ALGEBRAIC PROPERTIES OF CERTAIN RINGS OF CONTINUOUS FUNCTIONS by LI PI SU B.Sc. Taiwan Normal University, i960 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF! PHILOSOPHY in the. Department of Mathematics We accept this thesis as conforming to the required standard. THE UNIVERSITY OF BRITISH COLUMBIA August, I966. In presenting 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 of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y s h a l l make i t f r e e l y a v a i l a b l e f o r reference and study, I f u r t h e r agree that permission., f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s representatives. I t i s understood that copying or p u b l i c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission. The U n i v e r s i t y of B r i t i s h Columbia Vancouver 8, Canada Supervisor: Dr. J 0 V. Whittaker. i i . L I PI SU. ALGEBRAIC PROPERTIES OF CERTAIN RINGS OF CONTINUOUS FUNCTIONS. ABSTRACT We study the r e l a t i o n s between a l g e b r a i c p r o p e r t i e s of ' c e r t a i n r i n g s of f u n c t i o n s and t o p o l o g i c a l p r o p e r t i e s of the spaces on which the f u n c t i o n s are defined. We begin by c o n s i d e r i n g the r e l a t i o n between i d e a l s of r i n g s of f u n c t i o n s and z»filters. Let C m(X) be the r i n g of a l l m-times d i f f e r e n t i a b l e f u n c t i o n s on a C — d i f f e r e n t i a b l e n-manifold X 3 L_(X) the r i n g of a l l L c - f u n c t i o n s on a met r i c space X ' , and OtiX) - the r i n g of a l l a n a l y t i c functions, on a subset X of the complex plane. I t i s proved t h a t two m~(resp. Lc-) realcompact spaces X and • Y are cP-diffeomorphic (resp. Lc-homeomorphic) i f f C^Vx) and C m(Y) (resp. L„ ( X ) and L„(Y)) are r i n g isomorphic. Again* i f X and Y are m»(resp. Lc~) realcompact spaces, then X can be C m»(resp.Lc~) embedded as an open [resp. closed] subset i n Y i f f (^(X) (resp. L_(Xj) i s a aG-f>esp. 6 F - ] homomorphic image of C m(Y) (resp. L ^ Y ) ) . The subrings of C m (resp. L ) which determine the d®-dlffeombrphism (resp. Lc~homeomorphism) of the spaces are studi e d . We a l s o e s t a b l i s h ' a r e p r e s e n t a t i o n f o r a tr a n s f o r m a t i o n , more general than homomorphlsm, from a r i n g of Cm»differentiable f u n c t i o n s t o another r i n g of C ^ d i f f e r e n t i a b l e f u n c t i o n s . F i n a l l y , we show t h a t , f o r a r b i t r a r y subsets X and Y i i i . of the complex p l a n e , I f t h e r e i s " a r i n g isomorphism from (%(X) onto OiXY) which i s the i d e n t i t y on the constant f u n c t i o n s , then X and Y are c o n f o r m a l l y e q u i v a l e n t . i v . TABLE OF CONTENTS page INTRODUCTION 1 PART I. Rings of B i f f e r e n t i a b l e -Functions li Rings of d i f f e r e n t i a b l e f u n c t i o n s and i d e a l s 5 2. Zero-sets and m-completely r e g u l a r spaces 12 3. F i x e d , f r e e i d e a l s , and compact spaces 17 4. Real i d e a l s , m-realcompact spaces 23 5. The l o n g l i n e 26 6. Homomorphi'sms, C m~mapping, and (f-diffeomorphism 34 7. The embedding theorems 45 8. A r e p r e s e n t a t i o n theorem f o r transformations of r i n g s of C m ~ d i f f e r e n t i a b l e f u n c t i o n s 53 9. The r i n g s of Cm<»differentiable f u n c t i o n s on spaces which are not m-realcompact, and some a l g e b r a i c p r o p e r t i e s of C not a p p l i c a b l e i n (f 64 PART I I . The 'Rings of Lc{ (or L - ) : F u n c t i o n s 10. Rings, i d e a l s and some p r o p e r t i e s of L i p s c h i t z i a n or Lc»functions 72 11. L~complete r e g u l a r i t y and L°normality 80 12. F i x e d , f r e e i d e a l s and compact spaces 84 13- The Banach space L(X) and Le-realcompactness 86 14. Lc, L°mapping and Le, ]>homeOmorphisms 9^ V. page 15. Embedding theorems 99 16. The rings of Lc-functions defined on the metric spaces other than Lc-realcompact spaces, and rings of L i p s c h i t z i a n function on compact subsets of E n 102 PART III. The Rings of Analytic Functions IT. Rings of ana l y t i c function on any subset of complex plane 106 BIBLIOGRAPHY 117 v i . ACKNOWLEDGEMENTS I am greatly indebted to Professor J. V. Whittaker f o r suggesting the topic of t h i s thesis, f o r allowing me a generous amount of h i s time and for h i s many constructive comments during the preparation of t h i s thesis. I also wish to thank Professor D. Bures f o r h i s c r i t i c i s m of the dra f t form of t h i s work, and Miss S a l l y Bate f o r typing i t . The f i n a n c i a l support of the National Research Council of Canada and the University of B r i t i s h Columbia i s g r a t e f u l l y acknowledged. 1. INTRODUCTION Let (^(X) and C m(Y) be the rings of C m - d i f f e r e n t i a b l e ... i functions defined on the C^-differentiable n 1 ( n 2 ) , = m a n i f b i d s X, and Y, respectively, L(X 1) and L(Y 1) the rings of LipSchitzian functions defined on metric spaces X^ and Y^ , respectively, LC(X2.) L c ^ Y l ^ t t l e r i n & s o f L c-functions" defined on metric spaces X^ and Y^ , respe c t i v e l y (see ( 10 .l)), and C^(X^ and (^(Yg) the r i n g of anal y t i c functions defined on domains of the complex plane (orRiemann surfaces) Xg and Y 2 , respectively, where 0 m <, » i f the s p e c i f i c case i s not mentioned. During the l a s t twenty years, the r e l a t i o n s between the algebraic properties of C m(X) and (^(Y) ,' L(X j J and L(Y X) , a n d ^ ( X 2 ) and °c_(Y 2) and the topological properties of X and Y , X-^ and Y-j^ and X g and. Yg , r e s p e c t i v e l y / have been investigated. Hewitt (1948) [8] showed that two realcompact (or Q-) spaces X» and Y' (see §7 [8]) are homeomorphic i f f C(X') and C(Y') (that i s , when m =0) are isomorphic (Theorem 57"[8], Theorem (8.3) [7]) by means of the Structure space (Stone topology) (4.9) [7]). Myers (1954) [16] proved that two compact C m— d i f f e r e n t i a b l e a-manifOlds X and Y (1_< m < ») provided with a Riemanniari metric tensor of class C™"1 are C^-diffeomorphic i f f (^(X) and (^(Y) are i s o -morphic. P u r s e l l (1955) [20] established a stronger r e s u l t : two ( ^ - d i f f e r e n t i a b l e n-manifolds X and Y with neighbor-hood-finite covering of coordinate'neighborhoods are C m-diffeomorphic i f f C m(X) and C m(Y) are isomorphic, where 2. 1 < m < » . Nakai (1959) [17], showed again by u s i n g the Riemannlan m e t r i c tensor a stronger r e s u l t than Myers' w i t h 1 _< m <^ oo o L a t e r , i n i 9 6 0 , i n "Rings of Continuous Functions" [7] Gillman and J e r i s o n gave a systematic study of the r i n g C(X) on an a r b i t r a r y t o p o l o g i c a l space X . They study the r e l a t -i o ns between a l g e b r a i c p r o p e r t i e s of C(X) and t o p o l o g i c a l p r o p e r t i e s of X by examining the s p e c i a l f e a t u r e s of the f a m i l y of zero-sets (1.6) of an i d e a l of f u n c t i o n s . The method, used i n the book, w i l l p l a y the most important r o l e of t h i s work. For L(X) , Sherbert (1963) [26] has shown t h a t two com-pact m e t r i c spaces X.^ and Y 1 are L-homeomorphic {Ik. 2) i f f L(X) and L(Y) are isomorphic (Theorem 5.1 [ 2 6 ] ) . I t a l s o has been known f o r some time t h a t the confornial s t r u c t u r e of a domain i n the complex plane i s determined the a l g e b r a i c s t r u c t u r e of c e r t a i n r i n g s of a n a l y t i c f u n c t i o n s on i t . Bers (1948) [3] proved t h a t i f X 2 and Y 2 are two plane domains, then C M(X 2) and J l ( Y 2 ) are isomorphic i f f X 2 and Y 2 are conformally e q u i v a l e n t . Rudin (1955) [25] and Royden (1956) [22] have extended t h i s theorem t o the case i n which X 2 and Y 2 are a r b i t r a r y open Riemann surfaces. Rudin (1955) [25] a l s o showed t h a t i f X 2 and Y 2 are two plane domains w i t h no AB-removable p o i n t s , then they are con-f o r m a l l y e q u i v a l e n t i f the r i n g s & t * ( X 2 ) and CK~*(Y 2) of bounded a n a l y t i c f u n c t i o n s on them are i s o m o r p h i c (*) (*) Bers and Rudin do not assume a p r i o r i t h a t the complex constants are preserved under the given isomorphism. Royden, however, has t h i s a p r i o r i assumption on the complex constants. 3 . Later,- Ozawa and Mizumoto (1959) [18] proved that when X 2 and Y 2 are two plane domains whose complementary sets have p o s i t i v e capacity, respectively, i f there exists a d i r e c t r i n g isomorphism cp of Log" (X 2) onto Log^ (Y 2) such that cp(_c) = ;_c f o r every complex constant c j, then X 2 and Y 2 are conformally equivalent. In other aspects, we are also interested i n the represent-ation of the transformations of rings of certain kinds of functions. Whittaker (1961) [29.] and Kohls (1962) [12] gave a representation f o r the transformations of rings of continuous functions on d i f f e r e n t classes of topolog i c a l spaces. In 1965, M a g i l l [14] has obtained the algebraic conditions r e l a t i n g C(X) and C(Y) which are both necessary and s u f f i c -i e nt f o r embedding Y' i n X' , where X' and Y' are two realcompaet spaces. The primary aim of t h i s thesis i s to u t i l i z e the method of Gillman and Jerison f o r i n v e s t i g a t i n g the algebraic properties of C m(X) (§f.l- 4 and § 6 ) , and L C ( X 1 ) or L(X X) {\§ 10 -1*0 , and how they are r e l a t e d with the t o p o l o g i c a l properties of X and X-^ , respectively. Secondly, we generalize Magill's r e s u l t s to the rings (^(X) (§7) and L C ( X 1 ) ( § 1 5 ) . The other objective of t h i s work i s to es t a b l i s h a representation f o r the transformations of rings of G^-differentiable functions ( I 8) . In § 9 and § l 6 5 we also e s t a b l i s h that i f X and Y." (or X x and Y± ) are connected; then that C m(X) and C m(Y) (or L G ( X 1 ) and L C ( Y 1 ) ) are Isomorphic, implies X and Y (or X x and Y x) are C m-diffeomorphic (L c-homeomorphic). The subrings of C m(X) , C m(Y) (or L(X 1) and L(Y,)) which can determine the C^-diffedmorphism for L»homeomorphism) are studied. We also give some algebraic properties of C not applicable i n G m i n § 9. F i n a l l y , we discuss the r i n g pf a n a l y t i c functions defined on an a r b i t r a r y subset of the complex plane ( §17). 5. PART I RINGS OP DIFFERENTIABLE FUNCTIONS § 1. Rings• of i d l f f e r e n t i a b l e functions and ide a l s . In order to study the r e l a t i o n s betweenalgebraic prop-e r t i e s of (^(X) and topological properties of X , we s h a l l examine the special features of the family of zero-sets (see (1.6)) of an ideal,of C m(X) , where m w i l l be a f i x e d non-negative .integer or » . Hereafter we w i l l always r e f e r to m as an a r b i t r a r y integer such that 0 < m < '« . As we have learned i n Rings of continuous functions [73, such a family possesses properties analogous to those of a f i l t e r (see (1.8) to (1.15)). This f a c t w i l l play an important role i n the development of t h i s work. (1.1) Notation: We s h a l l write a l l equations involving n variables as i f there were a single variable present. For instance, we write f Q ( x ) f o r f ... Q (x^..., x n ) * y C x ' ) f o r ^ * \ f(xj_, x^), o X ^ 1... 3x^ x i (J) for ... (*«)•, 1 n etc. For any n-fold subscript k, we also l e t ak - \ + ... + k n , and 0 f c H = a k + o^ # By d(x,y) we s h a l l mean the distance between x and y f t , . ( x ) Note that f k ( x ' ) = £ •J| (x' - x ) * + R k(x',x) i s short f o r f . (x»,...,x«) = E • • • ' V ^ J * 1 ' *' I ' " ri x n 1^ + ... +ln < l n i r" m-(kT + . .. +k .) x l n V (x. - x x ) * l .,..(x n - ^ 5 ^ + ( x i , . . . . , ^ ; x 1,...,x n) (1.2) D e f i n i t i o n : Let f ( x ) = f Q ( x ) be defined i n the subset A of E n . ¥e say f(x) i s of class C m ( w i l l simply say C®) i n terms of the function f k i ^ ^ l 5 *' * ' x n U k l + * '' + k n — m ^ i f the functions f k k (x 1,...x n) are defined i n A for 1" * * vn a l l n -fold subscripts with k1+..'.+kn j< m , and f v . , ( x ) (1-1) f k(x») = E * ™ (x' -.x)** + R k(x'jx) °l: < m-ok f o r each f k ( x ) (o k.< m)* where R k(x'sx) has the following property. For any point x° i n A and any e > 0 , there i s a 6 > 0 such that i f x and x' are any two points of A with d(x°,x) < 6 and d(x',x°) < 6 , then (1-2) |R k(x*;x)| < d ( x , x ' ) m " o k e . Remarks: (!) I f m = 0 , then ( l - l ) and (1=2) state merely, that f ( x ) i s continuous. (2) For any i s o l a t e d point x° 3 and any £ > 0 , we may choose 6 > 0 so small that the only points x and x' s a t i s f y i n g d(x,x°) < 6 and d(x»-,x°)" <' 6 i s x° i t s e l f . Hence ( l - l ) becomes f k(x») = f k(x>) + R k(x';x !) so R k(x'jx) = 0 . Thus ( l - l ) and (1-2) are s a t i s f i e d even at the i s o l a t e d points 7'. (3) Moreover, i f A i s an open set, then f ( x ) ' i s • 0» i n the ordinary sense, and the f k ( x ) are the p a r t i a l derivatives of f(x) (see [30] § 3 ) . The converse i s true, by Taylor's Theorem. (1.3) D e f i n i t i o n ; Let X be any subset of E n . C m(X) = = (f : f i s a function which i s CP i n X}. C m*(X) = = { f 6 C m(X) : f i s bounded}. (1.4) Propositioni For any f and g e C m ( X ) , we denote the pointwlse addition, subtraction and m u l t i p l i c a t i o n of f and g by f+g , f - g , and f»g , respectively. Then C m(X) i s a commutative r i n g with unity demerit, denoted by U , (© i s the zero element) and C m*(X) i s a commutative Subring with U of C m(X). Proof i s obvious, (see Theorem 4 [31]). (1.5) D e f i n i t i o n : I f feC m ( x ) ( f e C m * ( X ) ) has a m u l t i p l i c a t i v e inverse i n Cm(X) {(fi*(X)) i s said to be a unit i n C m(X). (C m*(X)). (1.6) D e f i n i t i o n ; I f f£ C m(X) , then Z(f) = {x X : f(x) = 0} i s said to be the zero-set of r : Let Z(X) = (Z(f) : f6 (^(X)} . Remarks: ( l ) Note that i t i s clear that f£ C m(X) i s a unit i f f Z(f) = <|) (see Theorem 4 [31]'). Likewise, i f f i s a unit of C m*(X) , then Z(f) = § . But the converse need not hold, as the m u l t i p l i c a t i v e inverse f = 1 of f i n C m(X) may not 2 be a bounded function. For example: Let f(x) = e'"x . Then f € C m ( E 1 ) and Z(f) = <}) . But f " 1 ( x ) = e+* 2 , f " = = 1 e c r a ( E 1 ) - C ^ E 1 ) . (2) We s h a l l show that every closed subset of E 1 1 i s a zero-set of C^-differentiable function. 8. (1.7) D e f i n i t i o n : A nonempty subfamily gf of Z(X) i s said to be a z - f i l t e r on X , i f i t s a t i s f i e d the following condit-ions. (1) M s ; ( i i ) If Z1,-Z2e 3 , then Z1 n Zg e 3 ; and ( i l l ) I f Z £ 3 and Z« e Z(X) such that Z« z> Z , then Z'6 3? . (1.8) Proposition: I f I i s a proper i d e a l i n C m(X) / then the family Z[-l] = (Z(f) : f £ 1} i s a ' z - f l i t e r on X . Proof: ( i ) Since I contains no unit, ^ $,Z[l] . ( l i ) Let Z 1 3 Z 2 £ Z [ l ] , and ^ ^ f g g I such that Z 1 = Z^f^) , Z 2 = Z ( f 2 ) . Since I i s an i d e a l , ff + f | e i and z(f 1) n Z(f 2)= Z(f 2 + f | ) 6 Z [ l ] . That i s , Z X n Z 2 €Z[I] . ( i i i ) Let Z 6 Z[l] , Z'eZ(X) , f e l , and f'eC m(X) such that Z - Z(f) and Z8 - Z(f«) . Since I i s an i d e a l , ff» e I ,. However, Z' = Z we have Z» = Z» o"Z = Z(f') U Z(f) = Z ( f f ' ) e Z [ l ] . Q.E.D. Remark: The analogue of Prop. (1.8) with C m*(X) i n place of (^(X) i s f a l s e , i n general. For example, l e t us consider C m*(E^) and l e t f ( x ) = . Then f e . C m * ( E 1 ) . Set I = ( f ) , 1+x^ the i d e a l generated by f i n C m * ( E 1 ) . Then, i t i s clear that Z[I] s a t i s f i e s ( i i ) and ( i i i ) . However, Z(f) = <J)fiZ[l]. Note that " i d e a l " always means proper i d e a l , unless the contrary i s mentioned. (1.9) Proposition: If 3 i s a z - f i l t e r on X , then the family Z _ 1[??] = ( f e C m ( X ) : Z ( f ) e 3 ] i s an i d e a l i n (^(X) . Proof: ( l ) Let J = Z" 1^]-. Since <t> 3 , J must contain no unit so that J i s a proper subset of C m(X). Let f , g £ J , and h 6 C m ( X ) . Then Z(f.+ g) o Z(f) n Z(g)fi3 . By property ( i i ) of z - f i l t e r Z(f + g) e 3 . That i s , f + ge J - TTH*]. (2) Z[hf] = Z(h) U Z(f) O Z(f)£3 . By property ( i i i ) of z - f i l t e r again Z ( h f ) e 3 . Thus hf e J = Z" 1[5] . Hence Z""1^] i s a proper i d e a l i n ^ ( X ) . Remarks: ( l ) For f ;geC f f i(X) , the i d e a l (f,g) generated by f and g i s proper i f f Z(f) meets Z(g) . Equivalently f 2 + g 2 Is not a unit .of C m(X) . (2) ZlZ'Hv]] = 3 . For, l e t I = = { f e C m ( x T : Z ( f ) e 3 ) . Then ZlZ"1^]] = Z [ l ] = £Z(f) : f £1.} = (Z(f) : Z ( f ) ^ a f ) = 3 . Remark (2) shows that every z - f l i t e r i s of the form Z [ l ] for some i d e a l I i n C m(X) . (3) It i s clear that Z " 1 [ Z [ l ] ] r> I . The i n c l u s i o n may be proper. For instance, (a) we consider C (E ) , where m Is any p o s i t i v e integer. Evidently i e C m ( E 1 ) , where i (x) = x f o r a l l x f E 1 . I f I = ( i ) , then I consists of a l l functions f i n c f^E 1) such that f(x) = x»g(tf) f o r some g € C (E ) so that every function i n I vanishes at 0. Hence every zero-set In Z [ l ] contains the point 0. As a matter of f a c t , since Z [ l ] Is a z - f i l t e r that includes the set {0} , i t must be the family of a l l zero-sets containing 0. 1 0 . Let MQ = Z° 1[Z[I]] • Evidently i t consists of a l l functions i n C (E ) that vanish at 0 . Hence M c e r t a i n l y 3m+l contains I . However, Ml 4 I , for i ^ & M , and i f 3m+l 3m+l ° ° i " ^ € I , then i ~ ^ = g»i f o r some g g C ^ E 1 ) . But, 3m-r2 3m+l then g = i ~ ^ 4 C ^ E 1 ) . Hence i 5 £ M - I . Note that here M Q i s a maximal i d e a l . For, i f f t M Q , then Z(f) n Z ( i ) = <j> . From Remark ( 1 ) ( i , f ) = C ^ E 1 ) . Since (M Q,f) 2 C 1 ^ ) * (M Qf) = C ^ E 1 ) . In other words, M Q i s maximal. 171 1 (b) In case m = « , l e t us consider C (E ) , and l e t f 1 ( x ) = e f o r a l l x e E . Then f^e b m ( E 1 ) , and f ^ x ) | 0 f o r x f i E 1 and x + .0. Let I = ( f ^ ) .. This consists of a l l functions f i n C ^ E 1 ) such that f ( x ) = f-^x) g(x) f o r some g e'C^E1) so that every function i n I vanishes at 0 . Hence every zero-set i n Z[.l] contains the point 0 . As a matter of f a c t , since . Z [ l ] i s a z - f l i t e r that includes the set {0} , i t must be the family of a l l zero-sets containing G. Now, M Q = Z" 1[Z[I]] evidently consists of a l l functions i n (^(E 1) that vanish at 0 . Hence M Q c e r t a i n l y contains I . Moreover, i & M . I f i £ I , then x = g(x) f- L(x) for some g e C ^ E 1 ) . Consequently, g(x) = x/f^x) , i f x + 0 (as f x ( x ) = 0 only when x = 0 ) , and ^ J j . g(x) = - » , x _ 0 + g(x) = + » . Hence g cannot be continuous at the point 0 . This i s a contradiction. 11. In general, x n e M_ - I , for a l l p o s i t i v e integers n , hence f o r any f € (^(E 1) which Is 'analytic at the o r i g i n without a constant terra i s i n M„ - I . o Here, Z:[l] = Z[M Q] i n spite of the fac t that M Q contains I properly. Besides, M Q i s a maximal i d e a l i n C ^ E 1 ) . For, I f f £ MQ , then Z(f) i s d i s j o i n t from Z(f£) and so ( f - ^ f ) = G m(E 1) . Hence (M Q,f) = ( f - ^ f ) = C ^ E 1 ) . (1.10) Definitions A z - u l t r a f l i t e r on X i s a maximal z - f i l t e r . Note that every subfamily of Z(X) with the f i n i t e i n t e r s e c t i o n property, by Zorn's Lemma, i s contained i n some z ~ u l t r a f l i t e r on X . (1.11) Proposition: I f M i s a maximal i d e a l i n C m(X) , then Z[M] i s a z - u l t r a f l i t e r on X . Proof: It i s obvious that I f I ] ^ ^ & r t e ' f c w o a r b i t r a r y Ideals i n C m(X) such that I x e I 2 , then Z [ l J c Z[lg] . Also, i f 3^ and 3 2 are two a r b i t r a r y z - f l i t e r s In Z(X) such that 3^ c 3 2 , then Z " 1 ^ ] c Z " 1 [ 3 ? 