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UBC Theses and Dissertations

Functions with closed graphs. Leitch, F. Jonathan 1970

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FUNCTIONS WITH CLOSED GRAPHS • by F. JON LEITCH B.Sc. U n i v e r s i t y o f Guelph, 1968 THESIS SUBMITTED IN PARTIAL FULFILMENT THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the Department o f MATHEMATICS . We a c c e p t t h i s t h e s i s as c o n f o r m i n g t o the r e q u i r e d s t a n d a r d The U n i v e r s i t y o f B r i t i s h Columbia • March 1970 In presenting t h i s thesis i n p a r t i a l f u l f i l m e n t of the requirements for- an advanced degree at the University of B r i t i s h Columbia, I agree that the Library s h a l l make i t f r e e l y available for reference and study* I further agree that permission., for extensive copying of t h i s thesis f o r scholarly purposes may be granted by the Head of my Department or by his representatives. I t i s understood that copying or publication of t h i s thesis f o r f i n a n c i a l gain s h a l l not be allowed without ray written permission. Department of The Uni v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada Thesis Supervisor: J. V. Whittaker ABSTRACT This paper concerns i t s e l f mainly with those functions from one topological or metric.space to another that have closed graphs i n the product space. Their r e l a t i o n s h i p to closed, l o c a l l y closed, compact, continuous and subcontinuous functions i s studied i n order to determine the r e l a t i v e strength of the closed graph condition. The paper c o l l e c t s and i n some cases extends re s u l t s found i n papers by R. V. F u l l e r [ 2 ] , P. E. Long [7] P. Kostyrko ' a n d T. Shalat [ 4 ] , [5] and [ 6 ] . The main theorems deal with; l ) the characterization of continuous functions i n terms of subcontinuity and the closed graph property; 2) a proof that i f f has a closed graph then f i s the l i m i t of a sequence of continuous functions; and 3) a study of the operations under which the class of functions with closed graphs i s closed. I l l Table of Contents page Chapter 0: Notation and Def i n i t i o n s 1 Chapter 1: Introduction of closed graph 5 • property; open functions with closed- graphs Chapter 2: Subcontinuous functions; 10 r e l a t i o n s between functions with various properties Chapter 3: Characterization of continuous 17 functions; l i m i t theorem; statements about domain and range spaces Chapter 4: Points of i r r e g u l a r i t y ; l i m i t 29 numbers Chapter 5: Closed and l o c a l l y closed 35 functions; topologies determined by compact subsets Chapter 6: Analysis of the class of functions. 43 with closed graph; product spaces. i v Acknowledgements I should l i k e to thank Professor J.V. Whittaker for suggestion of the topic, his patient reading of the o r i g i n a l draft and for his hel p f u l c r i t i c i s m s . Also I would l i k e to acknowledge the f i n a n c i a l support of the National Research Council of Canada. CHAPTER 0 This chapter i s devoted to an explanation of the notation to be used and some d e f i n i t i o n s . D e f i n i t i o n s : , FILTER A class 5 of subsets of a nonempty set X i s a f i l t e r on X i f the following conditions are s a t i s f i e d : 1. i f A,B e 3 then ADB e 3 ' 2. i f A e and BsA then B e 5. 3. 0 k * A maximal f i l t e r on X i s c a l l e d an u l t r a f i l t e r . I f i s any f i l t e r on X and U. i s an u l t r a f i l t e r then either $<^lLor and IX. are not comparable. Maximal f i l t e r s always e x i s t since the class Q = {ACX|X q e A for a fi x e d X q e X} i s a p r i n c i p a l (fixed) u l t r a f i l t e r . I f X i s an i n f i n i t e set there e x i s t non-p r i n c i p a l (free) u l t r a f i l t e r s . TOPOLOGY A class 3" of subsets of a set X i s a topology for X i f the following conditions are s a t i s f i e d : 1. If A,B e then ADB e 3\ 00 2. i f A e ff n=l,2,... then u A e ff n n=i n 3. 0, X e if - 2 -The • s e t s b e l o n g i n g t o 3" .are c a l l e d the open s e t s o f X . I f . A e 3" t h e n the complement o f A i s s a i d t o be c l o s e d . To a v o i d p o s s i b l e c o n f u s i o n we p r e s e n t our d e f i n i t i o n of, the f o u r t o p o l o g i e s mentioned i n t h i s paper. FRECHET ( T 1 ) : HAUSDORFF.(T ) REGULAR: NORMAL: A t o p o l o g y 3" on a s e t X i s T^ i f and o n l y i f f o r each x 4 y, x,y e X, t h e r e e x i s t s M,N e 3" such t h a t x e M, y | M and y e N, x N . A t o p o l o g y J on a s e t X i s T^ i f and o n l y i f f o r each x =j= y, x3y e X, t h e r e e x i s t M,,N e 3" such t h a t x e M, y e N and MON = 0 A t o p o l o g y 31 on a s e t X i s r e g u l a r i f and o n l y i f f o r each x | A, where A i s a c l o s e d s u b s e t o f X , t h e r e e x i s t open M,N such t h a t ACN, x e M and MDN = 0 A t o p o l o g y 3" on a s e t X i s normal i f and o n l y i f f o r each p a i r o f c l o s e d s e t s A,BcX w i t h ADB =' $ t h e r e a r e open, d i s j o i n t s e t s M,N w i t h the p r o p e r t y t h a t .ACM and B<=N . NEIGHBOURHOOD N L e t x e X and l e t 3" be a t o p o l o g y on X . Then - 3 -N<=X i s a ^ - n e i g h b o u r h o o d o r s i m p l y a n e i g h b o u r h o o d o f x i f t h e r e i s a n 0 e J s u c h t h a t x e O5N D I R E C T E D S E T S AND NETS A b i n a r y r e l a t i o n R o n a s e t X i s c a l l e d a p r e o r d e r i f i t i s r e f l e x i v e a n d t r a n s i t i v e : i . e . 1. ( x , x ) e R f o r a l l x e X 2. i f ( x , y ) a n d ( y , z ) b e l o n g t o R t h e n ( x 3 z ) e R . A s e t X t o g e t h e r w i t h a d e f i n i t e p r e o r d e r i s c a l l e d a p r e o r d e r e d s e t . A d i r e c t e d s e t D i s a p r e o r d e r e d s e t w i t h t h e a d d i t i o n a l p r o p e r t y t h a t f o r e a c h a , b e D t h e r e e x i s t s a c e D s u c h t h a t a<c a n d b<c . A n e t i n a s p a c e X i s a m a p p i n g <p:D -» X o f some d i r e c t e d s e t D i n t o X . We w i l l d e n o t e n e t s e i t h e r b y { x o l a e A} o r , i f t h e d i r e c t e d s e t A n e e d n o t b e a m e n t i o n e d , , s i m p l y b y x Q . a A s u b n e t o f x w i l l be d e n o t e d b y x , - w h e r e a J Nb b i s a member o f t h e d o m a i n o f x ^ a n d N i s t h e a p p r o p r i a t e f u n c t i o n f r o m t h e d o m a i n o f x ^ t o t h e d o m a i n o f x a ' PRODUCT D I R E C T I O N L e t A a n d B be two o r d e r e d s e t s . I n A x B we d e f i n e t h e p r o d u c t d i r e c t i o n a s f o l l o w s ; (a-. ,b-.) _< ( a p , b ? ) - 4 -i n A x B i f and o n l y i f a 1<a i n A or a± = a g and b l — ^ 2 i n ^ * With "this d e f i n i t i o n A x B i s an ordered set, CONVERGENCE A net C x a l a e A} converges to a p o i n t x i n a space X i f and o n l y i f f o r each neighbourhood . N of x there e x i s t s an a„ e A such t h a t i f b>a.T , x, e N . N — N 3 b Throughout the remainder o f the paper we w i l l use the f o l l o w i n g n o t a t i o n : i f f i f and o n l y i f nbhd. neighbourhood X-A complement of A w i t h r e s p e c t to X A~ the c l o s u r e o f A G f the graph o f the f u n c t i o n f X x Y the C a r t