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DECOMPOSITION OF β -CONTINUITY IN IDEAL TOPOLOGICAL SPACES S. JAFARI β , K. VISWANATHAN, J.JAYASUDHA * College of Vestsjaelland South, Herrestrade 11, 4200 Slagelse,Denmark e-mail : πππ πππππππ ππ@πππππ.πππ Post-Graduate and Research Department of Mathematics, N G M College, Pollachi - 642 001, Tamil Nadu, INDIA e-mail : π£ππ π’πππ@π¦πβππ.πππ Abstract In this paper, we introduce and investigate the notions of βπ -continuous maps and βπ -irresolute maps in ideal topological spaces. Also we introduce the notion of π ππβ β πΌ sets and π ππβ β πΌ continuous maps to obtain a decomposition of β -continuity. 1 Introduction and Preliminaries The concept of ideals in topological spaces is treated in the classic text by Kuratowski [7] and Vaidyanathaswamy [13]. The notion of β -open sets in topological spaces was introduced by Jankovic and Hamlett [6]. Dontchev et al. [2] introduced and studied the notion of βπ -closed sets. Recently, Navaneethakrishnan and Paulraj Joseph [10] have studied further the properties of βπ -closed sets and βπ -open sets. An ideal β on a topological space (π, π ) is a non-empty collection of subsets of π satisfying the following properties: (1) π΄ β β and π΅ β π΄ imply π΅ β β (heredity); (2) π΄ β β and π΅ β β imply π΄ βͺ π΅ β β (ο¬nite additivity).A topological space (π, π ) with an ideal β on π is called an ideal topological space and is denoted by (π, π, β). For a subset π΄ β π, π΄β (β) = {π₯ β π : π β© π΄ β / β for every π β π (π₯)}, is called the local function [7] of π΄ with respect to β and π . We simply write π΄β in case there is no chance for confusion. A Kuratowski closure operator πΆπβ (.) for a topology π β (β) called the β -topology ο¬ner than π is deο¬ned by πΆπβ (π΄) = π΄ βͺ π΄β [13]. Let (π, π ) denote a topological space on which no separation axioms are assumed unless explicitly stated. In a topological space 0 0 2000 Mathematics subject classiο¬cation: Primary 54A05; Secondary 54D15, 54D30. keywords : βπ -closed set, βπ -open set, π ππβ β πΌ set 2 S. Jafari, K. Viswanathan, J.Jayasudha (π, π ), the closure and the interior of any subset π΄ of π will be denoted by πΆπ(π΄) and πΌππ‘(π΄), respectively. A subset π΄ of a space is said to be semi-open [9] if π΄ β ππ(πππ‘π΄) . A subset π΄ of a topological space (π, π ) is said to be g-closed [8] (resp. π -closed [12]) if πΆπ(π΄) β π whenever π΄ β π and π is open(resp.semi-open) in π; A subset π΄ of a topological space (π, π ) is locally closed (brieο¬y LC) if π΄ = π β© π where U is open and V is closed. A subset π΄ of a topological space (π, π ) is semi-locally closed (brieο¬y slc) if π΄ = π β© π where U is semi-open and V is semi-closed. A subset π΄ of a topological space (π, π ) is called π ππβ set if π΄ = π β© π where U is semi-open and V is closed. A function π : (π, π ) β (π, π) is g-continuous[8] (resp. π -continuous) if π β1 (π ) is g-closed(resp. π -closed) in (π, π ) for every closed set V of (π, π) . A subset π΄ of an ideal space (π, π, β) is β -closed [6] if π΄β β π΄ . A subset π΄ of an ideal space (π, π, β) is βπ -closed [2] (resp. βπ -closed [11]) if π΄β β π whenever π΄ β π and π is open(resp. semi-open). An ideal space (π, π, β) is called a πβ space [2] (resp. πβπ [11]) if every βπ -closed (resp. βπ -closed) subset of X is β -closed. Lemma 1.1. [6] Let (π, π, β) be a space with an ideal β on π, and π΄ is a subset of π. Then, 1. Every β -closed set is βπ -closed, 2. Every βπ -closed set is βπ -closed. We denote the family of all βπ -closed (resp. βπ - open) subsets of an ideal space (π, π, β) by βπ(π) (resp. βππ(π) ) . Deο¬nition 1.2. [5] A function f: (π, π, β) β (π, π) is said to be β -continuous if π β1 (π΄) is β -closed in (π, π, β) for every closed set A in (π, π) . 