THE STRUCTURE OF BN by RODNEY SEUNARINE RAMBALLY B.A. (Hons.), U n i v e r s i t y 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 t h i s thesis as conforming to the required standard THE UNIVERSITY OF BRITISH COLUMBIA August, 1970 In presenting an advanced the I Library further for degree shall agree scholarly by his of this written this thesis in at University the make tha it purposes for freely permission may representatives. thesis partial be It financial for of Columbia, British gain Depa r t m e n t Columbia for extensive by the understood permission. The U n i v e r s i t y o f B r i t i s h V a n c o u v e r 8, Canada of available granted is fulfilment shall Head be requirements reference copying that not the of agree and of my I this or allowed without that study. thesis Department copying for or publication my Supervisor: Professor T. Cramer ii ABSTRACT Our subject matter c o n s i s t s of a survey of the major r e s u l t s concerning the t o p o l o g i c a l space gN-N where N represents the space of n a t u r a l numbers with the d i s c r e t e topology, and Cech c o m p a c t i f i c a t i o n of gN the Stone- • N . We are mainly concerned with the r e s u l t s 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 w i t h an a r b i t r a r y d i s c r e t e space formulate our r e s u l t s i n terms of concerning . gN-N gN-N . X and finally In some cases, pre-1960 r e s u l t s are obtained as s p e c i a l cases of the r e s u l t s we derive using an a r b i t r a r y d i s c r e t e space X . The m a t e r i a l presented i s d i v i d e d i n t o four chapters. In Chapter I , we discuss c e r t a i n subsets of can be C*-embedded i n other subsets of gN-N conclusion that no proper dense subset of . gN-N gN-N which This study leads to the can be C*-embedded. In the second chapter we devise a general method of a s s o c i a t i n g c e r t a i n classes of points of P-points of gN-N gN-N w i t h c e r t a i n subalgebras of C(N) .. The form one of these c l a s s e s . The answer to R. S. Pierce's question, "Does there e x i s t a point of gN-N which l i e s s i m u l - taneously i n the closures of three pairwise d i s j o i n t open s e t s " 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 Hypothesis. , without the use of the Continuum Ill ACKNOWLEDGEMENTS I wish to express my appreciation and gratitude to my advisor, Dr. Timothy Cramer, f o r suggesting the t o p i c and . o f f e r i n g invaluable assistance i n the preparation of t h i s t h e s i s . I also wish to thank Dr. J . V. Whittaker f o r reading the thesis and Mrs. M. Wang f o r s k i l l f u l l y typing i t . To National Research Council of Canada go my thanks f o r t h e i r f i n a n c i a l assistance. iv TABLE OF CONTENTS Page INTRODUCTION CHAPTER CHAPTER 1 I. ON THE EMBEDDING OF CERTAIN SETS IN I I . AN APPLICATION OF B-SUBALGEBRAS OF BN-N CHAPTER I I I . CHAPTER BN . . . . . . C(N) TO . c-POINTS OF 4 19 gN-N 32 IV. TYPES OF ULTRAFILTERS AND THE NON-HOMOGENEITY OF BIBLIOGRAPHY BN-N 41 60 1 INTRODUCTION Our purpose i s to i n v e s t i g a t e the i n t e r n a l s t r u c t u r e of gN-N . We do t h i s w i t h the help of the major r e s u l t s of the l a s t ten years concerning t h i s t o p o l o g i c a l space. i n connection w i t h gN Most of the pre-1960 r e s u l t s can be found i n the Gillman and J e r i s o n t e x t , {4], and, when needed, we simply quote these r e s u l t s without proofs. Some of them w i l l be e a s i l y obtained from c e r t a i n theorems which we discuss here. the space In some cases we do not r e s t r i c t our considerations to N , simply because such a r e s t r i c t i o n does not decrease our work appreciably. I n such cases we discuss an a r b i t r a r y d i s c r e t e space and f i n a l l y apply our r e s u l t s to the space N . Much of the work i n the f r i s t three chapters depends h e a v i l y on the continuum'hypothesis. Thus i t i s expected that one of the major techniques used i n these chapters i s the i n d u c t i o n process, as i s indeed the case. The terminology and notations we use are those of the Gillman and J e r i s o n t e x t , [4]. However we would l i k e to make s p e c i a l mention of the f o l l o w i n g notations: D* = gD-D where D the continuum; i s a d i s c r e t e space); , for f continuous extension of f extension by N* = gN-N (or, more g e n e r a l l y , c w i l l denote the power of a f u n c t i o n on N , w i l l denote the to gN . We sometimes denote t h i s continuous f * . Unless otherwise stated we denote " c l o s u r e " by "c£" and " i n t e r i o r " by " i n t " . t o p o l o g i c a l space X The s e t of a l l continuous functions from the i n t o the t o p o l o g i c a l space numbers) i s denoted by R (space of r e a l C(X) and the s e t of a l l bounded functions i n 2 C(X) is denoted embedded in in . C(X) C*(S) Z(f) set {x of i f every be e X is . function to a function : 0} , and f(x) = in X a is itself. are not is set C(X) of points where A-points gX-X and types of a in X of I. We is of a 8N by to bs terms of a proved . We every H e r e we with then from subset point in a of of 2° C(N) which we 2 be X . Z(f) a • is said to be extended to a i f C*(X) Z(X) every For f The Cfunction function e zero-set C(X) of , X complement the necessary in we . of let The a certain here condition that and sufficient The r e s u l t s of t h e C*-embadded. introduce of regular . and that We dense the a defined subsets method subalgebra space. We the of show that conclude in the of N* the set that A of c a l l the these C*(N) known associating A-points P-points well for is of algebra points is one gN-N result the dense of these are that i n the gN-N P-points. III. gN-N type prove cardinality subalgebras dense II.- completely Chapter that c a l l can in investigate finally gX-X is gX-X C*(N)-points, has we in X C*-embedded. Chapter a C(S) of cozero-set. formulated:in It S C*-embedded denoted conditions f o r subsets of chapter in extended Chapter are subspace is called chapter A S zero-sets zero-set C*(X) Similarly can = X by This is is a contained short chapter simultaneously in i n which the we prove closures 3 of c disjoint open sets Chapter introduce a set ultrafilters the producing continuum on IV. T , the N . We relation. hypothesis of we BN-N We discuss elements then Using prove . of which define this the types a we of c a l l certain concept ultrafilters and / types relation without non-homogeneity \ the of and of on (j) assuming BN-N . the T free called the 4 CHAPTER ON 1.1. THE EMBEDDING The in gN prove that that i f from which we gN-N are assumed Fine to and be p e of be able to C*-embedded. be this then 3N-N spaces OF CERTAIN C*-embedded shall not dimensional 1.2. can purpose i n completely of SETS chapter in other conclude that our to not The construct, certain of We gN proper A l l results . C*-embedded concepts proofs. regular. gN subsets is the IN is gN-N-{P} We u s e most I of dense spaces here shall in and of zero- considered are due also 3N-N subsets F-spaces sets are mainly to Gillman. Definition. X is called an F-space It is proved in [4], i f every cozero-set i n X is is - C*-embedded. locally compact F-space. N and is an a-compact example We n o w he S families there exists define are the n+1 partition of U e order of such *LL- members of "IL of X we S C We to say that to be i f . a space then X gX-X is a compact a-'space. that V ^ U zero-dimensional. ~V refines . Let largest non-empty a that compact define the with mean not such proceed sets. "It of but 14.27, finite i f H be integer a n collection of for cover for intersection. Let By and every of X which an disjoint V . e We there open-closed open and • 5 closed cover 1.3. subsets of S of of Definition. cover of whose order A X S a X that completely separated in such 0 _< f , x e B . It any two and closed is _< 1 proved completely sets is said refinement Recall that X zero-dimensional. S . a is 0 Thus a partition X zero-dimensional open-closed A of a l l x that are B is in e the A , a and S is every in X a . X f C ' (X) in are are 5 C = 1 for disconnected;that a l l equivalent: complementary 6.10, basic space f(x) following totally [4], of function contained ; proved a i f p a r t i t i o n of and exists 16.17, also be for sets space It an there = [4], to subsets i f f(x) in by two separated for is zero. space has union gN open X is is totally disconnected. For p in gY such intersects 1.4. Y that If In realcompactness two Z e space every Y , let zero-set rY i n denote gY that the set of contains p a l l points also . Definition. exists any rY Chapter as Z(gY) definitions = Y are , II, follows: such then that Y we Y shall is p equivalent. is e said find to i t realcompact Z <Z Y* Let the . We f i r s t be realcompact. more convenient i f for shall every now definition show be to p e define Y* that I there these and the 6 / second be II. I implies II : p £ rY . so f o r some z e r o - s e t Z c Y* y and so £ BY - Y that p e Z c Y* p e Y* , then containing II implies p , I : p £ Y . then o b v i o u s l y . Hence Let y e rY . Z n Y = <j> , a c o n t r a d i c t i o n . every z e r o - s e t Thus containing y Now intersects Y . lemma i s needed both i n t h i s chapter If zero-set in Remark. BY-Y Y BY-Y such if y e Y Hence zero-set i n t e r i o r ; f o r then i f t h e r e e x i s t s p e Z^ c l o s e d under c o u n t a b l e and zero s e t has non-empty Since . in Z^ , d i s j o i n t Since from zero s e t s a r e zero-set, c o n t r a d i c t i n g the f a c t t h a t a non-empty interior. Y i s l o c a l l y compact then f o l l o w s from t h e ' f a c t t h a t Thus i n t ( Z ft Z^) = <j> Z p e Z - cl i n t Z, i n t e r s e c t i o n we then have a non-empty empty i n t e r i o r , l o c a l l y compact. The f o l l o w i n g i s the c l o s u r e o f i t s i n t e r i o r . has non-empty i n t Z , such t h a t subset Z i s l o c a l l y compact and realcompact, then, each then from l o c a l compactness t h e r e e x i s t s a z e r o - s e t Proof: Then and i n the n e x t . I t s u f f i c e s to prove t h a t a non-empty Z^ f) Z , w i t h Hence Y = rY . Hence the two d e f i n i t i o n s a r e e q u i v a l e n t . 