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129
Neutrosophic Beta Omega Closed Sets in Neutrosophic Topological Spaces
S. Pious Missier1,b), A. Anusuya2,a), A. Nagarajan3,c)
1Head & Associate Professor, Department of Mathematics, Don Bosco College of Arts and Science,
(Affiliated to Manonmaniam Sundaranar University,Tirunelveli) Keela Eral, Thoothukudi, Tamil Nadu-628 908, India.
2 Research Scholar(Reg.No-19222232092024), V.O.Chidambaram College, (Affiliated to Manonmaniam Sundaranar University,Tirunelveli), Thoothukudi-628 003,
India.
3Head & Associate Professor, V. O. Chidambaram College, Thoothukudi, Tamilnadu-628 008, India. nagarajan.voc@gmail.com
b)spmissier@gmail.com
a)anuanishnakul@gmail.com
c)nagarajan.voc@gmail.com
Article Info
Page Number: 129 β 137 Publication Issue:
Vol. 71 No. 3s (2022)
Article History
Article Received: 22 April 2022 Revised: 10 May 2022
Accepted: 15 June 2022 Publication: 19 July 2022
Abstract
We introduce and examine several applications of neutrosophic beta omega closed sets namely πππ½π βπ π, ππππ½π βπ π, π½πππ½π βπ π, ππΊπππ½π βπ π, πΊππππ½π βπ πin this study. We also introduce the concept of neutrosophic betaomega continuous mapping in neutrosophic topology. Furthermore, we study the relation between the neutrosophic betaomega continuous mapping with the already existing neutrosophic continuous mapping. In addition, we discuss the properties of neutrosophic betaomega continuous mapping.
Keywords:πππ½π βπ π, ππππ½π βπ π, π½πππ½π βπ π, ππΊπππ½π βπ π, πΊππππ½π βπ π,neutrosophic beta omega continuous mapping.
AMS Mathematics Subject Classification:18B30,03E72 I. INTRODUCTION
Fuzzy set theory introduced by Zadeh[11] has laid the foundation for the new mathematical theories in the research of mathematics. Later, Attanasov[2] developed the concept of intuitionistic fuzzy sets. Smarandache[7] was the first to coin the term "neutrosophic set.".
Neutrosophic operations and neutrosophic continuous functions have been investigated by Salama[10]. Here, we shall introduce the applications of neutrosophic beta omega closed set and neutrosophic continuity of neutrosophic-beta-omega sets. Also we present the implications of neutrosophic βbeta-omega-continuous mapping.
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II. PRELIMINARIES
Definition 2.1.[7] If βπ is a nonempty fixed set. A neutrosophic set (N-Set) GN is a form factor GN = {ο‘ο‘π, ΞΌG N(π), ΟGN(π), Ο GN(π)ο‘: πο‘βπ} where ΞΌ
G N(π), ΟGN(π) and Ο GN(π)represents the membership level, non-determination level and non-membership level respectively of each element πο‘ο‘βπto the set πΊπ. A Neutrosophic set πΊπ = {ο‘π, ΞΌ
G N(π), ΟGN(π), Ο GN(π)ο‘: πο‘βπ} can be recognized as a triple order ο‘ΞΌ
G N, ΟGN, Ο GNο‘ο‘in ]0,1+[on
βπ.
Definition 2.2. [1] For any two sets GNand HN,
1. GN βο‘HNο‘ο‘ΞΌG N(π) ο‘ο‘ΞΌH N (π), ΟGN(π) ο‘ο‘ΟHN(π) and Ο GN(π) ο‘ο‘Ο HN(π) ο‘ο‘ πο‘βπ 2. GNβ HN = ο‘π, ΞΌG N(π) ο‘ο‘ΞΌH N(π), ΟGN(π) ο‘ο‘ΟHN(π), Ο GN(π) ο‘ο‘Ο HN(π) ο‘
3. GNβ HN = ο‘π, ΞΌG N(π) ο‘ο‘ΞΌH N(π), ΟGN(π) ο‘ο‘ΟHN(π), Ο GN(π) ο‘ο‘Ο HN(π) ο‘ 4. GNC = {ο‘π, Ο GN(π), 1- ΟGN(π), ΞΌ
G N(π)ο‘: πο‘βπ} 5. 0N = {ο‘ο‘π, 0, 0, 1ο‘: πο‘βπ}
6. 1N = {ο‘ο‘π, 1, 1, 0ο‘: πο‘βπ}
Definition 2.3. [9] A neutrosophic topology (NT) is a family ππ of neutrosophic subsets ofβπwhose member satisfy the following axioms:
1. 0π, 1π βο‘ππ
2. βπ β© βπ βο‘ππο‘for any βπ, βπ β ππ 3. βͺ βπ π β ππο‘βο‘{ βπ π : i βο‘J} β ππ
Any neutrosophic set in ππ is a neutrosophic open set (NO-set). If a neutrosophic set GN is a neutrosophic open set,then its complement GNc is a neutrosophic closed set(NC-set) in
βπ.Also, neutrosophic topological space (NTS) is denoted by (βπ, ππ).
