# Characterizations of Functions with Closed Graph

Characterizations of Functions with Closed Graph The nature and usefulness of KC functions in topology have been well established in literature. In this paper, we introduce a new class of functions to be called inversely KC functions and emphasize its utility in characterization of functions with closed graph for the first time. We hereby study the relation between KC functions and Inversely KC functions and investigate various specific conditions under which these functions can be converted into functions having closed graph. Mathematics Subject Classification 2010: 54C10, 54D50. Key words: closed graph, KC function, inversely KC function, Frechet space, kspace. 1. Introduction A number of conditions on the functions or their inverses preserving closedness, openness or compactness of sets have been studied by Fuller [1]. Various important relationships between the class of functions having closed graph and the class of functions which map compact sets into closed sets have also been studied. Some interesting results which were proved by Fuller are the following: (i) For any function f : X Y a closed function with closed point inverses and domain as a regular space has a closed graph. (ii) A function into a Hausdorff space is continuous if and only if it is sub-continuous and has a closed graph. The functions which map compact sets into closed sets have been named as KC functions by Wilansky [6]. In literature, the nature and usefulness of KC functions in topology have been well established but the functions under which inverse image of a compact set is closed are not described so far which are to be called as inversely KC functions by us. It is well known that a function with closed graph is KC as well as Inversely KC. In this paper we study the relation between KC functions and Inversely KC functions and investigate the various specific conditions under which inversely KC functions or KC functions can be converted into functions having closed graph. It is also shown that a square Frechet, KC space is T2 and that the product of a Frechet space with a first countable space need not be Frechet. Some counter examples are also given. 2. Preliminaries We have used some of the standard notation and definitions. By a space we shall mean a topological space. Let X be a space. For H Y X, clY (H) denotes the closure of H in Y . A subset K of X is called limit point compact (see [4]) if every infinite subset of K has a limit point in K. For any space X, X will denote the one point compactification of X and Q will denote the set of all rationals with the usual topology. If A is a subset of X, we say that X is T1 at A (see [2]) if each point of A is closed in X. According to Wilansky X is said to be Frechet space (or closure sequential) if for each subset A of X, x cl(A) implies there exists a sequence {xn } in A converging to x while X will be called square Frechet if X × X is Frechet (see [6]). X is said to be a k-space if O is open (closed) in X whenever O K is open (closed) in K for every compact subset K of X and X is said to be KC if every compact set is closed (see [5]). Let X and Y are two spaces. A function f : X Y is said to be compact (compact preserving) if inverse image (image) of each compact set is compact. If graph of f that is, the set {(x, f (x)) : x X} is a closed subset of the product space X × Y then f is said to have closed graph (CG in short). The Theorems 2.1 to 2.4, which are rephrased here, will be used frequently. Theorem 2.3 follows from Theorem 1.4.5 of (see [3]) and Theorem 2.1. Theorem 2.1 ([1, Theorem 3.6]). For any function f : X Y , CG always implies KC as well as inversely KC. Theorem 2.2 ([1, Theorem 3.11]). If X is locally compact and regular, then any function f : X Y has CG if and only if f is KC with closed point inverses. Theorem 2.3 ([1, Theorem 3.11]). If Y is locally compact and regular, then any function f : X Y has CG if and only if f is inversely KC and Y is