UBC Theses and Dissertations

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UBC Theses and Dissertations

Orderable topological spaces Galik , Frank John 1971

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. ORDERABLE TOPOLOGICAL SPACES by FRANK JOHN GALIK B.Sc. (Hons.), U n i v e r s i t y o f B r i t i s h Columbia, 1 9 6 9 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE i n the Department o f ... MATHEMATICS ' We ac c e p t t h i s t h e s i s as conforming to the r e q u i r e d standard The U n i v e r s i t y o f B r i t i s h Columbia . May 1 9 7 1 In presenting t h i s t h e s i s in p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study. I f u r t h e r agree tha permission for extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s . I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. Department of The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada S u p e r v i s o r : Dr. T. Cramer ABSTRACT L e t (X , 3") be a t o p o l o g i c a l space. I f < i s a t o t a l o r d e r i n g on X , then (X , 3" , <) i s s a i d to.be an ordered t o p o l o g i c a l space i f a subbasis f o r 3" i s the c o l l e c t i o n o f a l l s e t s o f the form {x £ X | x < t} or [x f X | t < x} where t £ X . The p a i r (X , JJ) i s s a i d to be an.orderable t o p o l o g i c a l space I f there e x i s t s a t o t a l o r d e r i n g , < , on X such t h a t (X , 3" , <) i s an ordered t o p o l o g i c a l space. 1 - • • ' ) • . D e f i n i t i o n : L e t T be a subspace o f the r e a l l i n e {R . L e t • Q be the u n i o n of a l l n o n - t r i v i a l components o f T , both of whose end p o i n t s belong to C 1 R ( C 1 R ( T ) -T) . The f o l l o w i n g c h a r a c t e r i z a t i o n o f o r d e r a b l e sub-spaces o f IR i s due to M. E. Rudin. Theorem: L e t T be a subspace o f IR w i t h the r e l a t i v i z e d u s u a l topology. Then T i s o r d e r a b l e i f and o n l y i f T s a t i s f i e s the f o l l o w i n g two c o n d i t i o n s : (1) I f T - Q i s compact and (T-Q) n C l ^ Q ) = 0 then e i t h e r Q = 0 or T - Q = 0 i i i . (2) I f I i s an open i n t e r v a l o f IR and p i s an end p o i n t o f I and i f {p} U(I H(T-Q)) i s compact and {p} -Cl^ m o j n C 1 R ( I fl(T-Q)), then p T or {p} i s a component of T T h i s theorem enables us to prove a c o n j e c t u r e o f I.-.L. Lynn_, namely C o r o l l a r y : i f T c o n t a i n s no open compact s e t s then T i s t o t a l l y o r d e r a b l e . I f T i s a subspace o f an a r b i t r a r y ordered t o -p o l o g i c a l space a g e n e r a l i z a t i o n o f the theorem can be made. The g e n e r a l i z e d theorem i s s t a t e d and some exam-p l e s are g i v e n . ^ i v . TABLE OF CONTENTS PAGE INTRODUCTION • 1 NOTATION 3 BIBLIOGRAPHY . 62 ACKNOWLEDGMENTS I. wish to thank Dr. T. Cramer f o r su g g e s t i n g the t o p i c o f this t h e s i s and f o r h i s i n v a l u a b l e a s s i s t -ance d u r i n g the p a s t year. I would a l s o l i k e to thank Dr. L. P. B e l l u c e f o r h i s h e l p f u l c r i t i c i s m and Dr. M. E. Rudin f o r her sug g e s t i o n s and i n t e r e s t i n my work. The f i n a n c i a l a s s i s t a n c e o f the N a t i o n a l Research C o u n c i l o f Canada and the U n i v e r s i t y of B r i t i s h Columbia i s g r a t e f u l l y acknowledged. INTRODUCTION I n t h i s t h e s i s v/e examine a r e s u l t by Mary E l l e n Rudin which g i v e s a c h a r a c t e r i z a t i o n of o r d e r a b l e t o p o l o g -i c a l spaces. Dr. Rudin won the 1965 Netherlands Mathematical S o c i e t y p r i z e f o r t h i s work. Dr. Rudin has s t a t e d and proved the r e s u l t f o r the s p e c i a l case when the t o p o l o g i c a l space i s a subspace o f the r e a l l i n e . • S i nce the p r o o f o f the g e n e r a l r e s u l t f o l l o w s almost i d e n t i c a l l y t h a t o f the s p e c i a l case and s i n c e an attempt to prove the main theorem r e s u l t e d i n c o m p l i c a t e d unreadable n o t a t i o n , Dr. Rudin has omitted the p r o o f of t h i s theorem. We have p o i n t e d out, i n the t e x t of the t h e s i s , t h a t the g e n e r a l r e s u l t cannot be used i n many a p p l i c a t i o n s . I n the d i s c u s s i o n f o l l o w i n g Theorem (2) we g i v e our reasons. The paper by Dr. Rudin was not read by a r e f e r e e and there were some mistakes which we have c o r r e c t e d . We have been i n touch w i t h Dr. Rudin and she has acknowledged those c o r r e c t i o n s made to her paper. The minor e r r o r s i n Lemma (8) and the more s e r i o u s e r r o r i n Lemma (12) of Dr. Rudin's paper w i l l be d e s c r i b e d i n the t e x t o f the t h e s i s . There i s a l s o an e r r o r i n the statement of the theorem as i t appeared i n the " T r a n s a c t i o n s o f the American Mathematical S o c i e t y " , and Dr. Rudin has p o i n t e d t h i s out. As the p r o o f o f the r e s u l t f o r a subspace o f the r e a l l i n e i s a c o n s t r u c t i v e one, we have i n c l u d e d s u f f i c i e n t Page 2 examples I n t h i s , t h e s i s to g i v e the reader a good i d e a of the c o n s t r u c t i o n i n v o l v e d . Thus the reader can achieve a good grasp o f the p r o o f by r e a d i n g the s e c t i o n s l a b e l e d " D i s c u s s i o n ' I n . a d d i t i o n , we have r e l a t e d Dr. Rudin's paper to work done by I . L. Lynn and shown how this, f o l l o w s from Dr. R u d i n 1 s work. A l s o , some a p p l i c a t i o n s f o r the main theorem are g i v e n . NOTATION , Given a t o p o l o g i c a l space, T and a subset, X c T , then the c l o s u r e o f X i n •T i s denoted C1 T(X) . The s e t o f r e a l numbers i s denoted by R and the n a t u r a l . numbers are d e f i n e d by |N = {1 ,2,3...} . I f X c (R and a, b <= R d e f i n e a X + b = {a x + b | x <= X} . The usual, c o nventions are used i n den o t i n g i n t e r v a l s o f p . F o r example, (0,1) = {x £ R | o < x < 1} and [2,3] = {x g IR |" 2 _< x < 5 ) and (0,1) U[2,3] i s i n d i c a t e d i n d i a -grams as f o l l o w s . ( > E 3 • I t w i l l o f t e n be u s e f u l l to i n d i c a t e on a diagram a s e t whose elements are I n t e r v a l s . We use shading to i n -d i c a t e such a set.. For example, {[0,1] , [ 1 , 2 ] , [3,4]} i s denoted i n diagrams as f o l l o w s « » a. 3 H-D e f i n i t i o n : L e t (X,cT) be a t o p o l o g i c a l space. If. < i s a t o t a l o r d e r i n g on X , then (X,3",<) i s s a i d to be an ordered t o p o l o g i c a l space i f a subbasis f o r U i s the c o l l e c t i o n o f a l l s e t s o f t h e form {x 6 X | x < t} or {t £ X | t < x} where t € X Page 4 (X,U) i s s a i d to be an o r d e r a b l e t o p o l o g i c a l space i f there e x i s t s a t o t a l o r d e r i n g , A , on X such t h a t (X,3",A) i s an ordered t o p o l o g i c a l space. Example (1): L e t T = (0,1) U [ 2 , 3 ] . L e t "J be the u s u a l t o -p o l o g y f o r R r e l a t i v i z e d to T . Then T i s not o r d e r a b l e . p r o o f : Suppose f : T -» S i s an homeomorphism of T onto an ordered t o p o l o g i c a l space, S . Without l o s s of g e n e r a l i t y , f i s . o r d e r p r e s e r v i n g on both (0,1) and [ 2 , 3 ] . F o r example, assume " f i s n e i t h e r order p r e s e r v i n g nor order r e v e r s i n g on (0,1) . Then, without l o s s o f g e n e r a l i t y , there e x i s t a 3 b, c ^(0,1) such t h a t a <• c < b and f ( a ) < f ( b ) < f ( c ) . L e t B = {x 6 (a,b) | ' f ( x ) > f(*>)} • Since c <= B , B ^ 0 and l e t y = i n f ( B ) . There are 3 p o s s i b i l i t i e s . Hence we can assume t h a t f i s o r d e r p r e s e r v i n g on (0,1) and on [ 2 , 3 ] . f [ ( 0 , l ) ] i s an i n t e r v a l , f o r otherwise f [ ( 0 , l ) ] i s not connected. S i m i l a r l y f o r f [ [ 2 , 3 ] ] . Without l o s s o f g e n e r a l i t y , * f (a) < f ( b ) f o r a l l a 6(0,1) and b e [ 2 , 3 ] . Hence, s i n c e f [ ( 0 , l ) ] has no l a r g e s t element, f ( 2 ) i s an a c c u mulation p o i n t of f [ ( 0 , l ) ] which c o n t r a d i c t s the f a c t s t h a t f i s " a n homeomorphism and t h a t 2 i s not an accumu-Page 5 l a t i o n p o i n t "of (0,1) Example. (2): De f i n e I = [-1,0] , ' p _ r l 1 1 y ' r ~ ^3 3 ~5 3 y 3 • • • i i ~(~2 » » ^2 = %  3 ^  3 3^  3 ' ^  = ^TT 3 T o ^ 3 ' ' CO L e t X = I U Pjy^GjL • With the r e l a t i v i z e d t o p o l -ogy, X i s not o r d e r a b l e . p r o o f : Suppose g: X -• S i s an homeomorphism of X onto S , an ordered t o p o l o g i c a l space. Choose from each G^, Zt-€ G^ . CO _ CO Then {g(Z.)}._, converges to g(0) as does {g( J} Now g [ I ] i s an i n t e r v a l , f o r otherwise g [ I ] i s not connected and s i n c e g , without l o s s o f g e n e r a l i t y , p r e s e r v e s o rder on I i 'g(O) i s the maximum of g [ I ] . Now s i n c e the two sequences _ oo oo { g ( ^ P 3 - ) } n = 0 a n d C g ( z i ) } i = = 1 both converge t o g ( 0 ) a n d o n l y . f i n i t e numbers.of each sequence are l e s s than g ( 0 ) , v/e can choose m , r 6 IN'f such t h a t g ( z m ) •< g ( ^ r + ^ ) a n d n o p o i n t o f 1 0 0 1 t % n+ T ^ n = 0 i s b e t w e e n s ( z m ) a n d g( 2r+y ' ^ o w a n ^ ° P e n s e t c o n t a i n i n g g ( g r + ^ ) c o n t a i n s a p o i n t of g [ G r a ] but 2~r+3 i s not a c c u m u l a t i o n p o i n t o f G m and t h i s c o n t r a d i c t i o n e s t a b l i s h e s t h a t X i s not l i n e a r l y o r d e r a b l e . E — ) x Page 6 D e f i n i t i o n : L e t T be a subspace o f IR • Denote by Q the u n i o n o f a l l n o n - t r i v i a l components of T , both o f whose end p o i n t s belong to C 1 R ( C l ^ ( T ) - I ) . * • The f o l l o w i n g theorem g i v e s a c h a r a c t e r i z a t i o n o f o r d e r a b l e t o p o l o g i c a l subspaces o f IR . The remainder o f t h i s c h a p t e r w i l l be devoted to the p r o o f . Theorem (1): L e t T be a subspace of IR w i t h r e l a t i v i z e d u s u a l topology. Then T i s o r d e r a b l e i f and o n l y i f T s a t i s f i e s the f o l l o w i n g 2 c o n d i t i o n s : . , (1) I f T - Q i s compact and (T - Q ) p. C1 R ( Q ) = 0 then e i t h e r Q =. 0 or T - Q ="-0 . (2) I f I i s an open i n t e r v a l o f R and p i s an endpoint o f I and i f {p } U ( i n(T-Q)J i s compact and {p} = C l R ( i n Q ) n . C l ^ i n f C - Q ) ) , then- p | T or {p} i s a .component o f T . _ Notes: 1. I n example (1) above, Q = (0,1) , T - Q = [ 2 , 3 ] and hence T - Q i s compact and (T - Q ) n Clp ( Q ) = 0 . But Q * 0 and (T - Q ) * 0 and so c o n d i t i o n (1) i s not s a t i s f i e d . 