Definition 2.4. [8] A N-SetπΊπ of (βπ,ππ) is calledneutrosophic beta omegaclosed(ππ½ππΆ) if π½πππ(πΊπ) οππ whenever πΊποππ and ππ is Nππ in (βπ,ππ).
Definition 2.5. [8]For every set πΊπ β ο‘βπ, ππο‘, we define 1. π½ππππ(πΊπ)= β©{ππ: πΊπβ ππ, ππ β ππ½πCο‘βπ, ππο‘ο‘}.
2. π½ππππ‘π(πΊπ) = βͺ{ππ: ππ β πΊπ and ππ β ππ½ππο‘βπ, ππο‘}
Definition2.6. [6]A mappingπ: βπ, ππ β π€π, ππ is said to be
1. Neutrosophic-continuous(πππππ‘) if πβ1(πΊπ) is ππΆset in ο‘βπ, ππο‘ο‘for eachππΆ-set πΊπ in (π€π, ππ).
2. Neutrosophic-pre-continuous(ππππππ‘) if πβ1(πΊπ) is πππΆset in ο‘βπ,ππο‘ο‘for eachππΆ-set πΊπ in (π€π, ππ).
3. Neutrosophic-beta-continuous(ππ½ππππ‘) if πβ1(πΊπ) is ππ½Cset in ο‘βπ, ππο‘ο‘for eachππΆ-set πΊπ in (π€π, ππ)
4. Neutrosophic-alpha-generalized-continuous(ππΌπΊππππ‘) if πβ1(πΊπ) is ππΌπΊπΆset in ο‘βπ, ππο‘ο‘for eachππΆ-set πΊπ in (π€π, ππ).
III. APPLICATIONS OF NEUTROSOPHIC BETA OMEGA CLOSED SETS Definition 3.1.A neutrosophic topological space(βπ,ππ) is called a
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131 2. Pre-πππ½π-space(ππππ½πβπ π) if allππ½ππΆset is NPC.
3. Beta-πππ½π-space(π½πππ½πβπ π) if allππ½ππΆset is Nπ½πΆ.
4. Semi-generalized-πππ½π-space(ππΊπππ½πβπ π) if allππ½ππΆset is NSGC.
5. Generalized-semi-πππ½ π-space(πΊππππ½πβπ π) if allππ½ππΆset is NGSC.
Example 3.1.Let βπ = [0, 1], ππ = {0π, πΊππ Ξ» , π»ππ Ξ» , 1π},(βπ, ππ) = {0π, πΊπ(Ξ»), π»π Ξ» , 1π}. where
πΊπ(Ξ») =
< π, π, 1 β π > π€ππππ 0 β€ π β€1
2
< 1 β π, 1 β π, π >1
2β€ π β€ 1 and π»π(Ξ») =
< π, π, 1 β π > π€ππππ 0 β€ π β€1
2
< 1
2, 1
2, 1
2> 1
2β€ π β€ 1 . Then ππ is a NT. Here, (βπ, ππ) is πππ½πβπ π.
Example 3.2. Let βπ = {π1,π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.7,π2
0.6,π3
0.5 ,
π1 0.6,π2
0.6,π3
0.7 , π1
0.3,π2
0.5,π3
0.4 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ)-{ ππ / πΊπβππ β 1π} = πππΆ(βπ, ππ). Therefore,every ππ½ππΆset isπππΆ. Hence(βπ,ππ) is ππππ½πβπ π.