2 ] . Hence the r e s u l t follows immediately from Propositions (1.8) and (1.9) • (1.12) Proposition: I f ^ i s a z - u l t r a f l i t e r on X , then Z _ 1[>£] i s a maximal Ideal i n C^X) . Proof i s similar to that of Prop. ( l . l l ) . I t follows from Propositions ( l . l l ) and (1.12) that the mapping Z i s one-one from the set of a l l maximal ideals i n C m(X) onto the set of a l l z - u l t r a f l i t e r s on X . 12. (1.13) Proposition: Let M be a maximal i d e a l i n (^(X) . If Z(f) meets every member of Z[M] , then f £ M . Proof: Suppose that f M . Then, since M i s a maximal i d e a l , (M,f) = (^ (X)... We know that the unity UeC m (X) = (M,f). Hence ( X = h + g o f f o r some heM and geC m (X) . Hence Z(h + g.f) = Z(U) = <|) . But (j) = Z(h + g o f ) 3 Z(h) n z(g-f) = z(h) n [ z(g) u z ( f ) ] = [ 'z(h) n z(g)} u f z ( h ) n z ( f ) ] . By hypothesis Z(h) ft Z(f) 4= $ • Thus, we have <|) o [Z(h) n Z(g)] U [Z(h) 0 Z ( f ) ] < l > • This i s impossible. Hence f e M. (1.14) Proposition: Let ST be a z - u l t r a f l i t e r on X... If a zero-set Z meets every member of >f , then Z € ^ . Proof: Since Z meets every member of ?fr , & U {Z} has the f i n i t e i n t e r s e c t i o n property. By Zorn's Lemma, ^ U (Z) generates a z-ul t r a f l i t e r which contains ^f" . But •/4~ i s a z- u l t r a f l i t e r , y f ' = A'. Hence Z : ^ ^ = . / f . (1.15) Definition:' An id e a l I i n Cm(X) i s said to be a z-ideal i f Z(f) e Z [ l ] implies f £.1 . That i s , I = Z" 1 [Z[I]] . It i s obvious that every maximal i d e a l i s a z-idea l , while the i d e a l s I = ( i ) and 1^ = (f^) , given i n Remark (3) of Prop. (1.9) are not z-ideals. Zero-sets arid m-completely Regular Spaces. In t h i s section, we w i l l show that every closed subset of E n i s a zero-set. (2> 1) Lemma: Let f n ( x ) = x m e " n x f o r each x£ . [ 0 > . ) , where 13-m i s a ..-•'fixed p o s i t i v e integer. Then { f ' : n e N] converges uniformly to 0 on [ 0 , » ) , where N always denotes the set of a l l p o s i t i v e integers. Proof: f < (x) = m x m - V n x - n x m e - n x = x m- 1e- n x(m - nxj . n> Set " f ( x ) = 0 . We have x = 0 or x = ^ . n v n • f»(x) = (m- i)x m - 2 e " " n x ( m - nx) - nx m"" 1e n x(m - nx) -^ m - l ^ n x = xTO-2g-nx ( m2 _ m _ 2mnx +; n 2 x 2 ) . f " g ) = e ) m - 2 e - m ( m 2 - m - 2m 2 + m2) ^ e" m < 0 . n vn' v n v nm-^ Hence f n ( X i ) h&s i t s maximum at x « •£ with value ( B ) m e ~ m ..' Now, f o r any given e > 0 , take N Q to be the p o s i t i v e integer such that N Q > m/e-Je . Then (|) me" m _< ) m e ~ m <_ • o mm o. \ } „ 'o e**m = e , whenever n > N . That i s |f I < e m /e e whenever n _> N Q . Hence {g n : n € N} converges uniformly to 0 . Q.E.D. Let c i E n B r(a) = {x e E n : Ux - al! < r) and c t E n B r,(b) = {x € E n : ||x - b|! _< r'} , where a = ('ap..'. ,a ), b = ( b ^ . . . , b n ) , x = (x^,. .. , x n ) , and r / r ' = k: . Then, i t i s evident that c-tgn B r(a) i s C^-diffeomorphic with ci^n Br,(h) under '. 1 1 <t>(x) = (<^(x),... ,<j)n(x)) , where' • 4>±(x) = ^ (x± - a ±) + b ± , 1 1 1 i n • Now, i f f i s a function defined on cl^,n B r ( a ) , then g(x) f ( k ^ •--•'•b1) + a^..; ,k(x n - b n) -+ a n) = f . f ^ x ) i s a function defined on c t E n B r, (b) , where <j)"^ "(x) = (<$>j(x),..., <t>n(x)) , with tjj^ * k ( x i - \>^) + 'a 1.y I X i _< n . Moreover, 14. I f f'./e. C* , then g e C 8 9 , and i f | D i f ( x ) | i s hounded by M o i * then (D^g^x)! i s hounded by k mM c r i > f o r an a r b i t r a r y p o s i t i v e i n t e g e r m , where D ^ f ( x ) , D ig(x) and <ji ' are as shown i n (1.1). ( 2 , 2 ) Theorem: Let F be an a r b i t r a r y c l o s e d subset of E 1 1 . Then F i s a C % z e r o = s e t , I.e. t h e r e i s f € C ^ E 1 1 ) such t h a t Z ( f j = F . PrOof: We know t h a t E 1 1 - F B r i ( a i ) , where B r i ( a i ) denotes the open b a l l w i t h r a d i u s r i 4= Q , and center' 1 - ( a * , . ..,a^)... Let l±+1 = r j / r i + i f o r 1 > 1 • We now f i r s t - d e f i n e Then i+1 > O Choose 0 < r c < r i • ep(t) =|e 0 t > 0 t < 0 I t i s clear that cp e. ^ ( E 1 ) . Next define S( s> = p ^ - s T * tp(s-r Q) ' ^ S c C ^ E 1 ) f ^ x ) - g(j|x - all) . Then ^ e C*(E n) and f-Jx) = {1 f o r x e clm B r (a 1) 0 f o r x € E 1 1 - B r ( a 1 ) 0 < f x ( x ) < 1 elsewhere . Let f 2 ( x ) = t1(i2{Xia2.) + a j , . . . , t 2 ( x n - a 2 ) ni a*) . F i n a l l y , l e t Then we have f l ( x ) 1 f o r x e c ^ B ^ ^ a 2 ) G f o r x e E 1 1 -• B - ( a 2 ) r 2 G < f 2 ( x ) < 1 elsewhere, and f 2 e C * ( E n ) 15. I n g e n e r a l , we d e f i n e f ± ( x ) = - a j ) + a*" 1.,... , l ± ( x n - a * ) + a*" 1) . f o r i = 2 , 3 , . Then, we have f±(x) = 1 f o r xe oe^n B„ ,-,. . N ( a i ) 0 f o r x e - B _ ( a 1 ) r I 0 < f ^ ( x ) < 1 elsewhere By choice of ( f n ) , we know t h a t D^f-^ i s continuous f O r a l l k e N , and v a n i s h e s except on c - t E n B r ( a 1 ) - B r ( a 1 ) which i s '1 o compact1. Thus I s hounded on c-t En B r ( a 1 ) so i s bounded on E 1 1 . Let l ^ ^ l l - M a k f ° r a k € N * Then, we have | D k f 2 | <. if^^ - I n g e n e r a l , \\?±\ 1 •' ^ i ^ k ' f . ( X ) 4 ^ 4 J - t -? Let f ( x ) = E - T r — 3 where C, = e J x J , f o r i e N . i = l °i 1 We w i l l show t h a t f € G w ( E n ) . By Theorem (7.17) [23] , we Vi(x) then have t o show t h a t f o r each a v e N , 2 H converges K 1=1 °i u n i f o r m l y . However, i = l c i *~ 1=1 G i Q & 1=1 °i Thus we o n l y have t o show t h a t § -^-i—^- converges. Use i = l °i the " R a t i o t e s t " , (Jg - t j k / c i + i ) / (jli ^Vcj) = * i J 1 / e ( * B y V i r t U e o f L e m m a (2' 1) > iS * i V e ( i + l U i + l = 0 < B y c o r o l l a r y ([5], P.108), 16. 1=1 2 i s convergent. Therefore OS s =1 i converges uniformly for each ak s N , so that f e C (E ) . Moreover, Z(f) = ff Z(f.) , as each function f . i s a nGn-negative (2.35) Definitions Let X be a t o p o l o g i c a l space. X i s said to be m-completely regular i f f o r every closed subset F of X and x e X-F , there e x i s t s a function f e C m(X) such that f ( x ) = 1 and f ( F ) • [0} . (2.4) Theorems A topo l o g i c a l space i s m-completely regular i f f the family Z(X) = (Z(f) s f c C m(X)] i s a base f o r the closed subsets of X . Proof: Necessity: Suppose that X i s m~completely regular. For any closed subset F of X' and x e X-F , there e x i s t s an f e C m(X) such that f ( x ) =1 and f ( F ) = {0} . Hence Z(f) 3 F and x f Z(f) . Consequently Z(X) i s a b a s e f o r the closed subsets of X . (See [7]) . S u f f i c i e n c y : Suppose that Z(X) i s a base fo r the closed subsets of X . For each closed subset F of X and x e X-F , there i s g e (^(X) such that Z(g) o F and x 4 Z(g) . Let g(x) = r . Then r + P . • Let f = g o r " 1 , -1 - i where r , i s the constant function of. value r <. . Then f e C m(X) as f 1 and g e C m(X) . Moreover, f( x ) = 1 and f ( F ) = {0} Q.E.D. i = l 17. ( 2 . 5 ) D e f i n i t i o n : A topological space Is said to be m-normal If f o r any d i s j o i n t closed subsets F-^ and F 2 , there i s an f € C m(X) such that f (P^), = ( 0 } and f ( F 2 ) - {1} . Having proved that every closed subset of E 1 1 Is a zero-set, we can show that every subset of E 1 1 Is m-completely regular as well as m-normal. (2.6) Proposition: Every subset X' of E 1 1 i s m-normal. Proof: Let F.^ and F 2 be any two d i s j o i n t closed subsets of X . By property of the r e l a t i v e topology, we have two closed subsets F£ and F 2 of E 1 1 such that F| fl X = , i = 1,2. By Theorem (2.2), there are f ^ f g € C m ( E n ) such that Z(t±) = F j . Let f±\X = g ± . Then, g ± € C m(X) and Z ^ ) = Z ( f ± ) n X = F[ n X -. F so that Z(g±) n Z(g 2) = <j> . Now, set f = g x g . Since Z(g 1) n Z(g g) = <|) , f i s well-defined and f e G m(X) . Moreover Z(f) = Z ^ ) and f ( Z ( g 2 ) ) = (1) . Hence f ( F x ) = { 0 ) , and f ( F 2 ) = [1} . Q.E.D. (2.7) Corollary: Every subset X of E 1 1 Is m-completely regular. Proof: Since X i s Hausdorff and m-normal, X i s m-completely regular. Q.E.D. § 3 Fixed, free i d e a l s and compact spaces. In t h i s section we s h a l l see the characterization of fi x e d maximal Ideals of C^(X) (see ( 3 »3 ) ) and how they are 18. related to a compact space (see (3.11)). (3.1) Definition: Let I be any ideal in (^(X) or Cm*(X) . Then I is said to be a fixed ideal i f nz[l] is not empty; and I is said to be a free ideal i f nZ[I] is empty. Remarks: (l) In view of the definition of free ideal, we know that I is free i f f for each x e X , there is a function in I that does not vanish at x . (2) If I is a fixed ideal in C^X) , then the set S = flZ[I] is not empty, and the set I« = [f e (^(X) : f[S] • (0)} is a fixed ideal Containing I . Hence a fixed maximal ideal must be of this form. Furthermore, since I' can be enlarged by making S smaller, the only candidates for fixed maximal ideals are the ideals I' for which S contains just one point. The corresponding statements hold for Cm*(X) . (3) The ideal I' mentioned above evidently contains the z-ideal Z ° * 1 [ Z[l]] . In general, the two are not the samej the set S = nZ[l] need not be a member of Z[l] , even i f S is a zero-set. To see this, let us consider a space X c E 1 , where X contains at least one point which is not isolated, say a . Let 0,. denote the set of a l l f e Cm(X) for which Z(f) is a neighborhood of a . 0 is then a z-ideal. We claim that nz[0 1 = {a} . Suppose b € nZ[0_] and b 4s a • Since X is m-completely regular there is an f e Cm(X) such that Z(f) is a neighborhood of a but b 4 Z(f) . That i s , f e 0 Q so that b.| n z [ 0 j which is a EL cl contradiction. Hence nZ[O a ] «= {a} . Now, since {a} is a 19. closed set, by Theorem (2.2), i t i s a zero-set. However, "a" Is not an Isolated point i t cannot be a neighborhood of i t s e l f . Hence {a} i s not a member of Z[0 a] • Note that since nZ[l] i s a closed subset i n X , by (2.2), I t i s a zero-set. (Compare [73 P.55). (3.2) Notation: I f I i s an i d e a l i n C m(X), 1(f) w i l l denote the residue class of f , f+Ij r w i l l denote the constant function on X of value r f o r a l l r e /R, the f i e l d of r e a l numbers. (3.3) Theorem: (1) The f i x e d maximal ideals i n C m(X) are pr e c i s e l y the sets ^ = {f e C m(X) : f ( p ) > 0} . (2) The Ideals are d i s t i n c t f o r d i s t i n c t P • (3) For each p , (^(Xj/Mp i s isomorphic with the r e a l f i e l d fR. In f a c t , the mapping Mp(f) -• f(p) Is the unique isomorphism;.of Cm(X)/Mp onto fR , where p € X . Proof: ( l ) Consider the mapping cp from (^(X) to fR , defined by ep(f) = f(p) . Evidently i t i s a homomorphism, and i t s kernal i s Hp . Since f o r each r e fR. 9 r_(p)-= r , so that ep i s onto the f i e l d fR 1 Hence i t s kernal Is maximal. On the other hand, i f M i s any f i x e d i d e a l i n (^(X) , there e x i s t s a point p e flZ[M] . Clearly, M i s contained i n which has just been shown to be a maximal i d e a l . Hence i f M i s maximal, then we must have M = . (2) I f p 4s P' » hy the m-completely r e g u l a r i t y of X , there i s f e C^X) such that f(p') = 1 and f(p) = 0 20. (Prop. (2.3)). Hence f e Mp but f {Mp, . That i s , (3) Form; the proof of (1) , Mp is the kernal of a homomorphism of Cm(X) to , so. that Cm(X)/Mp is isomorphic with fk . The uniqueness of the isomorphism follows from the fact that the only non-zero automorphism of fR is the identity [7, -(0.23)3 . ! (3.4) Theorem: (l) The fixed maximal ideals in Cm*(X) are precisely the sets M*• « {f e C^X) : f(p) = 0} . (2) The ideals M* are distinct for distinct p. (3) For each p, Cm*(X)/M* is isomorphic with the real f ie ld fR . In fact, the mapping M*(f) -» f(p) is the unique isomorphism of Gm*(X)/M* onto , where p "e X . Proof is identical with that of Theorem (3.3) except for the notation. (3.5) Proposition: If X is compact, then every ideal I in Cm*(X) is fixed. Proof: We know that Z[l] is a family of closed subset with finite intersection property in the compact space X Hence nz[i] + (|) . (3.6) Propositions If X is a compact space, then the corres-pondence p Hp is one-one from X onto the set of a l l maximal ideals in C^X) . Proof: In view of Prop. (3.5) , every ideal is fixed. By Prop. (3.3), each fixed maximal ideal in Cm(X) is of the form 21. Mp for some p € X , and •« J ^ t Iff p = p« . Hence the assertion holds. Q.E.D. We now shift our emphasis from Ideal to z=fliter, (see (1.8) to (1.12)) . (3.7) Definition? Let 3 be a z-f i l ter in Z(X) . Then 3 is said to be a free or fixed z-f i l ter according as 03 = <j) or + 0 • The following property Is the immediate consequence of (3.7) . (3.8) Proposition: Every ideal of Gm(X) is fixed i f f every z-f i l ter Is fixed. (3.9) Lemma: A zero-set Z e Z(X) is compact i f f i t belongs to no free z-f l i ter . Proof: Necessity: Suppose Z Is compact and Z belongs to a z-f i l ter 3. Set 3' = {Z 0 P : F e 3) . Then (1) (|) 4 3' * since Z and F are in 3, Z fl P f f ; ( i i ) ( z n F) n . (z n F«) = z n (F n F«) e an as F n F' € 3 . Thus 3' Is a family of closed sets in Z with the finite intersection property. It follows that 038 4s $ • But H3 = 03' . That i s , 3 is fixed. Sufficiency: Let B be any family of closed subsets of Z with the finite intersection property. Since Z is closed In X , the members of £ are closed in X . Let 3 be the collection of a l l elements In Z(X) each of which contains 22, a f i n i t e i n t e r s e c t i o n of members of e . E v i d e n t l y , 3 i s a z - f l i t e r and Z i s i n 3 . (Since Z € B ) . By h y p o t h e s i s , flSP 4s 4^ • We know t h a t every c l o s e d s e t i s a z e r o - s e t and v i c e v e r s a . Hence nfl = A3 4 ty • T h e r e f o r e , Z i s compact. (3.10) P r o p o s i t i o n : L e t Mr be a z - u l t r a f l i t e r on X , and each of i t s members be noncompact. Then i s f r e e . P r o o f : Suppose t h a t p Q e fts4" • Then tp Q} i s a c l o s e d s e t so i s a z e r o - s e t . Now, (p Q} meets every member of at P G » "by Prop. (1.14) , {p Q} ejdf . T h i s i s a c o n t r a d i c t i o n ( 3 . 9 ) • Hence ^ i s f r e e . Q.E.D. Remark: N e v e r t h e l e s s , i t i s not t r u e t h a t i f every member of a z - f i l t e r 3 , i s noncompact, then 3 i s f r e e . For i n s t a n c e , c o n s i d e r C ^ E 1 ) , and l e t f (x) = S i n x . Then f e (^(E 1) , but z ( f Q ) 4 ty i s not compact (as Z ( f Q ) i s not bounded). Let I = ( f 0 ) > the p r i n c i p a l i d e a l i n C m(E^") generated by f . Hence every f u n c t i o n f e I can be w r i t t e n as f ( x ) = f 0 ( x ) g ( x ) f o r some g e G m ( E 1 ) . Then Z ( f ) c o n t a i n s Z ( f Q ) so t h a t Z ( f ) i s not bounded. Thus Z ( f ) i s not compact f o r any f e I . However, n z [ I ] = Z ( f Q ) 4 ty • In g e n e r a l , f o r any f € C ^ E 1 ) such t h a t Z ( f ) i s not bounded, the z - f i l t e r Z [ l ] , where I = ( f ) , i s f i x e d i n s p i t e of the f a c t t h a t each member of i t i s noncompact. ( J . - l l ) Theorem: In X c E 1 1 , the f o l l o w i n g are e q u i v a l e n t : (1) X i s compact. (2) Every i d e a l i n C m(X) i s f i x e d , i . e . every z - f i l t e r i s f i x e d . (2*) Every i d e a l i n Cra (X) i s fixed, (3) Every maximal i d e a l i n (^(X) i s f i x e d , I.e. every z - u l t r a f l i t e r i s fixed. (3*) Every maximal i d e a l i n C m*(X) Is fix e d . Proof: That ( l ) i s equivalent with (2) follows from Lemma (3-9) with Z = X and the f a c t X belongs to every z - f i l t e r i n X . Likewise, ( l ) implies (2*) because when X Is compact C m(X) = C (X) . Now, suppose that (2*) i s true; and 3 i s any family of closed subsets of X with f i n i t e i n t e r s e c t i o n property. Let A = {f e C m(X) : 0 _< f ^ 1 and Z(f) ^ F, f o r some F e 3} , and e = (Z(f) : f e A} . Then, i t i s clear that DB = 03 . Let I = (A) , the i d e a l i n C m*(X) generated by A . By hypothesis, 0Z[l] =f= $> • But Z [ l ] 3 8 , Ofi 2 n z[l] 4= • Hence 03 + $ • In other words X i s compact. This shows that (2*) implies ( l ) . Hence ( l ) Is equivalent with (2*). Consequently (2) and (2*) are equivalent. F i n a l l y , (2) i s equivalent to (3) and (2*) with (3*), because every free i d e a l i s contained i n a free maximal i d e a l . Q.E.D. §4 Real Ideals, m-realcompact Spaces. ' In 19^8, E. Hewitt defined r e a l maximal Ideals and r e a l -compact spaces (Q-spaces) (see [8], §7 and [7], Ch. 5). He also contributed many Interesting properties about r e a l maximal Ideals and realcompact spaces ( [8 ] and [7]). Unfortunately, those properties can not be carried to the rings of C m-d i f f e r e n t i a b l e functions, since they are not lattice-ordered rings. (See [7] (0.19) and Ch. 5). Indeed* i f f e (^(X) i s not nonnegative or nonpositive, then | f | 4 C m(X) . 24. Recently/(1964), R. Bkouche has shown that every para-compact Hausdorff dlfferentiable n-manlfold is m-realcompact (see (4.2) and [4] Theorem 2). Here, we wi l l show that every closed subset of E 1 1 is m-realcompact. Moreover, we know that every -differentiable n-manifold with countable basis can be C -^embedded in a closed subset of E 2 n + 1 (cor. 1.J2 [ l6 ] ) . We know that every residue class field of Cm(X) or Cm*(X) module a maximal Ideal contains a canonical copy of the real f ield fR : the set of Images of the constant functions under the canonical homomorphism. For, let M be a maximal Ideal In (^(X) , and r^ + r 2 be arbitrary constant functions. If M(r^) - M(r 2) , then 'M(rx - r g ) = M(r_1) - M(r 2) = 0 . That is - r 2 e M . This is impossible for x2 ~ l s a unit. Hence M(r_1) =f M(*2) * ' 6 1 1 ( 1 that the set K = {M(rJ : r e fR}-, forms a field is clear. We shall Identify this sub-field with fR . Thus, in Theorems (3.3) and (3.*0* we can write Mp(f) - f(p) for a l l f e Cm(X) , and M*(f) = f(p) for a l l f e Cm*(.X) . (4.1) Definition: Let M be any maximal Ideal in Cm(X) (or Cm*(X)). Then M is said to be a real ideal i f the canonical copy of fR is the entire f ield Cm(X)/M (respectively Cm*(X)/M) , land M is said to be a hyper-real ideal If the canonical copy of fR is not the entire f ield Cm(X)/M (respectively Cm*(X)/M). Remark: By Theorems (3.3) and (3 .4) , every fixed maximal Ideal in Cm(X) or Cm*(X) Is real . 25. (4.2) Definition: A topological space X Is said to be m-realcompact, i f every real maximal ideal in (^(X) is fixed. It is clear that i f X is compact, then X is m-realcompact. We Will show next that every closed subset of E 1 1 Is m-realcompact. (4.3) Lemma: An ideal in Cm(X) is free i f f for every compact subset A of X there exists an f e I having no zero in A . Proof: Necessity: Suppose I Is free and A is an arbitrary compact subset of X . If for each f e l , Z(f) n A + ty > then the family 3 = {F = Z(f) n A : for some f e 1} is a family of closed sets in A having the finite intersection property. Indeed,' for each P 1 , P 2 e 3 > we have F± = Z(t^) 0 A for some f ± e I , I = 1 ; 2 . Then P ^ P g = ( Z ^ ) n A) n z ( f 2 ) n A) - (z ( f x ) n z(f 2 )) n A = z(tf + t\) n A + ty , as f ? . + f f e I . Thus m \ ty . Hence nZ[l] = n Z(f ) 3 1 * f€l ~* n Z(f) n A = 03 + ty'• This is a contradiction. Hence there fe l is f e l with Z(f) n A = ty.. Sufficiency: Suppose for every compact set A c X there is an f e l such that Z(f) A A = ty . Then, for any {x} which is compact, there Is an f e l such that Z(f) n {x} m ty . That Is, x 4 Z(f) . Hence nz[l] = ty . Q.E.D. (4.4) Proposition: Let X be a closed (unbounded) subspace of E 1 1 . Then X is m-realcompact. 26. Proof: Suppose that M i s a free maximal Ideal and (^(Xj/M i s the r e a l f i e l d fR . Let g(x) =•-—j; . Then that M 2 * 1 g € C m(X) and g i s a unit i s clear. Hence g 4 M . That i s , M(g) 4 0 . For any p o s i t i v e number r and a s u f f i c i e n t l y small number e > 0 , g < x - £ f o r a l l but a compact subset of E 1 1 , say Ay . Then = A ' fi X i s compact i n X as X i s closed. Let A' = c-t'(X - B ) which Is closed i n X so i s closed i n E 1 1 . Thus, there i s an f e (^(E 1 1) c C m(X) such that Z(f) = A' . We w i l l show that Z(f) e Z[M] . I t i s enough to show that Z(f ) D Z for some Z e Z[M] . However, we know that B£ i s compact i n X . By Lemma (4.35), there Is t1 e M such that Z ( f 1 ) (1 B =• |l . In other words, Z(f ] L) c X .