1.5. Lemma. Thus Z 0 Y = <j> . f o r i f i t were then t h e r e would e x i s t a z e r o - s e t y e Z c Y* Y C rY . . Z If Y Y i s open i n i s open i n BY of a compact space, hence compact). BY BY-Y i s compact i f and o n l y i f Y thereby making Thus BY-Y BY-Y (this is a closed i s C*-embedded i n 7 3Y by [4], 6.9(b). Z = Z(f) - Y Thus f o r Z an a r b i t r a r y f o r some f e C*(3Y) . L e t p e Z . By realcompactness p £ rY . Thus there e x i s t s a f u n c t i o n but so g(y) ^ 0 f o r any p e Z(h) . Also zero-set i n 3Y-Y , g e C*(3Y) y e Y . L e t h = | f | + |g| Z(h) C Z . Thus g(p) = 0 such that . Then h(p) = 0 p e Z(h)C Z . We s h a l l show that t h i s non-empty set Z(h) contains a non-empty open s e t . L e t 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 n be such such that |h(y) - h(y ) l < — for y e V . Let u e n ' n n C*(3Y) that u(y^) = 1 u(y) = 0 f o r every n , y eY - V n (such a f u n c t i o n function on u e x i s t s because f o r every f ^ which i s equal to 1 . Hence q £ 3Y-Y 0 i s such that everywhere u(q) 4 0 . , This, means that every neighborhood of compact sets there e x i s t s a a t y^ and equal to Y-V . L e t u = \ f ). Suppose n n q meets i n f i n i t e l y many of the h(q) = 0 ; one may j u s t i f y t h i s as f o l l o w s : since every neighborhood of q meets i n f i n i t e l y many sets , then we can f i n d a net {S } such that each S is in V and {S } converges n n n n ° to q . Since h i s continuous we have l i m h(S ) = h(q) . Now n-*° |h(S ) - h(y )| < — ' n n n w 1 . So l i m h(S ) = 0 . Thus n h(q) = 0 . Hence f o r n N any point q e 3 Y - Y such that (BY-Y) - <£ Z(h) ,. Also u(q) $ 0 , we have (3Y-Y) - Z(u) h(q) = 0 . Thus i s a non-empty open subset 8 of BY-Y . Since 1.6. Remark. Z(h)<r Z we conclude that has non-empty i n t e r i o r . Local compactness i s a c r i t i c a l property of the space i n the above lemma. Y We can prove t h i s 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 . R Z Define a real-valued f u n c t i o n h on as f o l l o w s : , X _< TT , X _> TT h(x) = Let h have an extension (1) Z(g)/J Q = (j> (2) Z(g) =f ij) g to . BQ • Then (there e x i s t s a sequence such that We would l i k e to show that Int { } converging to IT a n a e Q . Then n Cl„Au 3Q ~Z(g) = <j> . Assume n }^Z(g)) . x e Int BQ-Q Then there e x i s t s some b a s i c open set Z e Z ( Q ) such that Q BQ - ci Z) . Since i s dense i n BQ then s > 0 i n BQ where (See remark at end (3Q - cSL Z) (BQ - cl Z) /? Q 4 <j> . Hence there e x i s t s an element y f i n d some (3Q - cl Z) x e (3Q - c i l Z) - Q C Z(g) . for the existence of such that such that y \ Z y e (y-e, y+e) but and i s open and since Q - Z f cj> so y £ Q . Thus we can (y-e,y+e) f) Z = (j> Note that t h i s implies that cA .(y-e,y+e) P c% Z = cfi Q pQ From above g{ [cH g(z) = 0 (y-e,y+e) ] - Q} f o r every = 0 ~Z(g) . 3Q-Q z E BQ - cA Z - Q . Hence since we'have shown that 9 c£ _(y-e,y+£) f) c£ Z = <j> contradiction. such that . But by d e f i n i t i o n of h (For y ^ ir since ir sj: (y-S,y+6) this i s a y e Q . Hence we can f i n d a 6 >0 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 c o n t r a d i c t i o n since we s a i d that g{[c£ (y-£,y+£)] - Q} = 0 ) . 3Q Remark. We would l i k e to show that a basic open s e t i n gQ i s the complement of a b a s i c closed set. From [4], 6.E(1), we know that basic closed sets i n gQ are of the form Now l e t O be an open set and p £ O ment of O . Then O a closed s e t F and O' c F X c F = &gQ c z where z . Let Q 1 i s closed i n gQ such that ^gg denote the comple- and p £ Q' f ° some Z e Z(Q) . SO there e x i s t s , Z e Z(Q) and p £ F r ( t h i s i s by d e f i n i t i o n of basic closed s e t s ) . and so we have, since Thus O D / _t p ^ F , p £ gQ - c^^^ Z c 0 . Hence the complements of b a s i c closed sets are basic open s e t s . 1.7. Remark. then If Y i s l o c a l l y compact and realcompact but not compact, 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 s e t S has open closure then the complement of S , Z and so Z say, has open closure. intZ i s closed 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 know that Thus gY-Y i s a P-space by [ 4 ] , 4 J . We already gN-N'' i s not b a s i c a l l y disconnected (see [4], 6W) and gN-N i s not a P-space. i 10 Given a space X , let SC X and p e X - S . Our other results' w i l l be formulated i n terms of the f o l l o w i n g c o n d i t i o n which we denote by (p, S) . a cozero-set HC Remark. (p, S) : There e x i s t a neighborhood S (S D V) - H such that V of p and has empty i n t e r i o r . This c o n d i t i o n holds i f p ^ & S , and i f S is a c cozero-set, then the c o n d i t i o n holds f o r a l l p \ S . 1.8. Lemma. Let F be a compact set i n X holds f o r a l l p e F . Then Proof: {V.,, i intZ C F c Z such that (p, X - F) f o r some zero-set Z . We s h a l l show that there e x i s t a f i n i t e open cover V } n such that of intZ fc F and zero-sets - F C X - V holds then f o r every and a cozero-set p e F H C x - F P Consider a l l such sets can l e t {V : p e F} P P subcover, l V p n V containing f o r k = 1, ..., n . Since fc such that i n t [ (X - F) n V P F (p, X - F) there e x i s t s a neighborhood of p , V p , - H ] = <}> . P f o r p e F . Without l o s s of g e n e r a l i t y we be an open cover of P , V Z , Z„, .... Z 1 2 ' n F so there e x i s t s a f i n i t e n . We s h a l l now c a l l t h i s f i n i t e k subcover P V^, V n . Similarly l e t H complement of H^ H C X - F . Also k be = , k = 1, n . Let the , k = 1, .. ., n . Then Z^.- F 5 i n t [ ( X - F) 0 V O Z ] = <J> . Hence no element fc fc V, i s an element of i n t Z, - F , i . e . X - ( i n t k k i n t Z. - F C (X - V, ) . Let Z = D Z, . Since k k , k k then Z i s a zero-set containing F and i n t Z v 0 since 0 Q of Z, - F) 3 V, and so k k Z, 3 F f o r a l l k k - F c X - (j V . k k 11 But X-i/V. k int Z F c X - F k . X - F . Hence - F c r X - l ' V 0 FC U v k since Thus c£(int Z Q - F) >') F = $ i s compact, then from [ 4 ] , 3.11, we conclude that c&(int Z int Z - F) 0 - F 0 0 containing f F 0 • Theorem. then are completely separated. f Let X intZ - F . Hence Q Z 0 be an F-space, Sc X V of and . p e cA S - S . I f p , and i f (p, S) H C S such that S fl holds S fi V i s open. a cozero-set V H c S p e cJl S . C l e a r l y p . Also int[S O V contains an open set U 0 (S Pi V) Pi O and non-empty since neighborhood V Also there e x i s t s a neighborhood such that 0 - H] = <j which contains Q /T V ^ <> j of p i s open and d i s j o i n t from i n t ( X - H) . We know that there e x i s t s a neighborhood since and We s h a l l show that there e x i s t s a neighborhood and a cozero-set holds. F i s C*-embedded i n S O {p} . Proof: that and So the required zero-set i s Z H i s open f o r some neighborhood S F This 'means that and d i s j o i n t from fi Z ) C F C ( Z P) Z ) . S fi V Since are contained i n d i s j o i n t zero-sets, i . e . there e x i s t s a zero-set •Z^ int(Z . k and O nV V of 0 since p p of such p and (p, S) and O /Q S ^ <> j i s a neighborhood of i s open, being the i n t e r s e c t i o n of two open s e t s , If) O Is a neighborhood of p . The required is V / ? Q Now l e t f e C*(S) and g e C*(H U {p}) be an extension 12 of f|H where f|H i s the r e s t r i c t i o n of f to an F-space so every cozero-set i s C*-embedded). on S 0 {p} H /? V i s dense i n , x e S , x = p (S U {p})/0 V f . Thus S 1.10. Theorem. S S G h X . . £/ {p} S C X Then i s a cozero-set i n i s C*-embedded i n S for a l l a S . We s h a l l use a well-known technique of i n d u c t i o n i n t h i s We can assume that S = S a . S and that Since every cozero-set i n J i s an F-space f o r a l l a . For given S C S c n 0 a < e C"(S^) assume that f o r every £ < a Let g e C*(G) g^ S a . ... 1 i s C*-embedded then Let g = g| (G /') S ) . has been extended to a f u n c t i o n , and that f 0 f ^ <£_ ... (note that we can make such an extension because (G H s ) i s a cozero-set i n so h , i s an F-space. Consider any t (follows be a union of • a S h i s a continuous extension be an F-space and l e t and i f G fl S^ Proof: proof. is S O' {p} . C*-embedded i n X in a if G C S then (b) is Let cozero-sets (a) X Define a f u n c t i o n = (S /"> ± from the f i r s t part of the p r o o f ) , then of (note that such that • r f(x) h(x) = \ ^ g(p) Since H (G H S ) which takes i s a cozero-set i n S-(G f~i S ) ^ S S and because there e x i s t s some f u n c t i o n to zero and - G) C" s - (G /) S ) . (S S ^ Thus 13 h takes (S _ - G) cozero-set i n to zero a l s o . Nov? h (S^ - G> . Now use the f a c t that the bounded f u n c t i o n g makes (G O S _) a i s an F-space to extend G /] S on the cozero-set Q> to f s on S s> s Also note that we are t r e a t i n g each f u n c t i o n f_^ as a set of ordered pairs). The f u n c t i o n U f [i e £ a £<a i s w e l l defined and continuous L S_ U (G /) S ) . £<a the cozero-set on To show that i t i s continuous ; we K 0 consider any open set i n the range of ( L- £ ;j g ) "*"(o>) = (V f £ a £ £<a £<a f and g £ (J g . Then s iJ g "*"(£>) which i s an open set since the a 1 functions O f £<a are continuous. L/' Also l) (G S a S ) a £ £<a cozero-set because i t i s the countable union of cozero-sets. i t i s a cozero-set i n the F-space Hence f a (U f n g ) >£ a £<a e C"(S ) . Now a S . Thus G In f a c t ( r e c a l l that S c S c a 1 has a continuous extension to a f u n c t i o n U f S 0 i s a continuous extension of a i s C*-embedded i n n g is a ... ). to a l l of S . \ (b). (a) since i f G That S i s an F-space follows q u i t e e a s i l y from i s any cozero-set i n S , then G f) S i s a cozero-set a in S for a l l a and so G i s C*-embedded which implies that S is an F-space. We now apply t h i s theorem to open subsets of".gN-N statement. are F-spaces. We can show that gN-N gN-N to conclude that a l l The f o l l o w i n g i s a proof of t h i s 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 w r i t t e n as an i n t e r s e c t i o n of a set of closures of zero-sets i n N . Thus every open set can be w r i t t e n as a union of cozero-sets. As already mentioned, gN-N Hence our theorem applies w i t h Remark. Let embedded i n gN such that f*(Z) = 0 sets i n gN-N and an open set i n gN-N . be a zero-set i n gN-N . Since and so Z ± Z , where i s a zero-set i n gN Z Let X Z^ S i s C*f * on also. gN By [ 4 ] , r zero- i n Z(N) . be an F-space having j u s t 2 be open and l e t p e c£ S - S , and suppose that Then gN-N e Z(N) . Hence there are A ± s i n c e there are jV^ sets 1.11. Theorem. S there e x i s t s a bounded continuous f u n c t i o n Z = f) ci lew 6E(3), S Z X = gN-N i s an F-space. zero-sets. Let (p, S) fails. i s not C*-embedded i n S U {p} . Proof: Let {S_}_ * £ £<w be a family of cozero-sets i n X whose 1 union i s S a function (such a family e x i s t s because f o r every f P such that Consider the set of f (p) = 1 P Z(f ) ) . Let and {V>.K there e x i s t s f o r a l l x e X - S. be a base of zero-set- 55<a) p neighborhoods of f (x) = 0 P p e S 1 p . We proceed i n d u c t i v e l y as f o l l o w s . For each a < a)^ , assume that cozero-sets been defined f o r every A and £ < a . Since i n t [ S f) V a - U. (A .?