FUNCTIONS WITH CLOSED GRAPHS • by F. JON L E I T C H B.Sc. University of Guelph, 1968 THESIS SUBMITTED I N P A R T I A L FULFILMENT THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the Department of MATHEMATICS . We a c c e p t t h i s to the r e q u i r e d The University t h e s i s as c o n f o r m i n g standard of B r i t i s h • M a r c h 1970 Columbia In presenting t h i s t h e s i s 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 U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study* I f u r t h e r agree that permission., f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s representatives. I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without ray w r i t t e n permission. Department of The U n i 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 T h i s paper concerns i t s e l f mainly w i t h those f u n c t i o n s from one t o p o l o g i c a l or metric.space to another t h a t have c l o s e d graphs i n the product space. T h e i r r e l a t i o n s h i p to c l o s e d , l o c a l l y c l o s e d , compact, continuous and subcontinuous f u n c t i o n s i s s t u d i e d i n order t o determine of the c l o s e d graph c o n d i t i o n . the r e l a t i v e s t r e n g t h The paper c o l l e c t s and i n some cases extends r e s u l t s found i n papers by R. V. F u l l e r [ 2 ] , P. E. Long [7] P. K o s t y r k o ' a n d The main theorems d e a l w i t h ; l ) T. S h a l a t [ 4 ] , [5] and [ 6 ] . the c h a r a c t e r i z a t i o n of continuous f u n c t i o n s i n terms o f s u b c o n t i n u i t y and the c l o s e d p r o p e r t y ; 2) f 3) a proof t h a t i f f graph has a c l o s e d graph then i s the l i m i t o f a sequence of continuous f u n c t i o n s ; and a study of the o p e r a t i o n s under which the c l a s s of f u n c t i o n s with c l o s e d graphs i s closed. Ill Table of Contents page Chapter 0: Notation Chapter 1: I n t r o d u c t i o n of c l o s e d graph p r o p e r t y ; open f u n c t i o n s w i t h closed- graphs 5 • Chapter 2: Subcontinuous f u n c t i o n s ; r e l a t i o n s between f u n c t i o n s with various properties 10 Chapter 3: C h a r a c t e r i z a t i o n of continuous f u n c t i o n s ; l i m i t theorem; statements about domain and range spaces 17 Chapter 4: P o i n t s of i r r e g u l a r i t y ; numbers limit 29 Chapter 5: Closed and l o c a l l y c l o s e d functions; topologies determined by compact subsets 35 Chapter 6: A n a l y s i s of the c l a s s of f u n c t i o n s . w i t h c l o s e d graph; product spaces. 43 and D e f i n i t i o n s 1 iv Acknowledgements I should l i k e for suggestion to thank P r o f e s s o r J.V. Whittaker of the t o p i c , h i s p a t i e n t r e a d i n g of the o r i g i n a l d r a f t and f o r h i s h e l p f u l c r i t i c i s m s . Also I would l i k e of the to acknowledge the f i n a n c i a l support N a t i o n a l Research C o u n c i l of Canada. CHAPTER 0 T h i s chapter i s devoted t o an e x p l a n a t i o n o f the n o t a t i o n t o be used and some d e f i n i t i o n s . Definitions: , FILTER A class filter on X 5 o f subsets o f a nonempty s e t X i f the f o l l o w i n g c o n d i t i o n s are s a t i s f i e d : 1. i f A,B e 3 2. i f A e 3. 0 k* i s any f i l t e r and either $<^lLor on and ADB e 3 then A maximal f i l t e r If BsA on X then X q ultrafilter. principal e A (free) X B e 5. and U. i s an u l t r a f i l t e r then IX. are not comparable. for a fixed If ' i s c a l l e d an u l t r a f i l t e r . Maximal f i l t e r s always e x i s t Q = {ACX|X isa X q e X} i s an i n f i n i t e since the c l a s s i s a principal (fixed) s e t t h e r e e x i s t non- ultrafilters. TOPOLOGY A class X 3" o f subsets o f a s e t X i s a topology f o r i f the f o l l o w i n g c o n d i t i o n s a r e s a t i s f i e d : 1. I f A,B e then ADB e 3\ 00 2. i f A e ff n=l,2,... n 3. 0, X e if then u A n=i e ff n - 2 - The • s e t s b e l o n g i n g of X . I f . A e 3" to 3" . a r e c a l l e d t h e n t h e complement o f t h e open A i s said sets t o be closed. 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 of, t h e f o u r t o p o l o g i e s m e n t i o n e d i n t h i s FRECHET (T ): A topology 1 3" exists and HAUSDORFF.(T ) M,N e 3" y e N, A topology x such t h a t on a s e t X M,,N e 3" and 3 topology only Let i s T^ i f and e X, there y e N and x e M 3" o n a s e t X ADB =' $ M,N where X , there ACN, sets i s regular i f x | A, i ff o r each p a i r with disjoint NEIGHBOURHOOD y | M x e M, on a s e t X 1 such t h a t A,BcX .ACM x e M, 3 such t h a t o n l y i f f o r each M,N A there xy x =j= y , a closed subset o f NORMAL: e X, N . J A topology and i f and MON = 0 and REGULAR: i s T^ x,y x 4 y, o n l y i f f o r each exist paper. on a s e t X o n l y i f f o r each our d e f i n i t i o n exist and A open MDN = 0 i s normal i f of closed there i s sets a r e open, with the property that B<=N . N x e X and l e t 3" be a t o p o l o g y on X . Then - 3 N<=X is there a is ^-neighbourhood or an 0 e J such simply a that x neighbourhood of x i f e O5N DIRECTED SETS AND NETS A binary preorder i f i t is set X (x,x) 2. i f additional c e R e D by such a f x a PRODUCT is called a i.e. e X (y,z) set D or, of D of the the xQ a x a belong to preorder is preordered each and a,b b<c X into i f is for space function R then called a is X . e D set with there the exists . a mapping We w i l l directed set <p:D -» X denote A nets need not of either be . a will domain o f from the be denoted x ^ and domain o f bJ y x,Nb N x ^ is the to the where domain ' DIRECTION Let we x definite a<c in a subnet member appropriate o a that simply by A is a l l and set that e A} mentioned,, b X and t r a n s i t i v e : for with property directed {xola a set . directed A net some on a set. A a e R (x,y) together preordered R reflexive 1. (x3z) A relation define the A and product B be two o r d e r e d direction as sets. follows; In (a-.,b-.) A x B _< (ap,b?) - 4 in b A x B l—^2 i n i f and only ^ * With i f a <a in 1 "this d e f i n i t i o n A a or = a ± A x B and g i s an o r d e r e d s e t , CONVERGENCE A net space there X Cxala i f and o n l y e x i s t s an following A } converges t o a p o i n t t h e r e m a i n d e r o f t h e p a p e r we w i l l u s e notation: neighbourhood X-A complement o f A~ the closure A of with respect the graph o f the f u n c t i o n the Cartesian product of x„ -• x a the net Z the p o s i t i v e 3 -» Y o X with Y f X converges t o the p o i n t x a the f u n c t i o n there such A' x to A X x Y f:X x 3 nbhd. + of T i f and o n l y i f f ina s u c h t h a t i f b>a. , x, e N . — N b iff G x i f f o r each neighbourhood . N a„ e A N Throughout the e integers f maps X into Y exists that the derived accumulation set of A , i . e . the s e t of a l l (cluster) points of A . - 5 CHAPTER 1 Following F u l l e r [ 2 ] , we introduce the notion of a function having a closed graph i n terms of l i m sup (Ls) and lim i n f (Li) of nets of sets. We also present a character- i z a t i o n of open functions and give necessary and s u f f i c i e n t conditions for an open function to have a closed graph. D e f i n i t i o n 1.1 A function to X x Y) iff topology of closed f: X -* Y has a closed graph [(x, f ( x ) ) | x e X} X x Y . (relative i s closed i n the product In terms of nets t h i s becomes i f f (x , f(x )) -« (p,q) in X x Y Gf is implies that q = f(p) • The following Lemma w i l l be used a great deal i n Chapter jj. Lemma 1.2 Let