e s i a n product of X w i t h Y x„ -• x the net x o converges to the p o i n t x a a Z + the p o s i t i v e i n t e g e r s f:X -» Y the f u n c t i o n f maps X i n t o Y 3 there e x i s t s such t h a t A' the d e r i v e d s e t of A , i . e . the s e t of a l l accumulation ( c l u s t e r ) p o i n t s o f A . - 5 -CHAPTER 1 Following Fuller [ 2 ] , we introduce the notion of a function having a closed graph in terms of l im sup (Ls) and lim inf (Li) of nets of sets. We also present a character-ization of open functions and give necessary and sufficient conditions for an open function to have a closed graph. Definition 1.1 A function f: X -* Y has a closed graph (relative to X x Y) i f f [(x, f (x)) |x e X} is closed in the product topology of X x Y . In terms of nets this becomes G f i s closed i f f (x , f(x )) -« (p,q) in X x Y implies that q = f(p) • The following Lemma w i l l be used a great deal in Chapter jj. Lemma 1.2 Let f :X -» Y be given. Then G f i s closed i f f for each x e X and y e Y , where y 4 f(x) , there exist open sets • U and V , containing x and y respectively, such' that f ( U ) n v = 0 . Proof: If the condition holds and (x,y) G^ then U x V is an open set in X x Y such that (U x V)nG^ = 0 which Implies G~ is closed. - 6 -I f G f i s c l o s e d and ( x , y ) G f , t h e n y ^ f ( x ) , so t h e r e e x i s t s a b a s i c o p en s e t o f t h e f o r m U x V where U a n d V a r e open^ c o n t a i n i n g x a n d y r e s p e c t i v e l y , a n d s u c h t h a t (U x V ) n G f = 0 . T h e r e f o r e f ( U ) H V = 0 . The t o p o l o g i c a l c o n v e r g e n c e o f a n e t o f s u b s e t s o f a t o p o l o g i c a l s p a c e X may be d e f i n e d i n t h e same manner a s th e t o p o l o g i c a l c o n v e r g e n c e o f a se q u e n c e o f s e t s . D e f i n i t i o n 1.3 I f C-An|n e D} i s a n e t o f s u b s e t s o f X , t h e n L i A ( L s An) i s d e f i n e d a s t h e s e t o f a l l x e X s u c h t h a t n v ' e v e r y nbhd o f x i n t e r s e c t s A n f o r a l m o s t a l l ( r e s p e c t i v e l y f o r a r b i t r a r i l y l a r g e ) n . We s a y t h a t a s t a t e m e n t S on e l e m e n t s o f a d i r e c t e d s e t D i s f u l f i l l e d f o r : 1. a l m o s t a l l n e D i f t h e r e i s an n e D s u c h o t h a t S i s f u l f i l l e d f o r a l l n>n — o 2 . a r b i t r a r i l y l a r g e n e D i f t h e s e t o f a l l n e D f o r w h i c h S i f f u l f i l l e d i s c o f i n a l i n . D . A n e t i s s a i d t o be t o p o l o g i c a l l y c o n v e r g e n t ( t o a s e t A) i f L i A = L s A (= A ) , i n w h i c h c a s e A = L i m A ' n n v '' n F o r f u r t h e r d e f i n i t i o n s a n d t h e o r e m s on n e t s o f s e t s see S. Mrowka [ 8 ] . The f o l l o w i n g lemma i s i n t e r e s t i n g i n i t s own r i g h t a n d i s n e e d e d i n t h e p r o o f o f t h e s u c c e e d i n g t h e o r e m . - 7 -Lemma 1.4 I f f:X -* Y i s a function, y ' i s a net i n Y , a, and p e Ls £~^(ya) > then there exists a net x ^ i n X with the property that x^ -* p and f ( x N t ) ) i s a subnet of Proof: Assume y i s a net i n Y and l e t p e Ls f ^(y-) J a. K a' Then f o r each nbhd U of p and each index a there i s an index N(a,U)>a such that f - 1 ( y ] \ r ( a T J ) ^ u ^ ^  • Direct the nbhds of p downwards by containment ( i . e . ^ o £ 2 • • ' ) > l e t the pairs (a,U) have the product d i r e c t i o n , and choose xN(a,U ) 6 f _ 1 ( % ( a , U ) ) f 1 U n > t h u s ° b t a i n i n S t h e n e t x N ( a , U ) i n X , Now we have . f ( x j j ( a TJ )) = yN(a U ) w i l i c t l l s a subnet of y a and x j j / a u ) ~* ^  s i n c e £°r any nbhd U of p , xN(b U ) 6 ^n P r o v i a e d only b>a and k>n . k ., Theorem 1.5 I f f:X -* Y i s a function the following are equivalent: a) f has a closed graph b) I f y a - q i n Y then Ls f _ 1 ( y a ) 5 f _ 1 ( q ) c) I f y - q i n Y then L i f - 1 ( y ) c f _ 1 ( q ) a a Proof: Assume • a) holds and l e t y 0 -• q i n Y . a Let p e Ls f ~ ^ ( y a ) • By Lemma 1.4 there i s a net x ^ i n X such that x^ -* p and f ( x N f e ) i s a subnet of y a . - 8 -Thus we have ( x N t l .> f ( x N b ^ "* ( p -> q ) a n d s i n c e G f i s c l o s e d Q = f ( p ) 5 i . e . p e f " " 1 ( q ) . T h e r e f o r e b) i s t r u e . I f b ) h o l d s t h e n c) h o l d s s i n c e f o r any n e t o f s e t s E we have L i E d L s E a a— a Assume t h a t c ) h o l d s and suppose (x f ( x )) -» ( p , q ) . a. a Then y = f ( x . ) - q and p e L i f~' 1"(y_) . S i n c e b y a s s u m p t i o n cl cl ci L i f _ 1 ( y a ) £ f _ 1 ( q ) we have p e f _ 1 ( q ) o r f ( p ) = q and hence a) h o l d s . D e f i n i t i o n 1 .6 I f a f u n c t i o n f : X -» Y i s s u c h t h a t U open i n X i m p l i e s f ( U ) i s open i n Y t h e n f i s s a i d t o be a n open f u n c t i o n . Theorem 1 .7 I f . f : X -* Y i s a f u n c t i o n , t h e f o l l o w i n g a r e e q u i v a l e n t : a) f i s open b ) y a - q i n Y i m p l i e s f _ 1 ( q ) c L i f _ 1 ( y a ) c ) y a - q i n Y , i m p l i e s f - 1 ( q ) 5 L s f 1 ( y a ) P r o o f : Assume a) h o l d s , l e t y q i n Y and l e t cl p e f _ 1 ( q ) . Suppose p d L i f _ 1 ( y ) . Then t h e r e e x i s t s a nbhd U o f p s u c h t h a t f r e q u e n t l y f _ 1 ( y )DU = Qf . S i n c e cl U must c o n t a i n an open s e t c o n t a i n i n g x we c a n t a k e U to. be o p e n . Now we have y i s f r e q u e n t l y o u t s i d e f ( U ) , b u t a s i n c e f ( U ) i s open a n d q e f ( U ) , f ( U ) must be a nbhd o f q and t h i s gives the desired contradiction. Thus b) holds. I f b) holds then c) holds since f - L ( q ) 5 L i f - 1 ( y a ) C L s f _ 1 ( y a ) . Assume c) holds, and suppose f i s not open. Then there i s an open UcX and a net y i n Y-f(U) such that — a y o -» q for some q e f(U) . By assumption f ~ 1 ( q ) c i J s f~ 1(y„) . Let p e f _ 1(q ) n u . Then U i s a nbhd of p and f _ 1 ( y a ) n u i s frequently nonempty since p e Ls f _ 1 ( y a ) • But t h i s means that y i s frequently i n f(U) a contradiction to the fac t that y i s a net i n Y-f(U) . Hence a) holds, a Using Theorems 1 .5 and 1 .7 and the d e f i n i t i o n of Lim E where E i s a net of sets we arri v e at the following a a theorem. Theorem 1 .8 The function f:X -» Y i s open and has closed graph i f f y a — q i n Y implies Lim f _ 1 ( y a ) = f _ 1 ( q ) Proof: By the theorems c i t e d above f i s open and has closed graph i f f y Q -• q i n Y implies • f _ 1 ( q ) e L I f - 1 ( y a ) C L s f - 1 ( y a ) c f _ 1 ( q ) Thus L i f _ 1 ( y a ) = Ls f " 1 ( y a ) = Lim f _ 1 ( y a ) = f _ 1 ( q ) and the theorem holds. - 1 0 -CHAPTER 2 .In t h i s chapter we introduce suboontinuous functions. We then study some rel a t i o n s between functions with closed graphs and closed, compact and subcontinuous functions. Throughout the chapter we w i l l denote by U(X,Y) the class of a l l functions f:X -» Y whose graphs, , are closed. The following theorem w i l l lead us to the d e f i n i t i o n of subcontinuity ( F u l l e r [ 2 ] ) Theorem 2 . 