«* 5 B , contained i n S , have (p, S) B 5 r f a i l s then U SJ] + K 15 (note that S i s also a cozero-set i n non-empty cozero-sets S n V - a c7 (A,. U >? G = A L> B . G 1 S 5 A (/ S J 5 B Thus we can choose d i s j o i n t . contained i n . Define A =.W A Hence, by c o n s t r u c t i o n , f o r each = (A L' B) /) S„ = I and S) . • K Theorem 1.10, G <--• (A a B ) /) a K a ± K i s C"-embedded i n d i s j o i n t since the sets a , B = £ < a l> B c , , which i s a cozero-set. S . Note that A and B So by are and the sets B are. Hence A and a a are complementary open sets i n G and by c o n s t r u c t i o n of the sets and B Thus a A G U {p} we conclude that both and B since B A and p G because A G ; hence A and B A and A' and G i s C"-embedded i n S). but not i n (they are B' C ' -embedded 5 c in i s not C*-embedded ( r e c a l l that / We now apply Theorem 1.9 and Theorem 1.11 3N-N. to the space to conclude the f o l l o w i n g : 1.12. Let C"-embedded i s S C (BN-N S V {p} p. are contained i n d i s j o i n t z e r o - s e t s ) . S 0 {p} i s not A A' /? B' = <j> i n the in S A i s contained i n the zero-set G Hence G B Thus from Urysohn's Extension Theorem, [ 4 ] , 1.17, G V {p} . B meet every neighborhood of i s i n the closures of both i s contained i n the zero-set subspace B are completely separated i n the subspace completely separated i n and A be open and i f and only i f p e ciL S - S . (p, S) holds. Then S is 16 1.13. Theorem. Let K be a compact F-space such that K = 3Y-Y where ..., Y i s l o c a l l y compact and realcompact and such that zero-sets. (a) (Such a space K i s 3N-N). K has j u s t o 2 Then The f o l l o w i n g are equivalent f o r an open s e t , (i) (p, S) holds f o r a l l p e K - S . (ii) S 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: zero-set ( i ) implies ( i i ) : Z By Lemma 1.5 we see that f o r any i n K , ck i n t Z = Z . Since p e K - S , then (p, K - (K - S)) holds f o r a l l p e K - S , i . e . (p, K - F) holds f o r a l l p z F where compact). in (p, S) holds f o r a l l F = K - S (note that F is We now apply Lemma 1.8 to conclude that f o r some zero-set K , i n t Z c F C Z . Hence Z C F c Z . Thus F ci i n t Z c. cZ F c cJl Z i s a .zero-set, i . e . which we conclude that S K - S Z and so i s a zero s e t , from 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 (b) ( i i i ) implies ( i ) follows from Theorem 1.11. Since no point of K i s i s o l a t e d , then no point i s a zero-set; f o r i f a s i n g l e t o n were a zero-set then, as already proved, t h i s 17 s i n g l e t o n would have non-empty i n t e r i o r c o n t r a d i c t i n g the f a c t that has no i s o l a t e d points. K 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 that D D be a proper dense subset of K . We must show i s not C*-embedded. Assume that D i s a proper subset there e x i s t s an element above that D D f , which i s a (K - {p}),i.e. f e C*(K - {p}). has no continuous extension to K . Consider continuous function on Since p e K - D . We have proved K - {p} i s not C*-embedded so there e x i s t s bounded continuous f u n c t i o n on f i s C*-embedded. such that f|D . This i s a 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 c o n t r a d i c t i o n 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 t h i s theorem to p e gN-N, subsets of gN-N-{p} gN-N gN-N we conclude that f o r i s not C*-embedded i n gN-N and that proper dense are not C*-embedded. Before ending t h i s chapter we make the f o l l o w i n g remarks. 1.14. Remark. Let X 2' card(X - S) < 2 be a compact F-space and l e t 0 . Then S i s pseudocompact. SC X be such that 18 Proof: Assume S i s not pseudocompact. unbounded real-valued f u n c t i o n and so (see [4], 1.20). C"-embedded so X - S D gN-N Hence S N Then S admits an i s a closed subspace of By [4], 14 N(5), every countable set i n an F-space i s N i s C*-embedded i n X . Hence A'" 2 ° and since card(gN - N) = 2 c£ N = gN . i s pseudocompact. 1.15. Each pseudocompact subset i s of c a r d i n a l i t y Proof: gN-N . P of gN which contains N onto . W , where <> j be a mapping from Let be extended continuously to a mapping then W C <}>*(?) • (j>* of i s closed i n gN into W i s the c> j can [0, 1] . Since Since the continuous image of a pseudocompact space i s pseudocompact - see [4], 1G - then <j>*(P) N j> 2 ° . set of r a t i o n a l s i n £0, 1] . By the c o m p a c t i f i c a t i o n theorem But Now we have a c o n t r a d i c t i o n . Again t h i s remark can be applied to N C P S <J>*0?) i s pseudocompact. [0, 1] (since pseudocompact i s equivalent to countably compact i n a normal, T^ space. But since [0, 1] is first countable then countably compact i s equivalent to compact and since [0, 1] i s Hausdorff, compact implies closed). c£ W C c£ <}>*(P) = <S>*(P) , i . e . and so card P > 2 0 Now since 10, 1] c «|>*(P) . Hence W c <}>-(P) then c [0, 1] = , 19 CHAPTER I I AN APPLICATION OF 2.1. g-SUBALGEBRAS OF C(N) TO gN-N In [10], W. Rudin has shown, using the continuum hypothesis, that gN-N has a dense subset of 2 P-points. now i s to prove a more general r e s u l t than t h i s . regular space called X associate a set of points i n r e s t r i c t i o n s on of 2 A and g-subalgebra A , we s h a l l c a l l e d A-points. we s h a l l show that gX-X C(X) , Then w i t h some has a dense subset A-points. Rudin's theorem will then be e a s i l y obtained by proving that C*(N) is a are the C*(N)-points. space X gX-X purpose For a completely we s h a l l define a c l a s s of subalgebras of g-subalgebras, and w i t h each Our N g-subalgebra and that the P-points of gN-N I t i s not necessary to r e s t r i c t our work to the since t h i s r e s t r i c t i o n does not s i m p l i f y our work much. we work with an a r b i t r a r y completely regular space Hence X . Most of the r e s u l t s i n t h i s chapter are due to D. Plank (see [ 8 ] ) . Note that under pointwise operations are algebras over and a l TR ( r e a l numbers). C(X) and C*(X) We note also that i f f e C(X) denotes the one-point c o m p a c t i f i c a t i o n of IR , then by the c o m p a c t i f i c a t i o n theorem there i s a unique continuous f u n c t i o n 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 x\ r e s p e c t i v e l y the s e t of xi prime i d e a l s and maximal i d e a l s i n A , where A C(X) i s a subalgebra of . We now define the h u l l - k e r n e l topology on )° xi and M . In xi general the h u l l - k e r n e l topology on a c o l l e c t i o n o7 of prime i d e a l s i n a commutative r i n g A w i t h 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. P r o p o s i t i o n . The sets E. (a) = {P ec7 : a e P} 7 where a e A , are o/ closed and c o n s t i t u t e a base f o r the closed sets i n vJ subalgebra of Proof: C(X)). (i) To show that the sets of the form closed i t s u f f i c e s to show that E^Ca) = {P z'3 a £ P . Since a E f] {P zrj E^(a) : Flfl E (a)J : a £ P} then E.(a) are C/ E^(a) C. E (a) . 7 . Let P e E. (a) . We must show that PO/TE. (a) =/HP a e P £ iJ : a £ P} and since and so P £ E. (a) . Hence the sets are closed. (ii) To show that these sets form a base f o r the closed sets consider S (A i s any i s closed then P 0 £J and S = ~S and so S closed such that P Q £ S . Since S =' {P e J : PO/0 S} . I f ; fl S = {a} , a £ A., then we are f i n i s h e d since there would then e x i s t an element C which i s closed and of the form E (a) (take C = S) such 21 that SC C I f fi S a' £ P and P D fj: C (note that |C 0 because {a} =/? S c|- ° ) . p i s not a s i n g l e t o n then consider one element 0 P O .OS 0 and so P would be an element of 0 such that thereby g i v i n g P i n ,// such that because a' £ P 0 and so D a' e Q i s closed. . The proof i s now complete. We s h a l l put the h u l l - k e r n e l topology on s h a l l introduce a family .ofy Now a' e P . C a l l t h i s Q e D . We have already shown that £D 0 P S , a contradiction. S C D since f o r any Q £ S , Q D f] S D . Then Clearly a' £/?S . Note that there must e x i s t one such element; otherwise consider the family of sets set P , and we of prime i d e a l s 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 s t a t e 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 (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 i d e a l i n C(X) associated with the point P e gX ( c a l l t h i s maximal i d e a l =• {f e C(X) : (fg)*(p) ). Then = 0 for all g e C(X)} . Gillman and J e r i s o n have also proved i n [ 4 ] , 7.2, that f o r all M L = {f e C*(X) : (fg)*(p) = 0 Li g e C*(X)} and that the * mappings p and p that the mappings p -> onto the maximal-ideal the mapping p -> M^ 1 M^. are one-to-one. and p -> spaces I t i s proved i n [5] are homeomorphisms of gX and >' ,. . We s h a l l prove l a t e r that 7 1 r i s continuous f o r any subalgebra A <^ C(X) . The 22 above characterizations of M and P M ,. 