Example 3.3. Let βπ = {π1,π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.6,π2
0.6,π3
0.6 ,
π1 0.7,π2
0.7,π3
0.7 , π1
0.3,π2
0.3,π3
0.4 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ)-{ ππ / πΊπβππ β 1π} = ππ½πΆ(βπ, ππ). Therefore,every ππ½ππΆset is NΞ²C. Hence(βπ, ππ) is π½πππ½πβπ π.
Example 3.4. Let βπ = {π1, π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.3,π2
0.4,π3
0.4 ,
π1 0.2,π2
0.3,π3
0.1 , π1
0.8,π2
0.8,π3
0.9 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ) = ππΊππΆ(βπ, ππ). Therefore every ππ½ππΆset is ππΊππΆ. Hence(βπ, ππ) is πΊππππ½πβπ π.
Example 3.5. Let βπ = {π1, π2, π3},ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.3,π2
0.3,π3
0.3 ,
π1 0.1,π2
0.3,π3
0.1 , π1
0.8,π2
0.8,π3
0.8 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ) = πππΊπΆ(βπ, ππ). Therefore every ππ½ππΆset is πππΊπΆ. Hence(βπ, ππ)is ππΊπππ½πβπ π.
Theorem3.1.
(a). Every ππππ½π βπ π is π½πππ½π βπ π. (b). Every πππ½π βπ π is ππππ½π βπ π. (c). Every πππ½π βπ π is π½πππ½π βπ π. (d). Every πππ½π βπ π is πΊππππ½π βπ π. (e). Every πππ½π βπ π is ππΊπππ½π βπ π.
Proof :
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(a). Let(βπ, ππ) beππππ½π βπ π. We know that, every πππΆ set is ππ½πΆ set and since every ππ½ππΆset is πππΆ, we get ππ½ππΆset is ππ½πΆ set.Hence(βπ, ππ) is a π½πππ½π βπ π.
(b), (c), (d) follow from the facts that every ππΆ set is πππΆ set, ππ½πΆ set, ππΊπC set.
πππΊπΆ set respectively.
Remark 3.1.The examples below show that the converse of the preceding theorems does not have to be true.
Example 3.6.Takeβπ = {π1, π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.3,π2
0.3,π3
0.4 ,
π1 0.3,π2
0.3,π3
0.3 , π1
0.6,π2
0.6,π3
0.6 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ) = ππ½πΆ(βπ, ππ). But πππΆ(βπ, ππ) = (βπ, ππ)-{ππ / πΊπ β ππ β πΊππΆ}-{ππ / πΊπ β ππ β πΊππΆ} .Hence(βπ, ππ)is π½πππ½πβπ πbut not ππππ½πβπ π.
Example 3.7. Let βπ = {π1, π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.7,π2
0.6,π3
0.5 ,
π1 0.6,π2
0.6,π3
0.7 , π1
0.3,π2
0.5,π3
0.4 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ)-{ππ / πΊπ β ππ β 1π} = πππΆ(βπ, ππ)and ππΆ(βπ, ππ)={0π, πΊππΆ, 1π}. Hence(βπ, ππ)is ππππ½πβπ πbut not πππ½π βπ π.
Example 3.8. Let βπ = {π1,π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = <ΞΆ, π1
0.6,π2
0.6,π3
0.6 ,
π1 0.7,π2
0.7,π3
0.7 , π1
0.3,π2
0.3,π3
0.4 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ)= (βπ, ππ)-{ππ / πΊπβ ππ β 1π} = ππ½πΆ(βπ, ππ) and ππΆ(βπ, ππ) ={0π, πΊππΆ, 1π}. Hence(βπ, ππ)is π½πππ½πβπ πbut not πππ½π βπ π.
Example 3.9. Let βπ = {π1, π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = <ΞΆ, π1
0.3,π2
0.4,π3
0.4 ,
π1 0.2,π2
0.3,π3
0.1 , π1
0.8,π2
0.8,π3
0.9 >.Then ππ is a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ)= ππΊππΆ(βπ, ππ) and ππΆ(βπ, ππ) ={0π, πΊππΆ, 1π}. Hence(βπ, ππ)is πΊππππ½πβπ πbut not πππ½π βπ π.
