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UBC Theses and Dissertations

Generalization of topological spaces Lim, Kim-Leong 1966

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A GENERALIZATION OF TOPOLOGICAL SPACES by  B.Sc,  KIM-LEONG LIM Nanyang U n i v e r s i t y , S i n g a p o r e , 1964.  A THESIS SUBMITTED I N PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n t h e Department of MATHEMATICS  We a c c e p t t h i s t h e s i s as conforming t o t h e required standard  THE UNIVERSITY OF BRITISH COLUMBIA August,  1966  In presenting  t h i s t h e s i s m p a r t i a l f u l f i l m e n t of the requirements  for an advanced degree a t 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,  1 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 my w r i t t e n permission.  Department of  Mathematics  The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8 , Canada Date  September  13, 1966.  (ii)  ABSTRACT Given a set subsets of  X .  X ,  let  P(X)  A nonempty s u b - c o l l e c t i o n  c a l l e d a generalized topology (X, tl  )  for  We  X  and  *U, of  P(X)  tC -open and  define  topology  Elements of  U  an  are s a i d  t h e i r complements are s a i d t o be  tc - c l o s u r e ,  the n a t u r a l way.  is  the o r d e r e d p a i r  i s c a l l e d a g e n e r a l i z e d t o p o l o g i c a l space or  a b s t r a c t space or s i m p l y a space. t o be  be the c o l l e c t i o n of a l l  tc-limit  point,  Most of the b a s i c n o t i o n s  are d e f i n e d a n a l o g o u s l y .  and  y_-closed. so on i n  i n point set  I t i s e x p e c t e d t h a t many  important r e s u l t s i n p o i n t set topology  w i l l not be c a r r i e d  over and a number of i n t e r e s t i n g p r o p e r t i e s w i l l be l o s t or weakened.  Nevertheless,  some of them w i l l s t i l l h o l d  d e s p i t e the absence of the f i n i t e  i n t e r s e c t i o n axiom and  a r b i t r a r y u n i o n axiom f o r the c o l l e c t i o n of The  true the  subsets.  p r i m a r y o b j e c t i v e of t h i s t h e s i s i s t o i n v e s t i g a t e  which theorems i n p o i n t s e t t o p o l o g y our more g e n e r a l s e t t i n g .  s t i l l remain v a l i d i n  A secondary o b j e c t i v e i s t o  provide  some counterexamples showing c e r t a i n b a s i c r e s u l t s i n p o i n t set topology  t u r n out t o be f a l s e i n the s e t t i n g .  be noted t h a t o t h e r b a s i c n o t i o n s which are not here a t a l l can be d e f i n e d s i m i l a r l y . a t t a i n d e s i r a b l e and c o n d i t i o n s must be  I t should  discussed  However, i n order  interesting conclusions, a d d i t i o n a l  imposed.  to  (iii)  TABLE OF CONTENTS  Page INTRODUCTION  1  PRELIMINARIES  3  SECTION 1:  Generalized  SECTION 2:  Mappings o f Spaces,  5  T o p o l o g i c a l Spaces Subspaces  and  Spaces I n d u c e d b y Mappings  SECTION 3 :  13  S e p a r a t e d S e t s , Connectedness and L o c a l Connectedness  SECTION 4:  19  Compactness, L o c a l Compactness and Countable Compactness  SECTION 5:  26  L i n d e l o f Spaces and Axioms o f Countability  35  SECTION 6:  Axioms o f S e p a r a t i o n  4l  SECTION 7:  I n v e r s e Image Spaces  54  BIBLIOGRAPHY  59  (iv)  ACKNOWLEDGEMENTS  A g r e a t debt o f g r a t i t u d e i s acknowledged t o Dr. D. Drake o f t h e Department a s s i s t a n c e and encouragement  o f Mathematics f o r h i s  during the p r e p a r a t i o n of  t h i s t h e s i s and t o Dr. H. A. T h u r s t o n f o r e x a m i n i n g t h e proofs. I a l s o w i s h t o thank t h e E x t e r n a l A i d O f f i c e o f Canada f o r t h e s c h o l a r s h i p w h i c h made i t p o s s i b l e f o r me to c a r r y out the necessary research.  1. Introduction As a g e n e r a l i z a t i o n o f the s e t o f r e a l numbers, a t o p o l o g i c a l space i s d e f i n e d  i n terms o f the f i n i t e  inter-  s e c t i o n axiom and the a r b i t r a r y union axiom f o r the c o l l e c t i o n of open s e t s . and  In order  t o l e t i t be as general  so t o have as few axioms as p o s s i b l e , we a r e l e d t o d e f i n e  the n o t i o n o f " g e n e r a l i z e d  t o p o l o g i c a l space".  Given a s e t X , of subsets o f for  as p o s s i b l e ,  X  X .  Then  l e t %L be a nonempty  (1 i s c a l l e d a g e n e r a l i z e d  and the ordered p a i r (X,  t o p o l o g i c a l space. be nonempty.  collection  £_ )  topology  i s called a generalized  Note t h a t we r e q u i r e only the c o l l e c t i o n  I n other words, f o r any a b s t r a c t s e t we may  suppose t h a t we know which subsets a r e t o be considered  as  open s e t s . The  most i n t e r e s t i n g f a c t i s t h a t we are able t o  d e f i n e most o f the b a s i c notions prove g e n e r a l i z e d  of point  s e t topology and  forms o f some o f the important theorems o f  p o i n t s e t topology although we do not r e q u i r e that t h i s t i o n be c l o s e d under f i n i t e  collec-  i n t e r s e c t i o n and a r b i t r a r y union.  I t i s expected t h a t many important theorems may not be  c a r r i e d over and some o f them can be weakened.  be a l s o noted t h a t other n o t i o n s , quotient defined  l i k e product space and  space, which a r e not d i s c u s s e d similarly.  I t should  However, i n order  here a t a l l can be  t o a t t a i n d e s i r a b l e and  i n t e r e s t i n g r e s u l t s , a d d i t i o n a l c o n d i t i o n s must be imposed.  2.  For i n s t a n c e , the r e q u i r e m e n t s t h a t the empty s e t and each ground s e t  X  be members o f  1A  a r e n e c e s s a r y f o r the  p r o j e c t i o n mapping t o be open i n t h e case o f p r o d u c t space. F i n a l l y we i n c l u d e here a s e c t i o n f o r i n v e r s e Image spaces.  The r e l a t i o n s between a space and i t s i n v e r s e image  space under a mapping a r e s t u d i e d . [  3.  Preliminaries We assume here the b a s i c p r o p e r t i e s o f s e t t h e o r y . Throughout the t h e s i s we w i l l use the c o n t r a c t i o n " i f f " f o r the phrase  " i f and o n l y i f " . Let  X  be a s e t and  be a c o l l e c t i o n o f s u b s e t s o f  A  a subset o f  X .  X .  Let  3  We w i l l adopt the f o l l o w i n g  conventions: cA = X •- A ,  1)  complement o f  A  with respect to  X  A - 35 = •{A - F:> F€3] iii)  A f l 5 = {A fl F: Fe3}  iv)  A l j 3 = {A (J F: Fe3} {A}10 = {B: B=A  v) vi)  or  Be3]  U 3 = U { F : Fe3}  vii)  n  3 = n{F:  viii)  {F  ix)  1 S  Fe3} .,F },  F ,.  {F : a  2  n  aeA) ,  where  where F  tx  F  c X  i  c X  f o r each  f o r each  a  i n an i n d e x s e t A .  Theorem. subsets of  X .  Let  Let 3  X  be a s e t and l e t A  and  B  be a c o l l e c t i o n o f s u b s e t s o f  be X .  Then i)  A 0. (U3F) = u(ATl3r) ;  ii)  A u (na) = n(Aua) ;  ill)  X - U3 = fl(X-3) ;  iv)  X - 03 = U(X-3f) ;  v)  U3 = 0  and  03 = X ,  i f 3  i s the empty c o l l e c t i o n :  4.  and vi)  A - B = An(X-B)  Theorem. mapping o f of  X  A c X  into  and l e t and  8  X  Y .  and Let  Y  be a c o l l e c t i o n o f s u b s e t s o f  B _ Y .  ii)  f ( n _ ) c n { f ( F ) : F€5] ;  v) vi) vii) viii)  be a  Let  Y  Then  f(U») = U { f ( F ) : Fe_} ;  iv)  be s e t s and l e t f  be a c o l l e c t i o n o f s u b s e t s  ~  i)  iii)  Note.  X  Let  #  f ( U 8 ) = U { f " ( B ) : Beft) ; _1  f  1  (na) = n[f  _ 1  f (Y-B) _ 1  - 1  (B):  = X - f  _ 1  Bee}  ;  (B) j  f" (f(A)) = A  f o r each  Ac  X  i f f f  is  f(f  f o r each  B c Y  i f f f  i s onto ;  1  f  _ 1  (B)) = B  (B) = 0  _ 1  We have adopted  i f f BHf(X) = 0 . the f o l l o w i n g c o n v e n t i o n s :  1)  f(A) = (f(a):  2)  f " ( B ) = {x: f ( x ) - b , 1  aeA}  ;  and  beB) .  1-1 ;  5.  SECTION 1: The  present  G e n e r a l i z e d T o p o l o g i c a l Spaces. s e c t i o n i s devoted t o the n o t i o n of  g e n e r a l i z e d t o p o l o g i c a l spaces and  other r e l a t e d notions  as open s e t s , c l o s e d s e t s , c l o s u r e and  so on; t o g e t h e r  such  with  some b a s i c r e s u l t s .  D e f i n i t i o n 1.1 P(X)  Let  X  be an a r b i t r a r y s e t and l e t  be the c o l l e c t i o n of a l l the s u b s e t s  a nonempty s u b c o l l e c t i o n of  P(X)  g e n e r a l i z e d topology  and  for  X  .  Then  of U  the ordered  X .  Let  V.  i s called pair  be  a  (X, IL  )  i s c a l l e d a g e n e r a l i z e d t o p o l o g i c a l space or s i m p l y a space. The  elements o f  if  li  we  lc  are s a i d t o be  <i_-open or s i m p l y open  i s c l e a r l y u n d e r s t o o d from the c o n t e x t ; i n t h i s case  can s i m p l y speak o f the space  X .  I t i s o b v i o u s from the d e f i n i t i o n t h a t each t o p o l o g i c a l space i s a space but the converse- need not be t r u e i n g e n e r a l . Thus any  statement which i s t r u e i n g e n e r a l i z e d t o p o l o g i c a l  spaces i s t r u e i n t o p o l o g i c a l spaces. D e f i n i t i o n 1.2 C c X  i s s a i d t o be -open.  i s no  We  confusion.  Let V.-closed  s i m p l y say t h a t  (X, li  )  be a space.  i f f the s e t E  X - C  A set is  i s c l o s e d whenever t h e r e  D e f i n i t i o n 1.3 Ac  X\  The s e t  % - c l o s u r e of  (X, 26  Let  A" = fl{ C: A c A  C ,  )  be a space and  X - C€ tt }  or s i m p l y the c l o s u r e o f  i s c a l l e d the  A .  An immediate consequence o f the above d e f i n i t i o n i s Theorem 1.4  For e v e r y s e t  A  i n a space  (X, ii ),  A rz A" . We now prove a theorem which i s o f g e n e r a l  utility  i n many o f the l a t e r theorems. Theorem 1.5 (X, U. ) ,  then  If  Since  £C: A c C , X - Ce U } , .  i s a c l o s e d s e t i n a space  A" = A .  Proof:  Theorem 1.4  A  Hence  A  i s an element o f the c o l l e c t i o n we have  AcA  .  But  A c A~ ,  A" = A .  The converse o f the p r e c e d i n g  theorem i s t r u e i n  t o p o l o g i c a l s p a c e s , and so i t might be c o n j e c t u r e d A  i s a s e t i n a space  A  i s closed.  