# Some Properties of(r,s)-T0 and (r,s)-T1 Spaces

Authors: S. E. Abbas; Biljana Krsteska
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

We define(r,s)-quasi-T0, (r,s)-sub-T0, (r,s)-T0 and (r,s)-T1 spaces
in an intuitionistic fuzzy topological space and investigate some properties of these spaces and the relationships between them. Moreover, we study properties of subspaces and
their products.


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## Body

## 1. Introduction and Preliminaries

Kubiak [1] and Šostak [2]
introduced the fundamental concept of a fuzzy topological structure, as an
extension of both crisp topology and fuzzy topology [3], in the sense that not
only the objects are fuzzified, but also the axiomatics. In [4, 5], Šostak gave some
rules and showed how such an extension can be realized. Chattopadhyay et al.
[6] have redefined the same concept under the name gradation of openness. A
general approach to the study of topological-type structures on fuzzy power
sets was developed in [1, 7–10].As a generalization of fuzzy sets, the notion of intuitionistic
fuzzy sets was introduced by Atanassov [11]. By using intuitionistic fuzzy sets, Çoker [12], and Çoker and Dimirci [13] defined the
topology of intuitionistic fuzzy sets. Recently, Mondal and Samanta [14]
introduced the notion of intuitionistic gradation of openness of fuzzy sets,
where to each fuzzy subset there is a definite grade of openness and there is a
grade of nonopenness. Thus, the concept of intuitionistic gradation of openness
is a generalization of the concept of gradation of openness and the topology of
intuitionistic fuzzy sets.In this paper, we define(r,s)-quasi-T0, (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1 spaces in an intuitionistic fuzzy topological
space and investigate some properties of these
spaces and the relationships between them.
Moreover, we study properties of subspaces and
their products.Throughout this paper, letX be a nonempty set, I=[0,1], and I0=(0,1] and I1=[0,1).
For α∈I, α¯(x)=α for all x∈X. A fuzzy pointxt for t∈I0 is an element of IX such that(1.1)xt(y)={t,ify=x,0,ify≠x.
The set of all fuzzy
points in X is denoted by Pt(X).
A fuzzy point xt∈λ if and only if t<λ(x).
A fuzzy set λ is quasi-coincident with μ,
denoted by λqμ,
if there exists x∈X such that λ(x)+μ(x)>1.
If λ is not quasi-coincident with μ,
we denote it by  λq¯μ.Definition 1.1 (see [14]).
An intuitionistic gradation of openness (IGO, for short) onX is an ordered pair (τ,τ*) of functions from IX to I such that
(IGO1)
τ(λ)+τ*(λ)≤1,
for all λ∈IX,(IGO2)
τ(0¯)=τ(1¯)=1, τ*(0¯)=τ*(1¯)=0,(IGO3)
τ(λ1⋀λ2)≥τ(λ1)⋀τ(λ2) and τ*(λ1⋀λ2)≤τ*(λ1)∨τ*(λ2),
for each λ1,λ2∈IX,(IGO4)
τ(⋁i∈Δλi)≥⋀i∈Δτ(λi) and τ*(⋁i∈Δλi)≤⋁i∈Δτ*(λi),
for each λi∈IX,i∈Δ.
The triplet(X,τ,τ*) is called an intuitionistic fuzzy topological
space (IFTS, for
short). τ and τ* may be interpreted as gradation of openness
and gradation of nonopenness, respectively.
An IFTS(X,τ,τ*) is called stratified if
(S)τ(α¯)=1 and τ*(α¯)=0 for each α∈I.Let(𝒰,𝒰*) and (τ,τ*) be IGOs on X.
We say (𝒰,𝒰*) is finer than (τ,τ*) ((τ,τ*) is coarser than (𝒰,𝒰*)) if τ(λ)≤𝒰(λ) and τ*(λ)≥𝒰*(λ) for all λ∈IX.Theorem 1.2  (see [3, 15]).
Let(X,τ,τ*) be an IFTS. For each r∈I0,s∈I1,λ∈IX,
an operator 𝒞:IX×I0×I1→IX is defined
as follows:(1.2)𝒞(λ,r,s)=⋀{μ∣μ≥λ,τ(1¯−μ)≥r,τ*(1¯−μ)≤s}.Then it satisfies the following
properties:
(1)
𝒞(0¯,r,s)=0¯,𝒞(1¯,r,s)=1¯, for all r∈I0, s∈I1;(2)
𝒞(λ,r,s)≥λ;(3)
𝒞(λ1,r,s)≤𝒞(λ2,r,s),
if λ1≤λ2;(4)
𝒞(λ∨μ,r,s)=𝒞(λ,r,s)∨𝒞(μ,r,s), for all r∈I0, s∈I1;(5)
𝒞(λ,r,s)≤𝒞(λ,r′,s′),
if r≤r′, s≥s′,
where r,r′∈I0, s,s′∈I1;(6)
𝒞(𝒞(λ,r,s),r,s)=𝒞(λ,r,s).Definition 1.3 (see [14]).
A functionf:(X,τ,τ*)→(Y,𝒰,𝒰*) is said to be as follows:
(1)
IF continuous ifτ(f−1(μ))≥𝒰(μ) and τ*(f−1(μ))≤𝒰*(μ),
for each μ∈IY;(2)
IF open ifτ(μ)≤𝒰(f(μ)) and τ*(μ)≥𝒰*(f(μ)),
for each μ∈IX;(3)
IF homeomorphism if and only iff is bijective and both f and f−1 are IF continuous.Definition 1.4 (see [16]).
Let0¯∉ΘX be a subset of IX.
A pair (β,β*) of functions β,β*:ΘX→I is called an IF topological base on X if it satisfies the following conditions:
(B1)
β(λ)+β*(λ)≤1, ∀λ∈ΘX,(B2)
β(1¯)=1 and β*(1¯)=0,(B3)
β(λ1⋀λ2)≥β(λ1)⋀β(λ2) and β*(λ1⋀λ2)≤β*(λ1)∨β*(λ2),
for each λ1,λ2∈ΘX.An IF topological base(β,β*) always generates an IGO, (τβ,τβ**) on X in the following sense.Theorem 1.5 (see [16]).
