SOUSLIN'S CONJECTURE AND SOME EQUIVALENCES by IAN ALISTAIR DANIEL B. Sc., University of B r i t i s h Columbia, 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the Department of MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA August, 1971 In p r e s e n t i n g an the thesis advanced degree at Library I further for this shall agree scholarly by his of this written the fulfilment of University of make i t f r e e l y that permission p u r p o s e s may representatives. thesis in p a r t i a l for be British available for extensive granted by the It i s understood financial gain shall •• K A "T The U n i v e r s i t y o f B r i t i s h V a n c o u v e r 8, C a n a d a ^ V\ M Columbia \ CS requirements Columbia, Head o f my be I agree r e f e r e n c e and copying of that not permission. D e p a r t m e n t of for the that study. this thesis Department copying or for or publication allowed without my Supervisor: P r o f e s s o r T. Cramer ABSTRACT T h i s t h e s i s d e a l s w i t h some e q u i v a l e n c e s to S o u l i n ' s h y p o t h e s i s . A l s o i n c l u d e d i s a consequence of i t s n e g a t i o n , the e x i s t e n c e o f a normal Hausdorff space which i s not c o u n t a b l y paracompact. Two e q u i v a l e n c e s to S o u s l i n ' s h y p o t h e s i s can be o b t a i n e d by c o n s i d e r i n g the c o u n t a b l e c h a i n c o n d i t i o n i n the product w i t h i t s e l f o f an o r d e r e d Peano maps on o r d e r e d c o n t i n u a . e x i s t e n c e o f a t r e e of antichain continuum and by c o n s i d e r i n g An e q u i v a l e n c e to a n e g a t i v e answer i s the i n which every c h a i n and cardinality, ( s e t o f p a i r w i s e incomparable o b j e c t s ) i s c o u n t a b l e . Another e q u i v a l e n c e to S o u s l i n ' s h y p o t h e s i s i s o b t a i n e d by assuming t h a t a continuum which i s t h e c o n t i n u o u s image of an o r d e r e d continuum i s m e t r i z a b l e i f and o n l y i f i t has the c o u n t a b l e c h a i n c o n d i t i o n . t h a t a compactum which i s the continuous We get y e t another by assuming image of an ordered compactum i s s e p a r a b l e i f and o n l y i f i t has the c o u n t a b l e c h a i n c o n d i t i o n . Finally, i f we r e s t r i c t o u r s e l v e s to compacta of the type j u s t mentioned we find t h a t S o u s l i n ' s h y p o t h e s i s i s e q u i v a l e n t to a g e n e r a l i z a t i o n of a m e t r i z a t i o n theorem o f S. Mardesic. ACKNOWLEDGEMENTS I would l i k e to thank Dr. T. Cramer f o r suggesting this topic and f o r giving me much help during the writing of the t h e s i s . I would also l i k e to thank Dr. J . V. Whittaker f o r reading i t . Thanks also goes to Dr. Z. A. Melzak f o r t r a n s l a t i n g some P o l i s h for me and to Dr. F. Burton-Jones f o r suggesting several papers f o r me to read. I would also l i k e to thank the National Research Council of Canada for t h e i r f i n a n c i a l assistance. TABLE OF CONTENTS , . Introduction Chapter 0 '• Terms, d e f i n i t i o n s and s i m p l e e q u i v a l e n c e s o f Souslin's hypothesis Chapter 1 Page — ° — ; S o u s l i n ' s problem and a normal H a u s d o r f f space which i s n o t c o u n t a b l y paracompact Chapter 2 The c o u n t a b l e c h a i n c o n d i t i o n i n E 1 5 E x E where i s an o r d e r e d continuum 31 Chapter 3 S o u s l i n Trees 36 Chapter 4 Continuous 43 images o f o r d e r e d compacta and c o n t i n u a . In 1923 it the o r d e r M. S o u s l l n [3] c o n s i d e r e d a t o t a l l y o r d e r e d s e t , gave topology and asked whether i t i s s e p a r a b l e i f i t i s connected and has the c o u n t a b l e c h a i n c o n d i t i o n , i . e . c o n t a i n s a t most a c o u n t a b l e number o f d i s j o i n t open s e t s . We s h a l l prove i n Chapter 0 that a p o s i t i v e p o s i t i v e answer i s e q u i v a l e n t to s a y i n g t h a t a t o t a l l y o r d e r e d s e t w i t h no first or l a s t element i s t o p o l o g i c a l l y the r e a l numbers i f i t i s connected and has the c o u n t a b l e c h a i n c o n d i t i o n i n t h e o r d e r t o p o l o g y . I t i s clear t h a t a n e g a t i v e answer i s e q u i v a l e n t to t h e e x i s t e n c e of a t o t a l l y s e t which i s connected, nonseparable when g i v e n t h e o r d e r t o p o l o g y . ordered and has t h e c o u n t a b l e c h a i n c o n d i t i o n Any such t o t a l l y o r d e r e d s e t w i l l be c a l l e d a Souslin l i n e . In Chapter 1 we assume the e x i s t e n c e o f a S o u s l i n l i n e and a r e a b l e to f i n d an open subsegment o f i t which has no s e p a r a b l e From t h i s we produce a t r e e - l i k e s t r u c t u r e , we take the p r o d u c t w i t h not c o u n t a b l y paracompact. i n o r d e r t o prove N ~ {0} T. , o f s u b i n t e r v a l s o f which t o get a normal H a u s d o r f f I n Chapter space which i s 2 we c o n s t r u c t a s e t s i m i l a r to T t h a t a continuum i s s e p a r a b l e i f and o n l y i f i t s product w i t h i t s e l f has t h e c o u n t a b l e c h a i n c o n d i t i o n . T h i s l e a d s t o two e q u i v a l e n c e s to S o u s l i n ' s h y p o t h e s i s , one o f which i n v o l v e s Peano maps. p a r t i a l l y order subintervals. T w i t h no u n c o u n t a b l e from Chapter I n Chapter 1 and get a t r e e o f c a r d i n a l i t y chains or a n t i c h a i n s . 3 we 1 The e x i s t e n c e o f such a t r e e i n t u r n i m p l i e s the e x i s t e n c e of a S o u s l i n l i n e . I n Chapter k we assume S o u s l i n ' s h y p o t h e s i s , c o n s i d e r c o n t i n u a and compacta .which a r e c o n t i n u o u s o r d e r e d c o n t i n u a and o r d e r e d compacta, r e s p e c t i v e l y , and prove images o f results concerning the s e p a r a b i l i t y o f the compacta and the m e t r i z a b i l i t y o f the continua. We also find theorem of M a r d e s i c ' s that we [8] can r e l a x the c o n d i t i o n s on a m e t r i z a t i o n by c o n s i d e r i n g the images o f o r d e r e d In both t h e s e cases we g e t e q u i v a l e n c e s to S o u s l i n ' s T h i s t h e s i s i s not a complete l i s t hypothesis. references Those w i s h i n g and f o o t n o t e s more should of Kurepa conjecture. of r e s u l t s concerning c o n s u l t M. E. E s t i l l [12]. compacta. Souslin's [4] and the TERMS, DEFINITIONS, AND SIMPLE EQUIVALENCES OF SOUSLIN'S HYPOTHESIS T h i s c h a p t e r i n t r o d u c e s d e f i n i t i o n s , n o t a t i o n and s e v e r a l s i m p l e e q u i v a l e n c e s to S o u s l i n ' s c o n j e c t u r e . The condition. while, a b b r e v i a t i o n , C.C.C., w i l l be, used The symbol, Q\ , will Q , will chain stand f o r t h e f i r s t u n c o u n t a b l e s t a n d f o r a l l the o r d i n a l s s t r i c t l y S m a l l Greek l e t t e r s w i l l ordinal l e s s than fi stand f o r members o f fi' w h i l e l e t t e r s such as i, j , k, Zy m, n, p, and q N . will stand f o r members o f the n a t u r a l numbers, A compactum w i l l be a compact H a u s d o r f f be a connected f o r the countable space w h i l e a continuum will compactum. We s h a l l o f t e n use the f a c t that a t o t a l l y ordered s e t has a connected o r d e r t o p o l o g y i f and o n l y i f i t i s o r d e r complete and has no gaps. A totally o r d e r e d s e t has a gap i f i t c o n t a i n s two p o i n t s w i t h no element between them. The f o l l o w i n g lemma c o n t a i n s s e v e r a l e q u i v a l e n c e s t o S o u s l i n ' s conjecture. Lemma 0.1 Let X be a t o t a l l y the o r d e r t o p o l o g y . 1. If o r d e r e d s e t which has the C.C.C. when g i v e n The f o l l o w i n g statements X i s connected are equivalent. i t i s separable 2. X i s separable 3. If X i s connected and has no f i r s t or l a s t element then i t i s homeomorphic to the r e a l numbers with the usual topology. Proof 1 X' implies . Notice that 2 X' Take the Dedekind completion of X" X . We wish to construct a t o t a l l y which i s a quotient space of what we are doing i s connecting removing a l l gaps. X To construct equivalence r e l a t i o n R(y) t o t a l l y ordered sets, (S(x) | x , ordinal} and c a l l i t with the order topology has the C.C.C. and that i t s s e p a r a b i l i t y implies that of set X on 1 we w i l l define f o r each o r d i n a l . S(x) y an We w i l l also define a c o l l e c t i o n of {W(x)| x , ordinal} where Intuitively, by making i t order complete and by X" X X'. and i s connected. ordered , and a c o l l e c t i o n of r e l a t i o n s i s a r e l a t i o n on W(x) . Let aR(o)b i f and only i f there exists a f i n i t e number of elements between them. Let W(o) be X' and l e t S(o) be ordinal and assume that Let way. W(x + 1) Let W(y) and be the quotient, c , d e W(x + 1) R(o) . Consider related i f and only i f there exists f o r u < w} for such that u < w} and a = , x i s an y < x and order i t i n the obvious S(x + 1 ) — r e l a t e d i f and only i f there i s only a f i n i t e number of points between them. A(u) £ A(w) where S(y) are defined for - a l l W(x) | S(x) be x + 1 Let a , b e X' be. R(x + 1)- (A(y) | y < x + 1 , A(y) e W(y) , {B(y) | y < x + 1 , B(y) e W(y) , B(u) e A(w) A(o) , b = B(o) and A(x + 1) and B(x + 1) are S(x + 1 ) - related. for some y < x obvious way If x i s a limit ordinal l e t a R(y) b and l e t . Let W(x) c , d e W(x) be be X'/R(x) ordinal. Order X/R and has the C.C.C. 2 and l e t be is 3 For A(n) onto f(m) x(j) more than [ Now r e l a t e d to - , n e N ~ {o} X . let . Let f(o) be x(o) f(£ - 1) . Suppose we have (j {w e B(n) such that k n-I k —rn-1 f(—r-) y e B(n) | w < y} J k+1 n-i f(—r-) i s the least natural number such that Using the same method we can extend i s a one to one, Z • x(i) f U{B(n). | n e N}, Consider to . Let such that (n > 1) I > 0 and f (£) X(j) is such that has been put into a. one to one k+1 —rn-1 < y < and an {k / n | k e N} (m , £ e B (1)) i s the l e a s t natural number has members between be . j B , X' from the nonnegative r a t i o n a l s , . x be Also l e t {x(i) | i e N} = such that