- B 6 e c t x ( X - B £) = Z(f) . Hence Z(f) € Z[M] . There-fore, g < r - 6 on the zero-set Z(f) , and r - g j> e . Let h± = (f - gJ2 on Z(f) . Then h± i s c"1 on Z(f) which i s closed i n E* . By Whitney's Analytic Extension Theorem (see [30]), we have a C m extension h . That i s o h|Z(f) = hj, . Hence h = r - g on Z(f) . Therefore, h s r - g (mod M). In other words M(h ) = M(r - g) = M(r) - M(g) = r - M(g) . But, since C m(X)/M i s r e a l o p M(h ) = (M(h)) J> 0 * so we have M(g) <_ r . Since r i s any p o s i t i v e number, M(g) i s I n f i n i t e l y small. This i s a contra-d i c t i o n . Q.E.D. §5 The Long Line. We now w i l l give an example to show that a non-paracompact space may not be an m-xealcompact space. • 2 7 . Let W be the set of a l l o r d i n a l 'members l e s s than the f i r s t uncountable ordinal, and {I : a e ¥} be a c o l l e c t i o n of a open i n t e r v a l s indexed by the set W , that i s , . I Is an open vX i n t e r v a l paired with an element a of W . We i n s e r t the open Interval I between a and a + 1 , where a +1 denotes the immediate successor of a (and a -1' w i l l denote the immediate predecessor of a i f a i s not a l i m i t o r d i n a l ) , and L denote the union of W and a l l the Inserted open in t e r v a l s . Now, we order L by the following f i v e conditions. Let x and y be any points of L . Then x < y I f (1) x and y are i n W and x <, y In W , (2) x i s i n W , and y i s i n an i n t e r v a l I , and x = a or x < a i n W ? (3) x i s i n i n t e r v a l I , and y Is i n W , and a < y i n W y (4) x i s i n i n t e r v a l I , and y Is i n an i n t e r v a l Ip , and a < 0 i n W ) or (5) x and y are both i n the same i n t e r v a l I and and x < y In X * a We then topologlze L by means of the order topology and the r e s u l t i n g space Is the "long l i n e " . 1 [9] . We s h a l l show now f i v e properties of the long l i n e . (5.1) Proposition: L s a t i s f i e s the f i r s t axiom of count-a b i l i t y (5.2) Proposition: L i s a Hausdorff space. The proofs of these two propositions are straightforward 28. from the d e f i n i t i o n of the long l i n e . (5.3) D e f i n i t i o n : An isotone mapping i s a mapping which i s order preserving. An isotone-homeomorphism f i s a mapping which i s a homeomorphism and both f and Its inverse f " 1 are isotone. And two ordered spaces are ca l l e d isotohe-homeomorphic i f there Is an isotone-homeomorphism from one onto the other. (5.4) Proposition: For each a e L , a + 1''.* [ l * a ] Is isotone-homeomorphic to the unit i n t e r v a l [0,1], consequently-each point of L , not the f i r s t element 1 , has an open neighborhood which i s homeomorphie to an open i n t e r v a l . Proof: We w i l l show t h i s proposition/by t r a n s i n f i n i t e induc-t i o n . We know that [1,2] i n L i s Isotone-homeomorphic to [0,1] In TS?~ . Now, assume that [ l , p ] In L i s Isotone-homeomorphic to [0,1] f o r each p < a . We have to show that [l, a ] i s also isotone-homeomorphic to [0,1] . Case 1, i f a i s a non-limit o r d i n a l , then by the Induction hypothesis [ l , a - l ] i s 1 Isotone homeomorphic to [0,1] . However, we know that [a-l,a] and [0,1], [0,1] and [O,-^],, and [0,1] and [•2*1] are isotone-homeomorphic. Thus, [ l , a ] i s isotone-homeomorphic to [0,1] . Case 2, I f a i s a l i m i t o r d i n a l , then there exists a sequence [ a n ] which converges to a . F i r s t of a l l , we s h a l l show that there exists a sequence of functions [ f n 3 such that f n i s an isotone-homeomorphlsm of [ l , a n ] onto [0, -JJ-^ -] , and f n \ l 1 » a n a i ^ f n - l * f o r a 1 1 n € N o 29. By our induction hypothesis., there e x i s t s [g^] such that i s an isotone-homeomorphism of [ l , a n ] onto [1,0] . Let e n map [0,1] to [0, -~j^;] such that e n(x) = x fo r each n e N . Evidently e i s an Isotone-homeomorphism of [0,1] XX onto [0, -—^j-] . Let h n he a mapping defined as follows: h n = e n o g n on [ l , a R ] f o r each n e N . Then ^ i s an isotone-homeomorphism of [ l , a ] onto [0, •=§_] . Now, l e t f^(x) = h^(x) f o r x € [ 0 , 0 ^ ] and f 2 ( x ) = g2<>h2(x) f o r each x e [a^,a 2] > where g 2 i s defined as follows; 1 1 y ~ h 2 ^ a i ) e^y) — g +s i , h 2 ( a i ) f o r y € * h 2^ a 2) ] -[ h 2 ( a 1 ) , 2/3] . Since g 2 Is a l i n e a r transformation, i t i s 1 2 an isotone-homeomorphism of [h^a-jO, 2/3] onto [•£, ^ ] . Therefore f 2 i s an isotone-homeomorphism of [0,a2] onto [0,2/3] such that t^[03a^] = f-^ . And so on. In general, f o r each n , l e t f n ( x ) = ^ n „ 1 ( x ) f o r x e [°» a n-il » f n ( x ) = g ^ h ^ x ) f o r x e [ a n = > 1 , a n ] . where ^(y) = ^ + -n - l x y h n ( a n - l ^ •=-—=•) • — — i s an Isotone-homeomorphism of n n u /„ \ n+I " h n < a n - l } ^ h n ^ a n - l ^ nTX^ o n t o ^ n r * nTT^ Thus, f n i s an isotone-homeomorphism of [ l , a ] onto [0,-^j] such that ^ n K 1 * 0 ^ ! ^ = f n - l . Therefore, we have obtained the required sequence of functions [ f n 3 . Next, we s h a l l show that the sequence [ f n ] has a l i m i t f , say, such that f i s an isotone-homeomorphism of [ l , a ] onto [0,1] . Define f ( x ) = f°n(x) f o r each x € [ l , a ) , and f ( a ) = 1 . Then the function f i s we l l -30. defined. Indeed, f o r each x € [ l , a ) , there i s an n such o that n > n Q implies x < an so that n > n Q implies f n ( x ) i s constant, and hence the l i m i t always exists. Then the following f i v e properties follow Immediately from the d e f i n i t i o n of f and the choice of {f 1 s (1) f i s one-one . (2) f " 1 i s well-defined„ (By (1))'. (3) f ( a n ) = f n ( a n ) = f o r each n , so that i f f f ( 0 - 1 , n-«* v n' (4) f and f = 1 are Isotone mappings, and f i n a l l y (5) f and f " 1 are continuous at each point of [l , a ) and [ 0 , 1 ) respectively. To show that f i s continuous at a , we have to show *1 J yy. that f o r each sequence CPn} converging to a , f (P n ) = f ( a ) (= 1) . (By Prop. (5.1) and theorem (5.3*0 [19] )• Without loss of generality, we may assume that |3 n < P n + 1 f o r each n £ N . Since f i s an isotone mapping {f(£„)} i s a n monotonic, nondecreasing sequence. It i s clear that (f ( P n)3 i s hounded above by 1 . Thus f ( ^ n ^ exists- I»et b = JJjJ f ( P n ) • The* b 1 1 • I f b < 1 / t h e n by (5)> we know that f Is continuous at b , so that f(p„) = b which implies ^ 6 n « f = 1 ( b ) t But f°1(b)=^Cand P n = a . This i s impossible. Hence b = 1 , or £jj f O n ) = 1 = f ( a ) • Consequently, f i s continuous at a . Now, we want to show that f " 1 Is continuous at 1 . We know that the space [0,1] 51. s a t i s f i e s the second axiom of countability, hence again i t i s enough to show that f o r any sequence [b n] which converges to ^'"^"^n^ = = a * W e m a y also assume that {b n) i s a monotonic non-decreasing sequence converging to 1. Since f°°^ i s an isotone mapping {f~^"(b n)3 i s a monotonic non-decreasing sequence. I t i s also clear that {f" 1(b )} Is bounded above by 1 . Thus J^JJ f°°1(b)^ exists . Let 0 = i i f j f" 1(b„) . Then 0 < a . I f 0 < a , then by (5) as n~*°» xi —• f " 1 Is continuous at 0 , ^ J J J f " 1 ( h n ) = 0 implies Urn b = lim --I/*. N „ f(0) . So that we have f(0) + 1 > and J^ JJ b R = 1 . This i s impossible. Hence 0 - a , or ^ f ~ 1 ( b n ) = a = f " x ( l ) . That i s , f"" 1 i s continuous at 1 . Therefore f i s an ISotone-homeomorphlsm of [ l , a ] onto [0,1] . By t r a n s f i n i t e induction, the f i r s t part of the proposition i s proved. Let a e L , a 4 I • Then [l,a+l] Is isotone-homeomorphic to [0,1] , so that (l,a+l) an open neighborhood of a , i s isotone-homeomorphic to (0,1) . Q.E.D. (5.5) Proposition: L Is countably compact but Is not para-compact hence i t Is not a compact space. Proof: Let A be any countably i n f i n i t e subset of L . Then, A w i l l be contained i n the union of [I. : a e A c W] , where a A has at most countably many elements. Let a_ be the l e a s t upper bound of A . Then [ l , a Q ] i s homeomorphic to [0*1] . Hence [ l , a Q ] i s compact.. How A c [ l , a D ] , A must have a l i m i t point i n [ l , a Q ] c L . That Is, L, i s countably compact. 32. Now, we w i l l see that L i s not paracompact. In view of Prop. (5 .4) , every point a e L, a 4 1 has an open neighbor-hood home omorphic to E 1 . Moreover, [1,2) i s an open set of L which Is a h a l f open i n t e r v a l i n E 1 . Thus, L i s l o c a l l y metrizable (see [9] p. 80) . But L i s not metrizable. By Theorem 2-68 ([9], p. 8l) L i s not paracompact. Consequently, i t i s not compact. Q.E.D. (5.6) Proposition: Gf any two d i s j o i n t closed sets i n L , one i s bounded. Proof: I f F^ and F 2 are two cofi r i a l closed sets, we -can choose ah increasing sequence {a : n e N] where a € F, i f n -1 n 1 n i s odd, and a n e F 2 i f n i s even. Then, since- F^ and F 0 are closed f a = i * " * a e F, n F 0 . This i s a contra-d nsN n n^» n 1 2 d i c t i o n . Q.E.D. From Prop. (5.4), we know that each point a"+ 1 o f the long l i n e has an open neighborhood which i s homeomorphic to an open i n t e r v a l , while each open neighborhood of 1 i s homeo-morphic to the ha l f open i n t e r v a l [0,1) , say. The long l i n e then can be considered as a 1-dimensional manifold with a boundary point 1 . Hence we can define the d i f f e r e n t i a b l e functions on L . (5.7) Proposition: Every function f e C(L) i s a constant on a t a i l L - L(a) , where a depends on f , and L(a) = [a e L : a < a} . Proof: It i s clear that every t a i l L - L(a) i s countably compact, ( i n f a c t , i t i s homeomorphic to L i t s e l f ) . Thus, 33. each image set! f [ L - L(a)] i s a countably compact subset of E 1 and hence compact, (as E 1 has a countable base)*- so that the i n t e r s e c t i o n 0 f [ L . - L(a)] of the nested family i s non-aeL empty. Choose a number r belonging to t h i s i n t e r s e c t i o n . Then the closed set f° 1(r) i s c o f i n a l i n L . Now, f o r each n e N the closed set = {x e L : |f(x) - r | >. |j} i s d i s j o i n t from ,f (r) . Hence, by Prop. ( 5 . 6 ) , Is bounded. That i s , there Is a e L such that a_ < a f o r n n — n a l l e A . Let an e L , and a„ > f ^ E a„ . We have n n o o — nsN n f [ L - L(a Q)3 = {r} . Q.E .D. Let L* be the union space of L and the point 0 , the f i r s t uncountable ordinal. Then, i t i s clear that L* i s a compact 1-dimensional manifold. For each f e C m(L) , we extend f to a function f * on L* by defining that f*(fi) i s the f i n a l constant value of f . Evidently, f * e C m(L*) , and f * i s the unique d i f f e r e n t i a b l e extension of f . On the other hand, f o r each g e C m(L*) , the r e s t r i c t -ion of g to L belongs to C m(L) . I t follows that C m(L) i s isomorphic with (^(L*) , under the mapping f -» f * . Since L* Is compact, we already have a complete des-c r i p t i o n of the maximal ideals i n C m(L*) t every i d e a l i s f i x e d , and the maximal Ideals assume the form Mff = ( f * e C m(L*) : f*(g) = 0} , where cr e L* . (By Theorem (3.3). ) By v i r t u e of ' the Isomorphism of C m(L*) with C m(L) , the maximal id e a l s i n C m(L) are i n one-one correspondence with those of C m(L*) . Moreover, the fix e d maximal ideals i n C m(L) correspond to the Ideals M^ i n C m(L*) f o r each a e L , leaving just one free 34. maximal i d e a l i n C ^ L ) , namely, M Q = {f £ C m(L) : f * £ MF I) , the one that corresponds to MF I . Though M Q Is fr e e , i t i s not hyper-real, f o r C m ( L ) / M Q ~ C m ( L * ) / M N . Hence L Is not m-realcompaet. § 6. Homomorphisms, Cm-mappings and G^-diffeombrphisms. In t h i s section we w i l l describe the r e a l t i o n between any Cm-mappings (see (6.1)) from I c E ^ 1 into Y c and homomorphisms from C m(Y) to G m(X) . We s h a l l f i n d that, i n a sense, every homomorphism from one r i n g of C m - d i f f e r e n t i a b l e functions i n t o another i s induced by a C m-mapping (see (6.20)). We w i l l also show that any two m-realcompact spaces X and Y are C m-diffeomorphic (see (6.1)). i f f C m(X) and G m(Y) are isomorphic (see (6.19)). (6.1) D e f i n i t i o n : Let X c E**1 and Y c E^ 2 . A mapping T : X -* Y Is said to be a C m-mapping at a point p , i f each component of T(X) = ( T 1 ( X 1 , . . . , x n ^ ) , j . . , T n 2(x 1,...x n^)) i s C m at p . I f T i s C m at each point of X , then T i s said to be a C m-mapping on X . And T i s C m-diffeomorphism i f T i s a Cm-mapping, one-one, onto and i t s inverse mapping T - 1 i s also a (f-mapplng. We w i l l say then X and Y are C m-diffeomorphic. Note that by def. (6.1), i t i s clear-that X and Y are C m-diffeomorphic implies = n 2 . (6.2) D e f i n i t i o n : An f e C m(X) i s daid to be a l o c a l i - t h projection at a point a i f there exists a neighborhood U of a such that f |U~= Jj , where M w i l l always denote the i=th projection of the space E n . 35. (6.3) Lemma: Let X be any subset of E 3 1 . For each a e X and r > G , there are ^ , ( 1 <. i < n) , h^ e C**(X) suclv that h ^ x ) = x i f o r a l l x e clx B r(a) . Then we c a l l h ^ ( l _< I _< n), the bounded l o c a l projections at a . Proof: Choose r' >.r . As shown i n Theorem (2.2) , there exists - g e C*^E n) such that g(x) = f l i f x e ci^n B r(a) G i f x e E n - B r,(a) V 0 < g(x) < 1 , elsewhere . Let -0(x) = x ± 3 1 <_ 1 <_ ri , the projections of E 1 1 . Set b-±(x) = ^ ( x ) o g(x) . Then, i t i s clear that h±e C**^) c C (X) and s a t i s f i e s the required condition. Q.E.D. Let C m be a subset of C m(Y) (or C m*(Y)) , and T be n T h o a mapping from X c E x to . Y .c E c . Then we w i l l see what C m should be i n order that g«T e C m(X) (or C m*(X)) f o r a l l g € C^ implies T i s a C m-mapping from X int o Y . (6.4) Theorem: Let T be a mapping from X to Y and C m be a subset of (^(Y). Then (1) T i s a C m-mapping implies g«r £ C m(X) f o r a l l (2) I f g.T € C m(X) f o r each g e C m , and C m includes a l l projections of X , then T i s a C m«mapping on X . Proof: ( l ) It i s clear that g«T e C m(X) f o r each g e C™ . (2) Since g«T € C m(X) for each g e C m which includes a l l projections on X , we have, i n p a r t i c u l a r , >£ OT(X) = T i ( x ) e C m(X) f o r 1 < i < n g . Hence, by Def. (6.1) T i s a Cm-mapping. 56. (6.5) Theorem: Let T be a mapping from X to Y and C m be a subset of C m*(Y) . Then ( 1 ) T i s C m-mapping implies g . T j C m*(X) f o r a l l g € C m , (2) If g . T € C m(X) for each g .€ C m , and C m includes a l l l o c a l projections defined as i n Lemma (6.3) , then T i s a C m-mapping on X . Proof: ( 1 ) I t Is clear that g<»T e,C m(X) f c ^ each g e.C m . Tfl 9 ^ yy. Moreover, g e G (Y) and r[X] c Y , hence g o T e C (X) . (2) The proof i s s i m i l a r to that of Theorem (6.4) (2) with the projections replaced by the bounded l o c a l projections. (6.6) D e f i n i t i o n : Let ep be given mapping from a set A into a set B . For each mapping g from B i n t o D , the composi t i o n g«© c a r r i e s A into D . Thus, ep induces a mapping ep' : D 5 - D*" , e x p l l e i t y ep'(g) = g»ep , and ep» Is said to be an induced mapping of cp . There i s a d u a l i t y between the properties one-one and onto (provided that D has more than one element); cp» i s one one I f f cp i s onto, and cp' Is onto i f f cp i s one-one. We are concerned with a C a p p i n g T of X into Y ., where the r o l e of D i s taken by E*~ . The appropriate sub-set of E 1 Y w i l l be either C m(Y) or C ^ Y ) . Evidently, the induced mapping T«, defined by t'(g) = g.T e C m(X) f o r each g s C m(Y) [resp. C m*(Y)] i s a homomorphism from C r a(Y) into C m(X) [resp. C m*(Y) in t o C m*(X)3 . Indeed, f o r any g,g" e C m(Y) and any x e X . (g+g')(T(X)) = go T(x)+g« »(T(X)) = (g»t)(x)-r(g».T)(x) = (g.T+g 'o 37. SO that T'(g+g') = (g+g' ) »T = goT+g' oT = T'(g) + T'(g) . \ S i m i l a r l y , T'(g.g*) = (g.T). (g'j »T) « (T« (g)) »(T ' (g)) . (6.7) Proposition: The homomorphism T' carr i e s the constant functions onto the constant functions of (^(ftX]) 1 i d e n t i c a l l y . Proof: For any x € X and r £ [R / we have T'(r)(X) = r(T(x)) = r . Hence T'(_r) = r on T[X] . Q. E. D. (6.8) Proposition: The homomorphism t 1 ' determines the mapping T uniquely. Proof: I f CT i s also determined by t ' » then a' = T' . Thus for each x e X , g(a(x)) - g ( T(x)) f o r a l l g e C m(Y) . By m-complete r e g u l a r i t y of X , CT(X) = T(X) . Hence a = T . Q.E.D. We now examine the d u a l i t y r e l a t i o n between T and T1 . (6.9) D e f i n i t i o n : A subset A of X i s C m [resp. C m * ] -embedded i n X i f for each f e C m(A) [resp. C m*(A )J, there i s g e C m(X) [resp, C ^ X ) ] such that g | A = f . (6.10) Theorem: Let T be a C m-mapping from X e E**1 into n c Y c E ^ , and T' be the induced homomorphism g -» g«T from C m(Y) in t o C m(X) [resp. C m*(Y) in t o C m*(X)] . (1) f Is an isomorphism (into) i f f T[X] i s dense i n Y . ( 2 ) T' i s onto i f f T i s a C m-diffeomorphism whose image i s Cm-emhedded [resp. Cm*-embedded]. Proof: ( l ) T' i s an isomorphism i f f T'(g) =9 implies g = © J But the 'latter means ( T ' ( g ) ) ( x ) = 0 f o r a l l x e X . 38. implies g = Q on , Y . That i s , T 5 i s an isomorphism i f f g —-© on T [X ] implies g = Q on Y - ; . By m-complete r e g u l a r i t y of Y , we have that T< i s an isomorphism i f f T [ X ] i s dense i n Y . ( 2 ) -Necessity: By hypothesis, f o r each f € ( ^ ( X ) , . there e x i s t s g e G m(Y) s u c h that T(g) = f . We s h a l l show that T i s one-one. If T ( X ^ ) = T ( X 2 ) , then g(T(x^)) = g { T ( x 2 ) ) for a l l g e C m ( Y ) . That i s , ( T»(g))(x 1) = ( T ! ( g ) ) ( x 2 ) for a l l g s C m(Y) . Since T ! i s onto, _f(x :) = f ( x 2 ) f o r a l l f e C M ( X ) . By m-complete r e g u l a r i t y of X , = x 2 o Hence, T i s one-one. Therefore, the inverse mapping T = 1 of T i s well defined as a mapping from T [ X ] to X . We know that f » t _ 1 i s the 'function g j T[X ] e C m ( T [ X ] ) f o r a l l f € C M ( X ) , and G M ( X ) includes a l l pro-jections, i t follows from Theorem (6ik)(2) that T" 1 i s a Cm-mapping. Hence T i s a C m-diffeomorphism from X onto T[X] . Moreover, f o r any h e C M ( T [ X ] ) , h«T s C M ( X ) , i n other words, there exists f € C ^ X ) such that h<>T = f or h = f O T = 1 . The l a t t e r i s g | T[X] ' for some g € C m(Y) such that T ! ( g ) = f. Hence T[X] i s (^-embedded In Y . Sufficiency: By hypothesis, T ~ i s a C -mapping from T[X ] onto X . Consider any f e C M ( X ) , the function fOT" 1 € ( ^ ( T C X ] ) , by hypothesis, has an extension g € C M ( Y ) such that g | T [ X ] = f O T " 1 . That i s f s g»T = T»(g) . Hence T ' i s onto. m# The proof for C i s exactly the same, since we know, by Lemma (6„3) that C M * ( X ) contains bounded l o c a l projections. 5 9 . (6.11) Corollary: I f T i s a C m-diffeomorphism from X onto then T' i s an isomorphism of C m(Y) onto C M ( X ) . Proof: Since T [ X ] = Y , by the theorem, T i s both i s o -morphism and onto. (6.12) Corollary: I f T i s a C m-diffeomorphism of a compact n, no space X . c E -1 to Y c E- , then the induced mapping T' i s onto. Proof: By hypothesis, T [ X ] i s compact i n Y , so i t i s compact no no i n E , hence i t i s closed i n E *• . By Whitney's Analytic Extension Theorem [30], f o r each g e C M ( T [ X ] ) there exists f € C m ( E n 2 ) c C m(Y) such that f | T [ X ] = g . Hence T [ X ] ' m i s C -embedded i n Y . Moreover, T i s a C -diffeomorphism hence T' Is onto. (By Theorem (6.10)). Next, we examine the Inverse problem of determining when a given homomorphism of C m(Y) into C M ( X ) i s induced by some C m-mapping from X into Y . Here, we s h a l l f i r s t consider the homomorphisms from C r a(Y) into fR , that i s , the case i n which X consists of just one point. (6.13) Proposition: Any nonzero homomorphism cp from C m(Y) [or C m*(Y)3 Into fR i s onto fR . In fa c t cp(r) = r f o r a l l r € fR . Proof: We know that cp(g) = cp( g°u) = cp(g)<>cp(u) fo r a l l g e C m(Y) , and cp i s not i d e n t i c a l l y zero.