Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 718 https://internationalpubls.com (1, 2)*-D**SpOpen Sets in Bitopological Spaces P. Devi Prabha1, R. Asokan2 1&2 Department of Mathematics School of Mathematics Madurai Kamaraj University Madurai-625021, Tamil Nadu, INDIA Email: 1deviprabhaponnusamy@gmail.com, 2asokan.maths@mkuniversity.org Article History: Received: 08-04-2024 Revised: 20-05-2024 Accepted: 07-06-2024 Abstract This article explains the concept of (1, 2)*D** semi- pre-open sets based on the concepts of semi- preopen sets and semi- pre- continuity in topological space. In addition to that the concept of (1,2)*D**Sp generalized continuous maps and generalized homeomorphisms are also discussed. Keywords: (1, 2)*-Open map, (1, 2)*-D**SpOS, (1, 2)*-D**SpContinuous map. 1. Introduction Bhattacharya and Lahiri [1] introduced a new class of semi generalized open sets by means of semi open sets introduced by Levine [5]. In view of that we introduce a new class of open sets namely (1, 2)*-D**Spopen sets and their properties are also studied.. Also (1, 2)*-D**SpContinuous maps, irresolute maps, (1, 2)*-D**SpConnected sets, homeomorphism are also studied with their characterizations. 2. Preliminaries Entire area of this paper, (X, 1, 2) X will denote bitopological space (briefly, BTPS). Definition 2.1: Let H be a subset of X. Then H is said to be 1,2-open [7] if H = A  B where A  1 and B   The complement of 1,2-open set is called 1,2-closed. Notice that 1,2-open sets need not necessarily form a topology Note; 1,2-open sets need not necessarily form a topology. Definition 2.2 [7]: Let H be a subset of a bitopological space X. Then (i) the τ1,2-closure of H, denoted by τ1,2-cl(H), is defined as {F : H  F and F is τ1,2-closed} (ii) the τ1,2-interior of H, denoted by τ1,2-int(H), is defined as {F : F  H and F is τ1,2-open} mailto:1deviprabhaponnusamy@gmail.com mailto:2asokan.maths@mkuniversity.org Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 719 https://internationalpubls.com Definition 2.3: A subset H of a BTPS X is called: (i) (1, 2)*-semi-open set [8] if H  1,2-cl(1,2-int(H)); (ii) (1, 2)*-preopen set [6] if H  1,2-int(1,2-cl(H)); (iii) (1, 2)*--open set [3] if H  1,2-int(1,2-cl(1,2-int(H))); (iv) regular (1, 2)*-open set [6] if H = 1,2-int(1,2-cl(H)). (v) (1, 2)*-gsp-closed [10, 11] if (1,2)*-cl(A)  U whenever A  U and U is (1, 2)*-open. Then complement of (1, 2)*-gsp-closed set is called (1, 2)*-gsp-open set. The complements of the above-mentioned open sets are called their respective closed sets. 3. Definition 2.4: [2] A subset H of a space (X, 𝜏) is called -cld if it contains all its condensation points. The complement of -cld set is called -open. 4. Definition 2.5. A bijection