2 βπ -continuity and βπ -irresoluteness Deο¬nition 2.1. A function f: (π, π, β) β (π, π) is said to be βπ -continuous if π β1 (π΄) is βπ -closed in (π, π, β) for every closed set A in (π, π) . Remark 2.2. If β = {β } in the above deο¬nition then the notion of βπ -continuity coincides with the notion of π -continuity. Decomposition of β -continuity in ideal topological spaces 3 Deο¬nition 2.3. A function f: (π, π, β) β (π, π, π₯ ) is said to be βπ -irresolute if π β1 (π΄) is βπ -closed in (π, π, β) for every βπ -closed set A in (π, π, π₯ ) . Theorem 2.4. For a function f: (π, π, β) β (π, π) , the following hold: 1. Every continuous function is βπ -continuous; 2. Every β -continuous function is βπ -continuous; 3. Every π -continuous function is βπ -continuous; 4. Every βπ -continuous function is βπ -continuous. Proof. (1) Let f be a continuous function and V be a closed set in (π, π) . Then π β1 (π ) is closed in (π, π, β) . Since every closed set is β -closed and hence βπ -closed, π β1 (π ) is βπ -closed in (π, π, β) .Therefore, f is βπ -continuous. (2) Let f be a β -continuous function and V be a closed set in (π, π) . Then π β1 (π ) is β -closed in (π, π, β) . Since every β -closed set is βπ -closed, π β1 (π ) is βπ -closed in (π, π, β) .Therefore, f is βπ -continuous. (3) Let f be an π -continuous function and V be a closed set in (π, π) . Then π β1 (π ) is π -closed in (π, π, β) . Since every π -closed set is βπ -closed [11], π β1 (π ) is βπ -closed in (π, π, β) .Therefore, f is βπ -continuous. (4) Let f be a βπ -continuous function and V be a closed set in (π, π) . Then π β1 (π ) is βπ -closed in (π, π, β) . Since every βπ -closed set is βπ -closed, π β1 (π ) is βπ -closed in (π, π, β) .Therefore, f is βπ -continuous. β‘ Remark 2.5. The relationships deο¬ned above, are shown in the following diagram: continuity β π -continuity β π -continuity β β β β -continuity β βπ -continuity β βπ -continuity None of these implications is reversible as shown by the following examples. Example 2.6. Let π = {π, π, π}, π = {β , {π}, {π, π}, π} , β = {β , {π}} and π = {β , {π}, π} . Deο¬ne f: (π, π, β) β (π, π) as f(a)=b, f(b)=a, f(c)=c. Then for the set 4 S. Jafari, K. Viswanathan, J.Jayasudha π = {π, π} , π β1 (π ) = {π, π} is βπ -closed but not closed. Hence f is βπ -continuous but not continuous. Example 2.7. Let π = {π, π, π}, π = {β , {π}, {π, π}, π} , β = {β , {π}} and π = {β , {π, π}, π} . Deο¬ne f: (π, π, β) β (π, π) to be the identity map. Then for the set π = {π} , π β1 (π ) = {π} is βπ -closed but not β -closed. Hence f is βπ -continuous but not β -continuous. Example 2.8. Let π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} , β = {β , {π}} and π = {β , {π, π}, π} . Deο¬ne f: (π, π, β) β (π, π) to be the identity map. Then for the set π = {π, π} , π β1 (π ) = {π, π} is βπ -closed but not βπ -closed. Hence f is βπ -continuous but not βπ -continuous. Theorem 2.9. A function f: (π, π, β) β (π, π) is βπ -continuous if and only if π β1 (π ) is βπ -open in (π, π, β) for every open set V in (π, π) . Proof. Let f: (π, π, β) β (π, π) be βπ - continuous and V be an open set in (π, π) . Then π π is closed in (π, π) and π β1 (π π ) is βπ -closed in (π, π, β) . But π β1 (π π ) = (π β1 (π ))π and so π β1 (π ) is βπ -open in (π, π, β) . Conversely, Suppose that π β1 (π ) is βπ -open in (π, π, β) for each open set V in (π, π) . Let F be a closed set in (π, π) . Then πΉ π is open in (π, π) and by hypothesis π β1 (πΉ π ) is βπ -open in (π, π, β) . Since π β1 (πΉ π ) = (π β1 (πΉ ))π , we have π β1 (πΉ ) is βπ -closed in (π, π, β) and so f is βπ -continuous. Deο¬nition 2.10. For A function f: (π, π, β) β (π, π), the subset {(π₯, π (π₯)) : π₯ β π} β π × π is called the graph of f and is denoted by G(f). Lemma 2.11. The following properties