Let(β,β*) be an IF topological base for X.
Define the functions τβ,τβ**:IX→I as follows: for each μ∈IX,(1.3)τβ(μ)={⋁{⋀i∈Jβ(μi)},ifμ=⋁i∈Jμi,μi∈ΘX,1,ifμ=0¯,0,otherwise,where ⋁ is taken over all families {μi∈ΘX∣μ=⋁i∈Jμi}, (1.4)τβ**(μ)={⋀{⋁i∈Jβ*(μi)},ifμ=⋁i∈Jμi,μi∈ΘX,0,ifμ=0¯,1,otherwise,where ⋀ is taken over all families {μi∈ΘX∣μ=⋁i∈Jμi}.
Then
(1)
(X,τβ,τβ**) is an IFTS;(2)
A mapf:(Y,τ,τ*)→(X,τβ,τβ**) is IF continuous if and only if β(λ)≤τ(f−1(λ)) and β*(λ)≥τ*(f−1(λ)),
for all λ∈ΘX.Lemma 1.6 (see [17]).
LetX be a product of the family {Xi∣i∈Γ} of sets, and for each i∈Γ, πi:X→Xi a projection map. For each λ∈IX, i,j∈Γ,
and λi∈IXi,
the following properties hold:
(1)
πi(πi−1(λi)⋀λ)=λi⋀πi(λ);(2)
if⋁xi∈Xiλi(xi)=αi for i∈F with each finite index subset F of Γ−{j} and put α=⋀i∈Fαi,
then
(a)
⋁x∈X(⋀i∈Fπi−1(λi))(x)=α;(b)
πi(⋀i∈Fπi−1(λi))=α¯.Definition 1.7.
Let(X,τ,τ*) be an IFTS, μ∈IX, xt∈Pt(X), r∈I0, and s∈I1.
It holds that(1.5)𝒬(xt,r,s)={μ∈IX∣xtqμ,τ(μ)≥r,τ*(μ)≤s}.A fuzzy set μ∈𝒬(xt,r,s) is called (r,s)-𝒬 open neighborhood of xt.


## 2. Some Properties of Product Intuitionistic Fuzzy Topological Spaces

Theorem 2.1.
Let{(Xi,τi,τi*)}i∈Γ be a family of IFTSs, let X be a set and for each i∈Γ, fi:X→Xi a map. Let (2.1)ΘX={0¯≠μ=⋀j=1nfkj−1(νkj)∣τkj(νkj)>0,∀kj∈K},for every finite set K={k1,…,kn}⊂Γ.
Define the functions β,β*:ΘX→I on X by
(2.2)β(μ)=⋁{⋀j=1nτkj(νkj)∣μ=⋀j=1nfkj−1(νkj)},β*(μ)=⋀{⋁j=1nτkj*(νkj)∣μ=⋀j=1nfkj−1(νkj)}, where ⋁ and ⋀ are taken over all finite subsets K={k1,k2,…,kn}⊂Γ.
Then,
(1)
(β,β*) is an IF topological base on X;(2)
the IGO,(τβ,τβ**) generated by (β,β*) is the coarsest IGO on X for which each i∈Γ, fi is IF continuous;(3)
a mapf:(Y,τ1,τ1*)→(X,τβ,τβ**) is IF continuous if and only if for each i∈Γ, fi∘f is IF continuous.Proof.
(B1) It is trivial.
(B2) Sinceλ=fi−1(λ) for each λ∈{0¯,1¯}, β(1¯)=β(0¯)=1 and β*(1¯)=β*(0¯)=0.
(B3) For all finite subsetsK={k1,…,kp} and J={j1,…,jq} of Γ such that (2.3)λ=⋀i=1pfki−1(λki),μ=⋀i=1qfji−1(μji),we have(2.4)λ⋀μ=(⋀i=1pfki−1(λki))⋀(⋀i=1qfji−1(μji)).Furthermore, we have for each k∈K∩J,(2.5)fk−1(λk)⋀fk−1(μk)=fk−1(λk⋀μk).Put λ⋀μ=⋀mi∈K∪Jfmi−1(μi)(ρmi), where (2.6)ρmi={λmiifmi∈K−(K∩J),μmiifmi∈J−(K∩J),λmi⋀μmiifmi∈(K∩J).
We have(2.7)β(λ⋀μ)≥⋀j∈K∪Jτj(ρj)≥(⋀i=1pτki(λki))⋀(⋀i=1qτji(μji)),β*(λ⋀μ)≤⋁j∈K∪Jτj*(ρj)≤(⋁i=1pτki*(λki))∨(⋁i=1qτji*(μji)).
Then,β(λ⋀μ)≥β(λ)⋀β(μ) and β*(λ⋀μ)≤β*(λ)∨β*(μ).
(2) For eachλi∈IXi,
one family {fi−1(λi)}, and i∈Γ,
we have(2.8)τβ(fi−1(λi))≥β(fi−1(λi))≥τi(λi),τβ**(fi−1(λi))≤β*(fi−1(λi))≤τi*(λi). Thus, for each i∈Γ, fi:(X,τβ,τβ**)→(Xi,τi,τi*) is IF continuous. Let fi:(X,τ∘,τ∘*)→(Xi,τi,τi*) be IF continuous, that is, for each i∈Γ and λi∈IXi, τ∘(fi−1(λi))≥τi(λi) and τ∘*(fi−1(λi))≤τi*(λi).
For all finite subsets K={k1,…,kp} of Γ such that λ=⋀i=1pfki−1(λki),
we have (2.9)τ∘(λ)≥⋀i=1pτ∘(fki−1(λki))≥⋀i=1pτki(λki),τ∘*(λ)≤⋁i=1pτ∘*(fki−1(λki))≤⋁i=1pτki*(λki).