2 . i n example (2) above, Q = ( | , 1) U (| , \) u ( ^ , j) [}..:, C1 R(Q) = [| , 1] U [ | , ^] U ...U{0} , and T - Q = [-1 , 0] -U{| , £ , £ , ...} . Page 7 Thus T - Q i s compact and (T -Q) fi C l ^ Q ) = {0} and so c o n d i t i o n (1) holds v a c u o u s ly. However, c o n s i d e r the i n -t e r v a l I = (0,1) and l e t p = 0 . ' * Then [0 , ^ , ^ , ^ -j {p} U ( i n ( T - Q ) ) i s compact and {0} = C l R ( i n Q ) a C 1 R ( I n(T - Q ) ) hut 0 g T and {0} i s not a component o f T hence con-d i t i o n (2) does not h o l d . We prove now t h a t c o n d i t i o n s (1) and (2) are n e c e s s a r y N e c e s s i t y . o f c o n d i t i o n (1): Suppose t h a t T - Q i s compact, t h a t (T - Q ) n Clj R ( Q ) f 0 t h a t q g Q and p 6 T - Q' , and t h a t A i s a t o t a l o r d e r o f T g i v i n g T i n t e r v a l topology. Without l o s s .of g e n e r a l i t y , assume p A q . L e t p' =' sup [x £ T - Q | x A q} . Then s i n c e T - Q i s compact, p'g T - Q . Since p'^ C L ^ Q ) and T has i n t e r v a l topology, there e x i s t s r , : q ' e T such t h a t p'g {x £ T | r A x. A q'} c T - Q . By d e f i n i t i o n o f p' , q £ Q. and q ' i s an end p o i n t of a component of Q . But q ' ^ C 1 R ( C 1 R ( T ) -T) and t h i s c o n t r a d i c t i o n e s t a b l i s h e s the • r e s u l t . N e c e s s i t y of c o n d i t i o n ( 2 ) : Suppose I i s an open i n t e r v a l o f R , p i s an end-p o i n t o f I , {p} U(TO(T-Q)) i s compact, and {p} i s the i n t e r -s e c t i o n of C l R ( l . n Q ) and C l R ( i n ( T - Q ) ) . Suppose, also., t h a t Page 8 . p f T , t ^ p , and t i s a member of C , the component o f T to which p belongs. Suppose t h a t A i s a t o t a l o r d e r i n g on T, f o r which T has i n t e r v a l topology. There i s no p o i n t i g I f! T such t h a t i e C f o r C c T - Q or e l s e C c Q . I f v/e assume there e x i s t s i £ I , i g C and C c T - Q then p i s not an accumulation p o i n t of I R Q . S i m i l a r l y f o r C. c Q , p i s not an accumulation p o i n t o f I D(T-Q) . Hence there i s no p o i n t i £ I fl T between p and t i n A . For otherwise {x g C'| x A i ) and {x e C | i A x} i s a s e p a r a t i o n f o r C . Without l o s s of g e n e r a l i t y , assume t < p and t A p .. Choose i g I A' T where p A i . Then {x £ T | t < x < i ] i s an open s e t c o n t a i n i n g .p. and so there e x i s t c and i ^ such t h a t c , i 0 6 {x ^ T | t < x < i ) and p e { x € T | c A x A i Q } c { x € T | t < x < i } , and f u r t h e r , {x £ T | p A x A i Q } <= I . S i n c e p i s an a c c u m u l a t i o n p o i n t of I R Q , there e x i s t s q € I fl Q such t h a t q 6 [x e T | c A x A 1Q} and p A q . S i m i l a r l y , s i n c e p i s an a c c u mulation p o i n t o f I'n(T-Q) and {p} u{l'n(T-Q)) i s compact, there e x i s t s p'g I n(T-Q) where p-A p'. A q and '{x £ T | p' A x A q} c o n t a i n s no p o i n t o f T - Q . Since p' 4 p , p' |' C 1 | R ( I N Q ) A N D hence there e x i s t q' and r where p' g {x $ T | q' A x A r} c T - Q . Then q' € Q and. q' i s an end p o i n t o f a component o f Q but q' { C l R ( C l e ( T ) -T) . T h i s c o n t r a d i c t i o n e s -t a b l i s h e s the r e s u l t . Page 9 Define IR*. ' to be the c o m p a c t i f i c a t i o n of IR by-adding the -points -co and co where -oo < x < m f o r a l l x £ B . Nov; we r e p l a c e tR by S , which i s obt a i n e d by s u b s t i t u t i n g an i n t e r v a l homeomorphic to [0 , 1] f o r ,each t r i v i a l component of IR* - T . When t h i s i s done T i s a subspace of S and S can be g i v e n a t o t a l o r d e r A which p r e s e r v e s the order o f T i n R . C l e a r l y i f the theorem i s shovm w i t h R r e p l a c e d by S ,. i t can be shown as s t a t e d . D e f i n i t i o n : L e t H be the unio n o f a l l _ o p e n i n t e r v a l s I of S such t h a t C l s ( l ) n(T-Q) i s compact and I n(T-Q) fl C l g ( Q ) 0 . ' ; Lemma (0): I f S = H . then T - Q i s compact and (T-Q) n C l g ( Q ) 0 . ' \ • ' Proof: Since H i s compact there e x i s t 1^ , 1^ , I , open I n t e r v a l s o f H such t h a t H c =iy1 Ij_ a n d C 1 s ^ " [ i ^ ft(T-Q) i s compact and I, n(T-Q) n C1 Q(Q) = 0 f o r 1 = 1 , 2 , . n Then T - Q = (T-Q) n H .n = U (T-Q) n C l G ( I i ) and hence T - Q i s compact i = l b Page 10 A l s o , ( T - Q ) n C l ( Q ) = u [ ( T - Q ) n c i G ( Q ) n I.] i = l Lemma (1): I f S = H and c o n d i t i o n (1) h o l d s , then T i s t o t a l l y . o r d e r a b l e . -Proof: Suppose S = H . Then c o n d i t i o n (1) and the p r e v i o u s p r o p o s i t i o n imply T c Q or Q = 0 . I f Q = 0 , then s i n c e S = H , T - Q = T ' i s compact. T h i s c l e a r l y i m p l i e s t h a t T"~ i s l i n e a r l y o r d e r a b l e under the o r d e r A . F o r l e t X Q £ T be g i v e n a n d ° { x g T | a A x A b} where a, b g S . Then s i n c e T i s compact, a' = sup{x £ T | x A a or x = a} and b 7 = i n f {x e T | x A b or x = b} are both i n T and {x £ T | a A x A b} = {x <= T | a 7 A x A b 7 } . I f T c Q then a l s o T i s l i n e a r l y o r d e r a b l e under A . For l e t x Q€ T be g i v e n and x Q £ {x £ T | a A x A b} where- a , b £ S . I f x Q i s an i n t e r i o r p o i n t of Q , c l e a r l y there e x i s t a 7 , b' g T where x Q € i> € T I a 7 A x A b 7} c {x g T | a A x A b} . I f XQ i s not an i n t e r i o r p o i n t o f Q then x^ i s an end p o i n t o f a component o f T and s i n c e x^g C l g ( C l g ( T ) -T) there e x i s t a7-, b 7 £ T such t h a t x Q € { X £ T | a 7 A x A b 7} c {x e T | a A x A-.b). • Page 11 •We now complete the p r o o f of the theorem by showing t h a t i f S - H f 0 , c o n d i t i o n (2) i m p l i e s t h a t T i s l i n e a r l y o r d e r a b l e . Lemma (1): Suppose c o n d i t i o n (2) holds and l e t I and J be two open i n t e r v a l s o f S such t h a t I c H , J c H I fl J = 0 , and x i s a common end p o i n t of I and J where x \ H . I f J c T then J c T - Q, and under t h i s assumption, x <J C l s ( i n ( T - Q ) ) D i s c u s s i o n : A method of p r o o f can be seen i f one t r i e s to c o n s t r u c t a counterexample w i t h the r e q u i r e d c o n d i t i o n s . In a l l cases l e t x = 0 I = (-1,0) \ J = (0,1) . Case (1): Suppose x | T , Since the components of S - T are n o n - t r i v i a l , our example might be the f o l l o w i n g . T = (-1 , ~) U (0 , 1) . Here, c l e a r l y , x e H . Case ( 2 ) : Suppose x £ T"". There are 3 d i s t i n c t ways to make x s a t i s f y the c o n d i t i o n x £ C 1 S ( C 1 S ( T ) -T) (a) L e t A = [| , 1) T = (| A - 1) U(jj A -|) U ( i ^ A -^) U... U [ 0, 1) I n t h i s case there are p o i n t s o f I not i n H . (b) L e t A = , 1) , and T = ( J A - I ) U ( J A - J ) U ^ A ^ ) U . . . U [ 0 , 1 ) I n t h i s case x g H . Page 12 T = ( i A - 1) u ( i A - h U ( i r - A - h U ... U[0 , 1) (c) L e t A"= (| , |) U{1} , and 1 U ( j - |) ^ n - ? j ! I n t h i s case, c o n d i t i o n (2) does not h o l d . The ot h e r p a r t o f the lemma i s e a s i l y <seen. [ ) ' [ ) r ) r f [ — — — — ) T -I -k ~\ 'I -> o I — c ) c ) i r r ) T -\ -> -\ -> - v • o • » c ) »—e-v——e^-^-H-«-£— : — ) T Proof: We will-now f o r m a l i z e the arguement i n the d i s c u s s i o n . S i n c e J c T , e i t h e r J c Q or J c T - Q . Assume to the c o n t r a r y that- J c Q . Then there e x i s t s an open i n t e r v a l I ' c I such t h a t x i s " an e n d p o i n t . o f i ' and {x] U (i'fl (T-Q) j i s compact. For otherwise there i s a p o i n t o f I which i s a l s o i n H . Hence i f x € T , by c o n d i t i o n (2) , x i s not i n both Clg(inQ) ' and C l g (ID (T-Q) J. I f x $ T , then x $ C l g (III (T-s i n c e the component of S -. T c o n t a i n i n g x i s not t r i v i a l . Hence i f x $ T or x $ Clg(in(T-Q)) then c l e a r l y x e H , a c o n t r a d i c t i o n . I f x ^ T and x cj Clg(inQ) then , CO by d e f i n i t i o n o f Q , there i s a sequence {x ] o u t s i d e of H n n=l Page 1J> and c o n v e r g i n g to x . T h i s c o n t r a d i c t i o n e s t a b l i s h e s t h a t J c T - Q . • - .• Now suppose x 6 Cls(m(T-Q)). Then x £ T s i n c e the components of . S - T are n o n - t r i v i a l and from above, x 4 Clg(inQ) . Hence there e x i s t s an open i n t e r v a l , I ' c o n t a i n i n g x such t h a t I ' fl(T-Q) fl Clg(Q) = 0 and C l (i') n(T-Q) i s compact. (For, otherwise, l ' c o n t a i n s o a sequence o f p o i n t s not i n H and co n v e r g i n g to x) Hence x £ H and t h i s c o n t r a d i c t i o n completes the p r o o f . Lemma (2): Suppose c o n d i t i o n (2) holds and l e t X be an open i n t e r v a l o f S whose end p o i n t s are not i n T . Suppose C1 Q(X) - H = {x}'. Then there i s a t o t a l o r d e r i n g , 9 f o r X n T such t h a t X n T has i n t e r v a l topology. F u r t h e r , 9(XDT) has no f i r s t element but has a l a s t element. D i s c u s s i o n : We c o n s i d e r 2 ca s e s . Case (1): Suppose x i s trie c l o s u r e o f some n o n - t r i v i a l component o f T . Example ( 3 ) : x = 0 , T = (0 , ^] , X = (-1 , 1) . In t h i s example x 3 T and x | H because f o r any open i n t e r v a l , I c o n t a i n i n g x , C l g ( l ) P. (T-Q) i s not compact. There can be no i n c r e a s i n g sequence from T co n v e r g i n g to x because S - T c o n t a i n s no n o n - t r i v i a l components, hence l o c a l l y about x t h i s example i s t y p i c a l f o r x ^ T . Page 14 O b v i o u s l y the induced order from IR s a t i s f i e s the h y p o t h e s i s . I • •- • . 1 i Example (4):: L e t A = (^ , 1) , x = 0 . Define T = A - 1) U ( ^ A , \) U ( ^ - A - \) U... U[0 , i ] , X = (-1 , I n t h i s example x £ T and x ^ H because x i s i n the c l o s u r e o f T - Q and i n the c l o s u r e o f Q . There can be no i n c r e a s i n g sequence from T - Q c o n v e r g i n g to ' x because o f c o n d i t i o n (2) , hence, l o c a l l y about x t h i s example i s t y p i c a l f o r x £ T . A g a i n the induced order from IR s a t i s -f i e s t h e - h y p o t h e s i s . -<- ) - ( ) •-<-• H f [ T Case (2): Suppose x i s not i n the c l o s u r e of some non t r i v i a l component of T . .' Example (5): l e t x = 0 , T = ( | , | ) . u ( - | , - | ) u ( £ , | ) u ( - - | / - £ ) u(|4) u(4,-| u nfo 1 1 1 1 1 1 Here, x £ T and x | H because x i s i n the c l o s u r e o f Page 15 T -. Q and i n the c l o s u r e o f Q . An o r d e r , 8 which s a t i s -f i e s the h y p o t h e s i s i s as f o l l o w s : To complete the d e f i n i t i o n of 0 , we add t h a t 0 o r d e r s each i n t e r v a l which l i e s to the l e f t of 0 as the induced order from |R and each i n t e r v a l which l i e s to the r i g h t of 0 as the r e v e r s e of the. induced order from IR . —= ( ) t-khc-U^—4)*-B-«—(—)—— 1 T -j- _A -> -y -> * -x J- x > J- I x 1 1 . 