k E N for some A(n) order preserving correspondence with a subset of f ^ b a , b e X' We w i l l define an order preserving, i s defined for a l l m < L u{ (m) | in < n}] exists let and notice that i f i s connected ~ U{A(m) | m < n} ( x ( i ) | X ( i ) > X(o)} suppose be f ordered i n the I t must therefore be separable as i s be a countable dense subset of one-to-one map, R(x) i n the obvious way implies B(n) a , i f and only i f S(x) - r e l a t e d i f and only i f there i s only a f i n i t e number of points between them. R- r e l a t e d i f and only i f a R(x) b Z . Notice that there 1 • Since Let, f(y) Z i s dense i n be x(i) ( x ( i ) | x ( i ) < x(o)} onto, order preserving map from where k f ( — r - ) and n—X i s between Z X it i k+1 f(—r) n-1 so that to the r a t i o n a l s . . Since each member of X is a sequential limit of members of Z f can be extended to a l l of X 3 implies and rememb er that X 1 If X has a f i r s t or last element remove them i s connected. Q. E. D. A totally ordered set which has the C.C.C. and is nonseparable when given the order topology is called a Souslin space. It i s clear from Lemma 0.1 that the existence of a Souslin space, like that of a Souslin line, i s equivalent to a negative answer to Souslin's conjecture. We have stated Souslin's conjecture i n terms of certain ordered spaces which are topologically the reals. Our conjecture can also be stated in terms of ordered space which are homeomorphic to the unit interval, [0,1]. Lemma 0.2 The following are equivalent to Souslin's hypothesis 1. An ordered continuum with the C.C.C. i s separable 2. An ordered compactum with the C.C.C. is separable 3. An ordered continuum with the C.C.C. i s homeomorphic to [0,1] Proof Follows from the statement and proof of Lemma 0.1. SOUSLIN'S PROBLEM AND A NORMAL HAUSDORFF SPACE WHICH IS NOT COUNTABLY PARACOMPACT A normal space i s said to have property descending sequence of closed sets, ( C ( i ) | i e N} there exists a sequence of open sets for every i and n { D ( i ) | i e N} D {D(i) = 0 . i f • for every , with n u l l i n t e r s e c t i o n | i e N} such that C(i) C D(i) In 1951 C. H. Dowker [2] proved that a normal space i s countably paracompact i f and only i f i t has property D . H e then used this r e s u l t to produce a normal space which i s not countably paracompact. This space i s not Hausdorff and Dowker asked whether there existed a normal Hausdorff space which was not countably paracompact. In 1955 M. E. Rudin [3] assumed Souslin's hypothesis to be f a l s e and used Dowker's characterization of countable paracompactness i n normal spaces to reply i n the a f f i r m a t i v e . This chapter involves the construction of her normal Hausdorff space which i s not countably paracompact. Assume there exists a Souslin l i n e , s h a l l do i s produce an open i n t e r v a l i n S subsegments. i s nonvoid. Assume that M S . The f i r s t thing we which contains no separable , the c o l l e c t i o n of open separable segments, By the Hausdorff Maximal P r i n c i p l e every member of M i s contained i n a maximal chain of open separable segments. chain i s countable. while S's By Lemma 5.5 this This implies that the union of t h i s chain i s separable connectedness implies that i t i s an open segment. c e r t a i n l y contained i n no other open.separable segment. This union i s A l l t h i s means that every open separable segment i s contained i n a maximal one. I f we consider the c o l l e c t i o n of maximal open separable segments we see that i t i s pairwise d i s j o i n t and therefore countable. We can also see that i t s union, which i s separable, must have a complement which contains an open segment, This segment, s , contains no separable subsegments since every separable • open segment i s contained i n a maximal one. let s be s we s h a l l construct for each countable c o l l e c t i o n of open i n t e r v a l s , i s countable. R(3) In the case where Let R(o) be {s} R(a) has been defined f o r a l l 3 < a exists. E(ct) i s (c(k) | k e A} I f R(a-l) s ~ {t | t e R(a)} then surely Define , such that , R(3) disjoint s ~ y { t | t e R(a)} a , assume and l e t R(a) be the set of . Since R(a) w i l l be S Suppose now that i s defined and that i s connected q } and {(a(k) , c(k))|k e A} . I f s ~ y {t | t e R(ct-l)} w i l l be also. and that, f o r g < a a nonvoid 1 i s {(a(k) , b ( k ) ) | k e A , A < ^ y {(c(k) , b(k)) | k e A} ^{ i s void . Let E(a) be a set made up maximal segments i n y { t | t e R(a-l)} ~ E(a) R(a) a e Q . Given a successor o r d i n a l , of one point from each member of R(a-l) countable. M S . Using that s a i s countable then i s a l i m i t ordinal s ~ {t [ t e R(3)} is R(a) to be a l l maximal open segments i n y { t | t £ R(3)} | 8 < a} countable complement, = V(a) . Since this i n t e r s e c t i o n has a y ( s ~ y { t | t e R(3)) Suppose we have a point, y | 3 < a} , R(a) , i n V(a) which i s not i n R(a) exists. . Since i t i s i n V(a) i t must be contained i n the i n t e r s e c t i o n of (t(3) | t(3) E R(3) descends as -v 3 , 3 < a} increases. which i s e a s i l y seen to be a chain which Since of the i n t e r s e c t i o n of t h i s chain. t i s not i n R(a) i t must be an endpoint There can only be a countable number of such chains and therefore only a countable number of V(a) and not i n R(a) . Since Please note that i f x s ~ V(a) t's which are i n i s countable so i s i s contained i n R(a) , where e R(3)> s ~ {t|teR(a)}. a is a l i m i t o r d i n a l , then: x where t(3) e R(3) Int( {t(3) = | 3 < a , x e t(3) n i s unique. Notation R w i l l be y{R(a) | a e fi'} , the unique o r d i n a l such that For every H(x) if x = y and i = j n £ N such that R(3) such that implies that we must n o t i c e several facts about (i) R(a) If 3 «5(x) w i l l be R we w i l l produce f(x,i) and i f x G y (y e R(a) k > n , x e R(^(x)) (3 , l i m i t ordinal) x e R(3) { f ( x , i ) | i e N} 9 = and i f x e R f(x,k) ^ = , a < 3) y . f(y,j) i f and only then there exists To produce this and ft' i s a l i m i t o r d i n a l and a < 3 then every element of contains a countably i n f i n i t e number of element from R(3) • sees this by n o t i c i n g that every subsegment of x e R(a) and that the subsegments of have a structure very s i m i l a r to R R(3) (iii) in R(a) that are i n R One i s nonseparable . (ii) y x H(x) has c a r d i n a l i t y For every such that x e R(3) x Q y Q if 3 and for every i s a limit ordinal a < 3 there i s a unique (iv) a(i) < 3 where (v) from each 3 If i s a l i m i t o r d i n a l then i t i s and If R(a) x i s in R(3) {x(l) , x(2), R(3) a > 3 } (3 | 1 e N , a(i) x(l) ^ of that . y(a(l) , x(l)) x's from R(3) f(x(l),j) . a(j) Since s h a l l write i t as H(x(i)) . i f and only i f i = j}) {f(x(l),i) j < i f(x(l),l) . (a < 3) y £. R(a) j > n implies that f(x(i),l) exists y(.l) . = j .We such that f(x(l),j) assume that define i C y z e R(y(l)) x(j) j < i x ( j ) ' s (j < i ) their we can pick Assume now Consider the unique . such that This R(3) and c a l l i t • Pick f(x(l),i) f(x(l),i) there exists i s defined f o r a l l From the d e f i n i t i o n of do contain these . = n such that y H(x(j)) . such that also have the condition that for x(l) £ fi' and for (j < i ) | i e N} shall contains an i n f i n i t e number x(l) C y(a(i) , x(l)) f(x(l),j) We y ( a ( l ) , x ( l ) ) e R(a(l)) contains an i n f i n i t e number of members from every H(x(j))'s w e y(a(l) , x(l)) such that i f and only i f not contain = has been defined for a l l By induction we can pick Now o Consider the unique one that i s not equal to any f(x(l),j) See the discussion i n (i) above. one can be picked and c a l l e d y ( a ( i ) , x ( l ) ) e R(a(i)) y(a(i) , x(l)) . and use induction to define H(x(l)) y{a(i) then i t contains at l e a s t one element has c a r d i n a l i t y ^ f i r s t produce = | oc(i) e fi',i e N} a(i) i a(j) where Since ^{a(i) such that R(3) z we j < i . First can see that there contains x By considering the members of but does R(a) which and by remembering the d e f i n i t i o n s of f(x(i),l) which i s a subinterval of z and which i s not equal to had to remember that f(x(j),£) z 3 = where n f(x(i),n) { y ( i ) I Yd) = Y(j) i f and only i f i x(i) e y(y(m) , x(i)) is a subinterval of which contain the z f(x(j) , £) n < m R(3) for j < i T be For each member of R x N T and notice j , y(x) < 3> y(y(m) , x(i)) . Remembering the definition of the y(y(ra.) we can pick a and £ e N {o} . T > x(i)) contains an infinite f(x(i) , m) which i s not equal and not equal to H(x(i)) f(x(i) , n) for for a l l x(i) e R(3) shall be the space we are after. we shall define a neighbourhood system. The topology obtained from these neighbourhood systems w i l l make T into a no.rmal Hausdorff space which i s not countably paracompact. For (x,n) where ^(x) i s not a limit ordinal the neighbourhood system is the family of a l l subsets which contain Consider 3 < <S(x) A(3(2)) l e t A(3) (x,l) where = {(y,D A(max(3(D ,3(2))) = ^(x) <£(x) (x,n) i s a limit ordinal. | 3 < tf(y) < tf(x) , x Q y> For every . A(3(D) n . Let the neighbourhood system for (x,l) be the family of a l l subsets which contain some If . Consider y(y(ni) , x(i)) e R(y(m)) . In this manner we can obtain our Let = R(3) and therefore disjoint from the element of R(y(m)) H(x(j))'s (j < i ) and the fact that to . To.do this we . It is easy to see that x(j)'s (j < i ) number of members of Z e N i s defined for a l l * n < m y(l) is from the paragraph above. such that and contains an infinite number of members from Now assume that that for j < i i s a limit ordinal and n A(3) is greater than 1 we shall use induction, assuming that neighbourhoods have been defined for (y,m) with $(y) and < m < n induction process s t a r t s with R(o)) x {n}'s the . , and for (y,n) with R(co) x {2} and goes through a l l the I t then continues to R(2w) x {n}'s R(2to) x {2} . The and proceeds through and so f o r t h . Consider (x,n) (^(x), a l i m i t o r d i n a l , and n > 1) and i n the following manner F((x,n),i) G((x,n),a) F((x,n),i) = { ( f ( x , j ) , n - l ) | j > i} G((x,n), ) = {(y,n) | x c a K(i) y J(a) y and y A neighbourhood system f o r where ?5(y) <.