our g e n e r a l  since  such t h a t  A~ = A ,  then  setting, for  i s a space w i t h 0  (X, tt )  that i f  T h i s , however, i s e a s i l y seen t o be f a l s e i n  Example 1.6 (X, U )  by  Let  X ^ 0" and  X" = X .  But  fl = {X} . X  Then  i s not c l o s e d ,  i s not open. The above example a l s o shows t h a t the c l o s u r e o f i  a s e t I n a space  (X, <IL )  need not be  closed.  7.  Theorem 1.7  Por e v e r y s e t A  i n a space  (X, £_ ),  A" - A" . Proof: that  A  A" c A~ _  By Theorem 1.4,  c A" .  =  I f there are closed sets containing set containing  A .  c C ,  A  and  A" c C  X - Ce  and  Thus e v e r y s e t .  let C  be any c l o s e d  %  by the d e f i n i t i o n o f c l o s u r e .  ii)  ii)  A ,  That i s :  A c C  i) Then  we s h a l l prove  C  Therefore  X - Ce «• .  satisfying  i ) also  satisifes  Hence t h e i n t e r s e c t i o n o f a l l s e t s s a t i s f y i n g  contained  i n the i n t e r s e c t i o n of a l l sets s a t i s f y i n g  That i s t o say ,  A~ c A" ,  Then c l e a r l y  A  =  = X  (X, li )  1.8  such t h a t Proof:  If  A  and  A c B ,  then  B  then  n[C: B c  theorem.  a r e s e t s i n a space  A" c B" .  A" = D{C: A c C , c  A ,  also.  We now prove a s i m p l e b u t i m p o r t a n t Theorem  i) .  by the d e f i n i t i o n o f c l o s u r e .  I f t h e r e a r e no c l o s e d s e t s c o n t a i n i n g A" = X .  i i )is  C ,  X - Ce IL } X - C€ H  ]  = B As an a p p l i c a t i o n o f t h e above theorem, we have C o r o l l a r y 1.9 s e t s i n a space  (X, IL  Let ) .  A  Then  and  B  be any p a i r o f  8.  i) ii)  (A U B ) ~  A" U B" c  Since  and  A c  B" c  (A U B ) ~  and  c A" 0 B" .  (A fl B ) " Proof:  A" c  c (A U B ) ~ j  A" U B"  .  A U B  and  (A U B ) ~ ,  The  proof  B c A U B ,  b y Theorem  of  i i )  1.8  need  in  i i )  not hold  need  c a n be  not hold.  easily  Example 1.10 {c,a} and  , {a,b}} , B" = {b}  implies which  that  ,  and  so t h a t  ,  .  that  the  following ,  H  B = {b}  .  Then  = {a,b}  .  But  Thus t h e r e  i)  = ({b,c} A" =  {a}  A U B =  Proof:  e x i s t spaces i n  If  (X, IL )  i s a space  such  that  F o r any  A c  X  ,  X - A" = X - 0  [B: AcB  = U{X-B: A c B  This notion  X - 0" = U{X  i s perhaps  - B:  X - Befl] = U % =  , X-Bett}  , X-Be  called  a  x  X  1.12  the p r o p e r time t o i n t r o d u c e  -neighborhood i f f there  exists  A  set  or simply Ue U  N  the  i n a space  (X, IL  a neighborhood  of a  such t h a t  V. }.  X , i . e . 0~ = 0 .  neighborhood. Definition  of  {a,b}  <f - 0 .  then  In p a r t i c u l a r ,  ,  (A U B ) " .  Theorem 1.11 X = (JU  ,  A" U B"  (A u B ) ~ = X  A" U B" ^  X = {a,b,c}  A = {a}  Hence  That the e q u a l i t y i n  seen from the  Let  have  is similar.  I t h a s b e e n shown i n t o p o l o g i c a l s p a c e s equality  we  xeU e  N .  )  is  point  9.  An immediate consequence o f the d e f i n i t i o n i s the following Theorem 1.13 (X,  it )  f  U  If  i s an open s e t i n a space  t h e n i t i s a neighborhood o f each o f i t s p o i n t s . Proof;  F o r each p o i n t  x  of  U ,  xeU c U .  The converse o f the above theorem need not be t r u e . Example 1.14 and N  N = X .  Then  X - {a,b} ,  Let  a e {a} c  N  and  <2L = {{a) , {b}}  b e {b) c N .  Hence  i s a neighborhood o f each o f i t s p o i n t s but not open. B e f o r e we c o n t i n u e w i t h our s t u d y of g e n e r a l i z e d  t o p o l o g i c a l s p a c e s , i t w i l l be u s e f u l t o have the f o l l o w i n g definition. D e f i n i t i o n 1.15 (X, H  ) .  Then a p o i n t  Let  A  be a s e t i n a space  of  X  i s called a  x  p o i n t or simply a l i m i t p o i n t of of  x  contains a point of  of a l l l i m i t p o i n t s of  A  (X, IL ) ,  A' c A .  Proof:  Since  d i s t i n c t from  A If  then  i f f each neighborhood  i s c a l l e d the  s i m p l y the d e r i v e d s e t o f Theorem l . l 6  A  A  A  A ,  a l i m i t point of  Hence  A .  x .  and i s denoted by A  The s e t  t_-derived set or A' .  i s a c l o s e d s e t i n a space  i s closed,  X - A  so i s a neighborhood o f each o f i t s p o i n t s . c o n t a i n s no p o i n t s o f  it-limit  i s open and But  so t h a t no p o i n t o f A» fl (X - A) = 0  X - A X - A i.e.  is A'cA.  10.  C o n s i d e r t h e f o l l o w i n g example the  converse o f the p r e c e d i n g Example  Then 0  A'=  0 ,  1.17  theorem i s f a l s e .  Let  X = {a}  A' c  A  l.l8  If  A  A c  B ,  so t h a t  o f a space i n which  .  ,  But  U  A  = {X}  and  A = X .  i s not closed,  since  i s n o t open.  Theorem (X, U  )  such that Proof:  neighborhood and  hence  A c  B  .  Let  of  x  x  point  sets  be a p o i n t point  of  B  i s a point  Corollary i n a space  B  1.19  of  of  of  i n a space  B' . A'  A  .  Then e a c h  distinct  distinct  from  x ,  from  x  since  B» .  Let  (X, i t ) .  are sets  A' c  then  contains  contains Thus  x  and  A  and  Then  B  be any p a i r o f  A' U B' c  i)  (A U B ) ' ; and /  ii) Proof: same r e a s o n i n g  This  Example  Let  1.10  A = {a}  A' U B' = Hence  0.  ;  1.20  B = {b)  But  A' U B' ^  A  .  and  Then  U B = {a,b}  (A u B ) ' .  c  A '  fl  B«  .  i  A' U B' ^ (A U B ) ' .  (X, it )  i . e . X = {a,b,c} and  '  .  spaces i n which Let  n B)  f r o m t h e above theorem by t h e  a s i n C o r o l l a r y 1.9  There e x i s t  Example  follows  (A  be t h e s p a c e g i v e n i n U  = {{b,c}, {c,a},{a,b}}.  A' = 0 = B' ,  and s o  ,  so  that  (A U B ) ' = {c} .  11.  A n o t h e r a p p l i c a t i o n o f Theorem 1.18 Theorem 1.21 A  U  F o r every s e t A  i s the f o l l o w i n g  i n a space  (X, U. ),  A' c A' . Proof:  set i n X  Suppose t h a t t h e r e does n o t e x i s t a c l o s e d  containing  A u A' c A"". containing  .  A~ = X , C  A' c C  Then  so t h a t  i s a closed set i n X  c C ,  by Theorem l . l 8 and  A' c A" ,  Therefore  A U A' c A" ,  Hence  Then  Now suppose t h a t  A .  Theorem 1.16  A .  by Theorem 1.4  since  C  .  We make t h e f o l l o w i n g o b s e r v a t i o n s : spaces i n w h i c h  i ) A U A' ^ A" ,  i s arbitrary.  There e x i s t  f o r some s u b s e t  A  of  the space; o r ii) subset  A  t h e s e t A U A'  o f t h e space. Example 1.22  and  A = (b) .  Then  A U A' = fa,b,c} .  i n Example 1.10;  L e t X = £a,b,c,d) , % = A" = X  Let  A' = {a,c} ,  [ {a,b} ,{b ,c}} so t h a t  A U A' ^ A" . (X, ft )  i . e . X = {a,b,c)  L e t A = {'a,b) .  A U A' = X ,  and  Therefore  Example 1.23  Ca,b)} .  i s n o t c l o s e d , f o r some  Then  be t h e space  and  IL = { { b , c ) , {c,a}  A' = [ c] ,  which i s n o t c l o s e d s i n c e  given  0  so t h a t  i s n o t open.  To end t h i s s e c t i o n , we w i l l i n t r o d u c e t h e n o t i o n o f "a base f o r t h e g e n e r a l i z e d t o p o l o g y " and g i v e an a l t e r n a tive characterization f o r i t .  12. D e f i n i t i o n 1.24 subcolleetion  fl  each  Ue 11 ,  that  U = Ufl* .  of  IC  (X, it )  be a space.  i s c a l l e d a base f o r ll  there e x i s t s a s u b c o l l e e t i o n fl.  fl  of  and each p o i n t  x  U  Let  (X, it )  of  U ,  i f f for  be a space.  i s a base f o r u  A subUe 11  i f f f o r each  there e x i s t s  A  o f fl. such  1  Theorem 1.25 collection  Let  V € fl such  that  xeV„CU . Proof:  L e t Ue ti  f o r some ft' c fl .  U = Ufl' , i . e . xeV  Necessity.  cU  f o r some  n o t h i n g t o prove. there e x i s t s  L e t Ue It  Thus assume  V e fl such t h a t x  U V xeU  .  I f U = (J ,  U ^ 0 xeV  and l e t  V eft',  c U .  ,  so t h a t  U =  U V xeU  there i s  xeU .  Then  (J V c U . xeU V e fl f o r  Hence  x  x  each  f o r some  Then  X  Sufficiency.  U c  xeV  xeU .  V efl .  X  But  Hence  and  x  ,  x  where x  xeU . Remark:  "interior",  One can a l s o d e f i n e  " e x t e r i o r " and "boundary"  the n o t i o n s o f  o f a s e t i n a space.  13.  SECTION 2:  M a p p i n g s o f Spaces,. S u b s p a c e s Induced by  rest new  section deals with  space  i n t o another; w h i l e the  s p a c e s f r o m g i v e n ones.  simply  2.1  (Y, t r  i n t o a space  A mapping  )  i s said  c o n t i n u o u s i f f Ve V  f  o f a space  t o be  ii-V  implies  Theorem 2.2 i n t o a space set  F ,  f  _ 1  (Y, V (F)  Y f  _ 1  .  Then  (Y) - F  )  )  - 1  (V)€  i s the  o f a space  il  following (X, lc ) _r-closed  ^-closed.  Necessity.  Y - F  _ 1  f  f  i s continuous i f f f o r each  is  Proof; in  A mapping  (X, U.  continuous or  that  A characterization of continuity  Let  F  v-open.  is  (F) = X - f  _ 1  (F)  be any Hence  is  <_r-closed s e t  f ~ ( Y - F) 1  __-c-pen,  so t h a t  f  _ 1  (F)  u-closed.  is  Sufficiency. Then  Y - V  f  - V) = f  - 1  various  o f i t d e a l s w i t h s e v e r a l b a s i c methods f o r c o n s t r u c t i n g  Definition  -  Spaces  Mappings.  The b e g i n n i n g o f t h i s t y p e s o f mappings f r o m one  and  (Y  so t h a t  is _ 1  V  <y"- l° cL c  1  is  H  (V) = X - f  it-open.  Theorem 2.3 (X,  _ 1  )  be any  If  «_r-open s e t i n  Y .  By h y p o t h e s i s ,  se  (Y) - f  f" (V)  from a space  Let  f  _ 1  Hence  and  i n t o a space  g  (V) f  is  l_-closed,  i s continuous.  a r e c o n t i n u o u s mappings (Y,  tr  )  and f r o m  (Y,  tr)  14.  Into a space f«g  ( Z , <xJ )  r e s p e c t i v e l y , then the composite mapping  i s continuous. Proof:  (f'g)"  = g-^f"  1  T h i s f o l l o w s from the f a c t .  1  D e f i n i t i o n 2.4 i n t o a space  (Y, V  open i f f Ue it  )  A mapping  implies that  (Y, V  )  set  f(C)  is  of a space  IL- <lr  f(U)e V  A mapping  C  is  (X, tL )  open or simply  .  f  o f a space  i s s a i d t o he  simply c l o s e d i f f the s e t the  f  i s s a i d to he  D e f i n i t i o n 2.5 i n t o a space  that  (X, u, )  tc- <\j c l o s e d or  ^.-closed i m p l i e s  that  'y-closed.  The' next theorem i s i n analogy t o Theorem 2 . 3 Theorem 2.6 mappings from a space from  (Y, V  )  (X, It )  f  and  (X, ti )  fog  g  a r e open ( r e s p .  i n t o a space ( Z ,<AJ )  i n t o a space  the composite mapping Proof:  If  (Y, IT  )  . closed) and  r e s p e c t i v e l y , then  i s open (resp. c l o s e d ) .  