1 'o0„ .10 1 . H • $ X Proof: L e t X' =, (p,q) „ L e t I = (p,x) , J = (x,q) . Then the order f o r X ITT i s as f o l l o w s : J (l(T-Q) has the t o t a l o r d e r i n g induced by A . We d e f i n e 8 '(Jn(T-Q) J = A ( J n ( T - Q ) J . S i m i l a r l y , , 0(in(T-Q)J =-A(in(T-Q) J , • G(JflQ) — A(JHQ) , e (inQ) = A (inQ) . Hence to d e s c r i b e a t o t a l o r d e r i n g , 0 f o r X D T we need o n l y s p e c i f y an o r d e r 0 f o r the s e t A = { (JH(T-Q ) J , in(T-Q) , JDQ , I HQ. , {x} DT } . Case (1): There e x i s t s a component, J / of JOT such t h a t Page 16 -x € ' C l g ( J 7 ) . Then we d e f i n e the f o l l o w i n g o r d e r f o r A , ( j n q ) G(inQ) e ( {x } nT) e ( J n ( T - Q ) j e ( i n(T-Q)) We no'w show t h a t XDT has an i n t e r v a l t opology . I f x_6 T. , l e t us denote by 3" the f a m i l y of 0 x 0 s e t s c o n t a i n i n g x ^ which are open i n the t o p o l o g y o f S r e l a t i v i z e d to T . L e t £ denote s e t s c o n t a i n i n g x n x 0 . which are open i n the i n t e r v a l t opology determined by the o r d e r 0 . L e t y £ X n T be g i v e n . •.(a) Suppose y = x . By lemma (1) , J 7 a T - Q and x ^ C l ( i n ( T - Q ) J . C l e a r l y , s i n c e ' x £ T and x e C l g ( J 7 ) then x £ T - Q . Since x £ T - Q , and because x <$ H we must have x a C l g ( m Q ) . By assumption, x c Cl g ( < i n(T - Q ) J and so 3^ c zx. Since x g C l g ( i n ( T - Q ) J and x 3 C l g ( J n Q ) ~ we have £ cr 3" x x (b) Suppose . y £ Q and y * x . I f y i s an i n t e r i o r p o i n t of Q , c l e a r l y £ = 3" so assume without l o s s o f y y g e n e r a l i t y t h a t y i s a- r i g h t hand end p o i n t of a component o f Q . Since y g H , l e t I 7 be an open i n t e r v a l c o n t a i n -i n g y such t h a t C l s ( I 7 ) n' (T - Q ) i s compact and I 7 n ( T - Q ) , n C l s ( Q ) = 0 . Then y $ C l g ( T - Q ) f o r otherwise C l g ( 1 7 ) n ( T - Q ) i s not compact. Hence £ c 3" . A l s o , y i s a l e f t hand y y• end p o i n t o f a monotone sequence from Q s i n c e y e C l g ( C l g ( T ) and y i s c l e a r l y a r i g h t hand end p o i n t of such a sequence Page 17 and so 3" c £ . (c) Suppose y € T - Q and y M . Note t h a t {t € T _'Q I t 6 y } t 0 s i n c e j ' c T - Q . L e t a = sup ( t € T - Q |. t 9 y ] . Then a € T - Q s i n c e a * x and a e H . I f {t £ T - Q | y 0 t} * 0 , l e t b = i n f ( t ? T - Q I y fi t ] , Then b * x s i n c e x $ I P. C l g ( T - Q ) and so b e (T-Q) . Hence (1) a £ T - Q and (2) b £ T - Q when {t £ T - Q | y 0 t} = 0 .. imply 3" cz-£ . Since y £ T - Q and y £ H , y <{ C l g ( Q ) and so £ c 3" . Hence by (a) (b) and (c) , X P T has • i n t e r v a l topology. Now from the argument i n (b) above, i f (IUJ) DT c o n t a i n s a p o i n t o f Q , 9(XPT) has no f i r s t element. So assume (IUJ) P. Q = 0 . Then x 5 Q and s i n c e x $ H , x l T - Q . Hence, by 3-ssumption, J 7 has no f i r s t element and so 9(XPT) has no f i r s t element. To show t h a t 0(XPT) has a l a s t element, l e t y £ J 7 and b = i n f [ t . £ T - Q | y 9 Then b ¥ x and so b £ T - Q . and b i s the l a r g e s t e l e -ment o f 0(XPT) . _ • Ca.se (2): For each component J 7 of TPJ , x $ C l g ( J 7 ) and, wit h o u t l o s s o f g e n e r a l i t y f o r each component I 7 o f I P T , x $ C 1 ~ ( I 7 ) . Then there e x i s t monotone sequences, Page 18 co co •*n=0 and ^n^n= 0 such t h a t (1) i o = p and CO { i n ] n = l c I - T , ( { i n } i s i n c r e a s i n g ) (2) J 0 = q and CO c J - T , (ld n 3 - ^ d e c r e a s i n g ) (3) l i m n-*» ^ = x D e f i n e the f o l l o w i n g o r d e r , 0 f o r X CI T . [ ( iQ, ! - ] / ) HQ] 0 E ( J X J J 0 ) HQ] 9 [ ( i - L ^ i g ) HQ] 0 [ ( j ^ ^ ) HQ] 0 . . . e[[x) DT] ... e [ ( i 1 3 i 2 ) ' n (T-Q)] 0 [ ( J 1 , J 0 ) n (T -Q ) ] 0 [ ( i o * - ^ To show 0 (XflT) has i n t e r v a l topology, l e t y e X n be g i v e n . (a) Suppose x = y . C l e a r l y x $ Q and so x £ T - Q and s i n c e x £ H , e i t h e r 012(1') fl (T -Q) i s not compact .for some open i n t e r v a l l ' c o n t a i n i n g x or x £ Clg(Q) . I f C l g ( l / ) n (T-Q) i s not compact then )^ c o n t a i n s a p o i n t not i n H ot h e r than x . Hence~"^x <= Clg(Q) . I f A = { t e T | x 0 t } * 0 , l e t b = i n f A . I f b = x , then-s i n c e x c C 1 Q(T-Q) , *J = Z . I f b * x , b e T - Q s i n c e O _ X X b e H a n d s i m i l a r l y 3" = £ . The same holds i f A = 0 . X X (b) Suppose y € Q . The argument f o r .(b) above h o l d s . (c) Suppose y g T - Q . The same argument f o r (c) above a p p l i e s s i n c e ( t 6 T - Q j t 0 y} * 0 . For i f x { T - Q then x e C l q ( T - Q ) s i n c e x <£ H . Page 19 D e f i n i t i o n : Denote by H' the u n i o n o f a l l open i n t e r v a l s of S which c o n t a i n not more than one p o i n t o u t s i d e H . Denote by G the s e t whose elements are the c l o s u r e s o f components from H 7 . L e t G' be the subset o f G con-s i s t i n g o f elements o f G which have an i n t e r i o r p o i n t not i n H . . •' We w i l l want to d i v i d e the terms of G 7 i n t o i n t e r v a l s , each o f which has one p o i n t o u t s i d e H and then use the p r e v i o u s lemma to or d e r each of the i n t e r v a l s . I n o r d e r i n g the terms of G' i t w i l l be convenient i f the p o i n t s of d i v i s i o n are not i n T . Hence the f o l l o w i n g lemma. Lemma ( 3 ) : Assume c o n d i t i o n (2) holds and th a t X g G' and J - i s a component of H Cl X . Then there e x i s t s x <= J such t h a t x { T . / Proof : Since X € G' , there e x i s t s an i n t e r v a l , I of S such t h a t I f H , I 0. J = 0 , and x i s a common end p o i n t o f I and J where .x 3 H . Assume t h a t J c T . Then by lemma (2) , J c T - Q and has an end p o i n t j such t h a t j $ C l g ( C l g ( T ) - T ) . Hence the r e e x i s t s an open bounded i n t e r v a l , J 7 , c o n t a i n i n g j and such t h a t J 7 c o n t a i n s no p o i n t s o f . C l g ( T ) - T . Hence Clo ( J 7 ) fl T i s c l o s e d s i n c e i t c o n t a i n s a l l i t s accumula-Page 20 t i o n p o i n t s and, b e i n g bounded i s compact. Since j ' con-t a i n s no p o i n t s o f C l g ( T ) - T , i t c o n t a i n s no end p o i n t s o f Q and hence no p o i n t s o f Q . Hence j ' n (T-Q) n Clg(Q) and j ' ; a H . T h i s i s c l e a r l y a c o n t r a d i c t i o n s i n c e J i s a component of H . We d e s c r i b e now, a t o t a l o r d e r i n g f o r a subset, Y* n T , of K f o r each K e G', f o r each component, J of K D H , by lemma ( 3 ) , s e l e c t a p o i n t Z j $ T . L e t Z be the s e t c o n s i s t i n g - o f a l l such z f o r J a component of K fl H . Note t h a t by d e f i n i t i o n o f G', the accumulation p o i n t s o f Z occur o n l y a t the end poin-ts o f K ' i f Z has any a c c u m u l a t i o n p o i n t s . Note, a l s o , t h a t the components o f K - Z have a n a t u r a l o r d e r induced from A , the order i n S . So, define. Y to be the c o l l e c t i o n of a l l com-ponents o f K - Z e x c l u d i n g the f i r s t component, i f i t e x i s t s and e x c l u d i n g the l a s t component, i f i t e x i s t s . L e t Y* = U {Clg(X) J X € Y] . Then f o r each X g Y , assuming c o n d i t i o n (2) h o l d s , the h y p o t h e s i s o f Lemma (2) holds so there i s a t o t a l o r d e r i n g 9 o f X n T where X n T has i n t e r v a l t o p o l o g y w i t h r e s p e c t to the order 9 and X (1 T has no f i r s t element but does have a l a s t element. C l e a r l y X n T w i l l have a t o t a l o r d e r i n g , namely - 0 , which y i e l d s i n t e r v a l t o p o l o g y and f o r which X fl T has a f i r s t element but no l a s t element . Now {X fl T I X € Y} has a n a t u r a l order induced Page 21 by S and so Y* n T = U {X n T | X £ Y} has a number o f t o t a l o r d e r i n g s where f o r each X £ Y , X fl T has i n t e r v a l t o pology. The f o l l o w i n g o r d e r s , X and x' w i l l assure t h a t , i n f a c t , Y* n T has the o r d e r topology. I f Y has a f i r s t element i n the -order induced by S , L e t Y Q be the f i r s t element. Otherwise, chose YQ a r b i t r a r i l y . Then s e t : (a) \ ( Y 0 n T) = e ( Y 0 n T) , (b) For a g i v e n X £ Y , X (X n T) = 0 (XflT) i f there i s an odd number o f terms o f Y between X and YQ , . (c) X (XHT) = -9 (XflT) o t h erwise. (Since the accumulation p o i n t s of Z occur o n l y a t the end p o i n t s o f K , there i s o n l y a f i n i t e number o f terms of Y between any such terms). "To complete the d e s c r i p t i o n o f the t o t a l o r d e r i n g on Y* 0. T , f o r X^, X2,.€ Y , s e t ( X i n T) X ( X £ n T), i f and o n l y i f (X ] L n T) A ( X 2 n T) The d e s c r i p t i o n of the t o t a l o r d e r i n g X* i s i d e n -t i c a l to the above wi t h 9 r e p l a c e d by -0 and -0 r e p l a c e d by 0 . I t i s an immediate v e r i f i c a t i o n t h a t Y* fl T has i n t e r v a l t o p o l o g y under X and x' when i t i s noted t h a t : (1) under X and x' i f x i s a l a s t p o i n t of X H T f o r X £ Y , then x has an immediate s u c c e s s o r and s i m i l a r l y i f Page 22 x i s a f i r s t p o i n t of X D T then x has an immediate p r e -decessor and (2) under X and x ' , X fl T has i n t e r v a l t o p o l o g y f o r each X g Y . Example (6): L e t T = (| , |] U , i ] U(| 3 , U ... G = ([0 , » ] , [-05 J 0]} and G' = {[0 , »]} . Choose Then Z = and then Y Since Y has no f i r s t element i n the 1 "5 o r d e r induced "by R , choose Y Q = , ^ j-] Then X and are ""defined as f o l l o w s : (1) . . . (1 , 1] x , | ] X (| a |] where X orders (jl > 2^ ' 5 B-] 5 ^15 5 9 " ' a s t h e i n d u c e d o r d e r f r ° m IR and X orders o t h e r i n t e r v a l s as the r e v e r s e of the i n -duced o r d e r from (R .. (2) * (?) 