^(x) . a < tf(y) < « K x ) } . (x,n) consists of sets of the form {(x,n)} K ( i ) i s a neighbourhood of each point of a neighbourhood of each point of and define G((x,n),a) sets containing something of the form . F((x,n),i) and J(a) Also i n t h i s system are a l l K(i) y J ( a ) y {(x,n)} We must now confirm that this i s a neighbourhood system. now (K(i) y J ( a ) intersection, y{(x,n)}) and L ( i , j , a , 3 ) , contains Without loss of generality assume F((x,n),j) 0 point of F((x,n),i) F((x,n),j) . and Also K(j) y J(B) y ^ (x,n)} n j < i neighbourhoods of (x,n) and 3 < a . n Consider Their y{(x,n)} . J ( a ) ) . Then K ( i ) ^ K(j) i s a neighbourhood of every G((x,n),ot) C i s a neighbourhood of every point of the neighbourhood system of K(i)) y(J(3) (K(j) is (x,n) G((x,n),3) G((x,n),a) . . and Thus J(3) n J(a) L(i,j,a,3) is in Thus the i n t e r s e c t i o n of any two i s again a neighbourhood of (x,n) Taking a l l sets which are neighbourhoods f o r each of t h e i r elements, we get a topology. I t w i l l be useful to prove that our o r i g i n a l neighbourhood system for (x,n) i s the neighbourhood system for (x,n) relative to this topology. That i s , we should prove that for each of our earlier defined neighbourhoods that AC B B of and A (x,n) there i s a neighbourhood A of (x,n) such i s in the neighbourhood system of each of i t s elements. To prove this we shall divide T into the three subsets we used to define the neighbourhood systems, that i s ; (i) a l l (x,n)'s such that (ii) a l l (x,l)'s ( i i i ) _11 and If B such that ^(x) is a limit ordinal (x,n)'s such that ^(x) i s a limit ordinal n >1 i s a neighbourhood of then the singleton, {(x,n)} y$(x) i s not a limit ordinal (x,n) (<^(x) , not a limit ordinal) has a l l the qualities of the neighbourhood we desire. If B i s a neighbourhood of i t must contain some A ( 3 ) = (y,l) and Therefore (x,l) (?5(x), a limit ordinal) then {(y,l) | 3 < ?5(y) < ?5(x) (z;l) are contained in A ( 3 ) and ^(y) < f5(Z) then . if z y . A ( 3 ) i s our desired neighbourhood of (x,l) Consider (x,n) , with $J(x) , a limit ordinal, and Assume that what we wish to prove i s true for a l l ^(y) < $(x) If , and that B m < n where (y,m) n > 1 such that (y,n) such that $(y) < «5(x) (x,n) i t contains a set of the form K(i) i s a neighbourhood of J(a) i s a neighbourhood of By our assumption and for a l l i s a neighbourhood of K(i) (j J(a) y{(x,n)} and , x C y} {(y,n) | x <~ y K(i) contains for every , and {(f(x,j), n-l)|j > ^(x) < <f>(y) < a} . j > i a set D(j) which contains J(a) (f(x,j), n-1) and i s in our topology. contains, for every neighbourhood of topology. a < 3 < $(x) (y,n) (x S y and Also by our assumption , a set, P(3) 5 $(y) = 3) which i s a and like J(ct) i s i n our Therefore our desired neighbourhood i s : ( U * ( J ) I J > i>) U<U^ (3) I « < 3 < «J(x)}) yCCx.n)} D p The following facts w i l l be used later. Lemma (1.1) C(n) = {(x,m) | m > n} i s closed for a l l n e N Proof Assume k < n n i s more than 2 and that . We w i l l prove that for every a neighbourhood of assumption (x,m) Notice also that (x,m) e ~ C(n) there exists , such that N(x,m) exists for a l l N(x,m) C. ~ C(n) (x,m). where exists for a l l (y,n-l) such that hood of (y,n-l) (y,n-l) with ^(y) < ^(x) and . By our m. i s less than n-1 , a limit ordinal, assume o < $(y) < $(x) . ,For every x £=: y there is P(y) which i s contained i n ~ C(n) has a neighbourhood contained i n ~ C(n) [U^(y) I ' ^ y N(x,m) , N(x,n-1) exists i f ^(x) i s not a limit ordinal. To prove the existence of N(x,n-1) for $(x) N(y,n-1) C(k) i s closed for a l l . Every , a neighbour- (f(x,j) , n-2) . This means that ,rf(x)> *5(y) > 1>J y I u ( K ( j ) | j e N}] <j{(x,n-i)} is a neighbourhood of (x,n-l) contained i n ~ C(n) The case when n is 2 can be handled i n a s i m i l a r manner. Lemma 1.2 . D(3) {(x,n) | <£(x) < 3 ) = i s countable open and closed. 5 Proof To prove that (y,m) e ~ D ( 3 ) If i s closed we s h a l l exhibit f o r each a neighbourhood, N(y,m), which i s a subset of ~ D ( 3 ) ^(y) i s a nonlimit o r d i n a l we w i l l l e t N(y,m) i n the case of be D(3) ( y , l ) where «5(y) , for a l l (x,n) and a l i m i t o r d i n a l , and (x,n) i n ~ D ( 3 ) . • and n = a neighbourhood which does not i n t e r s e c t of these neighbourhoods then members, D(3) • (x,l) be m D(3) and D(3) . B(x,n) To prove the existence of o r d i n a l and n > 1 «Ky) < ^00 and assume that m < n F((y,m),l) G((y,m),3) If A has i s the union i s our N(y,m) j5(x) , a l i m i t o r d i n a l , l e t . and f o r a l l Consider the sets , which i s contained i n I f ^(x) i s a nonlimit o r d i n a l l e t B(x,n) B(x,l) be defined i s open we w i l l exhibit f o r each of i t s (x,n) , a neighbourhood, with n < m F((y,m),l) and A y{(y,m)} (y,m) we w i l l l e t N(x,n) 0(x) < $(y) Each member of both To prove that while If we are considering m > 1 such that such that tf(x) < $(y) G((y,m),3) {(y,m)} ^(y) i s a l i m i t ordinal we w i l l l e t N(y,l) {(x,l) I y C x , «5(y) > j4(x) > 3 } with be and that B(y,m) B(y,n) be {(x,n)} {(y,l) | x ^ y B(x,n) where exists f o r a l l exists when and . For 1 < jj(y) < 3 ) $(x) i s a l i m i t (y,m) such that ^(y) i s l e s s than tf(x) . By our assumption each member of G((x,n),o) a neighbourhood which is contained in D(3) . these neighbourhoods then A y{(x,n)} If and F((x,n),l) has A is the union of a l l is what we are after. Lemma (1.3) If x and y are contained in R"(y) where u is a limit ordinal then there exists x C w , y 9i z a < \i and w , z e R(a) such that and w ^ z Proof From the definition of R(u) we can see that: x = lnt( {t(8) | 3 < V , t(B) e R(B) , n and x e t(B)l) y = l n t ( { u(3) | 3 < V , u(B) e R(3) , y e u(B)» n Our desired a must exist, otherwise x is equal to y Lemma (1.4) K(x,B) = ((y,l) | B < l*(y) < ?Kx) , x C y} is both open and closed. Proof If (y,l) is contained in K(x,3) ordinal then (y,l) e {(y,l)> C of If tf(y) is a limit ordinal and (y,l) (y,l) . K(x,3) and ^(y) is not a limit where {(y,l)> is a neighbourhood is contained in K(x,g) then Therefore ( y , l ) e {(w,D K(x,g) for each This i s t r i v i a l i f D((y,n)) $(y) K(x,g) (y,n) in f5(y) ~ K(x,g) (y,n)'s If D(y,l) n > 1. . $(y) < g . If then t e R(^(y)) that t C W j y C - Z j W ^ z Consider then To prove the existence of D((y,l)) exists when . and and w , z e R(a) v , a < «5(v) < <£(y)} . If {(v,l) | y C v , $(y) < $(v) < tf(x)} be our ($(y) , a l i m i t o r d i n a l and existence of D(w,m) for with $(w) < $(y) not contain a member of G((x,n),0) and where m = n K(x,g) . . w,z . D(x,l) D(y,n) i $(y) and Our w i l l be such such w i l l be then l e t . To prove the n > 1) assume the m < n such that and for F((x,n),i) By our assumption each member of has a neighbourhood which does not intersect The union of these neighbourhoods unioned with is is a limit a < ^(y) D(y,l) $(w) < <f>(y) and Pick $(y) t e K(x,g) «5(y) > «5(x) existence of (w,m) where o < $(w) < $(x)} then there exists From Lemma 1.3 above we have , (y,n) (y,l) where {(w,l) | y C w {(v,l) | y C and n we w i l l show that g < ?5(y) < $(x) that (w,m) D((y,n)) K ( x , g ) = 4 . such that i s a l i m i t o r d i n a l and then use induction for ordinal. . i s closed we must produce a neighbourhood i s a nonlimit o r d i n a l . for the other a l i m i t o r d i n a l and our w} E K ( x , p ) i s open. To prove that D((y,n)) | «Ky) < $(w) < g , y C {(y,n)} F((x,n), K(x,g) w i l l be what we want. Lemma (1.5) Consider such that xQ z (x,n) (n > 1 , $(x) > a l i m i t ordinal) . Then for every neighbourhood K does of and (x,n) z e R and for every m < n In fact, there exists K (y,m) e K such that i s also a neighbourhood of yC z 0(y) = and ^(x) . (y,m) Proof Let definition m = n-1 K and contains K be a neighbourhood of {(f(x,j), n-1) | j > i} such that j > k means that statement i s true for m+1 (m < n-1) y C z , «$(y) = for some j f(y,k)C K ^(x) and K f(x,j) C z . By for some i is a neighbourhood for the set. By the definition of k > i (x,n) H(x) and in fact there exists . Assuming that our there exists (y,m + 1) e K such that i s a neighbourhood for {(f(y,i),m)| i > j} . By the definition of . Therefore f(y,k) H(y) there exists k > j such that i s the point we are after. Lemma (1.6) Consider Then G((x,n),3) (x,n) where yUx.n)} = j£(x) is a limit ordinal and n > 1 {(y,n) | x C y , M rf(y) < rf(x)} i s closed. Proof For each point (y,m) e ~ [G((x,n),3) y{(x,n)}] we shall produce a neighbourhood which does not intersect G((x,n),3) y{(x,n)} existence of such neighbourhoods in obvious when $(y) or when m = 1 more than for each If m ^ n . In the case where $(y) . The i s nonlimit ordinal i s a limit ordinal and m is 1 we shall assume the existence of the desired neighbourhood (w,k) with k < m and for each (w,m) then i t i s clear that one can produce with (6(w) < $(y) F((y,m),i) and G((y,m),ct) such that n F((y,m),i) [G((x,n),3) y{(x,n)}] = <f> n and G((y,m),a) [ G ( ( x , n ) , 3 ) y{(x,n)}] = <f> . Each member of F((y,m),i) has a neighbourhood which does not intersect {(y,n)} [G((x,n),3) G((x,n),3) y{(x,n)} m = n (remember we are dealing with y{(x,n)}] such that R(y) (y - limit ordinal) rC^ . When vC w and , a < $(y) , and y C z as is F((y,m),l) then there exists u C z and w ^ z r e R($(y)) are contained in a < u and w,z e R(a) . If 3 < $(y) < $(x) , and w,z e R(a) . G((y,m),a) is disjoint from . Each member of G((y,m),a) neighbourhood disjoint from (y,m) e $(y) i s a limit ordinal and m > 1) we shall recall Lemma 1.3 which states that i f v,u exists G((y,m),a) is added to the union of these neighbourhoods we get what we desire. For the case where that and G((x,n),3) y{(x,n)} such then there such that v? ^ z , G((x,n),3) y{(x,n)} and F((y,m),l) . When has a {(y,m)} i s added to the union of these neighbourhoods we get what we wish. If $(y) < 3 then G((y,m),o) is disjoint from G((x,n),3) y{(x,n)} and i f «*(y) > $(x) G((y,m),$(x)) i s disjoint from G((x,n),3) y{(x,n)} . Since F((y,m),l) i s disjoint from F((y,m),i) both disjoint from G((x,n),3) we have a G((y,m),a) and G((x,n),3) y{(x,n)} and are able to produce our desired neighbourhood. Lemma (1.7) Consider Then (x,n) with F((x n),i) y{(x,n)} > $(x) , a limit ordinal and n > 1 is closed for any i Proof As may be suspected we shall produce for (y,m) e ~ lF((x,n),i) y{(x,n)}] a neighbourhood which does not