T1 at f (X). Theorem 2.4 ([3, Theorem 1.1.23]). For any function f : X Y , where X, Y are first countable, the following three conditions are equivalent: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses. 3. Characterizations of KC and inversely KC functions From the definition of KC and inversely KC functions, we immediately have the following theorem. Theorem 3.1. (a) A closed (continuous) function f : X Y is KC (inversely KC) if X(Y ) is KC; the converse holds if KC on the spaces is replaced by compact. (b) A compact preserving (compact) function is KC (inversely KC) if Y (X) is KC; the converse holds if KC on the spaces is replaced by compact. (c) A continuous function is KC as well as inversely KC if Y is KC. Theorem 3.2. For a function f : X Y , the following are equivalent: (a) f is KC. (b) f /H is KC for every subset H of X. (c) f /H is KC for every compact subset H of X. (d) f /H is closed for every compact subset H of X. Theorem 3.3. For a function f : X Y , the following are equivalent: (a) f is KC. (b) Given any subset S of Y and any subset U of X, with compact complement,containing f -1 (S), there exists an open set V containing S such that f -1 (V ) U . (c) Given any point y Y and any set U , with compact complement containing f -1 ({y}), there exists an open set V containing y such that f -1 (V ) U . Proof. First we prove (a) implies (b). Let S be a subset of Y , and U a subset of X, with compact complement, containing f -1 (S). Then f -1 (S) (X - U ) = . This implies that S f (X - U ) = . So S V = Y - f (X - U ). Since X - U is compact and f is KC, f (X - U ) is closed. Therefore V is open. Now V f (X - U ) = , this implies that f -1 (V ) (X - U ) = f -1 (V ) U. This proves (a) (b). (b) implies (c) is obvious. Finally we prove (c) implies (a). For any compact subset K of X, take y Y such that y f (K), then f -1 ({y}) K = implies that / f -1 ({y}) U = X - K. Therefore by (c), there is an open subset V of Y containing y such that f -1 (V ) U = X - K that is V f (X - U ) = or V f (K) = . This implies that y clY (f (K)). Therefore, f (K) is closed. / This proves (c) (a). Theorem 3.4. For a function f : X Y, the following are equivalent: (a) f is KC. (b) For each subset U of X with compact complement, the set G = {y Y : f -1 ({y}) U } is open in Y . (c) For each compact subset A of X, the set K = {y Y : f -1 ({y})A = )} is closed in Y . Proof. First we prove (a) implies (b). Let y G. Then f -1 ({y}) U y f (U ), this implies that y V = Y - f (X - U ). Therefore V is open in Y as f is KC. Also y V G. Thus G is open. This proves (a) (b). Now we prove (b) implies (c). Let A be a compact subset of X. Then G = {y Y : f -1 ({y}) X - A} is open. But G = Y - K. So K is closed. This proves (b) (c). Finally we prove (c) implies (a). Let A be a compact subset of X and yo f (A). Then f -1 ({yo }) A = . Let / -1 ({y}) A = )}. Then y Y - K and Y - K is open by K = {y Y : f o (c). Now f (A) (Y - K) = , this implies that yo clY (f (A)). Thus f (A) / is closed. This proves (c) (a). Theorem 3.5. For a function f : X Y , the following are equivalent: (a) f is inversely KC. (b) Inverse image of each set with compact complement is open. (c) For any point x X and any set V containing f (x), whose complement is compact, there exists an open set U containing x such that f (U ) V . Proof. First we prove (a) implies (b). Let K be a subset of Y such that Y - K is compact. Then by (a), f -1 (Y - K) is closed. So X - f -1 (K) is closed. Therefore f -1 (K) is open. This proves (a) (b). Now we prove (b) implies (c). Let x X and V be a subset of Y such that f (x) V, Y - V is compact. Then by (b), f -1 (V ) is open and x f -1 (V ). Therefore, there exists an open subset U of X such that x U f -1 (V ). Thus f (U ) V . This proves (b) (c). Finally we prove (c) implies (a). Let K any compact subset of Y and x (X - f -1 (K)). Take V = Y - K. Then f (x) V . So by (c), there is an open subset U of X containing x such that f (U ) V . Therefore x U f -1 (V ). This