It impliesτ∘(λ)≥β(λ) and τ∘*(λ)≤β*(λ) for each λ∈IX.
By Theorem 1.5(2), τ∘≥τβ and τ∘*≤τβ**.
(3) (⇒) Let f:(Y,τ1,τ1*)→(X,τβ,τβ**) be an IF continuous. For each i∈Γ and λi∈IXi,
we have(2.10)τ1((fi∘f)−1(λi))=τ1((f−1(fi−1(λi)))≥τβ(fi−1(λi)))≥τi(λi),τ1*((fi∘f)−1(λi))=τ1*((f−1(fi−1(λi)))≤τβ**(fi−1(λi)))≤τi*(λi). Hence, fi∘f:(Y,τ1,τ1*)→(Xi,τi,τi*) is IF continuous.
(⇐) For all finite subsets K={k1,…,kp} of Γ such that λ=⋀i=1pfki−1(λki),
since fki∘f:(Y,τ1,τ1*)→(Xki,τki,τki*) is IF continuous,
(A)τ1(f−1(fki−1(λki)))≥τki(λki),(B)τ1*(f−1(fki−1(λki)))≤τki*(λki).
Hence, we have(2.11)τ1(f−1(λ))=τ1(f−1(⋀i=1pfki−1(λki)))=τ1(⋀i=1pf−1(fki−1(λki)))≥⋀i=1pτ1(f−1(fki−1(λki)))≥⋀i=1pτki(λki),(by(A)),τ1*(f−1(λ))=τ1*(f−1(⋀i=1pfki−1(λki)))=τ1*(⋀i=1pf−1(fki−1(λki)))≤⋁i=1pτ1*(f−1(fki−1(λki)))≤⋁i=1pτki*(λki),(by(B)).
It impliesτ1(f−1(λ))≥β(λ) and τ1*(f−1(λ))≤β*(λ) for all λ∈IX.
By Theorem 1.5(2), f:(Y,τ1,τ1*)→(X,τβ,τβ**) is IF continuous.Definition 2.2.
Let(X,τ,τ*) be an IFTS and A⊂X.
The triple (A,τ|A,τ*|A) is said to be a subspace of (X,τ,τ*) if (τ|A,τ*|A) is the coarsest IGO on A for which the inclusion map i is IF continuous.Definition 2.3.
LetX be the product ∏i∈ΓXi of the family {(Xi,τi,τi*)∣i∈Γ} of IFTSs. The coarsest IGO, (τ,τ*)=(⨂τi,⨂τi*) on X for which each the projections πi:X→Xi is IF continuous, is called the product IGO of {(τi,τi*)∣i∈Γ} and (X,τ,τ*) is called the product IFTS.Lemma 2.4.
Let(Y,𝒰,𝒰*) be an IFTS and (β,β*) an IF topological base on X.
If f:(X,β,β*)→(Y,𝒰,𝒰*) is a function such that β(λ)≤𝒰(f(λ)) and β*(λ)≥𝒰*(f(λ)) for all λ∈ΘX,
then f:(X,τβ,τβ**)→(Y,𝒰,𝒰*) is IF open.Proof.
Suppose there existsμ∈IX such that(2.12)τβ(μ)>𝒰(f(μ))orτβ**(μ)<𝒰*(f(μ)),then there exists a family {λi∈ΘX∣μ=⋁i∈Γλi} such that(2.13)τβ(μ)≥⋀i∈Γβ(λi)>𝒰(f(μ))orτβ**(μ)≤⋁i∈Γβ*(λi)<𝒰*(f(μ)).On the other hand, since β(λ)≤𝒰(f(λ)) and β*(λ)≥𝒰*(f(λ))∀λ∈ΘX,
then we have(2.14)⋀i∈Γβ(λi)≤⋀i∈Γ𝒰(f(λi))≤𝒰[⋁i∈Γ(f(λi))]=𝒰[f(⋁i∈Γ(λi))]=𝒰(f(μ)),⋁i∈Γβ*(λi)≥⋁i∈Γ𝒰*(f(λi))≥𝒰*[⋁i∈Γ(f(λi))]=𝒰*[f(⋁i∈Γ(λi))]=𝒰*(f(μ)). It is a contradiction. Hence f is IF open.Theorem 2.5.
Let(X,τβ,τβ**) be a product space of a family {(Xi,τi,τi*)∣i∈Γ} of IFTS's. Then the following statements are
equivalent:
(1)
a projection mapπj:(X,τβ,τβ**)→(Xj,τj,τj*) is IF open;(2)
for everyμ=⋀i∈Γπi−1(λi) such that ⋁x∈Xiλi(x)=αi for each αi∈I and i∈Γ∘ such that a finite index subset Γ∘ of Γ−{j} and τi(λi)>0,
then  ⋀i∈Γ∘τi(λi)≤τj(α¯) and ⋁i∈Γ∘τi*(λi)≥τj*(α¯), where α=⋀i∈Γ∘αi.Proof.
(1)⇒(2): For every  μ=⋀i∈Γπi−1(λi) such that ⋁x∈Xiλi(x)=αi for each αi∈I and i∈Γ∘ such that a  finite index subset Γ∘ of Γ−{j}.
By Lemma 1.6(2(b)), we have, for α=⋀i∈Γ∘αi,(2.15)πj(μ)=πj(⋀i∈Γ∘πi−1(λi))=α¯.Since μ∈ΘX,
by Theorem 2.1, we have(2.16)⋀i∈Γ∘τi(λi)≤β(μ)≤τβ(μ),⋁i∈Γ∘τi*(λi)≥β*(μ)≥τβ**(μ).Furthermore, since πj is IF open, we have(2.17)τβ(μ)≤τj(πj(μ))=τj(α¯),τβ**(μ)≥τj*(πj(μ))=τj*(α¯).Hence, ⋀i∈Γ∘τi(λi)≤τj(α¯) and ⋁i∈Γ∘τi*(λi)≥τj*(α¯).