intersect F((x,n),i) y{(x,n)} . It is easy to see that such neighbourhoods exists for (y,m) is a nonlimit ordinal or when m where is 1 ? K y ) is a limit ordinal and m . To prove the same for (y,m) i s more than one we shall assume that our desired neighbourhoods exists for (z,k) with <f>(z) < j6(y) (z,m) with disjoint from F((y>m),l) . If tf(y) > H*) F((x,n),i) y{(x,n)} and G((y,m),$(x)) F((x,n),i) y{(x,n)} . When (Remember that of {(y,m)} f(z,j) = f(w,k) F((x,n),i) y{(x,n)} and for G((y,m) ,^(x)) is as i s F((y,m),l) . Each member of i s added to the union of these In the case where F((y,m),l) are disjoint from G((y,m),o) and {(y,m)} then k <m has a neighbourhood which Is disjoint from neighbourhoods we get what we want. G((y,m),o) and when f5(y) i f f z == w ^(y) < ? K x ) F((x,n),i) y{(x,n)} and j = k) . Each member F((y,m),l) has a nieghbourhood disjoint from . The sum of a l l these neighbourhoods unioned with i s the neighbourhood we want; We are now ready to prove,that T i s the space we desire. But f i r s t we need three definitions. Definitions If every Q i s a subset of R. then, x e R y C x (y e R) If x there exists is Q - full i s called z C y •such that then: L(x jg,Q) q-full if:for z e Q , i s the set of a l l y in q such that y C x , $(y) > g , and there does not e x i s t any Z such that ?5(z) > g , z C x and y C z From the d e f i n i t i o n of R and therefore countable. that $(y) < <5(x,g,Q) the members of L(x,g,Q) Let 3(x,g,Q) for every y in Q are d i s j o i n t denote the smallest o r d i n a l such i n L(x,g,Q) We can now proceed to prove that T . does not have property D , that i s , that i t i s not countably paracompact. Consider the c o l l e c t i o n of closed sets, C(n) i s {(x,m) | m > n} Suppose for (D(n) every | n e N} n (C(n) . Lemma 1.1 t e l l s us that i s a c o l l e c t i o n of open sets We s h a l l show that | n e N} , where C(n) i s closed. such that C ( n ) C D(n) i s non-null i ^{D(n) | n e N} Lemma 1.8 For Q(n) - full n e N l e t Q(n) be {y | (y,l) t D(n)} . No x e R is . Proof Suppose x i s Q(n) - f u l l {g(i) | i e N , g(i) e fi'} where Q(n)) and construct the sequence g(l) = «S(x) and g(i) = 6 ( x , g ( i - l ) . The d e f i n i t i o n of 6(x, g ( i - l ) , Q(n)) g(i+l) >^ g ( i ) . Therefore g = y{g(i) quarantees that | i e N> We f i r s t wish to prove that there exists yC x and (y,l) e D(n) s e p a r a b i l i t y of x D(n) . Therefore i s a limit ordinal. y e R(g) such that . From the d e f i n i t i o n of R(g) and the non- there exists if , n = 1 z c R(g) we are done. such that Assume zC x n > 1 . and Since (z,ri) e D(n) is open for (z,n) has a neighbourhood F( (z,n),i) where k > i such that j >k and c a l l i t y(n-l) D(n) i i s some natural number. Now there exists implies that . Notice that f(z,j)(^ x . Pick one such f(z,j) y(n-l) C x and that (y(n-l), n-1) e . It i s clear that this process w i l l produce a y(l) such that £ R(3) , y(l)C. x and (y(l),l) z D(n) y(l) Since w £ Q(n) i AC^D(n) which i s also a neighbourhood x i s Q(n) - f u l l . Because w . This y(l) w i l l be our y . there exists wC y such that i s contained in y , ^(w) > 3 > 3(i) for every . Therefore, by the definition of L(x,3(i) , Q(n)) , there exists for every since i e N , z(i) e L(x,3(i) , Q(n)) such that ^(z(i)) < 3(i+l) < 8,z(i) must also contain limit ordinal wC z(i) y . But . Because 8 is a (y,l) i s a limit point of {(Z(i),l) | i e N} (see the definition of the neighbourhood system of (y,l)) member of the open set, . Since D(n) , {(z(i),l) | i e N} This contradiction shows that no x (y,l) is a i s eventually i n D(n) . i s Q(n) - f u l l Corollary (1.9) For every z C y (z e R) then x £R there exists y C x (y E R) such that i f (z,l) e D(n) . Proof Immediate from the definition of Q - f u l l and Lemma 1.8. Q.E.D. For n £N l e t P(n) be is contained in D(n)} {x £ R | for every y ^ x(y £ R)(y,l) Theorem (1.10) The set, ^{D(n) | n e N} , i s non-null. Proof The preceding c o r o l l a r y shows that every x in This means that we can construct • {S(x,o,P(n)) | n e N} member of w . x . e P(n) Because , x , since $(z) > y z e P(n) with y y Thus > for every Is for every z e L(x,o,P(n)) is where n , n Pick , y e R(y) there exists $(w) > y and . y w£ which contains This places i s non-null and T is a such such that y such that This means that there i s In other words, f o r every as a subsegment. ^{D(n) | n e N} . P(n) - f u l l for every . n P(n)-full. x Since this set i s denumerable there exists 6(x,o,P(n)) < y that yC R(o) R w n and, i n fact, contains , there exists (y,l) i n every D(n) does not have property D Q.E.D. We w i l l now show that T i s a normal Hausdorff space. Lemma (1.11) Every point of T i s closed. Proof Consider $(y) (x,n) e T . i s a nonlimit ordinal then If (y,m) {(y,m)} f (x,n) i s also i n "separates" (y,m) T from and (x,n) . Consider now (y,l) where d(y) i s a limit ordinal. If $(x) > $(y) then {(w,l) | y C w , ^(y) < ^(w) < ^(x)> " separates" (y,l) from {(w,l) | y ^ w , o < ^(w) < j5(x)} prove that from (y,m) (x,n) (^(x) , a limit ordinal and k <m and every bourhood disjoint from {(x,n)} that neither contains (x,n) m > 1) (w,m) . Pick with G((y,m),a) has a neighbourhood which does not contain (w,k) and . To with has a neigh- F( (y,m),i) G((y,m),a) (x,n) . can be "separated" ?5(w) < $5(y) . Each member of these neighbourhood, A y{(y,m)} while does the trick when ^(x) < ^(y) we w i l l use induction, assuming that every ^(w) < j4(y) and (x,n) If A and so F((y,m),i) is the union of w i l l be what we want. Q.E.D. We w i l l now consider two disjoint closed sets H and K and prove that there are two disjoint open sets which "separate" them. Lemma (1.12) If all x's H with and K (x,n) are disjoint closed sets let H(n) in H in K . Suppose we have If x is H(i) - f u l l and K(n) yC x , y the set of a l l x's where both i s not K(j) - f u l l x and y for any be the set of with (x,n) are i n R i,j e N Proof Assume that we have H(i) - f u l l ordinals, and y x and is K(j) - f u l l (y(n) | n e N} . Let y in R such that y^ x , x is . We shall construct a sequence of y(l) be any ordinal more than j5(y) and let y(n) be if n is even. U^YCD R(y) S(y, y(n-l) , K(j)) i f n | i e N} = y is s t r i c t l y more than i s a limit ordinal. i t i s seen that we can pick subsegment of of y(n) Since y . Since y y( l) > for n n _ y(n) u(n) . there exists v(n) Since Now assume that z x u(n) = i = j such that H(i) - f u l l is a z is there is w' $(w') > y > y(n-l) for n of H(i) , for n {(w,i) |. z C » w'C , p < ^(w) < 6} k + Z there exists vC y wC v and . From H are both in and K . This i < j and must contain a set of the form ^(v) = such that (z,i) . such that . Assume now that (z,j) . A v(n) . A ,a neighbourhood of v(£) were mentioned above) i = j even, even. . Lemma 1.5 tells us that (v,i) where , i s less than . This means that and consider and , w contained in 3 < «5(w) < 6} . This means that there exists (u(k),j) e A for some point ^(y) 'u(n) ?5( u(n)) (z,i) i s in the closure of both is a neighbourhood of , z there is a member . But z C v(n) contradiction proves our lemma when {(w,j) | z C w such that y we can see that there exists . This means that A R(y) is H(i) - f u l l . Thus contains a set of the form, that , . The fact that tf(w) is larger . Because ( u(k),i) and (v(£),i) ( u(k) A w , H(i)) y(n-l) contained in L(x,y(n-1) , H(i)) such that «5(v(n)) < y(n) < y the definition of y(n-l) odd, implies that there exists which i s a subsegment of But z which i t s e l f is less than y contained in from <5(x, From the construction of is K(j) - f u l l L(y> Y(n-l) » K(j)) such that A z K(j) which i s a subsegment of than is odd and A y w E H(i) k eN is a neighbourhood . Since y is . We can easily see that f o r any n , v(n) i s i f v C v(n) contained . contains w and that the only way this can happen In other words f o r every i n v(n) . Remembering that A of v(m) f o r some ( z , j ) contains a point of H proves our lemma for i < j The case where m . even , v i s i s a neighbourhood of ( v , i ) {(u,i) | v C u , 3 < $(u) < y} . we know that i t contains a set of the form which must contain y(n) , n Therefore and a point of every neighbourhood K . This contradiction . j < i i s treated i n the same manner. Lemma (1.13) There exists If x i s contained for any n or x p e fi' such that i n R(u) then either contains no member of R(u) has the following x property. contains no member of H(n) K(m) f o r any m Proof Let of x P ( l ) be the set of a l l x's , in R , i s i n H(n) for any all x e R such that no subsegment of Let P(l) and and P(2) r e s p e c t i v e l y . 1 there exists w i K(m)} {a(m) Since both u E fi' such that and l e t in R P(2) be the set of i s i n K(m) f o r any m $(y) < u . I f I(m) then every subsegment of with P(l)' and . P(2)' are countable f o r every y in P(l) 1 or x e R(u) and assume that no_ subsegment of x i s for any m | m e N} such that no subsegment P(2)' be the c o l l e c t i o n of a l l maximal segments from P ( l ) Consider now K(m) - f u l l x n in R a(l) = u i s ly e R | x and i s I(m) - f u l l a(m) = y implies that . Construct 6(x , a(m-l) , I(m)) and P(2)' . let a be y{a(m) | m £ N} and consider K(m) there exists v(m)C z Notice that for any $(v(m)) . z e P(2) From the fact that x CI w x every . Let w ?5(w) < y i s K(m) - f u l l y y $(x) . x m Pick i s J(n) - f u l l and that $(v(n)) < y(n) H(n) some . Since our v E P(l) contains 1 . n y K(m) there exists $(y(m)) < a(m) was a r b i t r a r y we P(2)' which contains z& x z . we can conclude that for any . R(y) w(n) C If m z v(n) z belongs to J(n) = { z e R y where z and y y e R , H(n) l s be U^( ) n | n E N} . w(n) E J(n) . But z is P ( l ) and i s a subset of $(x) v y(n) and no subsegment of is in = |v C z and that we such that w(n) C. v(n) i t must happen that x But m y ( l ) be from i s arbitrary Therefore no subsegment of is in i s J(n) - f u l l Observe that tf(v) < y z (z C y C x) i s K(m) - f u l l . From Lemma 1.12 i t follows that f o r by l e t t i n g Thus x i n R(a) and assume that some subsegment, there exists < y < <f>(z) . x Therefore • i s i n K(m) v(n) E L(y, y(n-l) , J(n)) such that and . and since i s H(n) - f u l l z^y v(m) v(m) C y(m) we can see that (y(n) | n e N} 6(y, y(n-l) > J(n)) in = for some v i H(n)} can construct