-, where u i s the unity of C m(Y) . Hence cp(u) i s the unity i n fk . That i s , cp(u) = 1 . The mapping from fR to fR 9 defined be r -» cp(r) i s a nonzero homomorphism. By (0.22) i n [VJ. 4 0 . i t i s the i d e n t i t y . In other words, cp(r) = r . Hence cp i s onto. ( 6 . 1 4 ) Propositions The correspondence between the homomorphisms of Cm'(Y) [or Ci (Y)] onto ff( , and the r e a l maximal id e a l s i s one-one. Proofs The k e r n e l of a homomorphism, cp , of C m(Y) onto fR i s a maximal i n C m(Y) and a r e a l Ideal (as C m(Y)/ ker cp ~ ffl ) . On t h e other hand, each r e a l maximal i d e a l i s the kernel of such a homomorphism. Indeed, f o r each r e a l max-imal i d e a l M , l e t ? be the isomorphism from G m(Y)/M onto IH , and define cp s G m(Y) - / £ by cp(f) = f(M(f).) e fl> , f o r any f e C m(Y) . Clearly, cp i s onto, and ker cp ='{f.e C m(Y) : cp(f) = 0} = [f £ C m(Y) : j>(M(f)) = 0} = {f e C m(Y) : f € M} . Row, f o r any f ,g e C m(Y), cp(f+g) = J>(M(f+g)) = P(M(f) + M(g)) = ?(M(f)) + J 5 ( M ( g ) ) = cp(f) + cp(g) , and cp(f.g) = j>.(M(f . g ) ) ' -P ( M(f ) . M ( g ) ) = j H M ( f ) ) « ? ( M ( g ) ) . = cp(f)«cp(g) . Moreover, by (0.23) [7] , d i s t i n c t homomorphism onto fR have d i s t i n c t kernels. Hence i t i s one-one. The proof for (f1* i s simil a r . (6.15) Proposition: Y is.m-realcompact i f f to each homo-morphism cp from C m ( Y ) onto fR <= i . e . each nonzero homo-morphism into fR - there corresponds a unique point y Of Y such that q>(g) = g(y) f o r a l l g € C m(Y) . Proof: Necessity: Let cp be any homomorphism from C m(Y) onto fR . Then, by Prop. ( 6 . l 4 ) , ker. cp i s a r e a l maximal i d e a l i n (^(Y) . By m-realcdmpactness of Y , ker. cp = f o r unique y € Y . Hence cp(g) = cp(M (g)> = g(y) , 41. Sufficiency: The hypothesis says that the kernel of each homomorphism of (^(Y) onto fR i s a f i x e d i d e a l . In view of Prop. (6.14), each r e a l maximal i d e a l of (^(Y) Is then fixed. Hence Y is"m-realcompact. Q. E.D. Our f i r s t r e s u l t about homomorphisms from C m(Y) into C m(X) for X i s a generalization of Prop. (6.15). (6.16) Theorem: Let cp be a homomorphism from (^(Y) into C m(X) such that ep(u) = u . If Y i s m-realcompact, then there exists a unique C m-mapping T of X into Y such that T» '•= cp . Notice that the condition cp(u) = u i s necessary. Proof of the Theorem: For each point x e X , the mapping a : g -• (cp(g)-)(x) i s a homomorphism from C m(Y) i n JR. Since (cp(u))(x) - u(x) = 1 =J= 0 , a i s not zero homomorphism.' By Prop. (6.15), there i s a point y € Y -such that ep(g)(x) = g(y) for a l l g e C m(Y) . We set y = T(X) . Then that the mapping T from X into Y , thus defined,, s a t i s f i e s cp(g) = g°T for each g € C m(Y) i s clear. Since cp(g) € C m(X) for each g, e C m(Y) , Theorem (6.4) shows that T i s a C m-mapping. In view of Prop. (6.8), T i s a unique C^-mappIng for which T 1 = cp . Q.E.D. (6.17) Corollary; An in-real compact space Y contains a C m- image of X i f f C m(X) contains a homomorphic image of C m(Y) that includes the constant functions on X . Proof: Necessity: Let T be a C m-mapping from X into Y . Then T' , the induced mapping of T , i s a homomorphism from 42. C m(Y) into G m(Y) . By Prop. (6.7) C m(X) => T'(C m(Y)) which induces the constant functions on X . Sufficiency: T'(C m(Y)) contains the constant functions on X , there i s g e C m(Y) such that *r'(g) = u . Then, T ' (U ) = T ' ( U ) O U = T ' ( u ) o T ' ( g ) = T ' ( u o g ) = T ' ( g ) = u . Thus, the r e s u l t follows immediately from Theorem (6.16). Q.E.D. (6.18) Corollary; /Inm-realcompact space Y contains a dense (f1 image of X i f f (^(X) contains an isomorphic image of C m(Y) that includes the constant functions on X . (6.19) The Main Theorem: Two m-realcompact spaces X and Y are C^-diffeomorphic i f f C m(X) and C^Y) are isomorphic. Proof: The necessity follows from (6.11). Sufficiency: Let ep he an isomorphism of (^(Y) onto C m(X) . Then c p " 1 i s an isomorphism of C m(X) onto (^(Y) . By Theorem (6.16) there exist unique Cm-mappings T and T-^ from X Into Y and from Y into X , respectively, such that «p(g) '= g<»T 3 and c p - 1 ( f ) = feT-j^ , for each g £ (^(Y) and f - e C m(X) . Then, g(y) = c p ° 1 ( g o T ) ( y ) = (g.-T ).•»-.,_ ( y ) = g o ( T o T 1 ) ( y ) = g » ( T o T 1 ) ( y ) f o r a l l y e Y . That i s , T 0 ^ i s the i d e n t i t y mapping onto i t s e l f . S i m i l a r l y , T-^T i s the i d e n t i t y mapping of X onto i t s e l f . Thus T and -r^ are the inverse mappings of each other. Hence X and Y are C m-diffeomorphic. Q. E. B„ Remark: S. Bo Myers [l6], L„ E„ P u r s e l l [20] and M. Nakai [17] have dealt with C m - d i f f e r e n t i a b l e n-manifolds. This Theorem i s applicable to any closed subset of E 1 1 . 43. In spite of the remark made i n ; Theorem (6.l6), every homomorphism i s induced, i n essence, by a -mapping. (6.20) Theorem: Let cp be a homomorphism from C^Y) into C m(X), Y be m-realcompact. Then the set E = (x € X : cp(u)(x) = 1} i s open-and-^closed i n X . Moreover, there exists a unique C m-mapping T from E into Y , such that fo r a l l g e C m(Y) cp(g)(x) = g ( T ( x ) ) for a l l x e E , and cp(g)(x) = 0 f o r a l l x e X - E . Proof: As with any homomorphism, the element e = cp(u) i s an idenpotent i n C^X) . Hence i t i s the c h a r a c t e r i s t i c funct-ion of the set E = e _ : i"(l) . Since e i s continuous (as a matter of f a c t i t i s C m ) , E i s open-and-closed i n X . Moreover, again, as with any homomorphism, e i s the unity element of the image r i n g cp[Gm(Y)] . It follows that cp(g)[X - E] = cp(g)cp(u)[X - E] = cp(g)[X - E]ocp( u)[X - E] = LT.' {G} f o r a l l g € C m(Y) . If E I <| , consider the homomorphism a from C m(Y) into C m(E) , defined by a'(g). = cp(g) | E . It i s clear that a(u) = u e C m(E) f o r a(u) = e | E = u . By Theorem (6.16), there e x i s t s a unique C^-mapping T of E Into Y such that T' = a . We have then f o r any g e C m(Y) , cp(g)(x) = (a(g)(x)) = g ° T ( x ) f o r a l l x e E and cp(g)fx) = 0 f o r a l l x e X - E . Q.E.D. n„ (6.21) Theorem: Let Y be a compact subspace of E • , and cp be a homomorphism from C^Y) = C™*(Y) into C m*(X) . Then the set E - {x e X : cp(u)(x) = 1} i s open-and-closed i n X . Moreover, there exists a unique cP-mapping T from E into Y , 4 4 . such that f o r any g e C (Y) , cp(g)(x) = g(r(x)) f o r a l l x e E , and ep(g)(x) = 0 f o r a l l x e X - E . The proof i s similar to Theorem (6.20). (6.22) Corollary: Let cp be a homomorphism from C m(Y) i n t o a r i n g of d*1 functions. If Y i s m-realcompact, then there exists a unique closed subset F of Y such that the kernel of cp i s the z-ideal of a l l C m functions that vanish on F . Proof: Let ® be a homomorphism of C m(Y) onto ft which i s a subring of C m(X) , and E = {x e X : cp(u)(x) = l ] . By Theorem (6.20), there e x i s t s a (^-mapping T from E i n t o Y such that cp(g)(x) = g(*r(x)) f o r a l l x e E , and cp(g)(x) = 0 f o r a l l x is X - E , for any g e C m(Y). Let F = C ^ Y T [ E ] , and I = (g e (^(Y) : Z(g) D F] . Then I i s a z-ideal (see (1.15) )• On the other hand, ker cp = {g e C m(Y) : cp(g) = 0) . I t i s then clear that ker cp => I . .For any g Q e ker cp , cp(g 0) = © € ft c C m(X) so that cp(g Q)[X] = {0} . That i s , «P(S 0)( x) = g p(T(x)) = 0 f o r a l l x £ E , or g 0(-r[E]) = {0} . Thus Z(g Q)-5 f['E] so that Z(g Q) = a y Z ( g Q ) 5 c y [ E ] = F . Hence g e l . Consequently ker cp = I . The uniqueness of F follows from the choice of F . Q.E. D. (6.23) Proposition: An m-realcompact space Y contains a C m-embedded C m image of X i f f C r a(X) i s a homomorphic image of C m(Y). Proof: The necessity Is clear. For su f f i c i e n c y , l e t cp be a homomorphism of C m(Y) onto C m(X) , and. k € C m(Y) such that 45. cp(k) = u . Then cp(u) = cp(k)ep(u) = cp(kou) = cp(k) = u . By-Theorem (6.16), we have a unique C m~mapping of X into Y such that T ' = cp . Hence T ' i s onto. By Theorem (6.10)(2), then T [ X ] i s C^-embedded. (6.24) Proposition: A compact Space Y contains a C^-embedded C m image of X i f f C^IX) i s a homomorphic image of C m*(Y) . Proof Is simi l a r to that of Prop. (6.23). §7 The embedding Theorems. K. D. Ma g i l l , i n 1965, [14], established the algebraic conditions r e l a t i n g C(Y) and C ( X ) which are both necessary and s u f f i c i e n t f o r embedding X i n Y . ¥6 w i l l generalize hi s r e s u l t s to the rings of C m - d i f f e r e n t i a b l e functions. (7.1) D e f i n i t i o n : Let Ot be a c o l l e c t i o n of r e a l ideals i n a ring . Then f*l(X i s said to be a 6-real i d e a l . (7.2) D e f i n i t i o n : Let 0 be a subring of a r i n g A . Then we say that 0 i s 8-dense i n A i f f o r every p a i r I and I' of 6-real ideals of A with I - I ' + ^ j I - I ' contains an element of p . (7.3) D e f i n i t i o n : A homomorphism from a r i n g A1 into a r i n g A i s a 6-homomorphism I f i t i s n o n t r i v i a l and the image of A' i s 6-dense i n A. For example, a homomorphism onto i s a 6-homomorphism. (7.4) D e f i n i t i o n : A set of elements of a r i n g i s said to be a subreal, i f i t i s contained i n a r e a l i d e a l of the ri n g . 46. (7-5) D e f i n i t i o n : A 5-homomorphism i s a 5 F-homomorphism, i f the image of every r e a l i d e a l containing the kernel i s sub-r e a l . (7.6) D e f i n i t i o n : A 5-homomorphism from a r i n g A' in t o a r i n g A i s a 6 G-homomorphism i f for every r e a l i d e a l M of A' whose image i s subreal, there e x i s t s an element a ^ M such that the image of every r e a l i d e a l not containing - a i s subreal. Let cp be a homomorphism of a r i n g of functions A' into another r i n g of functions A . We w i l l say that cp has property (7-1) to mean that f o r every g e A and x 4 Z(g) , there i s an f e A' such that x 4 Z(cp(f)) , and Z(g) cr Z(cp(f)). . ni (7.7) -Theorem: Let Y be an m-realcompact space and X c E , and cp a homomorphism fr'om C m(Y) in t o G M ( X ) . Then cp has the property (7-1) i f f there i s a homeomorphism T from X in t o Y such that $ ( f ) = f o T f o r a l l f e C m(Y) and T i s a C m-mapping. In addition, i f ^ ( ^ ( Y ) ) contains a l l projections of X , th£n T i s a C m-diffeomorphlsm into Y . Proofs Sufficiency: Suppose . T i s a C m-diffeomorphism from X into Y such that 3>(f) = f « T f o r a l l f e C m(Y) and that x 4 Z(g) where g e C R A ( X ) . Then T(X) 4 c t Y T [ Z ( g ) ] . For i f T(X) e c,t Y «r[Z(g)] , then f o r each open neighborhood U of T(X), U H T[ Z( g) ]=)=({) . However, we know that x 4 Z(g) so that there exists an open neighborhood V . of x such that V n Z(g) = ({) . Thus T[V] Tl T[Z(g) ] = <|) (as T i s one-one), and T[V] i s open i n T[X] . Hence, there e x i s t s an ©penset G i n Y such that G n T[X] = T[V] . We know then that 47. t(x) e T[V] e G , and GO T[Z(g)] = G fl «r[Z(g)] n T[X] = (G n T[.X]) n T [Z(g)] = T[V ] n r [ Z(g)] = (j).'which i s a contra-d i c t i o n . Now, since r[x) 4 . C^TI Z(g) ] , by m-complete r e g u l a r i t y of Y , there i s an f e C^Y) such that Z ( f ) 3 ctyT[Z(g)3 and T(X) 4 Z ( f ) . Hence x 4 T" 1(Z ' ( f ) ) t and T " 1 [ z ( f ) n T[X]] 3 z(g) . ;But, ^ [ z f f ) n T[X]] = (x € x : f o r ( x ) = 0} = Z ( f o t ) = Z((P(f)). This proves the s u f f i c i e n c y . Necessity: Let (p be a homomorphism of C^Y) Into C M ( X ) s a t i s f y i n g (7-1). We define a mapping T from X onto Y as follows. Let x e X be given. 'Then the mapping (fx : C m(Y) - fR defined by <j^(f)'= q?(f) (x) i s a homomorphism. Moreover, x.4 Z(u)(=(j)) so by (7-1), there exists f e C m(Y) such that x 4 Z(q>(f Q )) . Hence <Px(tQ) - <P(f Q )(x) +-0 . That Is, '(p i s a non-zero homomorphism. By Prop, (6.15) since Y Is m-realcompact,. there exists a unique y .€ Y such that <j5• ( f) = f ( y ) f o r a l l f € C m(Y) . We then define T(X) = y . Then we have <p(f) = f o T f o r a l l f e C m(Y) . It remains fo r us to show that f i s a -mapping, T"1 e x i s t s and i s con-tinuous. Since f o T e C M ( X ) f o r a l l f e C m(Y) , by Theorem (6.4) T i s a Cm-mapping. Now, i f x =f= x' i n X , there i s a g e C ^ X ) such that g(x) = 0 and g(x>) = 1. By (7-1), there i s an f e C m(Y) such that x' 4 Z(<j?(f)) and x e Z(tp ( f ) ) . Hence f ( r ( x ' ) ) » [cp(f))(x'> + 0 , while f ( r ( x ) ) = (<p(f))(x) = 0. That i s , T(X) + T(X') so that T i s one-one. Therefore T - 1 i s well-defined. Now, T" 1 i s a mapping from T[X] onto X . In order to show that T"1 Is continuous, i t i s enough to show that T i s open. Let U be any open set In X , and x e ! U 4 8 . be a r b i t r a r y . Then x | Z(g) f o r some g € C M ( X ) .. By (7-1), there i s an f e C m(Y) such that x 4 Z(cp(f)) and Z(g) c Z(q>(f)). I t follows that T(X) e T [ X ] n [Y - Z(f)] c T [ X - Z ( g ) ] . That i s , T(X ) i s an i n t e r i o r point of T [ X ] (as Y --Z(f) i s open i n Y). Thus T [ U ] i s a set of Interi o r points. Hence i t i s Open so that T" 1 i s continuous. Now, we assume that (p(C m(Y)) contains a l l projections of X . Since T" 1 i s a mapping from T [ X ] onto X , and fo r every g e 9(C m(Y)), g s f . t f o r some f e C m(Y) so that g o T - 1 = ( f o T ) o T " 1 = f € C^Y) c C m ( T [ X ] ) . Since cp(c m(Y)) contains a l l l o c a l projections of X , by Prop. ( 6 . 4 ) , T" 1 i s a .Cm-mapping on T [ X ] . Q. E. D. Let F be a closed subset of X . Then Mp = (f e C (X) Z(f) 3 P} i s an i d e a l i n C m(X) . Moreover, Mp = n{M x : x e P} . Hence i t i s a 6 - r e a l i d e a l . I f X i s m-realcompact, then the converse i s also true. For, l e t be any 5 - r e a l i d e a l of C m(X) , then VL = D{M : M_ i s r e a l and a e A ) . However, f o r each a , there e x i s t s x e X such that M x = M Q . Thus, M, = n{M : x e E} , where E = {x e X : x corresponds to a for some a e A] . Let F = c t x E . We w i l l show that M E = n{M x : x e E} = n (M x : x e F] = Mp . Indeed, each f e Mp , f e Mg i s clear. Hence M E r> Mp . Take any g € Mg. We have Z(g) r> E . Thus c-t xZ(g) = Z(g) 3 = P . That i s , g € Mp . Therefore, ME = Mp . Since E i s uniquely deter-mined by A and P i s uniquely determined by E , P i s unique. Consequently, X i s m-realcompact i f f every 8 - r e a l i d e a l of (^(X) i s of the form Mp f o r some closed subset P of X . 4 9 . (7.8) Theorem: Let Y he an a r b i t r a r y topological space and X be an m-realcompact space. Then the following statements concerning a homomorphism (jp from C m(Y) into CM(X) are equivalent. (1) cp s a t i s f i e s property (7-1) . (2) cp i s a 6=homomorphism. (3) The image of c m(Y) separates points and closed sets and i s contained i n no 6-real i d e a l of C M(X). Proof: ( 1 ) implies (2). Let and M c be two 6-real ideals of C m(X) with Mp - Mp, 4 ty . Then F' e£ F and there i s an element x.,e F' - F and a g s C m(X) such that. Z(g) 3 F but x 4 Z(g) . According to ( l ) there exists an f e (^(Y) such that x 4 Z(cp(f)) and Z(g) c Z(cp(f)) . Hence cp(f) € Mp - Mp. . That i s , cp i s a 6-homomorphism. (2) implies ( l ) . For any two 5-real ideals Mp, Mp, i n (^(X) such that Mp - Mp, 4 ty , there exists an f € (^(Y) such that cp(f) e Mp - Mp, . In p a r t i c u l a r , for any zero-set Z(g) such that x 4 Z(g) , we have ^(g) M x ^ ^ * B y hypothesis, there e x i s t s an f €-CM(Y) such that cp(f) e M z ( g ) " M x ' T h u s > z( £p{ f)) 3 z ( g ) b u t * 4 Z(q>(f)) . (2) implies (3). Let Mp be an a r b i t r a r y 6-real i d e a l i n C m(X) . If F = X , then Mp ~ (0) and since cp i s nonzero, the Image of (^(Y) can not be contained i n Mp( = (0)) . Cn the other hand, i f F 4 X ,, choose x 4 F . Then M^ . - Mp 4 ty a n (2 hence must contain an element of cp(C m(Y)) . In either case, cp[C m(Y)] i s not contained i n Mp for any closed subset F of X . We w i l l now dhow that cp(CM(Y)) separates points and closed subsets. If F i s any 50. closed subset and x 4 F -> then Mp » M x 4= ty • By hypothesis there exists f £ C m(Y) such that cp(f) e M ? - M x . Hence cp(f)(x) + 0 and Z(cp(f)) => F . That i s , (cp(f))(x) 4 (cp(f))(F). (3>) implies ( 2 ) . Suppose (3) and cp i s not a 6-homomorphism. That i s , cp(C m(Y)) i s not 6-dense i n C m(X) . Hence, fo r some, Mp,Mp, we have , Mp-Mp 14s ty but (Mp-Mp, )H cp(C m(Y)) = ty . However, Mp, c M x f o r each x e F* . Thus Mp - Mp, r> Mp - M x for each x e F« and Mp-Mp, = M p - n{Mx : x € F 1} = U{M-p - M : x e F«} 4= • Hence, there exists at lea s t one x Q e F» such that Mp - M X Q 4 ty • But (Mp - Mx ) H cp(C m(Y)) c (Mp - Mp, ) n cp(C m(Y)) = ty . That i s , there does not exist any f e C m(Y) such that cp(f) £ Mp - M X q . In other words, there Is not any f e C m(Y) such that ep(f)(x) 4 ct xep(f)(F) . t h i s i s a contradiction. Q.E.D. ( 7 - 9 ) Theorem: Let X and Y be m-realcompact. A homo-morphism from C m(Y) into C m(X) i s a 6-homomorphism i f f there i s a homeomorphism T from X into Y such that cp(f) = f o T f o r a l l f € C m(Y) and T i s a Cm-mapping. Moreover, i f cp(C m(Y)) contains a l l projections of X , then T" 1 i s a Cm-mapping, that i s T i s a C r a^dlffeomoprhism. Proof follows from Theorems ( 7 - 7 ) and ( 7 - 8 ) . (7.10) Lemma: Let X and Y be any subsets of E 1 and E 2 , respectively, and T be a C m-mapplng from X into Y . Define a homomorphism cp from C m(Y) into C m(X) by ep(f) = f«T . Then for f i x e d i deals M and u of C m(Y) and C m(X) respect-y i v e l y , cp [ M l c M i f f T(X) = y . Proof: Sufficiency: Suppose T(X) = y . Then for f e M , Cr 5 1 . f ( y ) = 0 . On the other hand, cp ( f(x)) = f(«r(x)) = f(y) = 0 . Thus ep( f ) e M x . Hence cp(My) c M x . Necessity: Suppose ep(MTr) c Mv . Let f € C m(Y) , f ( y ) = r . Then f - r e M and cp ( f ) - r = cp ( f-r) e M_ . — y — — x Thus ep(f)(x) = r . Hence cp(f)(x) = fo<r(x )= f(y) f o r a l l f e C m(Y) which implies T(X) = y . (By m-complete r e g u l a r i t y ) . ( 7 . 1 1 ) Theorem: Let X c E^1 and Y c E**2 be m-realcompact. A homomorphism cp from C m(Y) into C M ( X ) i s a 6F-homomorphism i f f there i s a homeomorphism T from X into Y such that cp(f) = f o T for a l l f e C m(Y) , T[X] i s a closed subset of Y , and T i s a Cm-mapping. Moreover, i f cp(C m(Y)) contains a l l 1 in projections of X , then T~ i s also a C -mapping. Proof: Sufficiency: Let M be a r e a l i d e a l of C m(Y) which contains the kernel of. cp , ker cp . Then M = M for some y e Y . Moreover, since T[X] i s closed, y e T[X] . For otherwise, there would be an fe C m(Y) such that Z ( f ) => T [X ] and y 4 Z ( f ) . This would imply f € ker cp - M which i s a contradiction. Hence T(X) = y fo r some x e X , and by Lemma ( 7 . 1 0 ) , cp(M ) c M , that i s , the image M i s subreal y ^ i n C M ( X ) . Necessity: Suppose cp i s a 6F-homomorphism. By Theorem ( 7 . 9 ) > there i s a homeomorphism T from X into Y such that cp(f) = f o r for a l l f e C M ( X ) , and T i s a C m-mapping. We w i l l show that T[X] i s a closed subset of Y . Choose y e C ^ Y ( T [ X ] ) = F . Then, ker cp = Mp, er M . Hence, there i s a r e a l i d e a l M x of C M ( X ) such that ep(My) c M x . By Lemma ( 7 . 1 0 ) , we have y = T(X) . Thus y e T [ X] . That 52. i s , T[X] = c* y-r[X] . The l a s t part follows from that of Theorem (7.7). QoE„B„ n-. n o (7-12) Theorem: Let X c E 1 and Y c E 2 he m-realcompact. A homomorphism cp from C m(Y) into C M ( X ) i s a 6G-homomorphism i f f there exists a homeomorphism T from X into Y such that cp(f) = f «T and T i s a C m~mapping for a l l f e C m(Y) and T{X] i s an open subset of Y . Moreover, i f cp(C m(Y)) con-tains a l l projections of X / then T~ i s a C -mapping. Proof: Sufficiency: Let M be a r e a l i d e a l of C m(Y) whose image i s subreal i n C M ( X ) . Then, for some x € X , cp[M ] c M^ . From Lemma (7.10) again, T(X) = y . Then y 4 Y -T [ X] which i s a closed subset of Y and there i s an f e C r a(Y) such that Z(f) r> Y - T [ X] and y 4 Z ( f ) . Thus, f 4 M and i t f.olloi/s that the image of r e a l i d e a l not containing f i s subreal. Necessity: Suppose cp i s a 6 G-homomorphism. Then i t i s 6-homomorphism so that,by Theorem (7.9)s there exists a homeomorphism T from X into Y which i s a C m-mapping such that cp(f) = f o r . Let y € T [ X] . Then, y = T(X ) for some x e X . By Lemma (7.10), cp(My) c M x . Hence, there exists an f 4 My and the image of every r e a l i d e a l not containing f i s subreal. From t h i s i t follows that y € Y.