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Souslin's conjecture and some equivalences Daniel, Ian Alistair 1971

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SOUSLIN'S CONJECTURE AND SOME EQUIVALENCES by IAN ALISTAIR DANIEL B. Sc., University of British Columbia, 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS in 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 t h i s t h e s i s i n p a r t i a l f u l f i l m e n t o f the r e q u i r e m e n t s f o r an advanced degree at the U n i v e r s i t y o f B r i t i s h C o l u m b i a , I a g r e e t h a t t h e 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 r e f e r e n c e and s t u d y . I f u r t h e r agree t h a t p e r m i s s i o n f o r e x t e n s i v e c o p y i n g o f t h i s t h e s i s f o r s c h o l a r l y p u r p o s e s may be g r a n t e d by the Head o f my Department o r by h i s r e p r e s e n t a t i v e s . I t i s u n d e r s t o o d t h a t c o p y i n g o r p u b l i c a t i o n o f t h i s t h e s i s f o r f i n a n c i a l g a i n s h a l l not be a l l o w e d w i t h o u t my w r i t t e n p e r m i s s i o n . Department of • • K A "T ^ V\ M \ CS The U n i v e r s i t y o f B r i t i s h C olumbia V a n c o u v e r 8, Canada Supervisor: Professor T. Cramer ABSTRACT This thesis deals with some equivalences to Soulin's hypothesis. Also included i s a consequence of i t s negation, the existence of a normal Hausdorff space which i s not countably paracompact. Two equivalences to Souslin's hypothesis can be obtained by considering the countable chain condition i n the product with i t s e l f of an ordered continuum and by considering Peano maps on ordered continua. An equivalence to a negative answer i s the anti c h a i n (set of pairwise incomparable objects) i s countable. Another equivalence to Souslin's hypothesis i s obtained by assuming that a continuum which i s the continuous image of an ordered continuum i s metrizable i f and only i f i t has the countable chain condition. We get yet another by assuming that a compactum which i s the continuous image of an ordered compactum i s separable i f and only i f i t has the countable chain condition. F i n a l l y , i f we r e s t r i c t ourselves to compacta of the type j u s t mentioned we f i n d that Souslin's hypothesis i s equivalent to a g e n e r a l i z a t i o n of a metrization theorem of S. Mardesic. existence of a tree of c a r d i n a l i t y , i n which every chain and ACKNOWLEDGEMENTS I would like to thank Dr. T. Cramer for suggesting this topic and for giving me much help during the writing of the thesis. I would also like to thank Dr. J. V. Whittaker for reading i t . Thanks also goes to Dr. Z. A. Melzak for translating some Polish for me and to Dr. F. Burton-Jones for suggesting several papers for me to read. I would also li k e to thank the National Research Council of Canada for their financial assistance. TABLE OF CONTENTS Introduction , . '• Page — ° — Chapter 0 Terms, d e f i n i t i o n s and simple equivalences of Souslin's hypothesis ; 1 Chapter 1 Souslin's problem and a normal Hausdorff space which i s not countably paracompact 5 Chapter 2 The countable chain condition i n E x E where E i s an ordered continuum 31 Chapter 3 Souslin Trees 36 Chapter 4 Continuous images of ordered compacta and continua. 43 In 1923 M. Souslln [3] considered a t o t a l l y ordered set, gave i t the order topology and asked whether i t i s separable i f i t i s connected and has the countable chain condition, i . e . contains at most a countable number of d i s j o i n t open sets. 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 equivalent to saying that a t o t a l l y ordered set with no f i r s t 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 countable chain condition i n the order topology. I t i s c l e a r that a negative answer i s equivalent to the existence of a t o t a l l y ordered set which i s connected, nonseparable and has the countable chain condition when given the order topology. Any such t o t a l l y ordered set w i l l be c a l l e d a Souslin l i n e . able to f i n d an open subsegment of i t which has no separable su b i n t e r v a l s . 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 , T. , of subintervals of which we take the product with N ~ {0} to get a normal Hausdorff space which i s not countably paracompact. In Chapter 2 we construct a set s i m i l a r to T i n order to prove that a continuum i s separable i f and only i f i t s product with i t s e l f has the countable chain condition. This leads to two equivalences to Souslin's hypothesis, one of which involves Peano maps. In Chapter 3 we p a r t i a l l y order T from Chapter 1 and get a tree of c a r d i n a l i t y 1 with no uncountable chains or antichains. The existence of such a tree i n turn implies the existence of a Souslin l i n e . In Chapter k we assume Souslin's hypothesis, consider continua and compacta .which are continuous images of ordered continua and ordered compacta, r e s p e c t i v e l y , and prove r e s u l t s In Chapter 1 we assume the existence of a Souslin l i n e and are concerning the s e p a r a b i l i t y of the compacta and the m e t r i z a b i l i t y of the continua. We also f i n d that we can relax the conditions on a me t r i z a t i o n theorem of Mardesic's [8] by considering the images of ordered compacta. In both these cases we get equivalences to Souslin's conjecture. This thesis i s not a complete l i s t of r e s u l t s concerning Souslin's hypothesis. Those wishing more should consult M. E. E s t i l l [4] and the references and footnotes of Kurepa [12]. TERMS, DEFINITIONS, AND SIMPLE EQUIVALENCES OF SOUSLIN'S HYPOTHESIS This chapter introduces d e f i n i t i o n s , notation and several simple equivalences to Souslin's conjecture. The abbreviation, C.C.C., w i l l be, used f o r the countable chain condition. The symbol, Q , w i l l stand f o r the f i r s t uncountable o r d i n a l while, Q\ , w i l l stand f o r a l l the ordi n a l s s t r i c t l y l e s s than fi Small Greek l e t t e r s w i l l stand for members of fi' while l e t t e r s such as i , j , k, Zy m, n, p, and q w i l l stand f o r members of the natural numbers, N . A compactum w i l l be a compact Hausdorff space while a continuum w i l l be a connected compactum. We s h a l l o ften use the f a c t that a t o t a l l y ordered set has a connected order topology i f and only i f i t i s order complete and has no gaps. A t o t a l l y ordered set has a gap i f i t contains two points with no element between them. The following lemma contains several equivalences to Souslin's conjecture. Lemma 0.1 Let X be a t o t a l l y ordered set which has the C.C.C. when given the order topology. The following statements are equivalent. 1. I f X i s connected i t i s separable 2. X is separable 3. If X is connected and has no f i r s t or last element then i t is homeomorphic to the real numbers with the usual topology. Proof 1 implies 2 Take the Dedekind completion of X and c a l l i t X' . Notice that X' with the order topology has the C.C.C. and that i t s separability implies that of X . We wish to construct a totally ordered set X" which is a quotient space of X'. and i s connected. Intuitively, what we are doing i s connecting X by making i t order complete and by removing a l l gaps. To construct X" we w i l l define for each ordinal y an equivalence relation R(y) on X1 . We w i l l also define a collection of totally ordered sets, {W(x)| x , ordinal} , and a collection of relations (S(x) | x , ordinal} where S(x) i s a relation 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 R(o) . Consider x + 1 where x i s an ordinal and assume that W(y) and S(y) are defined for -all y < x Let W(x + 1) be the quotient, W(x) | S(x) , and order i t in the obvious way. Let c , d e W(x + 1) be S(x + 1)—related i f and only i f there is only a f i n i t e number of points between them. Let a , b e X' be. R(x + 1)-related i f and only i f there exists (A(y) | y < x + 1 , A(y) e W(y) , A(u) £ A(w) for u < w} and {B(y) | y < x + 1 , B(y) e W(y) , B(u) e A(w) for u < w} such that a = A(o) , b = B(o) and A(x + 1) and B(x + 1) are S(x + 1 ) - related. If x is a limit ordinal let a R(x) b i f and only i f for some y < x a R(y) b . Let W(x) be X'/R(x) , ordered in the obvious way and l e t c , d e W(x) be S(x) - related i f and only i f there is only a f i n i t e number of points between them. Now l e t a , b e X' be R- related i f and only i f a is R(x) - related to b for some x , an ordinal. Order X/R in the obvious way and notice that i f is connected and has the C.C.C. It must therefore be separable as is X' 2 implies 3 For n e N ~ {o} let A(n) be {k / n | k e N} and l e t B(n) be A(n) ~ U{A(m) | m < n} . Also l e t {x(i) | i e N} = Z be a countable dense subset of X . We w i l l define an order preserving, one-to-one map, f , from the nonnegative rationals, U{B(n). | n e N}, onto (x(i) | X(i) > X(o)} . Let f(o) be x(o) . Consider I > 0 and suppose f(m) is defined for a l l m < L (m , £ e B (1)) . Let f (£) be x(j) such that j is the least natural number such that X(j) i s more than f(£ - 1) . Suppose we have y e B(n) (n > 1) such that [ u{B(m) | in < n}] (j {w e B(n) | w < y} has been put into a. one to one order preserving correspondence with a subset of Z . Notice that there k k+1 1 exists k E N such that — r - < y < — r - • Since Z is dense in X i t n-1 J n-1 k k+1 has members between f (— r - ) and f (— r - ) • Let, f(y) be x(i) where i n-I n-i k k+1 is the least natural number such that x(i) is between f(— r - ) and f ( — r ) . n—X n-1 Using the same method we can extend f to (x(i) | x(i) < x(o)} so that f ^ i s a one to one, onto, order preserving map from Z to the rationals. 