1 Let f e U(X,Y) and l e t £ x n l n e b e a sequence of points i n X such that . x x , where x e X . Also l e t { f ( x n ) | x n e X} be compact i n Y . Then f ( x n ) - f ( x ) • Proof: Suppose that f ( x n ) does not converge to f(x) . Then there exists a nbhd N of f(x) with the property that f ( x n _ ) 4 N for i = 1 , 2 , 3 . . . ( 1 ) . Since '£f(xn)} i s compact i n Y there i s a sequence x n j = 1 , 2 , ... f o r which (f(x )}°° converges to some q e Y . Since f e U(X,Y) and (x , f ( x n )) - we must have 1 J - 1 J q = f(x) and hence (f(x )}c° -* f(x) . Therefore H 3=1 - 11 -{f(x )}°° converges to f(x) also, but t h i s contradicts (1) i i - 1 Hence f(x ) - f(x) . D e f i n i t i o n 2.2 The function f:X -. Y i s said to be subcontinuous i f f x n p i n X implies there i s a subnet f ( x ^ ) of f ( x a ) which converges to a point q e Y . S i m i l a r l y a function f:X -» Y i s said to be inversley subcontinuous i f f f(x ) -• q i n Y implies there i s a subnet a Xj^k of x^ which converges to some point p in. X . The concept of a subcontinuous function i s thus a generalization-of a function whose range i s compact (cf Theorem 2.1). I t i s also a generalization of a continuous function. Theorem 2.3 I f f:X -» Y i s a function and each point x i n X has a nbhd U such that f(U) Is contained i n a compact subset of Y then f i s subcontinuous. Proof: Let x be a net i n X such that x -» x . a a Since U i s a nbhd of x there i s an a Q such that for a l l a>a Q , x Q e U . Then f ( x a ) e f(U) f o r a l l a>a Q . Since f(U) i s contained i n a compact subset of Y there i s a sub-net f ( x N b ) of f ( x a ) that must converge to some q i n Y . Hence f i s subcontinuous. - 12 -The following theorem w i l l show that i f the range i s embedded i n a completely regular space then a subcontinuous function i s very nearly such that i t preserves compactness. Theorem 2.4 Let f:X - Y be a subcontinuous function and l e t Y be completely regular. Then f o r each compact subset K of X , f(K) i s compact. Proof: Let KcX be compact and l e t {Z |a e A} be a net in f(K) . Let B be a uniformity for Y . (see Dugundji [1] pp 200, 201) . Direct. B downwards by containment (cf Lemma 1.4) and l e t A x.B have the product order. For each (a,b) e A x B choose y ^ a b ^ e b [ Z a ] n f ( K ) and X(a ^ e Knf~" L(y^ a • Since K i s compact a subnet X(Mc, Nc) °^ x(a,b) converges. Since, f i s subcontinuous a subnet y ( K d ^d) of y ^ a ^  converges to a point q i n Y . C l e a r l y q e ~fJKJ Now consider the net ZT,, which i s a subnet of Kd Z & . We w i l l show that Z^ d -» q . Let b e B . By d e f i n i t i o n there i s a symmetric b e B such that b^ o b""1" = b^ o b^crb . By the choice of y ^ a b ^ we have that ( y ^ K d L d ) * Z K d ) i s eventually i n b± . Since L d ) - q , (q, y ( K d ? L d ) ) i s eventually in-,, b 1 . Since b-^  o b£b we have (q, Z K d ) i s eventually i n b^ and hence -» q . Consequently f.(K) i s compact. - 13 -Assuming the range i s embedded i n a Hausdorff space, the next theorem states that i f f e U(X,Y) then f handles compact sets with less success than continuous functions but i t s inverse does Just as well as the inverse of a continuous function. Theorem 2 .5 Let f e U(X,Y). I f K i s a compact subset of X [Y] then f(K) t f _ 1 ( K ) ] i s closed. Proof: Let K be a compact subset of X and assume that f(KJ Is not closed. Then there i s a net y a i n f(.K) such that y Q - q and q e Y-f(K) . Pick X q e K (1 f _ 1(y ) . Since a a a K i s compact there i s a subnet x N b of x & that converges to some point x e' K . Thus we have ( xJJ DJ ^ ( x N b ^ ~* ( X j ( l ) * Since f e U(X,Y) , q = f(x) and hence q e f(K) , but this contradicts the choice of q . Therefore f(K) i s closed. The second proof i s completely analogous and w i l l not be done. D e f i n i t i o n 2.6 Let f:X - Y be a function. I f f [ f _ 1 ] takes compact subsets of X [Y] onto compact subsets of Y[X] then f i s said to be compact-preserving [compact]. - Ik -D e f i n i t i o n 2.7 I f f [ f - 1 ] takes closed sets of X [Y] onto closed sets of Y [X] then f i s said to be closed [continuous]. In the following theorem we characterize compact and compact preserving functions i n terms of inverse sub-continuity and subcontinuity respectively. Theorem 2 .8 Let f:X -» Y be a function. a) The function f i s compact i f f f | f _ 1 ( K ) : f ~ 1 ( K ) - K i s inversley subcontinuous for every compact KcY . b) The function f i s compact-preserving i f f f |K:K -» f(K) i s subcontinuous f o r a l l compact KcY . Proof: The proof of a) i s analogous to b) so only b) i s proved. Assume that KcX i s compact and' that f i s compact preserving. Let x be a net i n K such that x -* x , c ° a a x e K . Since f(K) i s also compact the net f(x ) must cl have a subnet f ( x N ^ ) that converges to some q e f(K) . Hence f|K:K -• f (K) ' i s subcontinuous. Let f be subcontinuous. Let. KcX be compact and l e t y be net i n f ( K ) . - Pick x e K n f _ 1 ( y ). Then x i a ~ s a net i n K and hence there i s a subnet x ™ of x . that NB a converges, to some x e K . Since f.|K;K - f(K) i s subcontinuous we must have a subnet - 15 -f ( x M b ) of f ( x N b ) = y N b converging to a point q e f(K) . Since xMNTD^ = M^Nb l s a S U D n e - k °^ y a w e n a v e that f(K) i s compact. Corollary 2.9 a) I f f:X -» Y i s inversely subcontinuous and f~"!"(K) i s closed for each compact KcY then • f i s compact. b) I f f:X Y i s subcont inuous and f(K) i s closed for each compact KcX then f i s compa preserving. Proof: Again only b) i s proved. In any case we need only note that (using the notation of Theorem 2.8) since f(K) i s closed, the l i m i t point q of the net y ^ ^ must belong to f(K) . Corollary 2.10 I f f:X - Y and f e U(X,Y) and f i s sub-continuous .[inversely subcontinuous] then f i s compact-preserving [compact] Proof: By Theorem 2.5, • f(K) KcX and hence by Corollary 2.9 i s closed for every compact f • i s compact-preserving. - 1 6 -Corollary 2 . 1 1 • Let f: X -• Y be a function. Then:. a) i f Y i s Hausdorff and f i s both continuous and inversely subcontinuous then f i s compact. b) I f X i s Hausdorff and f Is both closed and sub-continuous then f i s compact preserving. Proof: We prove only b). Since X i s Hausdorff every compact K<=X i s closed and hence f(K) i s closed for every compact KcX . By Corollary 2 . 