implies that f -1 (V ) = f -1 (Y - K) = X - f -1 (K) is open. This proves (c) (a). Theorem 3.6. Let f : X Y is a function: (i) If f is inversely KC, then f /H : H Y is inversely KC for each compact subset H of X. (ii) If X is a k-space and f /H : H Y is inversely KC for each compact subset H of X, then f is inversely KC. Proof. We prove only (ii). Suppose f /H : H Y is inversely KC for each compact subset H of X. Then for any compact subset K of Y , H f -1 (K) = (f /H)-1 (K) is closed in H, for every compact subset H of X. This implies that f -1 (K) is closed in X as X is a k-space. Thus f is inversely KC. 4. Relationship between KC and inversely KC functions For arbitrary spaces, an inversely KC function need not be KC (even if the range space is T1 ) and a KC function with closed point inverses need not be inversely KC, as the following example shows. Example 4.1. Let X be an uncountable set with the countable complement topology and fix any point p X. Let f : X X be a function such that f (x) = x for x X and f () = p. Then f is inversely KC but not KC, although the range space is T1 . Also the inclusion function i : X X is KC with closed point inverses but not inversely KC. Theorem 4.2. Suppose f : X Y is inversely KC and Y is T1 at f (X). If Y is a Frechet space, then image of every limit point compact set is closed. Proof. Let H be a limit point compact subset of X and let y clY (f (H)) such that y f (H). There exists a sequence {xn } of points in H / such that f (xn ) y. Since H is limit point compact set, the sequence {xn } has a cluster point x, say, in H. Since Y is T1 at f (X), V = Y - {f (x)} is an open set containing y. Therefore there exists an integer no such that f (xn ) V for all n no as f (xn ) y. Let K = {f (xn ) : n no } {y}. Then K is compact. Since x (clX (f -1 (K)) - f -1 (K)), f -1 (K) is not closed in X. This is a contradiction as f is inversely KC. Thus f (H) is closed in Y . Corollary 4.3. Suppose f : X Y is inversely KC and Y is T1 at f (X). If Y is a Frechet space, then f is KC. Corollary 4.4. Suppose f : X Y is inversely KC and Y is T1 at f (X). If X is compact and Y is a Frechet space, then f is closed. Theorem 4.5. Suppose f : X Y is inversely KC. If Y is a k-space which is either regular and T1 at f (X) or T2 , then f is KC. Proof. Let H be a compact subset of X and K be any compact subset of Y . Since Y is a k-space, it is enough to prove that f (H) K is closed in K. Now f (H) K = f (H1 ) where H1 = H f -1 (K) is closed, compact in H. If f1 = f /H1 : H1 K, then f1 is inversely KC, where K is compact and is either regular and T1 at f1 (H1 ) or T2 . By Theorem 2.3, f1 has CG and so f1 is KC, by Theorem 2.1. Thus f (H) K is closed in K implying thereby that f (H) is closed in Y , since Y is a k-space. Hence f is KC, in this case also. Corollary 4.6. Suppose f : X Y is inversely KC. If X is compact and Y is a k-space which is either regular and T1 at f (X) or T2 , then f is closed. Theorem 4.7. Suppose f : X Y is KC and has closed point inverses. If X is Frechet, then inverse image of every limit point compact set is closed. Proof. Let K be any limit point compact subset of Y , and let x clX (f -1 (K)) such that x f -1 (K). Since X is a Frechet space, there exists / a sequence {xn } of points in f -1 (K) such that xn x. Then {f (xn )} is a sequence in K and K is limit point compact implies {f (xn )} has a cluster point, say y in K. Since y = f (x), U = X - f -1 ({y}) is an open set containing x. Therefore, there exists a positive integer no such that xn U for all n no as xn x. Then H = {xn : n no } {x} is compact, but f (H) is not closed as y (clY (f (H)) - f (H)). This gives a contradiction to the given condition. Thus f -1 (K) is closed in X. Corollary 4.8. Suppose f : X Y is KC and has closed point inverses. If X is a Frechet space, then f is inversely KC. Corollary 4.9. Suppose f : X Y is KC and has closed point inverses. If Y is compact and X is a Frechet space, then f is continuous. Theorem 4.10. Suppose f : X