5 suggest the f o l l o w i n g of C(X) P definition. 2.3. D e f i n i t i o n . M P For any subalgebra = {f E A : (fg)*(p) = 0 2.4. D e f i n i t i o n . A f o r a l l g E A} Let .*# =' {M P A : p e gX} and f o r p e gX l e t . . We now prove the following theorem i n order to show that .o^ may be given the h u l l - k e r n e l topology. mentioned we see that 2.5." Theorem. = M Proof: &i.' . =))','.,. 5 For p £ gX , M rJ • i s a prime i d e a l i n A P <j> f M For p £ gX , By the r e s u l t s already $ A P because OE M A I t i s c l e a r that M P f, g e A d e f i n i t i o n of. M P such that there e x i s t f \ M and P h, k s A fg £ M f e M 2.6. D e f i n i t i o n . of C(X) T a a g e M P : gX -> '-^ by A subalgebra if T or P A of C(X) i s a homeomorphism of g \ M such that (fghk)*(p) i 0 . Hence Define I i M A (gk)*(p) i 0 . Thus then e i t h e r and P . A i s an i d e a l i n A . To show that i t i s a prime i d e a l consider P P T (p) = M M . Then by (fh)*(p) ^ 0 fg \ M . Hence P and Thus i f P i s a prime i d e a l . P P A i s said to be a gX onto JTI . g-subalgebra 23 R e c a l l that we s a i d that of onto r,i gX and ))1 r , thereby making C(X) J( are homeomorphisms and C*(X) g-sub- 0*^ Ci algebras of t and C(X) . Before we introduce the concept of A-points we would l i k e to define a c e r t a i n class of sets and show that they are a base f o r the closed sets i n gX . Let T.^ denote the inverse map. For f e A let A = {p e gX : f e M^} A E^p (f) "A Now . because we have shown that <z$'^ by d e f i n i t i o n , {p £ gX : f £ M^} I t makes sense to speak of is = c o l l e c t i o n of prime i d e a l s . a Z((fg)*) where, as usual, geA Z denotes a zero-set. S (f) i s closed i n A 2.7. Theorem. (a) T Let : gX T. gX . A be a subalgebra of C(X) . A f s Then / i s a closed mapping i f and only i A Proof: gX , implies that i s continuous A (b) But zero sets being closed i n i a Hausdorff space. A (a) R e c a l l that we have shown that the sets of the form E.,-. (f) are b a s i c closed sets f o r Q5\ . Also , A A closed i n gX . Hence T i s continuous. t v A -1 T. A (f) = S (f) A is 24 (b) Then F,= Let be a closed mapping. {x £ gX : T ( X ) = T ( p ) } A are d i s j o i n t compact subsets of closed sets being zero sets i n g e A} so and A gX F = {x : (fg)*(:x) =0} A d i s j o i n t open neighborhoods and U = gX - ? V ( g X - U)] 0 A A i s closed, then q e V c V , 0 T.(U ) and 0 U U and 0 and V and K and V be T Hence ^ gX and p e U . Since T A U , 0 Then c l e a r l y of and A i s Hausdorff. i s Hausdorff then we use the f a c t i s continuous to prove that t A i s closed. Let B be a A closed subset of T are r e s p e c t i v e l y and l e t = ^[^(Vo)] c are A Conversely, i f Hence K are d i s j o i n t open neighborhoods 0 K F A respectively. that Now l e t U are open i n 0 and = gX - ^ [ ^ ( g X - V) ] . 0 = T^t^CUo)] , V 0 T (V ) A V F F , and therefore gX ). A = {f e A : (fg)*(p) = C> A closed subsets of a compact space of . K = {x e gX : T ( X ) = T ( q ) } (compact because gX ; M f Suppose ^^) -^ s gX . Since compact (T of a Hausdorff space T.(B) a gX i s compact then i s continuous). i s closed. B i s compact. Being a compact subset ( I t i s easy to show that a A compact subset of a Hausdorff space i s closed). algebra A , Remark. If T : gX A' i s a homeomorphism then -1 - ^A (f) {S ( f ) : f. e A} A forms a base f o r the closed sets i n f g-sub- i s a Hausdorff space. A V> Thus f o r a gX . This i s so because and the sets of the form E ( f ) , f £ A , form a 25 base for the closed sets i n A * / . > s already shown. a A Now l e t A that x be a 8-subalgebra of C(X) . This means i s a homeomorphism and so A {S.(f)' : f e A} closed sets i n 8X . For f e A , define S*(f) = S (f) /) X* , where A X* = 8X - X . The c o l l e c t i o n # A = {S*(f) : f e A} f o r the closed sets i n X* . Let S*(f) i s a base f o r the A fiS*(f) i s c l e a r l y a base denote the boundary of i n X* . 2.8. D e f i n i t i o n . Let A be a i s c a l l e d an A-point of X* 8-subalgebra of C(X) . A point p e X* i f , f o r a l l f e A , p { SS*(f) . Thus A the set of A-points i s the s e t P (X* - 6S*(f)) , x^hich i s an i n t e r feA A s e c t i o n of a family of card A , dense, open subsets of X* . We s h a l l now prove an existence theorem f o r A-points. F i r s t we need the f o l l o w i n g d e f i n i t i o n and propositions. 2.9. A space X non-void G -subset of X o has a non-void i n t e r i o r , r 2.10. Proposition. the Let Y Gg-property. i s said to have the G„-property i f every be a non-void, l o c a l l y compact, I f 73 i s a family of at most p o i n t s , then card l 0 space with co^ dense, open subsets of Y , then f ) D i s dense i n Y . Also i f Y u T has no i s o l a t e d 26 Let AJ = {U : a < co.. } a 1 Proof: i n Y , 0 Z > '1 Q, f cj) . We proceed an a r b i t r a r y non-void open set G i n d u c t i v e l y as f o l l o w s . a collection Let a < a>^ , and suppose that there i s defined {V^ : B < a} of non-empty open sets i n G y If a = y + 1 (i) V Y i if). We have x e C ] V„ 8<a Hence H x s for a l l x < t then we s h a l l show ^.V„ = V B<a B 0 Y 0 g B x x 3 .. I f B = Y then Y (ii) B<a V„ = V 0 B 3 = ^.c£ B<a Y . Hence we have V: D V 8<a V . Since 3 since y < V C cj> V Y Now i f x £ cJL, V„ Y x < 3 all T < a , i.e. X E V • But x < a 3 3 . Thus ^ 8<a a • for a l l n Y V„ = V 3 then . Y i s a l i m i t o r d i n a l then , then B<a fl V c O 3 B<a cJL. V Y ^ 3<a R 3 x £ V for x x < x + 1 < a and so f o r a l l 3 < a . Hence 3 V 3 V f o r a l l 3 < a , then implies that B < Y I R B Y 3 all 8 _< y • Y We s h a l l now show that i f a V for a l l Y B < a which implies that N x e V 0 3 Now l e t B < ct . This implies B , f o r a l l x < 8 . Let ' V„ c V B<a V.D V (and of course Y x e CJL. V f o r a l l B < a . Thus • , x < a which implies that x £ Y x < B. Hence V fl G V. C cJL, V c U O V c V . Then 6 U c G with compact B < ct closures such that f o r any c£ V . I t s u f f i c e s to show that f o r x £ V cl V 3 for x C f) V . 8<ct 3 27 Hence H V. 3<a being a.decreasing (that let {eft 3 < n V Y 3 : P 0 • 0 family 3 < ot} Then V v JL_ V . • Y V C 0 2 . has . But V Y and dense open a by set non-empty c L Y V hypothesis . open has case, And set <cU D i n V J C a a 3 a 3<a a V Si non-void since in Y n G U is Y compact and i n sets cJL r such with V a way such . the with that finite a, < Therefore ( c i , x V a : . a . 0 as a which G.-subset intersects compact is ' {V < follows: Hence non-void cl^. that , 6 locally : a w, } a is then there compact < co, } 1 a the exists and is defined collection of 1 intersection a„ V seen 3 interior is < a} 1 * n inductively c L is 0 3 intersection easily <r 0 : 3 1 3<a of V„ non-void family 1 either {cJL, Y sets, 0 But.' 3 decreasing ' in C compact a L Hence, 0 3<a of is c£ = 3 property (for consider cl^V^ Then 2 c L V C cJL, Y V a 0 U 2 a a C U V ) c 3<a H G a Si ( 2 for C V \ 3 2 a l l a (OH)n C cJL. l a . 1 the is dense i n Y Now i f Y construction closures. To see of the this , and such that Hence ^ o a<w T h u s /? ^ ) l a < to, G V c£ v + <j, . v Y a 1 . has sets no V consider isolated a. we the points can form two V that we a then V a have at 's each with already stage of disjoint constructed 28 and V an ,, ct+1 such pick ci V element y x that e V ,„ C cn-z V x - a . e By V ' .. crl-1 V' crt-1 . a e and Then ' V local and an V' 5" cri-1 fo. set serves arZ • v open ,„ compactness V as there Consider such cd-2 the exists V that other • V J y - a e that a a set V ,. a+1 j V ' a n d a+2 we mentioned, ^1 Hence now i f Y has no isolated then c a r d f)Z> _> 2 space X realcompact that p . The proof is complete. Recall there exists already 2.11. points Z discussed Proposition. the e that Z(BX) in If a such Chapter X is is e Z C X* . i f This for a l l p definition e X* was I. locally compact and realcompact then X* has Gg-property. Proof; proved We h a v e , that a non-empty every non-empty every point there exists in a G^ in X* Z must non-empty zero-set proved zero-set in this fact, which has contain G^ is this is a a subset in Lemma 1.5 non-empty a interior. compact compact of set, set) the where G. C and we Note , by that (e.g. [4], and w h i c h 3.11(b), contains 6 C has 2.12. . Now this non-empty zero-set has non-empty i n t e r i o r ,so every G^-subset of X* interior. Proposition. If X Proof: Suppose is realcompact p is an then isolated X* point has in no X* isolated . Let points. be a 29 zero-set neighborhood of set p in gX such that z ^ / 0 X* = {p} (such a Z^, e x i s t s 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 e x i s t s a zero-set Z^ £ Z(gX) p E Z C Z^ 0 Z Thus ' {p} = Z fi Z X* . 2 2 (Z fi Z C 2 such that Z' 0 X* = {p}) and e Z(gX) , which i s impossible by [ 4 ] , 9 . 6 . This completes the 2 proof. We can now prove very e a s i l y the existence theorem we are seeking. 2 . 1 3 . Theorem. Let compact. If A X be l o c a l l y compact and realcompact but not i s a g-subalgebra of has a dense set of isolated points. Now ^ . >A : 3X • * v ' / But since A X* A . c i s dense i n has no dense open subsets of X* i s a B-subalgebra of i s a homeomorphism and so X* X* . and has c a r d i n a l i t y at C(X) card X* <_ 2 then cs.x"d A. = 2 c Hence card fi& = 2 ° . The proof i s now complete s i n c e , as we have already seen, /?& X* From the l a s t two has the G^-property and i s a family of Hence by P r o p o s i t i o n 2 . 1 0 , fi ^ 2 card A = c , then 2 A-points. propositions we conclude that T with Let Z> = {X* - 6 S * ( f ) : f e A} Proof: least C(X) i s the set of A-points of X* . F i n a l l y we formulate our r e s u l t s i n terms of 3N . 30 2.14. Theorem. A point i n X* P-point .of C*(X)-point i f and only i f i t i s a X* . Proof: X is a (We f i r s t r e c a l l that a point i f any G^-subset of X containing p from the f a c t that every zero-set i s a conclude that p p i s a P-point of X i s a neighborhood of p .) p of X i s a P-point of i s a neighborhood of G^ Thus a point i n X* containing i s a P-point of i f and only i f i t i s not an element of the X*-boundary of any S ( f ) = ,{p e gX : (fg)*(p) = 0 Also, and from [ 4 ] , 3.11(b), we i f any zero-set of X Z e Z(X*) . R e c a l l that f o r a subalgebra p . Z X* where A , f o r a l l g e A} = (1 Z ( ( f g ) * ) . In the g£A case of C*(X) extension of a point p we get f of X* X*-boundary of X* is a to S ( f ) = Z(f ) 3 A gX) . (so p (Z fi X ) f o r any A C*(X)-point. p G.