9 3 f i s compact-preserving. - 17 -CHAPTER 3 In t h i s chapter, we give characterizations of con-tinuous and closed functions i n terms of the closed graph and subcontinuity properties. We then show that i f f i s a function with a closed graph then f i s the l i m i t of a sequence of continuous functions and point out the dependency of the closed graph property on the range space. The following two theorems w i l l show that the closed graph property complements the two subcontinuities i n i n t e r e s t -ing ways. Theorem 3 .1 I f f:.X -» Y i s a function then a s u f f i c i e n t condition for f to be. continuous i s that f have a closed graph and be subcontinuous. I f Y i s Hausdorff then the condition i s also necessary. Proof:. (S u f f i c i e n c y ) : Let x & be a net i n X such that x & -» x . Suppose that f ( x a ) does not converge to f(x) , i . e . f i s not continuous at x . Then f (x ) has a subnet, f(x„„ ) . v a' ' K Nb no subnet of which converges to f(x) . Since f i s sub-continuous some subnet ^ ( x ] y [ N C ) °^ ^^xNb^ converges to some point y e Y ... Hence we have (xmc» f ( x M N c ) ) - (x,y) and since G f i s closed y = f(x) . Therefore f( xM£j c) ~* f ( x ) which implies that the assumption was f a l s e . - 18 -(Necessity): I f f i s continuous then i t Is obviously sub-continuous. I f ( x a , f ( x a ) ) - (p,q) then x& - p and thus f ( x a ) -* f(p) since f i s continuous. I f Y i s Hausdorff. q = f(p) and hence f has a closed graph. The following example shows that Theorem 3.1 (Necessity) i s f a l s e i f Y i s only a space. Example 3.2 Let X = Y = Z +' and l e t both X and Y have the c o f i n i t e topology [ i . e . a set i s open i f f i t i s void or has f i n i t e complement]. Obviously the i d e n t i t y map i s continuous but i t does not have a closed graph, because the sequence {(n,n)|n e Z +] converges to every point (x,y) € X x Y . In every nbhd of a point (x,y) there i s an element of the base of the topology for X x Y of the form G^ x G^ where G-^G^ are open subsets of Z + and (x,y) e G^ x G^ . I f m = max [njn G x or n ^ Gg} then (k,k) e G 1 x G for a l l k X m . Corollary 3.3 Let the function f:X -• Y be su r j e c t i v e . I f U i s an u l t r a f i l t e r on X such that U x for some x e X then a s u f f i c i e n t condition for the image f i l t e r f(U) to converge to f(x) i s that G^ be closed and f be sub-continuous. I f Y i s Hausdorff the condition i s also necessa - 1 9 -Proof: ( S u f f i c i e n c y ) : By Theorem.3.1 f i s continuous and hence i f U - x then f(U) -» f(x) . (Necessity): As i n previous theorem. Corollary 3-i+ I f f:X - Y i s a function and Y i s compact and Hausdorff then f i s continuous i f f G~ i s closed. f Proof: Suppose i s closed. Since Y i s compact f i s subcontinuous. Hence by Theorem 3-l> f i s continuous. Let f be continuous and suppose there i s a net (xa> f ( x a ^ i n G f s u c h t h a t (xa> f ( x a ^ "* ( P ^ ) • T H E N x -» p and by continuity f ( x ) -* f(p) . Since Y i s cl cl Hausdorff q = f(p) and i s closed. The following example shows that the compactness condition on Y cannot be dropped. Example 3*5 Let X = (0,1] and Y = [ 0 , + » ) . Define f by - J l / x x e (0,») 1 10 x = 0 Then f i s c l e a r l y not continuous but G^ i s closed. The next Theorem from P.E. Long's paper [ 7 ] » gives a characterization of continuity i n terms of the closed graph - 20 -property and conditions on the domain and range spaces. Theorem 3 .6 . Let f:X -> Y be a function where Y i s countably compact and X i s f i r s t countable. I f G f i s closed then f • i s continuous. Proof: Suppose f i s not continuous. Then there i s an open VcY such that f _ 1 ( V ) i s not open i n X . Therefore f - 1 ( V ) contains a point x e X such that x i s a l i m i t point of X - f - 1 ( V ) . Since X i s f i r s t countable, there exists a sequence {xn} , with x^ e X - f~ 1(V) , that converges to x . In the countably compact space Y , the set ( f ( x n ) } has an accumulation point y | V . Then (x,y) ^ G^ , but (x,y) i s a l i m i t point of G^ since any open set i n X x Y containing (x,y) c l e a r l y contains points.of the form (x,f(x)) . This contradicts the assumption that G^ . i s closed, hence f i s continuous. The above example shows that t h i s theorem i s not true for Y l o c a l l y compact and X compact. Theorem 3 . 1 showed the r e s u l t of combining the closed graph.property with subcontinuity. I f instead we use inverse subcontinuity we might suspect that f _ 1 w i l l be continuous. The following theorem confirms t h i s f a c t . - 21 -Theorem 3-6 I f the function f : X -* Y has a closed graph and i s inversely" subcontinuous then f i s closed ( i . e . f - 1 . i s continuous). Proof: Let C be a closed subset of X and suppose that f(C) i s not closed. Then there exists a net x i n C and v a a point q e Y - f(C) such that f ( x a ) ~* Q • Since f is~ inversely subcontinuous there i s a subnet x,^ of x such J Nb a that X j ^ -* p and since C i s closed p e C . Hence we have (xNb> f ( x N b ) ) "* ( p> q) a n d q ^ f ( p ) s i n c e 0. e Y - f(C) and p e f(C) . This however, contradicts the assumption that G^ i s closed. D e f i n i t i o n 3 .7 • A function f : X -• Y belongs to Baire class a i f the set f - 1 ( G ) i s a Borel set of additive class A of Borel sets i n X f o r every open GcY . In p a r t i c u l a r , f e B 1 ( X 3 Y ) i f f _ 1-(G) i s an F c t subset of X for each open G5Y and f e B ( X 3 Y ) . i f f - 1 ( G ) i s open for each open GcY ; i . e . i f f i s continuous. The following theorem (from P. Kostyrko and T. Salat [ 5 ] ) shows that i f X and Y are metric spaces and f : X -» Y has a closed graph then f i s the l i m i t function of a sequence of continuous functions. - 2 2 -Theorem 3 . 8 Let (X,p) and (Y,a) be metric spaces where CO Y = U C with C compact n = 1 , 2 , 3 , . . . . Then . I l l II • U(X,Y)cB 1(X,Y) Proof:-Let f e U(X,Y) and l e t GcY be open. Define the set M = G fn(X x G) and note that M e P (X x Y) , i . e . M ^ CO i s an F set i n X x Y . Therefore M = U M where CT n=l n • . CO each M i s closed i n X x Y . Since Y = U C and each n=l C i s compact i t could be assumed that C ,, f o r n ^ n— n+1 n = l , 2 . , 3 j . . . • Let Z Q = (x *y ) e X x Y and l e t p be a p o s i t i v e number. Let S(x ,p) be a spherical nbhd of radius p -of X q . By assumption there i s an n Q € Z + such that y Q e c n > (and hence to C n for n>.n0) Define D n = C n +n-l f o r - n = 1>2>~5>-- > a n d % = S(x ,p) x D . o . ^ & Let M n p = MnnRp . Now each M i s closed (since both Mn and Rp are) and bounded (since R p i s ) i n X x Y and M n = U M n p . Let E N P = (x e x|2 y e Y • • (x,y) e M n p} and l e t E = { x e X | 3 y e Y - • (x,y) e M] . Then ro -1 E = U E ^ „ and E = f (G) . I f we can now show that n,p=l n p f o r any f i x e d n and p ,- E N P i s closed i n X , we w i l l be done. Let x e E ' [E'" i s the derived set of E 1 . np np np J - 2 > -Then there exists a net ( x a l a € A ) i n ^ np such that x -» x . From the d e f i n i t i o n of E this implies that a np ^ for each a e A there i s a y a e Y such that ( x a , y a ) e Now the elements y a l l belong to a compact D and hence there i s a subnet y N f e that converges to some point y e . Then we have ( x ^ , Y N F E ) - (x,y) e M n p since M n p i s closed. Hence x e E and therefore E i s closed set. Thus we np np have f - 1 ( G ) e F ( X ) , whence f e B 1 ( X A Y ) The following example, l i k e example 3-5 w i l l show CO that the condition that Y = U C where the C are n=l n compact cannot be dropped. The example a c t u a l l y i l l u s t r a t e s a function that has a closed graph but belongs to no Baire c l a s s . Example 3.8 Let X = [ 0 , 1 ] with the Euclidean metric and l e t Y be a set with c a r d i n a l i t y , the continuum, and the t r i v i a l metric. Let f :X -» Y be b i j e c t i v e . Let {(x ,y ) |a e A] a a • d be a net i n G> and assume that (x ,y ) -» (x ,y ). Then f v a ^ a ' v o30o' x o -» and y„ - y^. . Since Y has the t r i v i a l metric CL o a o y = y .for almost a l l a e A , i . e . there i s an a e A . a J o 3 o such that y = y for a l l a>a . From y = f ( x ) i t Ja Jo — o ^a K a' follows that yQ = f ( x & ) for a>.a0 a n d - since f i s i n j e c t i v e the net x "is stationary for a>a . From x -• x i t a — o a o