(2)⇒(1): From Lemma 2.4, we only show that β(μ)≤τj(πj(λ)) and β*(μ)≥τj*(πj(λ)) for all λ∈ΘX.
Suppose that there exists ν∈ΘX such that β(ν)>τj(πj(ν)) or β*(ν)<τj*(πj(ν)).
Then there exists a finite index subset Γ∘ of Γ−{j} with ν=πj−1(λj)⋀[⋀i∈Γ∘πi−1(λi)] (if necessary, we can take λj=1¯) such that(2.18)β(ν)≥τj(λj)⋀[⋀i∈Γ∘τi(λi)]>τj(πj(ν)),β*(ν)≤τj*(λj)∨[⋁i∈Γ∘τi*(λi)]<τj*(πj(ν)). On the other hand, by Lemma 1.6(2), we have(2.19)πj(ν)=πj[πj−1(λj)⋀(⋀i∈Γ∘πi−1(λi))]=λj⋀πj[⋀i∈Γ∘πi−1(λi)]=λj⋀α¯,where ⋁x∈Xiλi(x)=αi and α=⋀i∈Γ∘αi.
Since ⋀i∈Γ∘τi(λi)≤τj(α¯) and ⋁i∈Γ∘τi*(λi)≥τj*(α¯),
we have(2.20)τj(πj(ν))=τj(λj⋀α¯)≥τj(λj)⋀τj(α¯)≥τj(λj)⋀(⋀i∈Γ∘τi(λi)),τj*(πj(ν))=τj*(λj⋀α¯)≤τj*(λj)⋀τj*(α¯)≤τj*(λj)∨(⋁i∈Γ∘τi*(λi)).It is a contradiction.Theorem 2.6.
Let(X,τ,τ*) be a product space of a family {(Xi,τi,τi*)∣i∈Γ} of IFTSs and (Xj,τj,τj*) be stratified. Then, the following properties
hold:
(1)
(X,τ,τ*) is stratified;(2)
a projection mapπj:X→Xj is IF open.Proof.
(1) It is clear from the following: for allα∈I,(2.21)τ(α¯)≥β(α¯)=⋁{⋀i∈Γ∘τi(λi)∣α¯=⋀i∈Γ∘πi−1(λi)}≥τj(α¯)=1,τ*(α¯)≤β*(α¯)=⋀{⋁i∈Γ∘τi*(λi)∣α¯=⋀i∈Γ∘πi−1(λi)}≤τj*(α¯)=0.
(2) Sinceτj(α¯)=1 and τj*(α¯)=0 for all α∈I,
it satisfies the condition of Theorem 2.5(2).Theorem 2.7.
Let(X,τ,τ*) be a product space of a family {(Xi,τi,τi*)∣i∈Γ} of IFTSs and let (Xj,τj,τj*) be stratified. Then for every X˜j=Xj×∏{yi∣i≠j} in X parallel to Xj, πj|X˜j:X˜j→Xj is an IF homeomorphism.Proof.
LetX˜j=Xj×∏{yi∣i≠j}.
Since i:X˜j→X˜j and πj:X˜j→Xj are IF continuous, πj∘i=πj|X˜j is IF continuous. Moreover, πj|X˜j is bijective.
Now we only show thatπj|X˜j is IF open. Suppose there exists μ∈IX˜j such that(2.22)τ|X˜j(μ)>τj(πj|X˜j(μ))orτ*|X˜j(μ)<τj*(πj|X˜j(μ)).Then there exists ν∈IX with μ=i−1(ν) such that(2.23)τ|X˜j(μ)≥τ(ν)>τj(πj|X˜j(μ))orτ*|X˜j(μ)≤τ*(ν)<τj*(πj|X˜j(μ)).From the definition of (τ,τ*),
there exists a family {νk∈ΘX∣ν=⋁k∈Kνk} such that(C)τ(ν)≥⋀k∈Kβ(νk)>τj(πj|X˜j(μ))orτ*(ν)≤⋁k∈Kβ*(νk)<τj*(πj|X˜j(μ)).On the other hand, since each νk∈ΘX,
there exists a finite index Fk of Γ−{j} with νk=πj−1(λkj)⋀(⋀i∈Fkπi−1(λi)). Since πi−1(λi)(x)=yi for i≠j,
then for each x∈X˜j, ⋀i∈Fkπi−1(λi)(x)=(⋀i∈Fkλi)(yi).
Put αk=(⋀i∈Fkλi)(yi).
Let μk=i−1(νk) for  each k∈K.
Then,(2.24)πj|X˜j(μk)(xj)=⋁{μk(x)∣x∈X˜j,πj|X˜j(x)=xj}=⋁{i−1(νk)(x)∣x∈X˜j,πj(x)=xj(μk=i−1(νk))}=⋁{νk(x)∣x∈X˜j,πj(x)=xj}=⋁{πj−1(λkj)(x)⋀(⋀i∈Fkπi−1(λi)(x)∣x∈X˜j,πj(x)=xj)}=⋁{λkj(πj(x))⋀(⋀i∈Fkλi)(πi(x))∣x∈X˜j,πj(x)=xj}=λkj(xj)⋀(⋀i∈Fkλi)yi=λkj(xj)⋀αk=(λkj⋀αk¯)(xj).Hence, πj|X˜j(μk)=λkj⋀αk¯.
Thus,(2.25)τj(πj|X˜j(μk))=τj(λkj⋀αk¯)≥τj(λkj)⋀τj(αk¯)=τj(λkj)≥τj(λkj)⋀(⋀i∈Fkλi),τj*(πj|X˜j(μk))=τj*(λkj⋀αk¯)≤τj*(λkj)∨τj*(αk¯)=τj*(λkj)≤τj*(λkj)∨(⋀i∈Fkλi).From the definition of (β,β*),
it implies(2.26)τj(πj|X˜j(μk))≥β(νk),τj*(πj|X˜j(μk))≤β*(νk).Thus,(2.27)τj(πj|X˜j(μ))≥⋀k∈Kτj(πj|X˜j(μk))≥⋀k∈Kβ(νk),τj*(πj|X˜j(μ))≤⋁k∈Kτj*(πj|X˜j(μk))≤⋁k∈Kβ*(νk). It is a contradiction for (C).In an IFTS{(Xi,τi,τi*)∣i∈Γ}, X˜j=Xj×∏{yi∣i≠j} need not be homeomorphic to X from the following example.Example 2.8.