Y is KC and has closed point inverses. If X is a k-space which is either regular or T2 , then f is inversely KC. Proof. Let H be a compact subset of X. If f1 = f /H : H Y , then f1 is KC and has closed point inverses. By Theorem 2.2, f1 has CG, and so f1 is Inversely KC by Theorem 2.1. Now the theorem follows by Theorem 3.6(ii). Theorem 4.11. Suppose f : X Y is KC and has closed point inverses. If Y is compact and X is a k-space which is either regular or T2 , then f is continuous. Theorem 4.12. Let f : X Y be inversely KC, compact preserving, closed injection (KC, compact, continuous surjection). Then f is KC (inversely KC). Corollary 4.13. A homeomorphism f from a space X onto a space Y is KC if and only if it is inversely KC. 5. Characterizations of functions with closed graph Remark 5.1. Theorem 5.2 is a generalized version of Theorem 2.4 (without any change in the original proof) and Theorems 2.2 and 2.3 are jointly stated in Theorem 5.4. We shall generalize Theorem 5.4 by replacing the condition locally compact by k-space or locally compact, regular by Frechet on either of the spaces (Theorem 5.8 below). Theorem 5.2. Let X × Y be Frechet then for any function f : X Y , the conditions: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses of Theorem 2.4 are equivalent. Corollary 5.3. Every square Frechet, KC space is T2 . In particular, every first countable, KC space is T2 . Proof. Using the identity map i : X X in Theorem 5.2 and Theorem 1.1.8 (see [3]). Theorem 5.4. Let X and Y be both locally compact, regular. Then for any function f : X Y , the conditions: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses of Theorem 2.4 are equivalent. Remark 5.5. Example 5.6 below shows that the condition "Frechet" on X × Y in Theorem 5.2 cannot be replaced by "first countable" on one of the spaces and "Frechet" on the other. Moreover, part (a) of this example also shows that the generalized version of Theorem 5.4 i.e, Theorem 5.8 does not hold without "locally compact, regular" on at least one of the spaces. Part (b) of example 5.6 also shows that a KC and inversely KC function need not have closed graph, even if the range space is T2 . Example 5.6. (a) The inclusion map i : Q Q satisfies both (ii) and (iii) but not (i), since Q is KC but not T2 . Here none of the space is locally compact and regular, the domain space is first countable and the range is Frechet and T1 . (b) Also for any fixed p Q, the function f : Q Q which is identity on Q with f () = p, satisfies (ii) and (iii) but not (i). Here the domain space is Frechet and the range space is first countable and T2 . Using example 5.6, we get the following: Corollary 5.7. Q × Q is not Frechet, and so the product of a first countable space and a Frechet space need not be Frechet (since Q is Frechet). The following Theorem 5.8 which is a generalization of Theorem 5.4 above, follows from our Corollary 4.3, Theorem 4.5, Corollary 4.8, Theorem 4.10 and Theorems 2.1, 2.2 and 2.3. Theorem 5.8. Let f : X - Y be any function and let: (a) X be locally compact, regular, and Y be a k-space, which is either regular or T2 or Frechet, (b) Y be locally compact, regular and X be a k-space, which is either regular or T2 or Frechet. Then the conditions (i), (ii), and (iii) of Theorem 2.4 are equivalent. Remark 5.9. Theorem 5.2 requires in particular, X and Y both to be Frechet, while Theorem 5.4 requires X and Y both to be locally compact, regular. Our Theorem 5.8 requires in particular, one of the spaces to be Frechet, and the other locally compact, regular. Thus our Theorem 5.8 serves as a bridge between Theorem 5.2 and Theorem 5.4. Acknowledgement. The authors would like to express gratitude to referee for his valuable comments. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annals of the Alexandru Ioan Cuza University - Mathematics de Gruyter

# Characterizations of Functions with Closed Graph

, Volume 60 (2) – Nov 24, 2014
10 pages

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de Gruyter