follows that x„ = x for a>a . Hence (x ,y ) e G. and a o — o v o o' f therefore G^ i s closed. - 2h -Now l e t E be any non-Borel set. i n X and l e t G = f(E)cY . G i s open since in Y a l l sets are o p e n . Since f i s single valued E = f'^G) Hence f belongs to no Bore'l class. The previous theorem was l a t e r generalized as follows (Kostyrko [ 6 ] ) . Theorem 3.9 Let X be a normal topological space and l e t f e U(X,E^) where E^ = (-<», + «)• with the Euclidean metric, Then there i s a sequence of functions f n e B (X^E^)- , n = 1,2,3,... > such that |f n(x)|<n and lim f n ( x ) = f ( x ) for a l l x e X . Proof: Let P n = G f n(X x [-n,n]) n = 1,2,... . F n i s closed i n X x' E-^  and i t s projection X n to the set X i s also closed. ( X n = [x e X|3 y e (x,y) e F n ) . If X 4 0 then the function g = f|X i s continuous on X n °n > 1 n . n since i t has a closed graph (namely F^) and | g n( x)I<n ' f o r a l l x e X n . (cf Corollary J>.k). Since X i s a normal space there i s a continuous extension f of the function • r ft g n onto the whole space X such that |f (x)|<n for a l l x e X . I f ., X^ = 0 put f n ( x ) s 0 . The equality f(x) = lim f (x) f o r X e X follows n^co n - 2 5 -from the fac t that the sequence of sets f x n l n e z + ) i s CO increasing and X = U X „ I f x e X then there i s an n=l n . 4 -n e Z such that for a l l n>n , x e X and hence o — o 9 n f n ( x ) = f(x) . We now. consider the dependency of the closed graph condition on the space.in which the range Is embedded. Note that Theorem 3.1 and the following c o r o l l a r i e s require that the range space be Hausdorff. In general t h i s condition cannot be dropped, for i f i:X -» X i s the i d e n t i t y map on X , then G^ i s closed i f f X i s Hausdorff. Now l e t f:X -» Y be a function which has a closed graph.and i s not continuous. Let Y* be a compactification of Y . Then f :X - Y* is. subcontinuous since Y* i s compact I f f had a closed graph i t would be continuous (by Theorem 3. However continuity does.not depend upon the embedding space so that f:X -»-Y would be continuous, contrary to assumption. This shows that i f f e U(X,Y) but f £ B Q(X,Y) then f 4 U(X,Y*) . The following theorems (from P.E. Long's paper [7]) show that i f we know a function has a closed graph then we can make some statements about the domain and range spaces f . Theorem 3.1Q-, Let f:X -» Y be a function with G^ closed. Then f(X) i s T x . - 26 -Proof: Let y and w be d i s t i n c t points i n f(X) . Then there exists an x e X such that f(x) = w . Thus (x,y) | G^ , so. by Lemma 1.2 there e x i s t open sets U and V containing x and y respectively., such that f(U)nv = 0 Therefore w | V and Y i s T^ . Theorem 3-H Let f:X -* Y be any open surjection with G^ closed. Then Y i s T^ . Proof: Let y and w be d i s t i n c t points i n Y . Then there are d i s t i n c t points x and z i n X such that f(x) = y and f ( z ) = w . Since (x,w) ^ G^ and G^ i s clos there e x i s t open sets U and V containing x and w respectively, such that f(U)nV = 0 ; but f(U) i s open and contains y . Consequently Y i s T^ . Theorem 3.12 Let f:X - Y be i n j e c t i v e v/ith G f closed. Then • X i s T x . Proof: Let x and z be d i s t i n c t points i n X . Then f(x) ^ f ( z ) * so there e x i s t open sets U and V containing x and f(z.), respectively, such that f(U )nv = 0 . Thus z ^ U , implying X i s T, . - 27 -Theorem 3.13 I f f:X - Y i s b i j e c t i v e with G f closed, then both X and Y are T^ . Proof: Theorems 3.10 and 3.12 Theorem 3.14 Let f:X -» Y be i n f e c t i v e and subcontinuous with closed graph. Then X i s T^ . Proof: By Theorem 3.1 f i s continuous, hence i s an open surJection from f(X) onto X . Furthermore G^ i s isomorphic to G^-l, hence f - 1 has a closed graph i n Y x X . By Theorem 3.11, X i s T g . Theorem 3.15 Let f :X -» Y be a homeomorphism of X onto Y having G^ . closed. Then both X and Y are Tg . Proof: Theorems 3-H and 3.14 Theorem 3.16 Let. f:X -• Y be i n f e c t i v e , open, connected and have G^ closed. Then i f X i s l o c a l l y connected, X i s m 2 ° - 2 8 -Proof: Let x and z be d i s t i n c t points of X . Then f(x) ^ f ( z ) , so there exists an open connected set U containing x and an open set V containing f ( z ) such that f(U)nv = 0 . Since f(U) i s open, z | U . For otherwise UU{z] i s connected, so that f(Uu{z]) = f(U )u(f(z)} i s connected, since f i s connected. This i s an i m p o s s i b i l i t y . - 29 -CHAPTER 4 In t h i s chapter we review and extend some res u l t s obtained by Kostyrko and Salat i n [ 4 ] , Spaces X and Y. w i l l be Hausdorff unless otherwise stated. D e f i n i t i o n 4 . 1 Let f:X -» Y be a function. Define a e X to be a point of i r r e g u l a r i t y of F i f there exists a sequence x i n X with x -» a , x 4 a ; such that the sequence n. n 3 n ' ^ {f(x )} i s not compact i n Y „ We w i l l denote by the points of i r r e g u l a r i t y of f and by D^ the points of d i s c o n t i n u i t y of f . We note that for any,function f, N^cp^ • Theorem 4 . 2 I f f e U(X,Y) then N f = D f . Proof: Since for any function •N^cD^ we need only to show the reverse, i n c l u s i o n holds for f e U(X,Y). Assume on the contrary that D^ W^^ . . Let x e D^ be such that x | N^ . . Since x e D^ s x cannot be an i s o l a t e d point of X . Let (x |n e Z+} be a sequence of points of X such that x - x and x 4 x . Since x k N„ the sequence f f ( x )) n n . i ^ v n i s compact i n Y . From Theorem 2 . 1 i t follows that f ( x n ) - f ( x ) , which implies that x \ D f . Hence D fcN^ and the theorem Is proved. - 30 -Corollary 4.3 I f f e U(X 3Y) then f i s continuous i f f N f = 0 . Theorem .4.4 Let f e U(X,Y) and assume that Y i s l o c a l l y compact, i . e . that each point y e Y .has a nbhd whose closure i s compact i n Y , then D^ , i s closed. Proof: Since f e U(X,Y) we have D f = N f . Assume that 3\V i s not closed. Then there exists a net fx la e A} i n f . 1 a 1 • •* such that x -• x and x e X - N f , i . e . x i s a point of continuity of f . Since x e X - N. and x o -* x ' we have i a f(x ) - f(x) . Let K be a nbhd of f (x) such that K i s compact. By continuity there i s a nbhd V of x such that f (V)cKcK . Since x & -» x there i s an a Q e A such that for a l l a>a •, x & e V and hence f (*a) e f(V) . Since x e N„ for a l l a e A then for each a i a e A there i s a sequence {x a n.|n e Z +] of points i n X such that x -» x and {f(x „)|n e Z + l I s not compact . a, n a a, n i n Y . i . e . there i s no subsequence of {f(x „)|n e Z +} that converges i n Y . - ( l ) Now V i s a nbhd of x for a l l a>a . and since a — o x -» x for a l l a e A , therefore for a>a there exists a,n a . — o an n e Z + such that for a l l n>n , x„ e V . Hence for each a — a J a,n a>a the subsequence (f(x )ln>n , n e Z } i s a subset — o ^ ^ .x a.n' 1 — a * •* - 31 -of f(V)dK . Therefore f o r each a>. a 0 the sequence-f f ( x )ln>n . n e Z +} has a subsequence f f ( x ) | i e Z +] ^ v a,n / 1 — a 3 J ^ . v a,n^' 1 J that converges to a p o i n t i n K . But £f( x a n . ) l ^ - e i s a subsequence of £^( x a r , ) l n e Z +} a n ^ hence ( l ) has been c o n t r a d i c t e d . Therefore ' = . i s c l o s e d . In the f o l l o w i n g theorems Y w i l l be the r e a l l i n e Y = E^ = ( + co) w i t h the Eu c l i d e a n m e t r i c . D e f i n i t i o n 4 . 