-subset S o t h i s as f o l l o w s . 0 1 1 S*., (f) = Z ( f ) /I X* . Thus 3 ; Z e Z(gX) . Hence every P-point of v let p e C*(X)-point). of where Z^ e Z(X*) . We s h a l l 0 X* gX Since such that Z^ e Z(X*) gX then there S f) X* - Z^ . We j u s t i f y 1 CO From [4J, 1.10,.we can conclude that i s an open set i n X* then 0. = 0 i s the continuous i s then an element of the X*-boundary of a zero-set of w i l l not be a exists a So by d e f i n i t i o n 3 i s a C*(X)-point i f and only i f i t i s not on the Conversely show that by [ 4 ] , 7.2 ( f f o r every where £! X i . Since O. i s open i n gX . Hence Z = fl O 1 where m 1 i s open i n X* Z = f) (o f> X*) , 1 31 CO and i t i s c l e a r that O /I *) X = <f)0 )f) 1 Then S oo CO x i s the required X* . 1 G„-subset of Now l e t /? Q 1 X = S . gX . Now by complete r e g u l a r i t y there e x i s t s 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 zeroset neighborhood of that p o i n t ) . C*(X)-point of X* Hence i s a P-point of p £ S(Z^H X*) . Hence a X* . We can now f i n a l l y conclude that i f X is locally compact and realcompact but not compact and i f card(X) = c , then has a dense set of 2 P-points. Hence gN-N X* has a dense set of c 2 P-points. Remark. space Q* Local compactness i s an e s s e n t i a l property. Q (set of r a t i o n a l s ) ' s a t i s f i e s the property has no P-points (see [ 4 ] , 60.5). For the card C(Q) = c but 32 CHAPTER I I I c-POINTS OF 3.1. BN-N 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 c a r d i n a l greater than c then there does not e x i s t any a-point i n BN-N , because there are a t 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 e x i s t s 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 d i s c r e t e space of c a r d i n a l i t y m where 2 _< m n for a l l n < m . Following the n o t a t i o n of [ 4 ] we l e t , f o r any p e BN-N , A P be the free z - u l t r a f i l t e r on N 3.2. D e f i n i t i o n . cardinality Ap 0 0 p . i s c a l l e d uniform i f each element of AF has . Let uN = {p e BN : A 3.3. P r o p o s i t i o n . with l i m i t "* uN = B N - N P i s uniform} 33 Proof: , Let p e gN-N p e uN' . we must show that A and consider otherwise Thus A . In order to show that P has no element c o n s i s t i n g of a f i n i t e P subset of N . Suppose there e x i s t s P = {n^, . .. , n } where A P e A* 3 such that would not be a f r e e u l t r a f i l t e r . P P f) (N - {n » eA 1 {n^} ^ A n.^ £ N , 1 .1 i .1 j • Then P which means that Hence for P (N - {n^}) e A . P { n , . .., n } . e A . P 2 We proceed i n t h i s way u n t i l we can f i n a l l y conclude that {n_. } e A contradiction. gN-N C uN . p Hence no such element, i t i s easy to see that If p s N , then would mean set. A P {p} £ A P The f a c t that P , e x i s t s and uN C gN-N . Let p e uN , i . e . A = {Z : Z i s a zero set containing since any element i n N {p} e A N , then p £ gN-N can be made i n t o a zero A i s uniform. P P are of the form* ( ^gN Z) - N c where Z e A P . 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 card A = f o r every A £ £L , and cardCA^/O A^.) < ° A., A. £ , i ^ j . Then there i s a uniform u l t r a f i l t e r such that card {A zC-L : card(Z D A) = \} = card Proof: B = {x If c£.„ Z = {p £ gN : Z £ A } . Also b a s i c neighborhoods of gN 3 . 4 . Lemma. that J , For A £ CL p i c k A Now . But t h i s In the sequel we s h a l l use the f o l l o w i n g f a c t s . Z ,a i s uniform. p} i s a c o n t r a d i c t i o n since P P : A £ CL }. & AP f o r every N 0 for on c N P <f Q , then N Z £ A^ x^ £ ^ g A D uN ; l e t I f P, Q e CL such that such f x„ 34 because i f for A c N , cJl P fl Q s A Thus X 0 . v then A = A and since H A =' {P £ gN : A £ A }) card(P fl Q) = This means that . Since gN-N N P {Np : p £ uN} contains l e s s than so formed. set i s an open cover f o r p card Ct elements card CL elements of uN . Since uN i s compact, N , N l P Since each N then Also i f card(Z /) A) < for X x. £ c£ A fj yN for every Z £ A means P Consider there e x i s t s a , N 2 P , such that n contains l e s s than p. card B < card CL , a c o n t r a d i c t i o n . We claim i s the required u l t r a f i l t e r . P B . , N^ , Without l o s s of g e n e r a l i t y assume that t h i s B <^ uN . B p £ gN-N elements of P n uN C V N . Note that 1 p. and so , a contradiction. contains card fl f i n i t e number of elements from t h i s s e t , A r For i f not then p i c k a neighborhood of every such that that > 0 ( r e c a l l that i s compact, there e x i s t s a point such that every neighborhood of B . P £ A x P e A then P card B = card p e uN of = x A, £ A Z £ A A and f o r every I t i s uniform since then P X so Z % A x A A ). p e uN . fj: Z /) yN ( f o r Hence card(Z f) A) = J X 0 ke.fl. We digress a l i t t l e to prove the f o l l o w i n g theorem which w i l l be used i n the remainder of the chapter. 3.5. Theorem. Let m be a c a r d i n a l number such that n < m and l e t .X family Ct of subsets of every A £ CL and be a set of " c a r d i n a l i t y X such that card (A f) B) < m for m . 2 n _< m f o r every Then there e x i s t s a card Ci = 2 m , A, B £ Ct. and card A = m A r B . for 35 Proof: We can w e l l order X thereby l e t t i n g X = {x^ : a < a } Q X where a i s the i n i t i a l o r d i n a l such that 0 and order Y lexicographically. We would l i k e to show that where Z at every x for y > a 0 such that y(x )"= 0 • . v a} Q card Z = m . Indeed i s the subset of a a < a 0< a < = ra . Let Y = 2 0 Let Z = {y E Y : there e x i s t s for card a : a < a} a whose functions take on the value zero Y (i.e.Z Z = {Z 0 = {y £ Y : y(x ) = 0 y a f o r y > a} . y card ct For every- a there are 2 such functions and so card Z a = 2 ° ^ < m by hypothesis. Thus card Z = Y. — J J L a<a card Z a r < m-m = m . Thus card Z = m . For each of m A 0 y EY let be a sequence d i s t i n c t elements which converge to y {z^ : 0 < a } 0 and where z^ e Z ^ Y such that r y(x ) z (x ) = { , a < 0 a > 0 . a R U Now l e t f from {f[Sy] : y E Y} Z onto , X be a one-to-one map. Then i s the required family Note that theorem. •• N has the properties of X i n the above 36 3.6. Lemma. and L e t Ct be a c o l l e c t i o n o f subsets let A N i n Lemma 3 . 4 , the elements be as o b t a i n e d P of as g i v e n above Z of A? a being indexed by t h e o r d i n a l s l e s s than each a < to, 1 a subset card (X /) X ) a a < \ of a that f o r a < a and A that CT 4- A c a r d (J such elements) . s CT v < cr . f o r every •card{A z CI : c a r d A fl Z = a J Q cr j- A for a we have 1 second p r o p e r t y Let a < A A e tl such t h a t a f A a i ssatisfied Theorem. . Let card X Since collection {X } xa x<to^ T < to, and A (I , described. ct 1 gN then 0 Assume /Ik) a =' J 0 card \r card J Q card X = a \r <, and (A ^ A ) < \> % hence' the J a X' -'1 X = A a Y since this (A /) Z ) = a a L e t X = A ^ Z . Then a a a < ox.) . 1 so y = y ^ Z a r )A Y '"^ Z Y C A a Y A . T (X card = ( c Z a<to^ = {Z}' a a < to., n i s open i n and X C z a a be as a l r e a d y then a g a i n by Theorem 3 . 5 we have a 0 v TCI let U U T P o f subsets 1 T < to, J There e x i s t s a c - p o i n t i n gN-N Proof: every to^ . such t h a t card(Z CT Since a 3.7. c a r d Z^ fi A^ = s a t i s f y the r e q u i r e d p r o p e r t i e s s i n c e a < y < and so t h a t to., ( f o l l o w s from lemma 3 . 4 ) then 1 a f o r CT < ct ( r e c a l l X ot the s e t s A } = means t h a t we have an element A o card X = • a e /Z such t h a t we have chosen y so t h a t a P i c k an element ( r e c a l l t h a t we have and Z Then we can choose f o r i f a < Y < a, . ' • 1 J Proof: X to . of f~) X . ) < \ J such t h a t card X = Ta f o r T f CT . J 0 for Now f o r every CTOt „ X )^ gN t a gN-N . X a Since ci „ X gN Ta i s open i n We c l a i m t h a t {U } T T<(0^ i s a family uN . 37 of pairwise d i s j o i n t , open sets such that i.e. let p i s our c-point. q e U fl U a where x p e c£ a ^ x . Then f o r some a, , a~ we have 1 uN , q e c£„_ X f] yN . Hence 0N xc< q e (c£^ X fl 3N aa^ c£„, X ) f) yN . I n the case 3N x a oa^ O X xa /7 X ) a and since ) < \ . I f a, = ou J xa 1 2 properties of the sets {X } 2 that card (X fl aa so cra-^ A^ £ A q i s uniform. 2 ^ So i n q e c£„„ X f] c£^ X ^ 3N ac^ 3N xcu, J xa Xa ) < ''I then then we already know from ) < \, . But ra , X fl 1 X 1 q E yN 1 then 2 card (X ff X ) < *'t . aa xcu/ TO^ X card (X 2 X means that 1 0 ) C (X e i t h e r case a, 4 a„ T 2 card (X ca^ 2 2 7 (X T , That they are pairwise d i s j o i n t i s shown thus: q e c £ X • fl 3N aa^ D H f o r every T 2 e A and so q (X ^ X ) e A aa^ xa 2 q . Also 2 Hence card (X fl X ) = " aa^ xa l 0 , a contra- 2 diction. Hence our sets are d i s j o i n t . i s i n the closure of U f o r every T x We now proceed to show that p . Let Z E A ( i . e . a P p E c£ Z ). We know that ( ^ ^' yN) i s a basic neighborhood of p so we are f i n i s h e d i f we can show that f o r every x , c 2 flN C ^3N ^ ViN Z"' ^ <|> since every neighborhood of p V' card (X f> Z ) = ^ (X C Z x xa a xa a card X = % ) then there e x i s t s an element q i n •xa • c£ X f) c £ Z /) yN<r c £ Z' U ^ yN . Hence 3N xa 3« a 3N .a x c £3N Za /I U x /I H • intersect U . Since n 0 H OM o W y would then and 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 in P Lemma 3.4 was to show i n Lemma 3.6 that we can choose, f o r each a subset X w i t h those p r o p e r t i e s . a ^ we proved that Using t h i s c o l l e c t i o n ^ Z eA a {X} P a a<tu 6 1 p was a c-point of BN-N . Hence a reasonable a s s e r t i o n i s that every point of BN-N i s a c-point i f we can show that f o r every free u l t r a f i l t e r A there e x i s t s X C Z f o r every Z e A such a a a Y" ' 'T that card X = „ and card (X O X j < \ i f a T B • a a B P P J 3.8. Lemma. Let A be a free u l t r a f i l t e r on N P and order i t s elements by the o r d i n a l s l e s s than UJ_, . Then f o r each Z E A there 1 a card X = \ and card (X /? X j < * i f a a 8 X C. Z such that a a exists P J J a a ^ B • A'" P Proof: R e c a l l that every element i n A since X 1 £A A P i s uniform. P . F i x a < uj such that „ Let X^ be an i n f i n i t e subset of Z^ such that and assume that f o r each n X C Z has c a r d i n a l i t y card X = ^ 0 ,X iA P a < a we have chosen and card (X 0 X ) < K *r . L e t B = {CT < a : card (X fi Z ) = \} . 