UBC Theses and Dissertations

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UBC Theses and Dissertations

The structure of βN Rambally, Rodney Seunarine 1970

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THE STRUCTURE OF BN by RODNEY SEUNARINE RAMBALLY B.A. (Hons.), University of Saskatchewan, 1969 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER- OF ARTS i n the Department of MATHEMATICS We accept this thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA August, 1970 I n 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 t h e r e q u i r e m e n t s f o r a n a d v a n c e d d e g r e e a t t h e 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 m a k e 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 a n d s t u d y . I f u r t h e r a g r e e t h a 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 t h e H e a d o f my D e p a r t m e n t 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 n o t 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 . D e p a r t m e n t T h e 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 V a n c o u v e r 8, C a n a d a Supervisor: Professor T. Cramer i i ABSTRACT Our subject matter consists of a survey of the major results concerning the topological space gN-N where N represents the space of natural numbers with the discrete topology, and gN the Stone- • Cech compactification of N . We are mainly concerned with the results which were derived during the l a s t ten years. When there i s no advantage i n r e s t r i c t i n g our work to the space N we work with an arbitrary discrete space X and f i n a l l y formulate our results i n terms of gN-N . In some cases, pre-1960 results concerning . gN-N are obtained as special cases of the results we derive using an arbitrary discrete space X . The material presented i s divided into four chapters. In Chapter I, we discuss certain subsets of gN-N which can be C*-embedded i n other subsets of gN-N . This study leads to the conclusion that no proper dense subset of gN-N can be C*-embedded. In the second chapter we devise a general method of associating certain classes of points of gN-N with certain subalgebras of C(N) .. The P-points of gN-N form one of these classes. The answer to R. S. Pierce's question, "Does there exist a point of gN-N which l i e s simul-taneously i n the closures of three pairwise d i s j o i n t open sets" i s discussed i n Chapter I I I . F i n a l l y i n Chapter IV we present two proofs of the non-homogeneity of gN-N , without the use of the Continuum Hypothesis. I l l ACKNOWLEDGEMENTS I wish to express my appreciation and gratitude to my advisor, Dr. Timothy Cramer, for suggesting the topic and .offering invaluable assistance i n the preparation of this thesis. I also wish to thank Dr. J. V. Whittaker for reading the thesis and Mrs. M. Wang for s k i l l f u l l y typing i t . To National Research Council of Canada go my thanks for the i r f i n a n c i a l assistance. i v TABLE OF CONTENTS Page INTRODUCTION 1 CHAPTER I. ON THE EMBEDDING OF CERTAIN SETS IN BN . . . . . . 4 CHAPTER II. AN APPLICATION OF B-SUBALGEBRAS OF C(N) TO BN-N . 19 CHAPTER III. c-POINTS OF gN-N 32 CHAPTER IV. TYPES OF ULTRAFILTERS AND THE NON-HOMOGENEITY OF BN-N 41 BIBLIOGRAPHY 60 1 INTRODUCTION Our purpose i s to investigate the internal structure of gN-N . We do this with the help of the major results of the l a s t ten years concerning this topological space. Most of the pre-1960 results i n connection with gN can be found i n the Gillman and Jerison text, {4], and, when needed, we simply quote these results without proofs. Some of them w i l l be eas i l y obtained from certain theorems which we discuss here. In some cases we do not r e s t r i c t our considerations to the space N , simply because such a r e s t r i c t i o n does not decrease our work appreciably. In such cases we discuss an arbitrary discrete space and f i n a l l y apply our results to the space N . Much of the work i n the f r i s t three chapters depends heavily on the continuum'hypothesis. Thus i t i s expected that one of the major techniques used i n these chapters i s the induction process, as i s indeed the case. The terminology and notations we use are those of the Gillman and Jerison text, [4]. However we would l i k e to make special mention of the following notations: N* = gN-N (or, more generally, D* = gD-D where D i s a discrete space); c w i l l denote the power of the continuum; , for f a function on N , w i l l denote the continuous extension of f to gN . We sometimes denote this continuous extension by f* . Unless otherwise stated we denote "closure" by "c£" and " i n t e r i o r " by " i n t " . The set of a l l continuous functions from the topological space X into the topological space R (space of r e a l numbers) i s denoted by C(X) and the set of a l l bounded functions i n 2 C ( X ) i s d e n o t e d b y C * ( X ) . A s u b s p a c e S o f X i s s a i d t o b e C -e m b e d d e d i n X i f e v e r y f u n c t i o n i n C ( S ) c a n b e e x t e n d e d t o a f u n c t i o n i n C ( X ) . S i m i l a r l y S i s C * - e m b e d d e d i n X i f e v e r y f u n c t i o n i n C * ( S ) c a n b e e x t e n d e d t o a f u n c t i o n i n C * ( X ) . F o r f e C ( X ) , we l e t Z ( f ) = {x e X : f ( x ) = 0} , a n d we c a l l Z ( f ) a z e r o - s e t o f X . T h e s e t o f z e r o - s e t s i n X i s d e n o t e d b y Z ( X ) • T h e c o m p l e m e n t o f a z e r o - s e t i s c a l l e d a c o z e r o - s e t . C h a p t e r I . We i n v e s t i g a t e t h e n e c e s s a r y a n d s u f f i c i e n t conditions for subsets of 8N to bs C*-embadded. The results of t h e c h a p t e r a r e f o r m u l a t e d : i n t e r m s o f a c e r t a i n c o n d i t i o n d e f i n e d i n t h e c h a p t e r i t s e l f . I t i s f i n a l l y p r o v e d h e r e t h a t d e n s e s u b s e t s o f N * a r e n o t C * - e m b e d d e d . C h a p t e r I I . - H e r e we i n t r o d u c e a m e t h o d f o r a s s o c i a t i n g a s e t o f p o i n t s i n g X - X w i t h a t y p e o f s u b a l g e b r a A o f t h e a l g e b r a C ( X ) w h e r e X i s a c o m p l e t e l y r e g u l a r s p a c e . We c a l l t h e s e p o i n t s t h e A - p o i n t s o f g X - X . We t h e n p r o v e t h a t t h e s e t o f A - p o i n t s i s d e n s e i n g X - X a n d i s o f c a r d i n a l i t y 2° . We s h o w t h a t C * ( N ) i s o n e o f t h e s e t y p e s o f s u b a l g e b r a s o f C ( N ) a n d t h a t t h e P - p o i n t s o f g N - N a r e t h e C * ( N ) - p o i n t s , f r o m w h i c h we c o n c l u d e t h e w e l l k n o w n r e s u l t t h a t g N - N h a s a d e n s e s u b s e t o f 2 P - p o i n t s . C h a p t e r I I I . T h i s i s a s h o r t c h a p t e r i n w h i c h we p r o v e t h a t e v e r y p o i n t i n g N - N i s c o n t a i n e d s i m u l t a n e o u s l y i n t h e c l o s u r e s 3 o f c d i s j o i n t o p e n s e t s o f BN-N . C h a p t e r I V . We d i s c u s s t y p e s o f u l t r a f i l t e r s a n d i n t r o d u c e a s e t T , t h e e l e m e n t s o f w h i c h we c a l l t h e t y p e s o f t h e f r e e u l t r a f i l t e r s o n N . We t h e n d e f i n e a c e r t a i n r e l a t i o n (j) o n T c a l l e d t h e p r o d u c i n g r e l a t i o n . U s i n g t h i s c o n c e p t a n d w i t h o u t a s s u m i n g t h e c o n t i n u u m h y p o t h e s i s we p r o v e t h e n o n - h o m o g e n e i t y o f BN-N . \ / 4 C H A P T E R I ON T H E E M B E D D I N G O F C E R T A I N S E T S I N gN 1 . 1 . T h e p u r p o s e o f t h i s c h a p t e r i s t o c o n s t r u c t , c e r t a i n s e t s i n gN t h a t c a n b e C * - e m b e d d e d i n o t h e r s u b s e t s o f gN . We s h a l l a l s o • p r o v e t h a t i f p e 3N-N t h e n g N - N - { P } i s n o t C * - e m b e d d e d i n 3N-N f r o m w h i c h we s h a l l b e a b l e t o c o n c l u d e t h a t p r o p e r d e n s e s u b s e t s o f g N - N a r e n o t C * - e m b e d d e d . We u s e t h e c o n c e p t s o f F - s p a c e s a n d z e r o -d i m e n s i o n a l s p a c e s i n m o s t o f o u r p r o o f s . A l l s p a c e s c o n s i d e r e d a r e a s s u m e d t o b e c o m p l e t e l y r e g u l a r . T h e r e s u l t s h e r e a r e d u e m a i n l y t o F i n e a n d G i l l m a n . 1 . 2 . D e f i n i t i o n . X i s c a l l e d a n F - s p a c e i f e v e r y c o z e r o - s e t i n X i s C * - e m b e d d e d . I t i s p r o v e d i n [ 4 ] , 1 4 . 2 7 , t h a t i f . a s p a c e X i s -l o c a l l y c o m p a c t a n d a - c o m p a c t b u t n o t c o m p a c t t h e n g X - X i s a c o m p a c t F - s p a c e . N i s a n e x a m p l e o f s u c h a - ' s p a c e . We n o w p r o c e e d t o d e f i n e z e r o - d i m e n s i o n a l . L e t a n d S h e f a m i l i e s o f s e t s . We s a y t h a t ~V r e f i n e s H i f f o r e v e r y V e t h e r e e x i s t s U e "It s u c h t h a t V ^ U . L e t b e a c o v e r o f X . We d e f i n e t h e o r d e r o f *LL- t o b e t h e l a r g e s t i n t e g e r n f o r w h i c h t h e r e a r e n + 1 m e m b e r s o f "IL w i t h n o n - e m p t y i n t e r s e c t i o n . B y a n o p e n - c l o s e d p a r t i t i o n o f S C X w e m e a n a f i n i t e c o l l e c t i o n o f d i s j o i n t o p e n a n d 5 c l o s e d s u b s e t s o f S w h o s e u n i o n i s S . T h u s a p a r t i t i o n o f S i s a c o v e r o f S o f o r d e r z e r o . 1 . 3 . D e f i n i t i o n . A s p a c e X i s s a i d t o b e z e r o - d i m e n s i o n a l i f e v e r y b a s i c c o v e r o f X h a s a r e f i n e m e n t b y a n o p e n - c l o s e d p a r t i t i o n o f X . R e c a l l t h a t t w o s u b s e t s A a n d B o f a s p a c e X a r e c o m p l e t e l y s e p a r a t e d i n X i f t h e r e e x i s t s a f u n c t i o n f i n C 5 ' C ( X ) s u c h t h a t 0 _< f _< 1 , f ( x ) = 0 f o r a l l x e A , a n d f ( x ) = 1 f o r a l l x e B . I t i s p r o v e d i n [ 4 ] , 1 6 . 1 7 , t h a t t h e f o l l o w i n g a r e e q u i v a l e n t : a n y t w o c o m p l e t e l y s e p a r a t e d s e t s a r e c o n t a i n e d i n c o m p l e m e n t a r y o p e n a n d c l o s e d s e t s f o r a s p a c e X ; i s t o t a l l y d i s c o n n e c t e d ; - X i s z e r o - d i m e n s i o n a l . I t i s a l s o p r o v e d i n [ 4 ] , 6 . 