3 X ' ^ ' 5 ] X ' (5 ^ ^ w h e r e orders ,1 1,- ,1 In ,1 1 ( 3 ' 2"J 3 > B1 >[T5 3 I T J induced o r d e r from (R and X the induced o r d e r from !R . as the r e v e r s e of the o r d e r s o t h e r i n t e r v a l s as 4-JH 3-X ~» T 1 > 5 -5 o 10 > 1 '1 5 ' />/<»/ y / t I t t > t *t> I y / / / t 1 / / / i' t / i» V Page 2 3 Example ( 7 ) : F o r n = 1 , 2 , . .. , l e t f be the order p r e s e r v i n g homeomorphism of ( - 1 , 1 ) onto , • L e t T 7 be the subspace o f example ( 5 ) and l e t T = f- j j T ' ) U f g ( T 7 ) U f ^ ( T ' ) U' ... . L e t Q± be the order o f f i ( T / ) such t h a t 9 i ( f i ( T / ) ) i s order isomorphic to 9 ( T ' ) . Then G , G 7, Z , and YQ are as i n the p r e v i o u s example. The o r d e r s , X and \' are d e f i n e d as f o l l o w s : ( 1 ) f 5 ( T ' ) X f 2 ( T 7 ) X f - ^ T 7 ) where \ ( f i ( T 7 ) ) = e i ( f 1 ( T ) ) i f i i s odd X ( ^ ( T 7 ) = - ^ ( ^ ( T ) ) i f i i s even ( 2 ) \ 7 i s as i n ( 1 ) above w i t h odd and even i n t e r c h a n g e d . Lemma ( 4 ) : L e t K = [p , q] £ G 7 . Assume c o n d i t i o n ( 2 ) h o l d s and t h a t p £ P K but q $ P^ . . Then there e x i s t s x € (p i q) a n d a t o t a l o r d e r i n g , ir on [x , q] D T such t h a t (p , x] c H and TT orders T n i c e l y i n [ x , q] . D i s c u s s i o n : In the f o l l o w i n g examples l e t [p , q] = [ 0 , 5 ] . L e t (x , y) i n d i c a t e the ordered s e t {r £ R | x < r < y} w i t h order induced from R . L e t (x , y) i n d i c a t e the ordered s e t {r £ R | x < r < y) w i t h the r e v e r s e o f the *» o r d e r induced from SR . S i m i l a r l y f o r (x , y] e t c . Example ( 7 ) : D e f i n e A = ( y \ ) U (^ ,|) U ( ^ ) U ... U { ^,t&,...} U ( l , l | ] U . ( 2 , 2 ^ ] u'"(5,3|] U ( 4 , 4 ^ ] U ( 4 ^ 4 ^ L ) U 1 2 . L e t T-|_ be the s e t A t o g e t h e r w i t h an i n c r e a s i n g Page 2 it-sequence of d i s j o i n t s e m i - c l o s e d i n t e r v a l s c o n v e r g i n g to 0 and a descending sequence o f d i s j o i n t s e m i - c l o s e d i n t e r v a l s c o n v e r g i n g to 5 . Then [0 , 5] -H = { 0 , 1 , 2 , 3 , 4 , 5 } . Choose Z = { J , l j , 2 J , ^ , 4 ^ } Then Y = { [ | , l j ] , [ l j p 2 | ] >  [^> KT6 ]] '  N o w w e h a v e a procedure f o r o r d e r i n g T^ i n the u n i o n of the elements of Y which i s Y* - t ^ ^ J o " ] . I f x = ^ ( i n g e n e r a l , i f x i s the f i r s t element of Z) then we may d i s r e g a r d the I n t e r v a l [0 , j|] . I f ir^ i s to o rder n i c e l y i n , 5 ] i t w i l l have no f i r s t element and.no l a s t element. L e t X = [^jjr > 5 ] • ( l n g e n e r a l , X = (z , q] where z n i s the l a s t element of Z ) . The problem i s to o r d e r X P. T-^  and to i n s e r t t h i s ordered s e t i n t o the ordered s e t , X(y*fl T^) , i n such a way t h a t the r e s u l t i n g o r d e r has n e i t h e r f i r s t nor l a s t element. Since Y has an even number of elements d e f i n e 17^, on , 5 ] n ^ as f o l l o w s : . % 1 £2,11} T i ^ ( ^ ] ^ ( V ^ T T , (4|^) "•'( > • ( ) ( ] < r — l k 3 — £ — i - w T A [ V J O » X 3 H S * : 0 — -e> :—« * I o> si - H 2 ^Ji 11 n *4«t 4* 11 ' 11 4 4 t4* 4 ****** I t *^J4 * * 4 4 * * ' ^ Page 25 To see the procedure when Y has an odd number of elements l e t Tg. = T± - (4 , 4^ ]' . Then [0 , 5] - H = {0,1,2,3,5} • Choose Z = {| , l J , 2| , 3^ 3 . Then Y = t | '» l | ] * > 2|] [ 2 I > ^ ] } a n d X i s a s i n T-^  . Now i t i s not p o s s i b l e to choose x = ^ as b e f o r e , to o r d e r X fl Tg , and to i n s e r t t h i s ordered s e t i n t o the ordered s e t \ (Y*f) Tg) i n such a way t h a t the r e s u l t i n g o r d e r has n e i t h e r f i r s t nor l a s t element. Hence we choose x = ^ ( x i s the l a s t element i n (Tg-Q) n [0 , j£]) and attempt to o r d e r the s e t s ( i , )^ and X D Tg and i n s e r t these s e t s i n t o the ordered s e t \' (Y* 0 Tg) i n such a way t h a t the r e s u l t i n g ordered s e t has a f i r s t element and no l a s t element. Define i r p ( [ J , 5] H T ) as f o l l o w s : Page 2 6 v2(T^} r 2 ( 2 ~ 2 | ] ir2(3^) M4§>4^  ^ 2^1) ' "^cv ( ) ( 3 ( — 3 k—J 1 . T^nLo^I * J C o , 5 l - H 'Tjj * * ** t / *' ********* mtjt4ttt\^4***t*+/*' *''*'***''**** , r The above two examples are t y p i c a l o f the case when Y has a f i n i t e number o f elements and when q <f T or q g C l g ( X fl Q) Those which f o l l o w are two examples which are t y p i c a l of the case when Y has a f i n i t e number of elements and when q € T and q 4 C l g ( X H Q ) . 00 Example (8): L e t t ^ } ^ - - ^ b e a n i n c r e a s i n g sequence i n 1 2 1 2 [ ^ ' , 5 ) c o n v e r g i n g to 5 and l e t t = . Define T 3 = T l U ^ i I i-= 0,1,2,... } U { 5 ) . As the c o n s i d e r a t i o n s i n c o n s t r u c t i n g 7 f ^ ( T ^ ^ L X * a r e the same as those i n the p r e v i o u s examples, we w i l l d e f i n e ir-^^T^ fl [x , q] ) d i r e c t l y . The number o f terms o f Y i s even so l e t x = ^ and d e f i n e ir-, as f o l l o w s : j Page 2 7 { £ } T T 5 ( 1 A | ] T T 3 ( ^ | ) ^ ( 4 ^ ° , 4 ^ ) i r 5 ( 2 ^ | ] 7 T 3 ( 3 ^ | ] T r3( 4 , 4 i ] Tr 5,{t 03 Tr^{t±] T j { t 2 } . . . ir^{5) . To see the c o n s t r u c t i o n when t h e number of terms o f Y i s odd, d e f i n e T^ = T^ - (4., 4^] . L e t x = -| . D e f i n e -rr^([x,q] fl T^) as f o l l o w s : ^ ^ 4 ^ ( 4 I F ^ & V 1 ' 1 ^ ^ ( ^ 4 ^ M 5 ' 5 ^ 7 T 4 { t 0 ) 7 7 ^ } 7 T 4{t 2 3 7T. (53 F i n a l l y , v/e c o n s i d e r an example t y p i c a l of the case when Y has an i n f i n i t e number of terms. Example ( 9 ) : L e t A = (|,|) U (^,|) U U ... U { J , ^ , ^ , . U{ 0 ] . L e t A' be A together w i t h an i n c r e a s i n g sequence of s e m i - c l o s e d i n t e r v a l s , (a , b] , each c o n t a i n e d i n , 1) and c o n v e r g i n g to 1 . L e t T be the s e t A' together w i t h an i n c r e a s i n g sequence of semi-closed i n t e r v a l s con-v e r g i n g to 0 and a d e c r e a s i n g sequence of s e m i - c l o s e d i n t e r v a l s c o n v e r g i n g to 1 . I n t h i s case, l e t x = -g , the f i r s t p o i n t of Z and then \(T fl [-^  , 1]) , as des-c r i b e d i n example (6) o r d e r s T n i c e l y i n [-jj- , 1] . ... • «. " IW ) ' (- ) ( 3 C 1(3 • T A L o , v J Page 28 Proof o f Lemma ( 4 ) : . Let Z , Y , Y* , Y Q , X and x ' be as d e f i n e d p r e v i o u s l y f o r the element K of G' . Since p € P, . , there i s a f i r s t component , I o f K D H and hence i n the p r e v i o u s c o n s t r u c t i o n , YQ i s the f i r s t element o f Y . Ley y be the f i r s t p o i n t i n A o f Y* which i s a l s o the f i r s t p o i n t of Z . C a s e ( l ) : Suppose Y has a l a s t element. Then there i s a component X of K - Z whose end p o i n t i s q . Since the ac c u m u l a t i o n p o i n t s o f Z e x i s t o n l y a t the end p o i n t s o f K , Z has no accumulation p o i n t s and there are o n l y a f i n i t e number of components of' K fl H ' between I and X and hence o n l y a f i n i t e number of terms of Y . L e t TT(X D(T - Q ) ) = A(X D(T-Q)) and ir(X n Q) = A(X n Q) (a) Suppose q $ T o r q e C l g (X n Q.) . I f the number o f terms of Y i s even, l e t x = y , 7r(Y * n T) = X"(Y*n T) and TT(YQ[) T) TT(X D(T-Q)) TT((Y*- Y Q ) fl T)' TT(X fl Q) Tr((q} n T). Now we v e r i f y t h a t tr orders T n i c e l y i n [y , q] . As i n Lemma (2.) , i t i s v e r i f i e d t h a t i f X n(T-Q) has 2 elements, each p o i n t o f TT(X D(T-Q)) and each p o i n t o f ir(X n Q) has i n t e r v a l topology. To show t h a t ir({.x ^ q] fl T) has i n t e r v a l topology, we need o n l y v e r i f y the f o l l o w i n g : (1) TT(Y 0D T) has a l a s t p o i n t and i f X n(T-Q) i s non v o i d , TT(X n ( T - Q ) ) has a f i r s t and l a s t element. Since X c H and the l e f t hand.end p o i n t o f X i s i n H and not i n T , Page 29 TT(X fl (T - Q )) has a f i r s t element. I f q € C l G ( X n Q) , s i n c e q ^ 'PK , q f Clg (X n(T - Q ) ) and a g a i n , s i n c e X c H , TT(X n(T - Q ) ) has a l a s t element'. I f q $ T and q £ Clg ( x n(T - Q ) ) then q i s a t r i v i a l component o f S - T f o r any i n t e r v a l c o n t a i n i n g q c o n t a i n s a p o i n t not i n H . Hence q <£ C l g ( X n(T - Q )) and so u ( X n(T - Q ) ) has a l a s t element. (2) TT((Y* - Y Q ) n T) Has a f i r s t element and no l a s t element and 7r(X n Q) has no f i r s t element. Since the number o f terms o f Y i s even, (Y*- Y Q ) n T has a f i r s t element and no l a s t element. If. TT(X PI Q) has a f i r s t element t , then t i s a l e f t hand end p o i n t o f a c l o s e d component C , o f Q and hence i s an accumulation p o i n t o f T - Q . Hence t $ H and t h i s i s c l e a r l y a c o n t r a d i c t i o n . S i m i l a r l y we prove the l a s t f a c t (3) TT(X n Q) has no l a s t , element . Nov/, t h a t TT o r d e r s . [x , q] n TV n i c e l y i s imme-d i a t e . F o r y ^ T and [x , q] n T has no f i r s t element. I f q € T then q i s the l a s t p o i n t of [x , q] ( I T , otherwise [x , q] n T has no l a s t element. I f the number of terms of Y i s odd, then l e t z be the l a s t p o i n t o f (p , y) n(T - Q ) i n A . Since (p , y) c H and y $ T and p g P^ , the e x i s t e n c e of z i s guaranteed. L e t x = z and ir(Y * n T) = V (Y*n T) and l e t T T ( ( Z , y ) n Q) = A((z , y ) n Q) . As i n Lemma (2) i t i s shown t h a t T T ( ( Z , y ) n Q) Page 30 has i n t e r v a l topology. Then we s e t • ' i ' # • • ' • {z} TT(X P(T-Q)) TT(Y fl T) 7r((z , y) n Q) TT(X fl Q) Tr({q} ( I T ) . We can see by the same r e a s o n i n g as above t h a t T/(X P. (T-Q)) has a l a s t and a f i r s t p o i n t i f i t i s non v o i d , ir(Y*PT) has a f i r s t but no l a s t p o i n t , ( z , y ) P Q has n e i t h e r f i r s t nor l a s t p o i n t and ir(X P Q) has n e i t h e r f i r s t nor l a s t p o i n t . (b) Suppose q € T but . q $ C l g ( X P Q) . I f the number of terms of Y i s odd, l e t TT(Y*P T) = X(Y*PT) and l e t x = y . "' L e t (X P Q) 7T (Y*P T) TT (X P(T-Q)) IT {q} . I t i s v e r i f i e d , as above, t h a t ir ord e r s T n i c e l y i n [x , q] but i n t h i s case, 7 r ( X p(T-Q)) may have a l a s t element. I f the number of terms of Y i s even, l e t ir(Y*n T) = ) / ( Y * P T) and l e t x = z . L e t {z} T F(Y 0P T) TT((Z , y)n Q) i r ( X P Q) TT( ( Y * - Y Q ) P T) TT(X P(T-Q) )ir{q] Case ( 2 ) Y has no l a s t element. Then l e t TT(Y*P T) = X (Y*P T) and l e t x = y . Then c l e a r l y , v o r d e r s T n i c e l y i n [x , q] . -Lemma ( 5 ) : Suppose c o n d i t i o n ( 2 ) holds and t h a t [p , q] = K £ G . Suppose, a l s o , t h a t p , q e P K . Then there are p o i n t s , x , y £ (p , q) such t h a t x A y , ( p , x ] c H , and [y > l ) c H and there i s an or d e r TT on [x , y] P T t h a t o r d e r s T n i c e l y i n [x , y] . Page 3 1 Discussion: Since p , q £ we are concerned with cases where the number of elements of. Y i s f i n i t e . Again we consider 2 cases. Example ( 1 0 ) : The number of elements of Y i s odd. Let p = 0 and q = 5 • Let A - ( | , | ) U ( i , | ) U (|,^ -) U . . . U{J,ip^  Let A ' =-A U ( 5 - A) U ( l , l | ] U ( 2 , | ] U(3,3TJ] U { 0 , 5 } • F i n a l l y , define T to be the