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 1 If X has a first or last element remove them and rememb er that X is 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 is clear from Lemma 0.1 that the existence of a Souslin space, like that of a Souslin line, is equivalent to a negative answer to Souslin's conjecture. We have stated Souslin's conjecture in 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. is separable 2. An ordered compactum with the C.C.C. is separable 3. An ordered continuum with the C.C.C. is 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 is said to have property D if• for every descending sequence of closed sets, (C(i) | i e N} , with null intersection there exists a sequence of open sets {D(i) | i e N} such that C(i) C D(i) for every i and n { D ( i ) | i e N} = 0 . 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 result to produce a normal space which is not countably paracompact. This space is 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 false and used Dowker's characterization of countable paracompactness in normal spaces to reply in the affirmative. This chapter involves the construction of her normal Hausdorff space which is not countably paracompact. Assume there exists a Souslin line, S . The f i r s t thing we shall do i s produce an open interval in S which contains no separable subsegments. Assume that M , the collection of open separable segments, is nonvoid. By the Hausdorff Maximal Principle every member of M is contained i n a maximal chain of open separable segments. By Lemma 5.5 this chain is countable. This implies that the union of this chain i s separable while S's connectedness implies that i t i s an open segment. This union i s certainly contained in no other open.separable segment. A l l this means that every open separable segment i s contained in a maximal one. If we consider the collection of maximal open separable segments we see that i t is pairwise disjoint and therefore countable. We can also see that i t s union, which is separable, must have a complement which contains an open segment, s This segment, s , contains no separable subsegments since every separable • open segment is contained in a maximal one. In the case where M is void l e t s be S . Using s we shall construct for each a e Q1 a nonvoid disjoint countable collection of open intervals, R(a) , such that s ~ y { t | t e R(a)} is countable. Let R(o) be {s} . Given a successor ordinal, a , assume that R(3) has been defined for a l l 3 < a . Let E(a) be a set made up of one point from each member of R(a-l) and l e t R(a) be the set of maximal segments in y { t | t e R(a-l)} ~ E(a) . Since S is connected R(a) exists. If R(a-l) is {(a(k) , b(k))| k e A , A < ^ q} and E(ct) i s (c(k) | k e A} then surely R(a) w i l l be {(a(k) , c(k))|k e A} y {(c(k) , b(k)) | k e A} . If s ~ y {t | t e R(ct-l)} i s countable then s ~ {t | t e R(a)} w i l l be also. Suppose now that a is a limit ordinal and that, for g < a , R(3) is defined and that s ~ {t [ t e R(3)} is countable. Define R(a) to be a l l maximal open segments in ^{ y { t | t £ R(3)} | 8 < a} = V(a) . Since this intersection has a countable complement, y ( s ~ y { t | t e R(3)) | 3 < a} , R(a) exists. Suppose we have a point, y , in V(a) which is not in R(a) . Since i t is in V(a) i t must be contained in the intersection of (t(3) | t(3) E R(3) , 3 < a} which is easily seen to be a chain which descends as 3 increases. Since t is not in R(a) i t must be an endpoint -v of the intersection of this chain. There can only be a countable number of such chains and therefore only a countable number of t's which are in V(a) and not in R(a) . Since s ~ V(a) is countable so is s ~ {t|teR(a)}. Please note that i f x is contained in R(a) , where a is a limit ordinal, then: x = Int( n{t ( 3 ) | 3 < a , x e t(3) e R(3)> where t(3) e R(3) is unique. Notation R w i l l be y{R(a) | a e fi'} , and i f x e R , «5(x) w i l l be the unique ordinal such that x e R(^(x)) For every x e R(3) (3 , limit ordinal) we w i l l produce H(x) = {f(x,i) | i e N} 9 R(3) such that f(x,i) = f(y,j) i f and only i f x = y and i = j and i f x G y (y e R(a) , a < 3) then there exists n £ N such that k > n implies that f(x,k) ^ y . To produce this H(x) we must notice several facts about R and ft' (i) If 3 is a limit ordinal and a < 3 then every element of R(a) contains a countably i n f i n i t e number of element from R(3) • One sees this by noticing that every subsegment of x e R(a) is nonseparable and that the subsegments of x that are in R have a structure very similar to R . ( i i ) R(3) has cardinality Q i f 3 is a limit ordinal ( i i i ) For every x e R(3) and for every a < 3 there is a unique y in R(a) such that x Q y (iv) If 3 is a limit ordinal then i t is ^{a(i) | oc(i) e fi',i e N} where a(i) < 3 and a(i) i a(j) (v) If x i s in R(3) then i t contains at least one element from each R(a) where a > 3 . See the discussion i n (i) above. Since R(3) has cardinality ^ o w e shall write i t as {x(l) , x(2), } and use induction to define H(x(i)) . We shall f i r s t produce H(x(l)) . Consider the unique y(a(l) , x(l)) e R(a(l)) (3 = y{a(i) | 1 e N , a(i) = a(j) i f and only i f i = j}) such that x(l) ^ y(a(l) , x(l)) . Since y(a(l) , x(l)) contains an i n f i n i t e number of x's from R(3) one can be picked and called f ( x ( l ) , l ) . Assume now that f ( x ( l ) , j ) has been defined for a l l j < i . Consider the unique y(a(i) , x(l)) e R(a(i)) such that x(l) C y(a(i) , x(l)) . This y(a(i) , x(l)) contains an i n f i n i t e number of members from R(3) • Pick one that is not equal to any f ( x ( l ) , j ) (j < i) and c a l l i t f ( x ( l ) , i ) By induction we can pick {f(x(l),i) | i e N} such that f ( x ( l ) , i ) = f ( x ( l ) , j ) i f and only i f i = j .We also have the condition that for every y £. R(a) (a < 3) such that x(l) y there exists n such that j > n implies that f ( x ( l ) , j ) C y Now assume that H(x(j)) is defined for a l l j < i . F i r s t define f ( x ( i ) , l ) . From the definition of R(3) we can see that there exists y(.l) £ fi' and z e R(y(l)) such that z contains x but does not contain x(j) for j < i . By considering the members of R(a) which do contain these x(j)'s (j < i) and by remembering the definitions of their H(x(j))'s we can pick f ( x ( i ) , l ) which is a subinterval of z and which is not equal to f(x(j),£) for j < i and Z e N . To.do this we had to remember that z contains an infinite number of members from R(3) . Now assume that f(x(i),n) is defined for all* n < m and notice that 3 = n { y ( i ) I Yd) = Y(j) i f and only i f i = j , y(x) < 3> where y(l) is from the paragraph above. Consider y(y(ni) , x(i)) e R(y(m)) such that x(i) e y(y(m) , x(i)) . It is easy to see that y(y(m) , x(i)) is a subinterval of z and therefore disjoint from the element of R(y(m)) which contain the x(j)'s (j < i) . Remembering the definition of the H(x(j))'s (j < i) and the fact that y(y(ra.) > x(i)) contains an infinite number of members of R(3) we can pick a f(x(i) , m) which is not equal to f(x(j) , £) for j < i and £ e N and not equal to f(x(i) , n) for n < m . In this manner we can obtain our H(x(i)) for a l l x(i) e R(3) Let T be R x N {o} . T shall be the space we are after. For each member of T we shall define a neighbourhood system. The topology obtained from these neighbourhood systems will make T into a no.rmal Hausdorff space which is not countably paracompact. For (x,n) where ^(x) is not a limit ordinal the neighbourhood system is the family of a l l subsets which contain (x,n) Consider (x,l) where ^(x) is a limit ordinal. For every 3 < <S(x) let A(3) = {(y,D | 3 < tf(y) < tf(x) , x Q y> . A ( 3(D) n A(3(2)) = A(max(3(D ,3(2))) . Let the neighbourhood system for (x,l) be the family of a l l subsets which contain some A(3) If <£(x) is a limit ordinal and n is greater than 1 we shall use induction, assuming that neighbourhoods have been defined for (y,m) with $(y) < and m < n , and for (y,n) with ?5(y) <.