- Z(f) c T [ X ] . Indeed, for each y' € Y - Z(f) , f $ My, , cp(My, ) c Mx,.5 for some x' By Lemma -(7-10), y f = T ( X ' ) e T [ X] . Hence «r[X] i s open. The'last part follows frbm that of Theorem (7.7). QoEoDo 53-§8 A Representation Theorem fo r Transformations of Rings of C ^ d i f ^ erentiable Functions. Let X and Y he completely regular Hausdorff spaces, T he a transformation of C(Y) into C(X) which i s much more general than a homomorphism. We know that T can he represented by means of a continuous mapping from a subset of E 1 x X and a continuous mapping T from a subset of X into vY i f T s a t i s f i e s c e r t a i n conditions. (Theorem 1, [12]). We w i l l observe that i f T i s a transformation of C^Y) into C^X) , where Y i s m-realcompact, then T w i l l have a representation i f T s a t i s f i e s certain conditions (see Theorem (8.5)). We start with some generalizations of c l a s s i c a l r e s u l t s . (8.1) Lemma: Let f(x) = f (x- L,... ,x n) be def ihed i n an open subset G of E 1 1 , and the f i r s t p a r t i a l derivatives i n x. and x. exis t and be continuous. Then, i f f„ „ _ j x i x j ~ 2 2 a f • a f — 2 — t — e x i s t s and i s continuous i n G , then f = — -9Xj9x^ j i 9x^9Xj ex'ists and i s continuous In G . .(.Here we may assume that i < j ) . Proof: Let A be an abbreviation f o r the mixed second difference, Then A = f (x^,. . . iX^+h, •. . . ,Xj+k,. .. ,x n) - f (x^,. . . ,x.j+h . . . ,x R) - f(x 1,...,x ±,...,Xj+k,...,x n) + f ( x x , . . . , x n ) (8-1) We write P(x i) = f(x^,...,x^+k,...,x n) - f ( x 1 , . . . , x n ) . Then (8-1) becomes A = P(x i+h) - F ( x i ) . By the Mean Value theorem, we have A = hF ,(x i4-Gh) , where 0 < ©'< 1 . That i s , A = h [ f x . ( xl> * *•• > x± + @ ] s->' • - > xj + k> • • • > x n) ~ f X i ( x i * " *' > x ± + @ h> •'-ixn)]• 54. Again, by the Mean Value Theorem, we have A = hkf / ~(x,,. . . ,x.+©h,. . . ,x .+©'k, .. . ,x„ ) , 0 < 0 ' < 1 . Xj^x^j JL l j n = f (x-^,. .. ,x^+h,. . . ,x^+k,. . . ,x^) - f (x-^,. . . ,x^+h,. .. * x n ) - f ( x ± 3 . . . , x ± , . . . , X j + k , . . . , x n ) + f ( x 1 , . . . , x n ) (by (8-1)) (8-2). D i v i d i n g (8-2) by k , we have h f x , x . ' ( x l * * * ' J x i + € ) k > • • • 3 X j + © ' k , . . . ,x n) = -g [ f (x 1,. . .x±+h,. . . , X j + -. . . x n ) ~ f ( X 2 . 5 * * * ^ X j ^ h , • • • JX^) ] - [ f (x-^,.. . , x.^ ,. . . X j^-k, n. . 3 xn ) - f ( x 1 , . . . , x n ) ] . ' (8-3). Consider (x^,...,x n) as a ce r t a i n f i x e d point, and take the l i m i t of (8-3) as k -• G . Since f and f exist we X j X j X j have fx(x1>...,x±+hi...,xn) - f x (x- Li...,x n) = h f x ^ x (x^ ... ,x ±+©h,... ,Xy . . . ,x n) (8-4). Next, divide (8-4) by h , and take the l i m i t off i t as h - 0 . Since f i s continuous, we have x I x j h^O Ti J - f X j ( x l J * ' " >x±+h> • • • >xn^ " f X j ( x l * ' ' ' , x n ^ = f x ± x ^ ' Or f = f„ ^ We know that since f i s continuous X j X ± X ± X j X ± X j i n G so i s f „ . X j X i (8.2) Corollary: Let G be a d i f f e r e n t i a b l e n-manifold. a 2 f Then, f' = — exists and i s continuous on each coordinate x i x j B X j a x ± a 2 f neighborhood system implies that f = — exists and i s X < T a x i a x J 55. continuous on each coordinate neighborhood system. Proof: Note that the notations f(x, , . . . , x ), f , etc. refer only to the local coordinate neighborhoods. The proof is the same as that of Lemma (8.1). (8.3) Lemma: Let f(x) be defined in an open subset G of k,+. ..+k E , and f . , = —-— exist and be continuous n 1 in G . Then each partial derivative of f in one of the permutations of x- L , . . . ,x 1 X g , . . . j X ^ , ' x n > ' ' ' , x n ' k\ k 2 k n n (m = Z k. letters) exists, is continuous, and equal to 1=1 1 f x k l x k * ' 1 ' *' "> n Proof: We wil l f irst prove this in the case of f . Let x i x j x k P(x) = f (x) . Then, by assumption, F ._ exists and is con-x i x j x k tinuous. By Lemma (8.1), F v „ = F„ _ so i t is continuous, x k x j x J x k Thus f „ „ exists, is continuous and f = f „ . x i x k x j x i x k x j x i x j x k Since f v is continuous, fv „ is differentiable in x. . x i x j x k X ±Xj K By Lemma (8.1) f„ = f and is differentiable in x, . x j x i x i x j * Thus, f = f is continuous. The proof for the x j x i x k x i x J x k other permutation is similar. 56. Now, assume that i t i s true f o r a l l m _< p - I . We w i l l show that i t i s true f o r m = p . Without loss of generality, l e t us assume that f k' k f exists and i s continuous, x, i.-.x n 1 n where § k' = p . We know that every permutations i s a i = l product of transpositions ( I , j ) -* ( j , i ) . Moreover, the above argument applies to each transposition. Hence, each p a r t i a l derivative of f i s one of the permutations of x^,...,x^, n n* exists and i s continuous, and equal to f ki k l • Therefore the Lemma i s true- f o r a l l n . v - l V « ± n Notice that t h i s r e s u l t i s true In a d i f f e r e n t i a b l e n-manifold i n each l o c a l coordinate neighborhood. (8.4) Lemma: Let f ( t , x ) = f ( t , x 1 , — , x n ) be defined i n an open subset X c E , where X = E x X n , X n i s an open sub-set of E n . Let g(x) be a realvalued function defined on the open subset X_ , -2— , denoted by f , be non-zero for n at G a l l points of X , and f v i • v i v t = — - — — • • • • • • ' — • 1*1 • • • ^ . . . ^ i ^ i e x i s t and be continuous f o r each choice of k£,...,k^ such that 1 < m'« - § k» < 2 k. = m . Then 5 m f (g(x) ;x), e x i s t s and i n 57. i s continuous i f f k k exists and i s continuous. 9x, .. . Sx„ 1 n Proof: The s u f f i c i e n c y follows immediately from the Chain rule for p a r t i a l derivatives and the hypothesis. Necessity: By assumption and Lemma (8.3\ without loss of generality, we may consider the f i r s t p a r t i a l derivative of f(g(x),x) with respect to x1 . (Then we have: a f ( g ( x h x ) , h = f(g(x 1 4-h,.../x n), x 1+h 5...,x n)»f(g(x 1 V...,x n), Sx-^ x^,..-.,x ) + eh ; where e • 0 as h -• 0 . However, the r i g h t hand side can be written as [f (g(x^4li,.. . ,xn),x^+h,.. . ,x n) -f (g(x 1+h,.. . 5 x n ) , x 1 , . . . , x n ) ] + [f (g(x 1+h,. .. J,x n),x 1,. .. ,x n) -f (g (x 1 , . i , x n ) , x 1 , . : . , x n ) ] + eh = f 1(g(x 1+h,...,x n),x 1+©h,...,x n) «h+f Q ( g(x 1+©»h,...,x n),x x,...,x n) • [ g ( x x + h , . , x n ) - g(x x,...,x R)] + €h . where 0 <.©,.©' < 1 , by the Mean Value Theorem. Thus :3*(g(*?'*) = f 1(g(x : L+h,...,x n),x 1+©h,.,..,x n) • + ^(gCx^d'h,...^), 3x^ xl>''' >xn^° k [ g ( x 1 + n ^ - - l - 3 x n ) - g(x 1,...,x n)] + € . Let h - 0 . We have 3 f ( g ( x ^ x ) ? ,= f o ( g ( x ) , x ) lfitil-'+ f 1(g(x),x). 9x^ 9x^ By hypothesis, twe have B g ( x l = [ 9f(%(x)>x?? - f'^gCx),*) ]• / SX-^ SX-^ f 0 ( g ( x ) , x ) . ' (8-5) Hence d % ( x ) e x i s t s and i s continuous. We know that Ml£i2Lh*l .9 f (g(x),x) , and f (g(x),x) have continuous 1st ax. 58. p a r t i a l ^ derivatives bec'ause dg^J/dx-^ i s continuous. Hence by d i f f e r e n t i a t i n g (8-5) , we can show that g(x) has continuous 2nd p a r t i a l derivatives. Proceeding by induction, we conclude that, g(x) has continuous m-th p a r t i a l derivatives as we require. Q. E. D. I f , i n the lemma, i s replaced by a d i f f e r e n t i a b l e n-manifold, the r e s u l t i s also true i n each l o c a l coordinate neighborhood. Now, we are i n the p o s i t i o n to show the representation theorem. (8.5) Theorem: Let X be a d i f f e r e n t i a b l e n-manifold and Y an m-realcompact spaCe, T be a transformation of C m(Y) in t o C m(X) such that (-a) For x € X , i f f , g e C m(Y) . such that (Tf)(x) » (Tg)(x) then (T(f+h))(x) = (T(g+h))(x) and (T(f»h))(x) = (T ( g o h ) ) ( x ) f o r a l l h e G m(Y) . (b) For each x e X , the mapping Q : t -• (Tt)(x) from E 1 into E 1 i s G m i n t onto (T[ C m(Y) ]) (x) . (c) For each x e X , the mapping t - ( T t ) k ( x ) i s C m " k i n t , where 0 _< k < m , and since Tt e C m(X) , ( T t ) ( k ) i s a k-th p a r t i a l derivative of Tt . (d) C'(t) f t+ 0 f o r a l l t € E 1 and x € X , where C i s defined i n (b) . Set E = {x e X :, (T[C m(Y)])(x) eontaihs more than one point) . Then there i s a C m-mapping T of E into Y , and a continuous mapping oo from E 1 x E into E 1 such that tu(t,x 0) i s a homeomorphism onto (T[ C m(Y) ]) (x Q) f o r each x Q e E , i»(t,x) i s ; 59. C m i n t and x separately, and s a t i s f y i n g (1) (Tf)(x) = u)(f O T ( X ) , X ) i f x e E = (T©)(x) i f x € X - E for each f e C m(Y) Moreover, E i s an open subset of X , and ' T[E] i s dense i n Y i f f T i s one-one. In addition, i f T maps C m(Y) onto C m(X) , then T i s a C m-diffeomorphism of X onto a C^-embedded subset of Y . Proof: Let oo be the mapping from E 1 x E into E 1 defined by <u(t,x) = (Tt)(x) . F i r s t of a l l , we w i l l show that (a) implies that «)(t,x0) i s one-one fo r each x Q e E , where t i s variable, indeed, i f uo(r,x 0) = oo(s,x 0) f o r some r 4= s , then ( T r ) ( x Q ) = (Ts)(x Q) . We would have ( T ( r - s ) ) ( x Q ) = (T«)(x ), and (T(r-s . -A-))'(x ) = (Tu)(xJ (T(«.' -i-))(x ) = r-s ' r-s ( T © ) X q . Thus, f o r each h e C m(Y), (T(h»u))(x 0) = (T(h««))u Q) . That i s , (Th)(x Q) = (T©)(x Q) . This shows that x Q j E , which i s a contradiction. Thus' «)(t,x0) i s one-one, so that i t s inverse, denoted by a ( t , x Q ) , e x i s t s . By hypothesis (b) and the Brouwer Invariance of Domain Theorem, a ( t , x Q ) i s continuous. Since <»(r,x) = (Tr_)(x) where x i s variable, t«(r,x) i s C m i n x f o r each r e E 1 . By hypothesis (b) ut>(t,x0) = ( T t ) ( x Q ) i s C m i n t . By Lemma (P.675 [12]) both uu and a are j o i n t l y continuous. For each x e E , we define a mapping ep of (^(Y) i n t o fR by e p v(f) = a((Tf)(x),x) . We w i l l show that cp i s a homomorphism. Given any f,g i n C m(Y) , l e t «Px(g) = 8 > so that (Tf)(x) - u)(r,x) = (Tr)(x) and (Tg)(x) = ui(s,x) = 60. ( T s ) ( x ) . Then, by ( a ) , ( T ( f + g ) ) ( x ) = (T(r + g ) ) ( x ) = (T ( r + s ) ) ( x ) = «j(r+s,x) . Hence c? x(f+g) = a ( ( T ( f + g ) ) ( x ) , x ) = a ( ( T ( r + s ) ) ( x ) , x ) = r+s = cp x(f) + cp x(g) . A l s o ( T ( f e g ) ) ( x ) = (T(f«_s))(x) = ( T ( r . s ) ) ( x ) = aj(r.s,x) , hence ep x(f.g) = a ( ( T ( f o g ) ) ( x ) , x ) = r . s = c p x(f) ° c p x(g) . Moreover, cpx i s a non-zero homomorphism, f o r cx(t,x) i s one-one mapping, where x i s f i x e d . By Prop. (6.13) cp__ i s onto.- From. (6.15) and the m-realcompactness of Y , th e r e i s a unique p o i n t y e Y , say, such t h a t c p x(f) = f ( y ) f o r a i l f € C m ( Y ) . Thus, we can d e f i n e a mapping T of E i n t o Y as f o l l o w s : For each x € E , T ( X ) i s the unique p o i n t y of Y such t h a t ep ( f ) = x f ( y ) ( = f o T ( x ) ) f o r a l l f e C m ( Y ) . That i s , f(«r(x)).= . a ( ( T f ) ( x ) , x ) . I t f o l l o w s from the d e f i n i t i o n of a t h a t ( T f ) ( x ) = u)(f<>T(x),x) f o r each x € E . We now show t h a t ( T f ) ( x ) = (T©)(x) f o r x e X - E . We know t h a t f o r each x e X - E , ( T [ C m ( Y ) ] ) ( x ) i s o n l y one p o i n t . That i s , ( T f ) ( x ) (Tg)(x) f o r any f>g e C m ( Y ) . In p a r t i c u l a r , i f we l e t g = 0 , we have ( T f ) ( x ) = (T©)(x) f o r a l l f e (^(Y) . Thus, the e q u a t i o n ( l ) holds. From the d e f i n i t i o n of E , and the d i s c u s s i o n i n the f i r s t paragraph o f the p r o o f , we conclude t h a t the s e t (x e X : (Tu)(x) = (T©)(x)} which c o i n c i d e s w i t h X - E i s c l o s e d so t h a t E i s open i n X . Hence E 1 x E i s a d i f f e r e n t i a b l e (n+ 1)-manifold. Now, we are ready t o show t h a t T i s C m . From (c) and the d e f i n i t i o n of m , we have ( T t ) k ( x 0 ) = to k 2 *n+l^iX (see the n o t a t i o n i n Lemma (8.4)), i s c m" i c i n t f o r each 61, n±l i x_ € 'X , where 2 k. = k . That i s , i t s (m-k)-th derivative 1=2 1 e x i s t s and i s continuous i n t . On the other hand, f o r every f i x e d r e E 1 , (Tr)^ k^(x)'» w v. v (r,x) i s d i f f e r e n -2 i 2 . . . ( n + l ) ^ 1 + 1 t i a b l e i n x . By Lemma (P.675 [12]) (the f i r s t paragraph of the proof) m K k _(t,x) i s j o i n t l y continuous i n t 2 2... (n+1) ahd x . Thus, hy Lemma (8.1), a l l mixed p a r t i a l derivatives ex i s t and are continuous. From (d) we have (»1(t,x) =}= 0 , fo r (t,x) e E 1 x E . By Lemma (8.3), and the fac t that (Tf)(x) = uo(f.T(X),X) i s C m , we have f « T e C M ( E ) . Since f i s a r b i t r a r y i n C m(Y), by Theorem (6.4) T i s a Cm-mapping. On the other hand, by Lemma (8.3), and the continuity of ^ k« v (t,x) i n t and x , we have f o r each 2 2... (n+1)^i+l g e C m(X), t»(g(x),x) i s C m on E.-. t h i s f a c t w i l l be needed l a t e r . In v i r t u e of m-complete r e g u l a r i t y , we know that T [ E ] i s not dense i n Y i f only i f there exist f and g e C m(Y) that•coincide on T [ E ] hut not on Y . However, the l a t t e r i s equivalent to the condition that there e x i s t f and g e C m(Y) such that f + g but t»(f °T (X),X) = a)(g«T(x),x) fo r a l l x € E , that i s , (Tf)(x) = (Tg)(x) f o r a l l x e E . Hence T i s not one-one (as we know that (Tf)(x) = (T©)(x) = (Tg)(x) f o r a l l x € X - E ) . Therefore T [ E ] i s dense i n Y i f f T i s one-one. In addition, we assume now T i s onto. Then T[G m(Y)] = C m(X) which includes a l l constant functions. Thus, f o r each x e, X , (T[C m(Y)])(x) = C m(X)(x) 3 (r(x) = r : r e E 1 } = E 1 . 6 2 . On the other hand, f o r each f e C ( X ) , f ( x ) = r f o r some r € E 1 . Hence (T[ C^Y) ]) (x) = E 1 so that x c E . Consequently X = E . We w i l l show that T i s one-one. I f T ( X , ) = T ( X q ) = y Q , then f ( r ( x ) ) = f ( T ( x Q ) ) . Suppose x + x Q, Since T i s onto/ there e x i s t f and g e C m(Y) such that (Tf)(x) = (Tg)(x) but ( T f ) ( x Q ) 4= (Tg)(x Q) . Then a>(f .T(x),x) = (Tf)(x) = (Tg)(x) = uu(goT(x),x) and as i»(t,x) i s one-one f o r fi x e d x , f ( r ( x ) ) = g«T(x) so that T * T ( X g«T(x Q) . That i s , f ( y Q ) = g(y c) On the other hand, oo(f(y 0 l),x Q) = ( T f ) ( x Q ) + (Tg)(x Q) = tn(g(y 0),X 0) and eo(t,y 0) i s one-one, so we have f ( y 0 ) 4 g(y 0) which'is a contradiction. Hence x = x Q, I.e. T i s one-one. Therefore T " 1 IS well-defined on T [ X ] . Now, for each g e C M ( X ) , the mapping > f defined by f ( x ) = <»(g(x),x) f o r a l l x e E = ' X i s i n C M ( X ) (as shown above). Thus, g(x) = <x(f(x),x) f o r a l l x e_ X . Since T i n onto, there i s an f € (^(Y) such that T ( f Q ) = f . We then have g(x) = a ( ( T f Q ) ( x ) , x ) = f Q o T ( x ) for a l l x e X . In other words, C M ( X ) c {h«T : h e C m(Y)} . But, We know that h*T e C M ( X ) f o r a l l h e C m(Y) . Hence (^(X) = { h o T : h £ C m(Y)} . Now, f o r each f e ( ^ ( X ) , f = h»r f o r some h £ C m(Y) hence f ^ r " 1 - . -h | T [ X ] e C m ( T [ X ] ) . By Theorem ( 6 . 4 ) , T" 1 i s C m on T [ X ] . F i n a l l y , we w i l l show that T [ X ] i s Cm-embedded i n Y . For each g e C m ( * r [ X ] ) , the function f , defined by f(x) = ce (goT(x),x) f o r each x e T [ X ] , i s In C M ( X ) . Thus, there ex i s t s g Q € C m(Y) such that T g Q = f , or (Tg Q)(x) = f (x) t«(g 0 oT(x),x) = « ) ( g o T(x),x) f o r each x e X . Hence g ° T(x) = g Q o T(x) f o r each x e X , or g Q | T [ X ] = g , that i s g Q i s an extension of g . Q.E. D. 63. (8.6) Theorem: Let X and Y be any two d i f f erentiable n-i n 0 n l * N2" M A N I F , O L D S* o r subspaces of E and E , , respectively, T be a C^-mapping from X into Y , and uo a C m-mapping from E 1 x X into E 1 such that «j(t,xQ) i s a homeomorphism int o E"*" f o r each x Q € X . Then the transformation T defined f o r each f e (^(Y) by (Tf)(x) = t » ( f o T ( x ) , x ) i s into C M ( X ) , and T s a t i s f i e s conditions (a) and (b) of Theorem (8.5), T is one-one i f f T [ X ] i s dense i n Y . ' Moreover, i f T i s a C m-diffeomorphism of X onto a C m-embedded subset of Y , then T maps onto ( ^ ( X ) . Proof: That Tf i s i n ( ^ ( X ) follows immediately ftom the Chain rule, d e f i n i t i o n s of uo(t,x) and T , and the f a c t 1 that f e C m(Y) . Next, we w i l l show T s a t i s f i e s conditions (a) and (b) of Theorem (8.5). ,If f>g e C m(Y) are such that (Tf)(x) = (Tg)(x) f o r any x e X , then «j(f«>T(X),X) = uo(g«>T(x),x). But uj(t,x 0) i s one-one f o r f i x e d x Q e X , so we have fo-r(x) = g o T ( x ) . Now, f o r each h e (^(Y), (T(f+h))(x) = t»( (f+h )T(x),x) = t » ( f o T ( x ) + h«>T(x),x) = «j(goT(x) + h o T(x),x) = u3((g+h)T(x),x) = (T(g+h))(x) . S i m i l a r l y (T(f»h))(x) = (T(g»h))(x) . Hence T s a t i s f i e s (a) . That T s a t i s f i e s (b) i s a consequence of d e f i n i t i o n of T . The proof that T i s one-^one i f f T [ X ] i s dense i n Y is s i m i l a r to that T i s one-one i f f T [ E ] i s dense i n Y i n "Theorem (8.5). By hypothesis of ; uu , f o r x Q e X , l e t a ( t , x 0 ) be the inverse of «j(t,xQ) . Now, f o r each x e X , we have (Tf)(x) = oo(fOT(X),X) . Thus, f « T ( x ) = a((Tf)(x),x) on X . Hence a((Tf)(x),x) i s C01 on X , f o r each f e C m(Y) . 64. F i l i a l l y , assume that T i s a C m-diffeomorphism of X onto a .^embedded subset of Y . Given F €, C M ( X ) . , define, g hy g(y) = gor(x) - a ( F c T " 1 ( y ) , T " 1 ^ ) ) f o r a l l X ' = f - 1 ( y ) e X . As shown i n the1 l a s t paragraph, g e C M ( T [ X ] ) . We know that T [ X ] i s a CM-embedded subset of Y ', so there exists g Q € G M ( Y ) such that g Q | T [ X ] = g . We know that (Tg Q)(x) = t»(g«T(x), T^OT^X)) . By d e f i n i t i o n s Of g and a , we have. F o T _ 1 ( y ) = o)(g(y),x) f o r a l l x € X and y = T(X) e T . [ X ] . Hence (Tg Q)(x) = F «T"' 1(T(X)) = F(x) f o r each x ,e X . That i s , T g c = f or T maps C R A ( Y ) onto ( ^ ( X ) . Q.E.D. • Note that no extra condition i s needed f o r X or Y . i § 9 The Rings of C ^ d i f f erentiable Functions on Spaces which are not m-realcompact, and some Algebraic Properties of C not applicable i n C M . We have shown that i f X and Y are two m-realcompact spaces, then C M ( X ) and C M ( Y ) are isomorphic i f f X and Y are C m-diffeomorphic. We s h a l l make some observations about other cases i n t h i s section. We w i l l f i r s t of a l l construct a subring of (^(x) , where X c E n . Let •c(x 1,...,x n) « x± f o r 1 < 1 < n ,(as defined i n (6.2)), S-^ = {r : r i s the constant function defined on X with value r e /R), ahd S 2 = {^ 6-: 1 1 ..'<, h}. Let 3 = 3^^1)32 and 00 be the r i n g generated by S . Then, i t i s clear that Ot i s a commutative ring. Next, we embed Ot i n the quotient r i n g , R ( X ) , which consists of a l l quotients f/g* f , g € Ot and Z(g) = {x e X : g(x) = 0} = <}) . Evidently 65. ft(x) i s a commutative r i n g of r a t i o n a l functions on X with unity u , the constant function of value 1 , zero element © ,-the constant function of value 0 . In the following lemmas and theorems h-^ and Ag are rings of functions with the following properties: , ft(X) c ^ c C M ( X ) , R ( Y ) c Ag e C ^ Y ) and i f f e A± (or Ag) with Z(f) = $ , then f"" 1 e A x (or Ag) . ( 9 . 1 ) Lemma: There i s a function f e.M » {f e A x : f( a ) = 0 ] , f o r a e X such that Z(f) = {a] , and f belongs to no other free or f i x e d maximal i d e a l of A^ . . _n n 2 Proof: Let a = (a.