set A together with an increasing' sequence of d i s j o i n t semi-closed i n t e r v a l s converging to 0 and a decreasing sequence of d i s j o i n t semi-closed i n -tervals converging to 5 . Then [ 0 , 5 ] -H = { 0 , 1 , 2 , 3 , 5 } • Choose Z = {^,1^,2^,3^} . Let x = J * and y = ^ • ( l n general, x i s the f i r s t point of Z and y i s the f i r s t point of T-Q i n [z ,q] where z n i s the l a s t point of Z) . Then define T T ( [ X , y] fl T) as follows: (4>4"t) * ( 2 ^ T T { 4 | } . W H ^ H ( ] ( ] , T A t o . S ] O l a 3 ^ . f -Example ( 1 1 ) : The number of elements of Y i s even. Page 32 D e f i n e T' = T - (J^ ji] where T i s as, i n example (10) '2 above. Then [0,5] -H = {0,1/2,5} .' Choose Z = {J,l|.,2^ } . L e t x•= ^ and y = 2^  . ( I n g e n e r a l , x i s the f i r s t element o f Z and y i s the l a s t element o f Z . D e f i n e i r ( [ x , y ] fl T) as f o l l o w s : . » 4 i ( l , l i ] 7r(2,2^] . Note t h a t TT([X , y ] n T) i s j u s t X ( [ x , y ] n T) . P r o o f o f Lemma (5):' I f K $ G' i . e . (p , q) c H , t h e n s i n c e p g P^ , (p , q) -T 4= 0 and we may choose p o i n t s x , y € (p , q) so t h a t x and y b e l o n g t o the same components o f (p , q) -T . Then the lemma i s t r i v i a l l y s a t i s f i e d s i n c e [x , y] D T = 0 . Suppose (p , q) cp H ; t h e n K g G' and d e f i n e Z,Y,Y*,Y 0,\, and x' . S i n c e p , q £ P K , Z i s f i n i t e •.and hence Y has a f i n i t e number o f terms. L e t w be the f i r s t p o i n t ( w i t h r e s p e c t to A) o f Z and l e t z be the l a s t p o i n t o f Z .. Then w i s the f i r s t p o i n t o f Y* and z i s the l a s t p o i n t o f Y* . S i n c e w. , z | T , i f the number o f terms o f Y i s even, Y*fY T has no f i r s t o r l a s t element and i f x = w , y = z , t h e n X o r d e r s T n i c e l y i n Y* = [x , y] . I f the number o f terms o f Y i s odd, l e t y be the f i r s t p o i n t o f T-Q i n [z , q) . Such a y e x i s t s s i n c e q e P K . L e t Tr(Y*n T) = X(Y*fl T) Page 33 TT([-Z , y)n Q) = A ([z , y)n Q) and l e t ( [ z , y)n Q) ir(Y*n T) ir{y} . As shown p r e v i o u s l y , 7 r ( [ z , y)n Q) has the i n t e r v a l topology and i f [ z , y)D Q * 0 , i r ( [ z , y)D Q) has n e i t h e r f i r s t nor l a s t element. Then i r ( [ x , y]n T) has no f i r s t element, "but has a l a s t element and x $ T but y € T and so T i s ordered n i c e l y i n [x , y] . Lemma (6): Suppose c o n d i t i o n (2) holds and that K 6 G . Suppose one end p o i n t of K i s not i n P^ and the other i s not. the end p o i n t of any component of K n H . Then there i s a t o t a l o r d e r i n g , ir of T D K such that ir orders T n i c e l y i n K . D i s c u s s i o n : Since one end p o i n t of K i s not the end p o i n t of any component of K n H , i t i s c l e a r that Y has an i n f i n i t e number of elements. We consider 2 cases. I n each case l e t K = [0 , 5] . Case (L): The set Y has a f i r s t or l a s t element. Without l o s s of g e n e r a l i t y we may assume Y has a f i r s t element. Example (12): Let T be the union of {0,5}, (^ ,1) , and [2,3] together w i t h an i n c r e a s i n g sequence of d i s -j o i n t semi-closed i n t e r v a l s i n (4,5) converging to 5 , say {(a^, b^ ] | i £ IN} and together w i t h any i n c r e a s i n g sequence of d i s j o i n t semi-closed i n t e r v a l s converging to 0 . Page 34 Then ( j j ' l ) c Q , [2 , 3 ] C T - Q , and [0 , 5] -H = {0 , 5} U {^1 i € IN} . A t o t a l o r d e r , IT t h a t o r d e r s T n i c e l y i n [0,5] can be d e f i n e d as f o l l o w s : -> ^ {0} TT (|,1) TT ( a 1 , b 1 ] TT [2,3] Tr' ( a g , b 2 ] TT .(a^b ] IT ( a ^ b ^ ] . • • TT (5} I n t h i s example, p 6 Clg ( K n Q ) . -{ ) F -3 ' ( H I (4fr T A L o > ] V 2. 3 4 a, a t O j . . . 5" _| 0 b 6 9 \ 7. 3 * «i «JL V " i L o ^ I - H Example (13): Let A = {^ , . . . } U ( l , 2) . L e t T be the u n i o n of {0,5} and A t o g e t h e r w i t h an i n c r e a s i n g se-quence of d i s j o i n t , s e m i - c l o s e d i n t e r v a l s i n (2 , 5) converg-i n g to 5 , say {(a., b^] | i _€ IN} and t o g e t h e r w i t h any • i n c r e a s i n g sequence of d i s j o i n t s e m i - c l o s e d i n t e r v a l s converg-i n g to 0 . In t h i s case Q = (1 , 2) and [0 , 5] -H = {a^| i g IN} . A t o t a l o r d e r , ir t h a t orders T n i c e l y i n [0 , 5] can be d e f i n e d as f o l l o w s : {0} IT ... TT {•g} TT {^} Tfig-} TT ( a ^ b ± ] 7T (1 , 2) 7 7 " ^ , b g ] 7T (a^ ^ IT ( a , , b^] . . . ir {5} . I n t h i s example, p $ C l q ( K fl Q) . Page 35 '—©—«—o— OA i > 4 - 3 — 4 4 4 ^ T A t ^ S l Case (2): The set Y has n e i t h e r f i r s t nor l a s t element. Example (14): Define T7. = (T-A) U {(5-a±, 5-b±] | i € IN} where a , b^, A , and T are defined i n example (12) . C l e a r l y the order X ([0 , 5] f l T) s a t i s f i e s the hypothesis of the lemma. • • •  1 U C) -«—C 1 (10' T ALcyl Proof of Lemma (6): . Assume q i s not the end p o i n t of any component of K fl H , and that p P^ . and K = [p , q] . For the gi v e n K s since K g G7 , we can define Z which • has an i n f i n i t e number of elements and a l s o define YQJ Y , Y*, X i X 7 f o r the given K . Case (1): The set Y has a f i r s t or l a s t element, so assume without l o s s of g e n e r a l i t y t h a t Y has a f i r s t element. Then denote by X the component of (p , q) -Z whose end p o i n t i s p . Let TT (X n ( T - Q ) ) = A (X D ( T - Q ) ) and TT (X n Q) = A (X n Q) and v e r i f y immediately that TT (X fl ( T - Q ) ) Page j56 and TT(X n Q) has the i n t e r v a l topology. (a) I f p $ T or p £ C l G ( X n Q) d e f i n e the f o l l o w i n g o r d e r , TT on K n T . L e t 7r(Y * n T) = X ( Y * f l T) and ({p} n T) TT (x n Q) TT ( Y Q n T) TT (X n (T-Q)) TT ((Y*- Y Q ) n T TT(.{q} fl T) . From an argument i d e n t i c a l to t h a t of Lemma (4) rr o r d e r s T n i c e l y i n K . (b) I f p e T and p { C l g ( K n Q) d e f i n e the f o l l o w i n g o r d e r 7f on K fl T . L e t ir(Y*f) T) = \* (Y*0 T) and (X 0(T-Q)) TT(YQD T) TT (X fl Q)'TT((Y*- Y Q ) n T) 7r ( { q ] ' n T) . S i m i l a r l y , i t i s c l e a r t h a t TT orders T n i c e l y i n K . Case ( 2 ) : I f Y has n e i t h e r f i r s t nor l a s t element, then an o r d e r , TT which o r d e r s T n i c e l y i n K i s : ({p} n T) 7r(Y * n T) 7r({q] fl T) where i r ( Y * n T) = \ (Y * f l T) . I n p r o v i n g t h a t c o n d i t i o n (2) i m p l i e s t h a t T i s t o t a l l y o r d e r a b l e , we can prove the r e s u l t e a s i l y f o r the case when S £ G and S - H # 0 . P r o p o s i t i o n : I f c o n d i t i o n (2) holds and S £ G , S - H £ 0 , then T i s a t o t a l l y o r d e r a b l e t o p o l o g i c a l space. Page 37 Proof: Since S - H 0 but S g G , a l s o S g G' and so we can d e f i n e Z , Y , Y Q , Y * , and X f o r S £ G . Since +co and -co are c o n t a i n e d i n n o n - t r i v i a l components of S - T and the accummulation p o i n t s of Z occur o n l y a t +co and -co , Z has no ac c u m u l a t i o n p o i n t s and so Y has a f i n i t e number of terms. L e t ir ( Y * f l T) = X ( Y * n T) and ir ((T-Q) - Y * ) = A ( ( T - Q ) n Y * ) ' , TT(Q-Y*) - A ( Q - Y * ) . Define a t o t a l o r d e r TT on S R T by: ( Q - Y * ) 7 r ( Y o n T) T T ( ( T - Q ) - Y * J T T ( ( Y * - Y Q ) n T) . S i n c e Q - Y * c H i t f o l l o w s immediately t h a t ( Q - Y * ) has the o r d e r t o p o l o g y and. has n e i t h e r f i r s t nor l a s t element i f i t i s non-void. I t f o l l o w s a l s o t h a t ( ( T - Q ) - Y * ) has o r d e r t o p o l o g y aid has f i r s t and l a s t element i f i t i s non-v o i d . Hence T i s a t o t a l l y o r d e r a b l e t o p o l o g i c a l space. D e f i n i t i o n : I f I i s a c l o s e d i n t e r v a l o f S , $(J) i s a replacement of I i f (1) $ I s a - t o t a l o r d e r on the s e t J and J i s a topo-1- . l o g i c a l space w i t h the order topology, (2) T D I <= J and J n T i s homeomorphic to I n T , (3) J i s compact and connected, the end p o i n t s o f J are the end p o i n t s of I and J - T has no t r i v i a l com-ponents. We say §(J) i s a n i c e replacement of I i f $(J) i s a replacement of I and $ r e s t r i c t e d to J fl T or d e r s T n i c e l y i n I . Page 38 P r o p o s i t i o n : L e t I be a c l o s e d i n t e r v a l of S and assume TT o r d e r s T n i c e l y i n I . Then there i s a n i c e r e p l a c e -ment, §(J) of I . Proof: L e t I = [p , q] a n d ' l e t T' = (T n I) U{p , q} . Then TT Induces a n a t u r a l o r d e r .TT' on T 7 , namely x 7r'y i f and o n l y i f x = p or y = q or x i r y . L e t T'' be the s e t o b t a i n e d from T 7 by i n s e r t i n g between each a , b £ T 7 where b i s an immediate s u c c e s s o r of a , a s e t homeomorphic to (0 , 1 ) and l e t IT'' be the order induced from 7 r 7 on T 7 . Then l e t J be the s e t of a l l or d e r e d p a i r s (A , B) where ( 1 ) 0 > A c T" , 0 # B c T 7 7 , A U B = T " , and A n B = 0 , . ( 2 ) a 7r 7 /b Y a g A , V b £ B , (3 ) B has no s m a l l e s t element. Then d e f i n e an o r d e r , § on J by: (A , B) § ( A 7 , B 7 ) i f and o n l y i f A c A 7 . When J i s equipped w i t h the o r d e r topology, T D I i s ho-meomorphic to {(A , B) | sup' (A) £ T fl I} and J i s compact and connected and § r e s t r i c t e d to J n T o r d e r s T n i c e l y i n I . I Now f o r each elenie^it^_jC of.- & d e f i n e a replacement $ ( K Q ) and a s e t o? subsets of K , M K as f o l l o w s . Assume c o n d i t i o n i^'f h o l d s . Page 39 I f a n i c e replacement o f K e x i s t s , l e t $(KQ) be the n i c e replacement and l e t = 0 . I n (2) through (5) assume no n i c e replacement o f K i s p o s s i b l e . I f one endpoint o f K belongs to P^ and K <jj G 7 , l e t *(K 0 ) . = A(K) and 1^ = {K} . I f K <= G 7 and K = [p , q] , p € P K and q f P K then by lemma (4) there e x i s t s x g (p , q) and an o r d e r , TT , of [x , q] n T) such t h a t (p , x] c H and ir o r d e r s T n i c e l y i n [x , q] . By the p r e v i o u s p r o -p o s i t i o n there i s a n i c e 'replacement, § (B) of [x , q ] . L e t KQ= [p , x] U B and d e f i n e a t o t a l o r d e r & on K Q by $ [p , x] - A [p , x] and [p , x] $ (B) . L e t MK= {[p , x]} . I f K € G 7 and K = [p , q] and p , q 6 P K , by Lemma ( 5 ) there e x i s t s x , y 6 (p , q) such t h a t [p , x) c H , (y , q] c H and there e x i s t s a t o t a l o r d e r TT on [x , y] D T t h a t o r d e r s T n i c e l y i n [x , y] . L e t $(B) be-a n i c e replacement of [x , y] and l e t K Q = [p , x)u B U(y , q] . Define an o r d e r , § on K Q by $ [p , x] = A [p , x] , $ [y , q] = A [y , q] and [p , x) $ B $ [y , q] . L e t M K = {[p , x] , [y , q]}. . I f n e i t h e r end p o i n t o f K i s i n P^ , we f i r s t show t h a t there e x i s t s an o r d e r on K fl T f o r which o n l y a f i n i t e number o f p o i n t s o f K fl T do not have.