^(x) . The induction process starts with R(co) x {2} and goes through a l l the R(o)) x {n}'s . It then continues to R(2to) x {2} and proceeds through the R(2w) x {n}'s and so forth. Consider (x,n) (^(x), a limit ordinal, and n > 1) and define F((x,n),i) and G((x,n),a) in the following manner F((x,n),i) = {(f(x,j),n-l) | j > i} . G((x,n), a) = {(y,n) | x c y and a < tf(y) < « K x ) } . A neighbourhood system for (x,n) consists of sets of the form K(i) y J ( a ) y {(x,n)} where K(i) is a neighbourhood of each point of F((x,n),i) and J ( a ) is a neighbourhood of each point of G((x,n),a) . Also in this system are a l l sets containing something of the form K(i) y J ( a ) y {(x,n)} We must now confirm that this i s a neighbourhood system. Consider now (K(i) y J ( a ) y{(x,n)}) and K(j) y J(B) y ^ (x,n)} . Their intersection, L(i,j,a , 3 ) , contains (K(j) n K(i)) y(J ( 3 ) n J ( a ) ) y{(x,n)} . Without loss of generality assume j < i and 3 < a . Then F((x,n),j) 0 F((x,n),i) and K(i) ^  K(j) is a neighbourhood of every point of F((x,n),j) . Also G((x,n),ot) C G((x,n),3) and J(3) n J ( a ) is a neighbourhood of every point of G((x,n),a) . Thus L(i,j,a ,3) is in the neighbourhood system of (x,n) . Thus the intersection of any two neighbourhoods of (x,n) is again a neighbourhood of (x,n) Taking a l l sets which are neighbourhoods for each of their elements, we get a topology. It w i l l be useful to prove that our original neighbourhood system for (x,n) is the neighbourhood system for (x,n) relative to this topology. That is, we should prove that for each of our earlier defined neighbourhoods B of (x,n) there is a neighbourhood A of (x,n) such that AC B and A is in the neighbourhood system of each of its elements. To prove this we shall divide T into the three subsets we used to define the neighbourhood systems, that is; (i) a l l (x,n)'s such that y$(x) is not a limit ordinal (ii) a l l (x,l)'s such that ^(x) is a limit ordinal (i i i ) _11 (x,n)'s such that ^(x) is a limit ordinal and n > 1 If B is a neighbourhood of (x,n) (<^ (x) , not a limit ordinal) then the singleton, {(x,n)} has a l l the qualities of the neighbourhood we desire. If B is a neighbourhood of (x,l) (?5(x), a limit ordinal) then i t must contain some A ( 3 ) = {(y,l) | 3 < ?5(y) < ?5(x) , x C y} . i f (y,l) and (z;l) are contained in A ( 3 ) and ^(y) < f5(Z) then z y . Therefore A ( 3 ) is our desired neighbourhood of (x,l) Consider (x,n) , with $J(x) , a limit ordinal, and n > 1 Assume that what we wish to prove is true for a l l (y,m) such that ^(y) < $(x) , and that m < n and for a l l (y,n) such that $(y) < «5(x) If B is a neighbourhood of (x,n) i t contains a set of the form K(i) (j J(a) y{(x,n)} where K(i) is a neighbourhood of {(f(x,j), n-l)|j > and J(a) is a neighbourhood of {(y,n) | x <~ y , and ^(x) < <f>(y) < a} . By our assumption K(i) contains for every j > i a set D(j) which contains (f(x,j), n-1) and is in our topology. Also by our assumption J(a) contains, for every a < 3 < $(x) , a set, P(3)5 which is a neighbourhood of (y,n) (x S y and $(y) = 3) and like J(ct) is in our topology. Therefore our desired neighbourhood i s : (U* D(J) I J > i>) U<U^p(3) I « < 3 < «J(x)}) yCCx.n)} The following facts will be used later. Lemma (1.1) C(n) = {(x,m) | m > n} is closed for a l l n e N Proof Assume n is more than 2 and that C(k) is closed for a l l k < n . We will prove that for every (x,m) e ~ C(n) there exists N(x,m) , a neighbourhood of (x,m) , such that N(x,m) C. ~ C(n) . By our assumption N(x,m) exists for a l l (x,m). where m. is less than n-1 Notice also that N(x,n-1) exists i f ^(x) is not a limit ordinal. To prove the existence of N(x,n-1) for $(x) , a limit ordinal, assume N(y,n-1) exists for a l l (y,n-l) with o < $(y) < $(x) . ,For every (y,n-l) such that ^(y) < ^ (x) and x £=: y there is P(y) , a neighbour-hood of (y,n-l) which is contained in ~ C(n) . Every (f(x,j) , n-2) has a neighbourhood contained in ~ C(n) . This means that [U^(y) I ' ^ y , 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 in ~ C(n) The case when n is 2 can be handled in a similar manner. Lemma 1.2 . D ( 3 ) = {(x,n) | <£(x) < 3 ) is countable5 open and closed. Proof To prove that D ( 3 ) is closed we shall exhibit for each (y,m) e ~ D ( 3 ) a neighbourhood, N(y,m), which is a subset of ~D(3) If ^(y) is a nonlimit ordinal we w i l l l e t N(y,m) be {(y,m)} while in the case of (y,l) where ^(y) is a limit ordinal we w i l l let N(y,l) be {(x,l) I y C x , «5(y) > j4(x) > 3 } • If we are considering (y,m) with «5(y) , a limit ordinal, and m > 1 we w i l l l e t N(x,n) be defined for a l l (x,n) i n ~ D ( 3 ) such that 0(x) < $(y) and n < m and for a l l (x,n) such that tf(x) < $(y) and n = m Consider the sets F((y,m),l) and G((y,m),3) . Each member of both F((y,m),l) and G((y,m),3) has a neighbourhood which does not intersect D ( 3 ) . If A is the union of these neighbourhoods then A y{(y,m)} i s our N(y,m) To prove that D ( 3 ) is open we w i l l exhibit for each of i t s members, (x,n) , a neighbourhood, B(x,n) , which i s contained in D ( 3 ) • If ^(x) is a nonlimit ordinal let B(x,n) be {(x,n)} . For (x,l) with j5(x) , a limit ordinal, l e t {(y,l) | x ^ y and 1 < jj(y) < 3 ) be B(x,l) . To prove the existence of B(x,n) where $(x) is a limit ordinal and n > 1 assume that B(y,m) exists for a l l (y,m) such that «Ky) < ^ 00 and m < n and that B(y,n) exists when ^(y) is less than tf(x) . By our assumption each member of G((x,n),o) and F((x,n),l) has a neighbourhood which is contained in D(3) . If A is the union of all these neighbourhoods then A y{(x,n)} 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 a < \i and w , z e R(a) such that x C w , y 9i z and w ^  z Proof From the definition of R(u) we can see that: x = lnt( n{t(8) | 3 < V , t(B) e R(B) , and x e t(B)l) y = lnt( n { u(3) | 3 < V , u(B) e R(3) , y e u(B)» Our desired a must exist, otherwise x is equal to y Lemma (1.4) K(x,B) = ((y,l) | B < l*(y) < ?Kx) , xC y} is both open and closed. Proof If (y,l) is contained in K(x,3) and (^y) is not a limit ordinal then (y,l) e {(y,l)> C K(x,3) where {(y,l)> is a neighbourhood of (y,l) . If tf(y) is a limit ordinal and (y,l) is contained in K(x,g) then (y,l) e {(w,D | «Ky) < $(w) < g , y C w} EK(x,p) . Therefore K(x,g) is open. To prove that K(x,g) is closed we must produce a neighbourhood D((y,n)) for each (y,n) in ~ K(x,g) such that D((y,n)) nK(x,g) = 4 . This i s t r i v i a l i f f5(y) is a nonlimit ordinal. To prove the existence of D((y,n)) for the other (y,n)'s we w i l l show that D((y,l)) exists when $(y) is a limit ordinal and then use induction for (y,n) where $(y) is a limit ordinal and n > 1. . Consider then (y,l) where $(y) is a limit ordinal. If $(y) < g then {(w,l) | y C w and o < $(w) < $(x)} w i l l be our D(y,l) . If g < ?5(y) < $(x) then there exists t e K(x,g) such that t e R(^(y)) . From Lemma 1.3 above we have w,z and a < ^(y) such that t C W j y C - Z j W ^ z , and w , z e R(a) . Our D(y,l) w i l l be {(v,l) | y C v , a < «5(v) < <£(y)} . If «5(y) > «5(x) then l e t {(v,l) | y C v , $(y) < $(v) < tf(x)} be our D(x,l) . To prove the existence of D(y,n) ($(y) , a limit ordinal and n > 1) assume the existence of D(w,m) for (w,m) where $(w) < <f>(y) and m < n and for (w,m) with $(w) < $(y) and m = n . Pick i such that F((x,n),i) does not contain a member of K(x,g) . By our assumption each member of F((x,n), and G((x,n),0) has a neighbourhood which does not intersect K(x,g) The union of these neighbourhoods unioned with {(y,n)} w i l l be what we want. Lemma (1.5) Consider (x,n) (n > 1 , $(x) > a limit ordinal) and z e R such that x Q z . Then for every neighbourhood K of (x,n) and for every m < n there exists (y,m) e K such that y C z and 0(y) = ^(x) . In fact, K is also a neighbourhood of (y,m) Proof Let m = n-1 and K be a neighbourhood of (x,n) . By definition K contains {(f(x,j), n-1) | j > i} for some i and in fact is a neighbourhood for the set. By the definition of H(x) there exists k > i such that j > k means that f(x,j) C z . Assuming that our statement is true for m+1 (m < n-1) there exists (y,m + 1) e K such that y C z , «$(y) = ^(x) and K is a neighbourhood for {(f(y,i),m)| i > j} for some j . By the definition of H(y) there exists k > j such that f(y,k)C K . Therefore f(y,k) is the point we are after. Lemma (1.6) Consider (x,n) where j£(x) is a limit ordinal and n > 1 Then G((x,n),3) yUx.n)} = {(y,n) | x C y , M