5 I f f:X -•. E-^  i s a f u n c t i o n then f i s unbounded at a p o i n t x e X i f there i s a nbhd N of x i n which i t o o i s impossible to f i n d a s i n g l e r e a l M>0 such t h a t |f(x)|<M f o r a l l x e N . We note that i n the case t h a t Y = E-^  , po i n t s of i r r e g u l a r i t y of f and p o i n t s of unboundness c o i n c i d e , D e f i n i t i o n 4 . 6 The subset D .of E^ 4s s a i d to be of the f i r s t CO category i f D = u D where each D i s nowhere dense n = i •'' M i n E^ ( i . e . the clo s u r e of D^ contains no nonempty i n t e r v a l ) . I f D i s not of the f i r s t category we say D i s of the second category. Theorem 4 . 7 Let f e U(X,E 1) and assume that each nbhd K of x e X i s a second category s e t . Then D^  i s nowhere dense i n X . - 32 -Proof: Since f e U(X,E^) we have D f = . By Theorem 4 . 4 , i s closed and hence 0^ = X - 'is open x ° f - i ° x e X and 3 M>0 and a nbhd K of x o o • 3- x e K , |f (x) | <M j I t i s s u f f i c i e n t to show that 0 ,^ i s dense. Assume on the contrary that 0 f i s not dense i n X . Then there exists an x e X and a nbhd K of x such that CpfiK = $ . o o f Hence there does not e x i s t a single r e a l M>0 such that |f(x)|<_M for a l l x e K [although i t may well be true for some x e K . ] . Let T n = [x e K| |f(x)|<n} , n = 1 ,2 ,3 • CO K = U T and i t i s impossible that a l l the T are nowhere -i n L • n n=l dense i n K since then K would be a f i r s t category set, contrary to assumption. Hence for some n, T i s dense i n some open K^cK . Since f i s unbounded at x 3 a z e K ' ^ — o o such that |f(z )I>n . Since T i s dense i n there 1 N o 1 n i s a sequence fx, }m such that x, e T and x, -» z k-1 ° Since |f(x^)|<n for k = 1 , 2 , . . . there exists a sequence of indices [k } m such that {f(x, )} -*v and . s=l S |v|<n . Consequently we'have (x^ , f ( x ^ )) e G f and . s s (x f c , f(x )) - (z ,v) i G f since' |f(z ) |>n and |v|_<n.. S s This contradiction proves, the theorem. Corollary 4.8 Let I be a ( f inite or inf ini te) interval on E^ . Then for any f e U(I,E^) then the set i s nowhere dense in I . Definition 4.9 Let f be a real valued function determined on a par t ia l ly ordered set (X5<) and let x e X . We w i l l c a l l t a l imit of f at x on the right [left] i f there exists a sequence [ x n l n € Z } of points in X such that x n > x [x <x] for a l l n ; x -» x : and t = lim f(x ) . n • ' n ' v n' n-<» Denote by L [L ] the set of l imit numbers of f at a point x on the right [left] and let L = L UL be the set of a l l l imi t numbers of f at x . Theorem 4.10 Let f e .U(XjE^) and pick x e D^ , then i) L n = } 4 0 and i i ) L n [ ( - . , « ) - f ( x ) ] = 0 . A x Proof: i) This follows from Theorem 4.2 and the fact that in this case N f = [x e X | f (x) | i s unbounded] . i l ) Assume the contrary and le t t e L D [ ( - « , « > ) - f ( x ) ]. Then t e L and - f ^ f(x) . • Evidently there is a sequence [x In s Z"1"} in X such that x -» x and f(x ) - t . Now n J n v n we have (x , f(x )) e G„ and (x ., f(x ) - (x.t) k G_. v n* v n ' ' . f v n ^ n y v 5 * f - 34 _ This contradicts the fa c t that f e V(X3E^) 3 hence the theorem, i s proved. \ CHAPTER 5 This chapter discusses conditions for closed and l o c a l l y closed functions to have closed graphs.. We also look at spaces whose topologies are determined i n some measure by thei r compact subsets. Having already discovered conditions under which continuous and open functions have closed graphs we proceed to f i n d such conditions for a closed function. Example 5 . 1 Let X = ( 0 , 1 ] c.E 1', Y = { 0 , 1 } and.let f be the ch a r a c t e r i s t i c function of X mapping E^ into Y . Nov/ f i s a closed function but f does not have a closed graph. We note that f does not have closed point inverses. Following F u l l e r [ 2 ] we now define a l o c a l l y closed function. D e f i n i t i o n 5 . 2 . • , c We c a l l a function f:X-»Y l o c a l l y closed i f for every nbhd. U of each x <= X there i s a nbhd V of x such that V c U and f ( V ) . i s closed i n Y . I f the domain of a closed function f i s regular then f i s l o c a l l y closed. Also i f f:X-*Y i s such that X i s l o c a l l y compact and regular and f takes compact sets onto closed sets then .f i s l o c a l l y closed. - 36 -Example 5.3 Let X be the reals with the discrete, topology and Y the reals with the usual (Euclidean) topology, and l e t i:X-»Y be the i d e n t i t y . Then i i s l o c a l l y closed and i s i n f a c t continuous, but i t i s not closed. Theorem 5.^ I f f:X-*Y i s a l o c a l l y closed function then y a -+ q i n Y implies that Ls f "*"(y )cf~'I"(q) [cf. Theorems 1.5 and 1.7] Proof: Let y a - q i n Y and p e Ls f _ " L ( y a ) . Assume _____ X such that x M h-» p and f (x w > i) =. y ^ i s a subnet of that p ^ f (q) . By Lemma 1.4 there exists a net x ^ i n y a . Now X - f 1 ( c l ) i s a nbhd of p and since f i s l o c a l l y closed there i s a nbhd U of p with the properties: i ) UcX - f - 1 ( q ) and i i ) f (U) i s closed i n Y . Since xNb ~* p xNb i s e v e n i ; u a ^ y ^ a n ( ^ hence f ( x ^ ) i s eventually i n f(U) . This means that f ( x ^ ) i s eventually i n the complement of a nbhd of q , namely Y-f(U) . This contradicts the f a c t that y,„ = f(x, T, ) -» q . Corollary 5*5 If~„ f:X -» Y i s a l o c a l l y closed function and has closed point inverses then f has a closed graph. - 37 -Proof: By assumption f _ 1 ( q ) = f~"I"(q) for every q e Y . I f y -» q i n Y then L s f - 1 ( y ) c f _ 1 ( q ) which by Theorem 1.5 implies that f e U(X,Y). Corollary 5.6 I f the function f:X -» Y i s closed with closed point inverses and X Is regular then f has a closed graph. Corollary 5.7 I f the function f:X -» Y i s closed and subcontinuous with closed point inverses and X i s regular, then f i s continuous. Proof: Apply Corollary 5 .6 then Theorem 3.1 Corollary 5 .8 I f the function f:X Y i s l o c a l l y closed and inversley'subcontinuous with closed point inverses then f i s closed. Proof: Apply Corollary 5 .5 then Theorem 3 .4 Theorem 5 .9 ^ I f '.f:X-» Y i s a function where X i s regular - 38 -and l o c a l l y compact, the following are equivalent. a) f maps compact sets onto closed sets and has closed point inverses. b) f i s l o c a l l y closed and has closed point inverses c) f has a closed graph. Proof: We have already commented that (a).implies (b) and by Corollary 5>5s (t>) implies ( c ) . Assume then that G f i s closed. By Theorem 2.5 f maps compact sets onto closed sets. Also since points are compact, the same theorem yie l d s that f has closed point inverses. We now define some topological spaces which are, i n a sense, determined by t h e i r compact sets, and prove two theorems concerning these .spaces and compact preserving functions. •. . Defin i t i o n . 5 . 