are equivalent for a graph G(f ) of a function f: 1. G(f ) is βπ -continuous; 2. For each (π₯, π (π₯)) β π × π βπΊ(π ), there exists an βπ -open set ππ of X containing π₯ and an open set V containing π (π₯) such that π(ππ ) β© π β= π. Theorem 2.12. Let f: (π, π, β) β (π, π) be a function and g: π β π × π the graph function of f. If g is βπ -continuous then f is βπ -continuous. Decomposition of β -continuity in ideal topological spaces 5 Proof. Suppose that g is βπ -continuous. Let π₯ β π and V be any open set of Y containing f(x). Then π × π is open in π × π and by βπ -continuity of g, there exists π β βππ(π, π ) containing π₯ such that π(π ) β π × π . Therefore we obtain π (π ) β π. This shows that f is βπ -continuous. Theorem 2.13. Let (π, π, β) be an ideal topological space. If π΄ β βπ(π) and π΄ β π0 β βπ(π) then π΄ β βπ(π0 ). Proof. Let π΄ β βπ(π) and π΄ β π0 β βπ(π) then ππβ (π΄) β© π0 β π whenever π΄ β π β (π΄) = ππβ (π΄) β© π (Lemma 3.2 [1]).Hence π΄ β βπ(π ). and U is semi-open. Also πππ 0 0 0 Theorem 2.14. If f: (π, π, β) β (π, π) is an βπ -continuous function and π0 β βππ(π) . Then the restriction π /π0 : (π0 , π /π0 , β/π0 ) β (π, π) is βπ -continuous. Proof. Let V be any open set of (π, π) . Since f is βπ -continuous, π β1 (π ) is βπ -open in (π, π, β) and π β1 (π ) β© π0 = (π /π0 )β1 (π ) β βππ(π) . Moreover by Theorem 2.11 we have (π /π0 )β1 (π ) β βππ(π0 ) . This shows that π /π0 is βπ -continuous. Remark 2.15. The composition of two βπ -continuous maps need not be βπ -continuous as seen from the following example: Example 2.16. Let π = π = π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} and π = {β , {π}, π }, π = {β , {π, π}, π}, β = {β , {π}, {π}, {π, π}}, π₯ = {β , {π}}. Deο¬ne f: (π, π, β) β (π, π, π₯ ) and g: (π, π, π₯ ) β (π, π) to be the identity map. Then the map f is βπ -continuous, g is βπ -continuous but π β π is not βπ -continuous. Theorem 2.17. Let f: (π, π, β) β (π, π, π₯ ) be βπ -continuous and g: (π, π, π₯ ) β (π, π) be continuous. Then π β π : (π, π, β) β (π, π) is βπ -continuous. Proof. Let g be a continuous function and V be any open set in (π, π) . Then π β1 (π ) is open in (π, π, π₯ ) . Since f is βπ -continuous, π β1 (π β1 (π )) = (π β π )β1 (π ) is βπ -open in (π, π, β) . Hence π β π is βπ -continuous. Theorem 2.18. A function f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute if and only if the inverse image of every βπ -open set in (π, π, π₯ ) is βπ -open in (π, π, β) . Proof. This is a consequence of Theorem 2.9. 6 S. Jafari, K. Viswanathan, J.Jayasudha Theorem 2.19. Let f: (π, π, β) β (π, π, π₯ ) and g: (π, π, π₯ ) β (π, π, π¦) be βπ -irresolute. Then (π β π ) : (π, π, β) β (π, π, π¦) is βπ -irresolute. Proof. Let g be a βπ -irresolute function and V be any π¦π -open set in (π, π, π¦) . Then π β1 (π ) is π₯π -open in (π, π, π₯ ) . Since f is βπ -irresolute, π β1 (π β1 (π )) = (π β π )β1 (π ) is βπ -open in (π, π, β) . Hence π β π is βπ -irresolute. Theorem 2.20. If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and g: (π, π, π₯ ) β (π, π) is β - continuous then π β π : (π, π, β) β (π, π) is βπ - continuous. Proof. Let V be any closed set of (π, π) . Then π β1 (π ) is β -closed in (π, π, π₯ ) . Therefore, π β1 (π β1 (π )) = (π β π )β1 (π ) is βπ -closed in (π, π, β) , since every β -closed set is βπ - closed. Hence π β π is βπ -continuous. Theorem 2.21. Let (π, π, β) be an ideal topological space, (π, π, π₯ ) be a πβπ -space and (π, π) be a topological space. If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and g: (π, π, π₯ ) β (π, π) is βπ - continuous then π β π is βπ - continuous. Proof. Analogous to the Proof of Theorem 2.20. Theorem 2.22. Let (π, π, β) be an ideal topological space, (π, π, π₯ ) be a πβ -space and (π, π) be a topological space. If f: (π, π, β) β (π, π, π₯ ) is βπ -irresolute and g: (π, π, π₯ ) β (π, π) is βπ - continuous then π β π is βπ - continuous. Proof. Analogous to the Proof of Theorem 2.20. 