Example 3.10.Let βπ = {π1, π2, π3}, ππ = {0π, πΊπ, 1π} where πΊπ = < π, π1
0.3,π2
0.3,π3
0.3 ,
π1 0.1,π2
0.3,π3
0.1 , π1
0.8,π2
0.8,π3
0.8 >.Then ππis a NT. Then ππ½ππΆ(βπ, ππ) = (βπ, ππ)= πππΊπΆ(βπ, ππ) and ππΆ(βπ, ππ) ={0π,πΊππΆ, 1π}. Therefore every ππ½ππΆ set is πππΊπΆ. Hence(βπ, ππ) is ππΊπππ½πβπ πbut not πππ½π βπ π.
IV. NEUTROSOPHIC BETA OMEGACONTINUOUS MAPPING
Definition4.1. A mapping π: βπ, ππ β π€π, ππ is calledneutrosophic betaomegacontinuous(ππ½πππππ‘) if πβ1(πΊπ) is ππ½ππΆset in (βπ, ππ)for allππΆ-set πΊπ in (π€π, ππ).
Example 4.1. Let βπ = {π1, π2, π3}, ΞN= {πΏ1, πΏ2, πΏ3},ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, ππ, 1N} where πΊπ = < π, π1
0.6,π2
0.7,π3
0.6 , π1
0.7,π2
0.8,π3
0.7 , π1
0.4,π2
0.3,π3
0.4 >, π»π = <
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133 π, πΏ1
0.7,πΏ2
0.7,πΏ3
0.7 , πΏ1
0.8,πΏ2
0.8,πΏ3
0.7 , πΏ1
0.3,πΏ2
0.2,πΏ3
0.2 >, ππ =
< π, πΏ1
0.8,πΏ2
0.7,πΏ3
0.8 , πΏ1
0.9,πΏ2
0.8,πΏ3
0.7 , πΏ1
0.2,πΏ2
0.2,πΏ3
0.2 >.Then ππ and ππ are NTs. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3. Then π β1(ππ) is NΞ²ΟC in (βπ, ππ) for every ππΆset ππin (π€π, ππ). Henceπ is ππ½πππππ‘.
Theorem 4.1.
1. Every πππππ‘ mapping is ππ½πππππ‘mapping.
2. Every NΞ²cont mapping is ππ½πππππ‘mapping.
3. Every ππΊβππππ‘ mappingis ππ½πππππ‘mapping.
4. Every ππππππ‘ mapping is ππ½πππππ‘mapping.
5. Every ππGβππππ‘ mapping is ππ½πππππ‘ mapping.
6. Every ππππππ‘mapping is ππ½πππππ‘ mapping.
7.
Proof:
1. Takingπas aπππππ‘mappingandππ be any ππΆ-setin (π€π, ππ). Since,π is πππππ‘ mapping, we get πβ1(ππ) is ππΆ-set in (βπ, ππ) implies πβ1(ππ) is ππ½ππΆset. Hence π is ππ½πππππ‘mapping in (βπ, ππ).
2. Let π be any NΞ²ππππ‘mapping and ππ be any ππΆ-set in (π€π, ππ). Since π is NΞ²ππππ‘mapping, we get πβ1(ππ) is ππ½πΆ in βπ, ππ .Since everyππ½πΆset isππ½ππΆ,πβ1(ππ) is ππ½ππΆ. Hence π is ππ½πππππ‘mapping in (βπ, ππ).
3. Let π beππΊβππππ‘ mapping and ππ be any ππΆ-set in (π€π, ππ). Since π is ππΊβππππ‘mapping, we get πβ1(ππ) is ππΊ*C in (βπ, ππ). Therefore, πβ1(ππ) is ππ½ππΆ, because every ππΊβπΆ set is ππ½ππΆ set. Hence π is ππ½πππππ‘mapping in (βπ, ππ) 4. Let π be ππππππ‘mapping and ππ be any ππΆ-set in (π€π, ππ). Since π is ππππππ‘mapping, we
get πβ1(ππ) is πππΆin (βπ, ππ). Since allπππΆ set is ππ½ππΆ,πβ1(ππ) is NΞ²ΟC. Hence π is ππ½πππππ‘mapping in (βπ, ππ).