a(m-l) be the member of x e R(y) no subsegment of implies that Since such that and that no subsegment of n Since no subsegment of z C y(m) e I(m) Consider again of . be a subsegment of must be more than This means that f i n d that m z such that no subsegment of y(m) e L(x,a(m-1) , I(m)) < a . Let . In order that also contains for any x v . n Q.E.D. We s h a l l now construct two sequences of closed sets, ( A ( i ) | i e N}, and ( B ( i ) | i e N} y { A ( i ) | i e N} produce V [U^ 0) I 3 B and and W X y { B ( i ) | i e N} "separate" w be H i We s h a l l then [ y ( A ( i ) | 1 z N}] y V and and K {(x,n) | p5(x) < y} . . From Lemma 1.2 we know that i s open and closed and can be written as a sequence, Let A(o) be H X n and B(o) be closed and A(o) ^ B(o) that and B(m-l) and A(m-l) A(m-l) and B(m-l) p(m) t B(m-l) y{p(m)} from = K 'x 4> . T o construct Consider and B(m) be B(m-l) • Since 3 < *5Cy) < $(x)} p(m) = (x,n) and suppose V be equal to one. p(m) there exists J = Since n > 1 there exists A B(m-l) which contain and B(m) Assume the $(x) i s a l i m i t i s closed and does not contain , a neighbourhood of p(m). such that . By the d e f i n i t i o n of neighbourhood there exists A , , which does not i n t e r s e c t , that i s , assume that . Since B(m-l) {(y,l) | x Q y I t i s easy to see that these are what we want. ordinal and that G((x,n),3) d A(m-l) (p(m)}(Lemma 1.11) i s closed and d i s j o i n t , a neighbourhood of p(m) third p o s s i b i l i t y for p(m) p(m) B(m-l) ^A(m-l) = 4> and i s closed by Lemma 1.4. Let A(m) be A(m-l) y J B(m-l) , assume and B(m) are both closed and their i n t e r s e c t i o n i s void. i s closed and does not contain be and A(m) . I f ^(x) i s not a l i m i t ordinal l e t A(m) be B(m-l)»A(m) B(m-l) B(m) have been defined and that are closed. (p(m) | m e H} . A(o) and B(o) are both r1 Now l e t $(x) be a l i m i t ordinal and l e t n $ , A ( i ) ^ B ( i ) ,._= 0 and are both open. , both open, so that U E Let X , such that for every , both of which do not i n t e r s e c t A ^ B(m-l) = F((x,n),i) B(m-l) , and both of (x,n) . By Lemmas 1.6 and 1.7, F((x,n),i) y{(x,n)} yG((x,n),3) is closed. Let A(m) y G((x,n),3) and let B(m) If of A(m-l) p(m) e B(m-l) and be be A(m-l) y{(x,n)} yF((x,n) ,i) B(m-l) then repeat these operations with the roles B(m-l) reversed. Lemma (1.14) The sets, C = y{A(i) | i z N} and D = y{B(i) | i £ N} are both open. Proof Consider there exists C and prove that for every N(x,n) In the case where Suppose A(i) we see that Now let n that the C $(x) i s a limit ordinal. A(i) C X (x,n) = A(m-l) y J (x,n) such that $(x) is not a limit ordinal let N(x,n) n = 1 and Therefore , a neighbourhood of (x,n) contained i n i t where p(m) J = = N(x,n) C. C = {(x,n)} From the construction of {p(l) , p(2) ,...'.} for every for some m . Since p(m) i B(m-l) , A(m) {(y,m) | 3 < «Ky) < $(x)} . Let J be be more than one and l e t $(x) be a limit ordinal. contains some (w,m)'s which do not have min({ m | there exists i . N(w,m)'s N(x,n) Assume . Let k b (w,m) £ C which does not have a N(w,m)}) and l e t 3 be the min({ a | there exists (w,k) £ C with no N(w,k) and with (y,k) E C with such that $(w) = a}) N(y,k) (w,m) E C with = . There must exist $(y) = 3 does not exist but that N(w,m) does exist for a l l ?5(w) < 3 or with m < k . This (y,k) must be p(m) £ for some m and must therefore not be contained in B(m-l) that A(m) i s A(m-l) and for some y y{p(m)} y F(p(m),i) y G(p(m),y) for some i . Our definition of (y,k) tells us that neighbourhood for each point of F(p(m),i) C . This means and G(p(m) ,y) C contains a . Therefore contains a neighbourhood, N(y,k), for (y,k) . This contradiction shows that N(x,n) exists for a l l (x,n) e C The same argument proves that D is open. Q. E. D. Define V' V' , W' , V and W = {x e R(u) in the following manner: | there exists y C x such that y e H(i) for some i} (Remember W' = {x e R(u) u from Lemma 1.13) | there exists y9 such that x y e K(j) for some j} V = W = {(y,n) e R | y ^ x for some x e V'} {(y,n) e R | y x for some x e W'} Lemma 1.13 tells us that V' and W' are disjoint and that are disjoint. Lemma (1.15) The sets, W and V , are both open. W and V Proof If ^(x) i s not a limit ordinal and is a neighbourhood of (x,l) e V $(x) {(x,n)} .. If ^(x) i s a . To prove that (x,n) has a neighbourhood i n V where i s a limit ordinal and n > 1 we shall assume such neighbourhoods for a l l where (x,n) which i s contained in V then {(y,l) | ?5(x) > ^(y) > y , x Q y} does the same thing limit ordinal then for (x,n) e V (y,m) e V such that $(y) < ^(x) $(y) < $(x) and m < n and for (y,n) . Consider G((x,n).,u) . . (Remember y Our definition of V t e l l s us that G((x,n),y) C from Lemma 1.13). v while our assumption about (x,n) t e l l s us that each member of G((x,n),u) has a neighbourhood in . Let A be the union of a l l these neighbourhoods. V is contained i n V there exists the existence of i such that F((x,n),i) C V B w e V' such that f(x,j) C w when j > i . This implies . Therefore and each of its members has a neighbourhood i n V be the union of a l l these neighbourhoods. neighbourhood of x% w Since (x,n) A y B y{(x,n)} . Let is a (x,n) contained in V The same method proves that W i s open. Theorem (1.16) The sets, C y V and D y W, "separate" H C are both disjoint. and K . Proof A(i) and D Suppose there exists and B(j) where without loss of generality j > i x contained . Then by the construetion of the A(n)'s and the B(m)'s x i s contained i n A(j) This i s a contradiction. Since V and and W are contained i n {(x,n) | ^(x) < u} and we can say that C y V are d i s j o i n t . If (x,n) i s contained i n H or (x,n) C and i n the second Therefore and D are contained i n {(x,n) | $(x) > u} D y W ^(y) < y) C H ^ {(y,m) | ^(y) > y} x^ y C y V then either for some y e R(y) (x,n) C {(y,m) | . In the f i r s t case where . The same reasoning puts y K (x,n) C i s obviously i n V' . in D y W We have proven i n Lemmas 1.15 and 1.14 that D y W and C y V are both open. Therefore D y W and C y V "separate" H and K . Q. E. D. Since H and K Lemma 1.11 t e l l s us that T were a r b i t r a r y closed sets i s also Hausdorff. C T i s normal. THE COUNTABLE CHAIN CONDITION IN E E x E WHERE IS AN ORDERED CONTINUUM In 1952 Djuro Kurepa [12] proved that an ordered continuum, E , i s homeomorphic to I(closed unit i n t e r v a l ) if E E x has the C.C.C. We s h a l l prove this and state the obvious equivalence to Souslin's conjecture along with another involving Peano maps. Consider an ordered continuum We s h a l l define f o r each from E E such that a e fi' a c o l l e c t i o n , E D(a) E x has the C.C.C. , of closed segments and, using these closed i n t e r v a l s , s h a l l produce a countable dense subset A look at the proof of Lemma 0.1 w i l l then convince one that Let D(o) be {E} each nondegenerate i n t e r v a l , . If a Now l e t E l s connectedness D(a) be (nondegenerate) {X(o) | X and l e t X(o) y X ( l ) = i n common only one point and this point nondegenerate X X(o) and and If a and be i n D(a) • Let X X(l) ( j ( X ( l ) |x i s a l i m i t ordinal consider a l l {X(g) | g < a} if g > Y X . If X i s X(o) and i s in.D(a-l)} chains of closed i n t e r v a l s of the form X(g)CX(y) X ( l ) have i n t e r i o r of implies the existence of . X ( l ) be X(o) and i s i n the (nondegenerate) i s i n D(a-l)} i s homeomorphic t i s a successor ordinal consider X, i n D(a-l) two closed i n t e r v a l s such that E where X(g) e D(g) i f and only i f i t i s the i n t e r s e c t i o n of a l l elements i n one of these chains. Since E is connected such an i n t e r s e c t i o n w i l l again be a closed i n t e r v a l . We w i l l now state some lemmas concerning of i n t e r v a l s . E and our c o l l e c t i o n s . Lemma 2.1 If exists , X Y e D(3) i s contained i n D(a) and such that X £. Y 3 i s less than a ' then there . Proof By induction and by the d e f i n i t i o n of D(3) Lemma 2.2 If X and Y are i n t e r v a l s i n D(a) then they i n t e r s e c t at most one point and this point i s an endpoint f o r both of them. Proof By induction and the d e f i n i t i o n of D(3) Lemma 2.3 If a < 3 , X e D(a) i n one of X(o) or X ( l ) or X and Y e D(3) and Y then either Y i s contained intersect at at most one point and this point i s an endpoint of both of them. Proof By induction, the d e f i n i t i o n of D{y) and Lemma 2.2 . Lemma 2.4 If a ^ 3 then D(a) ' and D(3) have no elements i n common. Proof By d e f i n i t i o n and induction. Lemma 2.5 Let and l e t A E be a l i n e a r l y ordered space which i s connected be a chain of i n t e r v a l s in E and has the C.C.C. such that any subset of i t contains a greatest element, that i s , an element which contains a l l the others i n the subset. This chain i s countable. Proof Notice that each B e A A e A which i s contained i n A which are i n A Therefore . Since E { Int (A~B) | A e A has a "successor", that i s , a unique and which contains a l l subintervals of A i s connected , B D be This makes y{D(a) | a e ft'} and l e t endpoints of nondegenerate i n t e r v a l s countable dense subset i n E w i l l contain an i n t e r v a l . , the successor of A} countable c o l l e c t i o n of d i s j o i n t open sets. Let A ~ B in D F .We . The fact that F A countable. s h a l l show that {a} = there exists n a(a) e ft' such that {X(a) | a < a (a)} F is a i s dense follows from Theorem 2.6 a e E Q.E.D. be the c o l l e c t i o n of a l l the following theorem. For every w i l l be a where X(a) i s the nondegenerate i n t e r v a l i n D(a) which has. a as an element. Proof Given a l e t P(a,a) be y { X e D(a) | a e X} . Lemma 5.2 t e l l s us that P(a,a) i s an i n t e r v a l ( i f i t i s nonvoid) while Lemma 2.1 t e l l s us that i f P(a,a) exists then a < $ then P(y,a) P(3,a) ^ p(a,a) exists for a l l y < a . Note that P(a,a) Therefore 3 e ft' such that for a l l a > 3 This means that there exists i s the n u l l set. exists. A i n Lemma 2.5 and therefore i s {P(a,a) | a e fi'} forms a chain l i k e countable. P(o,a) and that i f Let a(a) be the least o r d i n a l such that P(a,a) i s the n u l l set. It i s possible to use induction to pick a sequence of i n t e r v a l s such that X(3) C X(ot) e D(a) X(a) . Suppose be two) Z for either such that V C y W e D(y) Z D(3) or Y such that and Y Z a . Ii D(a) has (there can at most , such that for every such that a e WC Z a < y < a(a) . Suppose t h i s didn't happen This would mean that there exists . Since a e U 3 < a(a) such that show this to be a contradiction. a eV there must e x i s t and U C. T f Z , Y each contain' a then l e t i t be X(a) . I f there are two does not contain any V or V C Z T e D(a) a , i n D(a) which contain pick one, say i t i s there exists T ,Z and Y a < 3 and i f i s defined for a l l 3 < a X(3) only one i n t e r v a l which contains intervals, , a e X(cx) {X(a) | a < a(a)} a < 3 < a(a) and either U e D(3) and . This means that . The d e f i n i t i o n of D(a) and Lemma 2.2 A l i t t l e insight and a f a i r amount of w r i t i n g w i l l prove that our X(a) Now i s not exactly if ^{X(a) j a < a (a)} i s contained i n every To see this look at the d e f i n i t i o n of a l i m i t o r d i n a l and when n { X ( a ) | a < a(a)} a {Int then X(ct(a)) w i l l for the two-cases when i s a successor o r d i n a l . 