1 0 , t h a t gN i s t o t a l l y d i s c o n n e c t e d . F o r a n y s p a c e Y , l e t r Y d e n o t e t h e s e t o f a l l p o i n t s p i n gY s u c h t h a t e v e r y z e r o - s e t i n gY t h a t c o n t a i n s p a l s o i n t e r s e c t s Y . 1 . 4 . D e f i n i t i o n . I f r Y = Y , t h e n Y i s s a i d t o b e r e a l c o m p a c t . I n C h a p t e r I I , w e s h a l l f i n d i t m o r e c o n v e n i e n t t o d e f i n e r e a l c o m p a c t n e s s a s f o l l o w s : Y i s r e a l c o m p a c t i f f o r e v e r y p e Y * t h e r e e x i s t s Z e Z ( g Y ) s u c h t h a t p e Z <Z Y* . We s h a l l n o w s h o w t h a t t h e s e t w o d e f i n i t i o n s a r e e q u i v a l e n t . L e t t h e f i r s t d e f i n i t i o n b e I a n d t h e 6 / second be I I . I implies II : I f p e Y* , then p £ Y . Thus p £ rY . so for some zero-set Z containing p , Z 0 Y = <j> . Hence Z c Y* and so p e Z c Y* . II implies I : Let y e rY . Then y £ BY - Y for i f i t were then there would e x i s t a zero-set Z such that y e Z c Y* . Hence Z n Y = <j> , a cont r a d i c t i o n . Now i f y e Y then obviously every zero-set containing y i n t e r s e c t s Y . Hence Y C rY . Thus Y = rY . Hence the two d e f i n i t i o n s are equivalent. The following lemma i s needed both i n t h i s chapter and i n the next. 1.5. Lemma. I f Y i s l o c a l l y compact and realcompact, then, each zero-set i n BY-Y i s the closure of i t s i n t e r i o r . Remark. I t s u f f i c e s to prove that a non-empty zero-set Z i n BY-Y has non-empty i n t e r i o r ; f o r then i f there exists p e Z - cl i n t Z, then from l o c a l compactness there e x i s t s a zero-set Z^ , d i s j o i n t from i n t Z , such that p e Z^ and i n t ( Z ft Z^) = <j> . Since zero sets are closed under countable i n t e r s e c t i o n we then have a non-empty zero-set, Z^ f) Z , with empty i n t e r i o r , contradicting the f a c t that a non-empty zero set has non-empty i n t e r i o r . Proof: Since Y i s l o c a l l y compact then BY-Y i s compact (this follows from the'fact that Y i s open i n BY i f and only i f Y i s l o c a l l y compact. Thus Y i s open i n BY thereby making BY-Y a closed subset of a compact space, hence compact). Thus BY-Y i s C*-embedded i n 7 3Y by [4], 6.9(b). Thus for Z an arbitrary zero-set i n 3Y-Y , Z = Z(f) - Y for some f e C * ( 3 Y ) . Let p e Z . By realcompactness p £ rY . Thus there exists a function g e C*(3Y) such that g(p) = 0 but g(y) ^  0 for any y e Y . Let h = |f| + |g| . Then h(p) = 0 so p e Z(h) . Also Z(h) C Z . Thus p e Z(h)C Z . We s h a l l show that this non-empty set Z(h) contains a non-empty open set. Let be a sequence of d i s t i n c t points i n Y on which h converges to zero. By l o c a l compactness we can choose d i s j o i n t compact neighborhoods of y such that |h(y) - h(y ) l < — for y e V . Let u e C * ( 3 Y ) n n ' n n b e s u c h t h a t u(y^) = 1 for every n u(y) = 0 , y e Y - V n (such a function u exists because for every n there exists a function f^ which i s equal to 1 at y^ and equal to 0 everywhere on Y-V . Let u = \ f ). Suppose q £ 3Y-Y i s such that u(q) 4 0 . n , This, means that every neighborhood of q meets i n f i n i t e l y many of the compact sets . Hence h(q) = 0 ; one may j u s t i f y this as follows: since every neighborhood of q meets i n f i n i t e l y many sets , then we can find a net {S } such that each S i s i n V and {S } converges n n n n ° to q . Since h i s continuous we have lim h(S ) = h(q) . Now n-*° |h(S ) - h(y )| < — . So lim h(S ) = 0 . Thus h(q) = 0 . Hence for ' n w n 1 n n n N any point q e 3 Y-Y such that u(q) $ 0 , we have h(q) = 0 . Thus (BY-Y) - <£ Z(h) ,. Also ( 3 Y-Y) - Z(u) i s a non-empty open subset 8 of BY-Y . Since Z(h)<r Z we conclude that Z has non-empty i n t e r i o r . 1.6. Remark. Local compactness i s a c r i t i c a l property of the space Y i n the above lemma. We can prove this by i l l u s t r a t i n g a non-empty zero-set i n 3 Q - Q with empty i n t e r i o r . Define a real-valued function h on R as follows: h(x) = , X _< TT , X _> TT . Let h have an extension g to BQ • Then (1) Z(g)/J Q = (j> (2) Z(g) =f ij) (there exists a sequence { a n} converging to IT such that a e Q . Then Cl„Au }^Z(g)) . n 3Q n We would l i k e to show that Int ~Z(g) = <j> . Assume x e Int ~Z(g) . B Q - Q 3 Q - Q Then there exists some basic open set (3Q - cl Z) i n BQ where Z e Z (Q) such that x e (3Q - c i l Z) - Q C Z(g) . (See remark at end for the existence of BQ - ci Z) . Since (3Q - cSL Z) i s open and since Q i s dense i n BQ then (BQ - cl Z) /? Q 4 <j> . Hence Q - Z f cj> so there exists an element y such that y \ Z but y £ Q . Thus we can find some s > 0 such that y e (y-e, y+e) and (y-e,y+e) f) Z = (j> Note that this implies that cAQ.(y-e,y+e) P c% Z = cfi pQ From above g(z) = 0 for every z E BQ - cA Z - Q . Hence g{ [cH (y-e,y+e) ] - Q} = 0 since we'have shown that 9 c£ _(y-e,y+£) f) c£ Z = <j> . But by d e f i n i t i o n of h this i s a contradiction. (For y ^  ir since y e Q . Hence we can fi n d a 6 > 0 such that ir sj: (y-S,y+6) and (y-5,y+S) C (y-£,y+e) . Now min {h(z) : z £ (y-<5,y+S)} > 0 by d e f i n i t i o n of h so 0 \ g{ [c£gQ(y-<5,y+S) ] - Q} . Hence our contradiction since we said that g{[c£3Q(y-£,y+£)] - Q} = 0). Remark. We would l i k e to show that a basic open set i n gQ i s the complement of a basic closed set. From [4], 6.E(1), we know that basic closed sets i n gQ are of the form c^gg z where Z e Z(Q) . Now l e t O be an open set and p £ O . Let Q1 denote the comple-ment of O . Then OX i s closed i n gQ and p £ Q' SO there exists, a closed set F such that F = c&gQ z f° r some Z e Z(Q) and p £ F _ t and O' c F (this i s by d e f i n i t i o n of basic closed sets). Thus O D / and so we have, since p ^  F , p £ gQ - c^^^ Z c 0 . Hence the complements of basic closed sets are basic open sets. 1.7. Remark. I f Y i s l o c a l l y compact and realcompact but not compact, then gY-Y i s not b a s i c a l l y disconnected, i . e . not every cozero-set has open closure. For i f every cozero set S has open closure then the complement of S , Z say, has open closure. Thus i n t Z i s closed and so Z i s open since by Lemma 1.5 Z = c£ i n t Z = i n t Z . Since every zero-set i s open, then gY-Y i s a P-space by [4], 4J. We already know that gN-N'' i s not b a s i c a l l y disconnected (see [4], 6W) and gN-N i i s not a P-space. 10 Given a space X , l e t S C X and p e X - S . Our other results' w i l l be formulated i n terms of the following condition which we denote by (p, S) . (p, S) : There exist a neighborhood V of p and a cozero-set HC S such that (S D V) - H has empty i n t e r i o r . Remark. This condition holds i f p ^ c& S , and i f S i s a cozero-set, then the condition holds for a l l p \ S . 1.8. Lemma. Let F be a compact set i n X such that (p, X - F) holds for a l l p e F . Then i n t Z C F c Z for some zero-set Z . Proof: We s h a l l show that there exist a f i n i t e open cover {V.,, V } of F and zero-sets Zn , Z„, .... Z containing F i n 1 2 ' n such that i n t Zfc - F C X - Vfc for k = 1, ..., n . Since (p, X - F) holds then for every p e F there exists a neighborhood of p , V p , and a cozero-set H P C x - F such that i n t [ (X - F) n V P - H P] = <}> . Consider a l l such sets V p for p e F . Without loss of generality we can l e t {V P : p e F} be an open cover of F so there exists a f i n i t e P l P n subcover, V , V . We s h a l l now c a l l this f i n i t e subcover P k V^, V n . S i m i l a r l y l e t H = , k = 1, n . Let the complement of H^ be , k = 1, .. ., n . Then Z^ .-5 F since H k C X - F . Also i n t [ ( X - F) 0 Vfc O Zfc] = <J> . Hence no element of V, i s an element of i n t Z, - F , i . e . X - (int Z, - F) 3 V, and so k k k k i n t Z. - F C (X - V, ) . Let Z 0 = D Z, . Since Z, 3 F for a l l k k v k , k k k then Z 0 i s a zero-set containing F and i n t Z Q - F c X - (j V . k k 11 But X - i / V . c X - F since F C U v . Thus k k i n t Z 0 - F c r X - l ' V X - F . Hence c£(int Z Q - F) >') F = $ . Since k k F i s compact, then from [4], 3.11, we conclude that F and c&(int Z 0 - F) are completely separated. This 'means that F and i n t Z 0 - F are contained i n d i s j o i n t zero-sets, i . e . there exists a zero-set •Z^ containing F and d i s j o i n t from i n t Z Q - F . Hence i n t ( Z 0 fi Z f) C F C (Z 0 P) Z f) . So the required zero-set i s Z 0H Z . • Theorem. Let X be an F-space, S c X and p e cA S - S . I f S fi V i s open for some neighborhood V of p , and i f (p, S) holds then S i s C*-embedded i n S O {p} . Proof: We s h a l l show that there exists a neighborhood of p and a cozero-set H C S such that S fl i s open and d i s j o i n t from int(X - H) . We know that there exists a neighborhood V of p such that S fi V i s open. Also there exists a neighborhood V 0 of p and a cozero-set H c S such that int[S O V 0 - H] = <j since (p, S) holds. V 0 contains an open set U which contains p and O /Q S ^ <j> since p e cJl S . Clearly Q /T V ^ <j> and O n V i s a neighborhood of p . Also (S Pi V) Pi O i s open, being the intersection of two open sets, and non-empty since If) O Is a neighborhood of p . The required neighborhood i s V / ? Q Now l e t f e C*(S) and g e C*(H U {p}) be an extension 1 2 of f|H where f|H i s the r e s t r i c t i o n of f to H (note that X i s an F-space so every cozero-set i s C*-embedded). Define a function h on S 0 {p} such that • r f(x) , x e S h(x) = \ ^ g(p) , x = p . Since H /? V i s dense i n (S U {p})/0 V± = (S /"> £/ {p} (follows from the f i r s t part of the proof), then h i s a continuous extension of f . Thus S i s C*-embedded i n S O' {p} . 