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 . Consider a set a CT X' i A . Then N - X e A . Since Z e A then (N - X ) fl Z A a a a a a i . e . Z - X e A . Thus i t s u f f i c e s to show that O { X :CTe B} i A . a a Since X i A f o r every cr e B then N - X e A and so for a l l y < a J P P P P P £ P p . ' CT P p J CT 0 {N - X^ :CT£ B} s A P CT , i . e . N - ^{x^ :CTe B} e A P . Thus 39 V {X : a e B} i A cr - L {X (Z . P Now : a e B}) ! let X such that card (X 0 X ) < \ have X £ A a . P Then by d e f i n i t i o n of I f on the other hand B (X o\ k k e N we can f i n d an element f) Z ) - 6 ' { X : j < k} a a. j j < k card (X cr, k n X Consider the set {x oo }, . ot^ k=l a = a, k {x \ . (X ) < -\ Thus CO }, . . ct^ k=l Let X a, k f) Z cr, k n fl Z ) = ~K a ot while f o r - 6/{X : j < k}) ^ <b . a. 2 be an i n f i n i t e subset of a "n X c Z ) . a a X" i A^ (note that a T k e N CO B = {<? }^ • in card (X 3 such that f o r some x because J o. we i s i n f i n i t e , then i t i s i n f i n i t e - l y countable by the continuum hypothesis so we w r i t e every B a < a f o r every J a be an i n f i n i t e subset of a Now f o r any a e B, A<" card (X' rt X ) = card (X rt X ) < k < " a cr a a, — k so CO 0 I }, , . I f however a I B then a k=l * card (X rt X ) < card (X H Z ) < \ by d e f i n i t i o n of B . Thus we from the way we have chosen {x k a a — a a have shown that f o r every card X = I J and ot 3.9. Theorem. Proof: card X^ = '\ card (X ^ a For every Let A P ) < ) g there e x i s t s a c o l l e c t i o n if J x p e gN-N C4 a « D ^ a X c a there e x i s t s = {Z } f o r every J a 0 , p E A P such that a a / g . i s a c-point of , X <Z Z a Z a gN-N as i n Lemma 3.8. . Since then we can use our Remark 3.5 to show that {X' \ } of subsets of X such that 40 card X xa = ' J Q now form sets and such that card (X xa /> X ) < aa J, 0 if x ^ a . We p r e c i s e l y as we d i d i n Theorem 3.7 and we proceed exactly as we d i d 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 r e s u l t s of t h i s 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 t h i s concept we s h a l l prove the non-homogeneity of gN-N ( t h i s 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 s e t s . 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. I f -rr i s a permutation on N Remark. on N then TT(^) = {TT(E) and Q i s an u l t r a f i l t e r : E e fi} I s an u l t r a f i l t e r on N . We show t h i s as f o l l o w s . (a) Let TT(E^) , TTCE^) e ir(^) . We would l i k e to show that TT(E ) 1 fl TT(E ) 2 e Tr(n) . We claim that For l e t y £TT(E^) and f o r some one-to-one. Obviously T\(JLA d £ E^ , Then f o r some . Tr ( 2^ e So y £ TT(E f> ;L TT(E ) = EA v • ^ u t e ]_ fl e 2 . , Tr(e^) = y s i n c e fr is . Thus TT(E ) fl ir(E ) C TT(E ^' E ) . 2 ir(E ^ E ) C TT(E ) ^ TT(E ) . Hence £ 2 e^ e = E ) ir(E ) = TT(E 2 42 TT(E ) 1 {) ]L Let = TT(E ) E 2 T T ( E ) -0 (b) TT(E ) 2 TT(E) C E Since E . We must show that E T:(E /'I 0 N . ) ^ Ec A Then 0 Q A <£* N then f o r some E c N) . A E ft . Hence e Q then e Tr(fi) . Since E 2 TT(A) = E Q IT i s a (we are 0 TT(E) <r i r ( A ) . Thus since TT(A) e Tr(fi) , i . e . E an u l t r a f i l t e r on 4.3. Theorem. 2 TT(O) . permutation on assuming that E ;L E ir(fi) . Hence -rr(fi) i s N . I f two free u l t r a f i l t e r s type then there i s a permutation ft^ TT of Q. and are of the same such that ' ^ N = ir(ft^) . 2 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 ultrafilters. Proof: Since ft^ and an isomorphism f(A) C. f(B) sets H^ f Q,^ axe of the same type then there exists taking ft^ to and that f(N) = N) . ^ 2 (note that i f A C B Let H are maximal proper subsets of N =' {q : q E N , q ^ n } T fl {E : E Eft^}f <f> then' H n E E £L 1 There must exist £ & such that f operates on H^ Now for any n i E T , a contradiction since and so Let E 0 E i s free. . Since H^ , Since E c- H n to give a set of the form are the only elements i n H^ =/7{f(H ) : n E F} n f . since otherwise s i m i l a r to H , H say. Define ir such that -rr(n) = m n m Now l e t E E and l e t F = N - E . Then E =/7{H : n 1 n the elements of . The which are permuted by That they are maximal proper subsets' i s obvious. n i H . n then E c H n £ f o r every f o r every n i f f(H ) = H . n m F} because n £ F . then 43 f(E) c f _ 1 f (H ) (E ) C 0 f (E) = E 0 -rr(n) = m f o r every E . . Thus Now if n f(f N - E _ 1 H m For . denote, complements) . required and r e c a l l E <= f ( E ) . 0 that m have Hence : n e F} because i s the o n l y element of N Ac N we have TT(A') TT(N - E ) = N - TT(E) Hence get f ( E ) = TT(E) (where primes [TT(A)]' = . . Thus, u s i n g the IT Hence fact i s the permutation. two u l t r a f i l t e r s N By symmetry we N - f ( E ) = TT(N - E) '. Thus We have now on . = N - / ? { f (H ) : n s F} =' {Tf(n) N - f ( E ) = TT(N - E ) , we that Q 0 0 any f (E) c: E ( E ) ) C. f ( E ) , i . e . f(H ) = H n m that i s not i n ir and so established that a necessary c o n d i t i o n f o r to be of the same type i s t h a t t h e r e e x i s t a p e r m u t a t i o n t a k i n g the one u l t r a f i l t e r to the o t h e r . But t h e r e a r e o n l y A 1 2 = c permutations free u l t r a f i l t e r s . on We 2 Hence no type can c o n t a i n more than can,now c o n c l u d e t h a t t h e r e a r e the s e t o f u l t r a f i l t e r s which has N . elements o f the d i f f e r e n t 2 types c because types forms a p a r t i t i o n o f and each type has o n l y c gN-N ultrafilters. F i n a l l y we must show t h a t every type c o n t a i n s p r e c i s e l y c ultrafilters. c infinite subsets of intersection. N can f i n d a family such t h a t any two members o f L e t • 0, be any f r e e u l t r a f i l t e r and the complement o f ultrafilter From Theorem 3 . 5 we A is infinite. containing E . For each E e 3 c o n s i s t i n g of 3 A e 0, let have finite be such be Then t h e r e e x i s t s a p e r m u t a t i o n on an N that 44 Q. carrying to ( t a k e the' p e r m u t a t i o n which c a r r i e s S i n c e we have c same type as such s e t s c E then we other u l t r a f i l t e r s . t h a t no type can c o n t a i n more than 4.4. We now of gN-N '. with p e P onto itself. We let I f we can say t h a t denotes f P denote consider P then l h from the s e t of a l l permutations f g(^ ) = 1 ft 2 a ^ p N). gN g : iP(N) Now ^(N) s i n c e an u l t r a f i l t e r then we onto and ft (= A 2 gN can say t h a t i n such a way , p^ e gN) i f and o n l y i f h(p ) = p 1 and o n l y i f p*^) and fi^(= = a.^ A 2 = A , a2 e gN) f o r some then p e P . we N (N) can be induces a that i f on p e P a and e a r l i e r remarks- i n t h i s s e c t i o n we ~' , a^ E gN) ^± l > Theorem 4.3 a a permuta- N , t h a t i f two u l t r a f i l t e r s (hence l ^ N* (where P on f p* . 2 p(fi^) = ^2 £2^(= A p*(a^) = ^ and N are u l t r a f i l t e r s a a N (i.e. ->£\N) of the same type then t h e r e e x i s t s an element e of elements to a homeomorphism of i s a s e t isomorphism We have a l r e a d y proved &2 i s of the as o p e r a t i n g on s u b s e t s o f P , p^ e gN) fi, ultrafilters). induces a s e t isomorphism the power s e t o f homeomorphism E) . take a c l o s e r l o o k a t the s e t of types of c o n s i d e r e d as a subset of fi^(= A free the continuous e x t e n s i o n o f N). to (Note t h a t we have a l r e a d y shown c Suppose _ f : N -> N t i o n on can conclude t h a t A ^2 = A and such that 2 ^" T ^ conclude u s f r o m that a r e o f the same type i f 45 Let that call T the elements then mapping T T a mapping o f i f and o n l y i f p * ( a ) = 8 TO = T6 t = T0 be a s e t and t of T f o r some the types o f u l t r a f i l t e r s 0 i s the type o f . type c o n t a i n s ultrafilters c Note a l s o ultrafilters and t h a t t h e r e a r e -1 we now have card T (t)= c and X i f X c£ X i s homeomorphic filter . {N f on N : N ] T„ Z = T . N z N with M is infinite. z e cl X - X X e x . Thus the form an u l t r a f i l t e r on of N H z J z z Then we g e t o f type ' T^Z . C o n s i d e r any s X = N i s a neighborhood ^ N = B'-. z t o be r e l a t i v e to X , T y the case when i s an u l t r a f i l t e r N* with We now d e f i n e the type of T In hood : x e X} gN-N and z c 1=2 9 T, Z , to be X T of types o f f r e e z . We know t h a t t h e r e e x i s t s a s e t X k : ^ ( X ) ->? (N) . Thus z i s taken by k to an u l t r a - C ^(N) N gN 2 each this u l t r a f i l t e r isomorphism Let to a c o u n t a b l e d i s c r e t e subset of i n t e r s e c t i o n o f the neighborhoods i.e. if that of u l t r a f i l t e r s We know t h a t written such Vie s h a l l N , i.e. card {M Call on t h a t s i n c e we have proved d i s c r e t e i f there e x i s t s a d i s j o i n t f a m i l y X . p e P . T Note t h a t we have d e f i n e d our We now d e f i n e a c o l l e c t i o n X onto SO as to be c o n s i s t e n t 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 b e i n g o f the same type. for N* N , where z e B such t h a t of and j^z . where We s h a l l now show that z} = A N z Z z^ = A , , thereby'giving i s a neighborhood of z . f o r o t h e r w i s e t h e r e e x i s t s some n e i g h b o r - N* ft B = cb , i . e . ' N* f) (N T) N) = A z z^ r z z r , i.e. 46 • (N f) N ) fl N = d) z z A N* /I N z z . But t h i s i s a c o n t r a d i c t i o n since neighborhood of z E CX, N-N = 0N-N and so (N* /) N ) /) N ^ (J> — z z e B and so B e A . Hence Now l e t B e A where But N z N z Hence when x,,z N = T . We would l i k e ' t o show that Z i s a neighborhood of - gN - c£(N - B) . Thus X . Thus z z„ C A N — B=N O N z B = {gN - c£(N-B)}rt N. i s a neighborhood of z . We know that '. gN - cft(N - B) is a z so we l e t B = N /? N . z =N and 2 Z E C £ X - X , Z N = A , and 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 z £ gN - N and X cr gN - N where X TZ = t' , T z = t f o r some i s d i s c r e t e and z E ci X - X . Thus we have <j)(t) = { T Z : z £ cZ X - X , T z = t , X C N*' and Hence the domain of when X discrete} < f > i s a l l of T , i t s range i s a subset of T and ( t , t') E <j> we say that t produces t ' or t ' i s produced by t . 