^,.. ,,a n) e X c r , and f(x) = X ( x ± - a^) . Then, i t i s clear that f e M and belongs to no other f i x e d maximal. Suppose that M i s a free maximal i d e a l i n A^ such that f e M . Since M i s free, there i s g e M such that g(a) + 0 . Let h = f 2 + g 2 . We have Z(h) <f) , so that i t s inverse h " 1 e A-^ . Since M i s an i d e a l u = hh" 1 € M . That i s , M i s the whole r i n g which i s impossible'. Hence f 4 M , or f belongs to M -only . Q.E. D. n' Let Y be a subset of E where n« may be equal to n . Let R ( Y ) be the r i n g defined i n the same manner as a ( X ) with domain Y , the r i n g of r a t i o n a l functions on Y . ( 9 . 2 ) Lemma: I f cp i s an isomorphism from A^ onto Ag , then f o r any M & c A 1 , cp(M&) i s a f i x e d maximal i d e a l i n Ag . Proof: Since cp i s an isomorphism onto, cp(MQ) i s a maximal i d e a l i n Ag . By Lemma ( 9 . 1 ) there i s f 0(jc) = 2 ( x i - a.^) 2 66. such that Z(f ) = {a} belongs to M only. Consider z ( ^ ( f 0 ) ) • I f Z ( ^ ( f Q ) ) = <j) , then ep(f Q) i s a unit i n Ag so that «p(M&) i s the whole r i n g Ag which i s impossible. Hence Z(cp(f" 0)) + § . . Gn the other hand, i f Z ( ^ ( f Q ) ) contains more than one point, say b, b' e Y , then cp(f Q) e and c p(f Q) e M^ , so that f 0 would belong to at l e a s t two maximal ideals which again i s impossible f o r f belongs to only one maximal ideal. Thus Z ( c p(f G)) = b , say. Hence ^CW^) = . (9 .3) Lemma: Let B^ and Bg be subrings of C(X) and C(Y) respectively, which contain a l l constant functions, cp be an isomorphism from Bg into B^ , and X be connected. Then cp i s the i d e n t i t y on the constant functions. Proof: It i s clear that - cp(u) = u and ep(©)•=•:© . By property of isomorphism, we have cp(n) = n f o r a l l p o s i t i v e integer n . Now, cp(u-u) = cp(©) = © or cp(u) + cp(-u) = © . Thus cp(-u) = -cp(u) = »u , so that cp(-n) - -n f o r a l l p o s i t i v e integer n . Moreover, c p ^ ) = _g = cp(p_«3) = ja ^ ( j f ) (by property of isomorphism), That i s , ep(j^) = J[ f o r any p o s i t i v e integer p and any integer We w i l l show that f o r constant function k , cp(k) i s also a constant function. Indeed, i f k i s an i r r a t i o n a l number, k-r 4 G f ° r a H r a t i o n a l numbers r . Thus e x i s t s f o r a l l r a t i o n a l numbers r-... Moreover, cp(k^r) »cp( ) = cp(k-r » -j^p ) = q>(u) = u , we have cp ( - j £ ~ ) = - ^ r j j 1 <P(kJ-r 1 , f o r a l l r a t i o n a l numbers' r . Now, i f cp(k) i s not 67. a constant function on X , then since X i s connected and cp(k) i s continuous, we must have cp(k)(X) i s a connected sub-1 . x set i n E . Thus cp(k)(X) contains some r a t i o n a l numbers say r 6 . Then c p ( ^ ) -with cp(k)(x) = r Q . This i s a contradiction. Consequently, cp(k) must be constant, so that cp , r e s t r i c t e d to the subring of constant functions on Y , i s a non-zero homomorphism of fR i n t o i t s e l f . By Theorem ( 0 . 2 2 ) [ 7 ] , cp i s the i d e n t i t y . Q.E.D. (9-4) Theorem: Let X and Y be two a r b i t r a r y subsets of E n , and cp be an isomorphism from Ag onto A^ leaving a l l constant functions unchanged. Then, cp induces a mapping T : X.-*'Y defined by cp(g) = g«T and T i s a C m-diffeomorphism. Proof: Define T to be' a mapping from X to Y as follows: T(X) = nZ[cp" 1(M x)] . Since cp i s an isomorphism, and i t onto, i t s inverse mapping cp"1 i s an isomorphism of A^ ^ onto Ag . By Lemma ( 9 . 2 ) , cp"°1(Mx) i s a f i x e d maximal i d e a l i n Ag . Thus, T i s a single valued mapping. Evidently, M T ( X ) = cp~1(Mx) . Let x and x' be i n X and x | x' . Then, by Lemma ( 9 . 2 ) , again cp"1(Mx) = M y and cp°1(Mx, ) = My, f o r some y, y' e Y . If y - y« , then cp~1(Mx) = My = cp" 1(M x t ) which implies M x = Mx, . This i s impossible f o r x ^ x' . Thus y ^ y f . Hence T i s one-one. Let y Q be a r b i t r a r y i n Y . Then My i s a f i x e d maximal i d e a l i n Ag and ^(My ) = M x f o r s o m e X 6 6 X ' T h u s y o " nZ t tp^CM^J] . - T(X Q), 68. This shows that T i s onto, How, f o r each g € Ag and each x e X , l e t ep(g)(x) = r . Then ip'(g) - T 'e , g - cp" 1^) = cp_1(cp(g) - r ) e MT(x),g(r(x)) = ( q f 1 ^ )J(T(X) ) = r(T(x)) = r = (cp(g)Xx) • Hence cp(g) = g°T . S i m i l a r l y , cp" 1(f) s for"1 » where T"1 : Y - X i s defined by T ~ x ( y ) = n z M M y ) ] . know that f o r a l l f e Ag , f c T.e.A^ . In p a r t i c u l a r , i f f = M (as defined above) , 1 < i _< n , then i s C m f o r each 1 < i < n . S i m i l a r l y , each component of T~ i s G . That i s , T i s a C m-diffeomorphism. Q/-E.D. ( 9 . 5 ) Corollary: Let X and Y be two connected subsets of E 1 1 , and cp be an isomorphism of A 2 onto A 1 . Then cp induces a cP-diffeomorphism, T, from X onto Y such that cp(g) = g o T f o r each g e Ag . Proof: We know that both A^ and Ag s a t i s f y the condition In Lemma (9.3) . Hence cp i s the Identity on the constant -funct-ions between Ag and A 1 . Then the conditions i n Theorem (9.'4) are s a t i s f i e d . The r e s u l t follows immediately from Theorem ( 9 . 4 ) . & . E , D. (9.6) Theorem: Let X and Y he two subsets of E n , and T be a C^-diffeomorphism of X onto Y . Then the induced mapp-in g T» defined by T'(g) = g o T i s an isomorphism of Ag onto A which i s the i d e n t i t y on the constant functions. Proof: For each g e Ag , g ° T e A^ , and f o r each f € A 1 , f a t " * 1 e Ag are clear. Now, f o r any f e Ag , T ^ f o r " 1 ) = • • T " 1 ' ) T = f . Thus, T' i s onto. Now, i f T'(g) = S , then (g» T)(x) = 0 f o r each x e X or g ° T[X] = g[Y] = {G} We 69. as T i s onto. That i s , g = 0 . This shows that T i s one-one. F i n a l l y , f o r each constant function r , T 1 ( r ) ( x ) = r(-r(x)) = r , f o r each x . That i s T»(r) = r . Q.E.D. Remark: I f A-^ = ft(X) and Ag = & ( Y ) , then T and T"1 are not only C m , each of t h e i r components i s a r a t i o n a l function. We w i l l name t h i s mapping as rational-homeomorphlsm. We also know that there i s a non-linear rational-homeomorphism. That i s , l e t X = Y = E 1 1 - ( 0 , . . . , 0 ) and T ( X ) = (T-JCX), ... ,T ( X ) ) x. be defined as T.£K) = — ± — f o r 1 <: i < n . Then i t s x 1 +...... +x n inverse i s known to be T " 1 ( y ) = (<|)^(y),... , ^ n ( y ) ) with ^ (y) = -g—d ^ , 1 < j < n , d y x+. ..+y2 Next, we w i l l see some algebraic properties of the rings of continuous functions which are inapplicable i n the rings of C r a - d i f f erentiable functions, where 1. < m < «. (1) The f i r s t one i s that the rings of continuous functions are lattice-ordered. But the rings of C m - d i f f e r e n t i a b l e functions are not. For instance, l e t X = E 1 . Consider (^(X) . Then we know that I(x) = x , i e C m(X) but | i | 4 C m(X> . Thus neither f A 0 nor f v 0 , In general, i s i n (^(X) . (2) We know that,' i n the r i n g of continuous functions,, for a z-ideal I , 1(f) _> 0 i f and only i f f f i s non-negative on some zero-set of I (see [73(5.4)(a)). Also, I i s a z-i d e a l implies that I i s convex. Hence, by Theorem (5.2)[7], I ( f ) _> 0 i f there i s g e C(X) such that g j> 0 and g = f (mod I ) . In the r i n g of d i f f e r e n t i a b l e function such a g need not exist. For example: Consider X = E 1 and C 1(X). 70. Let I = {f e C^X) : Z(f) 3 [0,1]}. Then since I i s a z-i d e a l , i t i s convex. However i t i s not absolutely convex, f o r 1 2 G (X) i s not lattice-ordered. Now, l e t f Q ( x ) = x - x . I t i s clear that f _> 0 on a zero-set of I . But, i f g € C 1(X) so that g _> 0 and g agrees with f on [ Q ? l ] , then the derivative of g does not exi s t at 0 or 1 . Indeed, l i m + - g(0) = l i m Ax - Ax 2 = i but Ax ^ Ax J L l * . g(Ax ) . - . -s(0), , fil±m. < o , as g(x) > 0 f o r a l l x. Thus, g'(x) does not exi s t at x = 0 . S i m i l a r l y , g ( A x + i ) , - gp . ) . - = lim g(Ax+l) Q b t Ax-0 L x Ax-0 A x * l i m g(AXH-l) - ,g(l) = l i m 1 + AX - (1+Ax) _ lim, Ax-Ax -2AX * x - ° ~ Ax AX A X ^ ° " Ax = - 1. That i s , g does not e x i s t at x = 1 . Hence, there i s not any g e C 1(X) with g >_ 0 and f s g (mod I) though we know that f i s nonnegative on a zero-set of I . (3) If I and J are z-ideals i n C(X) , then I J = I fl J . This i s not true i n C m (X) . Let X = E 1 , consider i n . C 1 ( E 1 ) , I = J = M Q = [f e (^(E 1) : f ( 0 ) - 0} . Then I n J = M Q . But I J + I 0 J . Indeed, i e M Q = I fl J where i ( x ) = x . I f i = gh f o r gome g e l and h e J , then g(0) = 0 = h(0) ;, and g,h are C^-differentiable. Accordingly, we have g(x) = g(0) .+•'. ax + ex , h(x) = h(0) + a'x + e'x , where a and a' are constant and e, e' -• 0 as 2 • 1 x - 0 . Thus, g(x)h(x) = (a+e^a'+ejx . Now, i'(x) = 1 71. f o r a l l x e E 1 . However, (g.h)«(x)! x = 0 = g(xUh(x) - g (0)h (0 ) m l i m ^a+€.j(a»4:€°,)x2 = Q ; H e n c e l m Q which i s impossible. That Is,; i s ( I n J)» IJ. This example also shows that the following i s inapplicable to (^(X) or C m*(X). I f P and Q are prime ideals i n C (or C*), then PQ = P n Q . For ^ C M * ( X ) , we t a k e X = (°n,n) and use the i same argument as above. 72. PART II THE RINGS OP La-(OR L~) FUNCTIONS We now study the same sort of properties when (^(X) i s replaced by the r i n g of functions s a t i s f y i n g a L i p s c h i t z condit-ion on each compact- subset of a metric space. £10. Rings, Ideals and Some Properties _of L i p s c h i t z i a n or Lc° Functions (10.1) Definitions A r e a l <=valued function f defined on a metric space (X,d) i s said to be a function s a t i s f y i n g a Lips c h i t z condition on each compact subset of (X,d), i f f o r each compact A c X , there i s a p o s i t i v e number K A such that fOr any two points x 3x' € A* |f(x) - f(x«)| <_ K Ad(x,x»). For the sake of brev i t y , we w i l l c a l l such a function Le-f unction, (see uniform L i p s c h i t z condition on each compact subset P.35 [^6].) (10.2) Definitions A real-valued function f defined on a metric space (X,d) Is said to be a L i p s c h i t z i a n function (or to s a t i s f y a Li p s c h i t z condition) of constant K , i f there e x i s t s a p o s i t i v e number K such that f o r any two points x,x« € X, |f(x)_ - f ( x ' ) | < K d(x,x») . Note that i f the metric space i s compact, then an Lc-function i s , clearly/an L=function. Remarks ( l ) The Lips c h i t z condition implies uniform continuity. Indeed, i f f s a t i s f i e s a Li p s c h i t z condition of constant K on (X,d) , then f o r any given p o s i t i v e number e , choose 6 = e/K , we have f o r any x,x> e X, |f (x)-T(x».) | <_ Kd(x,x' )\< 73. K«e/K = € whenever d(x,x«) < 5 = c/K (2) However, a continuous function may not be a L i p s c h i t z i a n function. For example, f ( x ) = x V ^ iavcontinuous on A = { x 1 <. x < 1) . But |f (x) f ( x « ) | - Ix1/5 - x' 1/ 5! - |x - x« |/|x 2 / 3 + x 1 / 5 x'3-/3 + X » 2 / 5 | = ' d(x,x') 1 | x*/3 + x l / 3 X , V 3 + x !'^/3j * W h e r e 1^7-5 + X V 3 x , l / 3 + x , * / 3 | i s hot bounded. Hence f i s not a L i p s c h i t z i a n function. (3) It i s also,easy to see that an Le-function i s a continuous function. But, a continuous function need not be an Lc-function. Consider X = E 1 and f(x) = x1^ . Then, as i t was shown i n Remark (2) , f o r any compact subset containing the o r i g i n , there does not e x i s t any constant s a t i s f y i n g the required condition. 1 (10.3) Definitions Let (X,d) be a metric space. Lc(X) • Cf f i s a real-valued Lc-function on X). L(X) = [t s f i s a real-valued, bounded function on X, and i s "a L i p s c h i t z i a n function of some constant' K}. L-function means L i p s c h i t s i a n function. (10.4) Theorems. The family Lc(X) and L(X) are commutative rings with unity under pointwise addition, subtraction and m u l t i p l i c a t i o n . In such case, the unity i s u , the function with constant value 1 9 and Q the constant function with value 0 , i s the zero'element of them. Remarks The condition that f i s bounded on X can not be omitted i n L(X). For,' l e t (X,dO be an unbounded metric space. Then, f p ( x ) = d('p,x) i s a L i p s c h i t z i a n function with 7 4 . constant 1 . For If (x) - f (x ' ) l = |d(p,x)---- d(p,x« ) I _< & IT 2 d(x,x» ) . However, l e t g(x) = f (x)*f (x) = (d(p,x)) i s not L i p s c h i t z i a n of any constant. Indeed, |g(x) - g(x')| = |d(p,x)) 2 - (d(p,x)) 2| = [d(p,x) + d(p,x')].|d(p,x).- d(p,x)| > d(x,X')•|d(p,x) - d(p,x«)|. Here.we know that |d(p,x) -d(p,x')| i s unbounded. That i s , g i s not Li p s c h i t z i a n . The reason that we discuss t h i s p a r t i c u l a r r i n g L(X) i s that i n §13 , we s h a l l discuss that L(X) i s a Banach algebra^ and Lc-realcompactness. Proof of the Proposition: F i r s t consider L(X). Let f and g be a r b i t r a r y from L(X) of constants K^ and Kg , respect-i v e l y , and |f | _< K', |g| _< K" . Then, f o r any x,x' € X . |(f + g)(x) - ( f + g)(x«)| = |(f(x) - f(x«)) + (g(x) - g(x'))| < |f(x) - f ( x ' ) | + |g(x) - g(x«) l < Kjd(x,x».) + Kgd(x,x«) < (Kj^ + Kg)d(x,x> ) . Take K = + Kg . Moreover f + g i s bounded by K» + K" . Hence f + g e L(X) . Now, |fog(x) - f og(x ' ) l= |f(x)g(x) - f(x»).g(x»)| < |f(x)| l g(x) t ........ g(x')| + |g(x» ) | o |f (x) - f ( x ' ) | _< K 'oKgd(x,x«) + K "K ' od(x,x') = ( K ' K g + K"K 1)«d(x,x') . Also | fog| '« |f | . | g| < K ' . K " . . Hence f«g e L(X) . Evidently, u arid Q e L(X) . For L C(X) , take any f ? g € L C(X) and any compact subset A of X . Then, by the f i r s t part we know that f|A,g|A e L(A). Hence, (f + g)|A, (fg)|.A € L(A) . Since A i s an a r b i t r a r y subset of X , by d e f i n i t i o n of L C(X) , f + g , f g e L C(X) . ; It i s obvious that L(X) c L C(X) . Hence u, Q e L C ( X ) . Q.E.D. Remark: If f e L(X), then |f| e L(X) . For | | f ( x ) | - |f(x« ) l l _< |f(x) - f ( x ' ) l < Kod(x,x») f o r a l l x,x» e X , and | f | i s 75. bounded. Hence, i f f , g e L(X.) , t h e n |f - g| e L(X) so th a t f v g-- f"+ * + ' f ° g< and f A g - L±£^£^Ji. 2 2 e L(X) . That i s , L(X) i s a l a t t i c e - o r d e r e d r i n g (p.7 [ 7 ] ) . S i m i l a r l y , we can show t h a t Lc(X) i s a l a t t i c e - o r d e r e d r i n g . (10.5) P r o p o s i t i o n : I f f i s a L i p s c h i t z i a n f u n c t i o n on (X,d) w i t h constant K , then j f | A n <£ L(X) w i t h the same constant. Proof: We know t h a t | f | A n i s bounded by n . Both | f | f I .+ £ .-. I I f I - £1 and n are L i p s c h i t z i a n . Hence | f j A n = - - / 2 i s L i p s c h i t z i a n . (10.6) Lemma: Let f e L(X) and r _< | f | f o r some p o s i t i v e , number r . Then f " 1 ( = l / f ) e L(X) . Proof: We know that | f ( x ) - f(x«)l < K d(x,x«) f o r some p o s i t i v e number K . We a l s o have |-y (x) - ^ ( x ! ) | = = |f(x) - f(xv)| < • | f ( x ) - . f . ( x » ; ) | ; < K D ( X^ X, ) ; f ( x ) f ( x ' ) | f ( x ) | . | f ( x « ) l ~ T ~~ T Moreover l / f _< l / r . Hence 1/f e L(X) . (10.7) Lemma: Let f e L*(X) - {f e L C ( X ) : f is'bounded}, and r < | f | f o r some p o s i t i v e number r . Then f = 1 (= l / f ) e LJ(X) • Proof: Let A be any compact subset of X . Then, as shown i n Lemma (10.6), we know l / f |A € L(A) . Hence l / f e L£(X). (10.8) P r o p o s i t i o n : For any p e X , def i n e f (X) = d(p,x). Then f e L(X) i f f (X,d) i s bounded. We ; s h a l l c a l l such a f u n c t i o n t a d i s t a n t f u n c t i o n . 76. Proof: We have already Shown that f is a L i p s c h i t z i a n function i n the Remark of Theorem (10.4). Thus f € L(X) i f f f i s hounded. Hence the r e s u l t follows immediately from the boundedness of a metric space. Q.E.D. (10 .9) D e f i n i t i o n : The set Z(f) = {x 6 X : f ( x ) = 0} i s said -to be the zero-set of f . Let Z(X) = (Z(f) : f e L(X)} -[Z(f) : f e L(X)} ( s e e ( l l . S ) ) . (10.10) D e f i n i t i o n : I f f e LC(X),(L*(X) or L(X)) has a multi-p l i c a t i v e inverse i n L (X), (L* (X) or L(X)) i s said to be a unit i n L C ( X ) 5 (L*(X) or L(X)). Remarks: ( l ) It i s clear that f e L_(X) i s a unit i n L_(X) i f f Z(f) = ty and f e L* (X), (L(X)) is a unit i n L*(X) (L(X>) i f f |f | > r f o r some r e , r >,0. (2) Consider an unbounded metric space (X,d), and f ( x) = ' 1 1 • • . Then |f | < 1 , and |f (x) - f (x> )| -1.+ d(,p,x) ld(p>x) - d(p,x»)| < d(x,x») . Hence f e L(X) and (1 + d ( p , x ) ) ( l + d(p,x«)) Z(f) = ty but | f | } r f o r a l l p o s i t i v e number r . Consequently f " 1 4 L(X) so that f i s not a unit. (10.11) D e f i n i t i o n : A nonempty subfamily g of Z(X) i s said to be a z - f i l t e r on (X,d), i f i t s a t i s f i e s the conditions (1), ( i i ) and ( i i i ) of (1.7). (10.12) Proposition: I f I i s a proper i d e a l i n I»C(X) , then the family Z [ l ] - {Z(f) : f e l } i s a z - f i l t e r on X . Proof i s sim i l a r to (1.8). 77. Remark: If L_(X) i s replaced by L ( X ) or L*(X) , then the r e s u l t need not be true. Let (X,d) be unbounded, and f( x ) = - — 1 -' f o r a f i x e d point p e X . Then, we know d(x,p) +1 that f e L(X) (and L*(X)) but f " 1 4 L(X) (nor L*(X)). Set I = (f) , the ideal generated by f . We know then that <|) = Z(f) e Z[l] . That i s , Z [ l ] i s not a z - f i l t e r . (10.13) Proposition: I f 3? i s a z - f i l t e r on X , then the familys Z ^ 1 ^ ] = {f e L Q ( X ) : Z(f) € 3 } , Z £ i [ 3 ] = {f e L*(X): c c Z(f) € 3 } and Z^Es] = ffeL(X) :Z(f )e3?} are ide,als i n L (X), L*(X) and L(X) , respectively. Proof i s si m i l a r to (1.9). Notice that we w i l l use Z" 1 to denote the appropriate 1 -1 -1 „i notation of Zj~ , ZL# or Z^ . c c Remarks: ( l ) Ztz" 1 ^]] - 3 . This shows that every z - f i l t e r i s of the form Z[l] f o r some i d e a l i n L_(X) , L*(X) or L(X). ( 2 ) I t i s clear that Z " 1 ^ ! ] ] =>'I . The i n c l u s i o n may be proper.' For. example, l e t (X,d~) be a (bounded) metric o 2 space, and f Q ( x ) = ( f (x)) » [d(p,x)] . Then f Q e L e(X) (L*(X) or L(X)). Let I = ( f Q ) . This consists of a l l funct-ions f i n L(X) such that f = f »g f o r some g e L„(X) (L*(X) or L(X)). In p a r t i c u l a r , every function i n I vanishes at _p . Hence every zero-set i n Z[l] contains the point p . As a matter of f a c t , since Z [ l ] i s a z - f i l t e r that includes the set {p} = Z ( f Q ) , I t must be the family of a l l zero-sets contain-ing p . 78. The i d e a l M Q = Z'-^ Z f l ] ] evidently consists of a l l functions i n L ( X ) (L*(X) or L(X)) which vanish at p . Hence MQ p I . However, M Q 4= I • For instance, f p ( x ) vanishes at p and f € L C(X) (LJ(X) or L(X)). Thus f p e MQ. Suppose f € I . Then, f = f 0«g f o r some g e L C(X) (L*(X) or L(X)). But, then g(x) = [d ( p , x ) ] " 1 i s discontin-uous at the point p . Thus g 4 I v ( x ) (L*(X) or L(X)). This i s a contradiction. Note that Z[M Q] = Z [ l ] , i n spite of the fa c t that M Q + I • It i s also obvious that M Q i s a fi x e d maximal i d e a l ; (10.14) D e f i n i t i o n : A z - u l t r a f l i t e r on X i s a maximal z - f i l t e r on X . (10.15) Proposition: If M i s a maximal i d e a l iri - L C ( X ) , then Z[M] i s a z - u l t r a f l i t e r on X . Proof i s s i m i l a r to (1.11). (10.16) Proposition: If Mr-is a z - u i t r a f l i t e r on X , then • Z " 1 ^ i s a maximal i d e a l i n L (X) . Proof i s ' s i m i l a r to (1.12). (10.17) D e f i n i t i o n : An i d e a l I i n L C(X) i s said to be a z~ id e a l i f Z(f) e Z [ l ] implies f e l . That i s , I = Z ^ t Z t l ] ] . Example: Every f i x e d maximal i d e a l i s a z-ideal. Note that i f I i s an i d e a l i n L*(X) (or L(X)) such that there i s an f e l with Z(f) = (j) and f " 1 \ L*(X) (or L(X)), 79-then I is clearly not a z-ideal, (10.18) Proposition: If M Is a maximal ideal in L.(X) (L*(X), L(X)), and Z(f) meets every member of Z[M] , then f € M . Proof is similar to (1.13). (10.19) Proposition: If >fis a z-ultraf l i ter on X , and a zero-set Z meets every member of , then Z s Proof is similar to (1.14). (10.20) Definition: in ideal I in L (X) (L*(X) or L(X)) is said to be a prime ideal i f f <»g e I implies f e I or g e l . (10.21) Theorem: For any z-ideal I in LQ(X) (L*(X) or L(X)), the following are equivalent. (1) I Is prime. (2) I contains a prime ideal. (3) For a l l g»h e L e(X) (L*(X) or L(X)), g.h = S , then g e l or h € I . (4) For every f e L.