for each f : X -» Y be given. x e X and y e Y , where open sets • U and V , containing such' that Proof: Then i s closed i f f y 4 f(x) , there e x i s t x and y respectively, (x,y) G^ then f (U)nv = 0 . I f the condition holds and i s an open set i n Implies Gf X x Y such that G~ i s closed. ( U x V)nG^ = 0 U xV which - 6 G If so there U and such exists V (U x V ) n G that Definition A n v for nbhd and convergence defined of a y ^ f(x) , U x V y Therefore may b e where respectively, f(U)HV = 0 and . of a net of subsets of i n t h e same m a n n e r a s sequence of sets. i s a net of subsets i s defined of x arbitrarily D x convergence n We set X , then f s e t o f the form . C-A |n e D} ( L s An) ' every = 0 f G 1.3 If Li (x,y) containing space topological and open topological topological the a basic a r e open^ The a i s closed f as the s e t o f a l l intersects large) say that A x f o r almost n of X , e X such then that a l l (respectively n . a statement S on elements of a directed i s fulfilled for: 1. almost a l l n e D i f t h e r e i s an n e D such o that 2. set n e D A net i s said large f o r which S fora l l n e D i f the set of a l l t o be t o p o l o g i c a l l y v further n>n — o i ffulfilled A) i f L i A = Ls A (= A ) , i n w h i c h ' n n '' i s c o f i n a l i n .D . convergent case A d e f i n i t i o n s and theorems (to a = Lim A on n e t s n of sets S. M r o w k a [ 8 ] . The and i s fulfilled arbitrarily For see S i s needed following lemma i n the proof i s interesting of the succeeding i n i t s own theorem. right - 7 - Lemma 1.4 If f:X -* Y i s a function, y ' i s a net i n Y , a, p e Ls £~^(y ) and > then t h e r e e x i s t s a n e t x ^ a x^ w i t h the p r o p e r t y t h a t -* p and f(x ) N t ) in X i s a subnet o f Proof: Assume y J Then f o r each nbhd index N(a,U)>a nbhds o f p N(a,U ) 6 f _ 1 U (%(a,U ) of y N(b and U ) k 6 ^n x p e Ls f ^ ( y - ) a' K and each index f - 1 (y]\r( TJ)^ a a there i s an ^ ^ • u (i.e. ^ D i r e c t the o 2 ••') > £ (a,U) have the product d i r e c t i o n , and choose X , Now we have x p such t h a t in a of and l e t downwards by containment l e t the p a i r s x i s a net i n Y a. jj/ P ) f 1 U n > .f( jj( r o v i a e d h u ° s TJ )) x u ) ~* ^ a t = a s only i n c e y b t a i n i n N(a U ) £° t h e n and e t x N w i l i c t l any nbhd r b>a S l s ( ,U ) a subnet a U of p , k>n . ., Theorem 1.5 If equivalent: f:X -* Y i s a f u n c t i o n the f o l l o w i n g a r e a) f has a c l o s e d graph b) If y - q in Y then Ls f _ 1 c) If y - q a in Y then L if - 1 a (y )5f _ 1 (q) (y ) f a _ 1 (q) a c Proof: Assume • a) Let X p e Ls f ~ ^ ( y ) • a such t h a t x^ -* p holds and l e t y -• q i n Y . a By Lemma 1.4 there i s a n e t x ^ and f(x N f e 0 ) i s a subnet o f y a . in - 8T h u s we h a v e Q = f Ntl .> f ( Nb^ y 1 f _ 1 = f(x.) - q a n d s i . Therefore n c e G b) i s since f i closed s true. f o r any n e t o f c) h o l d s and suppose sets _ 1 a (x p e L i f~' "(y_) and f ( x ) ) -» ( p , q ) . a. a Since by assumption . 1 cl (y )£f hence q L i E dLs E a— a cl Li p p e f"" (q) Assume t h a t Then "* ( - > ) x b) h o l d s then c) h o l d s we h a v e a x (p) 5 i . e . If E ( ci (q) we h a v e p e f _ 1 (q) or f(p) = q and a) h o l d s . Definition 1.6 If implies a function f(U) f : X -» Y i s open i n Y i s such t h a t then f U open i n X i s s a i d t o be a n open function. Theorem 1.7 If. equivalent: f : X -* Y is a function, the f o l l o w i n g are a) f i s open b) y a - q in Y c) y a - q in Y , implies implies f _ 1 (q)cLi f (q)5Ls - 1 f f _ 1 1 (y ) a (y ) a Proof: Assume a ) h o l d s , let y q in Y and l e t cl p e f nbhd _ 1 (q) U . Suppose of p p d L i f such t h a t _ 1 (y ) . frequently Then t h e r e f _ 1 exists ( y )DU = Qf . a Since 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 be o p e n . since Now we h a v e f(U) y x we c a n t a k e U to. i s frequently outside f(U) , but a i s open a n d q e f ( U ) , f ( U ) must be a nbhd o f q and t h i s g i v e s the d e s i r e d c o n t r a d i c t i o n . I f b) holds f- (q)5Li f- (y )CLs L then c) holds f 1 a _ 1 since (y ) . a Assume c) h o l d s , and suppose there y i s an open -» q o f i s not open. Then UcX and a n e t y i n Y-f(U) such that — a q e f ( U ) . By assumption f ~ ( q ) c i s f~ (y„) . f o r some 1 1 J p e f (q)nu Let Thus b) h o l d s . . _ 1 Then i s f r e q u e n t l y nonempty U i s a nbhd o f since p e Ls f _ 1 p (y ) a and f (y )nu _ 1 a • But t h i s means that y i s f r e q u e n t l y i n f ( U ) a c o n t r a d i c t i o n t o the f a c t that y i s a n e t i n Y-f(U) . a Hence a) h o l d s , U s i n g Theorems 1 . 5 and 1 . 7 and the d e f i n i t i o n o f Lim E where a E i s a n e t of sets we a r r i v e a a t the f o l l o w i n g theorem. Theorem 1.8 The f u n c t i o n f:X -» Y y a — q in Y implies Lim f _ 1 i s open and has c l o s e d graph i f f (y ) = f _ 1 a (q) Proof: By the theorems c i t e d above c l o s e d graph i f f y -• q Q in Y • f (q)eLI f- (y )CLs _ 1 1 a Thus and Lif _ 1 f - 1 (y )cf _ 1 a 1 the theorem h o l d s . a i s open and has implies ( y ) = Ls f " ( y ) = Lim f a f (q) _ 1 (y ) = f a _ 1 (q) 10 - - CHAPTER 2 .In t h i s chapter we i n t r o d u c e suboontinuous We then study some r e l a t i o n s between f u n c t i o n s w i t h graphs and c l o s e d , compact and subcontinuous Throughout the chapter we w i l l the class of a l l functions f:X -» Y functions. closed functions. denote by U(X,Y) whose graphs, , are closed. The f o l l o w i n g theorem w i l l of l e a d us to the d e f i n i t i o n (Fuller[2]) subcontinuity Theorem 2 . 1 Let of f e U(X,Y) points i n let {f(x )|x n X such t h a t . x e X} n £ and l e t x x , be compact i n n l n b e where Y . e sequence a x e X . Then f( x Also )- f ( ) • x n Proof: Suppose t h a t f( Then there e x i s t s a nbhd that f( x n compact i n which 4 _) (f(x f e U(X,Y) N ) does not converge to of )}°° and and hence ( 1 ) . Since x converges to some (x , f ( J )) - x n f(x) . f ( x ) w i t h the p r o p e r t y there i s a sequence 1 q = f(x) n for i = 1 , 2 , 3 . . . N Y x n '£f(x )} n j = 1,2, q e Y . ... Since we must have - J 1 (f(x )} ° c H 3=1 -* f ( x ) . Therefore is for - 11 {f(x )}°° i-1 i Hence converges to 2.2 The f u n c t i o n x a l s o , but t h i s c o n t r a d i c t s (1) f(x ) - f(x) . Definition iff f(x) p n f(x ) in X f:X -. Y i s s a i d to be i m p l i e s there i s a subnet which converges to a p o i n t a subcontinuous Similarly a function f(x^) of q e Y . f:X -» Y i s s a i d to be i n v e r s l e y subcontinuous i f f f ( x ) -• q i n Y i m p l i e s there i s a subnet a Xj^k of x^ which converges to some p o i n t p in. X . The concept o f a subcontinuous f u n c t i o n i s thus a g e n e r a l i z a t i o n - o f a f u n c t i o n whose range i s compact 2.1). ( c f Theorem I t i s a l s o a g e n e r a l i z a t i o n of a continuous f u n c t i o n . 2.3 Theorem If has a nbhd of Y f:X -» Y i s a f u n c t i o n and each p o i n t U such t h a t then f f(U) x in X Is c o n t a i n e d i n a compact subset i s subcontinuous. Proof: Let Since a>a Q f(U) net Hence U , x x be a net i n a i s a nbhd o f Q e U . Then x f( X such t h a t t h e r e i s an x a ) e f(U) a x N b f ) of f ( x a ) -» x . a such t h a t f o r a l l Q f o r a l l a>a i s c o n t a i n e d i n a compact subset of f( x Y Q Since there i s a sub- t h a t must converge to some i s subcontinuous. . q in Y . - 12 The f o l l o w i n g theorem w i l l show t h a t i f the range i s embedded i n a completely r e g u l a r space then a subcontinuous f u n c t i o n i s v e r y n e a r l y such t h a t i t p r e s e r v e s compactness. Theorem 2 . 