Obvious.  Theorem 2.7  Let  onto a space  (Y, V  f  be a 1-1  ) .  Then  mapping of a space f  i s open i f f  f  is  closed. Proof: onto,  For each  A c X ,  since  f ( X - A) = f ( X ) - f ( A ) = Y - f ( A ) .  f  i s 1-1  and (l)  15.  Suppose t h a t closed set. is f  Then  f  X - C  ir-open, so t h a t  i s open and t h a t is  f(C)  C  is a  i^-open and hence is  V-closed,  f ( X - C)  hy ( l ) .  Hence  i s a c l o s e d mapping. Now «&-open.  is f  suppose t h a t  Then  X - U  V-closed.  f  is  i s c l o s e d and t h a t 1c-closed  In view o f ( l ) ,  and hence  f(U)  is  U  is  f ( X - U)  tr-open.  i.e.  i s an open mapping.  Some p r o p e r t i e s , which w i l l be d i s c u s s e d l a t e r s e c t i o n s , of spaces are not preserved open or c l o s e d mappings.  i n the  by continuous,  They a r e , however, preserved  by  the s o - c a l l e d "homeomorphisms".  D e f i n i t i o n 2.8 space a  IL-ir  f  and  (X, it )  Let  onto a space  (Y,V  homeomorphism or simply f"  1  are continuous.  a homeomorphic homeomorphisms,  be a 1-1  f  ) .  mapping o f a  Then  f  i s called  a homeomorphism i f f both  A property  o f a space i s c a l l e d  i n v a r i a n t i f f i t i s i n v a r i a n t under i . e . whenever  i s a homeomorphism o f  X  onto  X  has the p r o p e r t y  Y ,  then  Y  and  there  has the property.  As i n t o p o l o g i c a l spaces, we can d e f i n e a "subspace" of a g i v e n space as f o l l o w s : D e f i n i t i o n 2.9 Let  1L  E  '  Let/ E  = ( E n U : Ue IL } .  be a s e t i n a space  Then  U  E  i s c a l l e d the  (X,  It).  16.  w i t h r e s p e c t t o %L  r e l a t i v e g e n e r a l i z e d topology i n E the ordered p a i r  (E, & )  i s c a l l e d a subspace o f  E  and  (X, <26 ) .  Before p r o v i n g a p a i r o f u s e f u l theorems, we need the f o l l o w i n g lemma. L e t ( E , U^)  Lemma 2.10 space  (X, 26 ) .  It-  and  L e t "6 and 3  be a subspace o f a  be the c o l l e c t i o n s o f a l l  ^g-closed sets r e s p e c t i v e l y .  Proof:  U = E - E 0%L  Since  E - 3 = ft, = E fl % ,  = E0(X-^)=En^  Theorem 2.11  denote the  If A  A" ^ E = EOA"^ ,  E 0 A"  .  i s a s e t i n a subspace (E,#g) of where  A~ E U  ti-g- and the «6-closures o f A  Proof:  5 = E (l€ .  we have  E  = E -11  a space (X,ft,), then  Then  and  A~  respectively.  = E n(n{C : A c C , X - Ce<&}) _=J7»[E n C : A c C  }  X - Cett]  = 0{F : A c P , E - P€ U^} , by Lemma 2.10 . = A~^E .  Theorem 2.12 (E, &-) E  i.e. E c u A*U-  o f a space  Let A  (X, i i )  .  be a s e t i n a subspace If E  there e x i s t s a s u b c o l l e c t i o n 11' ,  then  A  l t t  denote the Ug-  respectively.  E = E 0 A'^  has a  it-open cover,  <U. o f i l }  ,  where  and the U.-derived  such t h a t  A » ^ E and  sets of A  u  17.  Proof:  ^ - n e i g h b o r h o o d of V„  may assume IL such t h a t  7  f_-,-open.  x ,  contains point of  and l e t V  be any  Without l o s s o f g e n e r a l i t y , we  = E (1 U X  d i s t i n c t from  V  x .  t o be X  A  xeE fl A ' f  Let  ,  Hence t h e r e e x i s t s  Then  U  U  in  x  contains point of X  since A  xeA'^  .  This implies that  d i s t i n c t from  x  since  A c E :  X  and so  xeA' , i . e . E D A ' ^ - c A ' <*E . Now l e t x e A ' ^ E and U be any ^.-neighborhood X of x . A g a i n we may assume U t o be ft-open . L e t X V = E fl U„ . Then V i s &-,-open and c o n t a i n s p o i n t o f A d i s t i n c t from x . Hence U, a l s o c o n t a i n s p o i n t x x  r  of  A  d i s t i n c t from  A'^-EcEDA'^  .  x ,  so t h a t  xeE H A  , i.e.  , %  A' *<-E = E 0 A <  Consequently  Two t y p e s o f spaces i n d u c e d by a space and a mapping  f  a r e d e f i n e d as f o l l o w s .  . (X,2£ )  One o f them, t h e  i n v e r s e image space, w i l l be s t u d i e d i n g r e a t d e t a i l s i n the l a s t s e c t i o n . D e f i n i t i o n 2.13 empty s e t 4*3  Z  onto a space  of subsets of  i n <OJ i f f W = f ordered p a i r (Xfit)  Let  - 1  Z  f  (X,ii)  .  as f o l l o w s :  (U)  f o r some  be a mapping o f a nonDefine a c o l l e c t i o n  A subset  W  of  Z is  Then -W ^ 0 .  Ue U .  The  (Z,-t<j) i s c a l l e d the i n v e r s e image space o f  under the mapping  g e n e r a l i z e d t o p o l o g y on  f . Z  C l e a r l y <6J i s t h e s m a l l e s t  such t h a t  f  i s continuous.  18.  Definition 2.l4 (X tc }  )  onto a s e t Y .  a subset V  B  o f subsets  tr i f f f ^ 0 . space o f  - 1  of  Y  of  (V) = U  Y  f ( B ) e IL . _ 1  as f o l l o w s : f o r some pair  A subset  U€«£. (Y, V)  under t h e mapping  l a r g e s t generalized topology Remark;  be a mapping o f a space  Suppose f u r t h e r t h a t t h e r e e x i s t s  such t h a t  The ordered (X,i6)  Let f  on  Y  Define a c o l l e c t i o n V  of  Y  is in  Then by h y p o t h e s i s , i s c a l l e d t h e image f .  Clearly V  such t h a t  f  i s the  i s continuous.  One can a l s o d e f i n e t h e n o t i o n s o f a  p r o d u c t space and a q u o t i e n t space.  However, i f one e x p e c t s  the i n v e s t i g a t i o n t o be f r u i t f u l , one s h o u l d impose some o t h e r c o n d i t i o n s on t h e g e n e r a l i z e d t o p o l o g i e s .  /  19.  SECTION 3:  Separated Sets, Connectedness and Local  Connectedness.  The p r e s e n t s e c t i o n i s mainly devoted t o some b a s i c p r o p e r t i e s o f separated s e t s and connected sets i n spaces. F i n a l l y the n o t i o n o f l o c a l connectedness  D e f i n i t i o n 3.1 (X, _.) iff  A  Two s e t s  are s a i d t o be  n B"  - 0 = A"  A  i s also  and  B  defined.  i n a space  ^ - s e p a r a t e d or simply separated  nB  .  The f o l l o w i n g theorem  i s an immediate  consequence  of the above d e f i n i t i o n . Theorem 3.2 i n a space B  (X, it ) .  r e s p e c t i v e l y , then Proof:  Let If C  A  C  and  and  and  B  D  D  be separated sets  are subsets o f  and  a r e separated.  The p r o o f i s s t r a i g h t f o r w a r d .  v i r t u e o f Theorem 1 . 8 , we have  A  CfiD"cAn  Indeed, by  B~ = 0  and  C " n D c A " n B = 0 . Theorem 3.3  If A  or both closed* i n a space and  B - A  and  (X,il )  B  are e i t h e r both open  then the s e t s  3  A-B  are separated. Proof:  c AfleBOB'n(cA)~ ,  Note t h a t by p a r t  (A - B) fl (B - A)" = ADcBfl(BO cA) ~ i l ) o f C o r o l l a r y 1.9 .  the' i n t e r s e c t i o n  (A - B) fl (B - A)"  A f)(cA)~  cB fl B~ .  and  If A  Hence  i s a subset o f both  i s open, then  cA i s  20.  A fl ( c A ) " = A 0 cA = 0 .  c l o s e d and so  cB fl B" = cB fl B = 0 .  then  i n e i t h e r case. = 0  i f B  If B  (A - B ) 0 (B - A ) ~ = 0  Hence  S i m i l a r l y , we can show t h a t  i s open o r  A  C o r o l l a r y 3.4  i s closed,  (A - B ) ~ n (B - A)  i s closed. If A  and  B  a r e d i s j o i n t and  e i t h e r b o t h open o r b o t h c l o s e d i n a space  (X,% ) ,  then  they are separated. Proof:  T h i s f o l l o w s i m m e d i a t e l y from t h e p r e c e d i n g  theorem. D e f i n i t i o n 3.5 9£-connected  A space  (X, ft.) i s s a i d t o be  o r s i m p l y connected i f f i t i s not t h e u n i o n o f  nonempty p a i r w i s e d i s j o i n t open s e t s . (X, ez.) is  E  i s s a i d t o be connected i f f t h e subspace  i n a space (E, H^)  connected. Theorem 3 . 6  empty s e p a r a t e d s e t s X  A set  If A  X ^ A U B  and  B  f o r any p a i r o f non-  i n a space  (X,<2£),  then  i s connected. Proof:  If X  exist a collection  were n o t c o n n e c t e d , t h e n t h e r e would  [U }  o f nonempty p a i r w i s e d i s j o i n t  Ct Cfc£/\  open s e t s such t h a t  X =  U U aeA  .  Let  a  f i x e d element  o f the c o l l e c t i o n .  U *  be an a r b i t r a r y  a  Then  X = U * U ( U U ) . a^a* a  Clearly  U * fl ( U  U ) " = U * fl ( U  U ) = 0 .  a  Furthermore,  21.  if  a ^ a* ,  then  which i s c l o s e d . c  n U  U~ CX  (X - U ) n U cx  ( cV_*  a  J  U  =  nonempty s e t s  U  so t h a t  = 0 ,  a  U * c X - U  Ct and so  i f a j£ a* .  , cx  Thus  cx  ( I*  U  U  rVa*  a  = 0  U~*Ct c X - U  CX Hence  cx n  U * D U  and  #  n U ) = 0 .  a  Consequently the  a  U  U  are separated.  This  c o n t r a d i c t s our h y p o t h e s i s t h a t  X  i s n o t t h e u n i o n o f two  nonempty s e p a r a t e d s e t s .  X  i s connected.  Hence  The n e x t theorem t e l l s us t h a t connectedness i s a c o n t i n u o u s i n v a r i a n t ( a p r o p e r t y p r e s e r v e d by c o n t i n u o u s onto mappings) Theorem 3.7 connected space  If  (X,iC)  f ' i s a c o n t i n u o u s mapping o f a  onto a space  (Y,ir) ,  then  Y  is  connected. Proof:  If Y  exist a collection U-open  {V  s e t s such t h a t  were n o t connected, t h e n t h e r e would o f nonempty p a i r w i s e ^ d i s j o i n t Y = UV_  .  Then  X = f  _ 1  ( Y ) = f" (UV ) 1  ry  ct  - Uf ^-,) - 1  CX  {f  - 1  .  ( V )}_._. Ot  f  i s a c o n t i n u o u s onto mapping,  i s a c o l l e c t i o n o f nonempty p a i r w i s e d i s j o i n t  CtfcA  2_-open s e t s . connected.  Since  ct  T h i s c o n t r a d i c t s our h y p o t h e s i s t h a t  Hence  Y  i s connected.  X  is  22.  Theorem 3.8  L e t G = {C^}  connected s e t s i n a space Then  C = UC^  (X, U)  be a c o l l e c t i o n o f  such t h a t  OC^ / 0 .  i s connected.  'Proof:  OC^ ^ 0 ,  Since  we may l e t  x  e r 0  lC^ .  £ V } be any c o l l e c t i o n o f p a i r w i s e d i s j o i n t  Let  2^-open  a  s e t s such t h a t  C = UVV . Then x eV f o r some a . a o a o be an a r b i t r a r y f i x e d element o f G . Then Q  Let C  \*  V C  \^ *  C^* =  C  \*  0  = C n U 0  V  r vGt =  U V  a)  , l\ *  C  =  (  U  C  \ *  where (  0  G  n  V  0  a  n  d  U e it , U  J Cc  =  C  l  *\  0  M Y .  € C  \*  U  «  €  #  all  p / a  Similarly  o  .  C. c U  and hence  C c V  = C  K #  a * o  e  f  °  r  e  a  C  h  '  0  y  .  Then t h e  #  \  fl U  or  C  #  = 0 for c U C = uC, c U '  Thus  o  K  .  i.e.  a  C  t  Then  fl  u V c V  a  a  o  we must have  for a l l a 4 a  rt  .  and so  U  a  V  / a a^a o  u V a^a a  = 0 .  c  V  a  o  Hence  Q  This implies that  u n i o n o f nonempty p a i r w i s e d i s j o i n t ly  a .  ^  Q  V_ = 0  i  a  V fl ( u V ) = 0 , °o a^a  But  r  o  f o r a l l \ ^ \* .  o  K  C  W  Q  p  x  implies that  Hence  a -  V  ^  o connectedness o f  n  Qt  = C. * 0 V  C. n U  But  o  ( C * n U ) fl (C^^ n U ) = 0  Q  X  x  f o r each  = U ( C * fl U ) and  Hence if  (  C  i s not the  2^-open s e t s .  i s connected.  \  \ \  Consequent-  23.  There i s an important p r o p e r t y o f connectedness which f o l l o w s from the next theorem. Theorem 3 . 9 ( X , V.) is  and  E  If  Proof;  where  i s a connected s e t i n a space  { rP V  Let  a€A  t  e  E - uV  E  f o r each  a .  Let  then  (CnVp) 0  (C0V ) Y  C n(E0U ) = C n U € U cx ct o  =  0.  V. = E fl u * , a  L X  Furthermore  f o r each  u (CflV ) = 0 a^a*  connected, we must have  E  U(Cnv ) = Cn(ijV ) - CnE=C.  Then  C X  p / y ,  C n V = cc  .  a  CX  If  then  c o l l e c t i o n of pairwise  a  & -open s e t s such t h a t  U e  C c E c C" ,  i s a s e t such t h a t  connected.  disjoint  C  a .  Since  f o r some  C is  a*eA .  a  Hence = C  C = C fl V * a* "  c  and so  C c V - . a*  V~#e> hy Theorem 2.11  Thus  E <= E fl C"^-  and Theorem 1.8  ; and so  CX  E = V~L*E  .  We c l a i m t h a t  t h e r e would e x i s t  a  n  ^ a*  U  V  = 0 .  E - V  isa  V  such t h a t  _L X J  since  4 0 • But  CX-j  2_-,-closed s e t c o n t a i n i n g  CX-|^  F o r i f not, then  V  #  ,  C X  V - fl V  = 0 .  Hence  E = V"^c E - V  1  CX  which i s a c o n t r a d i c t i o n . E  ,  Consequently  E = V * , i.e.  i s not the u n i o n o f nonempty p a i r w i s e d i s j o i n t  &_-open  s e t s and so i s connected. Corollary 3.10 space  (X,ii ) ,  then  C"  If  C  i s a connected s e t i n a  i s connected.  24.  Proof:  T h i s f o l l o w s from  C c C~ c C  and  Theorem 3 . 9 • D e f i n i t i o n 3.11 are  s a i d t o be  X  Points  o f a space  q  (X,  ^ - c o n n e c t e d o r s i m p l y connected i n X i f f  t h e r e e x i s t s a connected s e t i n X  containing a l l  x  a  .  In t h i s new t e r m i n o l o g y we may r e p h r a s e Theorem 3.8 as f o l l o w s : Theroem 3.12 (X,U ) . x  and  I f f o r each p o i n t  xo  X  ' Let  be a p o i n t o f a space  q  x  of  are connected, then  X  3  Proof:  F o r each p o i n t  a connected s e t c o n t a i n i n g  x  X ,  the p a i r o f p o i n t s  i s connected, x  of  and  X  a c o l l e c t i o n o f connected s e t s w i t h  q  X , .  let C  Then  n C xeX  be  ^^) eX X  ^ 0  i  and  x  X =  U C xeX  Then by Theorem 3 . 8 ,  .  x  D e f i n i t i o n 3.13 locally  x  of  X ,  (X, U)  each neighborhood o f  a connected neighborhood o f  (E, Ug)  A space  i s connected.  i s s a i d t o be  ^ - c o n n e c t e d o r s i m p l y l o c a l l y connected i f f f o r  each p o i n t  (X, it )  X  x .  A set  E  x  contains  i n a space  i s s a i d t o be l o c a l l y connected I f f t h e subspace i s l o c a l l y connected. I t i s e v i d e n t t h a t connectedness does not i m p l y  l o c a l connectedness and l o c a l connectedness does n o t i m p l y connectedness s i n c e we have counterexamples I n t o p o l o g i c a l spaces.  s  25.  Theorem 3.14 s e t s i n a space locally  (X, ft.) forms a base f o r U ,  x  of  xeV  .  contained i n  Let  X .  By h y p o t h e s i s , that  then  X  is  connected. Proof:  point  I f the c o l l e c t i o n of connected open  U  Then N .  N  be any neighborhood o f an a r b i t r a r y  Then there e x i s t s  Ue 11 such that  c o n t a i n s a connected open s e t V  V  i s a connected neighborhood of  xeU c N. such x  26.  SECTION 4:  Compactness, L o c a l Compactness and Countable Compactness.  The f i r s t h a l f o f t h i s s e c t i o n r e s u l t s on compactness. establish  c o n t a i n s the b a s i c  I n the second p a r t we s h a l l  the r e l a t i o n s h i p s  between compactness, l o c a l  compactness and c o u n t a b l e compactness. D e f i n i t i o n 4.1 space  (X, IL )  A c UG  .  G c il. £> of  G  G  of sets i n a  i s c a l l e d a cover f o r a s e t  A cover If G  A collection  G  for A  A c X i f f  i s c a l l e d an open cover i f f  i s a cover f o r A ,  i s c a l l e d a subcover o f  then a G  subcolleetion  for A  i f Jb i s a  cover f o r A .  D e f i n i t i o n 4.2  A space  (X, 2d)  i s s a i d t o be  <U -compact or s i m p l y compact i f f e v e r y open c o v e r f o r has a f i n i t e i s s a i d t o be (E,  subcover f o r  X .  A set  E  i s a space  (X,H )  «_-compact or s i m p l y compact i f f the subspace  i s compact. The f o l l o w i n g  r e s u l t i s an immediate consequence  of the above d e f i n i t i o n . Theorem 4.3 compact i f f e v e r y subcover.  X  A set  E  i n a space  id-open c o v e r f o r E  (X, U.)  has a f i n i t e  is  27.  Theorem 4 . 4 [X^] X^  Let  (X, il )  be a space and l e t  be a f i n i t e c o l l e c t i o n o f s u b s e t s o f i s compact, t h e n Proof;  UX  Let  X .  i s compact.  i  By Theorem 4 . 3 , i t  Y = \JX^ .  s u f f i c e s t o show t h a t every open cover f o r Y subcover. for  Y .  each  F o r t h i s purpose, l e t {U }  X^  theorem.  X^ .  fU J  i s compact i m p l i e s t h a t  f i n i t e subcollection  has a f i n i t e  be an open cover  Then i t i s an open cover f o r each  X^ ,  I f every  For  contains a  a  which i s a cover for X^, by the p r e c e d i n g  Hence the u n i o n o f these c o l l e c t i o n s i s a f i n i t e  subcover f o r Y .  I t might be c o n j e c t u r e d t h a t a c l o s e d s u b s e t o f compact space i s compact and t h a t compactness i s a c o n t i n u o u s invariant  i n our g e n e r a l s e t t i n g . Theorem 4 . 5  T h i s I s i n f a c t the case.  A c l o s e d s u b s e t o f a compact space  I s compact. Proof:  Let  X .  Then  X - C  C .  Then  G u{X - C)  C  be a c l o s e d s u b s e t o f a space  i s open.  Let  G  be an open cover f o r  i s an open cover f o r X .  i s compact, t h e r e i s a f i n i t e subcover ,©U{X-C}=X. for  C  and  C  Hence £  (Y, <U )  (X, U )  i s compact.  Let  G  f  X  such t h a t  i s a f i n i t e subcover o f  i s compact by Theorem 4 . 3  Theorem 4 . 6 a compact space  £> o f  Since  G  .  be a c o n t i n u o u s mapping o f  onto a space  (Y, V)  .  Then  28.  Y = U3 . is a  Proof;  Let  #  Define  G = f  be a tr-open cover f o r Y . ^ ) .  1  %-open cover f o r X  2_-compactness o f subcover <& o f  G  X  since  for X . 3  f  and so G  i s continuous.  «_r-compact, s i n c e  Let 8  be the  such t h a t  Y = f ( X ) = f(U,o ) = U f ( s ) is  X = UG ,  The  i m p l i e s that there e x i s t s a f i n i t e  f i n i t e s u b c o l l e e t i o n of  Y  Then  Then  = U8 , 8  corresponding  oO = f ( f t ) .  Then  - 1  since  f  i s onto.  Thus  is finite.  B e f o r e p r o v i n g two more i m p o r t a n t r e s u l t s c o n c e r n ing  compactness, we need t h e f o l l o w i n g d e f i n i t i o n . D e f i n i t i o n 4.7  t o have the f i n i t e  A collection  3  i n t e r s e c t i o n property  of sets i s s a i d  i f f the i n t e r s e c t i o n  o f t h e members o f each f i n i t e s u b c o l l e e t i o n o f  3  s  i s non-  empty. Theorem 4.8  A space  (X, it)  every c o l l e c t i o n of closed sets i n X  i s compact i f f which has the f i n i t e  i n t e r s e c t i o n p r o p e r t y has a nonempty i n t e r s e c t i o n . Proof:  Necessity.  Let %  be any c o l l e c t i o n o f  c l o s e d s e t s i n a compact space  X  s e c t i o n property.  H€ = 0 .  X = X-  Assume t h a t  0f=X-nfc=  U ( X - C ) .  w i t h the f i n i t e  Hence  inter-  Then {X - C : c e S }  ceg  i s an open cover f o r X  and so t h e r e e x i s t a f i n i t e  of s e t s  in  C ,C ,...,C 0  l  d  n  such t h a t  X =  number  U (X - C,) . i=l 1  29.  But then  0 = X - X  = X-  i=l c o n t r a d i c t s our hypothesis  Sufficiency. and  i=l  1  that %  n"6^  t i o n p r o p e r t y . Hence  S C  U (X - C ) =  .  1  has the f i n i t e  c o n s i d e r the c o l l e c t i o n  Q %  be any open cover f o r - {X - A : AcG]  We have  f l ^ = n{X - A : AeG}  since  G  finite  i n t e r s e c t i o n p r o p e r t y , i . e . there e x i s t ft C 1=1  such t h a t  X .  = 0 .  ±  Hence €  Hence  X  of c l o s e d  = X - u(A : AeG}  sets.  in t  intersec-  0 .  Let  i s a cover f o r  This  = X - UG = 0,  does not have the C^,...,C  X = X - ft C = (X 1=1 -1=1 n = U A.i, , where ±  1  l  A i Consequently  X  Let  c o l l e c t i o n & of sets i n  X  (X, 11 )  Let  sets with the f i n i t e each  }  with the f i n i t e  intersection  n{D~:  De&)  a  in  A  {C } _ cr aeA A  4 0  ^ 0 ,  be a c o l l e c t i o n of c l o s e d  i n t e r s e c t i o n property. and  D C~ ^ 0 aeA  Then  by h y p o t h e s i s .  a  D C aeA  I f any  i s compact. Proof:  for  , for  = 1,2, . . n .  be a space.  p r o p e r t y has the f u r t h e r p r o p e r t y t h a t X  1  i s compact.  Theorem 4 . 9  then  = X - C eG  1  and  X  i s compact, by Theorem 4 . 8  .  C~ = C Hence  C) ±  30.  J u s t as the n o t i o n o f connectedness was i n S e c t i o n 3, we may l o c a l i z e  localized  the n o t i o n o f compactness  as f o l l o w s : D e f i n i t i o n 4.10 locally x  of  A space  (X, <&)  i s s a i d t o he  ^.-compact or s i m p l y l o c a l l y compact a t a p o i n t X  i f f x  has a compact neighborhood.  X  i s said  t o be l o c a l l y compact i f f i t i s l o c a l l y compact a t each o f i t s points.  E c X  A set  i f f t h e subspace  i s s a i d t o be l o c a l l y  (E, U^)  compact  i s l o c a l l y compact.  Thus a l o c a l l y compact space has an open cover necessarily.  I t i s evident  not i m p l y compactness.  t h a t l o c a l compactness does  We show now t h a t under a s u i t a b l e  c o n d i t i o n i t i s i n f a c t t h a t case t h a t compactness i m p l i e s l o c a l compactness. Theorem 4.11  A compact space w h i c h has an open  c o v e r i s l o c a l l y compact. Proof:  Since  each o f i t s p o i n t s ,  X  Lemma 4.12 a space is  (X,U ) .  Then  X  i s a compact n e i g h b o r h o o d o f  i s l o c a l l y compact by Let  A  and  B  definition.  