(X) (L*(X) or L(X)), there is a zero-set in Z[l] on which f does not change sign. Proof: That (l) implies (2) is clear. If I contains a prime ideal P , and g<>h = £ , then g o h e P . Hence g e P c 1 or h e P c I . This shows that (2) implies (3). To see that (3) implies (4), for every f e L C W , consider (f v Q)°(t A Q) m Q e I . By (3) f v <9 € I or f A Q e I . Either case Implies (4). Finally, we wil l show that (4) implies ( l) . Given g»h e l , consider the function |gj - |h | . By 80. hypothesis, there i s a zero-set Z of Z [ l ] on which | g | | h i s nonpositive, say. Then every zero of h on Z i s zero of g . Hence Z(g) => Z n Z(g) = Z 0 Z(goh) e Z [ l ] so that Z[g] e Z [ l ] . Since I i s a z-idea l , g e l . Thus I i s prime. The proof f o r L*(X) or L(X) i s exactly the same with L*(X) or L(X) taking the place of L(X). %11 L^Oomplete Regularity and L-Normality. (11.1) D e f i n i t i o n : A metric space (X,d) i s said to be L-completely regular i f , f o r every closed subset F of X , and x e X - F , there i s a function f e L(X) such that f ( x ) = 1 and f ( F ) = {0} . (11.2) D e f i n i t i o n : A metric space (X,d) i s said to be L-normal i f , f o r any two d i s j o i n t closed subsets F , F' , there i s an f e L(X) such that f ( F ) = {0} and f(F>) = Cl] Note that a metric space i s a Hausdorff space so that L-normallty implies L-complete r e g u l a r i t y . (11.3) Theorem: A metric space (X,d) i s L-'completely regular i i f f the family Z(X) of a l l zero-sets of L(X) i s a base for the closed subsets of X . Proof i s si m i l a r to (2.4). (11.4) Lemma: Let A be a nonempty subset of a metric space (X,d) and f e L(A) . Then there i s a g e L(X) such that g|A = f . Proof: Suppose |f| _< n , and K i s the 'Lipschitzian constant of f . For each x e X , define f_(x) = Sup {f(x«) - Kd(x»,x 8 l . Then for each x Q e A , f o ( x Q 0 = J f | A * f ( x ' ) " Kd(x«x Q)} . We know that K > 0 and d(x',x) j> °' I f there ex i s t s , x^ e A such that f ( x x ) - K o d(x 1,x Q) > f ( x Q ) - Kd(x,x Q) = f ( x Q ) , then f ( x 1 ) - f ( x 0 ) > K«d(x 1,x G) which contradicts the f a c t that |f(x) - f(x«)l < Kod(x,x') f o r a l l x,x» e A . Thus, f ( x Q ) > f(x') - K>d(x«,x 0) f o r a l l x« e A . Hence x f A (f(x') - K«d(x,x')} = f ( x 0 ) f o r x Q e A . That i s , f Q | A = f . We w i l l show that f Q i s L i p s c h i t z i a n on X of constant K . For a r b i t r a r y x,x' e X , and x e A., f ( x Q ) - Kod(x 0,x«) + Kod(x,x») > f ( x D ) Kod(x Q,x) > f ( x Q ) -Kd(x o,x») - K«d(x',x) . Hence f Q ( x ' ) + K>(x«,x) = xeA { f ( X o ) " Kod(x G,x')} + K o d ( x , x « ) = n A { f ^ X J " K e d ( x o ^ ' ) + K o d ( x ^ x o O > x U ^ A t f ( x o ) - Kcd(x Q,x)} = f Q ( x ) . On the other hand, o f 0(x«) - K»d(x',x) = x U ^ A t f ( x Q ) - Kod(x o,x')} - K»d(xSx) = J A (f(x o) - Kod(x Q,xi) - K»d(x',x)} < X ^ A Cf(x Q) - Kod(x,x0)} = f 6 ( x ) . Hence | f Q ( x ) - f j x ' ) | < K«d(x,x>) . F i n a l l y , l e t g = f A n . By Prop. (10.5) g € L(X) . Hence g i s required. Q.E.D. (11.5) Theorem: ( l ) Every metric space (X,d) i s L-completely regular. (2) Every compact metric space (X,d) i s L-normal. Proof: (1) Let F be a closed subset of X and p € X - F . Then d(F,p) 4 0 . Let f be a function defined as follows: .82. f [ F ] - {1}, and f(p) = 0 . Let K = (d(F,p))° 1 . ¥e w i l l show that f i s L i p s c h i t z i a n on F u [p ] with constant K . If x,x» e F , then |f(x) - f(x ' ) l = 0 _< K>d(x,x? ) . I f x e F and x' » p , then |f(x) - f(x')|= 1 , and K>d(x,x») = d(x,x') d(F p) - 1 • Hence ' |f(x) - f(x')(<_ Kod(x,x«) for x,x' e F U {p} . It i s clear that f i s bounded by 1 . That i s , f e L(F U {p}). By Lemma (11.'4) there i s an f Q e L(X) such that f Q | F U {p} = f . Hence f Q [ F ] = {1} and f Q ( p ) = G . For (2), we only have to replace {p} by a closed set F' d i s j o i n t from F . a. E. D. Remark: The compactness can not be omitted i n (2). For instance 1 P p - p Let X = E , F = {(x^y) € E : xy = 1} and P» = {(x,y) e E : xy «= -1} . Then F and F' are two d i s j o i n t closed sets i n 2 2 E . However, i t i s clear that there i s not any f e L(E ) such that f [ F ] = {l) and f [ F ' ] = {0} . (11.6) Lemma: Let B r (x Q) = {x e X : d(x,x 6) < r ± 3 i = 1,2, where 0 _< r ^ < r 2 . Then, there i s an f e L(X) such that: f ( x ) = 0 f o r x e clx B r i ( x g ) (11-1) JO < f(x ) < 1 f o r x e Br (x Q) - cl^ (x Q) W(x') « 1 f o r x,e X - B r^(x' 0) Proof: Let cp(t) = J O . f o r t _< r 1 t - r n for r-, < t < r, r 2 " ^ l 1 - ~ 2 ^1 f o r t > r 2 , and f(x) = cp(d(x,x Q 83. Then, I t i s obvious that f s a t i s f i e s (ll°l). Moreover, f o r any x,x' e X , |f(x) - f(x«)| = |ep(d(x,x0)) - cp(d(x' ,x) )|_< rtFrT l d ( x ^ x o ) ' d ( x ' * x o ) l < r t f e " d ( x ^ 8 ) » Moreover, f i s bounded by 1 . Hence f € L(X) . Q.E. D. (11.7) Proposition: For every neighborhood U of a point x Q of (X,d) , there i s a zero-set which i s a neighborhood of x Q contained i n U . Proof: Since U i s a neighborhood of x , by r e g u l a r i t y , there i s an ^ > 0 such that C J ^ ^ r ( X Q ) c u • Choose r 2 > r 1 , by Lemma (11.6), there i s an f e L(X) s a t i s f y i n g ( l l - l ) . Hence Z(f) = ^ ^ r ( x 0 ) c u • Q.E.D. (11.8) Proposition: Every closed subset A of (X,d) i s an L-zero-set. Proof: Let f(x ) = d(A,x) f o r each x e X . Then,, that Z(f) = A i s clear. We;will show that f s a t i s f i e s a Lipsc h i t z condition with constant 1 . That i s , |f(x) - f (y) | _< d(x,y) f o r a l l x,y s X. . For any x,y e X , d(A,x) = i n f [d(a,x) : a e A] < i n f (d(a,y) + d(y,x) s a £ A} = i n f {d(a,y) a e A} + d(y,x) = d(A,y) + d(x,y) . Thus d(A,x) - d(A,y) _< d(x,y) . S i m i l a r l y , d(A,y) - d(A,x) < d(x,y) . Hence |f(x) - f ( y ) | _< d(x,y) f o r a l l x,y e X . Let g = f A l . By Prop. ( 1 0 . 5 ) * g e L(X) . Moreover, Z(g) = Z(f) = A . Q.E.D. 84. The f i x e d i d e a l s and f r e e i d e a l s are defined the same as i n (3.1). Moreover, the three Remarks of (*3. l ) are v e r i f i e d w i t h L (X), (L*(X) or L(X)) i n place of (^(X). We w i l l a l s o use 1 ( f ) t o denote the r e s i d u e c l a s s of f mod I . (12.1) Theorem? ( l ) The f i x e d maximal i d e a l s i n ^ ( X ) ^ (L*(X) or L(X)) are p r e c i s e l y the s e t s Mp = {f € L C ( X ) ^ (L*(X) or L(X)) s f ( p ) = 0} . (2) The I d e a l s M are d i s t i n c t f o r d i s t i n c t p P (3) For each p , L c(X)/Mp ? (L*(X)/Mp or L(X)/Mp) i s isomorphic w i t h the r e a l f i e l d fR . In f a c t , the mapping Mp(f) - f ( p ) I s the unique isomorphism of L c(X)/Mp^ (L*(X)/Mp or L(X)/Mp) onto 1R, where p e X . Proof i s . s i m i l a r t o (3.3). (12.2) P r o p o s i t i o n s I f (X,d) i s compact, then every i d e a l I i n M X ) = L(X) i s f i x e d . . Proof i s s i m i l a r t o (3.5). (12.3) P r o p o s i t i o n s I f X i s compact, then the correspondence p Mp I s one-one from X onto the set of a l l maximal i d e a l s i n L C ( X ) ( L ( X ) ) . Proof i s s i m i l a r t o (3.6). (12.4) Lemmas A zero-set Z s Z(X) , Z + $ i s compact i f f i t belongs t o fno f r e e z - f i l t e r . (See (3.7)). Proof i s s i m i l a r t o (3.9). .85. (12.5) Proposition: Let Ar be a 'z-ultraf l i t e r on (X,d) and eafih of I t s members be noncompact. Then Ar Is free. Proof: Suppose p e n . Then, consider f^C^) = d(p,x). We know that Z ( f p ) = {p} which i s compact and meets each member of Ar- at p . Hence Z(f ) & A r (by (10.19)). ' This i s a contradiction. Remark: As we have shown i n the Remark of (3.10), A must be a z - u l t r a f i l t e r . We consider X = E 1 and f Q ( x ) = s i n x . By the Mean Value Theorem, s i n x - s i n x« = (cos x 1)(x-x') where x^ i s between x and x«. Hence we have |f(x) - f ( x ' ) l = |cos x - Jo lx-x') _< d(x,x« ) . That i s , f 'e L C(X), (L£(X) or L(X)). Let I = ( f Q ) .:. Then Z [ l ] i s a z - f i l t e r but not a z - u l t r a f i l t e r . We know that each member of Z [ l ] i s unbounded so that i t i s noncompact. But flZ[l] = Z(f ) . (12.6) Theorem: For a metric space (X,d), the following are equivalent. (1) X i s compact. (2) Every i d e a l i n L C(X) i s f i x e d , i . e . every z - f i l t e r i s f i x e d . (2*) Every Ideal i n L*(X)•(©r L(X)) i s fixed. (3) Every maximal Ideal i n L C(X) i s f i x e d , i . e . every z - u l t r a f i l t e r i s fixed. (3*) Every maximal i d e a l In L*(XV (or L(X)) i s fix e d . Proof i s s i m i l a r to (3.11). 8 6 . §13 The Banaeh Space L(X) and Lc-realcompactness. As we have shown In §4, every residue f i e l d of L (X), L*(X) or L(X) modulo a maximal i d e a l contains.a canonical copy of the r e a l f i e l d fR . We s h a l l show l a t e r that for every max-imal i d e a l M of L*(X\(or L(X)J, L*(X)/M, (or L(X)/M) i s i s o -morphic with fk . (13.1) D e f i n i t i o n : For f e L(X) , Hfll^ - sup { | f (x) | : x e X] , l l f L = sup { l f ( x ? " f (*'•) 1 . X j ( X » € X , x + x'} , and a d ( x , x » ) lift! = Mm + Md • (13.2) Proposition: L(X) i s a Banaeh algebra under the above-defined norm. Proof: (,i) That ||f|j = G i f f f = 0 i s clear. ( i i ) For f,g e.L.(X), ||f+gll = ||f+gl!„ + !!f+glld < It fit. + llsll. + l!fll d + k ! l d = llfli + ltd • ( i i i ) HxflJ = ,|x|o||f|l for each X e fR Is clear, ( i v ) llfogjj = ||fogl|w + l|fogHd < llfl^Hgii. + !Ifgl)d . But, | |f . gL = sup { l f ( x > g ( x K ( x C ? ° g ( x 8 ) l s x + x« , x,x« 6 X} , and a d(x,x») |f(x)g(x) - f(x « ) g(x ' ) l < | f ( x ' ) h | g ( x ) : - g(x>)| + !g(x*) | • | f(x) - f(x>)| We have ||f.g||d <t|f|IJlgl1d + llg!ijlft!d / hence iif-gi! < i i f i i - i i g t + Hfiijgnd,+ ii gn.it * ii d < i i fM is i i . . (v) Let [ f n ] : be an a r b i t r a r y Cauehy sequence under the above^defined norm. Then, i t i s clear that {fn} i s a Cauchy sequence under the norm 11°II ->. We know that G*(X) i s a Banaeh algebra, hence there i s f € C*(X) such that ( f n 3 converges to f under j| »| . We w i l l show that f e L(X). We know that 8 7 . every Cauchy sequence i s uniformly bounded. Indeed, f o r each e > 0 , there exists a p o s i t i v e Integer N ^ such that IIfn - f j <.€ f o r a l l n,m> N | g j . Thus | | f ^ - f j < e Let = sup Cljfjj - f j l l i J < *• and K = max ( e . , ^ ) . (:€ ) Then, l l f N , ; ( £ ) - f j < K f o r a l l m . Hence \ \ f j - \\f^ |1 < K (by property ( I I ) ) , or 11 f J | < | | f w II + K f o r a l l m . i (<s) d(x,x') d(x,x') | f n(x») - f ( x ' ) l ' • — — • i s true f o r a l l x,x ? € X , and x f x' d(x,x«) and a l l n . Suppose f f L(X). That Is, f o r any p o s i t i v e number K , there ex i s t x,x' e X , x ^ x' such that lf(x) - tlx1 > K . In p a r t i c u l a r , f o r K = K + 1 , d ( x , x « ) ° | f ( x Q ) - f(x»)| • ' .-o , > KL + 1 f o r some x »x« € X , xn 4= x A • d(x 0,x») ° 0 ° ° ° However, on the other hand, f o r these two points XQ»XQ > d(x Q , x (J )) > 0 , there i s a p o s i t i v e integer 1 such that <*(xrt,x') l f C X J - f (x£) | | f - f J L < - — - • f o r a l l n > N . Thus ° — — — „ \*(*p)-- f n ( x o ) l , l f n ( x o ) - f n ( x o ) ' , l fn ( x o > " f ' d(xQ,x(!)) d ( * 0 , x i ) d ( * o * x b > jC 1 + K Q . This i s a contradiction. Hence f € L(X). F i n a l l y we have to show that flf ~ f|| -* 0 as n -» » . 88. Since ||f n - fj| «. ||f n - f|| w + | | f n - f|| d and | | f n - f f l . - 0 as n -» • we only have to show f | f n f|| d -* .0 .as n -* » Since [ f n 3 i s a Cauchy sequence under the norm tjofj = || ©H^ + ||e||d , giveh any € > 0 , there i s a p o s i t i v e integer N such that |'|.fm ~ f H < € whenever m,n H . This implies that II f f n l l d < € whenever m,n > I . That i s |———-—:~-~ •• •• | < e (13°1L f o r m,n > N and d(x,x«) x 4 x ? € ;X . Rewriting (13-1), we have | ° n n =| < s , f o r rn^n > J arid a l l d(x,x 8) d(x,x« ) x 4= x ! e X . Now, we hold m f i x e d , f o r any f i x e d p a i r of . f m W ~ f J x < ) x =f= x« € JL s we have .- i s fix e d . Let i t be d(x,x») f'(.x) ..- f n(x«) r , and r =.. ' .- • i <'•<-; •, „ we know, by the uniform d(x,x») convergence of {f^} to f under the sup norm, {r n} -£&2^r.Z.£Lxll as n °* » „ Moreover |r « r | < e f o r n > N . d(x,x») n Thus, we have |r - •• f X x ? ; - )| < e . Or { f m ^ ) ° f m ( x 8 ) _ d(x,x 5 ) '"" d(x,x«) f ( x ) " f ( x ' ) | < c f o r m > N . Since x,x« e X are a r b i t r a r y , d(x,x») ~ . ' - f j x ' ) we have | 111 m - f ( x ) ° x v A / [ < £ 5 f o r m > H d(x,x») ' d(x,x«) ~ ' ' and a l l x =f x 1 e X . That i s , ||f - f| ) d < e , f o r m > J . The proof i s then complete 89. We are now back to show that every maximal Ideal i n L ( X ) or L * ( X ) i s r e a l . We s h a l l discuss In the case of L * ( X ) . The proof f o r L ( X ) i s exactly the same except f o r the notation. In order to show that every L * { X ) / M i s isomorphic to the r e a l f i e l d , /R "by the theory of t o t a l l y ordered f i e l d s (see P. 209 [28])9 we f i r s t c l a s s i f y the elements of L*(X)/M as p o s i t i v e , • negative or zero i n such a way that f + g and f o g are p o s i t i v e and - f i s negative when f and g. are p o s i t i v e . (See [27]). (13.35) Definitions f i s said to be po s i t i v e I f f s j f | (mod M) and f 4 0 (mod M). f i s said to he negative i f - f i s p o s i t i v e . And f i s said to he zero i f f e M . Let (fa and 7PC denote the classes of po s i t i v e and negative elements of L*(X) (or L(X) , resp e c t i v e l y ) . In order to j u s t i f y t h i s c l a s s i f i c a t i o n , we s h a l l show the following lemmas. (13°4) Lemmas For each f € L * ( X ) • ( o r L(X)), one and only one of the three r e l a t i o n s f £ <P , f s M or f € TC ''holds. Proofs We know that (- f + | f | ) (f + | f | ) = • © s M . Since + f + |f| '€ L * ( X ) and L*(X)/M i s a f i e l d , a t .least one of the r e l a t i o n - f + |f| = 0 and f + | f j ' H 0 (mod M) i s v a l i d . If both hold, then 2f = ' ( f ' + | f | ) - ( - f + | f | ) H 0 (mod M). Thus f s 0 (mod M). Q.E.D. (13.5) Lemmas For any two elements f , g € L * ( X ) (or L ( X ) ) S i f 0 <_ f £ g and g e M, then f € M. Proof: If f 4 M , then f £ 0 (mod M) so that there exists an 90, element h e L*(X) such that, f°'h s i (mod M). By (13.4), and the f a c t that f»h = 1 , at leas t one of the re l a t i o n s ~h & |h| (mod M), and h s jhj (mod M) holds. If the f i r s t should hold we would have 1 + f|h| s 1 - f »h s .0 (mod M). However, hy hypothesis, f _> 0 , 1 + f |h| . >_ 1 so that l / l + f |h| e Lg(X) (by (10.7)). Thus u = J^LJLJL lM i € M . This Is a contra-1 + f|h| di c t i o n . I f the second should hold, we would have 1 - f | h | s i - f . h — 0 (mod M) or g|h| + (1 - f|h|) B 0 (mod M). But f <_ g , so. we have g|h| + (1 - f |h|) = i + ( g .- f ) | h | _>.l . Again, we have a contradiction. Hence f s 0 (mod M). Q.E.D. (13.6) Lemma: For any f, g <s L * ( X ) (or L(X)), i f f e P" , and f s g (mod M), then g e -Proof: Since f € p" , f a j f j and f ^ 0 (mod M). By hypothesis g's f* 0 (mod M). Thus, we have only either - g = jg| (mod M) or g s |g| (mod M). Suppose - g s |g| (mod M) holds, we would combine I t with f s | fj (mod M) to obtain |f| + |g| s 0 (mod-M). However, '0-<_ |g| <, |f| + |g| , by (13.5) |.g| H 0 (mod M) which i s a contradiction. Consequently g s |g| (mod M). That i s , g <s 7^ , as we showed above g + 0 (mod -M). Q.E.D. (13-7) Lemma: For any , f , g e L*(X) (or L(X>), I f f'e P* and g e ^ , then f + g, f og £ P" . Proof: By D e f i n i t i o n (13.3) f s | f | , f + 0 (mod M) and g s |g'U g k 0 (mod M). We then have that f g s | f | | g | = |fog| and f o g =}= 0 (mod M) as L * ( X )/M Is a f i e l d . That i s , 91. fog £ f J . Suppose .-(f •'+ g) s jf + g| 0 Then |f | + jg| + |f + g| s f + g - (f + g) = 0 (mod M). We have 0 < | f | < |f| + |g| + |f + g| , 0 < |gj < |f| + |gj + |f + g| which imply |f | -m 0 and |g| = 0 (mod M). . Thus, f = 0, g = 0 (mod M) . This i s a contradiction. Hence f + g B |f + gj (mod M) . I f f + g s 0 (mod M), then |.f |.+ |g| = f + g (mod M)o Again, from 0 < | f j < | f | + -|g| and 0._< |g|£ f f j + |g|. We have j f j = 0 and |g| E 0 (mod M) which i s a i contradiction. Therefore f + g = |f + gj and |f + g| f 0 (mod M). That i s , f + g e p" Combining these four Lemmas and the d e f i n i t i o n of a t o t a l l y ordered f i e l d (see P. 209, [ 2 8 ] ) , we haves (13.8) Propositions L*(X)/M (.or L(X)/M)'' I ® a t o t a l l y ordered f i e l d . (13.9) Lemmas Let M be a maximal i d e a l i n L*(X), (or L(X)) and L^^X^ (or L(X))' be. normed by the sup norm lUH^ . Then M i s "closed i n L*(X) (br L(X) respectively) under || oj|w Proofs In view of ( 2 M . 1 ) . [7'3, c£M Is either a proper i d e a l of L*(X> or L*.(X)"' i t s e l f . Suppose . q*M = L*(x)'. .Then u € GiM and for any neighborhood of u:, H e(u), H (u) •(! M + <j) . In p a r t i c u l a r , take € = 1/2 . Then ^ ^ ( u ) 0- M + $ • T n a' f c i s , there i s an f € M such that ||u - f||oo< 1/2 . Thus |u(x) ••- f ( x ) | < 1/2 , s© that |u(x) - f (x) | < 1/2 f o r each x e X or u(x) - |f(x)|'< 1/2 f o r each x € X . Hence 1/2 < |f(x) | f o r each x € X . By Lemma (10.7) l / f £ L*(X) . In other, words, M ha®, a unit so that M = L*(X) . This i s 92. Impossible. Hence atM Is a proper Ideal containing M . We must have M = c-tM . Q.E„D„ (13.10) Proposition; For each maximal i d e a l M i n L*(X)^ (or L(X)),. L*(X)/M i s an archimede&n ordered f i e l d . That i s , L*(l)/M,(or L(X)/M) = ^ . Proof: I t i s enough to show that f o r any f € such that f Is not a constant function, there i s a p o s i t i v e integer n such that f ~ 1/n € P" ';' Suppose that there does not exi s t such a po s i t i v e integer. Then we would have f => 1/n € 'TL f o r a l l n (as f i s not a constant function). That l s ^ (f - 1/n) + |f - 1/n I c M f o r a l l n . Consider now the sequence {g n = (f - 1/n) + |f - 1/n I ° n e N] which has f + j f | as the l i m i t under the norm IMI^ . Indeed, f + |f| - [(f - 1/n) + |f - l/n|3 = + 1/n - |f - 1/n I 0 • 0 1 1 t n e other hand, |f 1/n I >_ | f | - 1/n , hence we have 0 < f + | f | - [ (f ° 1/n) + |f, - 1/n I 3 <, 2/n . By (13-9) Lemma, f + |f | € clM = M . This show that =f s | f | (mod M). This i s a contradiction. We have proved that every maximal i d e a l i n L*(X) )) i s r e a l . In view of Theorem (12.6), we have that every r e a l maximal i d e a l i n L(X) i s f i x e d i f and only i f (X,d) i s compact. (13.11) D e f i n i t i o n : A metric space (X,d) i s said to he Lc-realcompact i f every r e a l maximal Ideal i s f i x e d . Now, we w i l l give an example to show that there i s an Lc-realcompact space which i s not compact. However, the 93. existence of non~Lc~realeompact 'spaces remains as an open question. (13.12) Lemma. An i d e a l i n Lin,(X) i s free i f f f o r every compact subset A of X there exists an f € I having n© zero i n A . Proof i s si m i l a r to (4.3). (13.13) Propositions Let X be any closed subspace of E 1 1 . Then X i s Le=>reale©mpaet, Proofs Suppose that L ~ /R for'a free maximal i d e a l M . Let g(x) =. -. ^ 4 . Then g <s.L«(X) and i s a unit/ i s clear. Hence g |.M and M(g) 4 0 » For any r e / ^ , r > 0 , g < r fo r a l l but a compact subset of ? , say A . Then B = A n X i s compact In X as X Is closed. Let A" = e£ x(X - B) which i s closed i n X so i s closed i n E 1 1 . Thus there i s an f €. L ( E n ) c L j E * 1 ) such that Z(f) = A 8 . We w i l l show that Z(f) e.Z(M) . I t Is. enough to show that Z(f) z> Z fo r some Z €-Z(M) . However, we know that B Is compact In X . By Lemma (13.12), there i s f 1 e M such that Z(f^) fl B = |) . That i s , Z{f±) e X ° B c c£ x(X - B) = Z(f) . ' Hence Z(f). £ Z(M) . g <, £ ®n the s» ere-set Z(f) arid r - g .0 . Hence M(r - g) > 0 . Or M(r) - M(g) > 0 „• That i s , r 2 M(g) . Since r i s any p o s i t i v e number, M(g) i s I n f i n i t e l y small. This Is a