4 Let f:X - Y be a subcontinuous f u n c t i o n and l e t Y be c o m p l e t e l y r e g u l a r . K of Then f o r each compact subset X , f ( K ) i s compact. Proof: Let in KcX f(K) . Let B ( c f Lemma 1.4) each X(a ^ (Mc, and l e t e Knf~" (y^ • L °^ y ( Clearly x K a (a,b) d be a n e t ^d) downwards by containment A x.B choose a subnet Y . Direct. B (a,b) e A x B Nc) {Z |a e A} be a u n i f o r m i t y f o r Y . (see Dugundji [1] pp 2 0 0 , 2 0 1 ) . X be compact and l e t have the product o r d e r . y^ ab Since ^ e b[Z ]nf(K) i s compact a subnet converges. of y^ a ^ and a K For Since, f i s subcontinuous converges to a p o i n t q in q e ~fJKJ Now c o n s i d e r the n e t Z ,, which i s a subnet o f Kd T Z & . We w i l l show t h a t there i s a symmetric Z^ ^ we have t h a t e v e n t u a l l y i n b± . Since . Since 1 e v e n t u a l l y i n b^ compact. Let b e B . e B y^ e v e n t u a l l y in-,, b -» q . b By the choice o f ab d and hence such t h a t L d ) b-^ o b£b By d e f i n i t i o n b^ o b""" = b^ o b^crb . 1 (y^ K dL d ) *Z - q , (q, y we have -» q . ( K d ? K d ) is L d ) (q, Z Consequently ) K d is ) is f.(K) i s - 13 Assuming the range i s embedded i n a Hausdorff the next theorem s t a t e s t h a t i f f e U(X,Y) then f space, handles compact sets w i t h l e s s success than continuous f u n c t i o n s but i t s i n v e r s e does Just as w e l l as the i n v e r s e o f a continuous function. 2.5 Theorem Let X [Y] f e U(X,Y). f(K) t f then _ 1 If K i s a compact subset o f ( K ) ] i s closed. Proof: Let K be a compact subset of f(KJ Is not c l o s e d . that y K a Q - q and Then there i s a net q e Y-f(K) . i s compact there i s a subnet to some p o i n t Since X x e' K . Thus N b in f(.K) _1 of we have q . a q x & that x x D q e f(K) , Therefore f(K) such Since converges ( JJ J ^ ( N b ^ f e U(X,Y) , q = f ( x ) and hence c o n t r a d i c t s the choice o f y X e K(1f (y ) . a a Pick x and assume t h a t ~* ( X j ( but t h i s i s closed. The second p r o o f i s completely analogous and w i l l not be done. Definition 2.6 Let f:X - Y compact subsets of then f be a f u n c t i o n . X [Y] If f [f _ 1 ] takes onto compact subsets o f i s s a i d to be compact-preserving l) * [compact]. Y[X] - Ik 2.7 Definition If sets o f f [f Y [X] ] - 1 takes c l o s e d sets o f then f X [Y] i s s a i d to be c l o s e d functions c o n t i n u i t y and s u b c o n t i n u i t y closed [continuous]. In the f o l l o w i n g theorem we c h a r a c t e r i z e and compact p r e s e r v i n g onto compact i n terms o f i n v e r s e sub- respectively. Theorem 2 . 8 Let a) f:X -» Y The f u n c t i o n f be a f u n c t i o n . i s compact i f f f | f ( K ) : f ~ ( K ) - K i s _ 1 1 i n v e r s l e y subcontinuous f o r every compact b) The f u n c t i o n f KcY . 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 o f a) i s analogous to b) so o n l y b) i s proved. Assume t h a t preserving. ° c x e K . Since have a subnet Hence y be a n e t i n K a f(x ^) N f i s compact x -* x , a f(x ) cl must q e f(K) . ' i s subcontinuous. f ( K ) . - Pick and hence there converges, to some f such t h a t that converges to some be subcontinuous. be n e t i n a net i n K i s compact and' t h a t f ( K ) i s a l s o compact the n e t f|K:K -• f (K) Let let Let x KcX x e K . we must have a subnet x L e t . KcX e Knf i s a subnet Since _ 1 be compact and ( y ). x™ NB Then of f.|K;K - f ( K ) x a i~ s x . that a i s subcontinuous - 15 f(x M b ) Since of x f(K) f( MNTD^ = x N b ) = y N b l a ^MNb s converging to a p o i n t S U D n e - k °^ y w e n a v q e f(K) . that e a i s compact. Corollary 2.9 a) If f:X -» Y f~" "(K) ! f b) If i s i n v e r s e l y subcontinuous and i s c l o s e d f o r each compact KcY then • i s compact. Y f:X i s subcont inuous and c l o s e d f o r each compact KcX then f(K) i s f i s compa preserving. Proof: Again o n l y b) i s proved. I n any case we need o n l y note t h a t ( u s i n g the n o t a t i o n of Theorem 2.8) since c l o s e d , the l i m i t p o i n t must belong to q of the n e t y^^ and f e U(X,Y) and f(K) i s f(K) . Corollary 2.10 If f:X - Y continuous . [ i n v e r s e l y subcontinuous] then preserving f f i s subi s compact- [compact] Proof: By Theorem 2.5, KcX • f ( K ) i s c l o s e d f o r every compact and hence by C o r o l l a r y 2.9 f • i s compact-preserving. - 16 - Corollary 2 . 1 1 • Let a) if Y f : X -• Y be a f u n c t i o n . i s Hausdorff and i n v e r s e l y subcontinuous b) If X i s Hausdorff and continuous then f f i s both continuous and then f Then:. f i s compact. Is both c l o s e d and sub- i s compact p r e s e r v i n g . Proof: We prove o n l y b ) . compact K<=X compact KcX . Since i s c l o s e d and hence By C o r o l l a r y 2 . 9 X i s H a u s d o r f f every f ( K ) i s c l o s e d f o r every 3 f i s compact-preserving. - 17 CHAPTER 3 In t h i s chapter, we give c h a r a c t e r i z a t i o n s o f continuous and c l o s e d f u n c t i o n s subcontinuity properties. i n terms o f the c l o s e d graph and We then show t h a t i f f f u n c t i o n w i t h a c l o s e d graph then sequence f isa i s the l i m i t o f a o f continuous f u n c t i o n s and p o i n t out the dependency of the c l o s e d graph p r o p e r t y on the range space. The f o l l o w i n g two theorems w i l l show t h a t the c l o s e d graph p r o p e r t y complements the two s u b c o n t i n u i t i e s i n i n t e r e s t i n g ways. 3.1 Theorem If for f f:.X -» Y i s a f u n c t i o n then a s u f f i c i e n t to be. continuous i s t h a t be subcontinuous. If Y f condition have a c l o s e d graph and i s H a u s d o r f f then the c o n d i t i o n i s also necessary. Proof:. (Sufficiency): Suppose t h a t not Let f ( continuous a t x a x & be a n e t i n such t h a t ) does not converge to x . Then f(x ) a' v no subnet o f which converges to continuous some subnet x y e Y ... Hence we have since G i s closed C (x » y = f(x) . mc Since ^^ Nb^ -» x . & M N c f f is f(x„„ ) . Nb K i s sub- converges to some x f(x x f(x), i.e. has a subnet, ' f(x). °^ ^( ]y[N ) point f X ) ) - (x,y) Therefore which i m p l i e s t h a t the assumption was f a l s e . f and ( M £ j ) ~* ( ) x f c x - 18 (Necessity): If f i s continuous then i t Is o b v i o u s l y sub- continuous. If ( x , f ( x ) ) - (p,q) a f ( x ) -* f ( p ) a since a f q = f ( p ) and hence x then i s continuous. f has a c l o s e d - p & If Y and i s Hausdorff. graph. The f o l l o w i n g example shows t h a t Theorem (Necessity) Example i s false i f Y i s only a 3.1 space. 3.2 Let X = Y = Z' + c o f i n i t e topology [ i . e . f i n i t e complement]. and l e t both X and {(n,n)|n e Z ] + O b v i o u s l y the i d e n t i t y map converges every nbhd of a p o i n t to every p o i n t (x,y) of the topology f o r X x Y are open subsets of Z G x Corollary Y or n ^ Gg} + have the a s e t i s open i f f i t i s v o i d or has but i t does not have a c l o s e d graph, because [njn thus the sequence (x,y) € X x Y . there i s an element G^ x G^ of the form and then i s continuous (x,y) e G^ x G^ x G In of the base where . I f (k,k) e G 1 f:X -• Y be s u r j e c t i v e . m = G-^G^ max for a l l k X m . 