rf(y) < rf(x)} is 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)} . The existence of such neighbourhoods in obvious when $(y) is nonlimit ordinal or when m = 1 . In the case where $(y) is a limit ordinal and m is more than 1 we shall assume the existence of the desired neighbourhood for each (w,k) with k < m and for each (w,m) with (6(w) < $(y) If m ^ n then i t is clear that one can produce F((y,m),i) and G((y,m),ct) such that F((y,m),i) n[G((x,n),3) y{(x,n)}] = <f> and G((y,m),a) n[G((x,n),3) y{(x,n)}] = <f> . Each member of F((y,m),i) and G((y,m),a) has a neighbourhood which does not intersect G((x,n),3) y{(x,n)} . When {(y,n)} is added to the union of these neighbourhoods we get what we desire. For the case where m = n (remember we are dealing with (y,m) e [G((x,n),3) y{(x,n)}] such that $(y) is a limit ordinal and m > 1) we shall recall Lemma 1.3 which states that i f v,u are contained in R(y) (y - limit ordinal) then there exists a < u and w,z e R(a) such that v C w and uC z and w ^  z . If 3 < $(y) < $(x) then there exists r e R($(y)) , a < $(y) , and w,z e R(a) such that v? ^  z , r C ^ , and y C z . G((y,m),a) is disjoint from G((x,n),3) y{(x,n)} as is F((y,m),l) . Each member of G((y,m),a) and F((y,m),l) has a neighbourhood disjoint from G((x,n),3) y{(x,n)} . When {(y,m)} is 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)) is disjoint from G((x,n),3) y{(x,n)} . Since F((y,m),l) is disjoint from G((x,n),3) we have a G((y,m),a) and F((y,m),i) both disjoint from G((x,n),3) y{(x,n)} and are able to produce our desired neighbourhood. Lemma (1.7) Consider (x,n) with $(x) , a limit ordinal and n > 1 Then F((x >n),i) y{(x,n)} is closed for any i Proof As may be suspected we shall produce for (y,m) e ~ lF((x,n),i) y{(x,n)}] a neighbourhood which does not intersect F((x,n),i) y{(x,n)} . It is easy to see that such neighbourhoods exists for (y,m) when f5(y) is a nonlimit ordinal or when m is 1 . To prove the same for (y,m) where ? K y ) is a limit ordinal and m is more than one we shall assume that our desired neighbourhoods exists for (z,k) with k < m and for (z,m) with <f>(z) < j6(y) . If tf(y) > H*) then G((y,m) ,^(x)) is disjoint from F((x,n),i) y{(x,n)} as is F((y,m),l) . Each member of F((y>m),l) and G((y,m),$(x)) has a neighbourhood which Is disjoint from F((x,n),i) y{(x,n)} . When {(y,m)} is added to the union of these neighbourhoods we get what we want. In the case where ^(y) < ? K x ) G((y,m),o) and F((y,m),l) are disjoint from F((x,n),i) y{(x,n)} (Remember that f(z,j) = f(w,k) i f f z =  w and j = k) . Each member of G((y,m),o) and F((y,m),l) has a nieghbourhood disjoint from F((x,n),i) y{(x,n)} . The sum of a l l these neighbourhoods unioned with {(y,m)} is the neighbourhood we want; We are now ready to prove,that T is the space we desire. But first we need three definitions. Definitions If Q is a subset of R. then, x e R is called q-full if:for every y C x (y e R) there exists z C y •such that z e Q , If x is Q - fu l l then: L(x jg,Q) is the set of a l l y in q such that y C x , $(y) > g , and there does not exist any Z in Q such that ?5(z) > g , z C x and y C z From the definition of R the members of L(x,g,Q) are disjoint and therefore countable. Let 3(x,g,Q) denote the smallest ordinal such that $(y) < <5(x,g,Q) for every y in L(x,g,Q) . We can now proceed to prove that T does not have property D , that i s , that i t is not countably paracompact. Consider the collection of closed sets, (C(n) | n e N} , where C(n) i s {(x,m) | m > n} . Lemma 1.1 t e l l s us that C(n) is closed. Suppose (D(n) | n e N} is a collection of open sets such that C(n)C D(n) for every n We shall show that ^{D(n) | n e N} is non-null i Lemma 1.8 For n e N l e t Q(n) be {y | (y,l) t D(n)} . No x e R i s Q(n) - f u l l . Proof Suppose x i s Q(n) - f u l l and construct the sequence {g(i) | i e N , g(i) e fi'} where g(l) = «S(x) and g(i) = 6(x,g(i-l) , Q(n)) . The definition of 6(x, g(i-l) , Q(n)) quarantees that g(i+l) >^  g(i) . Therefore g = y{g(i) | i e N> is a limit ordinal. We f i r s t wish to prove that there exists y e R(g) such that y C x and (y,l) e D(n) . From the definition of R(g) and the non-separability of x there exists z c R(g) such that z C x and (z,ri) e D(n) . Therefore i f n = 1 we are done. Assume n > 1 . Since D(n) is open (z,n) has a neighbourhood AC^D(n) which is also a neighbourhood for F( (z,n),i) where i is some natural number. Now there exists k > i such that j > k implies that f(z,j)(^ x . Pick one such f(z,j) and call i t y(n-l) . Notice that y(n-l) C x and that (y(n-l), n-1) e D(n) . It is clear that this process will produce a y(l) such that y(l) £ R(3) , y(l)C. x and (y(l),l) z D(n) . This y(l) will be our y . Since x is Q(n) - full there exists wC y such that w £ Q(n) . Because w is contained in y , ^(w) > 3 > 3(i) for every i . Therefore, by the definition of L(x,3(i) , Q(n)) , there exists for every i e N , z(i) e L(x,3(i) , Q(n)) such that wC z(i) . But since ^(z(i)) < 3(i+l) < 8,z(i) must also contain y . Because 8 is a limit ordinal (y,l) is a limit point of {(Z(i),l) | i e N} (see the definition of the neighbourhood system of (y,l)) . Since (y,l) is a member of the open set, D(n) , {(z(i),l) | i e N} is eventually in D(n) . This contradiction shows that no x is Q(n) - fu l l Corollary (1.9) For every x £ R there exists y C x (y E R) such that i f z C y (z e R) then (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 let P(n) be {x £ R | for every y ^  x(y £ R)(y,l) is contained in D(n)} Theorem (1.10) The set, ^{D(n) | n e N} , i s non-null. Proof The preceding corollary shows that every x in R is P(n)-full. This means that we can construct • {S(x,o,P(n)) | n e N} where x is a member of R(o) . Since this set i s denumerable there exists y such that 6(x,o,P(n)) < y > for every n . Pick y e R(y) such that y C x . Because x Is P(n) - f u l l there exists w £ y such that w e P(n) , for every n , and $(w) > y . This means that there is z e L(x,o,P(n)) , for every n , which contains w and, in fact, contains y since $(z) > y . In other words, for every n , there exists z e P(n) with y as a subsegment. This places (y,l) in every D(n) Thus ^{D(n) | n e N} is non-null and T does not have property D Q.E.D. We w i l l now show that T is a normal Hausdorff space. Lemma (1.11) Every point of T is closed. Proof Consider (x,n) e T . If (y,m) f (x,n) is also in T and $(y) is a nonlimit ordinal then {(y,m)} "separates" (y,m) from (x,n) . Consider now (y,l) where d(y) is a limit ordinal. If $(x) > $(y) then {(w,l) | y C w , ^ (y) < (^w) < ^(x)> " separates" (y,l) from (x,n) while {(w,l) | y^w , o < (^w) < j5(x)} does the trick when ^(x) < ^(y) . To prove that (y,m) (^(x) , a limit ordinal and m > 1) can be "separated" from (x,n) we will use induction, assuming that every (w,k) with (^w) < j4(y) and k <m and every (w,m) with ?5(w) < $5(y) has a neigh-bourhood disjoint from {(x,n)} . Pick G((y,m),a) and F( (y,m),i) so that neither contains (x,n) . Each member of G((y,m),a) and F((y,m),i) has a neighbourhood which does not contain (x,n) . If A is the union of these neighbourhood, A y{(y,m)} will be what we want. Q.E.D. We will 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 H and K are disjoint closed sets let H(n) be the set of a l l x's with (x,n) in H and K(n) the set of a l l x's with (x,n) in K . Suppose we have y C x where both x and y are in R If x is H(i) - ful l , y is not K(j) - f u l l for any i , j e N Proof Assume that we have x and y in R such that y ^  x , x is H(i) - f u l l and y is K(j) - f u l l . We shall construct a sequence of ordinals, (y(n) | n e N} . Let y(l) be any ordinal more than j5(y) and let y(n) be S(y, y(n-l) , K(j)) i f n is odd and <5(x, y(n-l) , H(i)) i f n is even. Since y(n) is strictly more than y(n-l) , U ^ YC D | i e N} = y is a limit ordinal. From the construction of R(y) i t is seen that we can pick z from R(y) such that z is a subsegment of y . Since y is K(j) - f u l l there is a member , w , of K(j) which is a subsegment of z . The fact that tf(w) is larger than y ( n _ l ) > for n odd, implies that there exists 'u(n) contained in L(y> Y(n-l) » K(j)) such that w u(n) . But ?5( u(n)) is less than y(n) which itself is less than y = ^(y) . This means that z is contained in u(n) . Since x is H(i) - f u l l there is w' of H(i) which is a subsegment of z . Because $(w') > y > y(n-l) , for n even, there exists v(n) contained in L(x,y(n-1) , H(i)) such that w'C v(n) . But «5(v(n)) < y(n) < y . Thus z C v(n) for n even. Now assume that i = j and consider A ,a neighbourhood of (z,i) . A contains a set of the form, {(w,i) |. z C » , p < (^w) < 6} . From the definition of y we can see that there exists k + Z such that ( u(k),i) and (v(£),i) ( u(k) and v(£) were mentioned