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1221-8421
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1221-8421
DOI
10.2478/aicu-2013-0050
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### Abstract

The nature and usefulness of KC functions in topology have been well established in literature. In this paper, we introduce a new class of functions to be called inversely KC functions and emphasize its utility in characterization of functions with closed graph for the first time. We hereby study the relation between KC functions and Inversely KC functions and investigate various specific conditions under which these functions can be converted into functions having closed graph. Mathematics Subject Classification 2010: 54C10, 54D50. Key words: closed graph, KC function, inversely KC function, Frechet space, kspace. 1. Introduction A number of conditions on the functions or their inverses preserving closedness, openness or compactness of sets have been studied by Fuller [1]. Various important relationships between the class of functions having closed graph and the class of functions which map compact sets into closed sets have also been studied. Some interesting results which were proved by Fuller are the following: (i) For any function f : X Y a closed function with closed point inverses and domain as a regular space has a closed graph. (ii) A function into a Hausdorff space is continuous if and only if it is sub-continuous and has a closed graph. The functions which map compact sets into closed sets have been named as KC functions by Wilansky [6]. In literature, the nature and usefulness of KC functions in topology have been well established but the functions under which inverse image of a compact set is closed are not described so far which are to be called as inversely KC functions by us. It is well known that a function with closed graph is KC as well as Inversely KC. In this paper we study the relation between KC functions and Inversely KC functions and investigate the various specific conditions under which inversely KC functions or KC functions can be converted into functions having closed graph. It is also shown that a square Frechet, KC space is T2 and that the product of a Frechet space with a first countable space need not be Frechet. Some counter examples are also given. 2. Preliminaries We have used some of the standard notation and definitions. By a space we shall mean a topological space. Let X be a space. For H Y X, clY (H) denotes the closure of H in Y . A subset K of X is called limit point compact (see [4]) if every infinite subset of K has a limit point in K. For any space X, X will denote the one point compactification of X and Q will denote the set of all rationals with the usual topology. If A is a subset of X, we say that X is T1 at A (see [2]) if each point of A is closed in X. According to Wilansky X is said to be Frechet space (or closure sequential) if for each subset A of X, x cl(A) implies there exists a sequence {xn } in A converging to x while X will be called square Frechet if X × X is Frechet (see [6]). X is said to be a k-space if O is open (closed) in X whenever O K is open (closed) in K for every compact subset K of X and X is said to be KC if every compact set is closed (see [5]). Let X and Y are two spaces. A function f : X Y is said to be compact (compact preserving) if inverse image (image) of each compact set is compact. If graph of f that is, the set {(x, f (x)) : x X} is a closed subset of the product space X × Y then f is said to have closed graph (CG in short). The Theorems 2.1 to 2.4, which are rephrased here, will be used frequently. Theorem 2.3 follows from Theorem 1.4.5 of (see [3]) and Theorem 2.1. Theorem 2.1 ([1, Theorem 3.6]). For any function f : X Y , CG always implies KC as well as inversely KC. Theorem 2.2 ([1, Theorem 3.11]). If X is locally compact and regular, then any function f : X Y has CG if and only if f is KC with closed point inverses. Theorem 2.3 ([1, Theorem 