3 < a . Therefore {a} exist. a is = . What we need now the d e f i n i t i o n of D(a) {a} X(3) ..for X(o) (X(o) x X ( l ) ) | X i s to show that and , X(l) for F i s countable. X e D consider the c o l l e c t i o n , nondegenerate i n t e r v a l i n Lemma 2.3 and the d e f i n i t i o n of X(o) and Remembering X(l) D} C E x E . From we can see that this i s a pairwise d i s j o i n t c o l l e c t i o n of open sets and therefore must be countable as must F . Therefore E i s separable. The obvious equivalence to Souslin's conjecture mentioned before i s : Consider an ordered continuum, when E E . Does E x E have the C.C.C. does? In 1960 Sibe Mardesic [13] considered an ordered continuum and proved that i f i t could be mapped continuously onto.its square then i t had the C.C.C. Souslin's hypothesis and Peano's theorem can prove the converse while the converse along with Kurepa's r e s u l t w i l l prove Souslin's hypothesis. CHAPTER 3 SOUSLIN TREES Assuming the existence of a Souslin line, collection R by inclusion. = S , consider the y{R(g) | g e ft'} from Chapter 1 and partially order i t Notice that for every x e R the set {y | y e R , x < y} is a well ordered set when given the reverse order from ordered set with such a characteristic i s called a tree. that this i s not the standard definition of tree. R . Any partially It should be noted Usually i t is defined as a partially ordered set in which the collection of predecessors of each point is a well-ordered set. Notice that our type of tree i s a standard type tree with the reverse order and visa versa. One may ask whether this tree, R , can be turned into a Souslin space by squeezing i t f l a t , that i s , by f i t t i n g each member of R(a + 1) These members of blocks of R(a) R(a + 1) into the member of R(a) from which i t came. can be looked upon as the .framework into which the are fitted. is possible for any tree, T This, i n fact, can be done with R and , with the following characteristics: 1. T has cardinality 2. Each chain i s countable 3. Each antichain i s countable An antichain i s a subset in which any two distinct elements are not comparable in the partial order. A tree which has the three characteristics above is called a Souslin tree. Our R with the partial ordering is a Souslin tree. Since S is connected and has the C.C.C. any chain is countable by Lemma 2.5. From the structures of the R(a)'s x C: y , y C x or x ^ y we can see that for = x and y in R either This means that any, antichain i n R is a collection of pairwise disjoint open entervals and as a result i s countable. R has cardinality ^ since each contains a countable number of elements. T(3) T(a) is nonempty and Remember that for a ^ g,T(a) and are disjoint. The purpose of this chapter is to show that a negative answer to Souslin's conjecture is equivalent to the existence of a Souslin tree. Notice that this equivalence has just been proven in one direction. Suppose we have a Souslin tree, shares several characteristics with our R T . We already know that T , above, namely those that mark i t as a Souslin tree. We shall add to T several other traits possessed by R and then squeeze i t down to form a Souslin space. for every a e fi' l e t T(a) be the set of a l l x First, in T whose successors, when given the reverse order, can be put into a one to one order preserving correspondence with (3 | 3 efl','• 3 < a} successors for each x . Remember that the set of is well-ordered when given the reverse order. shall now l i s t several characteristics of these We T(a)'s Lemma (3.1) Every x in T belongs to one and only one T(a) Proof For any x in T , i t s successors, when given the reverse order, i s a well-ordered set of c a r d i n a l i t y less than or equal to ^ \ Q and therefore can be put into a one to one order preserving correspondence with one and only one ordinal which i s s t r i c t l y less than ft Lemma (3.2) Consider x e T(a) and only one y e T(f3) . Then for every such that i s contained i n T(y) f o r some x <y 8< a there exists one . Also every successor of x y <a Proof This c l e a r l y follows from the d e f i n i t i o n of T(a) and the proof of Lemma 3.1. Lemma (3.3) For every a eft'T(a) i s a nonempty" antichain and therefore countable. Proof If there existed an a would show that T(8) d i c t i o n since each fact that such that T(a) were empty then Lemma 3.2 i s empty for a l l 8 > a T(y) i s countable and T . This would be a contra- has c a r d i n a l i t y ^ j . The T(a) i s an antichain comes from Lemma 3.2. Please note that a l l of the above mentioned c h a r a c t e r i s t i c s are held by the R(a)'s in R when R has the i n c l u s i o n order. R also possessed no short arms. We shall now give this characteristic to T . Let V be the collection of a l l x's which have a countable number of predecessors and let U V which have no successors. U countable. w e T(3) x e U such that y < w y < x be a l l y's in of course is an antichain and therefore By remembering that for every such that in T 3 < a z e T(a) and we can see that for every . This means that V y e V there exists there exists is countable and that T~V is a Souslin tree in which each member has uncountable number of predecessors. (Each uncountable subset of a Souslin tree is a Souslin tree) in such a Souslin tree every is more than a x e T(a) . Note that when 3 has a predecessor i n T(3) . To see this remember that each T(y) i s an antichain and that the set of successors of any point is a well-ordered set with the reverse order. Finally we must make sure that the members of do the members of R . T "fork" Suppose we have a Souslin tree, every member has an uncountable number of predecessors. i f and only i f there exists y T , as , i n which Let x which has the same successors as x course in this case the successors of x were countable then there would exist P e fi such that for every T(y) contains no points of contain a y form a chain. of T T' . T(3) shall not include x) be in 1 . If 1 (Of T 1 y >$ is nonempty and therefore must with an uncountable number of predecessors. This contradiction shows that T' and therefore a Souslin tree in which each the same successors. . T These predecessors is an uncountable subset x has a y which has We can therefore assume that each member of T "forks" and has an uncountable number of predecessors. We are now ready to squeeze Let x be related to y T , that i s , give i t a total order. i f and only i f they have the same successors. This is an equivalence relation which divides T into a collection of disjoint sets each of which i s countable since i t i s an antichain. This means that each equivalence class can be written as a sequence which w i l l give i t an obvious total order. For each x e T let {g(l)(x) , g(2)(x),... . ,} be i t s equivalence class with this order. We are defining a way in which the bricks of T(a + 1) are fitted into their framework in T(a) Now for the actual ordering, y e T(3) where (Lemma 3.2) If x such that and y . If y < x y < t . Consider x e T(a) and then there exists a unique . Let y < (T) x 1 , and m , n e N i f and only i f t = g(l)(t) . and m ^ n , a l l unique, such that t l e t z(y) and w(y) be the unique members of T(y)(y < a)(Lemma 3.2) which are more than respectively, and l e t K be null set l e t t be w(o) (z(y) | z(y) $ W(Y), y < a} y < g(m)(t) Note that t . y and x , If K while w(y) z(y) i s the unique i s the unique g(n)(t) g(m)(t) such that such that y < g(m)(t) , x < g(n)(t) such x < g(n)(t) . i s uniquely determined by the existence of the elements g(m)(t) , and g(n)(t) i s the , otherwise let t be the greatest element i n . If t i s contained i n T(y) then that y < g(m)(t) , . Let y < (T)x i f and only i f g(m)(t) < g(n)(t) . To prove the existence and uniqueness of K t e T(a + 1) are not related in the partial order then there exists y e ft , t e T(y) x < g(n)(t) 8> a <(T) t, and m ^ n To prove t r a n s i t i v i t y suppose that we have y < (T) z . ((i) x < y To prove that and y < z (ii) x < y p a r t i a l ordering ( i i i ) x while y and z are ordering nor are Now and T(y) T n y (iv) and If M M C t x . w y and and y are not related i n the p a r t i a l and apply the d e f i n i t i o n T , • Since each ? > 3 there exists a > y where T(£ +1) g(l)(w) < (T) s .We and t o t a l order we can see that g(l)(w) g(2)(w)) and s . such that g(2)(w) < (T) s contains no points from If x(a) < y(a) in (x(a) , y(a)) t t(a) , . If s e T(a) . i n the t o t a l order. where The same thing happens i f Therefore Let (g(l)(w) , t of open x(a) be a p a r t i a l order predecessor of are not related i n the p a r t i a l order. is then from the d e f i n i t i o n of our but does contain The same i s true f o r contains an (g(l)(w) , g(2)(w)) then a l l the p a r t i a l order predecessors of . such contains a that we have a d i s j o i n t c o l l e c t i o n {(x(a) , y(a)) | a e A} and l e t M 3 is then i t contains a Therefore M <(T) i s countable T(£) x e T(y) can show that g(l)(w) < s T(a) are not related i n the p a r t i a l order. Suppose now T(3(a)) of the t o t a l order. with the order topology from a nonempty i n t e r v a l containing no points from a < 3 are not related i n the are not related i n the p a