-inter-v a l topology. Assume t h a t K g G 7 and assume t h a t there Page kO are an i n f i n i t e number of p o i n t s o f A(K n T) t h a t do not have i n t e r v a l topology. Then there e x i s t s y 6 K such t h a t e v e r y open i n t e r v a l about y c o n t a i n s a p o i n t x £ K D T where x does not have i n t e r v a l topology. I f y i s an end p o i n t of K , c l e a r l y y $ C l g ( X n Q) or e l s e y $ C l G ( K n(T-Q)) and i f y i s an end p o i n t of K , s i n c e y £ H , y $ C l g ( K n Q) or e l s e y $ C l G ( K P(T-Q)) . Now i f y . | C l g ( K n Q) , by an argument i d e n t i c a l to t h a t i n Lemma (4), there i s an open i n t e r v a l , I , about y such t h a t e v e r y p o i n t o f (T-Q)P. I has the i n t e r v a l t o p o l -ogy. T h i s i s a c o n t r a d i c t i o n to the d e f i n i t i o n of y . S i m i l a r l y i f y ^ C l G ( K P.(T-Q)) there i s an o p e n ' i n t e r v a l , I , about y such t h a t e v e r y p o i n t of Q P I has i n t e r v a l t o p o l o g y . A g a i n t h i s c o n t r a d i c t s the d e f i n i t i o n of y . I f K £ G' then we can d e f i n e Y* and X f o r the g i v e n K and X(Y*P T) has i n t e r v a l t o p o l o g y and by an argument s i m i l a r to the above, there are o n l y f i n i t e number of membeis o f (K P T) -Y* t h a t do not have i n t e r v a l topology. Hence we have shown t h a t there e x i s t s an order on K P T f o r which o n l y a f i n i t e number o f p o i n t s o f K P T do not have i n t e r v a l topology. Now u s i n g the c o n s t r u c t i o n i n the p r e v i o u s p r o -p o s i t i o n , there e x i s t s a replacement o f K , §(KQ) , such t h a t $(KQP.T) f a i l s to have i n t e r v a l t o p o l o g y a t a t most f i n i t e l y many p o i n t s o f K N P T . Page hi Example (15-)-!- L e t A = ( - | , - | ] U  [~E> -JZ> ' ' ' 3 u ( - J ^ • U  U ^ ~ T 0 ^ ~ I T ^ U ' ' ' ' L e t B 1°e t l r i e s e t c o n s i s t i n g o f { 0 , 1 } t o g e t h e r w i t h the u n i o n of a s e t of d i s j o i n t s e m i - c l o s e d i n t e r v a l s , each c o n t a i n i n g one and o n l y one element of the. s e t p/ = { ^ i p ^ j p ' j p • • • 3 • F o r convenience we w i l l assume t h a t i f [a , t>) i s a component o f B then a has an immediate p r e d e c e s s o r i n B and b has no immediate s u c c e s s o r i n B . A l s o , assume t h a t the i n t e r v a l c o n t a i n -• 1 i n g i s c l o s e d on the l e f t and denote t h i s i n t e r v a l [ajj. , b^ j-) . Define' T = A U B . Then G = { [ - c o , 0 ] , [ 0 , 1 ] [ 1 , = 0 ] } and G' = {[-co , 0 ] , [ 0 , 1 ] ] F o r each element K f_ G , §( KQ) i s K w i t h order induced from R . A l s o Q-j= [ i , 0 ] , • M [ 0 , l ] = $ > M [ l , « ] - 0 • . • — ( — — 3 < r * - ^ > # [•)(•] I O — T -1 -i ~> -\ 0 \ \ >x \ > 1 iV^V/ / / / / / / f 11 s 11 ft ? >>)• it /// // / // /C / t / 11 t,' /v ,• / / / v,y tttftft+r+, ,X-AV^^- 6-L e t $(N) be the replacement of S ob t a i n e d by r e -p l a c i n g each term K of G by $(K Q) . L e t M = {Mgl K £ G} . Page 42 Lemma ( 7 ) : Suppose c o n d i t i o n (2) holds and S ^ G . Then $(N) and M s a t i s f y the f o l l o w i n g . (a) E v e r y p o i n t a t which $(T) f a i l s to have i n t e r v a l t o p o l -ogy i s a member o f some element o f M . (b) L e t H* be the un i o n of a l l open i n t e r v a l s of §(N) whose c l o s u r e s i n t e r s e c t T - Q , i n a compact s e t d i s j o i n t from Cl^(Q) . I f I <E M , there e x i s t s i $ I such t h a t C 1 N ( N - ( I U H * ) ) and i f the s e t of p o i n t s of (I fl T) a t which §(T) f a i l s to have i n t e r v a l t o p o l o g y i s i n f i n i t e , i t can be ordered i n a simple sequence, monotonic i n $ and with l i m i t p o i n t lT . (c) The s e t M i s co u n t a b l e and. f o r a l l g > 0 and. I e M there e x i s t s an open i n t e r v a l of f(N) c o n t a i n i n g tT such t h a t any two p o i n t s o f T - I are no f u r t h e r a p a r t than g on the o r i g i n a l l i n e (R . Proof: We show f i r s t t h a t M i s a d i s j o i n t f a m i l y . Let T,J,gM . I 6 Mj/ j J £ M j / a n c ^ there are 2 ' p o s s i b i l i t i e s , (1) I / = J ' and i n t h i s case c l e a r l y I fl J = 0 or e l s e I = J , ( 2 ) 1 7 * j ' . Assume . I n J * 0 . ' Then s i n c e i ' c I and J 7 c J , I 7 n J ' c I n J . F u r t h e r , I 7 and J 7 i n t e r -s e c t i n a t most one p o i n t so assume I fl J = {p} and p i s an end p o i n t of I 7 and J 7 . By d e f i n i t i o n o f Page 43 G y p i s not the end p o i n t o f n o n - t r i v i a l components of I'd II and j ' n H so assume p i s not the end p o i n t o f a n o n - t r i v i a l component of l ' n H . By Lemma (6) , the oth e r end p o i n t o f I 7 i s i n P^/ . Hence I was chosen by r u l e s (2) , (3) , (4) above and i n t h i s case p ^ I . T h i s c o n t r a d i c t i o n e s t a b l i s h e s the r e s u l t . (a) I f t e T and t t K € G then §(T)•clearly has i n t e r v a l t o p o l o g y a t t or e l s e t i s i n some term o f . I f t i s not i n some term K of G then t i s a 2 s i d e d a c c u mulation p o i n t o f p o i n t s o u t s i d e of. H and hence a 2 s i d e d ' a c c u m u l a t i o n p o i n t of p o i n t s o f T , and a l s o o f p o i n t s of B - T and so $(T) has i n t e r v a l t o p o l o g y a t t . (b) L e t I £ M and assume I g f o r K £ G . I f there e x i s t s p € ( P K n I) l e t = p . Otherwise, I i s a replacement o f K and s i n c e S £j G , K 4 S and there e x i s t s an end p o i n t o f I , d i f f e r e n t from -«> or +» . Le t ^ be such an end p o i n t . A g a i n by Lemma (6) , lzq C1 N ( S - (K U H)) and hence 1^ C 1 N ( N - (I U H * ) ) . I f the number of p o i n t s a t which $ (K) f a i l s to have i n t e r v a l t o p o l o g y i s i n f i n i t e , then by d e f i n i t i o n o f §(N) , t j € P K . Assume 1 i s a l e f t end p o i n t o f I and, l e t x be a p o i n t o f N such t h a t [l-^s x] = I and (l^,*.) c H and x $ P K . L e t V be the s e t con-s i s t i n g o f those p o i n t s v o f [ l r , x] such t h a t Page kk v g' T - Q or v i s between two p o i n t s \ t ^ and t g of T -and no p o i n t s o f Q are between t ^ and t ^ . Then V £ 0 s i n c e . P^ . and a l s o , there i s a p o i n t o f Q between e v e r y two components of V . Now s i n c e x ^ P^ .. , and (lr, x) c H , lj i s the l i m i t p o i n t o f a simple sequence which i s monotonic i n § c o n s i s t i n g o f components of V . But the p o i n t s a t which s(T) f a i l s to have the i n t e r v a l t o p o l o g y are the endpoints o f components of V hence the p r o o f o f (b) i s complete. (c)- To show,that M i s co u n t a b l e , we n o t i c e t h a t each term o f ; M corresponds to a non t r i v i a l c l o s e d i n -t e r v a l on the r e a l l i n e and hence the s e t M i s co u n t a b l e . To show the second p a r t o f (c) , note t h a t A p r e s e r v e s the order o f T i n (R and t h a t $ p r e s e r v e s the or d e r o f T i n A except w i t h i n terms o f G . Hence i f the o n l y term o f G which •t^ belongs to i s K then the r e s u l t i s c l e a r . I f l-^ belongs to some o t h e r term, L , of G .then tT£ P^ and /..j$(H fl L) by d e f i n i t i o n o f G and i n Lemma ( 4 ) or Lemma ( 6 ) the order ir on L i s d e s c r i b e d by case (2) . Now by d e f i n i t i o n of a replacement, the r e s u l t c l e a r l y h o l d s . Example ( 1 6 ) : L e t T be d e f i n e d as i n example ( 1 5 ) . Then M = §?r> > °]} a n d = • T n e lemma c l e a r l y h o l d s and i n t h i s case t? 1 N 1 = 0 . Page 45 Example ( 1 7 ) : Let T be d e f i n e d as above. Define f . ; [ - l , 1 ] - [ - - . , - - , . , ] f o r i - 0 , 1 , 2 , . . . to be the order 1 1 p r e s e r v i n g homeomorphism o f [ - 1 , 1 ] onto [- — -Le t T' = U {f j_(T) [ i = 0 , 1 , 2 , . . . } . Then M = { f ± [ - | , 0 ] | i = 0 , 1 , 2 , . . . } and i f ^ = f ± ( 0 ) f o r i = 0 , 1 , 2 , . . . . Lemma (8): I f c o n d i t i o n ( 2 ) holds and S $ G , then T i s t o t a l l y o r d e r a b l e . Proof: L e t § , M , N be as d e f i n e d i n Lemma ( 7 ) . . S i n c e M has a countable number of elements, l e t M = {M 1,Mg, . . . } . ' F o r each n £ |N , we w i l l c o n s t r u c t N n , a c o l l e c t i o n of compact i n t e r v a l s o f f(N) and a t o t a l o r d e r -i n g , A n ( T ) s a t i s f y i n g the f o l l o w i n g c o n d i t i o n s : 1 . The terms o f N n do not i n t e r s e c t i n T and i f n > 0 then every element o f N n i s a subset o f some element o t \-i : 2 . I f n > 0 and A n ( T ) f a i l s to have i n t e r v a l t o p o l o g y a t a p o i n t , t g T , then A ^(T) f a i l s to have i n t e r v a l t o pology a t t and. t'$ M . 3. I f A n ( T ) f a i l s to have i n t e r v a l t o p o l o g y at. a p o i n t , t g T , then there i s a term,, I £ M , c o n t a i n i n g t and t h e r e i s a term o f J n c o n t a i n i n g t and i n i t s i n t e r i o r . Page 46 4. (a) I f X c N n , thenA n (XflT) = $(XnT) or ^ ( X f l T ) = -I(XHT) (b) No p o i n t o f T - X i s between two p o i n t s of X fl T i n 5 . I f n > 0 and. x A n y and y A n -jX then x and y are i n the same term o f Nn_-^ . I f X g N n and X c o n t a i n s x or X c o n t a i n s y , then no two p o i n t s of X n T are a t a d i s t a n c e a p a r t , on the o r i g i n a l l i n e , g r e a t e r than i . 6. I f n > 0 and. x , y £ T and no p o i n t i s between x and" y i n A n ^ then no p o i n t of T i s between x and y i n A n We f i r s t v e r i f y t h a t the sequences {A } and co n = 0 {N 3 can be d e f i n e d . L e t A n = s f N n = {N} and. these n n = : 0 u u c l e a r l y s a t i s f y the h y p o t h e s i s . Assume we have d e f i n e d A n ( T ) and. N^ f o r a l l n < k . Then we d e f i n e A^ (T) and N^ as f o l l o w s . C l e a r l y i f \ _ ] _ ( T ) nas i n t e r v a l t o p o l o g y a t e v e r y p o i n t of M^n T then s e t \ ( T ) = A^ . 1 ( T ) and N R = \ ± and the c o n d i t i o n s are s a t i s f i e d . So assume there e x i s t s a p o i n t of M^n T where k^ l ( T ) fails to have i n t e r v a l topology. L e t t. denote and by h y p o t h e s i s (3) , l e t X be the term o f N k ^ c o n t a i n i n g , i n i t s i n t e r i o r , I and a l l p o i n t s where Page 47 A ( T O M ) f a i l s to have i n t e r v a l topology. Assume, K—1 k with o u t l o s s o f g e n e r a l i t y , t h a t I i s a r i g h t hand end p o i n t o f M .