1.10. Theorem. Let X be an F-space and l e t S C X be a union of cozero-sets S i n X . Then a (a) i f G C S and i f G fl S^ i s a cozero-set i n S for a l l a , then G i s C*-embedded i n S . (b) S i s an F-space. Proof: We s h a l l use a well-known technique of induction i n this proof. We can assume that S = S and that S 0 C Sn c . . . • a 1 Consider any S . Since every cozero-set i n S i s C*-embedded then a J a S i s an F-space for a l l a . Let g e C*(G) . Let g = g| (G /') S ) . For given a < assume that g^ has been extended to a function t e C"(S^) for every £ < a , and that f 0 f^ <£_ ... (note that we can make such an extension because (G H s ) i s a cozero-set i n S and so (G H S ) i s a cozero-set i n S because there exists some function h which takes S-(G f~i S ) to zero and (S - G) C" s - (G /) S ) . Thus ^ S ^ 13 h takes (S _ - G) to zero also. Nov? h (S^ - G> makes (G O S _) a cozero-set i n . Now use the fact that i s an F-space to extend the bounded function g on the cozero-set G /] S to f on S Q> s s s> Also note that we are treating each function f_^  as a set of ordered p a i r s ) . The function U f [i e i s well defined and continuous on £ a £<a the cozero-set L; S_ U (G /) S ) . To show that i t i s continuous we £<a K consider any open set 0 i n the range of O f (J g . Then £<a s ( L-1 £ ;j g ) "*"(o>) = (V f iJ g "*"(£>) which i s an open set since the £ a £ a £<a £<a functions f and g are continuous. Also L/' S l) (G S ) i s a £ a £ a £<a cozero-set because i t i s the countable union of cozero-sets. In fact i t i s a cozero-set i n the F-space S ( r e c a l l that S 0 c Sn c ... ). a 1 Hence ( U f n g ) has a continuous extension to a function >- £ a £<a f e C"(S ) . Now U f i s a continuous extension of g to a l l of a a a S . Thus G i s C*-embedded i n S . \ (b). That S i s an F-space follows quite easily from (a) since i f G i s any cozero-set i n S , then G f) S i s a cozero-set a i n S for a l l a and so G i s C*-embedded which implies that S i s an F-space. We now apply this theorem to gN-N to conclude that a l l open subsets of".gN-N are F-spaces. The following i s a proof of this statement. We can show that gN-N has 2 zero-sets (see remark at the end) and so by the continuum hypothesis i t has cozero-sets. 14 Every closed set i n gN-N can be written as an intersection of a set of closures of zero-sets i n N . Thus every open set can be written as a union of cozero-sets. As already mentioned, gN-N i s an F-space. Hence our theorem applies with X = gN-N and S an open set i n gN-N . Remark. Let Z be a zero-set i n gN-N . Since gN-N i s C*-embedded i n gN there exists a bounded continuous function f* on gN such that f*(Z) = 0 and so Z i s a zero-set i n gN also. By [4], 6E(3), Z = f) ci Z± , where Z ± e Z(N) . Hence there are Ar zero-lew sets i n gN-N since there are jV^ sets Z^  i n Z(N) . 1.11. Theorem. Let X be an F-space having just 2 zero-sets. Let S be open and l e t p e c£ S - S , and suppose that (p, S) f a i l s . Then S i s not C*-embedded i n S U {p} . Proof: Let {S_}_ * be a family of cozero-sets i n X whose £ £<w1 union i s S (such a family exists because for every p e S there exists a function f such that f (p) = 1 and f (x) = 0 for a l l x e X - S. P P P Consider the set of Z(f ) ) . Let {V>.K be a base of zero-set-p 5 5<a) 1 neighborhoods of p . We proceed inductively as follows. For each a < a)^  , assume that cozero-sets A and B , contained i n S , have been defined for every £ < a . Since (p, S) f a i l s then int[S f) V - U. (A B r U S J ] + a .?«* 5 5 K 15 (note that S i s also a cozero-set i n S) . Thus we can choose d i s j o i n t . non-empty cozero-sets A and B contained i n S n V - c7 (A,. U (/ S J . Define A =.W A , B = l> B , a >- ? 5 5 a c G = A L> B . Hence, by construction, for each £ < a , G 1 S = (A L' B) /) S„ = <-• (A B ) /) which i s a cozero-set. So by I • K a ± K a a K Theorem 1.10, G i s C"-embedded i n S . Note that A and B are d i s j o i n t since the sets A and the sets B are. Hence A and B a a are complementary open sets i n G and by construction of the sets A A and B we conclude that both A and B meet every neighborhood of p. a Thus A and B are completely separated i n the subspace G but not i n G U {p} since p i s i n the closures of both A and B (they are completely separated i n G because A i s contained i n the zero-set B' and B i s contained i n the zero-set A' and A' /? B' = <j> i n the subspace G ; hence A and B are contained i n d i s j o i n t zero-sets). Thus from Urysohn's Extension Theorem, [4], 1.17, G i s not C*-embedded in G V {p} . Hence S i s not C5'c-embedded i n S 0 {p} ( r e c a l l that G i s C"-embedded i n S). / We now apply Theorem 1.9 and Theorem 1.11 to the space 3N-N. to conclude the following: 1.12. Let S C (BN-N be open and p e ciL S - S . Then S i s C"-embedded i s S V {p} i f and only i f (p, S) holds. 16 1.13. Theorem. Let K be a compact F-space such that K = 3Y-Y where . . . , o Y i s l o c a l l y compact and realcompact and such that K has just 2 zero-sets. (Such a space K i s 3N-N). Then (a) The following are equivalent for an open set, S : (i) (p, S) holds for a l l p e K - S . ( i i ) S i s a cozero-set i n K . ( i i i ) S i s C"-embedded i n K . (b) No proper dense subset i s C*-embedded. Proof: ( i ) implies ( i i ) : By Lemma 1.5 we see that for any zero-set Z i n K , ck i n t Z = Z . Since (p, S) holds for a l l p e K - S , then (p, K - (K - S)) holds for a l l p e K - S , i . e . (p, K - F) holds for a l l p z F where F = K - S (note that F i s compact). We now apply Lemma 1.8 to conclude that for some zero-set Z i n K , i n t Z c F C Z . Hence ci i n t Z c. cZ F c cJl Z and so Z C F c Z . Thus F i s a .zero-set, i . e . K - S i s a zero set, from which we conclude that S i s cozero-set i n K . That ( i i ) implies ( i i i ) follows from the d e f i n i t i o n of F-space. That ( i i i ) implies ( i ) follows from Theorem 1.11. (b) Since no point of K i s isol a t e d , then no point i s a zero-set; for i f a singleton were a zero-set then, as already proved, this 17 singleton would have non-empty i n t e r i o r contradicting the fact that K has no isolated points. Thus the complement of any point i s not a cozero-set which implies by part (a) that the complement of any point i s not C*-embedded. Now l e t D be a proper dense subset of K . We must show that D i s not C*-embedded. Assume that D i s C*-embedded. Since D i s a proper subset there exists an element p e K - D . We have proved above that K - {p} i s not C*-embedded so there exists f , which i s a bounded continuous function on (K - {p}),i.e. f e C*(K - {p}). such that f has no continuous extension to K . Consider f|D . This i s a continuous function on D so i t has a continuous extension f to K . K But since D i s dense then f |(K - {p}) = f . This i s a contradiction since we are now saying that f has an extension to a l l of K . Thus no proper dense subset i s C*-embedded. Applying this theorem to gN-N we conclude that for p e gN-N, gN-N-{p} i s not C*-embedded i n gN-N and that proper dense subsets of gN-N are not C*-embedded. Before ending this chapter we make the following remarks. 1.14. Remark. Let X be a compact F-space and l e t S C X be such that 2' 0 card(X - S) < 2 . Then S i s pseudocompact. 18 Proof: Assume S i s not pseudocompact. Then S admits an unbounded real-valued function and so N i s a closed subspace of S (see [4], 1.20). By [4], 14 N(5), every countable set i n an F-space i s C"-embedded so N i s C*-embedded i n X . Hence c£ N = gN . Now A'" 2 ° X - S D gN-N and since card(gN - N) = 2 we have a contradiction. Hence S i s pseudocompact. Again this remark can be applied to gN-N . 1.15. Each pseudocompact subset P of gN which contains N i s of c a r d i n a l i t y j> 2 ° . Proof: Let <j> be a mapping from N onto . W , where W i s the set of rationals i n £0, 1] . By the compactification theorem cj> can be extended continuously to a mapping (j>* of gN into [0, 1] . Since N C P then W C <}>*(?) • Since the continuous image of a pseudocompact space i s pseudocompact - see [4], 1G - then <J>*0?) i s pseudocompact. But <j>*(P) i s closed i n [0, 1] (since pseudocompact i s equivalent to countably compact i n a normal, T^  space. But since [0, 1] i s f i r s t countable then countably compact i s equivalent to compact and since [0, 1] i s Hausdorff, compact implies closed). Now since W c <}>-c(P) then c£ W C c£ <}>*(P) = <S>*(P) , i. e . 10, 1] c «|>*(P) . Hence [0, 1] = , and so card P > 2 0 19 CHAPTER I I AN APPLICATION OF g-SUBALGEBRAS OF C(N) TO gN-N 2.1. In [10], W. Rudin has shown, using the continuum hypothesis, that gN-N has a dense subset of 2 P-points. Our purpose now i s to prove a more general result than t h i s . For a completely regular space X we s h a l l define a class of subalgebras of C(X) , called g-subalgebras, and with each g-subalgebra A , we s h a l l associate a set of points i n gX-X cal l e d A-points. Then with some re s t r i c t i o n s on A and X we s h a l l show that gX-X has a dense subset of 2 A-points. Rudin's theorem will then be ea s i l y obtained by proving that C*(N) i s a g-subalgebra and that the P-points of gN-N are the C*(N)-points. I t i s not necessary to r e s t r i c t our work to the space N since this r e s t r i c t i o n does not simplify our work much. Hence we work with an arbitrary completely regular space X . Most of the results i n this chapter are due to D. Plank (see [8]). Note that under pointwise operations C(X) and C*(X) are algebras over TR (real numbers). We note also that i f f e C(X) and a l denotes the one-point compactification of IR , then by the compactification theorem there i s a unique continuous function f* : gX -y a l such .that f * | x = f . By a subalgebra of C(X) we s h a l l mean a subalgebra i n the usual sense which contains the constant functions. 20 We s h a l l denote by. f' and Til respectively the set of x\ xi prime ideals and maximal ideals i n A , where A i s a subalgebra of C(X) . We now define the hull-kernel topology on )° and M . In xi xi general the hull-kernel topology on a c o l l e c t i o n o7 of prime ideals i n a commutative ring A with i d e n t i t y i s defined by taking a = {P e J : P 3 f) ft- } to be the closure of the subset 61 of <J i 2.2. Proposition. The sets E.7(a) = {P ec7 : a e P} where a e A , are o / closed and constitute a base for the closed sets i n vJ (A i s any subalgebra of C(X)). Proof: ( i ) To show that the sets of the form E.