r t i a l ordering We mentioned e a r l i e r that i f in z i s a countable set then there exists T(a) and consider the four p o s s i b i l i t i e s and <j> for a l l y > $ = predecessor i n each predecessor y z) T i s uncountable element, and l e t us show that nonseparable. that x < (T) z x < (T) y g(2)(t(a)) if intervals, be contained i n x(a) in T(3(a) + 1) . g(2)(t(a)) are x(a) and y(a) Notice that the g ( 2 ) ( t ( a ) ) ' s are a l l distinct from each other. We now have (g(2)(x(a)) | a e A} which is an anti-chain i n the partial ordering and which therefore i s countable. This makes our collection of open intervals countable. CONTINUOUS IMAGES OF ORDERED COMPACTA AND CONTINUA PART 1 Assuming Souslin's hypothesis to be true one is able to work with continuous images of ordered compacta and ordered continua and, in fact, formulate Souslin's hypothesis in terms of them. Mardesic and Papic [7J set the stage for this with these two lemmas: Lemma 4.1 If X has the C.C.C. and is the continuous image of an ordered compactum then this ordered compactum can be picked so that i t too has the C.C.C. ([2], Corollary 5, p. 18). Lemma 4.2 Consider compactum. If X X , a locally connected continuous image of an ordered is separable then i t has a countable base. ([.7.3'Theorem 8,p.17) Actually the results of Mardesic* and Papid were more general than those just stated but what we have here is sufficient for our purposes which are to prove the following two theorems, taken from the same paper, [7_]. Theorem 4.3 We have a p o s i t i v e answer to Souslin's conjecture i f and only i f a compactum which i s the continuous image of an ordered compactum i s separable when i t has the C.C.C. Proof Assume we have a positive answer to Souslin's conjecture and a compactum, X , which has the C.C.C. and which i s the continuous image of an ordered compactum. For our purposes we s h a l l assume that an ordered compactum with the C.C.C. i s separable. t e l l s us that X i s equal to ordered separable compactum. This fact along with Lemma 4.1 f(K) where X f i s continuous and K i s an then i s separable. The converse i s t r i v i a l . Theorem 4.4 A p o s i t i v e answer to Souslin's conjecture i s equivalent to the statement that a compactum which i s the continuous image of an ordered continuum i s metrizable i f and only i f i t has the C.C.C. Proof Every metrizable compactum i s separable and as a r e s u l t has the C.C.C. ([15], p. 187, Theorem 5.6). Assume we have a positive answer to Souslin's conjecture and a continuum, X , which has the C.C.C. and i s equal to an ordered continuum and f i s continuous. f(R) where Since a l l sets i n K K is of the form, {x | a < x < b , a, b e K} which i s connected we can say that compact and X A in K i s Hausdorff . Therefore , constitute a base each member of K i s locally connected. Theorem 1.3, p. 121 and Theorem 3.5, p. 125). countable base. Since X K is f(A) w i l l be closed for a l l closed sets f i s closed and X Souslin's conjecture makes Since X i s locally connected.([15], The affirmative response to separable so that Lemma 4.2 can give X a i s regular (Hausdorff and compact) i t i s metrizable ([16], p. 125, Theorem 17). Now for the converse. Suppose that in a compactum, which i s the continuous image of an ordered continuum, being metrizable and having the C.C.C. are equivalent. Then an ordered continuum with the C.C.C. i s metrizable and as a result separable ([15],p. 187, Theorem 5.6). question is answered i n the affirmative. Souslin's CHAPTER 4 PART 2 A compactum set X A/R X is said to have property i s metrizable where in the same component of A x R y y i f for every closed i f and only i f x . It i s said to have property and y are 0. i f and only i f there is a countable collection of open sets such that for every disjoint pair of closed sets, M which separates X disjoint sets, A A y B = S and between and N , there is a member, S M and N B ,such that and both A and . This means that there exists M^A B , of this collection , N^B , A^B are closed in A y B = . In & , 1967 Sibe Mardesic" [8] proved that a compactum i s metrizable i f and only i f i t has properties a whether property and y u . In this same paper he raised the question alone would imply metrizability for a compactum. There i s a lemma which shall be proven later which says that an ordered continuum with the C.C.C. has property y y . Therefore i f i n fact property does imply metrizability for a compactum we have a positive answer to Souslin's conjecture. Another lemma which too shall be proven later says that a compactum with property y has the C.C.C. Therefore i f we assume the Souslin hypothesis and remember the last section we can respond to Mardesic" that in a compactum which is the continuous image of an ordered continuum property y goes one step further. alone implies metrizability. However Mardesic [8] He shows that Souslin's hypothesis implies that in a compactum which i s the continuous image of an ordered compactum property y i s sufficient for metrizability. He also proves the easy converse. It is the purpose of this section to prove this equivalence. One of the things needed i s that i f a compact, separable Hausdorff space is the continuous image of an ordered compactum then i t possesses property a It i s easy to see at this stage that this fact plus the two lemmas mentioned above establish MardeVid's equivalence. We shall prove these three lemmas as well as Mardesic's metrization theorem involving properties u and a Preceding a l l this w i l l be the statements of several lemmas that we need. Lemma 4.5 A totally disconnected ordered compactum with only a countable number of gaps i s metrizable. Proof ([9J , Lemma 1, p. 867) Lemma 4.6 If X and only i f x i s a compactum then X/R i s a compactum where x R yif and y are in the same component of X Proof •[10] Lemma 4.7 The continuous image of a compact metrizable space In any Hausdorff space i s metrizable. Proof ([17], p. 102, Problem 8B). Lemma 4.8 Consider X , a separable compactum, which is the continuous image of an ordered compactum. sets which separates There i s a countable collection of closed X between any pair of disjoint closed sets. Proof (111] , Theorem 4) Lemma 4.9 In a separable continuous image of an ordered compactum each closed set i s a G(<5) and each open set i s a F(a) Proof ([7], Corollary 3). Here is the proof of MardeVic's metrization theorem. Lemma 4.10 A compactum X has properties u and a i f and only i f i t i s metrizable. Proof Since X i s a T^ , regular space (compact and Hausdorff) the existence of a countable base w i l l imply that Consider the countable collection a property y . For every n eN C = X is metrizable. ([16],p. 125). {S(n) | n e N) l e t x R(~S(n)) y are in the same component of ~S(n) . Property n ~S(n) / R(~S(n)) guaranteed by i f and only i f x and y says that for every i s metrizable. Since i t i s also the continued image of a compact space i t w i l l have a countable base, V(n) = {V(n,k) | k e N} ([15], p. 187). We w i l l now prove that G = y ;|{S(n) y-p(n)" is a countable base for our topology. onto ~S(n) / R (~S(n)) . Since 1 (V(n,k)) | k e N} | n e N;| (P(n) i s the projection of ~S(n) P(n) -1 (V(n,k)) is equal to 0(n,k) T ~ S ( n ) where that i s open in X S(n) y P(n) (V(n,k)) X w i l l show that for every point, y y tM x there exists i s T^ B eG there exists 0(n,k) i s open i s open i n X for every y e B and B ^ M and k . Because {y} and M such that (i) A y B = ~S(n) (ii) A (iii) y e A , M S. B (iv) A , B are closed i n A y B n . We , with = $ S(n) e C which separates between and B it . This assures , and closed set , M such that This means that there is A Each component of A y B n in ~S(n) B = < } > is connected and therefore cannot intersect both of these sets. This implies that ~S(n) / R(~S(n)) where P(n)(A) and closed. that n P(n)(B) Because = = ?5 and P(n)(A) P(n)(y) e P(n)(A) P(n)(y) e V(n,k) and Suppose now that X . A V(n,k) P ( n ) (B) X x R y for x , y e A component of A P(n)(B) $ . The member of G (V(n,k)) X A (X i s compact) . Therefore i s a closed subset and has property X and i s metrizable. Let 8 base together with a l l f i n i t e unions from i t . i s normal since i t i s compact and Hausdorff. and P ,such that M^ 0 covers ~(0yP) and y are i n the same Since = X i s a compact ( B ( i ) | i e N} be t h i s B I t i s easy to see that We s h a l l prove o This means that f o r any N ,there are two d i s j o i n t open sets and N the union of a l l elements of 8 8 A/R . Remember please i s the countable family of open sets necessary for property two d i s j o i n t closed subsets, M 0 is a u i s a countable base which i s closed under f i n i t e unions. X A/R Lemma 4.7 can t e l l us that i f and only i f x metric space i t has a countable base. B such . Suppose again that that are both open V(n,k) e V(n) i s metrizable and that i s metrizable with the r e s u l t that that = n i s a compact metric space Hausdorff space (Lemma 4.6) and P(n)(B) there exists we are looking f o r i s S(n) y P(n) of P(n)(A) y P . Since contained i n i t . which Is compact exists a f i n i t e c o l l e c t i o n , (B(i) | i ~ (My N) This c o l l e c t i o n from (X i s compact) = 1 i s open, i t i s . Therefore there n ) , from B such that y{B(i) | i = l,...,n} •= S e B contains M y N that . Let A be 0 V (>i MC A ,N9.B , A y'B="S A closed in. A y B i.e. and B be P :~S r we can see that "(OyP) but is disjoint from and A n . It i s easy to see B = i> . From their definitions and B are both open in A y B and therefore are both . Therefore X has property a S e B separates X between M and N , . These lemmas, needed for our major theorem, w i l l be stated along with their proofs. Lemma 4.11 Consider a compactum, X ordered compactum. If X , which is the continuous image of an i s separable i t has property er Proof By Lemma 4.8 there exists a countable collection (F(n) | n e N} of closed sets which separate between any two disjoint closed sets. Lemma 4.9 we know that each of these of open sets. is normal. F(n)'s Please notice that since X From i s a countable intersection is compact and Hausdorff i t This means that the closed neighbourhoods of a closed set form a neighbourhood basis for i t . n Now consider one of these F(n)'s which i s {0(n,j) | j e N , 0(n,j) open} The normality of X says that for each neighbourhood of F(n) , and S(n,j) j there is W(n,j) , an open set, such that F(n) C (n,j) ^ W(n,j) 9 l 0(n,j) S , a closed . It i s easy to see that F(n) = n {S(n,j) | j £ N} = n { C £ (S(n,j)) | j £ N} We w i l l show that {S(n,j) | n , j E N} 8 , the set of a l l finite intersections from , i s the collection we want.. If M disjoint closed sets there is an Remember that that F(n) = F(n) C. ~ (MyN) j (1) ,. . . , j (m) } in ~ (MyN) n F(n) which separates X n {C-£(S(n,j)) | j £ N} . Suppose not . then {MyN} N are two between them. {(S(n,j)) | j £ N} = {Ct(S(n,j)) . There is a f i n i t e collection from and | j £ N} and (C.