^ . By Lemma 7(b) and (c) there e x i s t s t e X fl T such t h a t A = [ y ? T I U y H ) s a t i s f i e s t g e ( T f l A ) I t ^ - t g l < ^  . Denote by d the endpoint o f M f^l X . Denote by (a , b) the s e t "{y € N | a $ y § b] and sim-i l a r l y denote [a , b] . Nov; l e t V' be the s e t of h a l f open i n t e r v a l s o f N which are components of X - T which have one and o n l y one end p o i n t i n [d , I] fl T . I f there i s an element v g V' such t h a t there e x i s t s x <= X n T where x i s the o n l y end p o i n t o f v - i n the i n t e r i o r o f X and there e x i s t s y €(T-X) such t h a t there are no p o i n t s o f T between x and y i n ^ _ ] _ ( T ) > then d e f i n e V = V / - { v Otherwise, V. = V'. . Now i t i s c l e a r from the d e f i n i t i o n o f V t h a t A k ^ ( ^ H T) f a i l s to have i n t e r v a l t opology a t x i f and o n l y i f x i s the end p o i n t o f some element o f V . Hence, from Lemma 7(b) , i f V has an i n f i n i t e number of members, these members can be ordered i n a sequence and t h i s sequence i s monotonic i n § and has l i m i t p o i n t l . I f .V has a f i n i t e number of members, l e t V =. {v,, v , v.} where v^? v i + i ^ o r 1 = 1 > • • • > J - 1 • L e t d.^  be any p o i n t o f v^ and l e t b^ be the end p o i n t o f v^ which i s i n [d , I] fl T . We w i l l d e f i n e an o r d e r , fi , on a super-s e t o f [d , 1] fl T . At p r e s e n t , l e t n([d , l] fl T) = $([d , 1] fl T) L e t U be the s e t o f a l l components o f (l , t ) -T having one and o n l y one end p o i n t i n T and Page 48 l e t U' be the unio n o f a l l such components. Case (1): Suppose I £ C l N ( U ' ) . I f u £ U l e t b(u) be the one end p o i n t o f u not in. T . Given u £ U , b(u) belongs to'some term I o f • M and a g a i n by Lemma 7(b) , of a l l terms w £ U s a t i s f y i n g b(w) £ I , there i s a f i r s t w i t h r e s p e c t to order '$ , • say u* . Define U* = {u* | u £ U} and since' l £ Cl^(U') and l belongs to o n l y one term of G , l e t {u^, u^,...} be a sequence o f teimis o f U* , monotonic i n $ and ha v i n g l as a l i m i t A p o i n t . L e t p^ be-any p o i n t o f u^ and l e t U be the sub-sequence o f U* th a t s a t i s f i e s , without l o s s o f g e n e r a l i t y , M U N ) -3 P n • Define L - U P ^ ^ ] I n = 1,2,...} . L e t L' be L together w i t h components of X ~ ( d , p^) and d e f i n e \ = (\_-j_- {X}) U L 7 . (a) Suppose V has o n l y a f i n i t e number of e l e -ments. We now extend the d e f i n i t i o n o f Q to • [d , p^] n T . L e t ( ( P ] _ , p Q ) n T) n ( ( d x , d 2 ) n T) Q ( ( P 2 , P l ) n T) Q ( ( d 2 , d ^ ) n n . . . n ( ( d j _ 1 ) n T) n ( ( P j , P J _ 1 ) n T) Q ( ( p J + 1 * P j ) n T) ; n . . . I f x , y € ( p n * P n _ i ) l e t y n x i f f y $ x and (1) . n £ j and d n § b n and b (u') $ P n or ( 2 ) n > j and n - j i s even. Now d e f i n e A, (T) as f o l l o w s , Page 4 9 x A^ . y i f f x A k y and not both of x and y are i n [d , p Q ] D T and • A k([d , p 0 ] n T) = a([d / p Q ] n T) i f A k ( x n T) and A k ( [ d , p Q ] fl T) = -Q([d , p Q ] n T) i f A k(X fl T). = -§(X n T) . We now check the hypotheses f o r A k and N k . C l e a r l y (1) and ( 5 ) are s a t i s f i e d . Since I f A k ^ ( T ) f a i l s to have i n t e r v a l topology a t t £ M k then t = b^ f o r some n and by d e f i n i t i o n of A , ( T ) , hypotheses (2) i s s a t i s f i e d . A By d e f i n i t i o n of U , h y p o t h e s i s ( 3 ) holds and ( 4 ) and ( 6 ) are immediate. (b) Suppose V i s i n f i n i t e . Define the f o l l o w -i n g o r d e r , Q on [d , p Q ] fl T , ( ( P l , p 0 ) n T ) Q ( ( d 1 , d 2 ) n T ) n ' ( ( ? . , P 2 ) n T ) Qt(%> D3 ) N T) n('(P 5, P 4 ) n T ) . . . . . . ( ( P i | , p 5 ) n T ) n ( ( P 2 * P]_) n T ) • : We. complete the d e f i n i t i o n of Q i n the u s u a l way. Suppose x , y £ (p , P n _ i ) ^ T a n d n i s o d d » Then y fi x i f f y $ x and b^ where n = 2 i + 1 . I f x , y'€(P n^ P n 2_) a n d n ^ s e v e n ^ then y fi x i f f y § x and ^ i s even or x $ y and ^ i s odd. A l s o , A K ( T ) and N K are d e f i n e d as b e f o r e and the hypotheses are s i m i l a r l y v e r i f i e d . ' Case ( 2 ) : Suppose 1 $ C l N ( u ' ) .. Then there e x i s t s a p o i n t t ' of (t , t ) such t h a t t 'e T and no component of (l, t') -T has an end p o i n t b e l o n g i n g to T .. For assume the c o n t r a r y . Page 50 Since I <f 01^ (11') and" e v e r y p o i n t a t which § (N) f a i l s to have i n t e r v a l t o p o l o g y i s i n some term of M , I i s the end p o i n t of M^g M where £ ^ k . I n the d e f i n i t i o n of §(N) , I i s the l e f t hand end p o i n t of a component of H* and a l s o the r i g h t hand end p o i n t of a component of H* . T h i s c l e a r l y c o n t r a d i c t s Lemma 7 (b) . Hence CO choose a sequence, [p } c ((£, t') -T) d e c r e a s i n g to n n=0 I . Between p n and P n_]_ there i s a p o i n t not i n H* . Hence between p and p there i s a p o i n t and s i n c e no *n * n - l ^ p o i n t of T i s the end p o i n t of a component of ((I, t')-T) , choose t so t h a t the component of T _ c o n t a i n i n g t i s CO CO t r i v i a l . So l e t {r } and {s ,} be sequences such. n n=0 n n=0 t h a t r ^ . T , s^^ T ., j. ='0,1,... and [ r n ) i n c r e a s e s to t n , fs } decreases to t and r . = p , , s_ = p 1 n J n 0 * n - l ' 0 • n . Then d e f i n e a t o t a l order §' on ((p , P n_-j_)fi T.) by: *' ( ( v T ) = §((so> ^ T ) * - ' . . ^ V r0)fl T ) = - $ ( ( t n ' r0)n , and ( ( s Q , S ; L ) n T) j ' ( ( r i , r Q ) n T) $' ( ( S l , s 2 ) n T) * ' ( ( r 2 , r 1 ) n T) ... <">'{tn} . Note § / ( ( p n , P n_]_) (" 1 T ) nas i n t e r v a l topology w i t h > no f i r s t element and a l a s t element. With §' r e p l a c i n g A $ , we can d e f i n e u' , U and we have the same case as case (1) . Define fi as i n case (1) . We note, however, t h a t h y p o t h e s i s (4) does not hold i f N^ i s d e f i n e d as i n case 1 . But s i n c e Ak( ( - 6 , t ' )n T) has i n t e r v a l t o p o l o g y Page 5 1 a t e very p o i n t , we can d e f i n e L = 0 and then N^ as be f o r e and hy p o t h e s i s (4) h o l d s . Now that we have d e f i n e d A n ( T ) and N n f o r each n , we w i l l show how to d e f i n e a t o t a l o r d e r , A (T) , such t h a t A (T) has i n t e r v a l topology. F i r s t n o t i c e that i f t £ T ', and f o r each k , there exusts X, 6 N, such t h a t t P X-, 6 N. then there i s an k k • ^ k k i n t e g e r ra such t h a t A n ( X n ) = A m ( X m ) f o r a l l n >y m . For, i f the above i t not t r u e , then by c o n s t r u c t i o n of N^ , the component of T c o n t a i n i n g t i s t r i v i a l and t h i s con-t r a d i c t s the f a c t t h a t |(N) i s a replacement of S . Now d e f i n e A (T) as f o l l o w s : CO ( 1 ) I f there e x i s t s a ( s m a l l e s t ) i n t e g e r , n , such t h a t x and y are i n d i f f e r e n t elements of N , l e t x A y n co i f and o n l y i f x A n y . (2) I f x and y s.re i n the same element, X of N f o r each n , then there e x i s t s m such t h a t f o r a l l n > m , A n ( X n f l T) = A m ( X n f l -T) . • Let x A^ y i f and o n l y i f x A m y We f i r s t show that A i s a t o t a l o r d e r i n g o f T . CO ° (a) I f x A ^ y by ( 1 ) and y A^ z by ( 1 ) then by hypothe-s i s (5) , x and z are i n di:f Cerent elements o f N^ where p = max (n , m) and x A p z so that x A^ z . (b) I f x A y by ( 1 ) and y A y 2 b y (2) then by hypothe-00 s i s (5) and 4(b) , x A z . Page 52 .(c) I f x Aro y by. (2) and y A ' z by (2) then c l e a r l y , x A z . CO We now show th a t A (T)' has i n t e r v a l topology.. Suppose A (T) f a i l s to have I n t e r v a l topology at.' t £ T . CO Then i f A (T) f a i l s to have i n t e r v a l t o p o l o g y a t t f o r a l l n , there e x i s t s M £ M such t h a t t £ M by ( 3 ) . 3 n n \ ^  1 But by (2) t f M n because A (T) f a i l s to have i n t e r v a l t opology. Hence there e x i s t s n such t h a t ^ n ( T ) has i n t e r v a l t o p o l o g y a t t . By (2) , A (T) has i n t e r v a l t o p o l o g y a t t f o r a l l m > n . To snow th a t A (T) has i n t e r v a l t o p o l o g y a t t , l e t t $ 0 = (z f IR | a < z < b] where a , b £ (R and 0 i s a b a s i c open s e t o f IR . Choose n Q such t h a t n Q > n , i < min { | t-a | , | t-b |} and n^ > k where k i s an i n t e g e r t h a t s a t i s f i e s (1 ) t i s , not i n any element o f N k or (2) f o r a l l m > k , there e x i s t s X £ XT such t h a t t c X v 1 — mv m ^ m and A M ( X m n T) = A k ( X k n T) . Then choose x , y £ T such t h a t t £ {u £ T I x A u A y] c 0 fl T . Then assume . 0 0 I x - t I < i i f t has no imr.iediate p r e d e c e s s o r In A and. n 0 n 0 I y - t I < i i f t has no immediate s u c c e s s o r i n A . I f n O n 0 t has an immediate s u c c e s s o r and p r e d e c e s s o r , say y and x then by (6) , t £ {u £ T | x u ^ y) c 0 (1 T . So assume t h a t x and y are chosen so t h a t | x - t | < — n o Page 53 and- | y - t | < — . By (5) and t h e ' d e f i n i t i o n of n , n 0 t c[u £ T | x A u A y } c o n T . ' CO CO Now l e t t f 0' = fu £ T I x i u} where x £ T 00 and 0' i s a subbasic open s e t of A (T) . L e t k be an CO i n t e g e r such t h a t (1) x and t are i n d i f f e r e n t elements of N k i f x and t are each i n an element of N^ or (2) f o r a l l m > k , there e x i s t . X £ N such t h a t x , t f X v ' — 3 _ m m 3 ^ \ and A f f i(X m n T) = \ ( X k n T) . L e t n Q be such t h a t n Q > k , A (T) has i n t e r v a l topology a t t. and — < i | t-x | . n 0 n 0 . S i n c e A (T) has i n t e r v a l topology a t t , l e t a £ IR be n 0 such t h a t t £{z e fR | z > a } n T c { u c T | x A u}. Then no by d e f i n i t i o n of n Q , t <={z £ JR | z > a} fl T c 0' . Example (18): I f T i s d e f i n e d as i n example (15) then we c o u l d d e f i n e as f o l l o w s an order § on T such t h a t T has the i n t e r v a l topology: f '< 4 § (4>4] - ( 4 4 } * [4>4 } $ *4 } * ( a i o - ^ i 6 - ] -A ^ ^ * (-j>-%) $ C ^ ' ^ S O $ C-|} . . . $ {0} § [4*4) $ ( a J'4 ] • $ [a^,b^) . . . * ( 1 } . C l e a r l y $ ( f ± ( T ) ) c o u l d be d e f i n e d s i m i l a r l y . Example ( 1 9 ) : I f T' i s d e f i n e d as i n example (17) then i t Page 54 should be c l e a r from the above example t h a t an o r d e r $' f o r T / can-be c o n s t r u c t e d so t h a t T 7 has the i n t e r v a l topology. F o r ' « / ( T / ) i s d e f i n e d as f o l l o w s : $ ' ( f . ( T ) ) = $ ( f i ( T ) ) f o r 1 = 0 , 1 , 2 , . . . and f- L(T) §' f & ( T ) $' f ^ ( T ) ... . . In f a c t , such an o r d e r , $' i s c o n s t r u c t e d i n the p r o o f o f lemma (8) and i s denoted by . In the n o t a t i o n of lemma (8) we have: M = { f . [ ~ ! , 0 ] I i = 0 , 1 , 2 . . . } , N0= OR*} , • N.