f : X → Y is called (1, 2)*-homeomorphism [4] if f is bijection, (1,2)*- continuous and (1,2)*-open. 5. Definition 2.6: A subset A of X is called (1, 2)*-D*-cld (briefly, (1,2)*-D*-cld) if (1, 2)*- scl*(A)  (1, 2)*-int(U) whenever A  U and U is (1, 2)*-ω-open. The complement of (1, 2)*-D*-cld set is called (1, 2)*-D*-open. 6. Definition 2.7: (1, 2)*-D**-closed (briefly, (1, 2)*-D**-cld) if (1, 2)*-spcl(A)  U whenever A  U and U is (1,2)*-D*-open. The complement of (1, 2)*-D**-closed set is called (1, 2)*-D**-open. The class of all (1, 2)*-D**-cld in X is denoted by (1, 2)*-D**-C(X). The complements of the above mentioned open sets are called their respective closed sets. 3 (1, 2)*-D**spOpen Sets Definition 3.1 For S ⊆ X , (1, 2)*-spcl**(S) = ∩{K/S ⊆ K , K is (1, 2)*gspClosed}. Remark 3.2 (1, 2)*-spcl**(S) = Kuratowski closure operator on X . Definition 3.3 S ⊆ X, (1, 2)*-D**spOpen iff there exists an τ1,2OS U Such that U⊆S ⊆ (1, 2)*- spcl**(U). in X. Example 3.4 Let G = {1, 2, 3}, τ1 = { G, φ, {1} } and τ2 = {G, φ, {1}, {1, 2}}. Then (1, 2)*-D**spOS of (X, τ1, τ2) are X, φ, {1}, {1,2} and {1,3}. Remark 3.5 If C ⊆ X, D ⊆ X ∋ C ⊆ D, then we have (1, 2)*-spcl**(C) ⊆ (1, 2)*-spcl**( D ). Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 720 https://internationalpubls.com Theorem 3.6 For S ⊆ X. we have S is (1, 2)*- D**spOpen iff S  (1, 2)*-spcl**(τ1,2Int(S)). Proof. Assume S is (1, 2)*- D**spOpen in X . U  S implies U ⊆ τ1,2int(S). hence from Remark 3.5 and by defn 3.3, (1, 2)∗-spcl**(U)  (1, 2)*spcl**(τ1,2-Int(S)). So S ⊆ (1,2)*-spcl**(τ1,2-Int(S)). To prove the converse let S ⊆ (1, 2)*-spcl**(τ1,2Int(S)). substitute U = τ1,2-Int(S). finally by defn 3.3. S in X is (1, 2)*- D**spOS . Theorem 3.7 In BTS X τ1,2-OS(R) ⇒ (1 ,2)*-D**spOS(R) Proof. Assume R be a τ1,2OS in X let R = τ1,2Int(R) ⊆ (1, 2)*-spcl**(τ1,2Int(R)). We have S is (1 ,2)*-D**spOS in X . Example 3.8 Reverse of theorem 3.7 is proved by this example S = {1,3} is a (1 ,2)*-D**sp-OS in X but not an τ1,2-OS in X. Definition 3.9 In BTS X (1, 2)*-D**spT1/2 space for every (1 ,2)*-D**spOS is τ1,2OS Remark 3.10 In (1, 2)*T1/2 space, Every (1, 2)*SpOS is (1 ,2)*-D**spOS. Remark 3.11 (1 ,2)∗-spcl**(A) ⊆ τ1,2spcl(A) for a subset A in X. Theorem 3.12 In BTS X. S is (1, 2)*-D**spOS ⇒ S i s (1, 2)*-spOS . Proof In X , Suppose S is (1, 2)*- D**spOS and By defn 3.3 and. By Remark 3.11, (1 ,2)*- spcl**(U) ⊆ τ1,2-spcl(U) . Hence U ⊆ S ⊆ τ1,2-spcl(U) ⇒ S is (1,2)*-spOS. Example 3.13 To prove the reverse of thm 3.12 is not true assume G = {1, 2, 3}, t1 = {G, φ, {2}} and t2 = {G, φ }. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 721 https://internationalpubls.com Then S = {2, 3} is ( 1, 2)*spOS but not a (1, 2)*-D**spOS. Remark 3.14 In BTS X Consider C ⊆ X, D ⊆ X we have (1, 2)*-spcl** (C∪D) = (1, 2)*-spcl**(C)∪ (1, 2)*-spcl**(D) . Theorem 3.15 In BTS X Consider C ⊆ X, D ⊆ X We have (1, 2)*-D**spOS ⇒ CunionD is also a (1, 2)*-D**spOS . Proof. In BTS X Suppose C and D are (1, 2)*-D**spOS in X. By defn & Remark3.14, (1, 2)*-spcl**(S) ⊆ (1, 2)*-spcl** (T) ⇒ (1, 2)*-spcl** (S ∪ T). ⇒ S ∪ T is also (1, 2)*-D**spOS . Example 3.16 In BTS X Suppose C , D are (1 ,2)*-D**spOS ⇒ C ∩ D may not (1, 2)*-D**spOS . Let X = {1, 2, 3, 4}, t1 = { X, φ, {1, 2}, {1 ,2, 3}, {1, 2, 4}} , t2 = { X, φ, { 1, 2 }, {3, 4} }. Then the set A = {1,2,3} and B = {3,4} are (1,2)*- D**spOS in X and A ∩ B = {3} is not a (1,2)*- D**spOS. Theorem 3.17 In BTS X Assume B be a (1, 2)*-D**spOS , B ⊆ C and B ⊆ C ⊆ ( (1, 2)*- spcl**(τ1,2Int(A)). we have C is a (1, 2)*-D**spOS Proof From statement B is (1, 2)*-D**spOS and By Thm 3.16 B ⊆ (1, 2)*-spcl**(τ1,2Int(B)), Also B ⊆ C Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 722 https://internationalpubls.com ⇒ τ1,2-Int(A) ⊆ τ1,2- Int(B). Hence, (1, 2)*-spcl**(τ1,2Int(B)) ⊆ (1, 2)*-spcl**(τ1,2Int(C)). We have C ⊆ (1, 2)*-spcl**(τ1,2Int(B)) ⊆ (1, 2)*-spcl**(τ1,2Int(C)) it proves C is (1, 2)*- D**spOS . Remark 3.18 A Map h : L → M is (1, 2)*gspContinuous ⇒ f((1, 2)*-spcl**(A))  1,2-spCl( f(A)). Theorem 3.19 In BTS X Suppose a map h : L → M be (1, 2)*gspContinuous and (1, 2)*Open ⇒ B is (1, 2)*- D**spOS ⇒ f(B) in Y is (1, 2)*-spOS . Proof From statement B is (1, 2)*-D**spOS in L. By defn 3.3 and. Remark 3.18, h((1, 2)*- spCl**(V)) ⊆ σ1,2-spCl(h(B)). We have h(B) ⊆ h((1, 2)*- spCl**(V)) ⊆ σ1,2-spCl(h(V)). Also given h is (1, 2)*Open Map h(V) in M is σ1,2-Open . it follows that h(B) in M is (1, 2)*- spOS. Theorem 3.20 in BTS X consider A Map h : L → M be a (1, 2)*homeomorphism. If B in L is (1, 2)*-D**spOS , then h(B) is (1, 2)*- D**spOS in M. Proof B is (1, 2)*-D**spOS in L.. By definition 3.3 h(V) ⊆ h(B) ⊆ h((1, 2)*-spCl**(V)). Also given h is (1, 2)*homeomorphism we have h((1, 2)*-spCl**(V)) ⊆ (1, 2)*- spCl**h(V)). it follows h(V) ⊆ h(B) ⊆(1, 2)*- spCl**( h(V)). so that h( B) in M is (1, 2) *-D**spOS Theorem 3.21 in BTS X Consider A Map h : L → M if h is (1, 2)*homeomorphism.and B in M is (1, 2)*-D**spOS then ∃ τ1,2-OS such that h−1(B) in M is (1, 2)*- D**spOS Proof From statement B in L is (1, 2)*-D**spOS . By defn 3.3 we have h−1(V) ⊆ h−1(B) ⊆ h−1((1, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 723 https://internationalpubls.com 2)*-spCl**( V)). and Since h is (1, 2)*homeomorphism ⇒ h−1((1, 2)*-spCl**(V)) ⊆ (1, 2)*-spCl**(h−1(V)). hence we have h−1(V) ⊆ h−1(B) ⊆ (1, 2)*-spCl**(h−1(V)) thus h−1(B) is (1, 2)*-D**spOS 4. (1, 2)*-D**spClosed and (1, 2)*-D**spOpen Mappings Definition 4. 