above) are both in A . This means that (z,i) is in the closure of both H and K . This contradiction proves our lemma when i = j . Assume now that i < j and that A is a neighbourhood of (z,j) . A must contain a set of the form {(w,j) | z C w , 3 < «5(w) < 6} . This means that there exists k e N such that (u(k),j) e A . Lemma 1.5 tells us that A is a neighbourhood for some point (v,i) where v C y and ^(v) = y . Since y is H(i) - f u l l there exists w C v such that w E H(i) . We can easily see that for any n , v(n) contains w and that the only way this can happen is i f v C v(n) . In other words for every y(n) , n even , v is contained in v(n) . Remembering that A is a neighbourhood of (v,i) we know that i t contains a set of the form {(u,i) | v C u , 3 < $(u) < y} . which must contain v(m) for some m . Therefore every neighbourhood of (z,j) contains a point of H and a point of K . This contradiction proves our lemma for i < j . The case where j < i is treated in the same manner. Lemma (1.13) There exists p e fi' such that R(u) has the following property. If x is contained in R(u) then either x contains no member of H(n) for any n or x contains no member of K(m) for any m Proof Let P(l) be the set of a l l x's in R such that no subsegment of x , i n R , is i n H(n) for any n and let P(2) be the set of a l l x e R such that no subsegment of x in R is in K(m) for any m . Let P ( l ) 1 and P(2)' be the collection of a l l maximal segments from P(l) and P(2) respectively. Since both P ( l ) ' and P(2)' are countable there exists u E fi' such that $(y) < u for every y in P ( l ) 1 or P(2)' . Consider now x e R(u) and assume that no_ subsegment of x is K(m) - f u l l for any m . If I(m) is ly e R | y implies that w i K(m)} then every subsegment of x is I(m) - f u l l . Construct {a(m) | m e N} with a(l) = u and a(m) = 6(x , a(m-l) , I(m)) and l e t a be y{a(m) | m £ N} . Let z be a subsegment of x in R(a) and consider K(m) for any m . Since no subsegment of x is K(m) - f u l l there exists v(m)C z such that no subsegment of v(m) is i n K(m) Notice that $(v(m)) must be more than a(m-l) . Therefore there exists y(m) e L(x,a(m-1) , I(m)) such that v(m) C y(m) • But $(y(m)) < a(m) < a . This means that z C y(m) e I(m) and since m was arbitrary we find that z e P(2) . Let w be the member of P(2)' which contains z . From the fact that ?5(w) < y = $(x) and that z & x we can conclude that x CI w and that no subsegment of x is in K(m) for any m Consider again x e R(y) and assume that some subsegment, y e R , of x i s K(m) - f u l l for some m . From Lemma 1.12 i t follows that for every n no subsegment of y is H(n) - f u l l . If J(n) = { z e R | v C z implies that v i H(n)} we can see that y is J(n) - f u l l and that we can construct (y(n) | n e N} by letting y(l) be y and y(n) be 6(y, y(n-l) > J(n)) . Pick z ^ y from R(y) where y l s U ^ ( n ) | n E N} . Since y i s J(n) - f u l l there exists w(n) C z such that w(n) E J(n) and v(n) E L(y, y(n-l) , J(n)) such that w(n) C. v(n) . But $(v(n)) < y(n) < y < <f>(z) . Thus z v(n) and no subsegment of z is in H(n) . Since our n is arbitrary z is i n P(l) and is a subset of some v E P ( l ) 1 . Observe that tf(v) < y = $(x) . In order that v contains z (z C y C x) i t must happen that v also contains x . Therefore no subsegment of x belongs to H(n) for any n Q.E.D. We shall now construct two sequences of closed sets, (A(i) | i e N}, and (B(i) | i e N} , such that for every i , A(i) ^ B ( i ) ,._= 0 and y{A(i) | i e N} and y{B(i) | i e N} are both open. We shall then produce V and W , both open, so that [ y ( A ( i ) | 1 z N}] yV and [ U ^ B 0 ) I 3 E U w "separate" H and K . Let X be {(x,n) | p5(x) < y} . From Lemma 1.2 we know that X is open and closed and can be written as a sequence, (p(m) | m e H} Let A(o) be H n X and B(o) be K r 1' x . A(o) and B(o) are both closed and A(o) ^  B(o) = 4> . T o construct B(m) and A(m) , assume that B(m-l) and A(m-l) have been defined and that B(m-l) ^A(m-l) = 4> and A(m-l) and B(m-l) are closed. Consider p(m) = (x,n) and suppose p(m) t B(m-l) . If ^(x) is not a limit ordinal l e t A(m) be A(m-l) y{p(m)} and B(m) be B(m-l) • Since (p(m)}(Lemma 1.11) is closed and disjoint from B(m-l)»A(m) and B(m) are both closed and their intersection i s void. Now let $(x) be a limit ordinal and let n be equal to one. Since B(m-l) is closed and does not contain p(m) there exists J = {(y,l) | x Q y , 3 < *5Cy) < $(x)} , a neighbourhood of p(m) , which does not intersect B(m-l) and is closed by Lemma 1.4. Let A(m) be A(m-l) y J and B(m) be B(m-l) V It is easy to see that these are what we want. Assume the third possibility for p(m) , that i s , assume that $(x) is a limit ordinal and that n > 1 . Since B(m-l) is closed and does not contain p(m) there exists A , a neighbourhood of p(m). such that A ^ B(m-l) = $ . By the definition of neighbourhood there exists F((x,n),i) , G((x,n),3) d A , both of which do not intersect B(m-l) and both of which contain (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) be A(m-l) y{(x,n)} yF((x,n) ,i) y G((x,n),3) and let B(m) be B(m-l) If p(m) e B(m-l) then repeat these operations with the roles of A(m-l) and 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 C and prove that for every (x,n) contained in i t there exists N(x,n) , a neighbourhood of (x,n) such that N(x,n) C. C In the case where $(x) is not a limit ordinal let N(x,n) = {(x,n)} Suppose n = 1 and $(x) is a limit ordinal. From the construction of A(i) we see that A(i) C X = {p(l) , p(2) ,...'.} for every i . Therefore (x,n) = p(m) for some m . Since p(m) i B(m-l) , A(m) = A(m-l) y J where J = {(y,m) | 3 < «Ky) < $(x)} . Let J be N(x,n) Now let n be more than one and let $(x) be a limit ordinal. Assume that C contains some (w,m)'s which do not have N(w,m)'s . Let k b the min({ m | there exists (w,m) £ C which does not have a N(w,m)}) and let 3 be the min({ a | there exists (w,k) £ C with no N(w,k) and with $(w) = a}) . There must exist (y,k) E C with $(y) = 3 such that N(y,k) does not exist but that N(w,m) does exist for a l l (w,m) E C with ?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) . This means that A(m) is A(m-l) y{p(m)} y F(p(m),i) y G(p(m),y) for some i and for some y . Our definition of (y,k) tells us that C contains a neighbourhood for each point of F(p(m),i) and G(p(m) ,y) . Therefore C 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' , W' , V and W in the following manner: V' = {x e R(u) | there exists yC x such that y e H(i) for some i} (Remember u from Lemma 1.13) W' = {x e R(u) | there exists y 9 x such that y e K(j) for some j} V = {(y,n) e R | y ^ x for some x e V'} W = {(y,n) e R | y x for some x e W'} Lemma 1.13 tells us that V' and W' are disjoint and that W and V are disjoint. Lemma (1.15) The sets, W and V , are both open. Proof If ^(x) is not a limit ordinal and (x,n) e V then {(x,n)} is a neighbourhood of (x,n) which is contained in V .. If ^(x) is a limit ordinal then {(y,l) | ?5(x) > ^ (y) > y , x Q y} does the same thing for (x,l) e V . To prove that (x,n) has a neighbourhood in V where $(x) is a limit ordinal and n > 1 we shall assume such neighbourhoods for a l l (y,m) e V such that $(y) < $(x) and m < n and for (y,n) where $(y) < ^ (x) . Consider G((x,n).,u) . . (Remember y from Lemma 1.13). Our definition of V tells us that G((x,n),y) C v while our assumption about (x,n) tells us that each member of G((x,n),u) has a neighbourhood in V . Let A be the union of a l l these neighbourhoods. Since (x,n) is contained in V there exists w e V' such that x % w . This implies the existence of i such that f(x,j) C w when j > i . Therefore F((x,n),i) C V and each of its members has a neighbourhood in V . Let B be the union of a l l these neighbourhoods. A y B y{(x,n)} is a neighbourhood of (x,n) contained in V The same method proves that W is open. Theorem (1.16) The sets, C y V and D y W, "separate" H and K . Proof C and D are both disjoint. Suppose there exists x contained A(i) and B(j) where without loss of generality j > i . Then by the construetion of the A(n)'s and the B(m)'s x is contained in A(j) This i s a contradiction. Since C and D are contained in {(x,n) | ^ (x) < u} and V and W are contained in {(x,n) | $(x) > u} we can say that C y V and D y W are disjoint. If (x,n) is contained in H then either (x,n) C {(y,m) | ^(y) < y) or (x,n) C {(y,m) | ^ (y) > y} . In the f i r s t case (x,n) C C and in the second x ^ y for some y e R(y) where y is obviously in V' . Therefore H ^ C y V . The same reasoning puts K in D y W We have proven in 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 were arbitrary closed sets T i s normal. Lemma 1.11 t e l l s us that T is also Hausdorff. THE COUNTABLE CHAIN CONDITION IN E x E WHERE E IS AN ORDERED