5. Let π be ππGβππππ‘mapping and ππ be any ππΆ-set in (π€π, ππ). Since π is ππGβππππ‘mapping, we get πβ1(ππ) is πππΊβπΆ in (βπ, ππ).We know that every πππΊβπΆ set is ππ½π set,πβ1(ππ) is ππ½ππΆ. Hence π is ππ½πππππ‘mapping in (βπ, ππ).
6. Let π be ππππππ‘mapping in (βπ, ππ). Let ππ be any ππΆ-set in (π€π, ππ)). Since π is πππππ‘mapping, we get πβ1(ππ) is πππΆin (βπ, ππ). Every πππΆ is ππ½ππΆ that impliesπβ1(ππ) isππ½ππΆ. Hence π is ππ½πππππ‘mapping in (βπ, ππ).
Example 4.2. Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3},ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, 1N} where πΊπ = < π, π1
0.2,π2
0.2,π3
0.1 , π1
0.7,π2
0.7,π3
0.7 , π1
0.7,π2
0.7,π3
0.7 >, π»π =
< π, πΏ1
0.2,πΏ2
0.2,πΏ3
0.1 , πΏ1
0.2,πΏ2
0.1,πΏ3
0.2 , πΏ1
0.7,πΏ2
0.8,πΏ3
0.8 >.Then ππ and ππ are NTs. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3. Then π is ππ½πππππ‘mapping but not ππ½ππππ‘ mapping.
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Example 4.3. Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3},ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, ππ 1N} where πΊπ = < π, 0.5π1,0.6π2,0.5π3 , 0.4π1,0.3π2,0.4π3 , 0.6π1,0.8π2,0.7π3 >, π»π = < π,
πΏ1 0.6,πΏ2
0.9,πΏ3
0.7 , πΏ1
0.8,πΏ2
0.8,πΏ3
0.7 , πΏ1
0.3,πΏ2
0.4,πΏ3
0.3 >, ππ =
< π, πΏ1
0.7,πΏ2
0.9,πΏ3
0.8 , πΏ1
0.8,πΏ2
0.9,πΏ3
0.7 , πΏ1
0.2,πΏ2
0.4,πΏ3
0.2 >.Then ππ and ππ are NTs. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3. Then π is ππ½πππππ‘mapping but not ππΊβππππ‘mapping.
Example 4.4. Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3},ππ = {0π, πΊπ, 1N} and ππ = {0π, HN, 1N} where πΊπ = < π, π1
0.3,π2
0.3,π3
0.1 , π1
0.2,π2
0.2,π3
0.2 , π1
0.8,π2
0.8,π3
0.7 >, π»π = < π,
πΏ1 0.6,πΏ2
0.6,πΏ3
0.6 , πΏ1
0.5,πΏ2
0.6,πΏ3
0.6 , πΏ1
0.4,πΏ2
0.4,πΏ3
0.4 >.Then ππand ππ are NT. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3. Then π is ππ½πππππ‘mapping but not ππGβππππ‘mapping.
Remark 4.1.The converse of the preceding theorem does not have to be true.
1. The example 4.1. shows that the converse of theorem 4.1(1) is not true.
2. Theexample 4.3. shows that the converse of the 4.1(4) does not exist.
3. The example 4.2 shows that ππ½πππππ‘need not beππππππ‘.
Remark 4.2. The following examples show that ππ½πππππ‘mapping set and ππΊπππππ‘ mapping are independent in (βπ, ππ).
Example 4.5. Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3}, ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, ππ 1N}, where πΊπ = < π, π1
0.9,π2
0.8,π3
0.9 , π1
0.8,π2
0.8,π3
0.7 , π1
0.3,π2
0.2,π3
0.2 >, π»π = < π,
πΏ1 0.7,πΏ2
0.7,πΏ3
0.8 , πΏ1
0.7,πΏ2
0.7,πΏ3
0.6 , πΏ1
0.4,πΏ2
0.3,πΏ3
0.7 >, ππ =
< π, πΏ1
0.6,πΏ2
0.6,πΏ3
0.8 , πΏ1
0.6,πΏ2
0.6,πΏ3
0.6 , πΏ1
0.5,πΏ2
0.4,πΏ3
0.7 >.Then ππ and ππ are NT. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3. Then π is ππ½πππππ‘mapping notππΊπππππ‘ mapping.