3.3 L e t the f u n c t i o n i s an u l t r a f i l t e r on X such t h a t U x f o r some If x e X then a s u f f i c i e n t c o n d i t i o n f o r the image f i l t e r f(U) converge be to continuous. f(x) If Y i s that G^ be c l o s e d and f U to sub- i s H a u s d o r f f the c o n d i t i o n i s a l s o necessa - 19 - Proof: (Sufficiency): if U - x By Theorem.3.1 then (Necessity): Corollary f i s continuous and hence f ( U ) -» f ( x ) . As i n p r e v i o u s theorem. 3-+ i If f:X - Y Hausdorff then f i s a f u n c t i o n and i s continuous i f f G~ f Y i s compact and i s closed. Proof: Suppose i s closed. i s subcontinuous. Let ( a> x x f ( -» p a ^ x i f n G Since Y i s compact Hence by Theorem 3-l> f i s continuous. be continuous and suppose there i s a net f s u c h t h and by c o n t i n u i t y cl Hausdorff f q = f(p) a ( a> t x f ( x a ^ "* ( P ^ ) • f ( x ) -* f ( p ) . Since T Y H E N is cl and i s closed. The f o l l o w i n g example shows t h a t the compactness c o n d i t i o n on Example Y 3*5 Let 1 - Jl/x 10 cannot be dropped. X = (0,1] and Y = [0,+») . Define f by x e (0,») x = 0 Then f i s c l e a r l y not continuous but The next Theorem from P.E. Long's paper G^ i s closed. [7]» a c h a r a c t e r i z a t i o n of c o n t i n u i t y i n terms o f the c l o s e d gives graph - 20 - p r o p e r t y and c o n d i t i o n s on the domain and range spaces. Theorem 3 . 6 . Let compact and f:X -> Y X be a f u n c t i o n where is first countable. If G Y i s countably i s c l o s e d then f f • i s continuous. Proof: Suppose open f - 1 VcY (V) f i s not continuous. such t h a t f _ 1 (V) contains a point p o i n t of X - f - 1 (V) . i s not open i n x e X Since such t h a t X x countable, there converges I n the countably compact space (f(x )} has an accumulation n but X x Y (x,y) f (x,y) x^ e X - f ~ ( V ) , t h a t 1 point i s a l i m i t p o i n t of containing (x,f(x)) . hence x . Therefore i s a limit {x } to with is first X . e x i s t s a sequence n , Then there i s an G^ y | V . Y , the s e t Then (x,y) ^ G^ , s i n c e any open s e t i n c l e a r l y c o n t a i n s p o i n t s . o f the form T h i s c o n t r a d i c t s the assumption that G^. i s closed, i s continuous. The above example shows t h a t t h i s theorem i s not true f o r Y l o c a l l y compact and Theorem 3 . 1 X compact. showed the r e s u l t of combining c l o s e d graph.property with s u b c o n t i n u i t y . I f i n s t e a d we use i n v e r s e s u b c o n t i n u i t y we might suspect t h a t be continuous. the f The f o l l o w i n g theorem confirms t h i s _ 1 will fact. - 21 - Theorem 3-6 f : X -* Y I f the f u n c t i o n i s inversely" subcontinuous then has a c l o s e d graph and f i s closed (i.e. f - 1 . is continuous). Proof: Let f(C) C i s not c l o s e d . v a point q e Y - f(C) such t h a t -* p ( Nb> ( N b ) ) x f x p e f(C) . is and s i n c e "* ( > ) p q a n C d f( x a i s closed ^ ( ) q f p s i n x ) ~* Q • i s a subnet J Xj^ and suppose Then there e x i s t s a net i n v e r s e l y subcontinuous there that X be a c l o s e d subset o f e C Since x,^ Nb of p e C . c in a x that and f is~ such a Hence we 0. e Y - f ( C ) have and T h i s however, c o n t r a d i c t s the assumption t h a t G^ closed. D e f i n i t i o n 3.7 • f : X -• Y belongs to B a i r e A function set f X in - 1 (G) i s a Borel f o r every open s e t of a d d i t i v e c l a s s subset o f ct f - 1 (G) X f e B1(X Y) 3 f o r each open i s open f o r each open The f o l l o w i n g theorem [5]) shows t h a t i f X A a i f the of B o r e l sets GcY . In p a r t i c u l a r , F class and has a c l o s e d graph then of continuous f u n c t i o n s . f Y G5Y GcY if and ; i.e. f -(G) i s an _1 f e B if f (X Y). 3 if i s continuous. (from P. Kostyrko and T. are m e t r i c spaces and i s the l i m i t f u n c t i o n of a Salat f : X -» Y sequence - 22 - Theorem 3 . 8 Let (X,p) and (Y,a) be m e t r i c spaces where CO Y = U C I with ll • C n = 1,2,3,... compact II . Then . U(X,Y)cB (X,Y) 1 Proof:Let f e U(X,Y) M = G n(X x G) set and l e t GcY CO ^ i s an set i n X x Y . F Define the M e P (X x Y) , i . e . M and note t h a t f be open. Therefore M = U M where n=l • . CT n CO each C M i s closed i n X x Y . = l,2.,3j... • Let a p o s i t i v e number. p -of X radius that D n = y M and Rp n n p = and e U q Z L e t S(x ,p) . n f o r - = M nR n p n > >~5>-- . > 2 Now each M n p . Let E N P n,p=l n a 0 n E ^ „ and E = f n n Q p € Z i s closed R p & ( s i n c e both M i s ) i n X x Y and • (x,y) e M] . (G) . such + . = (x e x|2 y e Y • • (x,y) e -1 be Define % = S(x ,p) x D . ^ d M let E = { x e X | 3 y e Y U i s an f o r n>. ) n a r e ) and bounded ( s i n c e ro E = C and l e t be a s p h e r i c a l nbhd o f By assumption there 1 = U C and each n=l C ,, f o r n— n+1 = (x *y ) e X x Y Q > (and hence t o c Q n +n-l o C Let M Y = i s compact i t c o u l d be assumed t h a t ^ n n Since M } np Then I f we can now show t h a t p f o r any f i x e d n and p ,- E N P i s c l o s e d i n X , we w i l l be done. Let x e E' np [E'" i s the d e r i v e d np set of E 1 . np J n - Then there x a 2> ( e x i s t s a net x - a l a € -» x . From the d e f i n i t i o n o f for each Now the elements there a e A there y i s a subnet Then we have Hence x e E have f - 1 is a y i ^ p n such t h a t n E this implies ^ np e Y a y such t h a t Y N F E ) - (x,y) e M E , whence (X) that (x ,y ) e a D a and hence t h a t converges to some p o i n t N f e and t h e r e f o r e (G) e F ) a l l belong to a compact (x^, np A since n p M f e B (X 1 A . i s closed. np i s closed set. np y e Thus we Y) The f o l l o w i n g example, l i k e example 3-5 w i l l show CO Y = t h a t the c o n d i t i o n t h a t compact cannot be dropped. U C n=l where the C are n The example a c t u a l l y illustrates a f u n c t i o n t h a t has a c l o s e d graph but belongs to no B a i r e class. Example 3.8 X = [0,1] Let Y w i t h the E u c l i d e a n m e t r i c and l e t be a s e t w i t h 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 t h a t (x ,y ) -» (x ,y ). Then f a^a' o o' x -» and y„ - y^. . Since Y has the t r i v i a l m e t r i c CL o a o v v 30 o y = y . a o J .for almost a l l a e A , i . e . there 3 such t h a t y a J follows that the net x = y Jo y Q f o r a l l a>a . — o = f(x ) & for " i s stationary for a a>. a From a n d 0 a>a . — o x„ = x f o r a>a . Hence a o — o G^ i s c l o s e d . e A K From v a o y = f(x ) i t ^a a' - since follows that therefore i s an f x a i s injective -• x o (x ,y ) e G. o o' f i t and 2h - Now l e t G = f(E)cY Since . f G E - be any non-Borel set. i n i s open since i n Y i s s i n g l e valued X and l e t a l l sets a r e o p e n . E = f'^G) Hence f belongs to no Bore'l c l a s s . The follows previous theorem was l a t e r g e n e r a l i z e d as (Kostyrko [ 6 ] ) . Theorem 3.9 Let f e U(X,E^) Then there X be a normal t o p o l o g i c a l space and l e t where E^ = (-<», + «)• w i t h the E u c l i d e a n i s a sequence o f f u n c t i o n s n = 1,2,3,... > such t h a t for a l l f |f (x)|<n n e B (X^E^)- , and l i m n metric, f ( ) = f( ) x x n x e X . Proof: Let P = G n ( X x [-n,n]) n = 1,2,... . c l o s e d i n X x' E-^ and i t s p r o j e c t i o n (X also closed. X n 4 0 n X = [x e X | 3 y e then the f u n c t i o n n (x,y) g = f|X °n > n x e X n space there • g . Since i s a continuous e x t e n s i o n f is n X If X n | g ( )I<n ' for x n i s a normal o f the f u n c t i o n ft r n e F ) . i s continuous on . 