4.6-. Remark. L e t { M ^ : n E u} be a countable decomposition of N and p i c k , f o r every - n , x ,y n n J E C £ M n - M n . Let 47 X = {x : n E to} n every n ,x Y = {y : n e to} . • •n and f Y , then n n ( i e to ) and a base neighborhood A 1 M. /) A i 1 H fl 2 . R e c a l l that a base f o r . Now assume that there x e cZ X /7 cZ Y . Consider a base neighborhood exists A. i cZ X f) cZ Y = <p i s {cZ A : A C N} the closed sets i n gN x_^ We s h a l l show that i f , f o r J M. = (M. /) A.) = d> and = M. . Then • ±2 = cJli^ML 1 1 1 1 ' (M. M. = M. U M. 1 1 1 : i e to} . Then closure of M. i U A for a l l i 1 containing 1 f) A ) . Let M. i i 1 f) . Let H, = c£t/{M. 1 E 0 H ± 2 = <j> A. = M. and i ^ : i e to} and 1 . But x e H containing y_^ such that 2 and so A^ contains the . Similarly x e EL . Thus 1 we have a c o n t r a d i c t i o n . 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 f o r each n , x 4 y n n J 4 . 7 . Lemma. card x <_ 2 Let y E gN - N . There e x i s t s a set x and whose elements are d i s c r e t e countable subsets of and such that i f Y y e cZ Y .- Y Proof: then i s any d i s c r e t e countable subset of Y 3 X' f o r some X' e x (and l e t X = {x } n p o s s i b l e , and consider a l l p o s s i b l e x = {X'} where For each countable decomposition x E cZ M - M n n n Let such that ) such that X'c X gN-N , and i f y e cZ X' - X" . {^ } of N n J gN-N choose an y e cZ X - X , i f ' such that y e cZ X' - X . so formed as we vary our decompositions of N . Now c e r t a i n l y c o n s i s t s of d i s c r e t e countable subsets of x gN - N . We s h a l l 48 use Remark 4.6 Y = {y '} to show that (where n construction of X . Let Y x then by Remark 4.6 number of n's . . J . has the asserted property. i s as in\the statement of the lemma). we have X = {x } n x Since {x } n y e cH X - X y and n J (see [ 9 ] , Lemma 2). X' discrete {x } • and ,n & X' be Y (1 X . because i t i s i n the closures of There are at most 2 Y Now y and X N and decompositions of K with each decomposition we get at most 2 sets X . A' }< card x _< 2 > 2 =2 . The proof i s now complete. Hence From the f o l l o w i n g theorem we s h a l l obtain a very proof of the non-homogeneity of 4.8. Theorem. gN-N 2^ Proof: 2 types and any type Let t types. We prove the second part f i r s t . <j)(t) = {TZ : z e cH X - X , R e c a l l that simple . Any type i s produced by at most produces {y } n have to agree f o r an i n f i n i t e So we l e t the required set i s i n the closure of By the f o r some countable i s i n the closures of {y } Let T z = t be any type. f o r some d i s c r e t e A. X c N*} . infinite. card (eft M Now let {M } be a decomposition of N where each n We know that since M i s i n f i n i t e then - M ) = 2 n n 2 conclude that there are where each such c£ X M is n f o r each n . Again using Remark 4.6 we y° o 2 2 d i s j o i n t sets cJl X = c£ {x : x e M } n n n & J contains a point z such that r z = t X and 49 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 then we can conclude that card <j)(t) = 2 2 such elements z . Before proving the f i r s t part we must prove the f o l l o w i n g . X C Y , Y If that both onto X X and d i s c r e t e , and z e c£ X - Y , then and Y x one-to-one map from a i n c l u s i o n map from are i n f i n i t e . X to Y . y X c: Y 6 Let Y T Z = T z . We assume V be a one-to-one map from onto N . Let Define the permutation i p N be the on N as indicated: p : N i.e. p = x ° i ° <5 • x —s- N We define z and z X that N z and X z Y are set isomorphic to u l t r a f i l t e r s i n such a way that Ta = T z xy = T z and A TZ = T z A Y V q*(a) = y as u s u a l . . A Ci claim that and A Y on In order to show that X i t s u f f i c e s to show that f o r some permutation .We We know i q X , on p , as defined above, i s the required permutation, i . e . p*(a) = y . Assume that neighborhoods B 0 N (j: A Y and A and B p ' (a) ^ y . 5 c A y e A , p*(a) e B , such that p ~ ( B ) f) N e A 1 . K Then we can f i n d d i s j o i n t where N^ 6 ( p * ( B ) / ? N) = (N n X) e 1 i s some neighborhood of a set isomorphism between P (N) and , Y Now [N n p''c (A)]/T [N/? p * " ^ ) ] = $ . Also -1 A f) N e A z . _1 This i s so because <P (X) which takes A 8 a induces to z X ; 50 (B) f) N e A " p* and so i s taken by t h i s isomrophism to some neighborhood 6 takes N^ of (p*~ (B)/'> N) Since z . But considered as a subset of to 1 _1 . We s h a l l now show that _1 [N /? ] /0 6[p" (A)/? N] = <)) . Assume t h i s i s f a l s e . _1 [ ( N / 1 Y)/7 6 ( p * ( A ) z = x &z r\ Now x( /0 Y) £ A N and 'f (N) : f (Y) X(6(P" (A) _1 n Y) f) ( 6 ( P " ( A ) fl N)) _ 1 X ( r e c a l l that Y which takes N)) = A/7 N z -1 (N /) Y) /I S ( p * ( A ) /) N) e z -1 where N' z to y y A) . Y Also A P N e A Y . Thus . Hence Y and so i s some neighborhood of [(N . induces a set isomorphism between (see remark l a t e r ) and ^ Y) /7 6 ( p * ( A ) /I N) ] E A xt(N x Then N) ] _1 X then L S[E fl p * ( A ) ] / ? <5[N f) p * ( B ) ] = (J) Y N (N /? X) . [N /? p * ( A ) ] / 7 [N /? p*" (B)] = <f> -1 N^/l X (N z . z Y) 0 6(p* """(A) N) = N^D Now ^ Y) n <5(p* (A) D N)] /IX _1 = (N /) X) fl 6 ( p * ( A ) fl N) z -1 = cKp* (B)/) N) /I o ( p * ( A ) / ? N) -1 ( r e c a l l that -1 5 ( p * ( B ) f) N) = N /7 X). _1 <5(p* (B)rt N) /I 6 ( p * ( A ) /) N) = <j) _1 [(N -1 But we have already shown that , i.e. 0 Y) P 6 ( p * ( A ) /) N) ] /) X = <j> . Hence _1 2 implies that N' /) Y /) X = <J> which 2 N' /? X = <j> . But t h i s i s a c o n t r a d i c t i o n since N' is 51 a neighborhood of z . and z e c£, X . Thus ( N rt Y) rt 6 ( p - ( A ) n N) = 4 -1 z Hence • x(N rt Y)/? (<5 ( p * ( A ) /> N)) = ^ -1 x x ( N / ) Y) z e A Thus we have x (N and Y X ( N Z x ^ Y (6 ( p A _ 1 (A)/} N)) = A/} N )x(S(p* - 1 ( A ) /? N)) = <f> n Y) f) x(6(P' (A) ft N)) s A c_1 z Thus A e . But (see remark l a t e r ) . Y and , a contradiction. Y p*(a) = y _1 Remark. We must now show that our d e f i n i t i o n of p X (<S(p* (A) rt N)) = A/? N . From we get X(<5(p* (A) 0 N ) ) = p ( p * ( A ) rt N) -1 _1 • = pt{y e = p{{y e = A f] N BN N : : e A} fl p*(y) p(y) e N] A}] . We now continue w i t h a proof of the f i r s t p a r t of the theorem. Consider any type For any <j)(t) t e <}> "'"(t ) 1 there e x i s t s a t ' . Then and f o r every Y e x X such that (^""'"(t') = {t : <j>(t) = t'} . appearing i n the d e f i n i t i o n of X ~2 Y and (follows from Lemma 4 . 7 and the above observation). x z = T Z = t v V Hence w i t h every 52 element t are at most elements in $ o 2 ( t ' ) we can associate an element of x . But there o elements of x ; so there are at most such 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 . f o r every p a i r a homeomorphism of N* x e N* , T onto N* Tx = Ty "' 2 N* taking x to y . L e t , f o r any f o r some homeomorphism f of N* T X = xy by e a r l i e r remarks (see s e c t i o n 4.4). Hence then i n t h i s case. card T = 2 cover N* = <f> "''(TX) . I f f ( x ) = y are equal f o r every and onto x, y e N* , there e x i s t s Thus i f N* x e N* . But i s homogeneous then a l l sets card Tx _< 2 (shown a l r e a d y ) . Tx by our l a s t theorem Also the sets T . This i s a c o n t r a d i c t i o n i f a l l sets Tx Tx f o r x e N* are equal. Hence i s non-homogeneous. The remainder of the chapter i s used to show that no homeomorphism of BN into N* has a f i x e d p o i n t . - a f a c t that w i l l enable us to give another proof of the non-homogeneity of using the continuum hypothesis. 4.10. Lemma. For a c l a s s other c l a s s such that classes for X , X, X Q ± i = 0, 1, 2 . 2 X fx T X BN-N without F i r s t l y we need two lemmas. let f be a mapping of f o r any such that • X = X X i n t o some x e X . Then there e x i s t d i s j o i n t Q U X 0 X ;L 2 and f (X ) D X =. <> j 53 Proof: We s h a l l associate w i t h every x e X a class C(x) defined as f o l l o w s : C(x) = {y e X : f x = f y m n We now show that any two classes are d i s j o i n t . f o r some m, n e N} C(x^) and C(x ) e i t h e r coincide or 2 Assume that they are not d i s j o i n t . Then l e t y £ CCx^ fl C ( x ) . Hence 2 (1) f x (2) f where n f y = f y x = f 2 y m, m', n, n' £ N . We can assume that n' f y u n t i l we achieve f x^ (while at the same time applying to maintain e q u a l i t y i n m m' f x Hence 2 2 x^ y f o r some m" £ N . 0 e C(x^) . We want to show mo Then by symmetry we s h a l l be f i n i s h e d . We have 0 Q f to m" = f Now consider any element m, n f to (1)). We then get ^irri-(n'-n) -^n' .m' f x^ = f y = f x (3) n' > n . Apply £ N . Assume m f y x^ = f > m" . With (3) i n mind we apply Q 0 n° £ C(x ) . 2 y where 0 f to m" f repeatedly u n t i l we get the f o l l o w i n g : _m + (m —m") f x 2 f "x 0 T 0 m = f °y Q n 2 0 where y f °x^ u n t i l we get a s i m i l a r equation). m 2 7T Q m" £ N Hence 0 £ C(x ) . _m _n = f x^j^ = f y (Note that i f m Q < m" . Hence . then we would apply By symmetry we get f to C(x^) = C(J 54 f(C(x)) C C(x) f o r x s X . We now show that C(x) = {y : f x = f y , m,n e N} m that f y e C(x) . Since f ( f x ) = f ( f y ) . Thus m n Thus . Let f y e f [ C ( x ) ] . We must show n y f C(x) m + 1 then x = f (fy) f x = f y and so m n f y E C(x) . which means that n f(C(x)) C C(x) . C(x) , x E X , i s X Since the union of the sets i s s u f f i c i e n t to prove the lemma f o r each class i n the hypothesis and y of the lemma be of the form n m EN smallest be the l e a s t m n such that n m m(x) = q . Then q integer such that f^a = f ^ ^ "*"(fx) of n(x) > 0 ) . n(x)-l n since l e t m(x) x with n(x) > 0 q i s the smallest (note that i t makes sense to speak x Thus n(b) = 0 such that b 1 and ^ b m(fx) = mx and n ( f x ) = n(x) - 1 . 