UBC Theses and Dissertations

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UBC Theses and Dissertations

Functions with closed graphs. Leitch, F. Jonathan 1970

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FUNCTIONS WITH CLOSED GRAPHS  •  by  F. JON 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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