ISSN: 1304-7981 Number: 8, Year:2015, Pages: 36-46 http://jnrs.gop.edu.tr Received: 17.06.2014 Editors-in-Chief : Bilge Hilal Cadirci Accepted: 01.01.2015 Area Editor: Serkan Demiriz Balanced and Absorbing Soft Sets Sanjay Roya;1 ([email protected]) b T. K. Samanta (mumpu¡[email protected]) aDepartment of Mathematics, South Bantra Ramkrishna Institution, Howrah-711101, West Bengal, India bDepartment of Mathematics, Uluberia College, Uluberia, Howrah-711315, West Bengal, India Abstract - The aim of this paper is to de¯ne the balanced soft Keywords - Soft set, balanced set and absorbing soft set over a linear space and construct some Soft set, absorbing Soft set. useful theorem depending upon these concepts. 1 Introduction Sometimes a few mathematical problems arise in economics, engineering and en- vironment which can not be solved successfully by use classical methods because of various type of uncertainties are present in these problems. To solve these problems, a few concepts have been constructed in several times such as theory of probability, the- ory of fuzzy sets, and the interval mathematics etc. Soft set is the most recent notion of these concepts. The concept of soft set was ¯rst introduced by D. Molodtsov [7] in 1999. In his work, he de¯ned the operation on soft sets like union, intersection, cartesian product etc. Also he constructed some applications of soft sets in his ¯rst paper of soft set theory. Thereafter so many research works[1, 2, 4, 5, 6, 8, 9, 10] have been done on this concept in di®erent disciplines of mathematics. In functional analysis, certain types of sets viz. balanced set, absorbing set and convex set are found to play pivotal roles. In this paper, we also try to de¯ne the concept of balanced soft set, absorbing soft set over a linear space to study the functional analysis. Then we establish some theorems concerning the said notions. 1Corresponding Author Journal of New Results in Science 8 (2015) 36-46 37 2 Preliminary In this section, U refers to an initial universe, E is the set of parameters, P (U) is the power set of U and A ⊆ E: De¯nition 2.1. [3] A soft set FA on the universe U is de¯ned by the set of ordered pairs FA = f(e; FA(e)) : e 2 E; FA(e) 2 P (U)g where FA : E ! P (U) such that FA(e) = Á if e is not an element of A. The set of all soft sets over (U; E) is denoted by S(U). De¯nition 2.2. [3] Let FA 2 S(U). If FA(e) = Á; for all e 2 E, then FA is called a empty soft set, denoted by ©. FA(e) = Á means that there is no element in U related to the parameter e 2 E. De¯nition 2.3. [3] Let FA;GB 2 S(U). We say that FA is a soft subsets of GB and we write FA v GB if and only if FA(e) ⊆ GB(e) for all e 2 E: De¯nition 2.4. [3] Let FA;GB 2 S(U). Then FA and GB are said to be soft equal, denoted by FA = GB if FA(e) = GB(e) for all e 2 E: De¯nition 2.5. [3] Let FA;GB 2 S(U). Then the soft union of FA and GB is also a soft set FA t GB = HA[B 2 S(U), de¯ned by HA[B(e) = (FA t GB)(e) = FA(e) [ GB(e) for all e 2 E. De¯nition 2.6. [3] Let FA;GB 2 S(U). Then the soft intersection of FA and GB is also a soft set FA u GB = HA\B 2 S(U), de¯ned by HA\B(e) = (FA u GB)(e) = FA(e) \ GB(e) for all e 2 E. De¯nition 2.7. Let U be an initial universe and f : X ! Y be a mapping, where X and Y are set of parameters. If FA be a soft set over (U; X), then f(FA), a soft set over (U; Y ), is de¯ned by ½ ¡1 [ ¡1 F (x) if f (y) 6= ©; f(F )(y) = x2f (y) A A © otherwise: De¯nition 2.8. Let U be an initial universe and f : X ! Y be a mapping, where X ¡1 and Y are set