be any two s e t s i n  i s U -compact i f f A 0 B A  A n B  UQ-compact.  Proof: Assume t h a t  The a s s e r t i o n i s o b v i o u s i f A n B = 0" .  A D B ^ 0  s  and l e t C = A 0 B .  i t s u f f i c e s t o show t h a t the  U -compactness h  By symmetry, of  C  implies  31.  the  U-Q-compactness  of  & - o p e n cover f o r  C .  B  for  some  C c  (IJG)  c %L .  G  C .  let  Cc  U ( G O B )  Therefore  =  ( U G )  Cc  since  C  is  B ,  G fl B , so t h a t  ,  f o r some  1  -compact.  1  Thus  n (J A . i=l  Cc  A  and hence is  compact space prove.  Every c l o s e d  locally  compact.  Let  E  be a c l o s e d  (X,U)  .  If  Proof:  Hence assume t h a t  p o i n t of  E .  Since  X  E 4 0 is  Then  since  N - E  is  2^-open,  N - E = N - M ,  hence  M  is  in  X  N  is  M is a  that  E  E  locally  M  locally  locally  be an  X .  p , Let  M is  N  there  Is  M = N fl E .  ^-open  .  and  W ^ - e o m p a c t , by By Lemma 4.12,  i s a neighborhood  i s a neighborhood of  p  ^ - c o m p a c t neighborhood of p .  to  arbitrary  N - E = N fl ( X - E )  ^-compact.  l o c a l l y compact at compact.  in  N - M is  Thus is  p  compact at  p  Furthermore,  implies that  Consequently  is  since  tig-compact.  of  so t h a t  ^-closed.  Theorem 4.5 ,  p  that  there i s n o t h i n g  and l e t  locally N  subset of a  subset of a  E = 0 ,  a compact neighborhood of  is  This implies  compact.  compact space i s  M  .  1  Theorem 4.13  But  i  C c ( u A . ) fl B = U ( A . f l B ) i=l 1=1 1  C  fl  U (A,OA)  1=1 A . €G , 1 = l , . . . , n  Jb be an  Then & can be w r i t t e n as  Hence  0 A = U(GflA).  To do t h i s ,  Since  p  Is  in p ,  of E . so  arbitrary,  32.  D e f i n i t i o n k.lk be c o u n t a b l y  E  i s said to  ^-compact o r s i m p l y c o u n t a b l y compact i f f  every i n f i n i t e A set  (X, it )  A space  subset o f  X  has a l i m i t p o i n t i n  (X, U-)  i n a space  compact i f f the subspace  X .  i s s a i d t o be countably/  (E,  i s c o u n t a b l y compact.  I t i s o b v i o u s t h a t c o u n t a b l e compactness does not i m p l y compactness s i n c e we have counterexamples i n t o p o l o g i cal  spaces.  To prove t h a t c o u n t a b l e compactness i s weaker  t h a n compactness i f the g i v e n space has an open c o v e r , we a p p e a l t o the f o l l o w i n g lemma. Lemma 4.15 (Y,ir ) E  o f a space  If  E  i s a subset o f a subspace  (X, it) ,  then  Suppose t h a t  E  E  i s <_r-compact i f f  i s ^.-compact. Proof:  i s any 'Ed-open cover f o r  E .  i s V-compact and  i^ ^ae, a  Then the c o l l e c t i o n o f s e t s  {YflU ] , i s a IT-open cover f o r E , s i n c e E c Y . cr aeA Thus the 'W-compactness o f E i m p l i e s t h a t t h e r e e x i s t s a f i n i t e subcover  {YCm  }, ,  {U  }, , ct  X—x  i s a f i n i t e subcover of  i s any V-open = Y fl U Ct  of  (X, %L).  _  A  for  E .  ctcxcA  cover f o r f o r some  Ct  and hence  {Ul  ^ • • •^n  Now suppose t h a t  V  for E  y •••} n  oc ^ X—x  E  i s li-compact  E .  and  Then f o r each  U £ 2_ ,  since  (Y, V )  ^ 3 a  a  6  A  a , i s a subspace  Ct  The c o l l e c t i o n  {U } Ct  .  Ctfe/N  is a  ?_-open cover f o r  33.  E  and hence t h e r e e x i s t s a f i n i t e subcover  of  {U )  for E ,  since  E  a  is  }. ,  i s ^-compact.  Then  ( 5 U ) = 5 (YHU^ ) = U i=l i i = l i 1=1  E = Yfl E c YH E  [U  a  A compact s u b s e t , which has an  open c o v e r , o f a space i s c o u n t a b l y Proof:  compact.  I n v i e w o f Lemma 4.15 » i t s u f f i c e s t o  show t h a t i f a space  (X,tt)  i s compact and has an open  c o v e r , t h e n every i n f i n i t e s u b s e t o f X .  For t h i s purpose, l e t E  which has no l i m i t p o i n t s i n X  i s not a l i m i t p o i n t o f  set  U  Hence  Hence  i  V-compact. Theorem k.l6  in  . a  1 = 1 , • • . ,n  containing E n U  x  has a l i m i t  Then each p o i n t  such t h a t  [U } X  compact space  X ,  v  X x  of  i . e . t h e r e e x i s t s an open E 0 U  - {x} = 0 .  c o n t a i n s , a t most, the one p o i n t  S i n c e the c o l l e c t i o n  point  by any s u b s e t o f  X .  E ,  X  x  itself.  i s an open c o v e r f o r the  Xfc A  there e x i s t s a f i n i t e  subcollection  * )i-l,...,n ° x>xeX ^ c h that X = J U ^ . This i m p l i e s t h a t E = E (1 X = E n ( U U j = U ( E H U ) . i=l l 1=1 i u  f  [U  X l  v  x  Hence  E  i s f i n i t e , since i t i s a f i n i t e union of sets,  each c o n t a i n i n g a t most one p o i n t . subset of X  x  X  Thus e v e r y  infinite  must have a t l e a s t one l i m i t p o i n t and hence  i s countably  compact.  34. C o r o l l a r y 4.17  A compact space which has an open  cover i s c o u n t a b l y compact. Theorem 4.l8  A c l o s e d subset o f a c o u n t a b l y  compact space which has an open cover i s c o u n t a b l y compact. Proof: compact space  Let  E  (X,%)  be a c l o s e d s e t i n a c o u n t a b l y  and  A  B e i n g an i n f i n i t e s u b s e t of x  in  X .  Thus we have  and Theorem 1.16  .  E  A  X ,  A  has a  x e A ' ^ c E'Wc  Hence  This implies that  an i n f i n i t e subset o f  xeE 0 A  has a  E .  <&-limit p o i n t  E ,  by Theorem  1.8  = A ' « E by Theorem  , u  l i m i t point i n  E .  2.12.  Consequently  i s c o u n t a b l y compact. We end t h i s s e c t i o n w i t h a couple o f i m p o r t a n t  examples which show t h a t l o c a l  compactness and  countable  compactness are independent n o t i o n s . Example 4.19 11  Let  (X, H.)  is locally  compact, f o r i f X  s e t and l e t  X .  Example 4.20  E  the s e t  X  i s not c o u n t a b l y  X .  is infinite. A 0 E  and so  x  of  x£E'  Let  P  Let  E = X - F  X ,  {x}  be a nonempty f i n i t e and  Por e v e r y i n f i n i t e s u b s e t  i s i n f i n i t e and each p o i n t o f  X  compact s i n c e i t i s not l o c a l l y  P .  Hence  X  subset  11= {E}  A .  at n o i n t s of  i s an  .  a l i m i t p o i n t of i s not l o c a l l y  X ,  has no l i m i t p o i n t  0  I n f a c t , f o r each p o i n t  of an i n f i n i t e s e t  But  E = X - l* ]  open s e t c o n t a i n i n g x ,  Clearly  compact.  X .  i s an a r b i t r a r y f i x e d p o i n t o f  q  t h e n the i n f i n i t e s e t  X ,  be an i n f i n i t e  c o n s i s t o f those and o n l y those s i n g l e t o n s o f  Then  in  X  A E  i s c o u n t a b l y compact.  . of is But  compact  35.  SECTION 5:  LindelSf  Spaces and Axioms o f  I n t h i s s e c t i o n , we g e n e r a l i z e Lindelof  Countability.  the n o t i o n s o f  spaces and axioms o f c o u n t a b i l i t y i n t o p o l o g i c a l  spaces t o t h i s g e n e r a l s e t t i n g and prove some o f the i m p o r t a n t theorems c o n c e r n i n g them. D e f i n i t i o n 5-1 Llndelof cover.  A space  (X,^.)  i f f e v e r y open cover f o r A set  E  i n a space  i f f the subspace  (E, tig)  X  i s s a i d t o be  has a c o u n t a b l e  (X,It)  i s s a i d t o be  sub-  Lindeldf  i s Lindelof.  Thus e v e r y compact space i s L i n d e l o f , f i n i t e subcover i s c o u n t a b l e .  since  a  The n e x t theorem I s a l s o  trivial. Theorem 5.2  A set  E  I n a space  L i n d e l o f i f f e v e r y open cover f o r Theorem 5.3  Every closed  E  (X, tt)  is  has a c o u n t a b l e subset of a  subcover.  Llndelof  space i s L i n d e l o f . Proof: space  (X,il)  prove.  .  Let If  Suppose t h a t  G U [X - E}  E E  be a c l o s e d  subset of a  has no c o v e r , t h e r e i s n o t h i n g t o G  i s an open c o v e r f o r  i s an open c o v e r f o r  X .  Since  X .  for  E  Hence  and  E  G - {X - E}  E . X  JO o f  L i n d e l o f , t h e r e e x i s t s a c o u n t a b l e subcover for  Llndelof  Then is G U {X - E}  i s a c o u n t a b l e subcover o f  i s L i n d e l o f by Theorem 5.2  .  G  36.  The next p a i r o f theorems a r e i n analogy t o Theorem 4 . 4 and Theorem 4 . 6 f o r compactness. Theorem 5.4 {X } i  Let  (X,li)  be a space and l e t  be a c o u n t a b l e c o l l e c t i o n o f s u b s e t s o f  every  X  i s L i n d e l o f , then  i  Proof;  Let  UX  1  Y = UX^ .  X . If  i s LindelSf. By Theorem 5.2 , i t  s u f f i c e s t o show t h a t e v e r y open cover f o r Y c o u t a b l e subcover.  Por t h i s purpose, l e t {U_,}  open cover f o r Y . For each  X. , i  has a  X. i  be an  Then i t i s an open cover f o r each i s Lindelof implies that  {U } ct  contains  a c o u n t a b l e s u b c o l l e e t i o n which i s a c o v e r f o r X^ , Theorem 5.2  .  X^ .  by  The u n i o n o f these c o l l e c t i o n s i s a c o u n t a b l e  subcover f o r Y . Theorem 5 . 5 a L i n d e l o f space (Y,V )  Let  (X, li )  f  be a c o n t i n u o u s mapping o f  onto a space  (Y, V)  .  Then  i s Lindelof. Proof;  Theorem 4 . 6  The p r o o f i s s i m i l a r t o t h a t o f  .  Theorem 5.6  A space  (X,$£)  i s s a i d t o be  second c o u n t a b l e i f f t h e r e e x i s t s a c o u n t a b l e base f o r %L . A set E  i s a space  a b l e i f f the subspace  (X, li) (E, li„)  i s s a i d t o be second  count-  i s second c o u n t a b l e .  37.  Theorem 5 . 7 a second for  E  (Lindelof)  c o u n t a b l e space has a c o u n t a b l e Proof:  t o prove.  3  Then  Then e v e r y open cover  subcover. has no open c o v e r , t h e r e i s n o t h i n g  i s c o u n t a b l e and  In 3 ,  A  be t h e s e t o f a l l such  choose an  E c U 3 c UJD . Setting  Ap  8  such t h a t  Ap's .  Consequently  F o r each  F c Ap . L e t  Thus £> c G  &  a  f o r some  E c UG = 1)3 .  in G  E = X  and  3 = {BeB : B c A  Define  F  and  be a s e t i n  be an open cover f o r E  countable base f o r 11 . AeG} .  (X, Ui) .  If E  Let G  Let E  i s countable  i s the required  subcover.  i n t h e p r e c e d i n g theorem, we g e t  the f o l l o w i n g r e s u l t . C o r o l l a r y 5.8  Every second c o u n t a b l e space I s  Lindelof. Theorem 5.9 space i s second  (X, tc)  Let E  fig .  exists a 8  %-open s e t U ,  V = E n ( U S ) = U(EH8») 1  be a  ^ - o p e n set.  such t h a t  Then  U = UB' and  Then t h e r e  V = E 0 U .  f o r some  Since  % c ft . 1  E 0 8' c E d 8 .  