3.11]). If Y is locally compact and regular, then any function f : X Y has CG if and only if f is inversely KC and Y is T1 at f (X). Theorem 2.4 ([3, Theorem 1.1.23]). For any function f : X Y , where X, Y are first countable, the following three conditions are equivalent: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses. 3. Characterizations of KC and inversely KC functions From the definition of KC and inversely KC functions, we immediately have the following theorem. Theorem 3.1. (a) A closed (continuous) function f : X Y is KC (inversely KC) if X(Y ) is KC; the converse holds if KC on the spaces is replaced by compact. (b) A compact preserving (compact) function is KC (inversely KC) if Y (X) is KC; the converse holds if KC on the spaces is replaced by compact. (c) A continuous function is KC as well as inversely KC if Y is KC. Theorem 3.2. For a function f : X Y , the following are equivalent: (a) f is KC. (b) f /H is KC for every subset H of X. (c) f /H is KC for every compact subset H of X. (d) f /H is closed for every compact subset H of X. Theorem 3.3. For a function f : X Y , the following are equivalent: (a) f is KC. (b) Given any subset S of Y and any subset U of X, with compact complement,containing f -1 (S), there exists an open set V containing S such that f -1 (V ) U . (c) Given any point y Y and any set U , with compact complement containing f -1 ({y}), there exists an open set V containing y such that f -1 (V ) U . Proof. First we prove (a) implies (b). Let S be a subset of Y , and U a subset of X, with compact complement, containing f -1 (S). Then f -1 (S) (X - U ) = . This implies that S f (X - U ) = . So S V = Y - f (X - U ). Since X - U is compact and f is KC, f (X - U ) is closed. Therefore V is open. Now V f (X - U ) = , this implies that f -1 (V ) (X - U ) = f -1 (V ) U. This proves (a) (b). (b) implies (c) is obvious. Finally we prove (c) implies (a). For any compact subset K of X, take y Y such that y f (K), then f -1 ({y}) K = implies that / f -1 ({y}) U = X - K. Therefore by (c), there is an open subset V of Y containing y such that f -1 (V ) U = X - K that is V f (X - U ) = or V f (K) = . This implies that y clY (f (K)). Therefore, f (K) is closed. / This proves (c) (a). Theorem 3.4. For a function f : X Y, the following are equivalent: (a) f is KC. (b) For each subset U of X with compact complement, the set G = {y Y : f -1 ({y}) U } is open in Y . (c) For each compact subset A of X, the set K = {y Y : f -1 ({y})A = )} is closed in Y . Proof. First we prove (a) implies (b). Let y G. Then f -1 ({y}) U y f (U ), this implies that y V = Y - f (X - U ). Therefore V is open in Y as f is KC. Also y V G. Thus G is open. This proves (a) (b). Now we prove (b) implies (c). Let A be a compact subset of X. Then G = {y Y : f -1 ({y}) X - A} is open. But G = Y - K. So K is closed. This proves (b) (c). Finally we prove (c) implies (a). Let A be a compact subset of X and yo f (A). Then f -1 ({yo }) A = . Let / -1 ({y}) A = )}. Then y Y - K and Y - K is open by K = {y Y : f o (c). Now f (A) (Y - K) = , this implies that yo clY (f (A)). Thus f (A) / is closed. This proves (c) (a). Theorem 3.5. For a function f : X Y , the following are equivalent: (a) f is inversely KC. (b) Inverse image of each set with compact complement is open. (c) For any point x X and any set V containing f (x), whose complement is compact, there exists an open set U containing x such that f (U ) V . Proof. First we prove (a) implies (b). Let K be a subset of Y such that Y - K is compact. Then by (a), f -1 (Y - K) is closed. So X - f -1 (K) is closed. Therefore f -1 (K) is open. This proves (a) (b). Now we prove (b) implies (c). Let x X and V be a subset of Y such that f (x) V, Y - V is compact. Then by (b), f -1 (V ) is open and x f -1 (V ). Therefore, there exists an open subset U of X such that x U f -1 (V ). Thus f (U ) V . This proves (b) (c). Finally we prove (c) implies (a). Let K any compact subset of Y and x (X - f -1 (K)). Take V = Y - K. Then