LetX={x1,x2,x3}, Y={y1,y2}, and Z={z1,z2} be sets and W=X×Y×Z a product set. Let π1:W→X, π2:W→Y, and π3:W→Z be the projection
maps. Define τ1,τ1*:IX→I by(2.28)τ1(λ)={1ifλ=0¯,1,¯12ifλ=λ1,0otherwise,τ1*(λ)={0ifλ=0¯,1,¯12ifλ=λ1,1otherwise,where λ1(x1)=0.5, λ1(x2)=0.2, and λ1(x3)=0.3.
Also, X˜j={(x,y2,z2):x∈X}, define τ|X˜j,τ*|X˜j:IX˜j→I by
(2.29)τ|X˜j(μ)={1ifμ=0¯,1,¯12ifμ=μ1,23ifμ=0.1,¯14ifμ=0.7,¯0otherwise,τ*|X˜j(μ)={0ifμ=0¯,1,¯12ifμ=μ1,13ifμ=0.1,¯34ifμ=0.7,¯1otherwise,where μ1(x1,y2,z2)=0.5, μ1(x2,y2,z2)=0.2, and μ1(x3,y2,z2)=0.3.
Then the projection map πj|X˜j:X˜j→X is bijective IF continuous, but πj|X˜j is not IF open, because(2.30)23=τ|X˜j(0.1¯)≰τ1(π1|X˜j(0.1¯))=0.Hence, X˜j and X are not homeomorphic.Theorem 2.9.
Let(X,τ,τ*) be a product space of a family {(Xi,τi,τi*)∣i∈Γ} of IFTSs. Then, the following properties
hold:
(1)
𝒞τ,τ*(∏i∈Γλi,r,s)≤∏i∈Γ𝒞τi,τi*(λi,r,s),∀λi∈IXi,r∈I0,s∈I1;(2)
if𝒞τi,τi*(λi,r,s)=λi,∀λi∈IXi,r∈I0,s∈I1,
then 𝒞τ,τ*(∏i∈Γλi,r,s)=∏i∈Γλi.Proof.
(1) Suppose𝒞τ,τ*(∏i∈Γλi,r,s)≰∏i∈Γ𝒞τi,τi*(λi,r,s).
Then there exist x∈X and t∈(0,1) such that(D)𝒞τ,τ*(∏i∈Γλi,r,s)(x)≥t>∏i∈Γ𝒞τi,τi*(λi,r,s)(x).Since ∏i∈Γ𝒞τi,τi*(λi,r,s)<t,
there exists j∈Γ such that ∏i∈Γ𝒞τi,τi*(λi,r,s)≤πj−1(𝒞τi,τi*(λi,r,s))<t.
Put πj(x)=xj.
It implies 𝒞τj,τj*(λj,r,s)(xj)<t.
From the definition of 𝒞τj,τj*,
there exists μj∈IXj with λj≤μj and τj(1¯−μj)≥r, τj*(1¯−μj)≤s such that(2.31)𝒞τj,τj*(λj,r,s)(xj)≤μj(xj)<t.On the other hand, we
have(2.32)λj≤μj⇒πj−1(λj)≤πj−1(μj)⇒∏i∈Γλi=⋀i∈Γπj−1(λi)≤πj−1(μj)⇒𝒞τ,τ*(∏i∈Γλi,r,s)≤πj−1(μj),because τ(1¯−πj−1(μj))=τ(πj−1(1¯−μj))≥τj(1¯−μj)≥r and τ*(1¯−πj−1(μj))=τ*(πj−1(1¯−μj))≤τj*(1¯−μj)≤s.
Hence,(2.33)𝒞τ,τ*(∏i∈Γλi,r,s)(x)≤πj−1(μj)(x)=μj(xj)<t.It is a contradiction for (D).
Hence,(2.34)𝒞τ,τ*(∏i∈Γλi,r,s)≤∏i∈Γ𝒞τi,τi*(λi,r,s).
(2) It is clear from the following:(2.35)∏i∈Γλi≤𝒞τ,τ*(∏i∈Γλi,r,s)≤∏i∈Γ𝒞τi,τi*(λi,r,s)=∏i∈Γλi.


## 3. Some Properties of(r,s)-T0 and (r,s)-T1 Spaces

Definition 3.1.
An IFTS(X,τ,τ*) is said to be as follows.
(1)
(r,s)-quasi-T0 space if for each xt,xm∈Pt(X) and t<m,
there exists λ∈𝒬(xm,r,s) such that xtq¯λ.(2)
(r,s)-sub-T0 space if for each x≠y∈X,
there exists t∈I0 such that there exists λ∈𝒬(xt,r,s) such that ytq¯λ,
or there exists μ∈𝒬(yt,r,s) such that xtq¯μ.(3)
(r,s)-T0 space if for each xt,ym∈Pt(X),
there exists λ∈𝒬(xt,r,s) such that ymq¯λ,
or there exists μ∈𝒬(ym,r,s) such that xtq¯μ.(4)
(r,s)-T1 space if for each xt,ym∈Pt(X) such that xt≰ym,
there exists λ∈𝒬(xt,r,s) such that ymq¯λ.Theorem 3.2.
Let(X,τ,τ*) be an IFTS. Then the following statements are
equivalent:
(1)
(X,τ,τ*) is (r,s)-T0 space;(2)
for eachxt,ym∈Pt(X), 𝒬(xt,r,s)≠𝒬(ym,r,s);(3)
for eachxt,ym∈Pt(X),
then  xt∉𝒞τ,τ*(ym,r,s) or ym∉𝒞τ,τ*(xt,r,s).Proof.