CONTINUUM In 1952 Djuro Kurepa [12] proved that an ordered continuum, E ,is homeomorphic to I(closed unit interval) i f E x E has the C.C.C. We shall prove this and state the obvious equivalence to Souslin's conjecture along with another involving Peano maps. Consider an ordered continuum E such that E x E has the C.C.C. We shall define for each a e fi' a collection, D(a) , of closed segments from E and, using these closed intervals, shall produce a countable dense subset A look at the proof of Lemma 0.1 w i l l then convince one that E is homeomorphic t Let D(o) be {E} . If a is a successor ordinal consider each nondegenerate interval, X, in D(a-l) and let X(o) and X(l) be two closed intervals such that X(o) yX(l) = X and X(o) and X(l) have in common only one point and this point i s i n the interior of X . If X i s nondegenerate Els connectedness implies the existence of X(o) and X(l) . Now let D(a) be {X(o) | X (nondegenerate) is in.D(a-l)} (j(X(l) |x (nondegenerate) i s in D(a-l)} . If a is a limit ordinal consider a l l chains of closed intervals of the form {X(g) | g < a} where X(g) e D(g) and X(g)CX(y) i f g > Y • Let X be in D(a) i f and only i f i t is the intersection of a l l elements in one of these chains. Since E i s connected such an intersection w i l l again be a closed interval. We w i l l now state some lemmas concerning E and our collections of intervals. Lemma 2.1 , ' If X i s contained in D(a) and 3 i s less than a then there exists Y e D(3) such that X £. Y . Proof By induction and by the definition of D(3) Lemma 2.2 If X and Y are intervals in D(a) then they intersect at most one point and this point is an endpoint for both of them. Proof By induction and the definition of D(3) Lemma 2.3 If a < 3 , X e D(a) and Y e D(3) then either Y i s contained in one of X(o) or X(l) or X and Y intersect at at most one point and this point is an endpoint of both of them. Proof By induction, the definition of D{y) and Lemma 2.2 . Lemma 2.4 If a ^  3 then D(a) ' and D(3) have no elements in common. Proof By definition and induction. Lemma 2.5 Let E be a linearly ordered space which i s connected and has the C.C.C. and l e t A be a chain of intervals in E such that any subset of i t contains a greatest element, that i s , an element which contains a l l the others in the subset. This chain i s countable. Proof Notice that each A e A has a "successor", that i s , a unique B e A which i s contained in A and which contains a l l subintervals of A which are in A . Since E is connected A ~ B w i l l contain an interval. Therefore { Int (A~B) | A e A , B , the successor of A} w i l l be a countable collection of disjoint open sets. This makes A countable. Q.E.D. Let D be y{D(a) | a e ft'} and let F be the collection of a l l endpoints of nondegenerate intervals in D .We shall show that F is a countable dense subset in E . The fact that F is dense follows from the following theorem. Theorem 2.6 For every a e E there exists a(a) e ft' such that {a} = n{X(a) | a < a (a)} where X(a) is the nondegenerate interval in D(a) which has. a as an element. Proof Given a let 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) is an interval ( i f i t is nonvoid) while Lemma 2.1 t e l l s us that i f P(a,a) exists then P(y,a) exists for a l l y < a and that i f a < $ then P(3,a) ^  p(a,a) . Note that P(o,a) exists. Therefore {P(a,a) | a e fi'} forms a chain l i k e A in Lemma 2.5 and therefore is countable. This means that there exists 3 e ft' such that for a l l a > 3 P(a,a) is the null set. Let a(a) be the least ordinal such that P(a,a) is the null set. It i s possible to use induction to pick a sequence {X(a) | a < a(a)} of intervals such that X(ot) e D(a) , a e X(cx) and i f a < 3 then X(3) C X(a) . Suppose X(3) is defined for a l l 3 < a . Ii D(a) has only one interval which contains a let i t be X(a) . If there are two intervals, Z and Y , in D(a) which contain a (there can at most be two) pick one, say i t is Z , such that for every a < y < a(a) there exists W e D(y) such that a e WC Z . Suppose this didn't happen for either Z or Y This would mean that there exists a < 3 < a(a) such that D(3) does not contain any V such that a e V and either V C y or V C Z . Since 3 < a(a) there must exist U e D(3) and T e D(a) such that a e U and U C. T f Z , Y . This means that T , Z and Y each contain' a . The definition of D(a) and Lemma 2.2 show this to be a contradiction. A l i t t l e insight and a f a i r amount of writing w i l l prove that our X(a) is contained in every X(3) ..for 3 < a . Now i f ^{X(a) j a < a (a)} is not exactly {a} then X(ct(a)) w i l l exist. To see this look at the definition of D(a) for the two-cases when a is a limit ordinal and when a is a successor ordinal. Therefore {a} = n{X(a) | a < a(a)} . What we need now i s to show that F i s countable. Remembering the definition of X(o) and X(l) for X e D consider the collection, {Int (X(o) x X(l)) | X , nondegenerate interval in D} C E x E . From Lemma 2.3 and the definition of X(o) and X(l) we can see that this is a pairwise disjoint collection of open sets and therefore must be countable as must F . Therefore E is separable. The obvious equivalence to Souslin's conjecture mentioned before i s : Consider an ordered continuum, E . Does E x E have the C.C.C. when E 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 result w i l l prove Souslin's hypothesis. CHAPTER 3 SOUSLIN TREES Assuming the existence of a Souslin line, S , consider the collection R = y{R(g) | g e ft'} from Chapter 1 and partially order i t by inclusion. 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 R . Any partially ordered set with such a characteristic is called a tree. It should be noted that this is not the standard definition of tree. 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 is 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 flat, that is, by fitting each member of R(a + 1) into the member of R(a) from which i t came. These members of R(a) can be looked upon as the .framework into which the blocks of R(a + 1) are fitted. This, in fact, can be done with R and is possible for any tree, T , with the following characteristics: 1. T has cardinality 2. Each chain is countable 3. Each antichain is countable An antichain is 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 we can see that for x and y in R either x C: y , y C x or x ^ y = This means that any, antichain in R is a collection of pairwise disjoint open entervals and as a result is countable. R has cardinality ^ since each T(a) is nonempty and contains a countable number of elements. Remember that for a ^  g,T(a) and T(3) 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, T . We already know that T shares several characteristics with our R , 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. First, for every a e fi' let T(a) be the set of a l l x in T whose successors, when given the reverse order, can be put into a one to one order preserving correspondence with (3 | 3 e fl' ,'• 3 < a} . Remember that the set of successors for each x is well-ordered when given the reverse order. We shall now l i s t several characteristics of these T(a)'s Lemma (3.1) Every x in T belongs to one and only one T(a) Proof For any x in T , its successors, when given the reverse order, is a well-ordered set of cardinality 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 is s t r i c t l y less than ft Lemma (3.2) Consider x e T(a) . Then for every 8 < a there exists one and only one y e T(f3) such that x < y . Also every successor of x is contained in T(y) for some y < a Proof This clearly follows from the definition of T(a) and the proof of Lemma 3.1. Lemma (3.3) For every a e ft' T(a) is a nonempty" antichain and therefore countable. Proof If there existed an a such that T(a) were empty then Lemma 3.2 would show that T(8) is empty for a l l 8 > a . This would be a contra-diction since each T(y) is countable and T has cardinality ^ j . The fact that T(a) i s an antichain comes from Lemma 3.2. Please note that a l l of the above mentioned characteristics are held by the R(a)'s in R when R has the inclusion order. R also possessed no short arms. We shall now give this characteristic to T . Let V be the collection of all x's in T which have a countable number of predecessors and let U be a l l y's in V which have no successors. U of course is an antichain and therefore countable. By remembering that for every z e T(a) and 3 < a there exists w e T(3) such that y < w we can see that for every y e V there exists x e U such that y < x . This means that V 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) . Note that in such a Souslin tree every x e T(a) has a predecessor in T(3) when 3 is more than a . To see this