Example 4.6. Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3},ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, 1N}, where πΊπ = < π, π1
0.9,π2
0.8,π3
0.9 , π1
0.8,π2
0.7,π3
0.7 , π1
0.2,π2
0.2,π3
0.2 >, π»π = < π,
πΏ1 0.1,πΏ2
0.1,πΏ3
0.1 , πΏ1
0.1,πΏ2
0.1,πΏ3
0.1 , πΏ1
0.9,πΏ2
0.9,πΏ3
0.9 >.Then ππ and ππ are NT. Define a mapping π: π₯π, ππ β π€π, ππ by π(π1) = πΏ1, π(π2) = πΏ2 and π π3 = πΏ3.Then π isππΊπππππ‘mapping but not ππ½πππππ‘mapping.
Remark 4.3. Consider the examples4.5. and 4.6., these examples supports the fact thatππΊππππ‘ and ππΌπΊππππ‘ mappings are independent with ππ½πππππ‘in (βπ, ππ).
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135 Figure.1(The implications of π΅π·πππππ mapping)
V. SOME THEOREMS AND PROPERTIES OF NΞ²Ο-CONTINUOUS MAPPING Theorem5.1.If π: π₯π, ππ β (π€π, ππ)is ππ½πππππ‘andπ: (π€π, ππ) β (β¦π, ππ) ππ πππππ‘ then their composition πππ: (βπ,ππ)β (πΊπ, ππ) is ππ½πππππ‘.
Proof : Let πΊπ be any N-set of (πΊπ, ππ). Since π is πππππ‘, πβ1(πΊπ)is NC-set in (π€π, ππ).
Since π is ππ½πππππ‘, πβ1(πβ1 πΊπ ) = (πππ)β1 πΊπ isππ½ππΆin (βπ,ππ). Therefore πππ is ππ½πππππ‘.
Remark 5.1. If two mappings areππ½πππππ‘, Then theircompositionneed not be ππ½πππππ‘. Example 5.1.Let βπ = {π1, π2, π3}, π€π = {πΏ1, πΏ2, πΏ3}, β¦π = {πΎ1, πΎ2, πΎ3}, ππ = {0π, πΊπ, 1N} and ππ = {0π, π»π, 1N}, ππ = {0π, πΌπ, 1N}, where πΊπ =
< π, π1
0.7,π2
0.8,π3
0.7 , π1
0.8,π2
0.7,π3
0.8 , π1
0.3,π2
0.2,π3
0.3 >, π»π =
< π, πΏ1
0.9,πΏ2
0.9,πΏ3
0.9 , πΏ1
0.9,πΏ2
0.9,πΏ3
0.9 , πΏ1
0.1,πΏ2
0.1,πΏ3
0.1 >, πΌπ =
< π, πΏ1
0.2,πΏ2
0.1,πΏ3
0.2 , πΏ1
0.1,πΏ2
0.1,πΏ3
0.1 , πΏ1
0.8,πΏ2
0.8,πΏ3
0.8 >.Then ππ and ππ are NTs. Define the mappingsπ: π₯π, ππ β π€π, ππ andπ: (π€π, ππ) β (β¦π, ππ)by π(π1) = πΏ1, π(π2) = πΏ2,π π3 = πΏ3, π(πΏ1) = πΎ1, π(πΏ2) = πΎ2 and π πΏ3 = πΎ3. Then πand π areππ½πππππ‘ mapping but π β π is notππ½πππππ‘ mapping.
Theorem 5.2. A map π: π₯π, ππ β (π€π, ππ)is ππ½πππππ‘ if and only if πβ1(πΊπ) is ππ½ππ in π₯π, ππ , for every ππ-set πΊπ in (π€π, ππ).
Proof : Let π: π₯π, ππ β (π€π, ππ)beππ½πππππ‘ and πΊπ be aππ-set in (π€π, ππ). Then πβ1(πΊππΆ) is ππ½ππΆin (βπ,ππ). But πβ1 πΊππΆ =(πβ1(πΊπ)πΆ and πβ1(πΊπ)is NΞ²ΟO in (βπ,ππ).
Converse is similar.
Theorem 3.3.3. Let π: π₯π, ππ β (π€π, ππ)be a map. Assume that NΞ²ΟO π₯π, ππ is ππΆ-set under any union. Then the following are equivalent :