(1)⇒(2): It is trivial.
(2)⇒(3): Let λ∈𝒬(xt,r,s) and λ∉𝒬(ym,r,s).
Since λ∉𝒬(ym,r,s),
we have(3.1)ym≤1¯−λ,τ(λ)≥r,τ*(λ)≤s.By Theorem 1.2, we have 𝒞τ,τ*(ym,r,s)≤1¯−λ.
Since xtqλ and λ≤1¯−𝒞τ,τ*(ym,r,s),
then xtq[1¯−𝒞τ,τ*(ym,r,s)].
Hence, xt∉𝒞τ,τ*(ym,r,s).
(3)⇒(1): Let xt,ym∈Pt(X) and xt∉𝒞τ,τ*(ym,r,s).
Then t>𝒞τ,τ*(ym,r,s)(x) implies xtq[1¯−𝒞τ,τ*(ym,r,s)].
Since 𝒞τ,τ*(ym,r,s)=⋀{μ∣μ≥ym,τ(1¯−μ)≥r,τ*(1¯−μ)≤s}.
Since τ(⋁(1¯−μ))≥⋀τ(1¯−μ) and τ*(⋁(1¯−μ))≤⋁τ*(1¯−μ),
we have τ(1¯−𝒞τ,τ*(ym,r,s))≥r and τ*(1¯−𝒞τ,τ*(ym,r,s))≤s.
Hence, [1¯−𝒞τ,τ*(ym,r,s)]∈𝒬(xt,r,s).
Since ym∈𝒞τ,τ*(ym,r,s) and ymq¯[1¯−𝒞τ,τ*(ym,r,s))].
Thus, (X,τ,τ*) is (r,s)-T0 space.We can prove the following corollaries as a similar
method as Theorem3.2.Corollary 3.3.
Let(X,τ,τ*) be an IFTS. Then the following statements are
equivalent:
(1)
(X,τ,τ*) is (r,s)-quasi-T0 space;(2)
for eachxt,xm∈Pt(X), 𝒬(xt,r,s)≠𝒬(xm,r,s);(3)
for eachxt,xm∈Pt(X),
then  xt∉𝒞τ,τ*(xm,r,s) or xm∉𝒞τ,τ*(xt,r,s).Corollary 3.4.
Let(X,τ,τ*) be an IFTS. Then the following statements are
equivalent:
(1)
(X,τ,τ*) is (r,s)-sub-T0 space;(2)
for eachx≠y∈X,
there exists t∈I0 such that 𝒬(xt,r,s)≠𝒬(yt,r,s);(3)
for eachx≠y∈X,
there exists t∈I0 such that xt∉𝒞τ,τ*(yt,r,s) or yt∉𝒞τ,τ*(xt,r,s).Example 3.5.
LetX={x,y} be a set. We define an IGO (τ,τ*) on X as follows:(3.2)τ(λ)={1ifλ=1¯or0¯,12ifλ=x0.7,0otherwise,τ*(λ)={0ifλ=1¯or0¯,12ifλ=x0.7,1otherwise.For each x≠y∈X,
there exists 0.4∈I0 such that x0.7∈𝒬(x0.4,1/2,1/2) and y0.4q¯x0.7.
Hence, (X,τ,τ*) is (1/2,1/2)-sub-T0 space. On the other hand, since 𝒬(y0.5,1/2,1/2)=𝒬(y0.6,1/2,1/2)={1¯},
by Corollary 3.3(2), (X,τ,τ*) is not (1/2,1/2)-quasi-T0 space.Theorem 3.6.
Let(X,τ,τ*) be an IFTS. Then the following statements are
equivalent:
(1)
(X,τ,τ*) is (r,s)-T1 space;(2)
for eachxt∈Pt(X), xt=𝒞τ,τ*(xt,r,s);(3)
for eachλ∈IX, λ=⋀{μ∣λ≤μ,τ(μ)≥r,τ*(μ)≤s}.Proof.
(1)⇒(2): We only show that 𝒞τ,τ*(xt,r,s)≤xt.
Let ym∈𝒞τ,τ*(xt,r,s).
Suppose that ym≰xt.
Since (X,τ,τ*) is (r,s)-T1 space, there exists λ∈𝒬(ym,r,s) such that xtq¯λ.
It implies xt≤1¯−λ with τ(λ)≥r and τ*(λ)≤s.
Hence, 𝒞τ,τ*(xt,r,s)≤1¯−λ.
Since ym∈𝒞τ,τ*(xt,r,s)≤1¯−λ,
we have λ∉𝒬(ym,r,s).
It is a contradiction. Hence, ym≤xt.
Since ym∈𝒞τ,τ*(xt,r,s) implies ym≤xt,
then 𝒞τ,τ*(xt,r,s)≤xt.
(2)⇒(3): Let ρ=⋀{μ∣λ≤μ,τ(μ)≥r,τ*(μ)≤s}.
We only show that ρ≤λ.
Suppose there exist x∈X and t∈(0,1) such that(3.3)ρ(x)>1−t≥λ(x).Then, λ≤1¯−xt.
Since xt=𝒞τ,τ*(xt,r,s),(3.4)τ(1¯−xt)=τ(1¯−𝒞τ,τ*(xt,r,s))≥r,τ*(1¯−𝒞τ,τ*(xt,r,s))≤s.Hence, ρ≤1¯−xt.
It is a contradiction.
(3)⇒(1): For each xt,ym∈Pt(X) such that xt≰ym, 1¯−xt≱1¯−ym.
From (3), since 1¯−ym=⋀{μ∣1¯−ym≤μ,τ(μ)≥r,τ*(μ)≤s},
there exists μ=1¯−ym∈IX such that(3.5)τ(1¯−ym)≥r,τ*(1¯−ym)≤s.Moreover, since 1¯−xt≱1¯−ym,
we have xtq[1¯−ym].