2 m(fb ) r m(b ) + 1 . Now since 2 m(b ) f. " a = b 2 1 = f°b 1 b £ X m(fb) T m(b) + 1 . Assume that we have , nO^) = 0 , n ( b ) = 0 2 m(fx) = mx . i s the smallest integer such that We now show that there e x i s t s a t most one that and be the f x = f ^ ^a . which i s the same as saying that X such that n We assert that f o r x E X For assume that x f a = f x f o r some n E N . Then l e t n(x) such that n £ N f^a = f ^ ^ x C(x) . Thus f o r any a e X . For x E X f x = f y . Now pick an element m C(x) . So we l e t X we may assume that there are m,n E N in X then i t n(b^) = 0 and d such ^J 2 D e ^ m(fb ) r mOt^) + 1 , ;L then m(b )+l f a = f ft^) . Since m(fb ) f m(b ) + 1 ;L 1 55 then we have a number p = m(fb ) n(fb ) f a = f (fb ) m(b )+l f = f b ^ then such that 0 _< p < m(b^) + 1 where (since we already have m(b )+l p < m(b^) + 1 ) . Hence f(b^) = f a = f a where m(b^) + 1 > p >_ 0 . m(b )+l S i m i l a r l y f (b^) = f a = f a where ra(b^). + 1 > q _> 0 . Assume 1 2 q without l o s s of g e n e r a l i t y that because i f m(b ) = m(b )+l 1 1 2 d e f i n i t i o n of m(b^) 2 m(b^) > m(b^) + 1 m(b ) m(b )+l b^ = f a = f a = f a . But then 2 m(b ) = m(b ) + 1 > q m(b^) > m(b ) . Then means that m(b^) > q . This c o n t r a d i c t s the since i t must be the smallest number such that m(b ) b^ = f a . Thus m(b ) > m(b ) + 1 . Hence m(b-) m(b )-m(b„) m(b„) m(b )-m(b ) \= f a=f ( f a) = f (b ) = f 1 2 X Z 1 1 2 m(b, )-m(b,)-l (fb 1 m(b )-m(b ) - l m(b )-(mb +l-q) = f • (f a) = f (a) . But m(b ) > m(b )+l-q > q This c o n t r a d i c t s the d e f i n i t i o n of m(b^) . Thus there e x i s t s at most one b e X such that n(b) = 0 and m(fb) 4 m(b) + 1 . b e x i s t s then put X I f such an element not then l e t X 0 = <j> . For every as we have already shown, x e X - X 1 0 : m(x) + n(x) i s odd} X Q : m(x) + n(x) i s even} i = 0, 1, 2 are d i s j o i n t and y = fx^ then m(y) + n(y) i s odd. and, , and . Obviously the sets X = X^ U X^ U a s s e r t i o n of the lemma we f i r s t consider Thus n(x) > 0 . Let X^ = {x £ X - X = {x £ X - X = {b} . I f m(fx) = m(x) , n ( f x ) = n(x) - 1 . Hence m(fx) + n(fx) = m(x) + n(x) - 2 we have Q Q X , . To prove the second i = 1 . Let y £ f(X^) . f o r some x^ £ X^ . We must show y £ X^ . I f y £ X^ But m(fx^) + n(fx^) = mx^ + nx^ - 1 , and 0. 56 m(fx^) + n(fx^) since m(x^) + n(x^) f ( X ^ ) f) X ^ = <j) . hypothesis. i s also odd. i = 0 The case Let f : K -> N and N - X = XQ U X c£ X U cl X 0 i ± UX 2 a net } the sequence and so b r fb by the continuous extension of f and in X 2 = gN c£ X_^ 0 c£ X i s the same as the closure . Then . Then by Lemma 4 . 1 0 = <f> f o r i = 0 , 1 , 2 . Also = <j> A C cl X . Let x converges to D 0 A = cZ X = <j> f o r i r j • Now l e t • Since f f x . But Q E c£ X . Q Consider i s continuous then f(S ) = S^ 0 {f(S )} = {S } which converges to x n n x . £ A . Thus c£ X 0 c£ X and obviously converging to x (fCS^)} f f . f ( X ) fl X where V cH X U ca X A = {x s BN : f x = x} so f Let X = {x e N : f x = x} Proof: f o r every Similarly i s t r i v i a l because BN . Then the set of f i x e d points of of the set of f i x e d points of n y ^ X^ . Thus The proof i s now complete. 4 . 1 1 . Lemma. to f x ^ e X^- . This i s a c o n t r a d i c t i o n i s odd because . Thus f o r every fx = x Q and 0 and t h i s ends the proof. The following theorem i s required f o r our second proof of the non-homogeneity of N* . 4 . 1 2 . Theorem. Let f has no f i x e d point. be a homeomorphism of BN into N* . Then f 57 Proof: Decompose (i) "^ i ^ (ii) k £ G^ (iii) {fk} U G G i s f a c a m i N i n t o sets G l y °f d i s j o i n t sets \j {G^.} = N f o r any k e N i s a neighborhood of fc satisfying f k . i n N L> f(N)' . (At the end of t h i s proof we s h a l l construct such a decomposition.) For l e t hn = f k i f n E G, . Thus n e N h isa k mapping from from N N to N onto f(N) . Let g = f ^ ° h . Hence such that gn = f ''"(hn) = f "'"(fk) (note that gn =f n f o r any f o r some k E N c o n d i t i o n ( i i i ) we have f k £ G. i s a mapping n because such that k E gn). Hence by Lemma 4.11 g g k £ G^ but n £ G^ has no f i x e d point. 3 i n N U f (N) and From hn = f k f o r every k n £ G^ (hence one such element n h ( f k ) = f k . We now have h x = x 3 3 Suppose there e x i s t s x £ 3N then x E ci N h ( f k ) = f k and so f o r x £ c£ f(N) . x e gN such that f x = x . Since 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 Hence from i s f k ) . Thus f x £ f(ci N) above h ( f x ) = f x = x . Now 3 c o n t r a d i c t i n g the f a c t that g 3 implies that f x e ci f(N) . g x = f ^"(h^x) = f ^(x)'= x , 3 has no f i x e d point. Thus f has no f i x e d point. We now construct a decomposition ( i ) , ( i i ) and ( i i i ) . Consider any decomposition {G, } 'of N {Hn} satisfying of N . Then we 58 know that f o r j e N , f i s i n the closure of n . Thus we l e t f. e cl H. . f o r some 1 i one 1 then consider H. -{1} (place 1 1 Hn f o r one and only- i . e N . I f 1 e H. 1 x , x i n some other component of the decom- 1 position). Let G = H. \ 1 For every -'{1} . I f 1 i H. \ then l e t G • 1 f_. e cl G. j e N , repeat t h i s process to get Thus we have UG, = N , k \ G ' f o r any k e N , and f k U G^ We want to show that = H. . \ and j £ G. . f k e cl G . i s a neighborhood of f k fk in N o f(N) . Since i N - G k f k e cl G^ then f k ij: c£(N - G > and so k open neighborhood of f k i n BN . fc j = cl = [cl G C\ k (f(N) U . Thus BN - cl (N - G ) i s an Now [BN - cS,(N - G )] rt (f(N) every which implies that fk ij: A (Note that G^ e A O N) N) G 0 N] L? [c£ G rt f (N)] k = G, 0 f k . ' cl G,k f (N) = f k because, by construction, f . e c£ G. 3 3 k and j 5* k) . Thus 4.13. Remark. cl G^ rt c£ G^ = t> j fk U G k i s a neighborhood of f £ cl G^ f o r f k i n N C f(N) . 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 contradiction. f o r i ^ j ; so f or f o r no type t . We prove t h i s by 59 Assume there e x i s t s some type Then where t = xx = x^x and X <J)(t) = t) . f o r some ( t , t) e ty such that x e gN - N and .X c: gN - N , ( t h i s i s a l l by d e f i n i t i o n of As before we l e t a neighborhood of x} = & an u l t r a f i l t e r on X for some x i s d i s c r e t e and x e c£ X - X {N f) X : N x x N . Then t = x x t x, , X ' r which i s s e t isomorphic to A x^.x = T y . But TX i s the type of A p £ P . Thus, whenever X , an u l t r a f i l t e r on y . Hence p*x = y T x = T y , we have an element p £P A such that p*x = y . Since T X = TX then f o r some p e P , p*x = x , V A. a contradiction. 4.14. We are now prepared to give another simple proof of the non-homogeneity of N* . As already mentioned ty "''(xx) = ty "'"(xy) if h Remark 4 . 1 3 , —1 then hx = y i s a homeomorphism of N* xx \ ty (xx) so N* . From —1 —1 —1 h onto xx £ tj> (xy) . Thus whenever xx £ <$> (xy) . Hence i f xx e ty (xy) then homeomorphism implies that hx £ y hx = y f o r any 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 f o r no homeomorphism t h i s concludes the proof since t h e i r always exist' x i= y , such that the type of x some x,y e gN - N , i s a r e l a t i v e type of y we have proved i n Theorem 4 . 8 that f o r any x £ gN - N , xx jC 2 2 ° types). h . But ( r e c a l l that produces 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. J e r i s o n , Rings of continuous f u n c t i o n s , Van Nostrand, P r i n c e t o n , 1960. [5] I . Gelfand and A. Kolmogoroff, On rings of continuous functions on t o p o l o g i c a l 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 a p p l i c a t i o n s 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 c o m p a c t i f i c a tions, Ill] Duke Math. J . V o l . 23 (1955), 409-420. A. T a r s k i , 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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The structure of βN Rambally, Rodney Seunarine 1970
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Title | The structure of βN |
Creator |
Rambally, Rodney Seunarine |
Publisher | University of British Columbia |
Date Issued | 1970 |
Description | Our subject matter consists of a survey of the major results concerning the topological space βN-N where N represents the space of natural numbers with the discrete topology, and βN the Stone-Čech compactification of N . We are mainly concerned with the results which were derived during the last ten years. When there is no advantage in restricting our work to the space N we work with an arbitrary discrete space X and finally formulate our results in terms of βN-N . In some cases, pre-1960 results concerning βN-N are obtained as special cases of the results we derive using an arbitrary discrete space X . The material presented is divided into four chapters. In Chapter I, we discuss certain subsets of βN-N which can be C*-embedded in other subsets of βN-N . This study leads to the conclusion that no proper dense subset of βN-N can be C*-embedded. In the second chapter we devise a general method of associating certain classes of points of βN-N with certain subalgebras of C(N) . The P-points of βN-N form one of these classes. The answer to R. S. Pierce's question, "Does there exist a point of βN-N which lies simultaneously in the closures of three pairwise disjoint open sets" is discussed in Chapter III. Finally in Chapter IV we present two proofs of the non-homogeneity of βN-N , without the use of the Continuum Hypothesis. |
Subject |
Linear topological spaces. |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-06-14 |
Provider | Vancouver : University of British Columbia Library |
Rights | For non-commercial purposes only, such as research, private study and education. Additional conditions apply, see Terms of Use https://open.library.ubc.ca/terms_of_use. |
DOI | 10.14288/1.0080516 |
URI | http://hdl.handle.net/2429/35409 |
Degree |
Master of Arts - MA |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
Aggregated Source Repository | DSpace |
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