1 0 Let X be a topological space and p. e X . a) X i s said to have property k^ at point p i f f for each i n f i n i t e AcX having p as an accumulation point., there i s a compact subset K of AU{p] such that, p e K and p i s an accumulation point of K . b) X has property k^ at point p i f f for each set A having p as an accumulation point, there i s a - 39 -subset B of A and a compact K_>BU{p] such that p i s an accumulation point of B . c) X has property k^ at point p i f f U i s an- open set i n X p r e c i s e l y whenever UDK i s open i n K for each compact set K i n X . X i s said to be. a k^ space i f X has property k^ at each of i t s points. We have as rel a t i o n s between the k^ spaces that k-^  implies kg implies k_. . Theorem 5*11 Let f:X -» Y be a compact function and l e t Y be a k-, space. A s u f f i c i e n t condition that f be closed i s 3 that G f be closed. I f X i s regular, T^ the condition i s also necessary. Proof: . (Suf f i c i e n c y ) : Let C be a closed subset of X . To show that f ( ' C ) i s closed we need to prove that f ( C)HK i s closed for each compact Key . Let KcY be compact. Then f~"1"(K) i s compact and hence so i s cnf _ 1(K). By Theorem 2,5* f[Cnf _ 1(K)] i s closed i n Y . Since f[Cnf _ 1(K)] = f(.C)DK. we have that f ( C ) D K i s closed i n Y and hence i n K for every compact KcY . Hence f i s closed. - 40 -(Necessity): Since X i s and f i s compact, point inverses are closed. Thus by Corollary 5«6, f has a closed graph.. Using techniques similar to the predeeding theorem • we can arrive at another characterization of continuity. Theorem 5.12 Let f:X -» Y be an i n f e c t i v e , compact-preserving function and X a space. A s u f f i c i e n t condition that f be continuous i s that Q be closed. I f Y i s Tg then the condition'is also necessary. Proof: ( S u f f i c i e n c y ) : Let f e U(X,Y) and l e t CcY be closed. To show that . f~"*"(C) i s closed i t i s necessary to show that f - 1 ( C ) r i K Is closed for each compact KcX . L e t KcX be compact, then so i s f(K) . f (.Cnf(K)) i s closed by Theorem 2.5. Since f i s i n f e c t i v e f - 1 [ f ( K ) ] = K so f - 1 ( c n f ( K ) ) = f _ 1 ( C ) n K i s closed i n X and hence i n K for each compact KcX . Hence f _ 1 i s closed and therefore f i s continuous. (Necessity): Let f be continuous and Y be Tg . Points are closed i n Y and since f-"'" i s closed f has closed point inverses.. Also f maps compact sets onto closed sets. Therefore by Theorem 5.9 C^ i s closed. - 41 -Example 5.13 We w i l l now construct a function which i s compact and has closed graph but i s not closed. Let X be an uncountable set. Let C-^  be the cocountable topology for X ; C-^  = {AcX|X-A i s countable] \J0 . Let Cg be the discrete topology for X . The only compact sets i n either (X, C-jO or (X,Cg) are the f i n i t e subsets of X . That t h i s i s true for (X,Cg) i s obvious, to see that i t i s true for (X,C-^) l e t A. = {a-^ag,. .. } be a countable subset of X . Take as an open cover of A the c o l l e c t i o n [ M n | M n = (X-A)U{an}} for n = 1,2,3,... - X - M n = X-[(X-A)u { a n ) ] - Afl(X-{a n}) = A-{a n) , so the Mn are legitimate open sets whose union contains A . C l e a r l y no f i n i t e subcollection pf the M w i l l cover A , so A i s not compact. . Consider the i d e n t i t y map. i:(X,C 2) -» -(X,Cj) . The function i has closed graph, for l e t ( x a? 1 ( x a ) ) ~* ( x,y) • Then x & must eventually be constantly x i n (X,C^) and so must i ( x ) = x o . I f i ( x ) converged to y ^ x then we would have a contradiction to the f a c t that points are closed In (X,C-^) • The inverse of I carr i e s compact ( f i n i t e ) sets of (X,^) onto compact ( f i n i t e ) sets of (X,Cg) and thus i i s compact. F i n a l l y since C^pC-, , i i s not closed. ( - k2 -Theorem.5.14 Let f :-X -» Y be a function that i s compact preserving and has closed point inverses. Let X and Y be' T 2 and l e t X have property kg at a point p . Then f i s continuous at p . Proof: Suppose f i s not continuous at p . Then there i s a nbhd V of f(p) such that for each nbhd U of p there i s a point x u e-U such that f ( x u ) k v . The c o l l e c t i o n A = C x u l u i s a nbhd of p] has p as an accumulation point. Thus A has a subset B , for which there i s . a compact K3BLl{p} . A l s o p i s an accumulation point of B . Consider the function f JK:K -* Y . f |K i s closed since Y i s Tg . As ( f | K ) " 1 ( y ) = f _ 1 ( y ) n K f o r a l l y e Y , f|K has closed point inverses. F i n a l l y , since K is.compact, Tg i t i s also regular and hence by Corollary 5.7 f|K i s continuous. However i f we pick a net x i n BcK such that • a — x a -* p then f ( x a ) i s never i n the nbhd V of f(p) . This contradiction proves the theorem. - kj> -CHAPTER 6 This chapter i s devoted to an analysis of the class U(X,Y) . We w i l l follow Kostyro and Shalat i n our e f f o r t to determine under what operations U(X,Y) i s closed. Composition The following example w i l l show that i f f , g e U(X,Y) then i t does not follow that fog e U(X,Y) . Example 6.1 Let X = [ 0 , 1 ] , Y - both with the Euclidean metric. Let f(x) = { 1 / x, * f [° > 1 ] and l e t g(y) '= { ^ g J f °Q Then g i s continuous on f(X) and f has a closed 0 does not have a closed graph In XxY . graph. However g[f(x)] = {^/2X ^ ° We now give some conditions for the closedness of graphs of a composition of functions. Theorem 6.2 Let X,Y,Z be topological spaces. Let f:X -» Z be continuous and l e t g:y -» Z have a closed graph. Then g.f e U(X,Y) . Proof: Let G^ „ be the graph of g°f and l e t (x ,z ) be a net In G „ such that (x , z ) -» (x .z ) . Then g o t v a - a / v o 5 o' - 44 --> X q and by the continuity c o n t i t i o n , ^ ( x a ) ~* ^ (-^o^ ' Hence (f(x ),z ) e G . Hence z = p;(f(x )) and therefore K K o o g o <=*\ \ o' ' G ,. i s closed. cr o i Theorem 6 .3 Let X.Y,Z be topological spaces such that f(X) i s compact and Z i s ' Tg . Let f e U(X,Y) and l e t g:Y -• Z be continuous, then g°f e U(X,Y) Proof: Since f(X) i s compact, f i s subcontinuous. Hence by Theorem 3 .1 that f i s continuous. Since Z, i s T^ the same , theorem y i e l d s that g has a closed -graph. Hence by Theorem 6 .2 g°f e U ( X , Y ) . . . Addition and M u l t i p l i c a t i o n We s h a l l show that, i n general, U(X,Y) i s not closed with respect to either addition or m u l t i p l i c a t i o n . . Example 6 .4 . Let X = [0,1] and Y = , both with the Euclidean metric. Define functions f:X -> Y and g:X -. Y a s f o l l o w s : f(x) - * t ° -g(x) - { " ^ * t % Then f and g both have closed graphs but (f+g)(x) = ^ - 0 does not have a closed graph. Since i f x e X , x -* 0 , a a x 4 0 for any a then . (f+g)(x ).= 0 for every a and hence (x a,(f+g)(x a)) - ( 0 , 0 ) f G f + g . . - 45 -Example 6 .5 Let X = [ 0 , 1 ] and Y = (-«>,+») and define f and g as follows: f(x) = { J / x * t °Q > g<x) = X for a l l x e X . Then f(x)-g(x) = | J * = 0 xsrhich does' not have a closed graph i n XxY . Although f(x)«g(x) need not belong to U(X,Y) for f,g e U(X,Y) i t i s true that U(X,Y) i s closed with respect to m u l t i p l i c a t i o n by a constant i f Y = (-«>,+<») . Theorem 6 .6 I f f e U(X 3E 1) and c e E± then cf e U(X,E-L) Proof: For c = 0 the theorem i s true, so assume that c e E 1 , c 4 0 and f e U(X,Y) . Suppose that cf k U(X,E 1) . Then there i s a net (x ,y ) e G „ (where y = cf(x )) such that (x ,y ) -* (x ,y ) I G » . i . e . y ^ cf(x ) and hence v a 5 J a / ^ o , J o Y cf * o. o' f(x ) ^  y / . But we then have that ( x , f ( x )) -» (x ,y y ) A v o ^o/c v a* v a'' K o i J o / c ' Y which contradicts the fac t that G^ i s closed. Hence c-f e U(X,E X) . Maximum- and Minimum I f f,g e e(X,Y) i t does not follow that either max (f,g) or min(f,g) need belong to U(X,Y) . Example 6 .7 Let X = [ 0 , 1 ] and Y = (-»,+») . Define f and g - 46 -as follows: f(x) = ( 1 / / x X f o°' 1"' a n d s(x) = 0 f o r a 1 1 t - 1 x - 0 x e X . Then min (f,g) = - [ ^ ° £ t ^Q'1^ which does not . have a closed graph i n XxY . Theorem 6 . 