Thus 1¯−ym∈𝒬(xt,r,s) such that ymq¯[1¯−ym].
Hence, (X,τ,τ*) is (r,s)-T1 space.Theorem 3.7.
Let(X,τ,τ*) be a stratified IFTS. Then (X,τ,τ*) is (r,s)-quasi-T0 space for all r∈I0,s∈I1.Proof.
Letxt,xm∈Pt(X) such that t<m.
Then there exists α∈I0 such that(3.6)t≤1−α<m.Since (X,τ,τ*) is stratified IFTS, we have τ(α¯)=1 and τ*(α¯)=0.
Hence, α¯∈𝒬(xm,r,s) such that xtq¯α¯.Theorem 3.8.
(1) Every(r,s)-T0 space is both (r,s)-quasi-T0 and (r,s)-sub-T0.
(2) Every(r,s)-T1 space is (r,s)-T0.Proof.
(1) For eachxt,xm∈Pt(X) such that t<m.
By (r,s)-T0 space, there exists λ∈𝒬(xt,r,s) such that xmq¯λ.
Since t<m,
we have xtq¯λ.
So, X is (r,s)-quasi-T0.
(2) For eachxt,ym∈Pt(X),
if xt≰ym,
by (r,s)-T1 space, there exists λ∈𝒬(xt,r,s) such that ymq¯λ.
Also, if ym≰xt,
by (r,s)-T1 space, there exists μ∈𝒬(ym,r,s) such that xtq¯μ.
Hence, X is (r,s)-T0.The converse of Theorem3.8 is not true from the
following examples.Example 3.9.
LetX={x,y} be a set. We define an IGO (τ,τ*) on X as follows:(3.7)τ(λ)={1ifλ=α¯forα∈I,13ifλ=μpq,0otherwise,τ*(λ)={0ifλ=α¯forα∈I,23ifλ=μpq,1otherwise,where for each 0<p≤0.4, μpq(x)=p and μpq(y)=q, 0≤q<p.
Since (X,τ,τ*) is a stratified IFTS, by Theorem 3.7, (X,τ,τ*) is (r,s)-quasi-T0 space for each r∈I0 and s∈I1.
Ifr>1/3, s<2/3,
and t∈I0,
then for each xt,yt∈Pt(X),
we have(3.8)𝒬(xt,r,s)=𝒬(yt,r,s)={α¯∣1−t<α≤1}.By Theorem 3.2(2) and
Corollary 3.4(2), (X,τ,τ*) is neither (r,s)-sub-T0 nor (r,s)-T0 for r>1/3 and s<2/3.
If0<r≤1/3, 2/3≤s<1, and x≠y∈X,
there exists 0.7∈I0 such that there exists μ(2/5)0∈𝒬(x0.7,r,s) with y0.7q¯μ(2/5)0.
Hence, (X,τ,τ*) is (r,s)-sub-T0 for 0<r≤1/3 and 2/3≤s<1.
Forx0.3,y0.3∈Pt(X), 0<r≤1/3,
and 2/3≤s<1,
we have 𝒬(x0.3,r,s)=𝒬(y0.3,r,s)={α¯∣0.7<α}.
Hence, it is not (r,s)-T0 for 0<r≤1/3 and 2/3≤s<1.
For 0<r≤1/3 and 2/3≤s<1, (X,τ,τ*) is both (r,s)-quasi-T0 and (r,s)-sub-T0,
but not (r,s)-T0.Example 3.10.
LetX={x,y} be a set. We define an IGO (τ,τ*) on X as follows:(3.9)τ(λ)={1ifλ=α¯forα∈I,12ifλ=μpq,0otherwise,τ*(λ)={0ifλ=α¯forα∈I,12ifλ=μpq,1otherwise,where for each 0<p<1, μpq(x)=p and μpq(y)=q, 0≤q<p.
Let zt,zm∈Pt(X) with t≠m for z=x or y.
We have(3.10)𝒬(zt,12,12)≠𝒬(zm,12,12).For xt,ym∈Pt(X),
for p>1−t,
we have μp0∈𝒬(xt,1/2,1/2) with ymq¯μp0.
Hence, (X,τ,τ*) is (1/2,1/2)-T0 space. On the other hand, let y0.5≰x0.5.
For each μpq∈𝒬(y0.5,1/2,1/2),
since q+0.5>1 and p>q,
we have x0.5qμpq,
that is, 𝒬(y0.5,1/2,1/2)⊂𝒬(x0.5,1/2,1/2).
Thus (X,τ,τ*) is not (r,s)-T1 space.Theorem 3.11.
Every subspace of(r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space.Proof.
Let(X,τ,τ*) be (r,s)-T1 space. Let at,bm∈Pt(A) such that at≰bm.
Then, at,bm∈Pt(X) such that at≰bm.
Since (X,τ,τ*) is (r,s)-T1,
there exists λ∈𝒬(at,r,s) such that bmq¯λ.
Since τA(i−1(λ))≥τ(λ)≥r and τA*(i−1(λ))≤τ*(λ)≤s,
we have i−1(λ)∈𝒬τ|A,τ*|A(at,r,s) such that bmq¯i−1(λ).
The others are similarly proved.We can prove the following theorem as a similar method
as Theorem3.11.Theorem 3.12.
Every IF homeomorphic space of(r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space.Theorem 3.13.
Let{(Xi,τi,τi*)∣i∈Γ} be a family of (r,s)-quasi T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space. Let (τ,τ*) be the product IGO on X=∏i∈ΓXi.
Then (X,τ,τ*) is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) space.Proof.
Letxt,ym∈Pt(X) such that xt≰ym.
Then there exists i∈Γ such that (πi(x))t≰(πi(y))m.
Since (Xi,τi,τi*) is (r,s)-T1 space, there exists λ∈IXi such that(3.11)λ∈𝒬τi,τi*((πi(x))t,r,s),(πi(y))mq¯λ.Since πi(xt)=(πi(x))tqλ if and only if xtqπi−1(λ),
we have(3.12)πi−1(λ)∈𝒬(xt,r,s),ymq¯πi−1(λ).Therefore, (X,τ,τ*) is (r,s)-T1 space. The others are similarly proved.Theorem 3.14.