I s a c o u n t a b l e base f o r ILg and so  countable.  countable  We c l a i m t h a t i t i s a c o u n t a b l e base  Indeed, l e t V  i s a base f o r U  E 0 8  be a subset o f a second  and ft a c o u n t a b l e base f o r 11 .  E D iB i s c o u n t a b l e . for  countable  countable.  Proof: space  E v e r y s u b s e t o f a second  E  Hence  Consequently  I s second  38.  D e f i n i t i o n 5.10  A space  (X, tL)  separable i f f there e x i s t s a countable s e t that  A" = X .  A set  E  i n a space  be separable i f f the subspace  i s s a i d t o be A c X  (X,u)  (E, ^_ )  such  i s said to  i s separable.  E  Although i t i s shown i n t o p o l o g i c a l spaces  that  second countable axiom i m p l i e s s e p a r a b i l i t y , i t might be true i n our case u n l e s s the space c o n s i d e r e d has  not  an  open cover. Theorem 5.11  (X,3d)  If  i s a second countable (X,G_ )  space which has an open cover, then Proof? base f o r ti . a point such  x.^  x^'s  from each  1  f o r each  i .  Let  there e x i s t s x ^ x^ ,  Let  U  A  A" = X .  Let  Then  xeX  so assume t h a t  xeB  i  c U .  But  thus each open s e t c o n t a i n i n g  whence .  of  x) c o n t a i n s a p o i n t of  xeA'  .  I t follows that  Therefore  Theorem 5.12  ;  be an open s e t c o n t a i n i n g  such t h a t  A" = X  choose  be the s e t of a l l  , 1=1,2,...  xeA c A" ,  B^e  each neighborhood  by Theorem 1.21  .  = {x^  ±  We prove t h a t then  x ,  B  A 0 B  i ,  from  be a countable  In view of the Axiom of Choice, we may  f o r some  and  i=l,2,...}  L e t ft = { B ^  i.e.  countable.  i s separable.  and  A  if  is x =  x  i  x ^ x^ x , ^eB^  then c  U  x  (and so  A  distinct  X c A u A ' c A " , X  i s separable.  Every subset, which has an open  cover, o f a second countable space i s s e p a r a b l e .  39.  I t i s a consequence of Theorem 5.9  Proof: Theorem 5.11  .  5.13  Definition  A point  x  of a space  i s c a l l e d a condensation p o i n t of a s e t i f f every open s e t c o n t a i n i n g number o f p o i n t s o f  Let  E  8 = {B^  space  uncountable  subset of a  E ,  : 1=1,2,...}  countable space  i s a subset o f  Then f o r each p o i n t  x  c o n t a i n s an  Every uncountable  base f o r tl i n a second  of  i n the  countable space has a condensation p o i n t . Proof:  that  x  E  (X,ll)  E .  Theorem 5 . l 4 second  and  X  xeE ,  be a countable  (X, il) .  and has no condensation p o i n t s . x  i s not a condensation p o i n t U  and hence there e x i s t s an open s e t  such t h a t  f o r t6,  E n U  Suppose  i s countable.  Since  1  there e x i s t an i n t e g e r  8  containing i s a base  such t h a t  and so  E 0. B.  E = U{x  : xeE) c ufETIB. : xeE} x  x  i s a l s o countable.  union of countable s e t s , so  E  ,  c U ,  xeB.  x  x  x  But which i s a countable  i s countable.  This  completes  the p r o o f of the theorem. Definition (X, t l ) .  5.15  Let  x  be a p o i n t of a  A c o l l e c t i o n 71 of neighborhoods  a l o c a l base a t a member of 72 .  x  i f f each neighborhood  of of  x x  space  i s called contains  40.  D e f i n i t i o n 5.16 be f i r s t  (X,ti)  i s s a i d to  countable i f f there e x i s t s a countable l o c a l base  a t each p o i n t of s a i d to be f i r s t first  A space  X .  A set  E  i n a space  (X,U)  countable i f f the subspace  is  (E, U^)  is  countable.  The p r o o f of the f o l l o w i n g theorem i s t r i v i a l . Theorem 5«17  Every second countable  which has an open cover i s f i r s t  Theorem 5.18 space i s f i r s t  countable.  Every subset of a f i r s t  countable  countable.  Proof: a first  space  Let  x  countable space  l o c a l base a t  x  in  X .  l o c a l base a t  x  in  E .  be a p o i n t of a subset (X,IL) Then  and  {B (x)} n  {B (x)0E} n  E  of  a countable  i s a countable  4l.  SECTION 6:  Axioms .of  Separation.  As we have seen i n the p r e v i o u s s e c t i o n s , the concept, o f a space, i f u n r e s t r i c t e d , i s t o o g e n e r a l purposes.  To o b t a i n c e r t a i n h i g h l y d e s i r a b l e  f o r many-  conclusions  about some matters, some r e s t r i c t i o n s must be imposed on the  spaces.  I n t h i s s e c t i o n s e v e r a l o f these  c a l l e d the axioms o f s e p a r a t i o n , tions are b r i e f l y  Q  there  not  y .  e x i s t s an open s e t U A set E  the subspace  The of  (E, U^)  and  which c o n t a i n s  y x  of but  is  T  Q  T  Q  .  f o l l o w i n g theorem i s a simple  characterization  TA -spaces. o  closures  points U  A space i s a T - s p a c e i f f the Q  of d i s t i n c t points are d i s t i n c t . Proof:  Suppose  of a T -space Q  containing  x  and  (X,IX.) .  y  a r e two d i s t i n c t  Then there e x i s t s an open  one o f them b u t not the other.  l o s s o f g e n e r a l i t y , we may assume t h a t It follows not  x  (X, ft.) i s s a i d t o be  i n a space  Theorem 6 . 2  set  (X, ft.) i s s a i d t o be  A space  i f f f o r every p a i r o f d i s t i n c t p o i n t s  X ,  iff  and some o f t h e i r i m p l i c a -  considered.  D e f i n i t i o n 6.1 T  restrictions,  x .  x^{y}~.  that  X - U  i s a closed  Furthermore, we see that But  xeU  but  set containing £y)~ " X - U .  xe{x}~ , by Theorem 1 . 4 ,  c  so t h a t  Without y£U . y  but  Thus  {x}~ ^ £y}~.  42.  Conversely, points not  T  and x  o f a space  y or  y.  ,  o  suppose t h a t t h e c l o s u r e s  (X,4L )  then there  of  X  y ,  are d i s t i n c t .  I f t h e s p a c e were  w o u l d e x i s t two d i s t i n c t p o i n t s  or  ( i i ) e v e r y open s e t c o n t a i n s b o t h  contains both  every closed  x  and  y .  s e t does n o t c o n t a i n  c a s e , we o b t a i n Therefore  x  s u c h t h a t e i t h e r ( i ) no open s e t c o n t a i n s  Case ( i ) i m p l i e s t h a t f o r e a c h o p e n s e t  X - U  of d i s t i n c t  X  {x}~ = {y}~ ,  is  T  x  U , t h e c l o s e d set  Case ( i i ) i m p l i e s x  or  and  y .  that  In either  which i s a c o n t r a d i c t i o n .  . o  Theorem 6.3  Every subset of a T -space i s  D e f i n i t i o n 6.4 iff  there  y  and t h e other  ,  containing  ( X , ll )  is  T  Q  i s s a i d t o be x  and  but not  x  x .  y  of  but not  A set  E  i f f t h e subspace  I n a space, i f every subset c o n s i s t -  o f e x a c t l y one p o i n t i s c l o s e d , t h e n t h e s p a c e i s  of a space  .  .  x  Proof;  one  y  i s s a i d t o b e 1^  T h e o r e m 6.5 ing  ( X , ll)  e x i s t two o p e n s e t s , one c o n t a i n i n g  I n a space ( E , Ug)  A space  f o r every p a i r o f d i s t i n c t points  X ,  T  Q  (X,ll)  Let  x  and  y  be two d i s t i n c t  i n which subsets  point are closed.  Then  .  points  consisting of exactly  X - [x]  i s an open s e t  containing  y  but not  x ,  while  X - {y}  containing  x  but not  y .  T h i s means t h a t  i s a n open s e t X  is  T, .  43.  A l t h o u g h t h e converse o f the above theorem i s t r u e i n t o p o l o g i c a l s p a c e s , i t i s f a l s e i n our g e n e r a l i z e d t o p o l o g i c a l s spaces.  The f o l l o w i n g i s a  Example 6.6 Then  X  is  T  .  1  Let  counterexample.  X = {a,b,c} , H=  But f o r each p o i n t  x  {{a], {b), {c}}.  of  X ,  {x}  is  not c l o s e d . Theorem 6 . 7  E v e r y s u b s e t o f a T^-space i s  Theorem 6.8  The i n t e r s e c t i o n o f a l l open s e t s  c o n t a i n i n g the p o i n t  x  i f f (X, H)  T  {x}  is  Proof: points  x  and  o f a space x  of  i s the s e t  .  Necessity. y  (X,tc)  T^ .  X  I f t h e r e e x i s t two d i s t i n c t  such t h a t each open s e t  U .A.  containing  x  a l s o contains  which i s a c o n t r a d i c t i o n . points  x  and  containing set  U  y  x  y  of  b u t not  containing  y  y ,  then  ye  0 U xeU etc  = {x},  Thus f o r e v e r y p a i r o f d i s t i n c t  X ,  t h e r e e x i s t s an open s e t  y .  U  S i m i l a r l y t h e r e e x i s t s an open  but not  Sufficiency.  Obvious.  Theorem 6.9  Every  x .  Hence  T,-space i s  X  T  is  T^ .  D e f i n i t i o n 6.10 be Hausdorff, or points  x  and  T  y  A space  of  X ,  there e x i s t two d i s j o i n t open  x  A set  (X,W)  i n a space  the subspace  i s said to  i f f f o r every p a i r o f d i s t i n c t  2  s e t s , one c o n t a i n i n g E  (X,&. )  (E, IL^)  and the other c o n t a i n i n g  y .  i s s a i d t o be Hausdorff i f f  i s Hausdorff.  The f o l l o w i n g two theorems f o l l o w  immediately  from the above d e f i n i t i o n .  is  Theorem 6.11  Every Hausdorff space i s  Theorem 6 . 1 2  Every subset of a H a u s d o r f f  space  Hausdorff. Proof;  Let  p o i n t s i n a subset  E  x  and  that  x€U and  1  and  and  x  V  2  yeU  2  2  .  be any p a i r o f d i s t i n c t  .  tC-open s e t s Let  are two d i s j o i n t yeV  y  of a Hausdorff space  t h e r e e x i s t two d i s j o i n t  xeV  .  V  ±  (X,tt). and  = E fl V  ±  Ug  , 1=1,2  Then such  .  Then  W^-open s e t s such t h a t  I t follows that  E  i s Hausdorff.  I t i s n a t u r a l t o ask i f a compact subset of a H a u s d o r f f space i s c l o s e d . Example 6.13 and  E = {a} .  Then  compact subset o f  X .  The answer Is g i v e n by  Let (X,U )  X = {a,b,c}  ,#='P(X) - {{b,c}  i s Hausdorff and  However,  E  E  i s not c l o s e d .  is a Thus  a compact subset o f a Hausdorff space need not be c l o s e d .  45.  D e f i n i t i o n 6.14 {x  : n=l,2,...}  n  (X,%)  which has an open cover.  ft-limit  set  U  that  x  x  Then the sequence  w i t h r e s p e c t t o IL  containing  x  x n  )  i f ff°  r  X n  n  x  is  each open  there e x i s t s an i n t e g e r  3  implies that  The  f  {x 3  or  3  p o i n t o f the sequence  n _> N  be a p o i n t and l e t  be a sequence o f p o i n t s o f a space  i s s a i d t o converge t o a  Let  N  such  -U •  f o l l o w i n g theorem t e l l s us t h a t l i m i t s o f  sequences i n Hausdorff spaces which have open covers are unique. Theorem 6 . 1 5 a  sequence  {x 3  If  and  x  y  a r e two l i m i t s o f  i n a Hausdorff space  n  (X,li)  then  3  x = y . Proof:  If  x  and  y  were d i s t i n c t .  