3 Decomposition of β -continuity In this section, we obtain a decomposition of β -continuity in ideal topological spaces. In order to obtain the decomposition of β -continuity we introduce the notion of π ππβ β πΌ set and π ππβ β πΌ continuous mapping in ideal topological spaces and prove that a map is β -continuous if and only if it is both βπ -continuous and π ππβ β πΌ continuous. Deο¬nition 3.1. A subset π΄ of an ideal topological space (π, π, β) is called π ππβ β πΌ set if π΄ = π β© πΉ where U is semi-open and F is β -closed. Remark 3.2. Every β -closed set is π ππβ β πΌ set but not conversely. 7 Decomposition of β -continuity in ideal topological spaces Example 3.3. Let π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} and β = {β , {π}, {π}, {π, π}}. Then the set π΄ = {π} is a π ππβ β πΌ set but not a β -closed set. Remark 3.4. The notions of βπ -closed sets and π ππβ β πΌ sets are independent. Example 3.5. Let π = {π, π, π}, π = {β , {π}, {π, π}, π} and β = {β , {π}}. Then the set π΄ = {π} is βπ -closed but not π ππβ β πΌ set. Example 3.6. Let π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} and β = {β , {π}, {π}, {π, π}}. Then the set π΄ = {π} is π ππβ β πΌ set but not βπ -closed. Theorem 3.7. A subset of an ideal topological space (π, π, β) is β -closed if and only if it is both βπ -closed and an π ππβ β πΌ set. Proof. Necessity is trivial. To prove the suο¬ciency, assume that A is both βπ -closed and an π ππβ β πΌ -set. Then π΄ = π β© πΉ wher U is semi-open and F is β -closed. Therefore π΄ β π and π΄ β πΉ and so by hypothesis, π΄β β π and π΄β β πΉ . Thus π΄β β π β© πΉ = π΄ . Hence A is β -closed. Deο¬nition 3.8. A function f: (π, π, β) β (π, π) is said to be π ππβ β πΌ continuous if π β1 (π ) is an π ππβ β πΌ set in (π, π, β) for every closed set V in (π, π). Example 3.9. Let π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} , β = {β , {π}, {π}, {π, π}} and π = {β , {π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map. Then f is a π ππβ β πΌ continuous map. Remark 3.10. From the deο¬nitions it is clear that every β -continuous map is π ππβ β πΌ continuous. But the converse is not true. Example 3.11. Let π = {π, π, π, π}, π = {β , {π}, {π, π, π}, π} , β = {β , {π}, {π}, {π, π}} and π = {β , {π, π, π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map. Then f is π ππβ β πΌ continuous but not β -continuous, as for the closed set π = {π} in (π, π) π β1 (π ) = {π} is not β -closed. Remark 3.12. The concepts of βπ -continuity and π ππβ β πΌ continuity are independent as seen from the following examples. Example 3.13. The map f deο¬ned in Example 3.11 is π ππβ β πΌ continuous but not βπ continuous, as for the closed set π = {π} in (π, π) π β1 (π ) = {π} is not βπ -closed. 8 S. Jafari, K. Viswanathan, J.Jayasudha Example 3.14. Let π = {π, π, π}, π = {β , {π}, {π, π}, π} , β = {β , {π}} and π = {β , {π, π}, π}. Deο¬ne the map f: (π, π, β) β (π, π) to be the identity map.Then f is βπ -continuous but not π ππβ β πΌ continuous, as for the closed set π = {π} in (π, π) π β1 (π ) = {π} is not an π ππβ β πΌ set. Theorem 3.15. A function f: (π, π, β) β (π, π) is β -continuous if and only if it is both βπ -continuous and π ππβ β πΌ continuous. Proof. This is an immediate consequence of Theorem 3.7. References [1] A. Acikgoz, T. Noiri and S. Yuksel, On πΌ -I-continuous and πΌ -I-open functions, Acta Math. Hungar.,105(1-2)(2004), 27-37. [2] J. Dontchev, M. Ganster and T. Noiri, Uniο¬ed operation approach of generalized closed sets via topological ideals, Math. Japan, 49(1999), 395-401. 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