= {R - , - i . ) I i = 1,2,...} . A l s o , A Q i s the induced order from IR and A^(T') i s d e f i n e d by ' .' . A. ( f J . ( T ) ) = $ ( f j ( T ) ) i f J < i , and A . ( f . ( T ) ) = A n ( f . (T)) otherwise , and (t±(T)) A. ( f g ( T ) ) A ^ f ^ T ) ) . The f o l l o w i n g r e s u l t was shown by I . L. Lynn. L e t T be a subspace of IR and T)(T) denote the end p o i n t s of •the open ends -of components of fR - T which are h a l f open i n t e r v a l s . I f no open subset o f T i s compact and C l ^ r ^ T ) i s c o untable then T i s l i n e a r l y o r d e r a b l e . Lynn a l s o c o n j e c t u r e d the f o l l o w i n g r e s u l t which i s e a s i l y proven u s i n g the main theorem. Page 5 5 C o r o l l a r y : I f T c o n t a i n s no open compact s e t s , then T i s l i n e a r l y o r d e r a b l e Proof: Assume T c o n t a i n s no open compact s e t s and we w i l l show that c o n d i t i o n (1) and c o n d i t i o n (2) h o l d . (1) F i r s t assume t h a t Q i s d e f i n e d as i n the main theorem and t h a t T - Q i s compact and (T-Q)fl C1{R(Q) = 0 . Then T - Q = T f l ( lR - C1£R(Q).) and so T - Q i s open i n T s i n c e T c o n t a i n s no open compact s e t s , -T-- Q = 0 so c o n d i t i o n (1) h o l d s . (2) Assume I i s an open i n t e r v a l o f IR and p i s an end p o i n t of I and. [p] U(I fl(T-Q)) i s compact and { P } = c i ^ i n(T-Q)) n c i j R ( i n Q) . . Since p £ C 1 R ( I fl (T-Q)) , I fl(T-Q) *=' 0 and choose x £ I n(T L e t A = {t £ I n(T-Q) | t <_ x and (x , t ) n Q = 0} , B = {t € I fl (T-Q) \ t >_ x and (t , x) n Q = 0} . A i s hounded below by p and A f 0 so l e t t ^ = i n f (A) . B i s bounded above s i n c e {p} U(I fl(T-Q)) i s compact and B # 0 hence l e t t g = sup (B) . C l e a r l y t ^ £ p and tl> l2 € 1 n ( T _ Q ) s i n c e (P) U(I fl (T-Q)) i s compact. Hence t - ^ t 2 f C 1 ( R ( I N ^ ) ) A N D S O [t±> t 2 ] fl T i s open and i s compact. T h i s c o n t r a d i c t i o n e s t a b l i s h e s c o n d i t i o n (2) v a c u o u s l y . Page 56 D e f i n i t i o n s : L e t §(R) be an ordered t o p o l o g i c a l space and l e t T be a subspace. (1) I f p g T , l e t 'A(p) denote the s e t of a l l ' sequences o f p o i n t s of T , of type a f o r a an o r d i n a l , approaching p which are raonotonic i n § and which have no subsequence of s m a l l e r c a r d i n a l i t y approaching p . (2) L e t Q, denote the s e t of a l l p o i n t s q g T such t h a t : (a) I f p i s the f i r s t or l a s t p o i n t of R or any p o i n t of R not In the component of T c o n t a i n -i n g q , then the i n t e r v a l , [p , q'j of §(R) does not i n t e r s e c t T i n a compact s e t . and (b) e i t h e r (1) the component of T c o n t a i n i n g q i s t r i v i a l or (2) there are terms of A(q) of d i f f e r e n t c a r d i n a l i t y or « (3) there are terms, A' , A g A(q) such that e v e r y subsequence o f A' approaching q has a l i m i t p o i n t not i n C 1 R ( T ) . (3) Define A(S) , to be the Dedikind' c o m p a c t i f i c a t i o n of §(R); t h a t i s , , l e t p be i n S i f (a) p £ R or (b) p Is a n . i n i t i a l i n t e r v a l of f(R) having no l a s t p o i n t and.such that no f i r s t p o i n t of R f o l l o w s t h i s i n t e r v a l i n R . L e t the t o p o l o g y of S be the i n t e r v a l t opology induced by A d e f i n e d as f o l l o w s : Page 57 L e t x , y £ S . Then x A y i f ( a) x , y £ R and x § y or (b) x £ R and y £ S - R and an element o f y f o l l o w s x i n $ or (c) x , y £ S - R and some term o f y f o l l o w s every term o f x i n § or " (d) y d. R and x g S - R and some term of x f o l l o w s y i n § . (4) The s e t s H , H' and G can he d e f i n e d as b e f o r e , p r e -c e d i n g Lemma (0) and Lemma (3) . L e t F denote, the s e t of a l l terms [p , q] of G where p A q and there i s no t o t a l o r d e r i n g T of ([p , q] 0. T) such t h a t (a) T ([p , q], fl T) has the i n t e r v a l t o p o l o g y a t each p o i n t , and (b) i f p £ T , p i s the f i r s t p o i n t o f T ( [ p - , q] fl T) and (c) i f q £ T , q i s the l a s t p o i n t o f T ( [ p , q] fl T) and (d) i f p i T , there i s no f i r s t p o i n t of F ( [ p , q] n, T) and i f q $ T there i s no l a s t p o i n t o f P ( [ p , q] fi Theorem (2) L e t $(R) be an ordered t o p o l o g i c a l space and T a subspace. The f o l l o w i n g c o n d i t i o n s are n e c e s s a r y and s u f -f i c i e n t to i n s u r e t h a t T i s an o r d e r a b l e t o p o l o g i c a l space. Page 58 (1) I f T - Q i s compact and (T-Q) fl C1 T ( Q ) . = 0 then e i t h e r i Q = 0 or T - Q = 0 . .^ " ( 2 ) I f I i s an open i n t e r v a l o f §(R) and p g(T-I) and • I n(T - Q).U CP3 i s compact and C 1 T ( I fl (T-Q)) fl C l T ( I f l Q ) = [p] then (a) the component of T c o n t a i n i n g p i s t r i v i a l , and (b) no term of A(p) i s uncountable. (3) I f S $ G , then, f o r each term I of F , there i s a p o i n t f ( l ) 6 T - I such that i f X c F and X* i s the i n t e r s e c t i o n of T and the uni o n of terms of X and p e(T-(X*tT f ( X ) ) ) , then p e'ci T(X*) i f and o n l y i f p r- C l T ( f ( X ) ) . D i s c u s s i o n : Because of the way F i s d e f i n e d , the theorem cannot e a s i l y be used to prove t h a t a subspace, T , i s an o r d e r a b l e t o p o l o g i c a l space. E s s e n t i a l l y , i f [p , q] i s a term of G , i t i s as hard to determine whether [p , q] fl T i s o r d e r a b l e as i t ' i s to determine whether T i s o r d e r a b l e . Le t fi be the f i r s t uncountable o r d i n a l . , We con-s i d e r the s e t , 2 ^ = R ordered l e x i c o g r a p h i c a l l y . We f i r s t show that R i s o r d e r complete. Let A be a non-void subset o f R -and A c l e a r l y has an upper bound s i n c e R . has a l a s t element. I f a i s an o r d i n a l and a < fi de f i n e ' A and y as f o l l o w s : Page 59 . ( 1 ) A Q = .{a f A | a Q = 1 ] and y Q = 1 i f {a c A | a Q 1} £ 0 and A Q = A and y Q = 0 otherwise, ( 2 ) i f a = p -I- 1 f o r f3 an o r d i n a l •A = (a 6 A D l a = 1 } and v = 1 i f [a c Afl I a 1 } ^ 0 and A _ = A „ and y = 0 otherwise, (3) ' i f a i s a l i m i t o r d i n a l , A = (fl A ) n [a <E A |. a = 1 } and y = 1 i f . a B<a p • . . « a A • * 0 but y = 0 i f A = 0 . • a. a a Now c l e a r l y , e i t h e r there e x i s t s , an o r d i n a l , such t h a t A = fa} f o r some a g A and f o r a l l a > a„ or a . 0 e l s e w h e r e e x i s t s a l i m i t o r d i n a l a- such t h a t A = 0 . . - 0 a Q In t h i s f i r s t case, a i s the l a r g e s t element of A and a = sup(A) . I f A a = 0 f o r some l i m i t o r d i n a l a Q then y = {y a} a < n i s the supremum o f A . Hence R i s o r d e r -complete and so any c l o s e d bounded subset o f R i s compact. Then by d e f i n i t i o n o f Q , i f a , b g R and [a,b] i s a com-ponent .of T then [a , b ] c T - Q . I f a , b £ R and (a, i s a component of T then (a , b) c Q . A l s o , i f p 6 R then A(p) c o n s i s t s of those se-quences c o n v e r g i n g to p of type a where a i s an o r d i n a l and Q = a . Hence c o n d i t i o n ( 2 ) i s f a l s e u n l e s s i t holds v a c u o u s l y . F u r t h e r , A(S) p h i c a l o r d e r i n g on R . Page 60 = §(R) where $ i s the l e x i c o g r a -Example (20): L e t T = [a , b] U (c , d) where a,b,c,d £ R and a < b < c < d . Then [a , b] c T - Q and (c , d) c Q and c o n d i t i o n (1) does not h o l d so t h a t T i s n o t o r d e r a b l e . Example (21): Choose p 6 R and l e t [a } be a descend-i n g sequence o f type Q c o n t a i n e d i n R c o n v e r g i n g ' t o p , For each o r d i n a l 'a < Q , d e f i n e the s e t T c R as f o l l o w s , a F i r s t w r i t e a = X + n where X i s a l i m i t o r d i n a l and -n i s a f i n i t e o r d i n a l . D e f i n e T a = ( a a + 1 , a^) i f n € { 2 , 5 , 8 , . . . } and T q = {a a} i f n £ { 1 , 4 , 7 , . . . } and' T f l = 0 otherwise. Define T = g < Q T a . Then T i s not o r d e r a b l e f o r c o n d i t i o n (2) does not h o l d s i n c e every term o f A(p) i s uncountable. We now.examine an ordered t o p o l o g i c a l space, R i n which e v e r y subspace, T s a t i s f i e s c o n d i t i o n (3) . Let y be an i n f i n i t e o r d i n a l . L e t p be the f i r s t o r d i n a l num-ber such t h a t p = 2 Y and c o n s i d e r the s e t R = 2^ ordered l e x i c o g r a p h i c a l l y . L e t § be the l e x i c o g r a p h i c a l order f o r R . Then d e f i n e , f o r each a < (3 , A^ = {x £ R { x^= 0 f o r a l l Y > a}. Define A = U 0 A . Then c l e a r l y A i s dense i n R . A l s o , assuming the G e n e r a l i z e d continuum H y p o t h e s i s , A = U ~ \ < ^ 2 5 < p . p = p and so c l e a r l y A = p . Page 61 Since R has a dense subset of c a r d i n a l i t y p , A(p) c o n s i s t s p r e c i s e l y of those sequences of c a r d i n a l i t y p c o n v e r g i n g to p and hence A(p) i s uncountable i f and o n l y i f p > w . Hence i f p > w con-d i t i o n (2) holds i f and o n l y i f i t holds v a c u o u s l y . We have, of c o u r s e , a l r e a d y examined R = IR*= 2^ i n d e t a i l . A l s o , we can show that $(R) i s order-complete In the same way t h a t we have shown that 2^ i s ordered l e x i o c -g r a p h i c a l l y i s order complete. Hence A(S)'= §(R) and i f [a , b] i s a component of T , then [a , b] c T - Q and i f (a , b) i s a component o f T , then (a , b.) c Q . Hence c o n d i t i o n (3) holds f o r a l l subspaces T of R . For assume, a t f i r s t , t h a t F =. p .' Then F = [I | a < p } and assume I = [a , b 1 where a , b c R . For I £ F ' d e f i n e f ( I ) = x where x = b f o r a l l v i a ' and v a Y Y x / = 1 where a' i s the f i r s t o r d i n a l g r e a t e r than a such t h a t b / = 0 . C l e a r l y c o n d i t i o n (3) h o l d s . I f F < p then c o n d i t i o n (3) holds v a c u o u s l y . Page 62 BIBLIOGRAPHY Mary E l l e n Rudin, " I n t e r v a l t o p o l o g y i n subsets o f t o t a l l y o r d e r a b l e spaces", Amer. Math. Soc. Trans-a c t i o n s 118 (1965), 376. I. L. Lynn, " L i n e a r l y o r d e r a b l e spaces", Amer. Math. Soc. T r a n s a c t i o n s 113 (1964), 189. 

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