(a) are C/ closed i t suffices to show that E^(a) C. E 7(a) . E^Ca) = {P z'3 : Flfl E ( a ) J . Let P e E. (a) . We must show that a £ P . Since PO/TE. (a) =/HP £ iJ : a £ P} and since a E f] {P zrj : a £ P} then a e P and so P £ E. (a) . Hence the sets E^(a) are closed. ( i i ) To show that these sets form a base for the closed sets consider P 0 £ J and S closed such that P Q £ S . Since S i s closed then S = ~S and so S =' {P e J : PO/0 S} ;. I f fl S = {a} , a £ A., then we are finished since there would then exist an element C which i s closed and of the form E (a) (take C = S) such 2 1 that SC C and P D fj: C (note that P 0 | C because {a} =/? S c|- p°) . If fi S i s not a singleton then consider one element a' £/?S such that a' £ P 0 . Note that there must exist one such element; otherwise P 0O .OS and so P 0 would be an element of S , a contradiction. Now consider the family of sets P i n ,// such that a' e P . C a l l this set D . Then S C D since for any Q £ S , Q D f] S and so a' e Q thereby giving Q e D . We have already shown that D i s closed. Clearly P 0 £ D because a' £ P 0 . The proof i s now complete. We s h a l l put the hull-kernel topology on , and we s h a l l introduce a family .ofy of prime ideals i n A , which w i l l reduce to '))'") when A = C(X) and A = C*(X) (A i s again a subalgebra of C(X)). F i r s t we state the following characterization (which was f i r s t introduced by Gelfand and Kolmogoroff, [5]; i t i s also discussed i n [4], 7.3 and 7 D(l)) of the maximal ideal i n C(X) associated with the point P e gX ( c a l l this maximal i d e a l ). Then =• {f e C(X) : ( f g ) * ( p ) = 0 for a l l g e C(X)} . Gillman and Jerison have also proved i n [4], 7.2, that M L = {f e C*(X) : (fg)*(p) = 0 for all g e C*(X)} and that the L i * mappings p and p M^. are one-to-one. I t i s proved i n [5] that the mappings p -> and p -> are homeomorphisms of gX onto the maximal-ideal spaces r and >7'1 ,. . We s h a l l prove l a t e r that the mapping p -> M^ 1 i s continuous for any subalgebra A <^  C(X) . The 22 above characterizations of MP and MP5,. suggest the following d e f i n i t i o n . 2.3. D e f i n i t i o n . For any subalgebra A of C(X) and for p e gX l e t MP = {f E A : (fg)*(p) = 0 for a l l g E A} . 2.4. D e f i n i t i o n . Let .*#A =' {MP : p e gX} . We now prove the following theorem i n order to show that .o^ may be given the hull-kernel topology. By the results already mentioned we see that = M 5 &i.'rJ. =))','.,. • 2.5." Theorem. For p £ gX , MP i s a prime ide a l i n A Proof: For p £ gX , <j> f MP $ A because OE MP and I i MP . A A A I t i s clear that MP i s an i d e a l i n A . To show that i t i s a prime idea l consider f, g e A such that f \ MP and g \ MP . Then by d e f i n i t i o n of. MP there exist h, k s A such that (fh)*(p) ^ 0 and (gk)*(p) i 0 . Thus (fghk)*(p) i 0 . Hence fg \ MP Thus i f fg £ MP then either f e MP or g e MP . Hence MP i s a prime i d e a l . Define T a : gX -> '-^  by T A ( p ) = MP 2.6. D e f i n i t i o n . A subalgebra A of C(X) i s said to be a g-subalgebra of C(X) i f T a i s a homeomorphism of gX onto JTI . 23 Recall that we said that and t J( are homeomorphisms of gX onto r,ir and ))1 , thereby making C(X) and C*(X) g-sub-Ci 0*^  algebras of C(X) . Before we introduce the concept of A-points we would l i k e to define a certain class of sets and show that they are a base for the closed sets i n gX . Let T.^ denote the inverse map. For f e A l e t A A = {p e gX : f e M^ } . I t makes sense to speak of E^p (f) because we have shown that <z$'^ i s a c o l l e c t i o n of prime ideals. " A Now by d e f i n i t i o n , {p £ gX : f £ M^ } = Z((fg)*) where, as usual, g e A Z denotes a zero-set. But zero sets being closed i n gX , implies that S A ( f ) i s closed i n gX . 2.7. Theorem. Let A be a subalgebra of C(X) . Then (a) T : gX i s continuous A A / (b) T. i s a closed mapping i f and only i f i s a Hausdorff space. A A Proof: (a) Recall that we have shown that the sets of the form - 1 E.t,-. (f) are basic closed sets for Q5\ . Also T . , A A A closed i n gX . v Hence T i s continuous. A (f) = S A ( f ) i s 24 (b) Let be a closed mapping. Suppose f . Then F,= {x £ gX : T A(X) = T A(p)} and K = {x e gX : T A(X) = T A(q)} are d i s j o i n t compact subsets of gX (compact because F and K are closed sets being zero sets i n gX ; M A = {f e A : (fg)*(p) = C> g e A} so F = {x : (fg)*(:x) =0} , and therefore F and K are closed subsets of a compact space gX ). Now l e t U and V be d i s j o i n t open neighborhoods of F and K respectively and l e t U 0 = gX - ? A V A ( g X - U)] and V 0 = gX - ^ [ ^ ( g X - V) ] . Since T A i s closed, then U 0 and V 0 are open i n gX and p e U 0 U , q e V 0 c V , U 0 = T^t^CUo)] , V c = ^ [ ^ ( V o ) ] . Then c l e a r l y T.(U 0) and T (V 0) are d i s j o i n t open neighborhoods of and A A A A respectively. Hence ^ i s Hausdorff. Conversely, i f i s Hausdorff then we use the fact that T i s continuous to prove that t i s closed. Let B be a A A closed subset of gX . Since gX i s compact then B i s compact. Hence T^^) - ^ s compact (T a i s continuous). Being a compact subset of a Hausdorff space T.(B) i s closed. ( I t i s easy to show that a A compact subset of a Hausdorff space i s closed). Thus for a g-sub-algebra A , i s a Hausdorff space. Remark. I f T : gX A' i s a homeomorphism then {S (f) : f. e A} A A forms a base for the closed sets i n gX . This i s so because -1 V f> - ^A (f) and the sets of the form E (f) , f £ A , form a 25 base for the closed sets i n A * / . > a s already shown. A Now l e t A be a 8-subalgebra of C(X) . This means that x i s a homeomorphism and so {S.(f)' : f e A} i s a base for the A A closed sets i n 8X . For f e A , define S*(f) = S (f) /) X* , where A A X* = 8X - X . The c o l l e c t i o n # = {S*(f) : f e A} i s c l e a r l y a base for the closed sets i n X* . Let fiS*(f) denote the boundary of S*(f) i n X* . 2.8. D e f i n i t i o n . Let A be a 8-subalgebra of C(X) . A point p e X* i s called an A-point of X* i f , for a l l f e A , p { SS*(f) . Thus A the set of A-points i s the set P (X* - 6S*(f)) , x^hich i s an i n t e r -f e A A section of a family of card A , dense, open subsets of X* . We s h a l l now prove an existence theorem for A-points. F i r s t we need the following d e f i n i t i o n and propositions. 2.9. A space X i s said to have the G„-property i f every non-void G r-subset of X has a non-void i n t e r i o r , o 2.10. Proposition. Let Y be a non-void, l o c a l l y compact, T 0 space with the Gg-property. I f 73 i s a family of at most co^  dense, open subsets of Y , then f ) D i s dense i n Y . Also i f Y has no isolated u l points, then card 26 Proof: Let AJ = {U : a < co.. } . I t suffices to show that for a 1 an arbitrary non-void open set G i n Y , 0 Z > '1 Q, f cj) . We proceed inductively as follows. Let a < a>^  , and suppose that there i s defined a c o l l e c t i o n {V^ : B < a} of non-empty open sets i n G with compact closures such that for any B < ct c£ y V G c U G fl V t for a l l x < ( i ) I f a = y + 1 then we s h a l l show ^.V„ = V (and of course B<a Y V i if). We have V. C cJL, V 0 c U 0 O V c V , for a l l x < 8 . Let Y B Y g B x x ' x e C] V„ . Then x e CJL. V 0 for a l l B < a . Thus x e V for a l l 8<a 6 Y • 3 x < B. Hence x s , x < a which implies that x £ since y < a • Hence H V„ c V B<a 3 Y Now l e t B < ct . This implies 8 _< y • I R B < Y then V.D V .. I f B = Y then V„ = V . Hence we have V n 3 V for a l l B Y B Y B Y B < a which implies that 0 V: D V . Thus ^ V„ = V . 8<a Y 8<a 3 Y ( i i ) We s h a l l now show that i f a i s a l i m i t ordinal then N V = ^.c£ V . Since V C cj> V , then fl V c O cJL. V R B<a 3 B<a Y 3 3 Y 3 B<a 3 B<a Y 3 Now i f x £ cJL, V„ for a l l 3 < a , then x £ V for Y 3 x a l l x < 3 • But x < a implies that x<x+1 < a and so x £ V for x a l l T < a , i . e . X E V for a l l 3 < a . Hence ^ cl V C f) V . 3 3<a 3 8<ct 3 27 H e n c e H V . = 0 C J L _ V . . B u t . ' { c J L , V „ : 3 < a } , 3 < a 3 3 < a • Y 3 Y 3 b e i n g a . d e c r e a s i n g f a m i l y o f c o m p a c t s e t s , h a s n o n - v o i d i n t e r s e c t i o n ( t h a t {eft V 0 : 3 < o t } i s a d e c r e a s i n g f a m i l y i s e a s i l y s e e n a s f o l l o w s : Y P l e t 3 n < 3 0 • T h e n c £ v V 0 C V 0 . B u t V 0 <r c L V 0 . H e n c e L ' 2 1 1 1 * H e n c e , i n e i t h e r c a s e , Si V 0 i s a n o n - v o i d G . - s u b s e t 3 < a 3 6 o f Y a n d b y h y p o t h e s i s h a s a n o n - v o i d i n t e r i o r w h i c h i n t e r s e c t s t h e d e n s e o p e n s e t . A n d s i n c e Y i s l o c a l l y c o m p a c t t h e n t h e r e e x i s t s a n o n - e m p t y o p e n s e t i n Y s u c h t h a t cl^. i s c o m p a c t a n d c L V < c U D i n V J C U n G . T h e r e f o r e ' {V : a < co, } i s d e f i n e d Y a a n 3 a a 1 3 < a i n d u c t i v e l y i n s u c h a w a y t h a t ( c i , V : a < w, } i s a c o l l e c t i o n o f x a 1 c o m p a c t s e t s w i t h t h e f i n i t e i n t e r s e c t i o n p r o p e r t y ( f o r c o n s i d e r cl^V^ a n d c J L r V w i t h a , < a „ . T h e n 2 c L V C U 0 ( Si V c ) C V C cJL. V ) , a n d s u c h t h a t a2 a 2 3 < a 2 3 \ a l a l c J L , V C U H G f o r a l l a < to, . H e n c e Y a a 1 (OH)n G ^ o c £ v v + <j, . Y a a < w 1 T h u s /? ^ i s d e n s e i n Y . N ow i f Y h a s n o i s o l a t e d p o i n t s t h e n a t e a c h s t a g e o f t h e c o n s t r u c t i o n o f t h e s e t s V w e c a n f o r m t w o V ' s w i t h d i s j o i n t a . a c l o s u r e s . T o s e e t h i s c o n s i d e r t h e V t h a t we h a v e a l r e a d y c o n s t r u c t e d a 2 8 a n d a n e l e m e n t x e . B y l o c a l c o m p a c t n e s s t h e r e e x i s t s a s e t V , , s u c h t h a t x e V ' .. a n d V ' 5" v • C o n s i d e r V - V , . j ct+1 crl-1 cri-1 fo. a a+1 p i c k y e V - V ' a n d a n o p e n s e t V s u c h t h a t y e V ' a n d a crt-1 cd-2 J a+2 ci V , „ C V . T h e n ' V , „ s e r v e s a s t h e o t h e r V t h a t we m e n t i o n e d , cn-z a arZ • a ^ 1 H e n c e i f Y h a s n o i s o l a t e d p o i n t s t h e n c a r d f)Z> _> 2 . T h e p r o o f i s n o w c o m p l e t e . R e c a l l t h a t a s p a c e X i s r e a l c o m p a c t i f f o r a l l p e X * t h e r e e x i s t s Z e Z ( B X ) s u c h t h a t p e Z C X * . T h i s d e f i n i t i o n w a s a l r e a d y d i s c u s s e d i n C h a p t e r I . 2 . 1 1 . P r o p o s i t i o n . I f X i s l o c a l l y c o m p a c t a n d r e a l c o m p a c t t h e n X * h a s t h e G g - p r o p e r t y . P r o o f ; We h a v e , i n f a c t , p r o v e d t h i s i n L e m m a 1 . 