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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Functions with closed graphs. Leitch, F. Jonathan 1970
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Title | Functions with closed graphs. |
Creator |
Leitch, F. Jonathan |
Publisher | University of British Columbia |
Date Issued | 1970 |
Description | This paper concerns itself mainly with those functions from one topological or metric space to another that have closed graphs in the product space. Their relationship to closed, locally closed, compact, continuous and subcontinuous functions is studied in order to determine the relative strength of the closed graph condition. The paper collects and in some cases extends results found in papers by R. V. Fuller [2], P. E. Long [7] P. Kostyrko and T. Shalat [4], [5] and [6]. The main theorems deal with; 1) the characterization of continuous functions in terms of subcontinuity and the closed graph property; 2) a proof that if f has a closed graph then f is the limit of a sequence of continuous functions; and 3) a study of the operations under which the class of functions with closed graphs is closed. |
Subject |
Functions |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-06-14 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080505 |
URI | http://hdl.handle.net/2429/35407 |
Degree |
Master of Arts - MA |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
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UBCV |
Scholarly Level | Graduate |
AggregatedSourceRepository | DSpace |
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