.

130 Yong Chan Kim and S. E. Abbas
Theorem 3.4. Let (X, T , T ∗ ) and (X, U, U ∗ ) be ifts’s. Let
(T st , Tst) ∗ and (U st , Ust) ∗ be stratification for (T , T ∗ ) and (U, U ∗ ) respectively.
If f : (X, T , T ∗ ) → (Y, U, U ∗ ) is intuitionistic continuous, then
f : (X, T st , Tst) ∗ → (Y, U st , Ust) ∗ is intuitionistic continuous.
Proof. Suppose there exist ν ∈ I Y
and r ∈ I 0 , s ∈ I 1 such that
U st (ν) > r > T st (f −1 (ν)), U ∗ st(ν) < s < T ∗
st(f −1 (ν)).
Since U st (ν) > r and U ∗ st(ν) < s, by the definition of (U st , U ∗ st), there
exists a family {ν j | j ∈ J} with ν = ∨ j∈J (ν j ∧ α j ) such that
U st (ν) ≥ ∧ j∈J
U(ν j ) > r, U ∗ st(ν) ≤ ∨ j∈J
U ∗ (ν j ) < s.
On the other hand, since
f −1 (ν) = f −1 ( ∨ j∈J
(ν j ∧ α j )) = ∨ j∈J
f −1 (ν j ) ∧ α j ,
by the definition of (T st , T ∗
st), we have
T st (f −1 (ν)) ≥ ∧ j∈J
T (f −1 (ν j )), T ∗
st(f −1 (ν)) ≤ ∨ j∈J
T ∗ (f −1 (ν j )).
Since f : (X, T , T ∗ ) → (Y, U, U ∗ ) is intuitionistic continuous, that is,
T (f −1 (ν j )) ≥ U(ν j ) and T ∗ (f −1 (ν j )) ≤ U ∗ (ν j ) for each j ∈ J,
T st (f −1 (ν)) ≥ ∧ j∈J
T (f −1 (ν j )) ≥ ∧ j∈J
U(ν j ) > r,
Tst(f ∗ −1 (ν)) ≤ ∨ T ∗ (f −1 (ν j )) ≤ ∨ U ∗ (ν j ) < s.
j∈J
j∈J
It is a contradiction. Hence f : (X, T st , Tst) ∗ → (Y, U st , Ust) ∗ is intuitionistic
continuous.
□
The converse of the previous theorem is not true from the following
example.

Connectedness in intuitionistic fuzzy topological spaces 131
Example 3.5. Let X be a nonempty set. Define IGO’s (T , T ∗ ) and
(U, U ∗ ) on X as follows: for each λ ∈ I X ,
{ 1 if λ = 1, 0,
T (λ) =
0 otherewise,
⎧
⎪⎨ 1 if λ = 1, 0,
1
U(λ) =
3
if λ = 0.5,
⎪⎩
0 otherewise,
{ 0 if λ = 1, 0,
T ∗ (λ) =
1 otherewise,
⎧
⎪⎨ 0 if λ = 1, 0
U ∗ 2
(λ) =
3
if λ = 0.5,
⎪⎩
1 otherewise.
Since 0 = T st (0.5) < U(0.5) = 1 3 and 1 = T ∗ (0.5) > U ∗ (0.5) =
2
3 , then the idintity function id X : (X, T , T ∗ ) → (X, U, U ∗ ) is not an
intuitionistic continuous. On the other hand, for a family {1} with
0.5 = 0.5 ∧ 1, we have U st (0.5) ≥ U(1) = 1 and Ust(0.5) ∗ ≤ U ∗ (1) = 0.
Hence U st (0.5) = 1 and Ust(0.5) ∗ = 0. Thus
: (X, T st , T ∗
st) → (X, U st , U ∗ st) is intuitionistic continu-
Therefore, id X
ous.
{ 1 if λ = α, ∀α ∈ L
T st (λ) = U st (λ) =
0, otherewise,
{ 0 if λ = α, ∀α ∈ I
Tst(λ) ∗ = Ust(λ) ∗ =
1, otherewise.
Theorem 3.6. Let (X, T st , T ∗
st) be a stratification of an ifts
(X, T , T ∗ ). A fuzzy set λ is (r,s)-component in (X, T , T ∗ ) iff λ is (r,s)-
component in (X, T st , T ∗
st).
Proof. (1) Let λ (r,s)-component in (X, T st , Tst). ∗ Suppose that λ
is not (r,s)-connected in (X, T , T ∗ ). Then µ ≠ 0 and ρ ≠ 0 are (r,s)-
separated in (X, T , T ∗ ) such that λ = µ∨ρ. Since T ≤ T st and T ∗ ≥ Tst
∗
from Theorem 3.1,
C Tst ,Tst ∗ (µ, r, s) ≤ C T ,T ∗(µ, r, s), C T st ,Tst ∗ (ρ, r, s) ≤ C T ,T ∗(ρ, r, s).
Hence µ and ρ are (r,s)-separated in (X, T st , Tst). ∗ Thus, λ is not (r,s)-
component in (X, T st , Tst). ∗ It is a contradiction.
(2) We will show that if λ is (r,s)-component in (X, T , T ∗ ), then λ is
(r,s)-connected in (X, T st , Tst). ∗ Let λ be a (r,s)-component in (X, T , T ∗ ).
Then C T ,T ∗(λ, r, s) = λ from Theorem 2.12(1). Suppose that λ is not