5 w h e r e we p r o v e d t h a t a n o n - e m p t y z e r o - s e t Z h a s n o n - e m p t y i n t e r i o r . N o t e t h a t e v e r y n o n - e m p t y G ^ i n X * m u s t c o n t a i n a c o m p a c t s e t , C , ( e . g . e v e r y p o i n t i n t h i s n o n - e m p t y G ^ i s a c o m p a c t s e t ) a n d b y [ 4 ] , 3 . 1 1 ( b ) , t h e r e e x i s t s a z e r o - s e t w h i c h i s a s u b s e t o f t h e G . a n d w h i c h c o n t a i n s 6 C . Now t h i s z e r o - s e t h a s n o n - e m p t y i n t e r i o r , s o e v e r y G ^ - s u b s e t o f X * h a s n o n - e m p t y i n t e r i o r . 2 . 1 2 . P r o p o s i t i o n . I f X i s r e a l c o m p a c t t h e n X* h a s n o i s o l a t e d p o i n t s . P r o o f : S u p p o s e p i s a n i s o l a t e d p o i n t i n X * . L e t b e a 29 zero-set neighborhood of p i n gX such that z ^ / 0 X* = {p} (such a set Z^ , exists because every neighborhood of a point i n a completely regular space contains a zero-set-neighborhood of the point - [ 4 ] , p. 3 8 ) . Since X i s realcompact there exists a zero-set Z^  £ Z(gX) such that p E Z 2 C X* . Thus ' {p} = Z fi Z 2 (Z fi Z 2 C Z' 0 X* = {p}) and Z^  0 Z 2 e Z(gX) , which i s impossible by [ 4 ] , 9 . 6 . This completes the proof. We can now prove very e a s i l y the existence theorem we are seeking. 2 . 1 3 . Theorem. Let X be l o c a l l y compact and realcompact but not compact. I f A i s a g-subalgebra of C(X) with card A = c , then X* has a dense set of 2 A-points. Proof: Let Z> = {X* - 6 S * ( f ) : f e A} . From the l a s t two propositions we conclude that X* has the G^-property and X* has no isolated points. Now ^ i s a family of c dense open subsets of X* . Hence by Proposition 2 . 1 0 , fi ^  i s dense i n X* and has c a r d i n a l i t y at least 2 . But since A i s a B-subalgebra of C(X) then > A cs.x"d A. c T : 3 X • * / v ' A i s a homeomorphism and so card X* <_ 2 = 2 Hence card fi& = 2 ° . The proof i s now complete since, as we have already seen, /?& i s the set of A-points of X* . F i n a l l y we formulate our results i n terms of 3N . 30 2.14. Theorem. A point i n X* i s a C*(X)-point i f and only i f i t i s a P-point .of X* . Proof: (We f i r s t r e c a l l that a point p of X i s a P-point of X i f any G^-subset of X containing p i s a neighborhood of p . Also, from the fact that every zero-set i s a G^  and from [4], 3.11(b), we conclude that p i s a P-point of X i f any zero-set of X containing p i s a neighborhood of p .) Thus a point i n X* i s a P-point of X* i f and only i f i t i s not an element of the X*-boundary of any Z where Z e Z(X*) . Recall that for a subalgebra A , S (f) = ,{p e gX : (fg)*(p) = 0 for a l l g e A} = (1 Z((fg)*) . In the g£A case of C*(X) we get S A ( f ) = Z ( f 3 ) by [4], 7.2 ( f 3 i s the continuous extension of f to gX) . So by d e f i n i t i o n S*.,;(f) = Z ( f 3 ) /I X* . Thus a point p of X* i s a C*(X)-point i f and only i f i t i s not on the X*-boundary of (Z fi X A) for any Z e Z(gX) . Hence every P-point of X* i s a C*(X)-point. v Conversely l e t p e where Z^  e Z(X*) . We s h a l l show that p i s then an element of the X*-boundary of a zero-set of gX (so p w i l l not be a C*(X)-point). Since Z^ e Z(X*) then there exists a G.-subset S of gX such that S f) X* - Z^  . We j u s t i f y o 1 CO this as follows. From [4J, 1.10,.we can conclude that Z = fl Om where 1 1 0 i s an open set i n X* for every i . Since O. i s open i n X* then 0. = 0 1 0 X* where £!X i s open i n gX . Hence Z = f) (o f> X*) , 1 1 31 CO CO oo and i t i s c l e a r that /I O  x*) = <f)0 X)f) X* . 1 1 Now l e t /? Q X = S . 1 Then S i s the required G„-subset of gX . Now by complete regularity there exists an element Z 2 e Z(gX) such that p e Z- c: S (obtained from [4], p. 38 - every neighborhood of a point i n a completely regular space contains a zero-set neighborhood of that point). Hence p £ S(Z^H X*) . Hence a C*(X)-point of X* i s a P-point of X* . We can now f i n a l l y conclude that i f X i s l o c a l l y compact and realcompact but not compact and i f card(X) = c , then X* has a dense set of 2 P-points. Hence gN-N has a dense set of c 2 P-points. Remark. Local compactness i s an essential property. For the space Q (set of rationals)' s a t i s f i e s the property card C(Q) = c but Q* has no P-points (see [4], 60.5). 32 CHAPTER I I I c-POINTS OF BN-N 3.1. We s h a l l c a l l a point i n BN-N a c-point i f i t l i e s ' simultaneously i n the closures of c pairwise d i s j o i n t open sets of BN-N . We remark at once that i f a i s any cardinal greater than c then there does not exist any a-point i n BN-N , because there are at most c d i s j o i n t open sets i n BN-N . In the f i r s t part of the chapter we s h a l l prove that there exists a c-point i n BN-N . Then l a t e r on we prove that every point of BN-N i s a c-point, the proof of which i s independent of that of the f i r s t part. However we include the f i r s t part because the same proofs can be applied to draw the same conclusion for any discrete space of c a r d i n a l i t y m where 2 n _< m for a l l n < m . Following the notation of [ 4 ] we l e t , for any p e BN-N , A P be the free z - u l t r a f i l t e r on N with l i m i t p . 3.2. D e f i n i t i o n . Ap i s called uniform i f each element of AF has ca r d i n a l i t y 0 0 . Let uN = {p e BN : A P i s uniform} 3.3. Proposition. "* uN = BN-N 3 3 Proof: , Let p e gN-N and consider A P . In order to show that p e uN' . we must show that A P has no element consisting of a f i n i t e subset of N . Suppose there exists P e A*3 such that P = {n^, . .. , n } where n.^  £ N , 1 .1 i .1 j • Then {n^} ^ A P for otherwise A P would not be a free u l t r a f i l t e r . Hence (N - {n^}) e A P. Thus P f) (N - {n 1» e A P which means that {n 2, . .., n } . e A P . We proceed i n this way u n t i l we can f i n a l l y conclude that {n_. } e A P , a contradiction. Hence no such element, P , exists and gN-N C uN . Now p i t i s easy to see that uN C gN-N . Let p e uN , i . e . A i s uniform. I f p s N , then A P = {Z : Z i s a zero set containing p} . But this would mean {p} £ A P since any element i n N can be made into a zero set. The fact that {p} e A P i s a contradiction since A P i s uniform. In the sequel we s h a l l use the following facts. I f Z N , then c£.„ Z = {p £ gN : Z £ A P} . Also basic neighborhoods of gN p £ gN-N are of the form* (c^gN Z) - N where Z e A P . 3 . 4 . Lemma. Let Ci- be an i n f i n i t e c o l l e c t i o n of subsets of N such that card A = for every A £ £L , and cardCA^/O A^ .) < ° 0 for P such that card {A zC-L : card(Z D A) = J\} = card & for every Z £ A^  A., A. £ , i ^ j . Then there i s a uniform u l t r a f i l t e r A on N Proof: , For A £ CL pick x^ £ c ^ g N A D uN ; l e t B = {x A : A £ CL }. I f P, Q e CL such that P <f Q , then f x„ 34 because i f = x then A = A H and since P £ A ( r e c a l l that x for A c N , cJl A =' {P £ gN : A £ A P}) then P e A and so X0 >r P fl Q s A v . This means that card(P fl Q) = 0 , a contradiction. Thus card B = card . Since gN-N i s compact, there exists a point p e uN such that every neighborhood of p contains card Ct elements of B . For i f not then pick a neighborhood of every p £ gN-N , N^ , such that N contains less than card fl elements of B . Consider P {Np : p £ uN} so formed. Without loss of generality assume that this set i s an open cover for uN . Since uN i s compact, there exists a f i n i t e number of elements from this set, N , N , N , such that P l P2 P n n uN C V N . Note that B <^  uN . Since each N contains less than 1 p. p. card CL elements of B then card B < card CL , a contradiction. We claim that A P i s the required u l t r a f i l t e r . I t i s uniform since p e uN . Also i f card(Z /) A) < for Z £ A P then x A fj: Z /) yN (for XA XA X x. £ c£ A fj yN means A, £ A so Z % A ). Hence card(Z f) A) = J 0 for every Z £ A P and for every k e . f l . We digress a l i t t l e to prove the following theorem which w i l l be used i n the remainder of the chapter. 3.5. Theorem. Let m be a cardinal number such that 2 n _< m for every n < m and l e t .X be a set of "cardinality m . Then there exists a family Ct of subsets of X such that card Ci = 2 m , card A = m for every A £ CL and card (A f) B) < m for A, B £ Ct. and A r B . 3 5 Proof: We can well order X thereby l e t t i n g X = {x^ : a < a Q} X where a 0 i s the i n i t i a l ordinal such that card a0 = ra . Let Y = 2 and order Y lexicographically. Let Z = {y E Y : there exists a < a0 such that y(x )"= 0 • . v for a < 0 < a Q} We would l i k e to show that card Z = m . Indeed Z = {Z : a < a 0} a where Z i s the subset of Y whose functions take on the value zero a at every x for y > a ( i . e . Z = {y £ Y : y(x ) = 0 for y > a} . y a y card ct For every- a there are 2 such functions and so card Z = 2° a r^ < m by hypothesis. Thus card Z = Y. card Z a — J J L A a<a0 < m-m = m . Thus card Z = m . For each y E Y l e t be a sequence {z^ : 0 < a 0} of m d i s t i n c t elements which converge to y and where z^ e Z ^  Y such that r y(x ) , a < 0 z R ( x ) = { a U • • a > 0 . Now l e t f from Z onto , X be a one-to-one map. Then {f[Sy] : y E Y} i s the required family theorem. Note that N has the properties of X i n the above 3 6 3 . 6 . Lemma. Let Ct be a c o l l e c t i o n of subsets of N as given above and l e t A P be as obtained i n Lemma 3 . 4 , the elements Z of A? a being indexed by the ordinals less than to . Then we can choose f or each a < to, a subset X of Z so that card X = o and so that 1 a a • a card (X /) X ) < J \ i f a < Y < a, . a a ' • 1 Proof: Pick an element e /Z such that card Z^ fi A^ = J 0 ( r e c a l l that we have card (J such elements) . Let a < to^  . Assume that for a < a we have chosen A s such that card(Z /Ik) = J'0 CT CT a and that A 4- A for every v < cr . Since CT y •card{A z CI : card A fl Z = J Q} = to., (follows from lemma 3 . 4 ) then t h i s a 1 \r means that we have an element A e tl such that card (A /) Z ) = J Q a a a and A j- A for CT < ct ( r e c a l l a < ox.) . Let X = A ^ Z . Then cr a 1 a a a \r the sets X s a t i s f y the required properties since card X = <, and ot a for a < y < we have A f A so card (A ^ A ) < J\> % hence' the 1 a y a y second property i s s a t i s f i e d since X' -'1 X = A ^ Z r) A '"^  Z C A A . a Y a a Y Y a Y 3 . 