f (x) V . So by (c), there is an open subset U of X containing x such that f (U ) V . Therefore x U f -1 (V ). This implies that f -1 (V ) = f -1 (Y - K) = X - f -1 (K) is open. This proves (c) (a). Theorem 3.6. Let f : X Y is a function: (i) If f is inversely KC, then f /H : H Y is inversely KC for each compact subset H of X. (ii) If X is a k-space and f /H : H Y is inversely KC for each compact subset H of X, then f is inversely KC. Proof. We prove only (ii). Suppose f /H : H Y is inversely KC for each compact subset H of X. Then for any compact subset K of Y , H f -1 (K) = (f /H)-1 (K) is closed in H, for every compact subset H of X. This implies that f -1 (K) is closed in X as X is a k-space. Thus f is inversely KC. 4. Relationship between KC and inversely KC functions For arbitrary spaces, an inversely KC function need not be KC (even if the range space is T1 ) and a KC function with closed point inverses need not be inversely KC, as the following example shows. Example 4.1. Let X be an uncountable set with the countable complement topology and fix any point p X. Let f : X X be a function such that f (x) = x for x X and f () = p. Then f is inversely KC but not KC, although the range space is T1 . Also the inclusion function i : X X is KC with closed point inverses but not inversely KC. Theorem 4.2. Suppose f : X Y is inversely KC and Y is T1 at f (X). If Y is a Frechet space, then image of every limit point compact set is closed. Proof. Let H be a limit point compact subset of X and let y clY (f (H)) such that y f (H). There exists a sequence {xn } of points in H / such that f (xn ) y. Since H is limit point compact set, the sequence {xn } has a cluster point x, say, in H. Since Y is T1 at f (X), V = Y - {f (x)} is an open set containing y. Therefore there exists an integer no such that f (xn ) V for all n no as f (xn ) y. Let K = {f (xn ) : n no } {y}. Then K is compact. Since x (clX (f -1 (K)) - f -1 (K)), f -1 (K) is not closed in X. This is a contradiction as f is inversely KC. Thus f (H) is closed in Y . Corollary 4.3. Suppose f : X Y is inversely KC and Y is T1 at f (X). If Y is a Frechet space, then f is KC. Corollary 4.4. Suppose f : X Y is inversely KC and Y is T1 at f (X). If X is compact and Y is a Frechet space, then f is closed. Theorem 4.5. Suppose f : X Y is inversely KC. If Y is a k-space which is either regular and T1 at f (X) or T2 , then f is KC. Proof. Let H be a compact subset of X and K be any compact subset of Y . Since Y is a k-space, it is enough to prove that f (H) K is closed in K. Now f (H) K = f (H1 ) where H1 = H f -1 (K) is closed, compact in H. If f1 = f /H1 : H1 K, then f1 is inversely KC, where K is compact and is either regular and T1 at f1 (H1 ) or T2 . By Theorem 2.3, f1 has CG and so f1 is KC, by Theorem 2.1. Thus f (H) K is closed in K implying thereby that f (H) is closed in Y , since Y is a k-space. Hence f is KC, in this case also. Corollary 4.6. Suppose f : X Y is inversely KC. If X is compact and Y is a k-space which is either regular and T1 at f (X) or T2 , then f is closed. Theorem 4.7. Suppose f : X Y is KC and has closed point inverses. If X is Frechet, then inverse image of every limit point compact set is closed. Proof. Let K be any limit point compact subset of Y , and let x clX (f -1 (K)) such that x f -1 (K). Since X is a Frechet space, there exists / a sequence {xn } of points in f -1 (K) such that xn x. Then {f (xn )} is a sequence in K and K is limit point compact implies {f (xn )} has a cluster point, say y in K. Since y = f (x), U = X - f -1 ({y}) is an open set containing x. Therefore, there exists a positive integer no such that xn U for all n no as xn x. Then H = {xn : n no } {x} is compact, but f (H) is not closed as y (clY (f (H)) - f (H)). This gives a contradiction to the given condition. Thus f -1 (K) is closed in X. Corollary 4.8. Suppose f : X Y is KC and has closed point inverses. If X is a Frechet space, then f is inversely KC. Corollary 4.9. Suppose f : X Y is KC