remember that each T(y) is 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 T "fork" , as do the members of R . Suppose we have a Souslin tree, T , in which every member has an uncountable number of predecessors. Let x be in T1 if and only i f there exists y which has the same successors as x . (Of course in this case the successors of x shall not include x) . If T1 were countable then there would exist P e fi1 such that for every y > $ T(y) contains no points of T' . T(3) is nonempty and therefore must contain a y with an uncountable number of predecessors. These predecessors form a chain. This contradiction shows that T' is an uncountable subset of T and therefore a Souslin tree in which each x has a y which has the same successors. We can therefore assume that each member of T "forks" and has an uncountable number of predecessors. We are now ready to squeeze T , that is, give i t a total order. Let x be related to y if and only if they have the same successors. This is an equivalence relation which divides T into a collection of disjoint sets each of which is countable since i t is an antichain. This means that each equivalence class can be written as a sequence which will give i t an obvious total order. For each x e T let {g(l)(x) , g(2)(x),... . ,} be its 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, <(T) . Consider x e T(a) and y e T(3) where 8 > a . If y < x then there exists a unique t e T(a + 1) (Lemma 3.2) such that y < t . Let y < (T) x i f and only if t = g(l)(t) . If x and y are not related in the partial order then there exists y e ft1 , t e T(y) , and m , n e N , a l l unique, such that y < g(m)(t) , x < g(n)(t) and m ^  n . Let y < (T)x if and only if g(m)(t) < g(n)(t) . To prove the existence and uniqueness of t let z(y) and w(y) be the unique members of T(y)(y < a)(Lemma 3.2) which are more than y and x , respectively, and let K be (z(y) | z(y) $ W(Y), y < a} . If K is the null set let t be w(o) , otherwise let t be the greatest element in K . If t is contained in T(y) then z(y) is the unique g(m)(t) such that y < g(m)(t) while w(y) is the unique g(n)(t) such that x < g(n)(t) . Note that t is uniquely determined by the existence of the elements t, g(m)(t) , and g(n)(t) such that y < g(m)(t) , x < g(n)(t) and m ^  n To prove transitivity suppose that we have x < (T) y and y < (T) z . To prove that x < (T) z consider the four p o s s i b i l i t i e s ((i) x < y and y < z ( i i ) x < y and y and z are not related in the part i a l ordering ( i i i ) x and y are not related in the partial ordering while y and z are (iv) x and y are not related in the partial ordering nor are y and z) and apply the definition of the total order. Now let us show that T with the order topology from <(T) i s nonseparable. If M C T is a countable set then there exists 3 such that T(y) n M = <j> for a l l y > $ • Since each T(a) i s countable and T is uncountable there exists ? > 3 such that T(£) contains an element, t . We mentioned earlier that i f x e T(y) then i t contains a predecessor in each T(a) where a > y . Therefore t contains a predecessor w in T(£ +1) .We can show that (g(l)(w) , g(2)(w)) is a nonempty interval containing no points from M . If s e T(a) where a < 3 , g(l)(w) < (T) s and g(l)(w) < s then from the definition of our total order we can see that g(2)(w) < (T) s . The same thing happens i f g(l)(w) and s are not related in the partial order. Therefore (g(l)(w) , g(2)(w)) contains no points from M but does contain t Suppose now that we have a disjoint collection of open intervals, {(x(a) , y(a)) | a e A} , in the total order. Let x(a) be contained in T(3(a)) and l e t t(a) be a partial order predecessor of x(a) i n T(3(a) + 1) . If x(a) < y(a) then a l l the partial order predecessors of g(2)(t(a)) are in (x(a) , y(a)) . The same is true for g(2)(t(a)) i f x(a) and y(a) are not related in the partial order. 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 in the partial ordering and which therefore is 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 X , a locally connected continuous image of an ordered compactum. If X 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 positive answer to Souslin's conjecture i f and only i f a compactum which is the continuous image of an ordered compactum is 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 is the continuous image of an ordered compactum. For our purposes we shall assume that an ordered compactum with the C.C.C. is separable. This fact along with Lemma 4.1 t e l l s us that X i s equal to f(K) where f i s continuous and K is an ordered separable compactum. X then is separable. The converse i s t r i v i a l . Theorem 4.4 A positive answer to Souslin's conjecture i s equivalent to the statement that a compactum which is 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 is separable and as a result 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 is equal to f(R) where K i s an ordered continuum and f is continuous. Since a l l sets in K of the form, {x | a < x < b , a, b e K} , constitute a base each member of which is connected we can say that K is locally connected. Since K is compact and X is Hausdorff f(A) will be closed for a l l closed sets A in K . Therefore f is closed and X is locally connected.([15], Theorem 1.3, p. 121 and Theorem 3.5, p. 125). The affirmative response to Souslin's conjecture makes X separable so that Lemma 4.2 can give X a countable base. Since X is regular (Hausdorff and compact) i t is metrizable ([16], p. 125, Theorem 17). Now for the converse. Suppose that in a compactum, which is 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. is metrizable and as a result separable ([15],p. 187, Theorem 5.6). Souslin's question is answered in the affirmative. CHAPTER 4 PART 2 A compactum X is said to have property y i f for every closed set X A/R is metrizable where x R y i f and only if x and y are in the same component of A . It is said to have property 0. i f and only if there is a countable collection of open sets such that for every disjoint pair of closed sets, M and N , there is a member, S , of this collection which separates X between M and N . This means that there exists disjoint sets, A and B ,such that M^A , N^B , A ^ B = & , A y B = S and both A and B are closed in A y B . In 1967 Sibe Mardesic" [8] proved that a compactum is metrizable if and only if i t has properties a and u . In this same paper he raised the question whether property y alone would imply metrizability for a compactum. There is a lemma which shall be proven later which says that an ordered continuum with the C.C.C. has property y . Therefore if in fact property y 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 if 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 alone implies metrizability. However Mardesic [8] goes one step further. He shows that Souslin's hypothesis implies that in a compactum which is the continuous image of an ordered compactum property y is 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 is that i f a compact, separable Hausdorff space is the continuous image of an ordered compactum then i t possesses property a It is 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 will be the statements of several lemmas that we need. Lemma 4.5 A totally disconnected ordered compactum with only a countable number of gaps is metrizable. Proof ([9J , Lemma 1, p. 867) Lemma 4.6 If X is a compactum then X/R is a compactum where x R y i f and only i f x 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 is metrizable. Proof ([17], p. 102, Problem 8B). Lemma 4.8 Consider X , a separable compactum, which is the continuous image of an ordered compactum. There is a countable collection of closed sets which separates 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 is a G(<5) and each open set is 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 is metrizable. Proof Since X is a T^ , regular space (compact and Hausdorff) the existence of a countable base will imply that X is metrizable. ([16],p. 125). Consider the countable collection C = {S(n) | n e N) guaranteed by property a . For every n e N let x R(~S(n)) y i f and only if x and y are in the same component of ~S(n) . Property y says that for every n ~S(n) / R(~S(n)) is metrizable. Since i t is also the continued image of a compact space i t will have a countable base, V(n) = {V(n,k) | k e N} ([15], p. 187). We will now prove that G = y ;|{S(n) y-p(n)" 1 (V(n,k)) | k e N} | n e N;| is a countable base for our topology. (P(n) is the projection of ~S(n) onto ~S(n) / R (~S(n)) . Since P(n) - 1 (V(n,k)) is open in ~S(n) i t is equal to 0(n,k) T~S(n) where 0(n,k) is open in X . This assures that S(n) y P(n) X(V(n,k)) is open in X for every