£(S(n,j)) | j = whose intersection i s contained y{C£(S(n,j)) | j £ N} would be a collection of closed sets with the finite intersection property. Since X i s compact Let K e B contradiction. F(n) F(n) would have a point i n MyN separates between M£.A,N£.B,A B n in A y B K separates M = be X between ^{S(n,j) | j = j (1) , . . . , j (m)} and N there exists ^ ,A y B . If we consider M . This i s a = ~F(n) and A ^ ~ K and A and and A B and . Since such that B are closed B ^ ~ K we can see that N Lemma 4.12 An ordered continuum, X , with the C.C.G. has property u Proof If A i s a closed subset of X we w i l l show that A/R with the quotient topology i s a totally disconnected ordered compactum with only a countable number of gaps. metrizable. Remember that Lemma 4.5 w i l l then imply that x R y i f and only i f x and A/R is y are in the same component of A If set W X also. x and y are contained i n W , then every point of A If this were not so W , a component of the closed between x and would not be connected. to put a linear order on the set of components of A order topology of this set. A/R of and A/R {W | W where ^ Y Consider Y e A/R} which i s open in A of X Y . If P {x | x e X , x < b _1 and to consider the A onto ({W | W £ Y}) i s -1 = g . 1 . b . of Y} ^A follows from the connectedness* ({W | W ^ Y}) and {x | x e X , x < b} ^A r i s closed. This shows that has the order topology. The fact that with the order topology, i s Hausdorff implies that has this order topology. This enables us , the projection from . The existence of b follows from the fact that i s contained in , a subbasis element of the order topology while the equality of P when A/R P i s an element of A/R nonempty then i t i s equal to y A P P is continuous i s compact and A/R i s closed when , A/R Therefore this topology i s in fact the quotient topology ([16}, Theorem 8, p. 95). From the structure of A/R is connected i f and only i f C we can see that P (C) , C £ A/R A consists of a single point. , Remembering that a component i s closed, we can use Theorem 3.4, p. 124 of [15] to show * A linearly ordered space i s connected in the order topology i f and only i f there are no gaps and the space i s order complete. that A/R is totally disconnected. A l l we need to do now is to prove that number of gaps. Suppose we have component of between them. A E ^ F e A/R Since X A/R such that there is no is connected and both are closed in X , E contains g.£.b. Using again the connectedness of for i t s e l f . a £,u.b. for itself and between these two points in X in ~A has only a countable is an element of ~A F E and contains a X we can see that and i n fact an interval which contains i t . In this manner assign to each gap in A/R open interval of X The C.C.C. of proves that we have only a countable number of gaps. X F an and uotice that any two of these intervals are disjoint. Lemma 4.13 A compactum, X , which has property \i also has the C.C.C. Proof { U(X) | X s Let open sets. From each closure of {x(X) that X onto y Since | X e L} pick be a collection of pairwise disjoint x(X) e U (X) A/R then {P(x(X))} X {x(X)} i f and only i f x A and denote by A the . From the definition of closure and the fact X e L {x(X)} since for every to U(X) X is Hausdorff i t follows that that for each A L} £ is open in U(X) A n . A = {x(X)} If is open for every is i t s own component. and y P . This means is the projection of X e L . (Remember This is true x are in the same component of is a closed subset of a Hausdorff space and X i s related A) has property u A/R i s a compact metric space. This means that i t is separable and can only have a countable number of points fore { U(X) | X e L} y such that {y} i s open. There- i s countable. Here is the major theorem. Theorem 4.14 A positive answer to Souslin's conjecture i s equivalent to the statement that a compactum which i s the continuous image of an ordered compactum being metrizable i f and only i f i t has property u Proof Assume that a compactum which i s the continuous image of an ordered compactum is metrizable i f and only i f i t has property u . With the aid of Lemma 4.12 we can see that an ordered continuum with the C.C.C. is metrizable. Since a compact metric space i s separable we have a positive answer to Souslin's conjecture. We already know that any metrizable compactum has property u (Lemma 4.10). Assuming a positive answer.to Souslin's conjecture consider a compactum, X , which has property of a ordered compactum. u and which i s the continuous image Theorem 4.13 implies that enables Souslin's hypothesis to t e l l us that X has the C.C.C. This is separable. is needed now i s to look at Theorem 4.11 which gives as a result makes i t metrizable. X X A l l that property a and 1. M. Souslin, Probleme 3, Fundamenta Mathematical, Vol. 1(1920), p. 223. 2. C. H. Dowker, On Countably Paracompact Spaces, Can. J. Math., Vol. 3. 3(1951), 219-224. M. E. Rudin, Countable paracompactness and Souslin's problem, Can. J. Math., Vol. 7(1955), 543-547. 4. M. E. E s t i l l , Concerning a problem of Souslin's, Duke Math. Journal, Vol. 5. M. E. E s t i l l , Separation in non-separable spaces, Duke Math. Journal, Vol. 6. 19(1952), 629-640. 18(1951), 623-629. M. E. Rudin, Souslin's conjecture, American Mathematical Monthly, Vol. 76, (1969), 1113-1119. J 7. S. Mardesic and P. Papic, Continuous images of ordered compacta, the Souslin property and diadic compacta, Glasnik Mat. - Fiz. Astronom. , Vol. 8. 17(1962), 3-25. S. Mardesic, Images of ordered compacta are peripherally metric, Pacific Journal of Math., Vol. 23, No. 23, 1967. 9. L. B. Treybig, Concerning continuous images of compact ordered spaces, Proceedings of the Amer. Math. Soc, Vol. 15(1964), 866-871 10. V. I. Ponomarev, On continuous decompositions of bicompacta, Uspechi Mat. Nauk. 12(1957), 335-340. 11. S. Mardesic, Continuous images of ordered compacta and a new dimension which neglects metric subcontinua, Trans. Amer. Math. Soc. 121(1966), 424-433. 12. Djuro Kurepa, Sur une propri£te* carateristique du continu l i n e a r e et l e probleme de Suslin, Publications de 1' i n s t i t u t mathematique de 1' Academe Serbe des Sciences, V o l . 4(1952), 97-108. 13. S. Mardesic", Mapping ordered continua onto product spaces, Glasnik Mat. F i z and Ast., 15-2, 14. (1960). A. Lelek, On Peano functions, Prace Matematyczne Warsaw, V o l . 7, (1962), 127-139. 15. James Dugundji, Topology, A l l y n and Bacon, Inc., Boston (1968). 16. John L. Kelley, General Topology, American Book, Van Nostrand, Reinhold, New York, Toronto, London, Melbourne, 17. (1955). S. - T., Hu, Elements of General Topology, Holden-Day, Inc., San Francisco, London, Amsterdam, (1969).
- Library Home /
- Search Collections /
- Open Collections /
- Browse Collections /
- UBC Theses and Dissertations /
- Souslin's conjecture and some equivalences
Open Collections
UBC Theses and Dissertations
Featured Collection
UBC Theses and Dissertations
Souslin's conjecture and some equivalences Daniel, Ian Alistair 1971
pdf
Page Metadata
Item Metadata
Title | Souslin's conjecture and some equivalences |
Creator |
Daniel, Ian Alistair |
Publisher | University of British Columbia |
Date Issued | 1971 |
Description | This thesis deals with some equivalences to Soulin's hypothesis. Also included is a consequence of its negation, the existence of a normal Hausdorff space which is not countably paracompact. Two equivalences to Souslin's hypothesis can be obtained by considering the countable chain condition in the product with itself of an ordered continuum and by considering Peano maps on ordered continua. An equivalence to a negative answer is the existence of a tree of cardinality, [formula omitted] ₁, in which every chain and antichain (set of pairwise incomparable objects) is countable. Another equivalence to Souslin's hypothesis is obtained by assuming that a continuum which is the continuous image of an ordered continuum is metrizable if and only if it has the countable chain condition. We get yet another by assuming that a compactum which is the continuous image of an ordered compactum is separable if and only if it has the countable chain condition. Finally, if we restrict ourselves to compacta of the type just mentioned we find that Souslin's hypothesis is equivalent to a generalization of a metrization theorem of S. Mardeśic. |
Subject |
Set theory |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-05-25 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080502 |
URI | http://hdl.handle.net/2429/34814 |
Degree |
Master of Arts - MA |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
Download
- Media
- 831-UBC_1971_A8 D35.pdf [ 2.37MB ]
- Metadata
- JSON: 831-1.0080502.json
- JSON-LD: 831-1.0080502-ld.json
- RDF/XML (Pretty): 831-1.0080502-rdf.xml
- RDF/JSON: 831-1.0080502-rdf.json
- Turtle: 831-1.0080502-turtle.txt
- N-Triples: 831-1.0080502-rdf-ntriples.txt
- Original Record: 831-1.0080502-source.json
- Full Text
- 831-1.0080502-fulltext.txt
- Citation
- 831-1.0080502.ris
Full Text
Cite
Citation Scheme:
Usage Statistics
Share
Embed
Customize your widget with the following options, then copy and paste the code below into the HTML
of your page to embed this item in your website.
<div id="ubcOpenCollectionsWidgetDisplay">
<script id="ubcOpenCollectionsWidget"
src="{[{embed.src}]}"
data-item="{[{embed.item}]}"
data-collection="{[{embed.collection}]}"
data-metadata="{[{embed.showMetadata}]}"
data-width="{[{embed.width}]}"
async >
</script>
</div>
Our image viewer uses the IIIF 2.0 standard.
To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.831.1-0080502/manifest