1 ( c f C o r o l l a r y J>.k). n t o the s e t X i s s i n c e i t has a c l o s e d graph (namely F^) and all F f n onto the whole space x e X . I f ., X^ = 0 The equality X such t h a t |f (x)|<n for a l l put f ( x ) s 0 . n f ( x ) = l i m f (x) f o r X e X n^co n follows - - 25 from the f a c t t h a t the sequence o f sets f x n l n e z + ) i s CO i n c r e a s i n g and X = U X „ n=l . If x e X then there i s an n n 4- o e Z such t h a t f o r a l l n>n , x e X — o n and hence 9 f ( x ) = f(x) . n We now. c o n s i d e r the dependency o f the c l o s e d graph c o n d i t i o n on the s p a c e . i n which the range Is embedded. Note t h a t Theorem 3.1 and the f o l l o w i n g c o r o l l a r i e s r e q u i r e the range space be H a u s d o r f f . In general t h i s cannot be dropped, f o r i f i : X -» X on X , then G^ Now l e t graph.and of Y . If f i s closed i f f f:X -» Y f :X - Y* condition i s the i d e n t i t y map i s Hausdorff. be a f u n c t i o n which has a c l o s e d i s not continuous. Then X that L e t Y* be a c o m p a c t i f i c a t i o n i s . subcontinuous s i n c e Y* i s compact had a c l o s e d graph i t would be continuous (by Theorem 3. However c o n t i n u i t y does.not depend upon the embedding space so t h a t This f:X -»-Y would be continuous, c o n t r a r y shows t h a t i f f e U(X,Y) but f £ B (X,Y) Q to assumption. then f 4 U(X,Y*) . The f o l l o w i n g theorems (from P.E. Long's paper [7]) show t h a t i f we know a f u n c t i o n has a c l o s e d graph then we can make some statements about the domain and range spaces f . Theorem 3.1Q-, Let f(X) is T x f:X -» Y . be a f u n c t i o n w i t h G^ closed. Then - 26 Proof: Let y there e x i s t s an (x,y) V and w x e X be d i s t i n c t p o i n t s i n such t h a t f(x) = w . f(X) . Thus | G^ , so. by Lemma 1.2 there e x i s t open sets containing Therefore Theorem Then x and y respectively., such t h a t w | V and Y is U and f(U)nv = 0 T^ . 3-H Let closed. f:X -* Y Then Y is Let y and be any open s u r j e c t i o n w i t h G^ T^ . Proof: w be d i s t i n c t p o i n t s i n there are d i s t i n c t p o i n t s x f(x) = y Since and f(z) = w . there e x i s t open sets U and r e s p e c t i v e l y , such t h a t and contains Theorem is z V in Y such t h a t (x,w) ^ G^ and containing x f(U)nV = 0 ; but Consequently X is Then G^ i s clos and f(U) w i s open T^ . 3.12 Let X y . and Y . T x f:X - Y be i n j e c t i v e v/ith G f closed. Then • . Proof: Let x and z be d i s t i n c t p o i n t s i n f ( x ) ^ f ( z ) * so there e x i s t open sets U x f(U)nv = 0 . and f(z.), r e s p e c t i v e l y , such t h a t z ^ U , implying X is T, . and V X . Then containing Thus - 27 Theorem 3.13 If both X f:X - Y and Y i s b i j e c t i v e with are G f c l o s e d , then T^ . Proof: Theorems 3.10 and 3.12 Theorem 3.14 Let c l o s e d graph. f:X -» Y be i n f e c t i v e and subcontinuous w i t h Then is X T^ . Proof: By Theorem 3.1 f an open s u r J e c t i o n from i s isomorphic to f(X) G ^ - l , hence Y x X . By Theorem 3.11, Theorem 3.15 Let having G^. i s continuous, hence f :X -» Y closed. X onto f - is X . Furthermore g . be a homeomorphism of Then both X G^ has a c l o s e d graph i n 1 T is and Y are X onto Y Tg . Proof: Theorems 3 - H Theorem 3.16 Let. have m 2 ° G^ and 3.14 f:X -• Y closed. be i n f e c t i v e , open, connected and Then i f X i s l o c a l l y connected, X is - - 28 Proof: Let x and z be d i s t i n c t p o i n t s of X . Then f ( x ) ^ f ( z ) , so there e x i s t s an open connected s e t U containing such t h a t x f(U)nv = 0 . UU{z] and an open s e t Since f(U) f containing i s open, z | U . i s connected, so t h a t connected, since V f(z) For otherwise f(Uu{z]) = f ( U ) u ( f ( z ) } i s connected. is T h i s 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 obtained otherwise Let f:X -» Y in X {f(x )} with We f x i s not We of X and Define a e X Y. will stated. be a f u n c t i o n . a p o i n t of i r r e g u l a r i t y of n. Spaces 4.1 Definition x extend some r e s u l t s Salat in [ 4 ] , by Kostyrko and be Hausdorff unless review and and D^ i f there be e x i s t s a sequence n -» a , xn 4 a '; such t h a t the sequence ^ 3 compact i n will by F to Y „ denote by the p o i n t s of irregularity the p o i n t s of d i s c o n t i n u i t y of note t h a t f o r a n y , f u n c t i o n f, N^cp^ f . • 4.2 Theorem If f e U(X,Y) then N = D f . f Proof: Since f o r any function the reverse, i n c l u s i o n holds f o r contrary Since Let x n that x e D^ (x - x D^^W^. . x s |n e Z } + and i s compact i n x Let Y we need o n l y to show f e U(X,Y). x e D^ be Assume on such that cannot be an i s o l a t e d p o i n t of be a sequence of p o i n t s of 4 x . Since n •N^cD^ x k N„ .i . From Theorem 2 . 1 f ( x ) - f ( x ) , which i m p l i e s t h a t n the theorem Is proved. the f x | N^. . X . such t h a t sequence ^ i t follows x \ D X the f f ( x )) n v that . Hence D cN^ f and - 30 Corollary 4.3 If f e U(X Y) then 3 f i s continuous i f f N = 0 . f Theorem .4.4 Let f e U(X,Y) and assume t h a t compact, i . e . t h a t each p o i n t i s compact i n Proof: Y , then Since 3\V f x c o n t i n u i t y of f(x f . there such t h a t Y . x -» x an a>a — o n a -» x & a there e Z . Assume t h a t f , i.e. f (x) V a Q f (* ) i s a point of x -* x ' we o have x e A is such t h a t such t h a t of p o i n t s i n + „)|n e Z l a, n + i s no subsequence of K then f o r each n {f(x x in e f(V) . a .|n e Z ] a 1 such t h a t for a l l a e A {x fx l a e A} a • •* a i s an and hence e N„ i and 1 of I s X not compact {f(x „)|n e Z} + Y . -(l) f o r a l l a>a . and s i n c e a — o f o r a l l a e A , t h e r e f o r e f o r a>a there e x i s t s . — o a V f i s a nbhd there x -» x and . a, n a Now = N f x e X - N. i i s a sequence i.e. D be a nbhd of e V & t h a t converges i n a,n K x •, x Since x x e X - N Since Since f o r a l l a>a a e A we have By c o n t i n u i t y there f (V)cKcK . in and ) - f(x) . Let compact. .has a nbhd whose c l o s u r e Then there e x i s t s a net . -• x is locally i s closed. f e U(X,Y) i s not c l o s e d . such t h a t D^, y e Y Y i s a nbhd of x such t h a t f o r a l l n>n , x„ e V . Hence f o r each — a a,n the subsequence (f(x )ln>n , n e Z } i s a subset ^ ^ . a.n' — a * •* + J x 1 - 31 of f(V)dK . f^f ( x a , n )ln>n — a v / 1 Therefore f o r each . n e Z } >. J contradicted. £^( Therefore x a r ,)ln ' e But Z } + = co) £f( x a n )|i e Z ] + 1 .)l^- J e ^ hence ( l ) has been a n . i s closed. I n t h e f o l l o w i n g theorems Y = E^ = ( + 0 v t h a t converges t o a p o i n t i n K . i s a subsequence o f t h e sequence- a has a subsequence f(x ^ . f a,n^' + 3 a Y w i l l be t h e r e a l w i t h the Euclidean line metric. Definition 4.5 If at a point f:X -•. E-^ xo e X i s a f u n c t i o n then i f t h e r e i s a nbhd i s impossible to f i n d a single r e a l for all x e N . N M>0 f of f xo i n which i t such t h a t We n o t e t h a t i n t h e case t h a t points of i r r e g u l a r i t y of i s unbounded |f(x)|<M Y = E-^ , and p o i n t s o f unboundness c o i n c i d e , Definition 4.6 The subset D .of E^ 4 s s a i d t o be o f t h e f i r s t CO category if D = u n=i in E^ (i.e. interval). D where each i s o f the second i s nowhere dense M the c l o s u r e o f If D D •'' D^ c o n t a i n s no nonempty i s n o t o f t h e f i r s t c a t e g o r y we say D category. Theorem 4 . 