8 I f f,g e U(XJ,E1) and both f and' g are subcontinuous then max (f,g) and min (f,g) belong to U(X 5E 1) . . Proof: By Theorem J>.1 both f and g are continuous. Hence both max (f,g)- and min (f,g) are continuous. Since E^ i s Hausdorff then by Theorem J . l we have that both max (f,g) and min (f,g) belong to U(X aE 1) . Absolute Value • Theorem 6 . 9 I f f:X - E 1 and f e U(X,E-L) then |f | e U(X 3E 1) . Proof: Let f e U(X,E ) and l e t (x ,y ) be a net i n i. a a G | F | such that ( x & , y a ) ( x Q , y o ) . I f f ( x & ) > 0 for i n f i n i t e l y many a e A , i . e . for a = a^ , •k = 1 , 2 , 3 , . . . , then ' ya = ' f ( X a )!' = f-( xa ) ~ f ( x J = = ' l f ( x J I a n d h e n c e  a k a k a k ° ° ° ( x Q , y o ) e G| fj . I f on the other hand f ( x a ) < 0 f o r i n f i n i t e l y . - 4 7 -many a e A then y - |f(x ) | = - f ( x - - f(x ) = -y k k k = | f ( x Q ) | and again (x.Q,yQ) e G j f j . Therefore |f| e U(X,E 1) Convergence Pointwise The following example w i l l show that U(X,Y) need not be closed with respect to pointwise convergence. Example 6.10 Let X = [0,1] and Y = E-^  and define f n ( x ) = x n for n = 1,2,33... and x e X . Then f e U(X,E-L) for a l l n . However f(x) = lim f (x) does not belong to U(X,E^) n-*t» since f(x) = .{$> * f [°>^ Almost Uniform A sequence of functions { f " n l n € ^ } converges almost uniformly to f on a space X i f [f |n e Z +] converges uniformly to f on a l l compact subsets of X . Theorem 6.11 If f n e U(X,Y) , n ~ l , 2 , 3 , . . . , where (x,p) and (Y,a) are. metric spaces and f -» f almost uniformly on X then lim f = f e ,U'(X,Y) . n-»ro Proof: Assume that f f In e Z +] converges to f almost n 2 D uniformly on X and that f e U(X,Y) . Let % be the metric - 48 -f o r X XY , then # 2 = p 2 + a 2 . Since f.tt U(X,Y) there e x i s t elements ( x n ^ y n ) ^ n &f such, that ..(xn,y ) -» ( x 0 , y o ) 4 . Then there e x i s t p o s i t i v e numbers '&1 and 6 g such that S ( ( x Q . y o ) , ' 6 1 ) n S ( ( x Q . f ( x Q ) ) , 62) = 0 1 As . f (x ') _> f (x ) there e x i s t s a k, such t h a t f o r a l l ' n v o' x o' 1 k >_ k 1 , a ( f k ( x o ) , f ( x 0 ) ) < 6 2 a n d h e n c e f o r a 1 1 k 2_ k i ' • 6 ( ( x 0 . f ( x 0 ) ) , ( x 0 , f k ( x 0 ) ) ) = ^ a 2 ( f ( x 0 ) / , f k ( x 0 ) ) < S 2 . Therefore ( x o , f k ( x Q ) ) e S ( ( x o , f k ( x o ) ) , 6 2) f o r a l l k > ^  . 2 Since f -» f almost u n i f o r m l y on' X , then K' = [ x o , x - j X g j . . . ] i s a compact subset of X . Then f o r every i n t e g e r k there i s an n^ (n1<n2<n^<...) such that CT(f (x )..f(x )) < 1/k f o r p = 0,1,2,... . k y L I t f o l l o w s from the above th a t I f . p -> co .then >(( ( x o , y o ) , ( x p , f ( x Q ) ) ) < 1/k and hence there Is a k ' such that f o r a l l k>k„ (x , f (x )) 2 — 2 ^ o' n f c v o 1 ' e S ( ( x o , y o ) , o i ) . Hence from 1 (x -f (x )) 4 S ( ( x , f ( x Q ) ),'p 6 k • f o r k_>kg which c o n t r a d i c t s 2 . The f o l l o w i n g v / i l l show that the converse of the preceeding theorem Is not n e c e s s a r i l y true. Example 6.12 Let X = [0,1] and define f = -f 1 / = 0 n L x - 1 / n x e (0,1] f o r n - 1,2,3,... . Then l i m f (x) = f ( x ) = 1 f o r n-+<= a l l x e X . Now f e U(X,E 1) and f e U(X,E 1) f o r a l l n-_ 49 -But since sup I x ^ ^ - l | = +» i t follows that convergence i s xeX • . not almost uniform. • In what follows i f [X |a e A), i s any system of topological spaces, then X X will-, be the Cartesian product GeA a of the sets X and TT X w i l l be the topological product a . aeA a supplied with the product topology. The following i s a well known lemma for which no proof i s given. :: . 6 . 1 3 Let [X |a e A] be a system of topological spaces and. set X = TT x • The net {x b|b e B] of points i n . X aeA a where x D = i x a | a e A} converges to x = ( x a l a e A] i f f for each a e A the net [x^ jb e B, x^ e X&} converges to x & . Theorem 6.14 Let '(X Q|a e A] be a system of topological spaces and on a space. X define functions f :X -» X for each ^ o a o a a e X a . Let G f • = {(x,f (x))|x e X } be closed sets i n a .TT{X |s = 0,a ; a e A] . Then the function f:X -» X X s ° aeA a defined by f(x) = ( f ^ ( x ) , f 2 ( x ) , . . . } has a graph G f that i s closed i n X x(X X ) . aeA Proof": We s h a l l show that G^ cz G f . Let x e G^ , x = (x .fx la e A]) for some x e X '. There i s a net' v o <• a 1 ' ' o o - 50 -{x b|b e B, x b = (xj,{xj|a'e A})} of points i n G f - {x} such that x b -* x . Note that x° = f (x b) for a l l a e A a a.v o y By the lemma 6.13 a x b x for a l l a e A . Hence a a b b f (x ) •* x for a l l a e A . Also x -• x . Therefore av o . a • o o ( x b , f f ( x b ) | a e A}) - (x J x la e A}) = (x , f f (x )|a e A}) . ^ o ^ a ^ o - 1 , v o b a J x o ^ a o Since G„ i s closed for a l l a, (x , f f (x )|a e A}) f P V o ^ a v o J a = (x ,f(x ) ) . Hence G„ i s closed. v o v o I Theorem 6.15 Let [X la e A] be a system of topological spaces. Let X be a space and for each a e A define f :X -» X o ^ a o a "co be subcontinuous. Define f:X -» X X as m Theorem 6.14. o A a aeA Then f i s subcontinuous. Proof: Suppose f Is not subcontinuous. This implies that there i s a net fx b|b e B] i n x such that fx b1 -» x e X o 1 o L o J o o but ..o subnet of {f (x ) |b e B] converges i n x X . Since ° b b , a e A a for any fix e d b e B 3 f ( x Q ) = Cfa( x 0)la e A] thi s means that for at lea s t one a e A , f f (x b)|b e B] has no convergent a o subnet i n x a (by Lemma 6.13) . But thi s contradicts the fact that f i s subcontinuous for each a e A . a Combining Theorems 3»1* 6.14, and 6.15 we see that If f i s defined as i n the two theorems above then f i s continuous. - 51 -BIBLIOGRAPHY 1. Dugundj'i, J . , T o p o l o g y , A l l y n and Bacon I n c . , B o s t o n , .1966. 2. F u l l e r , R.V., R e l a t i o n s Among C o n t i n u o u s and V a r i o u s Non~Contifiuous F u n c t i o n s , P a c i f i c J o u r n a l of M a t h e m a t i c s , V o l . 25, No. 3 , (1968) , PP 495-509. 3 . . H u s a i n , T.,- Al m o s t C o n t i n u o u s Mapping, R o c z n i k i -P o l s k i e g o Towarzystwa Matematycznego, X, (1966), pp 1-7. 4. K o s t y r k o , P. and S a l a t , T., o &VHK.UHax., r P A $ b ! KQToPblX C a s o p i s Math., V o l . 89, (1964) , pp 426-4J1 . i 5 • Ci ^ V H K U H f l X . P P A $ b l ^OTOPblX flBTTjaiOCfl .^AMK.HYh M H O ^ E C T B A M M JL ? , A c t a f a c . r e r . n a t u r . U n i v . Comenian, Math 10, No. 3 , (1965), PP 51-61}. 6. K o s t y r k o , P., A Note On F u n c t i o n s ¥ith C l o s e d Graphs, C a s o p i s Math., V o l . 9"4, (1969) , PP 202-205. 7. Long, P.E. , F u n c t i o n s W i t h C l o s e d Graphs, Mat. Montbliy, .Oct. 1969, PP 9~30-932. 8. Mrowka, S., On The Convergence o f Nets o f S e t s , Fund Math. 45 (T958) , PP 237-2T6". : 

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