n and k . We will show that for every point, y , and closed set , M , with y t M there exists B e G such that y e B and B ^ M = $ . Because x is T^ there exists S(n) e C which separates between {y} and M This means that there is A and B such that (i) A y B = ~S(n) (ii) A n B = <}> (i i i ) y e A , M S. B (iv) A , B are closed in A y B Each component of A y B is connected and therefore cannot intersect both of these sets. This implies that ~S(n) / R(~S(n)) = P(n)(A) y P(n)(B) where P(n)(A) n P(n)(B) = ?5 and P(n)(A) and P(n)(B) are both open and closed. Because P(n)(y) e P(n)(A) there exists V(n,k) e V(n) such that P(n)(y) e V(n,k) and V(n,k) n P ( n ) (B) = $ . The member of G we are looking for is S(n) y P(n) (V(n,k)) Suppose now that X is metrizable and that A i s a closed subset of X . A is a compact metric space (X is compact) and A/R is a Hausdorff space (Lemma 4.6) . Therefore Lemma 4.7 can t e l l us that A/R is metrizable with the result that X has property u . Remember please that x R y for x , y e A i f and only i f x and y are in the same component of A . Suppose again that X is metrizable. Since X i s a compact metric space i t has a countable base. Let 8 = (B(i) | i e N} be this base together with a l l f i n i t e unions from i t . It is easy to see that B i s a countable base which is closed under f i n i t e unions. We shall prove that B is the countable family of open sets necessary for property o X is normal since i t i s compact and Hausdorff. This means that for any two disjoint closed subsets, M and N ,there are two disjoint open sets 0 and P ,such that M ^  0 and N P . Since ~ (My N) i s open, i t i s the union of a l l elements of 8 contained in i t . This collection from 8 covers ~(0yP) which Is compact (X i s compact) . Therefore there exists a f i n i t e collection, (B(i) | i = 1 n), from B such that y{B(i) | i = l,...,n} •= S e B contains "(OyP) but is disjoint from M y N . Let A be 0 rV and B be P (>i:~S . It is easy to see that MC A ,N9.B , A y'B="S and A n B = i> . From their definitions we can see that A and B are both open in A y B and therefore are both closed in. A y B . Therefore S e B separates X between M and N , i.e. X has property a . These lemmas, needed for our major theorem, will be stated along with their proofs. Lemma 4.11 Consider a compactum, X , which is the continuous image of an ordered compactum. If X is 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. From Lemma 4.9 we know that each of these F(n)'s is a countable intersection of open sets. Please notice that since X is compact and Hausdorff i t is normal. This means that the closed neighbourhoods of a closed set form a neighbourhood basis for i t . Now consider one of these F(n)'s which is n{0(n,j) | j e N , 0(n,j) open} The normality of X says that for each j there is W(n,j) , a closed neighbourhood of F(n) , and S(n,j) , an open set, such that F(n) C S(n,j) ^  W(n,j) 9 l 0(n,j) . It is easy to see that F(n) = n{S(n,j) | j £ N} = n{C£ (S(n,j)) | j £ N} We will show that 8 , the set of a l l finite intersections from {S(n,j) | n , j E N} , is the collection we want.. If M and N are two disjoint closed sets there is an F(n) which separates X between them. Remember that F(n) = n{(S(n,j)) | j £ N} = n{Ct(S(n,j)) | j £ N} and that F(n) C. ~ (MyN) . There is a finite collection (C.£(S(n,j)) | j = j (1) ,. . . , j (m) } from {C-£(S(n,j)) | j £ N} whose intersection is contained in ~ (MyN) . Suppose not . then {MyN} y{C£(S(n,j)) | j £ N} would be a collection of closed sets with the finite intersection property. Since X is compact F(n) would have a point in MyN . This is a contradiction. Let K e B be ^{S(n,j) | j = j (1) ,... , j (m)} . Since F(n) separates between M and N there exists A and B such that M £ . A , N £ . B , A n B = ^ , A y B = ~F(n) and A and B are closed in A y B . If we consider A ^ ~ K and B ^ ~ K we can see that K separates X between M and N Lemma 4.12 An ordered continuum, X , with the C.C.G. has property u Proof If A is a closed subset of X we will show that A/R with the quotient topology is a totally disconnected ordered compactum with only a countable number of gaps. Lemma 4.5 will then imply that A/R is metrizable. Remember that x R y i f and only i f x and y are in the same component of A If x and y are contained in W , a component of the closed set X , then every point of A between x and y is contained in W also. If this were not so W would not be connected. This enables us to put a linear order on the set of components of A and to consider the order topology of this set. Consider P , the projection from A onto A/R and {W | W ^ Ye A/R} , a subbasis element of the order topology of A/R where Y is an element of A/R . If P - 1 ({W | W £ Y}) is nonempty then i t is equal to {x | x e X , x < b = g . 1 . b . of Y} ^A which is open in A . The existence of b follows from the connectedness* of X while the equality of P _ 1 ({W | W ^  Y}) and {x | x e X , x < b} r^A follows from the fact that Y is closed. This shows that P is continuous when A/R has the order topology. The fact that A is compact and A/R , with the order topology, is Hausdorff implies that P is closed when A/R has this order topology. Therefore this topology is in fact the quotient topology ([16}, Theorem 8, p. 95). From the structure of A/R we can see that P A(C) , C £ A/R , is connected i f and only i f C consists of a single point. Remembering that a component is closed, we can use Theorem 3.4, p. 124 of [15] to show * A linearly ordered space is connected in the order topology i f and only if there are no gaps and the space is order complete. that A/R is totally disconnected. All we need to do now is to prove that A/R has only a countable number of gaps. Suppose we have E ^ F e A/R such that there is no component of A between them. Since X is connected and both E and F are closed in X , E contains a £,u.b. for itself and F contains a g.£.b. for itself. Using again the connectedness of X we can see that between these two points in X is an element of ~A and in fact an interval in ~A which contains i t . In this manner assign to each gap in A/R an open interval of X and uotice that any two of these intervals are disjoint. The C.C.C. of X proves that we have only a countable number of gaps. Lemma 4.13 A compactum, X , which has property \i also has the C.C.C. Proof Let { U(X) | X s L} £ X be a collection of pairwise disjoint open sets. From each U(X) pick x(X) e U (X) and denote by A the closure of {x(X) | X e L} . From the definition of closure and the fact that X is Hausdorff i t follows that U(X) n A = {x(X)} . This means that for each X e L {x(X)} is open in A . If P is the projection of A onto A/R then {P(x(X))} is open for every X e L . This is true since for every X {x(X)} is its own component. (Remember x is related to y i f and only if x and y are in the same component of A) Since A is a closed subset of a Hausdorff space and X has property u A/R is a compact metric space. This means that i t is separable and can only have a countable number of points y such that {y} is open. There-fore { U(X) | X e L} is countable. Here is the major theorem. Theorem 4.14 A positive answer to Souslin's conjecture is equivalent to the statement that a compactum which is 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 is 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 is 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 u and which is the continuous image of a ordered compactum. Theorem 4.13 implies that X has the C.C.C. This enables Souslin's hypothesis to t e l l us that X is separable. A l l that is needed now is to look at Theorem 4.11 which gives X property a and as a result makes i t metrizable. 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(1951), 219-224. 3. 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. 19(1952), 629-640. 5. M. E. E s t i l l , Separation in non-separable spaces, Duke Math. Journal, Vol. 18(1951), 623-629. 6. 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. 17(1962), 3-25. 8. 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 lineare et le probleme de Suslin, Publications de 1' institut mathematique de 1' Academe Serbe des Sciences, Vol. 4(1952), 97-108. 13. S. Mardesic", Mapping ordered continua onto product spaces, Glasnik Mat. Fiz and Ast., 15-2, (1960). 14. A. Lelek, On Peano functions, Prace Matematyczne Warsaw, Vol. 7, (1962), 127-139. 15. James Dugundji, Topology, Allyn and Bacon, Inc., Boston (1968). 16. John L. Kelley, General Topology, American Book, Van Nostrand, Reinhold, New York, Toronto, London, Melbourne, (1955). 17. S. - T., Hu, Elements of General Topology, Holden-Day, Inc., San Francisco, London, Amsterdam, (1969). 

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