132 Yong Chan Kim and S. E. Abbas
(r,s)-connected in (X, T st , Tst), ∗ then µ ≠ 0 and ρ ≠ 0 are (r,s)-separated
in (X, T st , Tst) ∗ such that λ = µ ∨ ρ. Since T ≤ T st and T ∗ ≥ Tst,
∗
C Tst ,Tst ∗ (λ, r, s) ≤ C T ,T ∗(λ, r, s) = λ. So, C T st ,Tst ∗ (λ, r, s) = λ. Since
µ ≤ λ, we have C Tst ,Tst ∗ (µ, r, s) ≤ λ. It implies λ = C T st ,Tst ∗ (µ, r, s) ∨ ρ.
Put C Tst ,Tst ∗ (µ, r, s) = ω. If x ∈ supp(ω), then x ∈ supp(λ). Since λ is
(r,s)-component in (X, T , T ∗ ), by Theorem 2.12(4), x 1 ∈ λ = ω ∨ρ, that
is, ω(x) ∨ ρ(x) = 1. Since ω ∧ ρ = 0, then ρ(x) = 0. So, ω(x) = 1. Hence
ω is a crisp set. Since T st (1 − ω) ≥ r and T ∗
st(1 − ω) ≤ s from Theorem
1.3, and Theorem 1.4, for any family {α i ∧ η i | 1 − ω = ∨ i∈J α i ∧ η i },
T st (1 − ω) = ∨ { ∧ i∈J
T (η i )} ≥ r, T ∗
st(1 − ω) = ∧ { ∨ i∈J
T ∗ (η i )} ≤ s.
Without loss of generality, we assume that α i ≠ 0. Since ω(x) = 1 for
x ∈ supp(ω),
(1 − ω)(x) = ∨ i∈J(α i ∧ η i )(x)
⇒ 1 = ω(x) = ∧ i∈J(1 − α i )(x) ∨ (1 − η i )(x).
Thus, (1 − ω)(x) = ∨ i∈J η i(x) for x ∈ supp(ω). If y ∉ supp(ω), then
1 = (1 − ω)(y) = (α i ∧ η i )(y) ≤ ∨ i∈J η i(y). Hence, for any family
{α i ∧ η i | 1 − ω = ∨ i∈J α i ∧ η i }, we have
1 − ω = ∨ i∈J
η i .
It implies
T st (1 − ω) = T (1 − ω) = ∧ i∈J
T (η i ) ≥ r
T ∗
st(1 − ω) = T ∗ (1 − ω) = ∧ i∈J
T ∗ (η i ) ≤ s.
So, C T ,T ∗(ω, r, s) = ω. It implies
C T ,T ∗(C Tst ,Tst ∗ (µ, r, s), r, s) = C T st ,Tst ∗ (µ, r, s).

Connectedness in intuitionistic fuzzy topological spaces 133
Similarly, we have C T ,T ∗(C Tst ,Tst ∗ (ρ, r, s), r, s) = C T st ,Tst ∗ (ρ, r, s). So, µ
and ρ are (r,s)-separated in (X, T , T ∗ ) from
C T ,T ∗(µ, r, s) ∧ ρ
≤ (C T ,T ∗(C Tst ,Tst ∗ (µ, r, s), r, s) ∧ ρ) = (C T st ,Tst ∗ (µ, r, s) ∧ ρ) = 0,
µ ∧ C T ,T ∗(ρ, r, s)
≤ (µ ∧ C T ,T ∗(C Tst ,Tst ∗ (ρ, r, s), r, s)) = (µ ∧ C T st ,Tst ∗ (ρ, r, s)) = 0.
Thus, λ is not (r,s)-component in (X, T , T ∗ ). It is a contradiction.
(3) Let λ be a (r,s)-component in (X, T st , Tst). ∗ From (1), λ is (r,s)-
connected in (X, T , T ∗ ). There exists a (r,s)-component µ in (X, T , T ∗ )
containing λ. From (2), µ is (r,s)-connected in (X, T st , Tst). ∗ Thus λ = µ.
Let ρ be a (r,s)-component in (X, T , T ∗ ). Similarly, ρ is a (r,s)-
component in (X, T st , Tst).
∗
□
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Yong Chan Kim
Department of Mathematics
Kangnung National University
Gangneung 210-702, Korea
E-mail: yck@kangnung.ac.kr
S. E. Abbas
Department of Mathematics
Faculty of Science
South Valley University
Sohag 82524, Egypt
E-mail: sabbas@yahoo.com