would e x i s t two d i s j o i n t open s e t s xeU  and  yeV .  i s an i n t e g e r and M  since  f  N  x n  such t h a t  3  n > L  Thus  x = y .  n  n ^ N  converges t o  y  x  £ n  U  V  such t h a t x  implies that  there  3  x  e  U  n  x n  eV  •  fl V = j? ,  Let  L = max(M,N).  which i s i m p o s s i b l e .  A homeomorphic image o f a H a u s d o r f f space i s Hausdorff.  ,  there i s an i n t e g e r  3  implies that  implies that  and  converges t o  x  such t h a t  n _> M  Then  C )  Now s i n c e  U  Then there  In f a c t we have the f o l l o w i n g r e s u l t .  46.  Theorem 6.16  If  of a Hausdorff space {Y,V)  p o i n t s of  f  and  x  e x i s t two 1  ,  1  of  2  X  and  .  since  1-1  is  y  .  X  Thus  and  uy  t(V^)  X  not' i n  containing  y^  open.  i n a space  (E, tig)  and  (X, 1A,)  x  Since  there e x i s t two xeV  (X,11) P  in  such t h a t disjoint  respectively,  2  Consequently  Y  i s Hausdorff.  i s s a i d to be X  and  each p o i n t  d i s j o i n t open s e t s ,  the other c o n t a i n i n g  x .  A set  i s s a i d to be r e g u l a r i f f the  If of  (X, ti) X  and  there e x i s t s an open s e t Proof;  that  y  since  there  are two  2  points  subspace  i s regular.  then f o r each p o i n t  x ,  then  i=l,2 ,  i n - X\  2  f(U" )  there e x i s t two  Theorem 6.l8  x ,  U  and  A space  F , F  and  and  r e g u l a r i f f f o r each c l o s e d set of  ,  ±  i s Hausdorff implies that  containing  Theorem 6.17  E  (Y, V ) ,  be any p a i r of d i s t i n c t  2  f(x^) = y  d i s j o i n t open s e t s 1=1,2  f  y^  such t h a t  1-1  Y  one  open mapping  Then there e x i s t a p a i r of d i s t i n c t  open s e t s i n  x  and  onto a space  s  Let  Y .  i s onto and  x €U  (X ti)  i s a 1-1  i s Hausdorff. Proof;  x^  f  and  V  X - U  i s a r e g u l a r space, each open s e t  such t h a t  i.e.  V  containing  xeV" c U .  i s a c l o s e d set not  d i s j o i n t open s e t s  X - U c W ,  U  and  W  V c X - W c U .  containing such By  47.  Theorem 1.8 Thus  and  V" c ( x -W)  Theorem 1.5 ,  xeV~ c U ,  = X - W c U .  as a s s e r t e d .  The converse o f t h i s theorem i s f a l s e .  L e t us  consider Example 6.19 xeX" c X (X,tL)  I f X = {a,b}  f o r each p o i n t  t  x  of  X  and  since  U=  fx} ,  then  X~ = X . But  i s not r e g u l a r , s i n c e the only open and the only  closed sets i n X  are X  and  0  respectively.  As a consequence o f Theorem 6.l8, we have the following C o r o l l a r y 6.20 r e g u l a r space P ,  (X 1i)  If P  and  }  x  i s a closed set i n a  any p o i n t of- X  then there e x i s t two open s e t s  F c U ,  xeW  and  Proof; d i s j o i n t open s e t s  U" fl W" =  and  U  and  X ,  W c V ,  implies that  such t h a t  xeV  U" c X - ¥" ,  so t h a t  Theorem 6.21 is  regular.  since  X - V  and  P c U .  containing  i.e. X - V c X - W " .  U~ c X - V ,  such t h a t  there e x i s t two  By Theorem 6.l8 , there i s an open s e t W such t h a t  ¥  0 .  By r e g u l a r i t y o f V  U  not i n  But  x  U c X - V  i s closed.  Thus  U" fl W" = 0 .  Every subset o f a r e g u l a r  space  48.  Proof: (X.ll)  . Let  x  Let  E  be a subset o f a r e g u l a r space  be a p o i n t o f  s e t not c o n t a i n i n g  x .  a  ft-closed  such t h a t  C  does not c o n t a i n  set  two d i s j o i n t C c U  and joint  2  .  C  x .  E fl  and ¥  ¥  Let  x  and  U"  exists  Furthermore,  such t h a t  2  and x  y  E fl U~  and  F  xeU.^  a r e two d i s -  2  respectively,  Q  T^ .  be any p a i r o f d i s t i n c t Since  X  is  T  x  but not  y .  i s a closed set containing  y  but not  x „  X ,  U  T -space i s  (X,li) .  o  xeV  and  X - U c ¥ .  Therefore  a r e two d i s j o i n t open s e t s c o n t a i n i n g because  yeX - U .  Theorem 6.24  Hence  A space  i f f i t i s b o t h r e g u l a r and i s s a i d t o be  ,  Q  t h e r e e x i s t two d i s j o i n t open s e t s  D e f i n i t i o n 6.23  (X, U)  , there  i s r e g u l a r , there e x i s t  and  T -space  such t h a t  respectively,  T^  2.10  ^,-closed  containing  By r e g u l a r i t y o f  and  X  Every r e g u l a r  t h e r e e x i s t s an open s e t X - U  be a  i s regular.  points of a regular  Thus  Lemma  Since  Therefore  Proof:  F  F = E 0 C .  U-^  ( E , £__,)  and  Then by  ^-open s e t s  Theorem 6.22  V  E  ^g-open s e t s c o n t a i n i n g  and hence  i  X  (X,li) .  A set  i f f the subspace  x  is  and T  2  V y  .  i s s a i d t o be E  i n a space  (E,£__)  is  E v e r y subset o f a T^,-space i s  Tj T-, .  49-  Proof: Theorem 6.21  T h i s i s a consequence o f Theorem 6 . 7  .  Theorem 6 . 2 5 Proof:  Every  T^-space i s  Tg .  By Theorem 6 . 9 and Theorem 6 . 2 2  Theorem 6.26 and  A space I s  T^  .  i f f i t i s regular  T o  Proof:  The n e c e s s i t y i s o b v i o u s .  f o l l o w s from Theorem 6 . 2 2 and Theorem 6.11 D e f i n i t i o n 6.27  A space  The s u f f i c i e n c y .  (X,It)  i s s a i d t o be  normal I f f f o r each p a i r o f d i s j o i n t c l o s e d s e t s B , taht  and  t h e r e e x i s t two d i s j o i n t open s e t s A c u  and  B c V .  A set  E  and  i n a space  i s s a i d t o be normal i f f t h e subspace Analogous t o Theorem 6 . 1 8 ,  U  (E,  A V  and such  (X, U) i s normal.  we have t h e f o l l o w i n g  theorem f o r n o r m a l i t y . Theorem 6.28 t h e n f o r any c l o s e d s e t F ,  If F  Note t h a t  d i s j o i n t closed sets I n  X .  e x i s t two d i s j o i n t open s e t s D c W .  Since  i s a normal space,  and open s e t  t h e r e e x i s t s an open s e t Proof:  and  (X, ti)  V  U  containing F c V"~ c U .  such t h a t  D = X - U  and  F  By n o r m a l i t y o f V  V c X - W ,  and  W  we have  a r e two X ,  there  such t h a t  F c V  50.  V"c(X-W)~ = Xs e t such t h a t  The  W c X - D  = U .  Therefore  V  i s an open  P c V~ c U .  converse of the above theorem i s f a l s e .  For  instance, Example 6.29 {{a}, is  {b},  {c},  X}  Let .  X = [a,b,c]  Then the  {{b,c}, {c,a}, {a,b), 0} .  and  c o l l e c t i o n of c l o s e d (X, u)  Clearly  satisfies  the  c o n d i t i o n i n Theorem 6.28  and  0  two  d i s j o i n t c l o s e d s e t s which are not contained  in  .  But  {a,b}  sets  are two  d i s j o i n t open s e t s .  N o r m a l i t y i s not a h e r e d i t a r y p r o p e r t y , we  have counterexamples i n t o p o l o g i c a l spaces.  it  i s i n v a r i a n t under homeomorphisms. Theorem 6.30  since  However,  Every homeomorphic image of a normal  space i s normal. Proof: space  (X, t i )  and  F  and  f (Fg) _ 1  f  disjoint  are two  and two  - 1  (Y, V)  normal.  f  By n o r m a l i t y ,  respectively.  i s 1-1  Suppose t h a t  and  , open and  Then  F^  f~ (F ) 1  1  ^.-closed s e t s , s i n c e  Ug  Hence  d i s j o i n t U-open s e t s c o n t a i n i n g  since  .  ir-closed sets.  <2£-open sets  f (Fg)  a homeomorphism of a normal  disjoint  i s a continuous mapping. disjoint  be  onto a space  are two  2  Let  onto.  there  exist  containing f ^ ) F.^  and and  f two  f~ (F ) 1  1  f(Ug) Fg  are  respectively,  This implies that  Y  is  51.  D e f i n i t i o n 6.31 iff  it  (X,U)  i s both normal and  i s s a i d to be  .  ( X , i s A set  E  s a i d to be i n a space  ( E , tU^  is  We have seen i n Theorem 6 . 9 » Theorem 6.11  and  Theorem 6 . 2 5 , for  A space  that:  i = 1,2,3  .  i f f the subspace  A space i s  T^  implies i t  is  T^ ^  A q u e s t i o n now a r i s e s as to whether  T^-ness i m p l i e s the T ^ - n e s s . g i v e n i n the negative  Unfortunately,  i n general,  T^ .  the  the answer  is  as can be seen from the  following Example 6 . 3 2  Let  X = {a,b,c,d} U=  {{a},  and  {bj,  {c},  {dj,  {a,cjf,  Cb,dj] . Then the c o l l e c t i o n of a l l c l o s e d sets = [{b,c,dj, (X,4_)  is  {a,c,d},  T^ .  But i t  [a,b,d],  is  {a,b,c},  i s not r e g u l a r ,  {b,d},  {a,c}}  s i n c e there do not  e x i s t two d i s j o i n t open sets c o n t a i n i n g the c l o s e d set and the p o i n t  c  Theorem 6 . 3 3  If  it i s c l o s e d under a r b i t r a r y  then  X  is  (X,<2_) union,  s u b c o l l e e t i o n of  true i s a T^-space  such t h a t  i n the sense t h a t  the  11 i s a member of 11,  T, .  Proof:  It  suffices  to show that  let  x  point  X ,  there e x i s t s an open set  of  be any p o i n t of  X  Por t h i s purpose, y ^ x  {a,b,dj  respectively.  However the f o l l o w i n g i s  union of a r b i t r a r y  and  X .  is  regular.  For U  each containing  52.  y  but not  that {x}  x .  X-{x}=  Thus UU y^x  i s closed.  X - {xj c ,  so t h a t  U U y^x  X - fx}  c  implies  y  X - {x} i s open and  hence  y  Therefore  X  i s r e g u l a r s i n c e i t i s normal  and every s e t c o n s i s t i n g o f one p o i n t i s c l o s e d .  D e f i n i t i o n 6.34  A space  (X, «£)  i s s a i d t o be  completely normal i f f f o r each p a i r o f separated s e t s and A  B ,  A  there e x i s t two d i s j o i n t open s e t s , one c o n t a i n i n g  and the other c o n t a i n i n g  B  .  A set  E  i n a space  i s s a i d to be completely normal I f f the subspace i s completely  (X,%)  (E,?^,)  normal.  The f o l l o w i n g r e s u l t i s a consequence  of the above  definition. Theorem 6.35  Every completely normal space Is  normal. Proof;  Let  A  and  B  be any p a i r of d i s j o i n t  c l o s e d s e t s i n a completely normal space AflB"=A"nB = AnB separated.  Since  X  = 0 ,  so t h a t  and hence  X  A  and  B  Then are  i s completely normal, there e x i s t  d i s j o i n t open s e t s , one c o n t a i n i n g B  (X, U.) .  A  two  and the other c o n t a i n i n g  i s normal.  Theorem 6.36 space i s completely  Every subset of a completely  normal.  normal  53.  Proof: space  (X,H)  .  sets i n E .  Let Let  Then  Theorem 2.11 , since  E A  and  be any p a i r o f s e p a r a t e d  W  and  A fl B"^E = A fl B ~  A " * fl B = 0 . u  w  B~  fl E = A 0  u  By  3  Ac E , A fl B~* = 0  complete  normality of  Hence  B  B" E = 0  AD  Therefore  IL-open  be any subset o f a c o m p l e t e l y normal  sets  U  U fl E  containing  and  and  A  and s i m i l a r l y  V  B  3  _ a  B = 0 .  