7 Let x e X in X . f e U(X,E ) 1 i s a second c a t e g o r y and assume t h a t each nbhd set. Then D^ K of i s nowhere dense - 32 - Proof: Since 4.4, i s c l o s e d and x x ° - f It f e U(X,E^) i e X o ° hence and an x e X o and a nbhd Hence there |f(x)|<_M i s not f for a l l CO K = U T -i n n=l dense i n f and 0^, dense i n K of x o K x e K [although Let and T L s i n c e then K c o n t r a r y to assumption. K^cK . — such t h a t | f ( z )I>n o 1 N i s a sequence fx, } By Theorem 'is open K i s dense. X . of x . |f(x^)|<n of i n d i c e s |v|<n (x . i t may M>0 S the exists such t h a t w e l l be true f o r t h a t a l l the f such t h a t n, T e T are nowhere i s dense i n x 3 a z o i s dense i n x, • category s e t , i s unbounded a t T n T n e K' o there and x, -» z ° k =1,2,... there e x i s t s a sequence [k } such t h a t {f(x, )} -*v and . s=l S Consequently we'have (x^ , f ( x ^ )) e G s s m f )) - (z ,v) i , f(x fc for j CpfiK = $ . f would be a f i r s t Since m Then there k-1 Since o Assume on such t h a t Hence f o r some Since 1 . = [x e K| |f(x)|<n} , n = 1,2,3 n i t i s impossible • some open ^ = a nbhd does not e x i s t a s i n g l e r e a l x e K .]. some D 0^ = X - 3 M>0 i s s u f f i c i e n t to show t h a t 0 have x e K , |f (x) | <M • 3- contrary that we G f s i n c e ' | f ( z ) |>n s T h i s c o n t r a d i c t i o n proves, the theorem. and and . |v|_<n.. Corollary 4.8 Let I Then for any be a ( f i n i t e or i n f i n i t e ) i n t e r v a l on f e U(I,E^) then the set E^ . i s nowhere dense in I . D e f i n i t i o n 4.9 Let f be a r e a l valued function determined on a p a r t i a l l y ordered set t a l i m i t of a sequence [x n <x] f [xnln at € for a l l • n ;' Denote by f at a point x (X5<) x Z } and l e t x e X . We w i l l c a l l on the r i g h t [ l e f t ] i f there exists of points i n x n -» x ': L [L and ] X t = such that lim n-<» x n >x f(x v n ') . the set of l i m i t numbers of on the r i g h t [ l e f t ] and l e t be the set of a l l l i m i t numbers of f at L = L UL x . Theorem 4.10 Let i) L n f e .U(XjE^) =} 4 0 and and pick ii) x e D^ , then Ln[ (-.,«)-f(x) ] = 0 . x A Proof: i) This follows from Theorem 4.2 and the fact that i n t h i s case N f = [x e X | f ( x ) | i s unbounded] . i l ) Assume the contrary and l e t t e L and [x In s Z"1"}J n we have t e L D [ ( - « , « > ) - f ( x ) ]. Then - f ^ f ( x ) . • Evidently there i s a sequence in X such that (x , f(x )) e G„ and v n* v n ' ' .f x -» x n v and f(x ) - t . v n (x ., f(x ) - (x.t) k G_. v 5 n ^ ny * f Now - 34 _ T h i s c o n t r a d i c t s the f a c t t h a t theorem, i s proved. \ f e V(X E^) 3 3 hence the CHAPTER 5 T h i s chapter d i s c u s s e s locally at closed functions f o r c l o s e d and to have c l o s e d graphs.. spaces whose t o p o l o g i e s their conditions We a l s o look are determined i n some measure by compact subsets. Having a l r e a d y discovered c o n d i t i o n s under which continuous and open f u n c t i o n s have c l o s e d graphs we proceed to f i n d such c o n d i t i o n s f o r a c l o s e d f u n c t i o n . Example 5 . 1 X = ( 0 , 1] Let characteristic f c.E ', Y = { 0 , 1 } 1 function of X i s a closed f u n c t i o n but We note t h a t f mapping f E^ [2] into f be the Y . Nov/ does n o t have a c l o s e d graph. does n o t have c l o s e d p o i n t Following F u l l e r and.let inverses. we now d e f i n e a l o c a l l y closed function. Definition 5 . 2 . We c a l l a f u n c t i o n every nbhd. such t h a t U o f each V c U and f:X-»Y x <= X f(V). there i s a nbhd of x i s r e g u l a r then f is locally i s such t h a t A l s o i f f:X-*Y then . f i s locally f closed. V i s closed i n Y . f compact and r e g u l a r and ,c l o c a l l y closed i f f o r I f the domain o f a c l o s e d f u n c t i o n closed. • X i s locally takes compact sets onto c l o s e d sets - 36 5.3 Example Let and Y let i:X-»Y X be the r e a l s w i t h the d i s c r e t e , t o p o l o g y the r e a l s w i t h the u s u a l be the i d e n t i t y . (Euclidean) Then i topology, and i s l o c a l l y c l o s e d and i s i n f a c t continuous, but i t i s not c l o s e d . 5.^ Theorem If y a -+ q in f:X-*Y Y i s a l o c a l l y closed function implies Ls f "*"(y )cf~' "(q) that then [ c f . Theorems I 1.5 and 1.7] Proof: Let y - q i n Y and p e Ls f " ( y ) . Assume _____ p ^ f (q) . By Lemma 1.4 there e x i s t s a net x ^ i n _ L a that X y such t h a t a . Now a x -» p and Mh X - f ( l) 1 i) UcX - f Nb ~* x p eventually x - 1 (q) Nb in i s i s a nbhd and e v e n i ) =. y ^ U p of i i ) f (U) ; f(U) . u a ^y ^ c o n t r a d i c t s the f a c t t h a t y,„ i s a subnet of and s i n c e p a n ( f is w i t h the p r o p e r t i e s : i s closed i n ^ hence T h i s means t h a t i n the complement of a nbhd of Corollary w > i i s a nbhd o f c l o c a l l y c l o s e d there f (x Y . Since f(x^) is f ( x ^ ) i s eventually q , namely Y-f(U) . This = f(x, , ) -» q . T 5*5 If~„ f:X -» Y closed point inverses i s a l o c a l l y c l o s e d f u n c t i o n and has then f has a c l o s e d graph. - 37 - Proof: By assumption If y -» q in 1.5 implies that Y f then _ 1 (q) Lsf - 1 = f~" "(q) f o r every I (y )cf (q) _ 1 q e Y . which by Theorem f e U(X,Y). 5.6 Corollary I f the f u n c t i o n p o i n t i n v e r s e s and X f:X -» Y i s closed with Is r e g u l a r then f has closed a closed graph. 5.7 Corollary I f the f u n c t i o n f:X -» Y with c l o s e d p o i n t i n v e r s e s and X i s c l o s e d and subcontinuous i s r e g u l a r , then f is continuous. Proof: Apply C o r o l l a r y 5.6 3.1 5.8 Corollary I f the f u n c t i o n f:X inversley'subcontinuous is then Theorem Y i s l o c a l l y closed and w i t h c l o s e d p o i n t i n v e r s e s then closed. Proof: Apply C o r o l l a r y 5.5 Theorem 5.9 then Theorem 3.4 ^ I f '.f:X-» Y i s a f u n c t i o n where X i s regular f - 38 and locally a) f compact, the f o l l o w i n g are e q u i v a l e n t . maps compact s e t s onto c l o s e d s e t s and has c l o s e d point inverses. b) f i s locally c l o s e d and has c l o s e d p o i n t c) f has a c l o s e d inverses graph. Proof: We have a l r e a d y by C o r o l l a r y 5>5s commented t h a t ( a ) . i m p l i e s (t>) i m p l i e s ( c ) . Assume then t h a t f (b) and G i s closed. f maps compact sets onto c l o s e d s e t s . By Theorem 2.5 Also compact, the same theorem y i e l d s t h a t f s i n c e p o i n t s are has c l o s e d point inverses. We now d e f i n e some t o p o l o g i c a l spaces which a r e , i n a sense, determined by t h e i r compact s e t s , and prove two theorems concerning these .spaces and compact f u n c t i o n s . •. preserving . Definition.5.10 Let a) X for b) X be a t o p o l o g i c a l space and i s s a i d to have p r o p e r t y each i n f i n i t e AcX k^ having i s a compact subset that, and p e K X has p r o p e r t y A having p p k^ at point p point., there p. e X . pi f f as an accumulation K of AU{p] such i s an accumulation p o i n t o f at point p i f f f o r each s e t as an accumulation p o i n t , there isa K . - 39 subset c) B of A and a compact p i s an accumulation p o i n t of X has p r o p e r t y set i n X k^ k^ p p r e c i s e l y whenever i s s a i d to be. a K k^ in such t h a t B . at point f o r each compact s e t X K_>BU{p] iff U UDK i s an- open i s open i n K X . space i f X has property a t each of i t s p o i n t s . We have as r e l a t i o n s between the k-^ implies kg implies k^ spaces t h a t k_. . Theorem 5*11 Let a k-, space. 