Let{(Xi,τi,τi*)∣i∈Γ} be a family of IFTSs. Let (τ,τ*) be the product IGO on X=∏i∈ΓXi.
If (X,τ,τ*) is (r,s)-sub-T0 space, then (Xi,τi,τi*) is (r−ϵ,s+ϵ)-sub-T0 space for each ϵ>0 and for each i∈Γ.Proof.
Letxj,yj∈Xj such that xj≠yj.
Then there exists xi∈Xi for all i∈Γ−{j} such that x≠y∈X and(3.13)πi(x)={xiifi∈Γ−{j},xjifi=j,πi(y)={xiifi∈Γ−{j},yjifi=j.Since (X,τ,τ*) is (r,s)-sub-T0 space, there exists t∈(0,1) such that(3.14)ρ∈𝒬(xt,r,s),ytq¯ρ.Let (β,β*) be a base for (τ,τ*).
Since τ(ρ)≥r and τ*(ρ)≤s,
by Theorem 1.5, for ϵ>0,
there exists a family {ρk∣ρ=⋁k∈Δρk} such that(3.15)τ(ρ)≥⋀k∈Δβ(ρk)>r−ϵ,τ*(ρ)≤⋁k∈Δβ*(ρk)<s+ϵ.Since xtq[ρ=⋁k∈Δρk],
there exists k∈Γ such that xtqρk, β(ρk)>r−ϵ and β*(ρk)<s+ϵ.
Then, there exists a family {λi∣ρk=⋀i∈Fπi−1(λi)} which F is a finite subset of Γ such that(E)β(ρk)≥⋀i∈Fτi(λi)>r−ϵ,β*(ρk)≤⋁i∈Fτi*(λi)<s+ϵ.Without loss of generality, we
may assume j∈F because we can take F1=F∪{j} such that λj=1¯, τj(1¯)=1, and τj*(1¯)=0,
if necessary.
Sincextqρk and ytq¯ρk,(F)t>[⋁i∈F−{j}(1¯−λi)(πi(x))]∨(1¯−λj)(xj),(G)t≤[⋁i∈F−{j}(1¯−λi)(πi(x))]∨(1¯−λj)(yj).If (⋁i∈F−{j}(1¯−λi)(πi(x))≥t,
it is a contradiction for (F) and (G). Thus(3.16)⋁i∈F−{j}(1¯−λi)(πi(x))<t.It implies(3.17)t>(1¯−λj)(xj),t≤(1¯−λj)(yj).Furthermore, by (E), we have τj(λj)>r−ϵ and τj*(λj)<s+ϵ.
Hence,(3.18)λj∈𝒬τj,τj*((xj)t,r−ϵ,s+ϵ),(yj)tq¯λj.Thus, (Xj,τj,τj*) is (r−ϵ,s+ϵ)-sub-T0 space.In the above theorem, if(X,τ,τ*) is (r,s)-T1 (resp., (r,s)-quasi-T0 and (r,s)-T0), it is not true from the following example.Example 3.15.
LetX={x} and Y={y} be sets. Define IGO (τ1,τ1*) on X as follows:(3.19)τ1(λ)={1ifλ=α¯forα∈I,0otherwise,τ1*(λ)={0ifλ=α¯forα∈I,1otherwise,and IGO (τ2,τ2*) on Y as follows:(3.20)τ2(λ)={1ifλ=0¯or1¯,12ifλ=y0.2,0otherwise,τ2*(λ)={0ifλ=0¯or1¯,12ifλ=y0.2,1otherwise.Let X×Y={(x,y)} be a product set and (τ1⨂τ2,τ1*⨂τ2*) the product IGO on X×Y.
Since (x,y)0.2=π1−1(0.2¯)=π2−1(y0.2),
by Theorem 1.5, we have(3.21)(τ1⨂τ2)(0.2¯)=τ1(0.2¯)∨τ2(y0.2)=1,(τ1*⨂τ2*)(0.2¯)=τ2(0.2¯)⋀τ2*(y0.2)=0.We can obtain the product IGO, (τ1⨂τ2,τ1*⨂τ2*) as follows:(3.22)τ1⨂τ2(λ)={1ifλ=α¯forα∈I,0otherwise,τ1*⨂τ2*(λ)={0ifλ=α¯forα∈I,1otherwise. Then, (X×Y,τ1⨂τ2,τ1*⨂τ2*) are (r,s)-T1, (r,s)-T0,
and (r,s)-quasi-T0 for all r∈I0,s∈I1.
But (Y,τ2,τ2*) is not (r,s)-quasi-T0 for all r∈I0,s∈I1.
Hence, it is neither (r,s)-T0 nor (r,s)-T1 for all r∈I0,s∈I1.Theorem 3.16.
Let{(Xi,τi,τi*)∣i∈Γ} be a family of IFTSs and (τ,τ*) be the product IGO on X=∏i∈ΓXi.
If (X,τ,τ*) is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1) and (Xj,τj,τj*) is stratified for j∈Γ,
then (Xj,τj,τj*) is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1).Proof.
Let(X,τ,τ*) and X˜j=Xj×∏{yi∣i≠j} of X parallel to Xj.
Since (X˜j,τ|X˜j,τ*|X˜j) is a subspace of (X,τ,τ*),
by Theorem 3.11, (X˜j,τ|X˜j,τ*|X˜j) is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1). Since (Xj,τj,τj*) is stratified, by Theorem 2.7, πj|X˜j:(X˜j,τ|X˜j,τ*|X˜j)→(Xj,τj,τj*) is IF homeomorphism. From Theorem 3.12, (Xj,τj,τj*) is (r,s)-quasi-T0 (resp., (r,s)-sub-T0, (r,s)-T0, and (r,s)-T1).



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