7 . Theorem. There ex i s t s a c-point i n gN-N Proof: . Let (I , A P = {Z}' and X C z be as already a a < to., a a described. Since card X = 0 then again by Theorem 3 . 5 we have a ct v c o l l e c t i o n {X } of subsets of X such that card X = J 0 f o r xa x<to^ a Ta every T < to, and card ( X f~) X ) < J\ for T f CT . Now for every 1 TCI . CTOt T < to, l e t U = ( c Z n „ X ) ^  uN . Since ci „ X i s open i n 1 T gN ta gN Ta a<to^ gN then U i s open i n gN-N . We claim that {U } i s a family T T T<(0^ 37 of pairwise d i s j o i n t , open sets such that p e c£ for every T , i . e . p i s our c-point. That they are pairwise d i s j o i n t i s shown thus: l e t q e U fl U where a ^  x . Then for some a, , a~ we have a x 1 2 q e c£ D H X • fl uN , q e c£„_ X f] yN . Hence 3N aa^ 0N xc<2 q e (c£^7 X fl c£„,T X ) f) yN . In the case a, 4 a„ then 3N aa^ 3N xa 0 1 2 (X O X ) C (X /7 X ) and since card (X fl X ) < ''I then oa^ xa 2 a 2 1 a2 card (X X ) < J\ . I f a, = ou then we already know from ca^ xa 2 1 2 ^ properties of the sets {X } that card (X ff X ) < *'t . So i n TO^ aa 1 xcu/ either case card (X fl X ) < J\, . But q e c£„„ X f] c£^T X aa 1 r a 2 ^ 3N ac^ 3N xcu, means that X £ A q , X e A q and so (X ^ X ) e A q . Also cra-^  xa 2 aa^ xa 2 q E yN so A^ i s uniform. Hence card (X fl X ) = "l0 , a contra-aa^ xa 2 d i c t i o n . Hence our sets are d i s j o i n t . We now proceed to show that p i s i n the closure of U for every x . Let Z E A P ( i . e . T a p E c£ Z ). We know that ( c^ f l N 2 '^ yN) i s a basic neighborhood of p so we are finished i f we can show that for every x , C^3N ^ ViN Z"' ^ <|> since every neighborhood of p would then V' intersect U . Since card (X f> Z ) = ^ (X C Z and x xa a xa a card X = %) then there exists an element q i n •xa • n c£ X f) c£ 0 H Z /) yN<r c£ o W Z' U ^ yN . Hence 3N xa 3« a 3N .a x c£ O M Z /I U /I H • 3N a x y 38 Thus p i s a c-point i n BN-N . The purpose of demonstrating such an u l t r a f i l t e r A P i n Lemma 3.4 was to show i n Lemma 3.6 that we can choose, for each Z e A P a a subset X with those properties. Using this c o l l e c t i o n {X} a ^ ^ 6 a a<tu1 we proved that p was a c-point of BN-N . Hence a reasonable assertion i s that every point of BN-N i s a c-point i f we can show that for every free u l t r a f i l t e r A P there exists X C Z for every Z e A P such a a a Y" ' 'T that card X = „ and card (X O X j < J\ i f a T B • a a B 3.8. Lemma. Let A P be a free u l t r a f i l t e r on N and order i t s elements by the ordinals less than UJ_, . Then for each Z E A P there 1 a exists X C. Z such that card X = J \ and card (X /? X j < J*a i f a a a a 8 a ^ B • P A'" Proof: Recall that every element i n A has c a r d i n a l i t y „ since A P i s uniform. Let X^ be an i n f i n i t e subset of Z^  such that X1 £ A P . Fix a < ujn and assume that for each a < a we have chosen X C Z such that card X = ^ 0 , X i A P and card (X 0 X ) < K * r for a l l y < a . Let B = {CT < a : card (X fi Z ) = J\} . We now show CT a that, i f B i s f i n i t e then (Z - U{X : cr e B }) e A P . Consider a set a CT X' i A P . Then N - X e A P . Since Z e A P then (N - X ) fl Z £ A P a a a a a i.e. Z - X e A P . Thus i t suffices to show that O { X : CT e B} i A p . a a . CT ' Since X i A P for every cr e B then N - X e A p and so CT J CT 0 {N - X^ : CT £ B} s A P , i. e . N - ^{x^ : CT e B} e A P . Thus 39 V {X : a e B} i A P . Now l e t X be an i n f i n i t e subset of cr a (Z - L!{X : a e B}) such that X £ A P . Then by d e f i n i t i o n of B we have card (X 0 X ) < J \ for every a < a a a I f on the other hand B i s i n f i n i t e , then i t i s i n f i n i t e -CO l y countable by the continuum hypothesis so we write B = {<?n}^ • every k e N we can find an element x i n \ (X f) Z ) - 6 ' { X : j < k} because card (X fl Z ) = ~K while for o\ a a. a, a k j k j < k card (X n X ) < -J\ . Thus (X f) Z - 6/{X : j < k}) ^ <b . cr, o. cr, ot a. k 3 k 2 CO Consider the set {x }, . . Let X be an i n f i n i t e subset of ct^ k=l a oo "n {x }, . such that X" i A^ (note that X c Z ) . Now for any a e B, ot^  k=l a T a a A<" a = a, for some k e N so card (X' rt X ) = card (X rt X ) < k < " 0 k a cr a a, — k CO I from the way we have chosen {x }, , . I f however a I B then a k k=l * card (X rt X ) < card (X H Z ) < J\ by d e f i n i t i o n of B . Thus we a a — a a have shown that for every a there exists X c Z E A P such that a a card X = J I and card (X ^ x ) < J ) 0 i f a / g . ot a g 3.9. Theorem. For every p e gN-N , p i s a c-point of gN-N . Proof: Let A P = {Z } , X <Z Z as i n Lemma 3.8. Since C4 a«D^ a a card X^ = '\ for every a then we can use our Remark 3.5 to show that there exists a c o l l e c t i o n {X' \ } of subsets of X such that 40 card X = J'Q and such that card (X /> X ) < J , 0 i f x ^ a . We xa xa aa now form sets precisely as we did i n Theorem 3.7 and we proceed exactly as we did there to show'that p i s a c-point of 3N-N . 41 CHAPTER IV TYPES OF ULTRAFILTERS AND THE NON-HOMOGENEITY OF 6N-N 4.1. The major results of this chapter are due to Z. Froli'k. Here we discuss the concept of types of u l t r a f i l t e r s and using this concept we s h a l l prove the non-homogeneity of gN-N (this w i l l be proved x^ithout using the Continuum Hypothesis) . 4.2. D e f i n i t i o n . We say that two u l t r a f i l t e r s are of the same type i f they are isomrophic as p a r t i a l l y ordered sets. Theorem 4.3 w i l l show that not every two u l t r a f i l t e r s on N are of the same type. Remark. I f -rr i s a permutation on N and Q i s an u l t r a f i l t e r on N then TT(^) = {TT(E) : E e fi} Is an u l t r a f i l t e r on N . We show this as follows. (a) Let TT(E^) , TTCE^) e ir(^) . We would l i k e to show that TT(E 1 ) fl TT(E2) e Tr(n) . We claim that TT(E ) fl ir(E 2) = TT(E E ) . For l e t y £TT(E^) d T\(JLA . Then for some e^ e , Tr(e^) = y and for some £ E^ , T r( e2^ = v • ^ u t e ] _ = e2 s i n c e fr i s one-to-one. So y £ TT(E ;L f> EA . Thus TT(E ) fl ir(E 2) C TT(E ^ ' E ) . Obviously ir(E ^  E £) C TT(E ) ^ TT(E 2) . Hence 42 T T ( E 1 ) {) T T ( E 2 ) = T : ( E ; L / ' I E 2 ) . Since E ^ E 2 e Q then T T ( E ] L ) -0 T T ( E 2 ) E TT(O) . (b) Let TT(E) C E 0 . We must show that E Q e Tr(fi) . Since IT is a permutation on N then for some A <£* N TT(A) = E 0 (we are assuming that E 0 c N) . Then E c A since TT(E) <r i r ( A ) . Thus A E ft . Hence TT(A) e Tr(fi) , i.e. E Q E ir(fi) . Hence -rr(fi) is an u l t r a f i l t e r on N . 4.3. Theorem. If two free u l t r a f i l t e r s ft^ and Q. are of the same type then there is a permutation TT of N such that ' ^ 2 = ir(ft^) . c There are 2 types of free u l t r a f i l t e r s on N ; each type contains c u l t r a f i l t e r s . Proof: Since ft^ and Q,^ axe of the same type then there exists an isomorphism f taking ft^ to ^ 2 (note that i f A C B then f(A) C. f(B) and that f(N) = N) . Let H =' {q : q E N , q ^ n } . The sets H^ are maximal proper subsets of N which are permuted by f . That they are maximal proper subsets' is obvious. Now for any H^ , n i H . There must exist E E £L such that n i E since otherwise T n 1 T fl {E : E E ft^} f <f> , a contradiction since is free. Since E c- H n then' H n £ & and so f operates on H^ to give a set of the form similar to H , H say. Define ir such that -rr(n) = m i f f(H ) = H . n m n m Now let E E and let F = N - E . Then E =/7{H : n £ F} because 1 n the elements of E are the only elements in H^ for every n £ F . Let E 0 =/7{f(H ) : n E F} . Since E c H for every n then n n 43 f(E) c f (H ) f o r every n and so f (E) c: E Q . By symmetry we have f _ 1 ( E 0 ) C E . Thus f ( f _ 1 ( E 0 ) ) C. f(E) , i . e . E 0 <= f(E) . Hence f (E) = E 0 . Now N - E 0 = N -/?{f (H ) : n s F} =' {Tf(n) : n e F} because -rr(n) = m i f f(H ) = H and r e c a l l that m i s the only element of N n m that i s not i n H . Thus N - f(E) = TT(N - E) '. m For any A c N we have T T ( A ' ) = [ T T(A)]' (where primes denote, complements) . Hence TT(N - E ) = N - TT(E) . Thus, using the f a c t that N - f ( E ) = TT(N - E ) , we get f ( E ) = T T ( E) . Hence IT i s the required permutation. We have now established that a necessary condition f or two u l t r a f i l t e r s to be of the same type i s that there e x i s t a permutation ir on N taking the one u l t r a f i l t e r to the other. But there are only A1 2 = c permutations on N . Hence no type can contain more than c free u l t r a f i l t e r s . We can,now conclude that there are 2 types because the set of u l t r a f i l t e r s of the d i f f e r e n t types forms a p a r t i t i o n of gN-N which has 2 elements and each type has only c u l t r a f i l t e r s . F i n a l l y we must show that every type contains p r e c i s e l y c u l t r a f i l t e r s . From Theorem 3 . 5 we can f i n d a family c o n s i s t i n g of c i n f i n i t e subsets of N such that any two members of 3 have f i n i t e i n t e r s e c t i o n . Let • 0, be any free u l t r a f i l t e r and A e 0, be such that the complement of A i s i n f i n i t e . For each E e 3 l e t be an u l t r a f i l t e r containing E . Then there e x i s t s a permutation on N 44 carrying Q. to (take the' permutation which c a r r i e s A to E) . Since we have c such sets E then we can conclude that fi, i s of the same type as c other u l t r a f i l t e r s . (Note that we have already shown that no type can contain more than c free u l t r a f i l t e r s ) . 4.4. We now take a closer look at the set of types of elements of gN-N '. We l e t P denote the set of a l l permutations of N and p* with p e P the continuous extension of p to a homeomorphism of N* onto i t s e l f . Suppose _ f : N -> N i s a set isomorphism ( i . e . a permuta-t i o n on N). I f we consider f as operating on subsets of N then we can say that f induces a set isomorphism g : iP(N) ->£\N) (where P (N) denotes the power set of N). Now since an u l t r a f i l t e r on N can be considered as a subset of ^(N) then we can say that f induces a homeomorphism h from gN onto gN i n such a way that i f P l P2 fi^(= A , p^ e gN) and ft (= A , p^ e gN) are u l t r a f i l t e r s on N , then g(^ 1) = ft2 i f and only i f h(p 1) = p 2 . We have already proved that i f two u l t r a f i l t e r s and &2 a ^ e of the same type then there exists an element p e P such that a l a2 p(fi^) = ^2 (hence p*(a^) = ^ ^ ^± = A > ^2 = A ^" T ^ u s f r o m Theorem 4.3 and e a r l i e r remarks- i n t h i s s e c t i o n we conclude that a l ~' a2 £2^(= A , a^ E gN) and fi^(= A , a2 e gN) are of the same type i f and only i f p * ^ ) = a.