1 A Map h : L → M is (1, 2)*-D**spO- map if h(V) in M is (1, 2)*-D**spOS ∀ τ1,2OS V in M. Theorem 4.2 A map h : L →M is (1, 2)*-Open map ⇒ (1, 2)*D**spOpen -Map Proof. From statement h : L →M is (1, 2)*-Open map .and G is τ1,2-OS in L. we have h(G) in M is σ1,2Open. From Theorem3.7, h(G) is (1, 2)*-D**spOS in M.. Henceforth h is (1, 2)*-D**spOpen . Example 4.3 Reverse ofTheorem 4.2 is not true. by this example Let L= M= {a1, ,b1 ,c1} , τ1 = {L, φ, {a1}}, τ2 = { L, φ, {a1, c1}}. Let σ1 = {L, φ, {a1}}, σ2 = {L, φ, {a1}, {a1,b1}}. Let h : L → M is an identity map. we have h is (1, 2)*- D**spOpen but h is not (1, 2)*-Open. map Definition 4.4 The map h : L → M is (1, 2)*- D**spClosed Map if For every τ1,2-CS V in L, h(V) in M is (1, 2)*- D**spClosed Remark 4.5 h : L → M is (1 ,2)*Closed ⇒ h is (1, 2)*- D**spClosed but conversely not true Proof. From Theorem 4.2. proof is clear 5. (1, 2)*-D**spContinuous Mappings Definition 5.1 A map h : K →L is called (1, 2)*-D**spContinuous ∀ σ1,2-OS in L its inverse image of h is (1, 2)*-D**spOpen in K.. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 724 https://internationalpubls.com Theorem 5.2 A map h : K →L h is (1, 2)*Continuous ⇒ h is (1, 2)*-D**spContinuous. Proof Assume R as a σ1,2OS in L. Also h is (1, 2)*Continuous, h−1(R) is τ1,2Open in X. From Theorem 3.7, h−1(R) is (1, 2)*-D**spOpen in X . Thus h is (1, 2)*-D**spContinuous. Example 5.3 This example proves reverse of thm 5.2 Consider A map h : K →L let the two sets L = K = { a1 ,a2 ,a3}, τ1 = { L, φ, {a1}}, τ2 = { L, φ, {a1}, {a1, a2}}, σ1 = {K, φ, {a1}} and σ2 = {K, φ, {a1 ,a3}}. Suppose h : K → L be the identity map we have h is (1, 2)*- D**spContinuous but h is not (1, 2)*Continuous . Remark 5.4 In (1, 2)*-D**sp-T1/2 space, every (1, 2)*-D**spContinuous map is (1, 2)*Continuous. Theorem 5.5 h : K → L is a map. we have the below implications are true. • 1 h is (1, 2)*- D**spContinuous. • 2. For each σ1,2CS in L its inverse image is (1, 2)*-D**spClosed in K . Proof. (1) ⇒ (2) Let R is σ1,2CS in L. Then L - R is σ1,2Open in L. Also h is (1, 2)*-D**spContinuous, f−1 (L - R) is (1, 2)*-D**spOpen in K. so that we have K / f−1(R) is (1, 2)*-D**spOpen in K ⇒ f−1(R) is (1, 2)*- D**spClosed in K. (ii) ⇒ (i) Let S is a σ1,2OS in L. Then L-S is σ1,2Open in L. ⇒ f−1(L \ S) is (1, 2)*- D**spClosed in K, ⇒L \f−1(S) is (1, 2)*-D**spClosed in L .So that h−1(S) is (1, 2)*- D**spOpen in L. Hence h is (1, 2)*-D**spContinous. Theorem 5.6 A s s u m e h : K → L is a map If h is (1, 2)*-D**spContinuous Map, Then h (τ1,2-D**spcl(B)) ⊆ σ1,2-spcl(h(B)). Proof. Given h(B) ⊆ σ1,2-spCl(h(B)), ⇒ B ⊆ h−1(σ1,2-spCl((B)). Then σ1,2-spCl(h(B)) is a σ1,2CS in L and Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 725 https://internationalpubls.com h is (1, 