and has closed point inverses. If Y is compact and X is a Frechet space, then f is continuous. Theorem 4.10. Suppose f : X Y is KC and has closed point inverses. If X is a k-space which is either regular or T2 , then f is inversely KC. Proof. Let H be a compact subset of X. If f1 = f /H : H Y , then f1 is KC and has closed point inverses. By Theorem 2.2, f1 has CG, and so f1 is Inversely KC by Theorem 2.1. Now the theorem follows by Theorem 3.6(ii). Theorem 4.11. Suppose f : X Y is KC and has closed point inverses. If Y is compact and X is a k-space which is either regular or T2 , then f is continuous. Theorem 4.12. Let f : X Y be inversely KC, compact preserving, closed injection (KC, compact, continuous surjection). Then f is KC (inversely KC). Corollary 4.13. A homeomorphism f from a space X onto a space Y is KC if and only if it is inversely KC. 5. Characterizations of functions with closed graph Remark 5.1. Theorem 5.2 is a generalized version of Theorem 2.4 (without any change in the original proof) and Theorems 2.2 and 2.3 are jointly stated in Theorem 5.4. We shall generalize Theorem 5.4 by replacing the condition locally compact by k-space or locally compact, regular by Frechet on either of the spaces (Theorem 5.8 below). Theorem 5.2. Let X × Y be Frechet then for any function f : X Y , the conditions: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses of Theorem 2.4 are equivalent. Corollary 5.3. Every square Frechet, KC space is T2 . In particular, every first countable, KC space is T2 . Proof. Using the identity map i : X X in Theorem 5.2 and Theorem 1.1.8 (see [3]). Theorem 5.4. Let X and Y be both locally compact, regular. Then for any function f : X Y , the conditions: (i) f has CG. (ii) f is inversely KC and Y is T1 at f (X). (iii) f is KC and has closed point inverses of Theorem 2.4 are equivalent. Remark 5.5. Example 5.6 below shows that the condition "Frechet" on X × Y in Theorem 5.2 cannot be replaced by "first countable" on one of the spaces and "Frechet" on the other. Moreover, part (a) of this example also shows that the generalized version of Theorem 5.4 i.e, Theorem 5.8 does not hold without "locally compact, regular" on at least one of the spaces. Part (b) of example 5.6 also shows that a KC and inversely KC function need not have closed graph, even if the range space is T2 . Example 5.6. (a) The inclusion map i : Q Q satisfies both (ii) and (iii) but not (i), since Q is KC but not T2 . Here none of the space is locally compact and regular, the domain space is first countable and the range is Frechet and T1 . (b) Also for any fixed p Q, the function f : Q Q which is identity on Q with f () = p, satisfies (ii) and (iii) but not (i). Here the domain space is Frechet and the range space is first countable and T2 . Using example 5.6, we get the following: Corollary 5.7. Q × Q is not Frechet, and so the product of a first countable space and a Frechet space need not be Frechet (since Q is Frechet). The following Theorem 5.8 which is a generalization of Theorem 5.4 above, follows from our Corollary 4.3, Theorem 4.5, Corollary 4.8, Theorem 4.10 and Theorems 2.1, 2.2 and 2.3. Theorem 5.8. Let f : X - Y be any function and let: (a) X be locally compact, regular, and Y be a k-space, which is either regular or T2 or Frechet, (b) Y be locally compact, regular and X be a k-space, which is either regular or T2 or Frechet. Then the conditions (i), (ii), and (iii) of Theorem 2.4 are equivalent. Remark 5.9. Theorem 5.2 requires in particular, X and Y both to be Frechet, while Theorem 5.4 requires X and Y both to be locally compact, regular. Our Theorem 5.8 requires in particular, one of the spaces to be Frechet, and the other locally compact, regular. Thus our Theorem 5.8 serves as a bridge between Theorem 5.2 and Theorem 5.4. Acknowledgement. The authors would like to express gratitude to referee for his valuable comments.

### Journal

Annals of the Alexandru Ioan Cuza University - Mathematicsde Gruyter

Published: Nov 24, 2014