^ f o r some p e P . 45 Let T be a set and T a mapping of N* onto T such that TO = T6 i f and only i f p*(a) = 8 for some p e P . Vie s h a l l c a l l the elements of T the types of u l t r a f i l t e r s on N , i . e . i f t = T0 then t i s the type of 0 . Note that we have defined our mapping T SO as to be consistent x^ith our d e f i n i t i o n of two u l t r a f i l t e r being of the same type. Note also that since we have proved that each type contains c u l t r a f i l t e r s and that there are 2 types of free -1 c u l t r a f i l t e r s we now have card T (t) = c and card 1=2 We now define a c o l l e c t i o n X of u l t r a f i l t e r s to be di s c r e t e i f there e x i s t s a d i s j o i n t family {M : x e X} with M e x . We know that c£ X i s homeomorphic to gN i f X i s i n f i n i t e . Thus for X a countable d i s c r e t e subset of gN-N and z e cl X - X the i n t e r s e c t i o n of the neighborhoods of z with X form an u l t r a f i l t e r on X . C a l l t h i s u l t r a f i l t e r z . We know that there e x i s t s a set X isomorphism k : ^ (X) ->?9(N) . Thus z i s taken by k to an u l t r a -f i l t e r C ^(N) . We now define the type of z r e l a t i v e to X , written T,TZ , to be T X y In the case when X = N we get and j^z . where z^ i s an u l t r a f i l t e r on N of type ' T^Z . We s h a l l now show that z^ = A , i . e . {N f] N : N i s a neighborhood of z} = A Z , thereby'giving T„TZ = T . Consider any N H N , where N i s a neighborhood of z . N z J z z Let N ^ N = B's-. Then z e B f o r otherwise there e x i s t s some neighbor-hood N* of z such that N* ft B = cb , i.e.' N* f) (N T) N) = A , i . e . z z r z z r 46 • (N A f) N ) fl N = d) . But this i s a contradiction since N* /I N i s a z z z z neighborhood of z E CX, N-N = 0N-N and so (N* /) N ) /) N ^  (J> . Thus — z z z e B and so B e A . Hence z„ C A N — Now l e t B e A Z . We would l i k e ' t o show that B = N O N z where N i s a neighborhood of z . We know that B = {gN - c£(N-B)}rt N. z '. But gN - cft(N - B) i s a neighborhood of z so we l e t N - gN - c£(N - B) . Thus B = N /? N . z z 2 Hence when X = N and Z E C £ X - X , Z N = A , and x , , z = T N z 4.5. D e f i n i t i o n . The producing r e l a t i o n <J> on T i s defined to be the set of a l l ( t , t') E T x T such that T Z = t' , T z = t for some z £ gN - N and X cr gN - N where X i s discrete and z E ci X - X . Thus we have <j)(t) = { T Z : z £ cZ X - X , T z = t , X C N*' and X discrete} Hence the domain of <f> i s a l l of T , i t s range i s a subset of T and when ( t , t') E <j> we say that t produces t ' or t' i s produced by t . 4.6-. Remark. Let { M ^ : n E u} be a countable decomposition of N and pick, for every n , x , y E C £ M - M . Let - n Jn n n 4 7 X = {x : n E to} and Y = {y : n e to} . We s h a l l show that i f , for n • • Jn every n , x f Y , then cZ X f) cZ Y = <p . Recall that a base for n n the closed sets i n gN i s {cZ A : AC N} . Now assume that there exists x e cZ X /7 cZ Y . Consider a base neighborhood A^ containing x_^  ( i e to ) and a base neighborhood A 1 containing y_^  such that A. fl A 1 = d> and M. = (M. /) A.) U (M. f) A 1) . Let M. f) A. = M. and i 1 1 1 1 ' i i i ^ M. /) A 1 = M. . Then M. = M. U M. . Let H, = c£t/{M. : i e to} and i ±2 • 1 1 1 12 1 1 H 2 = cJli^ML : i e to} . Then E± 0 H 2 = <j> . But contains the closure of M. for a l l i and so x e H . Sim i l a r l y x e EL . Thus i 1 we have a contradiction. Hence c% X cZ Y = A> Note that the converse holds as w e l l , i . e . i f ci X fl cl Y = 6 then for each n , x 4 y n Jn 4 . 7 . Lemma. Let y E gN - N . There exists a set x such that card x <_ 2 and whose elements are discrete countable subsets of gN-N and such that i f Y i s any discrete countable subset of gN-N , and i f y e cZ Y .- Y then Y 3 X' for some X' e x where y e cZ X' - X" . Proof: For each countable decomposition {^n} of N choose an x E cZ M - M (and l e t X = {x } ) such that y e cZ X - X , i f n n n n J ' possible, and consider a l l possible X'c X such that y e cZ X' - X . Let x = {X'} so formed as we vary our decompositions of N . Now x cer t a i n l y consists of discrete countable subsets of gN - N . We s h a l l 48 use Remark 4.6 to show that x has the asserted property. Let Y = {yn'} (where Y i s as in\the statement of the lemma). By the construction of x we have y e cH X - X for some countable discrete X . Let X = {x } . Since y i s i n the closures of {x } • and {y } n ,n n then by Remark 4.6 {x } and {y } have to agree for an i n f i n i t e J . n Jn & number of n's . So we l e t the required set X' be Y (1 X . Now y i s i n the closure of X' because i t i s i n the closures of Y and X (see [9], Lemma 2). There are at most 2 decompositions of N and K with each decomposition we get at most 2 sets X . Hence A' }< card x _< 2 > 2 =2 . The proof i s now complete. From the following theorem we s h a l l obtain a very simple proof of the non-homogeneity of gN-N . 4.8. Theorem. Any type i s produced by at most 2 types and any type produces 2^ types. Proof: We prove the second part f i r s t . Let t be any type. Recall that <j)(t) = {TZ : z e cH X - X , T z = t for some discrete A. X c N*} . Now l e t {M } be a decomposition of N where each M i s n n i n f i n i t e . We know that since M i s i n f i n i t e then 2 card (eft M - M ) = 2 for each n . Again using Remark 4.6 we n n y- & ° J o 2 conclude that there are 2 d i s j o i n t sets cJl X = c£ {x : x e M } n n n where each such c£ X contains a point z such that r z = t and X 4 9 o TZ e cj>(t) . Since at most 2 of these elements z could be of any o 2 one type (mentioned already) and since we have 2 such elements z then we can conclude that card <j)(t) = 2 . Before proving the f i r s t part we must prove the following. I f X C Y , Y discrete, and z e c£ X - Y , then T VZ = T z . We assume that both X and Y are i n f i n i t e . Let 6 be a one-to-one map from N onto X and x a one-to-one map from Y onto N . Let i be the inclusion map from X to Y . Define the permutation p on N as indicated: p : N y X c: Y x — s - N i. e . p = x ° i ° <5 • We define z and z as usual. We know X i Ci Y that z and z are set isomorphic to u l t r a f i l t e r s A and A on X Y N i n such a way that T a = T z and x y = T z . In order to show that A X T VZ = T z i t suffices to show that for some permutation q on X , A Y q*(a) = y .We claim that p , as defined above, i s the required permutation, i . e . p*(a) = y . Assume that p5'c(a) ^ y . Then we can f i n d d i s j o i n t neighborhoods A and B such that y e A , p*(a) e B , A f) N e A Y , B 0 N (j: A Y and p A~ 1(B) f) N e A K . Now [N n p''c-1(A)]/T [N/? p * " 1 ^ ) ] = $ . Also 6(p* _ 1(B)/? N) = (N n X) e where N^ i s some neighborhood of z . This i s so because 8 induces a set isomorphism between P (N) and <P (X) which takes Aa to z ; X 50 " p* (B) f) N e A and so i s taken by this isomrophism to N^/l X some neighborhood N^ of z . But considered as a subset of N 6 takes (p*~1(B)/'> N) to (N /? X) . Since [N /? p* _ 1(A)]/7 [N /? p*" L(B)] = <f> then S[E fl p* - 1(A)]/? <5[N f) p* _ 1(B)] = (J) . We s h a l l now show that [N /? Y ] /0 6[p" _ 1(A)/? N] = <)) . Assume this i s fa l s e . Then X[(N z/1 Y)/7 6(p* _ 1(A) N) ] = x&zr\ Y) f) X ( 6 ( P " _ 1 ( A ) fl N)) . Now x ( N /0 Y) £ A Y ( r e c a l l that x induces a set isomorphism between :f (Y) and 'f (N) which takes z y to A Y) . Also X(6(P" _ 1(A) n N)) = A/7 N (see remark later) and A P N e A Y . Thus xt(N ^ Y) /7 6(p* - 1(A) /I N) ] E A Y . Hence (N /) Y) /I S(p* - 1(A) /) N) e z y and so (N z Y) 0 6(p* """(A) N) = N^D where N' i s some neighborhood of z . Now z [(N ^ Y) n <5(p*_1(A) D N)] /IX = (N /) X) fl 6(p* - 1(A) fl N) z = cKp* - 1(B)/) N) /I o(p* - 1(A)/? N) ( r e c a l l that 5 (p* _ 1(B) f) N) = N /7 X). But we have already shown that <5(p* _ 1(B)rt N) /I 6(p* - 1(A) /) N) = <j) , i . e . [(N 0 Y) P 6(p* _ 1(A) /) N) ] /) X = <j> . Hence N' /) Y /) X = <J> which 2 2 implies that N' /? X = <j> . But this i s a contradiction since N' i s 5 1 a neighborhood of z . and z e c£, X . Thus ( N z rt Y) rt 6(p- - 1(A) n N) = 4 Hence • x(N rt Y)/? x(<5 (p* - 1(A) /> N)) = ^ . But x(N z/) Y) e A Y and x(6 ( p A _ 1 (A)/} N)) = A/} N e A Y (see remark l a t e r ) . Thus we have X ( N Z ^ Y ) x ( S ( p * - 1 ( A ) /? N)) = <f> and x ( N z n Y) f) x(6(P' c _ 1(A) ft N)) s A Y , a contradiction. Thus p*(a) = y _1 Remark. We must now show that X(<S(p* (A) rt N)) = A/? N . From our d e f i n i t i o n of p we get X(<5(p*-1(A) 0 N ) ) = p(p* _ 1(A) rt N) • = pt { y e BN : p * ( y ) e A} fl N ] = p{{y e N : p ( y ) e A}] = A f] N . We now continue with a proof of the f i r s t part of the theorem. Consider any type t' . Then (^ ""'"(t') = {t : <j>(t) = t'} . For any t e <}> "'"(t1) and for every X appearing i n the d e f i n i t i o n of <j)(t) there exists a Y e x such that X ~2 Y and x vz = T VZ = t (follows from Lemma 4 . 7 and the above observation). Hence with every 52 element t i n $ (t') we can associate an element of x . But there o o are at most 2 elements of x ; so there are at most 2 such elements t . This completes the proof. 4.9. We now prove the non-homogeneity of N* . Let us assume that N* i s homogeneous, i . e . for every pair x, y e N* , there exists a homeomorphism of N* onto N* taking x to y . Let, for any x e N* , T = <f> "''(TX) . I f f(x) = y for some homeomorphism f of N* onto N* then T X = xy by e a r l i e r remarks (see section 4.4). Hence Tx = Ty i n this case. Thus i f N* i s homogeneous then a l l sets Tx "' are equal for every x e N* . But card Tx _< 2 by our l a s t theorem and card T = 2 (shown already). Also the sets Tx for x e N* cover T . This i s a contradiction i f a l l sets Tx are equal. Hence N* i s non-homogeneous. The remainder of the chapter i s used to show that no homeomorphism of BN into N* has a fixed point.- a fact that w i l l enable us to give another proof of the non-homogeneity of BN-N without using the continuum hypothesis. F i r s t l y we need two lemmas. 