3 that f:X -» Y G A s u f f i c i e n t condition that be c l o s e d . f be a compact f u n c t i o n and l e t Y If X i s regular, f T^ be be c l o s e d i s the c o n d i t i o n i s a l s o necessary. Proof: . (Sufficiency): f('C) that Let f(C)DK KcY . Y . be compact. cnf (K). _1 f To show f(C)HK i s closed Then f~" "(K) 1 By Theorem 2,5* i s compact f[Cnf (K)] i s _1 f[Cnf (K)] = f(.C)DK. we have t h a t _1 Since i s closed i n Hence X . Key . KcY hence so i s closed i n be a c l o s e d subset of i s c l o s e d we need to prove that f o r each compact and C Let Y and hence i n i s closed. K f o r every compact - 40 (Necessity): Since are X is and f i s compact, p o i n t Thus by C o r o l l a r y 5«6, closed. f inverses has a c l o s e d graph.. U s i n g techniques s i m i l a r to the predeeding theorem • we can a r r i v e a t another c h a r a c t e r i z a t i o n of c o n t i n u i t y . Theorem 5.12 Let f u n c t i o n and f:X -» Y be an i n f e c t i v e , X a space. be continuous i s t h a t Q compact-preserving A s u f f i c i e n t condition that be c l o s e d . If Y is Tg f then the c o n d i t i o n ' i s also necessary. Proof: (Sufficiency): Let show t h a t . f~"*"(C) f (C)riK - 1 f e U(X,Y) I s c l o s e d f o r each compact f(K) . f Theorem 2 . 5 . f f (cnf(K)) - 1 Since = f each compact To (C)nK KcX . KcX . L e t (.Cnf(K)) i s infective f i s closed i n X Hence f _ 1 - 1 KcX be i s c l o s e d by [f(K)] = K so and hence i n K for i s c l o s e d and t h e r e f o r e f continuous. (Necessity): are _ 1 be c l o s e d . i s c l o s e d i t i s n e c e s s a r y to show t h a t compact, then so i s is and l e t CcY closed i n Let Y point inverses.. f be continuous and and s i n c e Also f T h e r e f o r e by Theorem 5.9 Y f "'" i s c l o s e d - be f Tg . P o i n t s has c l o s e d maps compact sets onto c l o s e d C^ i s closed. sets. - 41 Example 5.13 We w i l l now c o n s t r u c t a f u n c t i o n which i s compact and has c l o s e d graph but i s not c l o s e d . Let X be an uncountable cocountable topology f o r Let Cg X ; set. (X, C-jO or C-^ C-^ = {AcX|X-A be the d i s c r e t e topology f o r sets i n e i t h e r Let (X,Cg) be the i s countable] \J0 . X . The o n l y are the f i n i t e compact subsets of X . That t h i s i s true f o r t h a t i t i s true f o r subset o f (X,C-^) for n n whose union c o n t a i n s M will M i Then x must i(x ) =x & A , so A X-Mn = i(x ) subcollection i : ( X , C ) -» -(X,Cj) . 2 has c l o s e d graph, f o r l e t . If X-[(X-A)u{an)] i s not compact. . must e v e n t u a l l y be c o n s t a n t l y o the c o l l e c t i o n A . C l e a r l y no f i n i t e cover be a countable are l e g i t i m a t e open sets n Consider the i d e n t i t y map. The f u n c t i o n A n = 1,2,3,... - - Afl(X-{a }) = A - { a ) , so the n i s obvious, t o see l e t A. = {a-^ag,. .. } X . Take as an open cover of [ M n | M n = (X-A)U{a }} pf the (X,Cg) converged x to ( ? 1 ( x a in x ) ) ~* ( , y ) • x a (X,C^) y ^ x and so then we would have a c o n t r a d i c t i o n t o the f a c t t h a t p o i n t s are c l o s e d In The i n v e r s e of (X,^) I c a r r i e s compact ( f i n i t e ) s e t s of onto compact ( f i n i t e ) s e t s o f i s compact. Finally since (X,C-^) • C^pC-, , i (X,Cg) and thus i s not c l o s e d . ( i - k2 - Theorem.5.14 Let preserving be' f T f :-X -» Y and has c l o s e d p o i n t i n v e r s e s . and l e t X 2 be a f u n c t i o n t h a t i s compact have p r o p e r t y i s continuous a t kg Let X and Y at a point p . Then p . Proof: Suppose i s a nbhd V is a point C l x A = A f(p) e-U u i s not continuous a t such t h a t has a subset K3BLl{p} . A l s o p p] f( has f|K Tg Y is x u )k p . The v U As of p f JK:K -* Y . B . f |K 1 for a l l _ 1 Finally, i s closed since i t i s a l s o r e g u l a r and hence by C o r o l l a r y 5.7 K y e Y , is.compact, f|K is continuous. However i f we p i c k a n e t x • a x a -* p then f ( x a ) i s there i s . a compact (f|K)" (y) = f (y)nK has c l o s e d p o i n t i n v e r s e s . there collection i s an accumulation p o i n t of Tg . Then as an accumulation p o i n t . B , f o r which there C o n s i d e r the f u n c t i o n since p . such t h a t f o r each nbhd i s a nbhd of u Thus of x u f in never i n the nbhd T h i s c o n t r a d i c t i o n proves the theorem. BcK — V of such t h a t f(p) . - kj> CHAPTER 6 T h i s chapter i s devoted to an a n a l y s i s of the c l a s s U(X,Y) . We w i l l determine f o l l o w Kostyro and S h a l a t under what o p e r a t i o n s U(X,Y) i n our e f f o r t to i s closed. Composition The f o l l o w i n g example w i l l then i t does not f o l l o w t h a t f , g e U(X,Y) show that i f f o g e U(X,Y) . Example 6 . 1 Let metric. X = [0,1] , Y - Let f(x) = { and l e t g(y) '= { ^ g Then graph. g However 1/x J , both w i t h the E u c l i d e a n * f [° > f° Q i s continuous on g [ f ( x ) ] = {^/ X 2 does not have a c l o s e d graph In We now graphs 1] give f(X) ^ ° and f has a c l o s e d 0 XxY . some c o n d i t i o n s f o r the closedness of of a composition of f u n c t i o n s . Theorem 6 . 2 Let X,Y,Z be t o p o l o g i c a l spaces. Let continuous and l e t g:y -» Z have a c l o s e d graph. f:X -» Z be Then g.f e U(X,Y) . Proof: Let be a net In G^ „ G g o t „ be the graph of such t h a t v g°f and l e t (x , z ) -» (x .z ) . a - a o o' / v 5 (x ,z ) Then - 44 -> X Hence G and by the c o n t i n u i t y c o n t i t i o n , q ( f (ox ),zo ) e Gg K . K Hence zo ^( x a = <=*\ p;(f(x )) \ o' ' ) ~* ^(-^o^ ' and t h e r e f o r e ,. i s c l o s e d . cr o i 6.3 Theorem Let X.Y,Z i s compact and Z be t o p o l o g i c a l spaces such t h a t i s ' Tg . L e t be continuous, then f e U(X,Y) f(X) and l e t g:Y -• Z g°f e U(X,Y) Proof: Since by Theorem 3 . 1 f(X) that i s compact, f g°f i s subcontinuous. i s continuous. same , theorem y i e l d s t h a t Theorem 6 . 2 f g e U(X,Y) Since Z, i s has a c l o s e d -graph. T^ Hence the Hence by . . . A d d i t i o n and M u l t i p l i c a t i o n We s h a l l show t h a t , i n g e n e r a l , U(X,Y) i s not c l o s e d w i t h r e s p e c t to e i t h e r a d d i t i o n or m u l t i p l i c a t i o n . . Example 6 . 4 . X = [0,1] Let Euclidean as metric. follows: Then f Define f(x) and g - and Y = , both w i t h the functions * t ° f:X -> Y -g(x) - {"^ both have c l o s e d graphs but does not have a c l o s e d graph. Since 4 0 f o r any (x ,(f+g)(x )) a a a then . (f+g)(x - (0,0) f G f + g . ).= 0 . g:X -. Y * t % (f+g)(x) = e X , x f o r every a if x a x and ^ - 0 -* 0 , a and hence - 45 6.5 Example X = [0,1] Let f and g for a l l and Y = (-«>,+») f(x) = { J as f o l l o w s : / * t ° x xsrhich does' not have a c l o s e d graph i n f , g e U(X,Y) >< g Q = x) X f(x)-g(x) = | J * = 0 x e X . Then Although and d e f i n e f(x)«g(x) i t i s true t h a t XxY . need n o t belong t o U(X,Y) U(X,Y) f o r 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 If f e U(X E ) 3 and 1 c e E then ± c f e U(X,E- ) L Proof: c = 0 For c e E , c 4 0 1 and f e U(X,Y) . Then there i s a n e t that f(x v o the theorem i s t r u e , so assume t h a t (x ,y ) e G Suppose t h a t „ (x ,y ) -* (x ,y ) I G » . i . e . a a ^ o o cf * v 5 ) ^ y J / / ^o/c , J . Y But we then have t h a t which c o n t r a d i c t s the f a c t t h a t G^ (where y c f k U(X,E ) . 1 = c f ( x )) y ^ cf(x ) o. o' v (x,f(x a* v a'' i s closed. such and hence )) -» (x ,y K o i J o / cy ' )A Y Hence c - f e U(X,E ) . X Maximum- and Minimum If f , g e e(X,Y) i t does not f o l l o w t h a t e i t h e r max ( f , g ) or min(f,g) need b e l o n g t o Example U(X,Y) . 