By  t h e r e e x i s t two d i s j o i n t A c U  such t h a t  V fl E  and  X  A n  a r e two d i s j o i n t  respectively.  and  B c V „  & -open s e t s E  This implies that  E  i s normal. As an immediate consequence  6.35 and  of,Theorem  Theorem 6.36 we have C o r o l l a r y 6.37  Every subset of a completely  normal space i s normal. D e f i n i t i o n 6.38 T^  A space  (X,tL)  i f f i t i s b o t h c o m p l e t e l y normal and  i n a space (E,& ) £  (X,1L)  is T  5  i s s a i d t o be  T,-  i s s a i d t o be .  A set  E  i f f t h e subspace  .  I n view o f Theorem 6.36 and Theorem 6 , 7  a  we have  Theorem 6.39  E v e r y s u b s e t o f a T^-space i s  Theorem 6.k0  E v e r y T^-space i s T^ .  Proof;  T h i s f o l l o w s from Theorem  6.35.  T,_ .  54.  SECTION 1%  I n v e r s e Image Spaces.  The n o t i o n of the i n v e r s e image space of a space i  under a mapping has been i n t r o d u c e d i n S e c t i o n 2.  The main  purpose of t h i s s e c t i o n i s t o i n v e s t i g a t e which p r o p e r t i e s the Inverse Image space would i n h e r i t i f a g i v e n space has these properties. Theorem 7 . 1 of a space set i n  Z  X ,  be the i n v e r s e image space  then the s e t If  f  - 1  f .  (K)  let  Z  a  Z  such t h a t  f w e r e  = f  _ 1  _1  f ~ (K) = U  (K) D W '  v  Thus f o r each  a  a ,  3  ,  Z  = f  _ 1  ^ 0 ,  Z ^  since  Ct i  0 = z Hence  p  n z  a  = f  a  _1  p  We  v  1  f o r some  0 .  Furthermore, i f  V  1  if  B £  Y  and  have shown t h a t  K = f(f K  _ 1  (K)  p  Hence  f~ (K) 1  •  Also  K n U  ) = f ( u Z ) = ^( ) z  a  a  (X,u)  Ct  €  = u(KnTJ ) . a  i s the u n i o n o f nonempty p a i r w i s e K  disjoint i s connec-  i s connected.  C o r o l l a r y 7.2 connected space  and so  .  Y  U^-oven s e t s which c o n t r a d i c t s our h y p o t h e s i s t h a t ted.  The i n v e r s e image space under a mapping  .  > then  6 ^ y  n f ( K r t J ) = f" (Knu nKnu ) _ 1  eU  a  ct  i  a ,  U  _ 1  ot  fl  f o r each  Z .  ( K ) n f" ^-,) = f (KnU ) ,  (KnU ) n (KfiU ) = 0 , P  (U ) or  _ 1  ot  = f (Knu )  y  i s a connected  ~tU , -open s e t s f (K) each a , we may  •  z  Ct  K fl U  K  not connected, then there  W  where  If  i s connected i n  e x i s t nonempty p a i r w i s e d i s j o i n t  in  a  (Z,<0)  under a mapping  Proof; would  Let  f  (Zy«j)  i s connected.  of a  55.  Proof: get  Taking  the d e s i r e d  i n the p r e c e d i n g theorem,  we  result.  Theorem 7.3 locally  K = X  The i n v e r s e image space  connected space  (X,Cd)  under a mapping  (Z,*a) f  of a  i s locally  connected. Proof: N  of  z ,  Por each p o i n t  there e x i s t s  z  W £i/J  of  Z  and any  such t h a t  neighborhood  z e ¥ c N_  Z  x = f(z) local  and  f (ir) -1  ¥ =  connectedness o f  borhood  V  of  x  ¥  .  Hence  Z  ,  V _ e, p  Since  under a mapping  Let  we may  {¥ } _  F  $  By Theorem 7.1  ¥  = f 1  - 1  [ J r (  f^OO  ,  f  (Z/ta)  u  c  h  that  0 U ) = Q i=l i 1=1  of a  i s compact.  ( U ) , U etc. a a  Z, .  For  Then  )  •  This implies  X = f(Z) = U U  f  X = - 1  a  (U a  i  B  u  •  ) = tl ¥ i=l i  Hence .  Z = f  _ 1  {IJ }  0  (X) =  T h i s shows t h a t  Z  is  a  i  compact, s i n c e i  (W } . , i s a finite ct^ l — i , _ ! , . . . , n 0  (  . cx  i s compact, t h e r e e x i s t s a s u b c o l l e e t i o n s  contained  cx  cx  W _ A  By  v  = " (U f  z  be an open cover f o r  K  let  = U  X  .  The i n v e r s e image space  3  CX  i  .  connected.  a Z = u W  x e U  there e x i s t s a connected n e i g h -  i s locally  (X 4l)  Prbof: a  Then  Furthermore, i t i s a neighborhood o f  compact space  each  X ,  U £U.  X  Theorem 7 A  O  Let  X  connected.  in  f o r some  such t h a t  X  is  .  Z  s u b c o l l e e t i o n of  _  56.  Theorem 7 . 5  The I n v e r s e image space  l o c a l l y compact space  (Z,to)  (X,<^)  under a mapping  f  z e Z  and  Then  of a  i s locally  compact. Proof;  Let  compact neighborhood  N  neighborhood o f  Clearly  {W } _ a aeA  z .  U f (W ) = U U " , W c  ft U i=l  that  i  a  M - f  _ 1  Let  M = f  M  (NJ • x  M  - 1  (K )  Then  N  _  1  Then  ( ft U 1=1 a  i s compact.  ) = ft i 1=1  Since  f" ( U 1  a  z  I  M  Is a  Let  x  N  N  1  L  has a  = f(M) c f ( U W ) = ' a  x  = f ~ ( U ) f o r each a. S i n c e  f o r some f i n i t e s u b c o l l e c t i o n  c f  .  V  x  has an open c o v e r .  he an open c o v e r f o r M .  A  N  .  x = f(z) .  i s compact,  {U 1 . , „ „ o f fO] ^ a^* 1=1,2...,n o?a€/ 1  ) = ft W i=l i a  i s arbitrary,  Z  .  A  T h i s shows  i s locally  compact. The p r o o f o f t h e f b l l o w i n g theorem i s s i m i l a r t o t h a t of Theorem 7 A  .  Theorem 7.6 L i n d e l o f space  (X,4L)  Theorem 7.7 second c o u n t a b l e space  The i n v e r s e image spacce under a mapping  f  (Z,tj)  i s Llndelof.  The i n v e r s e image space ' (Z/<u) (X,<&)  of a  under a mapping  f  of a  i s second  countable. Proof; for  U. .  Then  Let f"^©)  = {B. ]. _,  0  be a c o u n t a b l e base  = { f " ( B ) : B^ e ©} 1  i  i s countable.  We  CL.  show t h a t i t i s base f o r -ft/.  To do t h i s , l e t W  be any element of  57.  n  and  so  W = f  U = u fi » Hence  (U)  a  f o r some  f~ (iR) 1  '  B .  c  The  c o u n t a b l e space  Since 1  and  inverse  (Xy&)  .  fi  i s a "base,  W = f" (ufe ) = U - " (fi- ) •  Then  i s a b a s e f o r -UJ  Theorem 7.8 first  U e U.  f o r some  Z  ,  l  i s second  image s p a c e  u n d e r a mapping  1  countable.  (Z/ta) f  1  of a  is first  countable. Proof: x = f(z) . It  Let  Then  x  z  Hence  Z  f^Cfi^.)  i s first  So f a r we preserves  axioms.  to enquire  The  them a r e n o t i n h e r i t e d We  now  with  i=0,l,2...,5  such that  whether  T^  Let  X =  Let  f  1  {x^  2  and  2  space  In fact,  a l l the m6st o f  of r e g u l a r i t y  and  shows t h a t  i s not  T  (Z/to)  an open s e t c o n t a i n i n g  z^  Thus the space  , ed = { { x } , 1  f(z^) = x  But  2  .  image  be a mapping o f  the space  z,  2  I t i s quite  which  x )  Clearly  not  D^l^si  the  -space under a mapping i s n o t  f ( z ) = f ( z ) = x^  t h e space  ©.^ =  i t preserves  the exceptions  <0 =» {{z ,z 3, {z-,}} . 1  and l e t  T^  .  {z^ZgjZ-,,} .  Z =  above.  g i v e a counterexample  E x a m p l e 7.9 and  the i n v e r s e  answer i s n e g a t i v e .  i n v e r s e image s p a c e o f a for  Z  countable.  have s e e n t h a t  natural at t h i s point  normality.  of  i s a countable l o c a l base a t  a l l the p r o p e r t i e s mentioned  separation  p6int  has a c o u n t a b l e l o c a l b a s e  i s e a s y t o see t h a t  z ,.  be a n a r b i t r a r y  ,  (X,U)  since there  but not (Z,4J)  2  z  2  i s not  Z  {x-,}}  onto  .  Then  is  T^  1=0,1,2.  does not  or containing T.  X  , 1=0,1,2 —  exist z  g  5 •  but  58.  Theorem 7.10 regular  space  (X/U)  Proof: Z  not i n  0* e 11.  P  .  Let  Let  .  which i s e_-ciosed.  Hence f  _ 1  z e f = f  (X-U)  Hence  1  f  _ 1  z  be a - ^ / - c l o s e d  ^ ) '  _ 1  f  _ 1  s e t and  f o r some  and  X , there x e _ 1  .  2  x i X - U ,  e x i s t two  and  (U )  so  disjoint  X - U c U  respectively.  Therfore  The i n v e r s e  .  ?  Since P c f "  a r e two d i s j o i n t -ftj-open s e t s  2  T h e o r e m 7.11  a point of  m Z - (Z-P) = P , so t h a t  (U)  f ~ (U" )  P  z  f (Z-P) => U  (X-U) c f  a  i s regular.  1  x e  of  (Z/0)  Z - P = f~ (U')  such that  2  and  and  and  Then  U  - f  (X)  write  space  f  By r e g u l a r i t y o f  and  (U^)  taining  P  T h u s we may  TJ^  image  under a mapping  x«f(z)  i_-open s e t s  The i n v e r s e  Z  1  ^ ) .  con-  i s regular.  image s p a c e  (ZyfO)  of a  /  normal  space  (X,4L)  Proof: -to-closed Z - P  sets.  = f" (U ) 1  f [x-(u uu )] _ 1  1  Let Then  1  ±  u n d e r a mapping  ,  = f  2  P^  Z - P^ where  1  and  ( X ) -  P and  that  X -U  ±  f" (X-U ) c f " 1  1  i  are  two d i s j o i n t  Therefore  Z  ^ ) ,  be any p a i r o f d i s j o i n t Z - P  f^C^uu^ ,  e x i s t two d i s j o i n t c T^i-1,2.  1=1,, 2  'W-open s e t s  a r e 'toJ-o'pen.  2  .  and so  1  1  Hence  ^ ) = f 1  1  P^^  2  and  f" (V )  containing  (Z-P )]=  .  2  91- o p e n s e t s  .  u  2  fl ( X - U ) = 0  (X-TJ^  1  Let  f " R $ C - U ) n (X-u* ) ]=  Then  Z - [(Z-P-_)  TnusPj^ = Z - f "  i s normal.  i s normal.  U j e « , i=l,2  Z - [ Z - (P-^HPg)] = Z - Z = 0 Hence t h e r e  2  f  - 1  (X)  1  i  f P  such  2  - f" (U ) =  and and  V  2  _ 1  (V ) 2  respectively.  59.  BIBLIOGRAPHY A l e x a n d r o f f , P. and H. Hopf: V e r l a g , B e r l i n , 1935.  T o p o l o g i e I. S p r i n g e r -  Baum, J . D.: Elements o f p o i n t s e t topology. PrenticeH a l l , Inc., Englewood C l i f f s , N.J., 1964. Bourbaki, N.: T o p o l o g i e Generale. P a r i s , 1939-48.  Hermann and C i e . ,  Bushaw, D.: Elements o f g e n e r a l topology. and Sons, New York, 1963. B a a l , S. A.: P o i n t s e t topology. York and London, 1964.  John Wiley  Academic P r e s s , New  H a l l , D. W., and Spencer, G. L.: Elementary topology. John Wiley and Sons, Inc., New York, 1955. Hocking, J . G., and Young, G. S.: Topology. Wesley, Reading, Mass., 1961. Hu,  Addison-  S. T.: Elements o f g e n e r a l topology. Holden-Day, Inc., Say F r a n c i s c o , London, Amsterdam, 1964.  K e l l e y , J . L.: General topology. D. Van Nostrand Co. Inc., P r i n c e t o n , N,J., 1955Kowalsky, H.-J.: T o p o l o g i s c h e Raume. B a s e l and S t u t t g a r t , 1961.  Birkhauser Verlag  Kuratowski, C : I n t r o d u c t i o n t o s e t theory and topology Pergamon P r e s s , New York, 1962. Mamuzic, Z. P.: I n t r o d u c t i o n t o g e n e r a l topology. P. Noordhoff L t d . , Gronlngen, 1963. Moore, T. 0 . : Elementary g e n e r a l topology. PrenticeH a l l , Inc., Englewood C l i f f s , N.J., 1964. P e r v i n , W. J . : Foundations o f g e n e r a l topology. Academic P r e s s , New York and London, 1964. S i e r p i n s k i , W.: General topology, U n i v e r s i t y o f Toronto P r e s s , 1952.  

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