2)*-D**spContinuous map ⇒ h−1(σ1,2-spCl(h(B)) is (1, 2)*-D**spClosed in L. Hence τ1,2-D**spcl(B) ⊆ h−1(σ1,2-scl(f(B).it proves h (τ1,2-D**spcl (B)) ⊆ σ1,2-spcl(h(B)). (1, 2)*-D**spContinuous and (1, 2)*-D**spIrresolute Mappings Definition 6.1 Consider a map h : K → L H is (1, 2)*-D**spirresolute if for every (1, 2)*-D**spOS of L its inverse image of h is (1, 2)*-D**spOpen in K Remark 6.2 Consider a map h : K → L For every (1, 2)*-D**spCS of L by defn of 6.1 (1, 2)*-D**spClosed in K.. Theorem 6.3 Consider a map Proof. h : K → L is (1, 2)*-D**spirresolute implies h is (1, 2)*-D**spContinuous . Suppose R is a τ1,2OS in K. Also h is (1, 2)*-D**spirresolute proves h−1(R) is (1, 2)*- D**spOpen in K. Thus h is (1, 2)*-D**spContinuous. Example 6.4 Reverse part of the Theorem 6.3 Can be proved by the following example to show it is not true Let X = Y = {a1 ,a2 ,a3}, τ1 = {X, φ, {a1}, {a2}, { a1,a2}}, τ2 = {X, φ, {a1,a2}} , σ1 = {X, φ, {a1}}, σ2 = { X, φ, {a1}, {a1,a2}}. Let f : X → Y be the identity map. Hence f is (1, 2)*- D**sp-Continuous but f is not (1, 2)*-D**spirresolute. Theorem 6.5 Consider a map h : K → L h is (1, 2)*Continuous and L is (1, 2)*-D**sp-T1/2-space implies h is (1, 2)*- D**spirresolute. Proof Assume B be (1, 2)*-D**spOS in L. Also L is (1,2)*-D**sp-T1/2-space, implies B is an σ1,2OS in L and also h is (1, 2)*Continuous proves h−1(B) is (1, 2)*-D**spOS in K. Thus h is (1, 2)*-D**spirresolute. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 726 https://internationalpubls.com Theorem 6.6 Consider a map h : K → L h is (1, 2)*-D**spirresolute and k : L → M be an (1, 2)*-D**spirresolute maps. Then h o k : K → M is (1, 2)*- D**spirresolute. Proof. Suppose V be a (1, 2)*-D**spOS in M. so h−1(V) is (1, 2)*-D**spOpen in L implies h−1(k−1(V)) is (1, 2)*-D**spOpen in K. Thus (hok)−1(V) is (1, 2)*-D**spOpen in K. Hence hok is (1, 2)*-D**spirresolute. 7. (1, 2)*-D**spConnected Sets Definition 7.1 A space X is (1, 2)*-D**spdisConnected if it is the Union of two disjoint non empty (1, 2)*- D**spOS otherwise it is said to be (1, 2)*-D**spConnected Theorem 7.2 IN BTS X, the following statements are true. • X is (1, 2)*-D**spConnected. • φ , X are the subsets which are both (1, 2)*-D**spOpen and (1, 2)*-D**spClosed . Proof. i ⇒ ii Let U ⊆ X which is (1, 2)*-D**spOpen & (1, 2)*-D**spClosed. Then X/U is also (1, 2)*-D**spOpen & (1, 2)*-D**spClosed by defn of 7.1 . U and X/U implies either U = φ or X/U = φ. ii ⇒ i Suppose A ,B in X such that AUB =X where A, B not equal to empty (1, 2)*-D**spOS . So A- X/B is (1, 2)*-D**spCS ⇒ A is(1, 2)*-D**spO⊆ X and (1, 2)*- D**spCS ⊆ X as we assumed A = φ or X proves X is (1, 2)*- D**spConnected. Theorem 6.3 suppose a mapping j : K → L is i) (1, 2)*-D**spContinuous and onto ,K is (1, 2)*-D**spConnected ⇒ L is (1, 2)*Connected. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 727 https://internationalpubls.com (ii) If j : K → L is (1, 2)*-D**spirresolute surjection map and K is (1 ,2)*-D**spConnected ⇒ L is (1, 2)*-D**spConnected. Proof. Assume L is not (1, 2)*Connected. We have L = C ∪ D is not empty where C and D are disjoint σ1,2-OS in L. Also j is (1, 2)*-D**spContinuous , onto K = f−1(C) ∪ f−1(D) where f−1(C) and f−1(D) are disjoint but not empty (1, 2)*-D**spOSs contradicts X is (1, 2)*- D**spConnected. So that L is (1, 2)*Connected. ii proof obvious from 7.1. 8. (1, 2)*-D**spHomeomorphisms Definition 8.1 f : X →Y is a bijection map called (1, 2)*-D**sphomeomorphism if the mapping is (1, 2)*-D**spContinuous ,(1, 2)*-D**spOpen. Remark 8.2 Every (1, 2)*homeomorphism is (1, 2)*- D**sphomeomorphism but conversely not true Theorem 8.3 consider The mapping h : X →Y, is 1-1 and onto we have the following statements are true. • (i)h−1: Y → X is (1, 2)*- D**spContinuous. • (ii)The mapping is (1, 2)*-D**spOpen . • (iii)the mapping is (1, 2)*- D**spClosed . Proof. • (i)⇒ (ii) Let K be any τ1,2-OS in X. Since f−1 is (1 ,2)*-D**spContinuous, f( K) in Y is (1, 2)*- D**spOpen . Thus the mapping is (1, 2)* D**spOpen. • (ii)Implies iii In X suppose F is τ12CS , Then X/F is τ12OS and Also the mapping is (1, 2)*-D**spOpen , f (X/F) in Y is (1, 2)*-D**spOpen in Y.. But in Y, f(X/F) = Y /f(F) where f (F) is (1, 2)*-D**spClosed . Thus the mapping is (1, 2)*-D**spClosed Map (iii) implies (i) Suppose R is τ1,2-CS in X , We have f (R) in Y is (1, 2)*-D**spClosed . Also the mapping f is (1, 2)*- D**spClosed its inverse mapping is (1, 2)*-D**spContinuous. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 2s (2024) 728 https://internationalpubls.com Theorem 8.4 Suppose the mapping f is 1-1 , onto and (1, 2)*-D**spContinuous Then the implications are true. To prove the mapping f is • i)(1, 2)*-D**spOpen . • ii)(1, 2)*-D**sphomeomorphism . • iii)(1, 2)*-D**spClosed Proof Assume i) f is (1 ,2)*-D**spOpen also the mapping is 1-1 and onto , (1, 2)*-D**spContinuous From the definition8.1, the mapping is (1, 2)*-D**sphomeomorphism. (ii) is proved. assume (ii) the mapping is (1, 2)*-D**spOpen , 1-1 and onto it is (1, 2)*- D**spClosed from thm 7.8 . (iii) proved assume iii f is (1, 2)*-D**spClosed and bijective. F is (1, 2)*-D**spOpen map. By Theorem 8.3 ( i), proved REFERENCES [1] Bhattacharya, P. and Lahiri, B. K., Semi-Generalized closed sets in a topology, Indian J. Math., 1987, 29(3), 375. [2] Hdeib, H.Z., -closed mappings, Rev. Colomb. Mat., 1982, 16(3-4) 65-67. [3] Lellis Thivagar, M., Ravi, O. and Abd El-Monsef, M. 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