6.7 Let X = [0,1] and Y = (-»,+») . Define f and g 46 - as f o l l o w s : f ( x ) = (t 1 / / x X -1 - "' x f- o°' 0 1 a n s( x ) d = 0 f o r a 1 1 x e X . Then min (f,g) = - [ ^ ° £ have a c l o s e d graph i n ^Q' ^ which does not . 1 t XxY . Theorem 6 . 8 If f , g e U(X ,E ) J subcontinuous then max U(X E ) 5 and 1 both f ( f , g ) and min and' g are ( f , g ) belong to .. 1 Proof: By Theorem J>.1 both Hence both max E^ (f,g)- f and min and g are continuous. ( f , g ) are continuous. i s Hausdorff then by Theorem J . l we have t h a t both (f,g) and min ( f , g ) belong to Since max U(X E ) . a 1 Absolute Value • Theorem 6 . 9 If f:X - E and 1 f e U(X,E- ) then L |f | e U ( X E ) . 3 1 Proof: Let G|F| f e U(X,E ) i. such t h a t (x ,y ) & i n f i n i t e l y many ' a k y ' ( a )!' k = f a X = a ( x , y ) e G| j Q o f and l e t (x ,y ) a Q . o a e A , i.e.for f -( a ) ~ k x a . f ( x J= ° (x ,y ) a a be a n e t i n f(x ) > 0 If a = a^ , •k = ° = 'l ( JI ° I f on the o t h e r hand f x f for & ( x a a n ) < 1 , 2 , 3 , . . . , d 0 h f e o n r c then e infinitely . many a e A then - y 47 - - |f(x ) | = - f ( x k = |f(x )| and a g a i n Q k (x. ,y ) Q e Gj j Q - - f(x ) = -y k f . Therefore | f | e U(X,E ) 1 Convergence Pointwise The f o l l o w i n g example w i l l show t h a t U(X,Y) need not be c l o s e d w i t h r e s p e c t t o p o i n t w i s e convergence. Example 6.10 Let X = [0,1] for n = 1,2,33... n . However and and Y = E-^ x e X . f ( x ) = l i m f (x) and d e f i n e Then f f (x)= x n n e U(X,E- ) L does not belong t o for a l l U(X,E^) n-*t» since f ( x ) = .{$> * f [°>^ Almost Uniform A sequence o f f u n c t i o n s uniformly to f on a space u n i f o r m l y to f on a l l X {f" l n ^ } € n i f [ f |n e Z ] + compact subsets of converges almost converges X . Theorem 6.11 If (Y,a) X then f n e U(X,Y) , n ~ l , 2 , 3 , . . . , where are. m e t r i c spaces and lim f n-»ro f -» f (x,p) and almost u n i f o r m l y on = f e ,U'(X,Y) . Proof: Assume t h a t f f In e Z ] converges to f almost n and t h a t f e U(X,Y) . L e t % be the m e t r i c + 2 u n i f o r m l y on X D - 48 for X Y , then # X e x i s t elements ..(x ,y ) -» ( x 0 n and 6 g '& 1 ( x ,y ) o = 2 n 2 p + ^y ) ^ n 4 . 2 a Since &f n such, t h a t S ( ( x . y ) , ' 6 ) n S ( ( x . f ( x ) ) , 62) Q o 1 As . f ( x ') _> f (x ) n o' o' there e x i s t s k >_ k x x , (f (x ), f( 1 a k o )) < 0 •6((x .f(x )),(x ,f (x ))) 0 0 0 k o Q Since K' = [ x o f , x - j X g j . . . ] every i n t e g e r CT k a n d k k y Q a k, 1 h e n c e Q f o r a 1 1 o -» f k < S 2 / 1 2_ i ' k 2 0 = 0 such t h a t f o r a l l ' = ^a (f(x ) ,f (x )) 0 k 0 for all . 2 Therefore . 2 k > ^ a l m o s t u n i f o r m l y on' X , then i s a compact s u b s e t o f X . Then f o r t h e r e i s an n ^ (n <n <n^<...) 1 (x )..f(x )) < 1/k (f 6 e S ( ( x , f ( x ) ) , 62) (x ,f (x )) o k there . Then t h e r e e x i s t p o s i t i v e numbers such t h a t v f.tt U(X,Y) such t h a t 2 f o r p = 0,1,2,... . L I t f o l l o w s from t h e above t h a t If. p -> co .then >(((x ,y ), ( x , f o o p (x ))) Q < 1/k there I s a and hence k ' such t h a t f o r a l l k>k„ ( x , f ( x )) 2 — 2 ^ o' n o ' e S ( ( x , y ) , o ) . Hence from 1 (x - f ( x )) 4 S ( ( x , f ( x ) ),' 6 k • f o r k_>kg w h i c h c o n t r a d i c t s 2 . v 1 fc o o i Q The f o l l o w i n g v / i l l show t h a t t h e converse o f t h e p r e c e e d i n g theorem I s n o t n e c e s s a r i l y t r u e . Example 6.12 Let X = [0,1] and d e f i n e f n for n - 1,2,3,... all x e X . Now = -f x L 1 / - 1 / n = 0 x e (0,1] l i m f (x) = f ( x ) = 1 f o r n-+<= f e U ( X , E ) and f e U ( X , E ) f o r a l l . Then 1 1 n- p _ 49 But since sup I x ^ ^ - l | = +» xeX • . not almost uniform. • I n what f o l l o w s t o p o l o g i c a l spaces, then i f [ X |a e A), i s any system o f X X GeA of the sets X i t f o l l o w s t h a t convergence i s will-, be the C a r t e s i a n product a and TT X w i l l be the t o p o l o g i c a l product . aeA s u p p l i e d w i t h the product topology. a a The f o l l o w i n g i s a w e l l known lemma f o r which no proof i s given. :: . 6 . 1 3 Let [X |a e A] and. set X = T T aeA where x each = i D a e A x a • x be a system o f t o p o l o g i c a l spaces The n e t {x |b e B] b of points i n . X a | a e A} the n e t converges to x = ( [x^ jb e B, x^ e X } x a la e A] i f f for converges to & x & Theorem 6.14 Let ' ( X | a e A] Q and on a space. X ^ o a e X a . Let G f a define functions f :X -» X a o a • = {(x,f ( x ) ) | x e X } |s = 0,a ; a e A] . .TT{X be a system o f t o p o l o g i c a l spaces f o r each be c l o s e d sets i n Then the f u n c t i o n f:X -» X X ° aeA has a graph G that s a d e f i n e d by f(x) = (f^(x),f (x),...} 2 i s closed i n X f x(X X ) . aeA Proof": We s h a l l show that G^ cz G . L e t x e G^ , x = (x . f x l a e A]) f o r some x e X '. There i s a net' o <• a ' ' o o f v 1 . - 50 {x |b e B, x b such that x b b = ( x j , { x j | a ' e A})} -* x . of points i n Note t h a t x x a a for a l l a e A . b - {x} f x° = f ( x ) f o r a l l a a. o f o r a l l a e A . Hence a e A b v By the lemma 6.13 a G y b b f (x ) •* x Also x -• x . Therefore a o. a •• o o ( x , f f ( x ) | a e A}) - (x J x la e A}) = (x , f f (x )|a e A}) . ^o^a^o, o a o ^ a o v b b 1 v b J Since x G„ i s c l o s e d f o r a l l a, (x , f f (x )|a e A}) f o ^ a o a = (x , f ( x ) ) . Hence G„ i s c l o s e d . o o I P v V v J v Theorem 6.15 Let Let X o [X la e A] be a space and f o r each ^ "co be subcontinuous. Then be a system of t o p o l o g i c a l spaces. f Define a e A f:X -» o define X X a aeA f :X -» X a o a as m Theorem 6.14. A i s subcontinuous. Proof: Suppose f Is not subcontinuous. T h i s i m p l i e s t h a t there i s a n e t f x | b e B] i n x such t h a t f x 1 -» x e X o o o o o but ..o subnet of {f (x ) |b e B] converges i n x X . Since b b 1 f o r any f i x e d b e B f o r a t l e a s t one subnet i n x f a c t that a f a f 3 ° b f(x ) = a e A , Q Cf ( b)l , a x 0 a e A] a e J A a t h i s means t h a t f f ( x ) | b e B] has no convergent a o (by Lemma 6.13) . But t h i s c o n t r a d i c t s the b i s subcontinuous f o r each Combining If L a e A . Theorems 3»1* 6.14, and 6.15 we see t h a t i s d e f i n e d as i n the two theorems above then continuous. f is - 51 - BIBLIOGRAPHY 1. Dugundj'i, J . , 2. Fuller, 3.. H u s a i n , T.,- 4. K o s t y r k o , P. a n d S a l a t , T., R.V., Topology, A l l y n .1966. and Bacon I n c . , B o s t o n , R e l a t i o n s Among C o n t i n u o u s a n d 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 o f M a t h e m a t i c s , V o l . 2 5 , No. 3 , ( 1 9 6 8 ) , PP 495-509. Almost Continuous Mapping, R o c z n i k i P o l s k i e g o Towarzystwa Matematycznego, X, ( 1 9 6 6 ) , pp 1-7. o &VHK.UHax., r P A $ b ! C a s o p i s Math., V o l . 89, KQToPblX ( 1 9 6 4 ) , pp 4 2 6 - 4 J 1 . i 5• Ci ^ V H K U H f l X . P P A $ b l .^AMK.HYh ^OTOPblX MHO^ECTBAMM JL flBTTjaiOCfl , ? A c t a f a c . r e r . n a t u r . U n i v . Comenian, M a t h 10, No. 3 , ( 1 9 6 5 ) , PP 51-61}. 6. K o s t y r k o , P., A N o t e On F u n c t i o n s ¥ith C l o s e d G r a p h s , C a s o p i s M a t h . , V o l . 9"4, ( 1 9 6 9 ) , PP 2 0 2 - 2 0 5 . 7. L o n g , 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 C o n v e r g e n c e o f N e t s o f S e t s , M a t h . 45 ( T 9 5 8 ) , PP 237-2T6". : Fund
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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 |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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