of parameters. If GB be a soft set over (U; Y ), then f (GB), a soft set over (U; X), is de¯ned by ¡1 f (GB)(x) = GB(f(x)). De¯nition 2.9. [11] Let U be a universal set and E be a usual vector space over R or n C and FA1 ;FA2 ; ¢ ¢ ¢ ;FAn be soft sets over (U; E) and f : E ! E be a function de¯ned by f(e1; e2; ¢ ¢ ¢ ; en) = e1 + e2 + ¢ ¢ ¢ + en. Then the vector sum FA1 + FA2 + ¢ ¢ ¢ + FAn is de¯ned by (F + F + ¢ ¢ ¢ + F )(e) AS1 A2 An = ¡1 fF (e ) \ F (e ) \ ¢ ¢ ¢ \ F (e )g (e1;e2;¢¢¢ ;en)2f (e) A1 1 A2 2 An n De¯nition 2.10. [11] If U be a universal set and E be a usual vector space over R or C and t be a scalar and g : E ! E be a mapping de¯ned by g(e) = te, then the scalar multiplication tFA of a soft set FA is de¯ned by tFA = g(FA). Journal of New Results in Science 8 (2015) 36-46 38 Proposition 2.11. [11] If FA is a soft set over the universal set U and the parameter set E, where E is an usual vector space over K(R or C) and t 2 K, then 8 ¡1 < FA(t e) if t 6= 0; tF (e) = © if t = 0 and e 6= 0; A : S p2E FA(p) if t = 0 and e = 0: 2.1 Balanced Soft set Throughout this work, we denote E as a vector space over the ¯eld K(R or C) and U as an initial universe. Also we denote 0 and 1 as the zero element and unity of the ¯eld respectively. Also the zero vector of the linear space is denoted by 0 which can be easily separated from the zero element of the ¯eld. De¯nition 2.12. A soft set FA over (U; E) is said to be balanced soft set if tFA v FA for all t 2 K with jtj · 1. Example 2.13. Let the universal set U= the set of all real numbers and E be a real vector space and 0 2 A ⊆ E: Let FA be a soft set de¯ned by ½ (jej; 1) if e 6= 0; F (e) = A U if e = 0; where e 2 A: Then obviously, FA is a balanced soft set. Theorem 2.14. If FA is a soft set over (U; E), then tj¸j·1¸FA is a balanced soft set. Proof: Let j®j · 1 and e 2 E. Case 1. 0 < j®j · 1. ®(tj¸j·1¸FA)(e) 1 = (tj¸j·1¸FA)( ® e) 1 = [j¸j·1¸FA( ® e) = (tj¸j·1®¸FA)(e) ⊆ (tj¸j·1¸FA)(e), since j®j · 1 and j¸j · 1; j®¸j · 1 Case 2. ® = 0. Subcase 1. If e 6= 0, then obviously, 0(tj¸j·1¸FA)(e) = © ⊆ (tj¸j·1¸FA)(e): Subcase 2. If e = 0, then 0(tj¸j·1¸FA)(0) = [x2Ef(tj¸j·1¸FA)(x)g = [x2Ef[j¸j·1¸FA(x)g = f0FA(0)g [ f[x(6=0)2E0FA(x)g [ f[x2Ef[0<j¸j·1¸FA(x)gg 1 = f0FA(0)g [ f[x2Ef[0<j¸j·1FA( ¸ x)g, as [x(6=0)2E0FA(x) = © = 0FA(0): Again, (tj¸j·1¸FA)(0) = [j¸j·1¸FA(0) = f[0<j¸j·1¸FA(0)g [ f0FA(0)g = f[0<j¸j·1FA(0)g [ f0FA(0)g Journal of New Results in Science 8 (2015) 36-46 39 = 0FA(0). Thus, 0(tj¸j·1¸FA)(0) = (tj¸j·1¸FA)(0). Hence, tj¸j·1¸FA is a balanced soft set. Theorem 2.15. Let FA be a balanced soft set over (U; E). Then i) ®; ¯ 2 K and j®j · j¯j ) ®FA v ¯FA ii) ® 2 K and j®j = 1 ) ®FA = FA Proof: i) Case 1. ® = 0. Subcase 1. If ¯ = 0, then clearly, ®FA = ¯FA. Subcase 2. If ¯ 6= 0, then for any non-zero e 2 E, ®FA(e) = © ⊆ ¯FA(e) and for zero element of E, we have ®FA(0) = 0FA(0) ⊆ FA(0) as FA is a balanced soft set. That is, 1 ®FA(0) ⊆ FA( ¯ 0) = ¯FA(0). Therefore in this case ®FA v ¯FA. ® ® Case 2. ® 6= 0. Since j ¯ j · 1, we have ¯ FA v FA. 1 ® Let e 2 E and ¯ e = e1. Now since ¯ FA v FA, then ® ¯ FA(e1) ⊆ FA(e1) ¯ or, FA( ® e1) ⊆ FA(e1) ¯ 1 1 or, FA( ® ¯ e) ⊆ FA( ¯ e) 1 1 or, FA( ® e) ⊆ FA( ¯ e) or, ®FA(e) ⊆ ¯FA(e) that is, ®FA v ¯FA. 1 ii) Let j®j · 1. Then ®FA v FA. Again let j®j ¸ 1. Then j ® j · 1 which implies 1 that ® FA v FA. So as the procedure of case 2 for proof (i), we get FA v ®FA. Thus, ®FA = FA if j®j = 1: Theorem 2.16. If fFA® : ® 2 ¤g is a collection of balanced soft sets over (U; E), then u®2¤FA® is balanced.