4.10. Lemma. For a class X l e t f be a mapping of X into some other class such that fx T X for any x e X . Then there exist d i s j o i n t classes X Q, X±, X 2 such that • X = X Q U X;L 0 X 2 and f (X ) D X =. <j> for i = 0, 1, 2 . 53 Proof: We s h a l l associate with every x e X a class C(x) defined as follows: C(x) = {y e X : f m x = f n y for some m, n e N} We now show that any two classes C(x^) and C(x 2) either coincide or are d i s j o i n t . Assume that they are not d i s j o i n t . Then l e t y £ CCx^ fl C(x 2) . Hence (1) f x = f y ( 2 ) f x 2 = f y where m, m', n, n' £ N . We can assume that n' > n . Apply f to n n' f y u n t i l we achieve f y (while at the same time applying f to f m x ^ to maintain equality i n (1)). We then get ^irri-(n'-n) -^n' .m' f x^ = f y = f x 2 Hence m' m" (3) f x 2 = f x^ for some m" £ N . Now consider any element y 0 e C(x^) . We want to show y 0 £ C(x 2) . mo n° Then by symmetry we s h a l l be finished. We have f x^ = f y 0 where m0, n Q £ N . Assume mQ > m" . With (3) i n mind we apply f to m" f repeatedly u n t i l we get the following: _mT + (m0—m") _mQ _n 0 7T f x 2 = f x^ j^  = f y Q . Hence f m " x 2 = f n°y 0 where m" £ N . Hence y 0 £ C(x 2) . (Note that i f mQ < m" then we would apply f to fm°x^ u n t i l we get a si m i l a r equation). By symmetry we get C(x^) = C(J 54 We now show that f(C(x)) C C(x) for x s X . C(x) = {y : f m x = f n y , m,n e N} . Let fy e f[C(x)] . We must show that fy e C(x) . Since y C(x) then f m x = f n y and so f ( f m x ) = f ( f n y ) . Thus f m + 1 x = f n ( f y ) which means that fy E C(x) . Thus f(C(x)) C C(x) . Since the union of the sets C(x) , x E X , i s X then i t i s s u f f i c i e n t to prove the lemma for each class C(x) . So we l e t X i n the hypothesis of the lemma be of the form C(x) . Thus for any x and y i n X we may assume that there are m,n E N such that f m x = f n y . Now pick an element a e X . For x E X l e t m(x) be the smallest m E N such that f m a = f n x for some n E N . Then l e t n(x) be the least n £ N such that f n x = f m ^ x ^ a . We assert that for x E X with n(x) > 0 m(fx) = mx . For assume that m(x) = q . Then q i s the smallest integer such that f^a = f n ^ X ^ x which i s the same as saying that q i s the smallest integer such that f^a = f n ^ x ^ "*"(fx) (note that i t makes sense to speak of n ( x ) - l since n(x) > 0 ) . Thus m(fx) = mx and n(fx) = n(x) - 1 . We now show that there exists at most one b £ X such that n(b) = 0 and m(fb) T m(b) + 1 . Assume that we have d ^ J D 2 e ^ such that b 1 ^ b 2 , nO^) = 0 , n(b 2) = 0 and m(fb;L) r mOt^) + 1 , m(fb 2) r m(b2) + 1 . Now since n(b^) = 0 then m(b ) m(b )+l f. " a = b 1 = f°b1 and f a = f ft^) . Since m(fb;L) f m(b1) + 1 55 n(fb ) then we have a number p = m(fb ) such that f a = f (fb ) where m(b )+l 0 _< p < m(b^) + 1 (since we already have f = fb^ then m(b 1)+l p < m(b^) + 1). Hence f(b^) = f a = f a where m(b^) + 1 > p >_ 0. m(b 2)+l Similarly f (b^) = f a = f q a where ra(b^). + 1 > q _> 0 . Assume without loss of generality that m(b^) > m(b2) . Then m(b^) > m(b^) + 1 m(b ) m(b )+l because i f m(b1) = m(b 2)+l then b^ = f a = f a = f a . But m(b1) = m(b2) + 1 > q means that m(b^) > q . This contradicts the d e f i n i t i o n of m(b^) since i t must be the smallest number such that m(b1) b^ = f a . Thus m(b ) > m(b2) + 1 . Hence m(b-) m(b )-m(b„) m(b„) m(b )-m(b ) m(b, )-m(b,)-l \= f a = f X Z ( f a) = f 1 1 (b 2) = f 1 (fb m(b )-m(b ) - l m(b )-(mb +l-q) = f • (f q a ) = f (a) . But m(b ) > m(b )+l-q > 0. This contradicts the d e f i n i t i o n of m(b^) . Thus there exists at most one b e X such that n(b) = 0 and m(fb) 4 m(b) + 1 . I f such an element b exists then put X Q = {b} . I f not then l e t X 0 = <j> . For every x e X - X Q we have n(x) > 0 and, as we have already shown, m(fx) = m(x) , n(fx) = n(x) - 1 . Hence m(fx) + n(fx) = m(x) + n(x) - 1 . Let X^ = {x £ X - X 0 : m(x) + n(x) i s odd} , and X 2 = {x £ X - X Q : m(x) + n(x) i s even} . Obviously the sets X , i = 0, 1, 2 are d i s j o i n t and X = X^ U X^ U . To prove the second assertion of the lemma we f i r s t consider i = 1 . Let y £ f(X^) . Thus y = f x ^ for some x^ £ X^ . We must show y £ X^ . I f y £ X^ then m(y) + n(y) i s odd. But m(fx^) + n(fx^) = mx^ + nx^ - 1 , and 56 m(fx^) + n(fx^) i s odd because f x ^ e X^- . This i s a contradiction since m(x^) + n(x^) i s also odd. Thus y ^ X ^ . Sim i l a r l y f ( X ^ ) f) X ^ = <j) . The case i = 0 i s t r i v i a l because b r fb by hypothesis. The proof i s now complete. 4 . 1 1 . Lemma. Let f : K -> N and f the continuous extension of f to BN . Then the set of fixed points of f i s the same as the closure of the set of fixed points of f . Proof: Let X = {x e N : fx = x} . Then by Lemma 4 . 1 0 N - X = X Q U X± U X 2 where f ( X ) fl X = <f> for i = 0 , 1 , 2 . Also c£ X U cl X 0 V cH X U ca X 2 = gN and obviously c£ X 0 c£ X = <j> for every i and c£ X_^ 0 c£ X = <j> for i r j • Now l e t A = {x s BN : fx = x} . Then A C cl X . Let x Q E c£ X . Consider a net } i n X converging to x D • Since f i s continuous then the sequence ( f C S ^ ) } converges to f x 0 . But f(S ) = S^ for every n and so {f(S )} = {S } which converges to x Q . Thus f x Q = x 0 and n n so x0. £ A . Thus A = cZ X and this ends the proof. The following theorem i s required for our second proof of the non-homogeneity of N* . 4 . 1 2 . Theorem. Let f be a homeomorphism of BN into N* . Then f has no fixed point. 57 Proof: Decompose N into sets G s a t i s f y i n g ( i ) "^ Gi c^ i s a f a m i l y °f d i s j o i n t sets \j {G^ .} = N ( i i ) k £ G^ for any k e N ( i i i ) {fk} U Gfc i s a neighborhood of fk. i n N L> f(N)' . (At the end of this proof we s h a l l construct such a decomposition.) For n e N l e t hn = fk i f n E G, . Thus h i s a k mapping from N onto f(N) . Let g = f ^ ° h . Hence g i s a mapping from N to N such that gn =f n for any n because gn = f ''"(hn) = f "'"(fk) for some k E N such that k £ G^ but n £ G^ (note that k E gn). Hence by Lemma 4.11 g 3 has no fixed point. From condition ( i i i ) we have fk £ G. i n N U f (N) and hn = fk for every k n £ G^ (hence one such element n i s f k ) . Thus h(fk) = fk and so h 3 ( f k ) = fk . We now have h 3x = x for x £ c£ f(N) . Suppose there exists x e gN such that fx = x . Since x £ 3N then x E ci N and i t i s very easy to j u s t i f y that f ( c i N) c ci f(N) . Thus fx £ f(ci N) implies that fx e ci f(N) . Hence from above h 3 ( f x ) = fx = x . Now g 3x = f ^"(h^x) = f ^(x)'= x , contradicting the fact that g 3 has no fixed point. Thus f has no fixed point. We now construct a decomposition {G, } 'of N s a t i s f y i n g ( i ) , ( i i ) and ( i i i ) . Consider any decomposition {Hn} of N . Then we 58 know that for j e N , f i s i n the closure of Hn for one and only-one n . Thus we l e t f. e cl H. . for some i . e N . I f 1 e H. , 1 i 1 1 x x then consider H. -{1} (place 1 i n some other component of the decom-1 1 position). Let G = H. -'{1} . I f 1 i H. then l e t G = H. . 1 \ \ • 1 \ For every j e N , repeat this process to get f_. e cl G. and j £ G. . Thus we have UG, = N , k \ G ' for any k e N , and fk e cl G . We want to show that fk U G^ i s a neighborhood of fk f k i n N o f(N) . Since fk e cl G^ then G^ e A which implies that i f k N - G k ij: A and so fk ij: c£(N - Gk> . Thus BN - cl (N - G ) i s an open neighborhood of fk i n BN . Now [BN - cS,(N - Gfc)] rt (f(N) O N) = cl G k C\ (f(N) U N) = [cl G k 0 N] L? [c£ G rt f (N)] = G, 0 fk . ' k (Note that cl G, f (N) = fk because, by construction, f. e c£ G. f k 3 3 every j and cl G^ rt c£ G^  = tj> for i ^ j ; so f £ cl G^ for j 5* k) . Thus fk U G k i s a neighborhood of fk i n N C f(N) . or 4.13. Remark. The above theorem implies that no type i s a r e l a t i v e type of i t s e l f , i . e . ( t , t) e <p for no type t . We prove this by contradiction. 59 Assume there exists some type t such that ( t , t) e ty Then t = xx = x^x and t = x xx for some x e gN - N and .X c: gN - N , where X i s discrete and x e c£ X - X (this i s a l l by d e f i n i t i o n of <J)(t) = t) . As before we l e t {N f) X : N a neighborhood of x} = x,r , x x & X ' an u l t r a f i l t e r on X which i s set isomorphic to A y , an u l t r a f i l t e r on N . Then x^ .x = Ty . But TX i s the type of A X . Hence p*x = y for some p £ P . Thus, whenever T x = Ty , we have an element p £ P A such that p*x = y . Since T VX = TX then for some p e P , p*x = x , A. a contradiction. 4 . 1 4 . We are now prepared to give another simple proof of the non-homogeneity of N* . As already mentioned hx = y implies that ty "''(xx) = ty "'"(xy) i f h i s a homeomorphism of N* onto N* . From —1 —1 Remark 4 . 1 3 , xx \ ty (xx) so xx £ tj> (xy) . Thus whenever hx = y —1 —1 then xx £ <$> (xy) . Hence i f xx e ty (xy) then hx £ y for any homeomorphism h of N* onto N* . In other words i f the type of x i s a r e l a t i v e type of y then hx = y for no homeomorphism h . But this concludes the proof since their always exist' some x,y e gN - N , x i= y , such that the type of x i s a r e l a t i v e type of y ( r e c a l l that we have proved i n Theorem 4 . 8 that for any x £ gN - N , xx produces jC 2 ° 2 types). 60 BIBLIOGRAPHY [1] N. J. Fine and L. Gillman, Extensions of continuous functions i n gN , B u l l . Amer. Math. Soc. 66 (1960), 376-381. [2] Z. F r o l i k , Fixed points of maps of gN , B u l l . Amer. Math. Soc. 74 (1968), 187-191. [3] , Sums of u l t r a f i l t e r s , B u l l . Amer. Math. Soc. 73 (1967), 87-91. [4] L. Gillman and M. Jerison, Rings of continuous functions, Van Nostrand, Princeton, 1960. [5] I. Gelfand and A. Kolmogoroff, On rings of continuous functions on topological spaces, Dokl. Akad. Nauk SSSR 22 (1939), 11-15. [6] N. Hindman, On the existence of c-points i n gN-N , Proc. Amer. Math. Soc. 21 (1969), 277-280. [7] M. Katetov, A theorem on mappings, Comment Math. Univ. Carolinae 8 (1967), 431-433. [8] D. Plank, On a class of subalgebras of C(X) with applications to gX-X , Fund. Math. LXIV (1969), 41-54. [9] M. E. Rudin, Types of u l t r a f i l t e r s , Topology Seminar Wisconsin, 1965, 147-151. [10] W. Rudin, Homogeneity problems i n the theory of Cech compactifica-tions, Duke Math. J. Vol. 23 (1955), 409-420. Ill] A. Tarski, Sur l a decomposition des ensembles en sous-ensembles presque d i s j o i n t s , Fund. Math. 12 (1928), 188-205. 

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