contradiction. Therefore M must be fi x e d . Q 0E„B, •9*. §14 Lc, L°Mappings and..Lc^ I^Homeomor^MsmSo (14.1) D e f i n i t i o n : A mapping T from ( X , d 1 ) to (Y,d 2) i s said to be an L -[resp. L-] mapping i f , f o r each compact subset A of (X,d^),' there i s a p o s i t i v e number such that d 2 ( T ( x ) , T ( X » ) ) <,. K A d(x,x 5) f o r a l l x,x« e A , f i f there i s -a p o s i t i v e number K such that d 2 ( i r(x), T ( X » ) ) < K d(x,x') f o r a l l x,x» € X ). (14.2) D e f i n i t i o n : A mapping -r from (X,d 1) to (Y,d 2) Is said to be an Lc ( L j - homeomorphism, i f T i s one-one, onto, and both T and i t s inverse are L c-[resp. I>3 mapping. (14.3) Lemma: Let (X,d 1) and (Y,d 2) be two compact metric spaces, and T be a mapping from X into Y such that f « T € L(X) f o r each f e L(Y) . Then T i s an L-mapping. Proof: Let cp be a mapping from L(Y) into L(X) defined by cp(f) = f CT . Then cp i s c l e a r l y a homomorphism. By Theorem (2.5,17) [21], cp Is continuous under the norm ||»|| = || eH^ +|<>|!d defined i n (13.1), so that cp Is bounded ( f l 3 l Theorem 7A). How, f o r q e Y l e t f q(Y) » d(;q,y) f o r each y e Y and D = {fq 's q 6 Y} . Then, ||fqJ1 --- H f ^ + Hf qll d = Diam (Y) + 1. .. Jd 2(q^y) - d2(q,y')) For | | f J L = sup ( s y , y ' e Y , y + y ' } q a2 d 2(y,y«) d 0(y*y') <^ gup {r-"~;;nn > ^ ,• n =1} . Hence the set D Is bounded by d 2(y,y') DIam(Y)+ 1 9 so that cp(D) i s a bounded subset i n L(X) , say by K . In other words ||f . (T(X))- f . (T(X»)) | | < K f o r a l l q e Y , 95. so that | |f . T ( X ) - f • T ( x » ) l i d < K'• Hence >L H 2 | d 2 ( q , T ( x ) ) - d 2 (a,TM)| • — . • < K f o r a l l x,x» s I, x =^ x . In d 1(x,x») d 2 ( T(x), T(k»)) p a r t i c u l a r , i f we take q = T ( X ) , then < K d 1(x,x !) . fo r a l l x^x* €.X-, x 4s X S . I f x"= x' , then we have d 2 ( T ( x ) j T ( x ' " ) ) = d 1(x ix»') = 0 . Thus d 2 ( ? ( X ) , T ( X « ) ) < K d 1(x,x 8 ) f o r a l l x>.x» e. X. Q. E.D. (14.4) Proposition: Let (X,&-} 3(Y,d2) he metric spaces and T be a mapping from (X,d^) to (Y,.dg) . (1) I f T i s an L -mapping, then f « T s L_(X) f o r a l l t € LC-(Y) . (2) I f f o r e L„(X) f o r a l l f € L_(Y) , then T i s an L -mapping of (X,d x) in t o (Yjdg)-. Proofs ( l ) Take any compact subset A of X . Then we have two constants and Kg such that d 2 ( T ( x ) , T ( x ' ) ) , . <_ K ^ d ^ X j X ' ) and |f(y) - f (y«) | _< 'Kgdg(y,y») f o r x,x« e A and y,y> e T [ A ] , where T [ A ] is' a compact subset of Y as T i s continuous and A i s compact.. -Hence, we have |f©<r(x). - f » T ( x ' ) l S Kgdg(T(3f:),T(x ! , J) < KgKjd-j^xyx' ) for any x,x«'e A . • Therefore, f . T e L_(X) . (2) Consider any.compact subset A 4 $ of X . We w i l l show that T I S an L-mapping on A . We know from Theorem (3.8) [7] that T Is continuous. Hence T [ A ] i s compact. Let cp be a mapping from L_(Y) to L_(X) defined by cp(f) -• f o r - f o r a l l f e L_(Y). Then, I t Is obvious that cp i s a 96. homomorphism of L (Y) Into L (X) . We r e s t r i c t ep to L . ( Y ) | T [ A ] = {f I T [ A ] : f € L (Y)3 , then cp Is in t o L (X) | A = £g|A i g e L C(X)3 . Since A i s compact,, L ( X ) |A e. L ( A ) . Also f o r each f € L(A) , hy (11.4), there, is•an f e L(X) c L C ( X ) such that f|A = f Q . Thus L ( A ) c L C ( X ) J A so that L(A) = L C(X)|A . S i m i l a r l y , we have L q ( Y ) | T [ A ] = L ( T [ A ] ) . By Lemma (14,3), T i s an remapping on A . Since A i s a r b i t r a r y , T i s then an Lc-mapping. Q.E.D. We w i l l investigate the r e l a t i o n between a r b i t r a r y Lc-mappings from (X,d 1) into (Y,dg) and the r i n g homomorphisms of L _ ( Y ) to L_(X). We s h a l l see that every homomorphism from a r i n g L C(X) J Into another L C(Y) , In I- same sense, IS induced by an Lc-mapping (see (6.1) and (14.13)). Let T s X -» Y he an Lc-mapplng. Then, the induced mapping T 1 , defined hy T«(g) - g « T € L_,(X) for each g € L „ ( Y ) i s a homomorphism from L ( Y ) i n t o L„(X) . I t ca r r i e s the constant functions onto constant functions i d e n t i c a l l y . I. I,. 1 • ; Moreover, i t also determines the mapping T uniquely. The 'proofs are similar to (6.7) and (6.8) respectively^ (14.5) Theorems Let T be an LC~mapplng from (Xjd.^) to (Y,d 2) and T ' be the Induced homomorphism g -* g«>T from L C ( Y ) into L E ( X ) . (1) T' i s an isomorphism (into) i f f T [ X ] i s dense i n Y . ( 2 ) T» i s onto i f f T i s an Lc-homeomorphism of X onto an LcTembedded subset of Y «, i . e . f o r each f €.• L ( T [ X ] . ) , there i s an f 0 e L C ( Y ) with f 0 | T [ X ] = f . ; ' Proof i s similar 1 to (6.10). 97. We how examine the inverse problem of determining when a given homomorphism of L (Y) into L.(X) is induced by some Lc-mapping from X into Y . We wi l l f irst consider the homo-morphisms from L (Y) into fR , that i s , the case In which X is a single point. (14.6) Proposition: Every nonzero homomorphism cp from LC(Y) into fR is onto fR . " In fact, cp(r) = r for a l l r e fR . Proof is similar to (6.13). (14.7) Propositions The correspondence between the homomorphisms of L-_(Y) onto fR, , and the real maximal Ideals is one-one. Proof is similar to (6.14). (14.8) Propositions Y Is Lc-realcompact i f f to each homo-morphism ' cp from L_(Y) onto fR - i .e . each nonzero homomorphism into fR - there corresponds a.unique point y of Y such that cp(g) - g(y) for a l l g e LC(Y) . Proof Is similar to (6.15). (14.9) Theorems Let cp be a homomorphism from L (Y) into L (X) such that ep(u) = u . If Y is Lc-realcompact, then there exists a unique Lc-mapping T of X into Y such that T ' = cp . Proof is similar to (6.l6). (14.10) Corollary: An Lc-realcompact metric space (Y,d 2) con-tains an image of ah Lc-mapping of (X d^.^ ) i f f LC(X) contains a homomorphic image of LQ(Y) that includes the constant functions on X . 98. Proof Is si m i l a r to (6.17). (14.11) Corollary? Ah Lc-realeompact space, (Y,d 2) contains a dense image of an Lc=mapping of (X,d 1) i f f L C('X) contains an isomorphic image of L (Y) that includes the constant functions on X . Proof follows Immediately from (l4 . 5 ) and (14 .9). (14.12) The Main Theorem: Two LC-realeompact metric fepaeeS (X , ^ ) and (Y,d 2) are Lc-homeomorphic i f f L C(Y) and L C(X) are isomorphic. Proof i s si m i l a r to (6.19). Note that, i n particular, that i f we l e t (X,d 1) and (Y,d 1) he compact metric spaces, then they are L c = L-homeomorphic i f f L (X) = L(X) and L (Y) = L(Y) are isomorphic. D.R. Sherbert has a s i m i l a r r e s u l t . (See Theorem 5.1 [26].) In spite of the Theorem (14.9), every, homomorphism i s induced, i n essence, by an Le-mapplng. (14.13) Theorem: For any metric space (X,d 1) and an Le° realcompact metric space (Y,d 2) , l e t cp be a homomorphism from L„(Y) into L„(X) . Then-the set E = {x e X : cp(u)(x) = 1) i s open-and-closed i n X . Moreover, there ex i s t s a Unique Lc^mapping T from E into Y , such that for a l l g e L.(Y) , cp(g)(x) = g°(T(x)) f o r a l l x e E , and <¥>(g)(x) = 0 f o r a l l x. e X - E . Proof i s s i m i l a r to (6.20). 99. (14.14) Theorem; Let (Y ,d 2 ) he a compact space and ep he a homomorphism from L*(Y) = L(Y) into L*(X) . Then the set E = {p e X s tp(u)(p) = 1} is open-and^clpsed in X . Moreover, there exists a unique Lc-mapping from E into Y , such that for any g e L(Y) cp(g)(x) = g(t(x)) for a l l x e E and cp(g)(x) = 0 for a l l x e X - E . Proof is similar to (6.20). (14.15) Corollarys Let cp be a homomorphism from L (Y) into a ring of Lc-functions. If Y is Lc-realcompact, then there exists a unique closed set P in Y such that the kernel of cp is the z-ideal of a l l Lc-functions that vanish on P . Proof is similar to (6.22). (14.16) Propositions An Lc-realcompact space (Y,d 2) contains an Lc-embedded image of an Lc-mapping of (Xid-^ ) i f f LC(X) i s a homomorphic image of L (Y) . PrOof is similar to (6.23). (14.17) Propositions A compact space (Y,dg) contains an L*- embedded image of an Lc-mapping of ( X ^ ) i f f L*(X) Is a homomorphic image of L*(Y) = L(Y) . Proof is similar to (6.23). §15 Embedding Theorems The definitions (7.1) to (7.6) are applicable In this section. IGO. (15.1) Theorem: Let (Y,dg) he an Lc-realcompact metric space, (X^c^) be any metric space,1 and cp a homomorphism from LC(Y) into LC(X) . Then, ep has the property (7^1) i f f there is a homeomorphism T from X into Y such that ep(f) = f ° T for a l l f e LC(Y) and T is an Lc-mapping. In addition, i f cp(L (Y)) = L (X) , then T is an Lc-homeomorphism. Proof: The proof of the f irs t part is the same as Theorem (7.7). Now, let ep° he a mapping from cp(L. (Y)) = L (X) into L.(Y), defined by cp°(f) = f o r " 1 . Since each f e cp(L_(Y)) has the form f => f Q o T for some f Q e LC(Y) , we have ep°(f) .=' ( f Q o T ) o T r = fQ . We then show that T is an Lc-mapping> by Prop. (14.4) Q.E.D. (15 .2) Theorem: Let (Y,d 2) be a metric space, (X,d 1) be ah Le-realcompact metric space. Then the following statements concerning a homomorphism cp from LC(Y) into LQ(X) are equivalent. (1) ep has the property (7-1).' (2) cp is a 6-homomorphism. (3) The image of L (Y) separates points and closed sets and i t is contained in no 6-real ideal of L_(X). Proof: The proof is similar to Theorem (7.8). (15-. 3) Theorem: Let (Xjd.^) and (Y,dg) be two Lc-real comp act metric spaces. Then a homomorphism, cp' , from L (Y) into L (X) is a 6-homomorphism i f f there is a homeomorphism T from X into Y such that cp(f) - f » T for a l l f c L_(.X) , and T is an Lc-mapping. In addition, i f cp(Lc(Y)) is LC(X) i tself . 101. then T i s an Lc-home ©morphism. Proof: The f i r s t part i s sim i l a r to Theorem (7.9). The second part i s the same as that i n Theorem (15.1). (15.4) Lemma: Let ( X ^ d ^ and (Y,dg) he any metric spaces, and T be an Lc-mapping from X into Y . Define a homo-morphism op from L C(Y) into L C(X) by cp(f) = foT . Then for any ide a l s M^ . and My of L C(X) and L C(Y) , respectively, cp[My] c i f f T ( x ) = y . Proof i s si m i l a r to Lemma (7.10). (15.5) Theorem: Let (X,d-j^) and (Y,dg) be Lc-realcompact metric spaces. Then a homomorphism, cp , from L c(Y) into L C ( X ) i s a 6F-homomorphism i f f there exists a homeomorphism^ from X i n t o Y such that cp(f) s f«T f o r a l l f e L C ( Y ) , T[X] i s a closed subset of Y^ and T i s an Lc-mapping. In addition, i f cp[L_(Y)] = L ( X ) , then T i s an Lc- homeomorphism. Proof i s si m i l a r to (7.11) and the l a s t part of Theorem (15.l). (15.6) Theorem: Let ( X , d 1 ) and (Y,d g) be two Lc-realcompact metric spaces. Then a homomorphism, ep , from L_(Y) into L C ( X ) i s a 6G-homomorphism i f f there exists a homeomorphism^ T; from X i n t o Y such that cp(f) = f O T f o r a l l f € L (Y) , T[X] i s an open subset of Y , and T IS an Lc-mapping. In addition, i f ep i s onto, then T. i s a homeomorphism. Proof i s si m i l a r to (7.12) and the l a s t part of Theorem (15.l). 102. 16 The Rings of Lc-Funetions Defined on the -Metric Spaces other than Lc-Realcompact Spaces, and Rings of L i p s c h i t z i a n Functions on Compact Subsets of E n. (16.1) Lemma: Suppose p e X . Then there exists f.e L Q(X) such that Z(f) = (p) and such that f belongs to no maximal i d e a l other than Mp. Proof: Consider f(x) = d(p,x) f o r a l l x.e X . We know that f belongs to Mp and not any other f i x e d maximal i d e a l . Suppose f belongs to a free maximal i d e a l M . Then by the property of free i d e a l s , we have g e M such such that g(p)'-4 °-p p Let h = f + g . We have Z(h) = ty . Thus i t s inverse 1/h exists. Take any compact subset A of X . For t h i s A , there e x i s t two p o s i t i v e numbers K A and in such that |h(x) - h(x')| < K Aod(x,x») and |h(x)| > m f o r a l l x,x« e A . Hence |l/h<*) '* l/h<*'> | .-/M*f- h(x» ) j < 1 / m 2 | h ( x ) _ h ( x , } , |h(X)oh(X')l X -4y d(x,x») . Therefore 1/h e L_(X) . Since M i s an m i d e a l , we would have u = h»l/h € M . That i s , M = L (X) . This i s impossible. Q„E.D 0 (16.2) Lemma: I f cp i s an isomorphism from L„(Y) onto L.(X), then f o r M q c L C(Y) , and cp(M^) = Mp fo r some p e X , that i s , cp(Mq) i s a f i x e d maximal i d e a l i n L C(X). Proof: Since cp i s an Isomorphism onto, cp(M^) i s a maximal i d e a l . By Lemma ( l6.l) f Q(,y) = d(q,y) f e M only. Con-slder Z(cp(f Q)) . I f Z(ep(f )) = ty , then as shown i n the l a s t lemma cp(f ') i s a unit so that ' cp(M ) i s the whole r i n g L (X) q q c which i s impossible. Hence Z(cp(f )) 4 0 • On the other hand, 103. i f Z(op(fq)) contains more than one point, say p-^ and pg ^ then tp(fq) e and Mp so that f q would belong to at ; l e a s t two maximal ideals which again i s impossible f o r f belongs to only one maximal i d e a l . Hence Z(ep(f_)) = {p}, say. Therefore cp(Mq) = Mp . Q.E. D. (16.3) Theorem: Let (X3d-^) and (Y,dg) be any two metric spaces, and cp be an isomorphism from L (Y) onto L (X) leaving a l l constant functions unchanged. Then cp induces a mapping T : X -* Y defined by cp(g) = g«T and T Is an Lc-homeomorphism. Conversely, i f T i s an Lc-homeomorphism of X onto Y , then the induced mapping cp , defined by cp(g) = go*r , i s an isomorphism from L C(Y) onto L C(X) . Proof i s s i m i l a r to (9.4) and the converse i s obvious. (16.4) Corollary: Let (X,d 1) and (Y^d 2) be two connected -metric spaces and cp be an Isomorphism of L (Y) onto L (X) . Then cp induces a Lc-homeomorphism, T , from X on Y such that «p(g) g°T f o r each g e L C ( Y ) . Proof i s s i m i l a r to (9.5). Now we discuss the s i m i l a r properties as we have done i n § 9 f o r (^(X) . Hereafter, we consider X and Y as compact subsets of E n . (16.5) Lemma: The projection functions ^ ( x 1 , . . . , x n ) = x i f o r 1 C i _< n belong to L(X) . Proof: We know that |X(x) -^o(x» ) I = Ix.^ - x£ | _< d(x,x') f o r 1 < i < n , and 4>(X) i s bounded. Hence L(X) f o r 104. 1 < 1 < n • Q . E . D. (16.6) Propositions Let L» be a subfamily ©f L ( X ) 3 and T be a mapping from X i n t o Y . (1) If T i s an L-mapping, then f o r € L(X) f o r a l l f e L(Y) '. (2) i f L' contains a l l projections and f»T e L ( X ) f o r each f € • L* , then T Is an L=mapping from X into Y . Proofs ( l ) i s obvious. (2) For any x,x' € X••, and any f € L« , |f<>T(x) -f o T ( x ' ) l \ < K f o T<"d(x,x») , where K^.. = l | f ° T | l d . In p a r t i c u l a r , i f f =Z f o r 1 C i X h , then |(T(X)') ±- (•r(x,))1\ = IZ(T ( X ) ) - > 1 ( T ( X ' ) ) I < Kj o T od(x,x> ) , where ( T ( X ) ) ± denotes the i - t h coordinate of. T ( X ) , f o r 1< I _< n . Hence .'•'.(• d ( T ( x ) , T(X»)) = (2 1 ( T ( X ) ) • - (T(X«))J 2 )*< ( 2 4 o T ) l / 2 d(X'X?)< i = l x x i = l ^ T Let K = ( E m ) 1 ^ 2 . Then K i s independent of x and 1=1 x' as each K i s . That i s , d{Tix),T(x' )) < K>d(x,x« ) f o r a l l x and x* £ X . Hence T i s an L-mapping. Q . E j D i ! ' A subring B ( X ) of L ( X ) i s said to have the property (16-1) i f B ( X ) i s a subring of L ( X ) 3 B { X ) z> R ( X ) . and f e B ( X ) with Z(f) = § implies f " 1 ( = l / f ) e B ( X ) , where R ( X ) i s defined as i n §9 > We know that such a B ( X ) e x i s t s . For instance, l e t B Q ( X ) = [ f € L ( X ) f e C 3 ( X ) } . Then, i t i s obvious that ft(X) c B Q ( X ) c L ( X ) and f e B Q ( X ) with Z(f) = <J) implies f " 1 6 B N ( X ) . 105. (16.7) Lemma; There i s a function f € = [f e B(X) : f(x ) = 0] such that Z(f) = (x) and f belongs to no other free or f i x e d maximal i d e a l . Proof i s si m i l a r to (9.1). (16.8) Lemm'a; I f cp Is an isomorphism from B(Y) onto B(X), then f o r ML c B(Y) , cp(M ) Is a f i x e d maximal i d e a l i n B(X), where B(X) and B(Y) have property ( l6-l). Proof Is si m i l a r to (9.2). (16.9) Theorems, /Let X and Y he two compact subsets of E n :9 and cp he an isomorphism from B(Y) onto B(X) leaving the constant functions unchanged, where B(X) and B(Y) have the property (l6»l). Then cp induces an L-homeomorphism, T , from X onto Y , defined by cp(g) = g o T . Proof, i s .Similar to (9.^). (16.10) Corollarys Let X and Y be connected compact subsets of E n and cp be an isomorphism from B(Y) onto B(X) , where B(X) and B(Y) have the property (16-.I). Then ep J induced an L-homeomorphism, T , from X onto Y , defined by ep(g) = g » T . Proof I s s i m i l a r t© (9.5). 106. PART I I I THE •RINGS OP ANALYTIC FUNCTIONS § !7 Rings of Analytic Function on any Subset of the Complex Plane A l l the functions considered i n t h i s section are complex single-valued. (£ w i l l denote the complex plane. (17.1) Definitions Let G be an open subset of (C. A function f o r i l G i s said^to be an an a l y t i c function on G. , i f f o r each p € G , there i s a power series i; a (z - p) which n=0 n converges on D = [Z 1 |Z - pj < R) and f ( z ) = 2 a _ ( z - p ) n n=0 n f o r z € B , where R > 0 and D c G , and a n i s a complex number f o r each n = 0,1,2,... (17.2) Definitions Let X be an a r b i t r a r y subset of (D. A function f on X i s said to be an anal y t i c function on X , i f for each p € X , there i s a power series E a_(z - p ) n which n=0 n converges f o r | z - p| < R , and f{z) - E a_(z - p ) n f o r a l l n=0 n z £ X and |z - p| < R s where R > 0 . (17.3) Definitions Let X and Y be two a r b i t r a r y subspaces of (£.. A mapping T from X to Y i s said to be an ana l y t i c mapping i f T i s an a n a l y t i c function on X and valued i n Y . T i s said to be a conf©rmal mapping i f T i s one-one, onto. (See §2 Ch. I I [2].) (17.4) Definitions Let X be an a r b i t r a r y subset of (£,, and Oi(x) = {f : f i s an an a l y t i c function on X] . 107. Let f and g be any two elements from (%(X) , and f f g , and fog be the pointwise sum, difference, and product of f and g , respectively. Then, f o r each p € X we have power series E a ( z - p ) n and § P,-(z - p ) n converging on n=0 n n=0 n |z - p| < R-^ , and |z - p| < R 2 , respectively, and f ( z ) = It a (z - p ) n , g(z) = 2 6 (z - p ) n f o r z e X and |z - p| < R,, n=0 n n=0 n 1 |z - p| < R 2 , respectively, where R x > 0 and R 2 > 0 . Thus E (a + 0 )(z - p ) n = 2 a f z - p ) n + I 6(z - p ) n , n=0 n n n=0 n n=0 n and 2 ( 2 a k p n _ k ) ( z - p ) n =( E ajz - p) n>( E B ( z - p ) n ) n=0 k=0 n=0 n n=0 n converges f o r |z - p| < R and (f + g)(z) = f(z).+ g(z) = " ( an ± P n H z ~ P ) n ^ f°g(z) = f(z)«g(z) = E ( E cuB " )• n=0 n n n=0 k=0 * n " K (z - p ) n f o r z € X and |z - p| < R , where R = minfR^Rg} . Hence f + g and fog are a n a l y t i c at p . Since p i s a r b i t r a r y , f + g and f g e 0[(X) . Suppose f e 0t{X) such that Z(f) = [z € X : f ( z ) = 0} = 0 . For p e X , there exists a power series § a „ ( z " P ) n which converges on |z - p| < R , n=0 n R > 0 and f ( z ) = 2 a (z - p ) n f o r z e X and |z - p| < R. n=0 n I t i s clear that an 4= 0 . Then, that --. 1 — - — ~ also I a n ( z - p ) n n=0 n has a power series which converges at l e a s t on |z - p| < R, (See P. 145 [1]), and - i — = / z « (z --p 1 1))" 1 f o r z e X f ( z ) ln=0 n y |z-p| < R i s clear. Hence 1/f € Moreover, © , and 108. u , the constant function with values 0 , and 1 , respectively, are two entire functions on the complex plane, so that ©, u e C%{X) • The operations of addition and m u l t i p l i c a t i o n thus defined are associative and commutative, and the d i s t r i b u t i v e laws hold. Hence we have the following theorem? (17.5) Theorem; The family 0t(X) , defined i n (17.4), forms a commutative r i n g with unity, (17.6) Lemma; For p e l , there i s an f ? ^ = [f e.#{X) • f(p) = 0} such that Z(f) = {p} 3 and f belongs to no maximal i d e a l other than Mp . ProOf; Let f ( z ) = z - p . Then that f e Mp and f belongs to no other f i x e d maximal Ideal i s clear. Now, suppose that M i s a free maximal i d e a l such that f € M . Since M i s free, there i s g e M eC%|X) such that g(p) + 0 • We know that g i s a n a l y t i c at p so there Is a power series S o t n ( z - P ) n which converges f o r \z - p| < R , and n=0 g(z) = , £ a_(z p ) n f o r z s X and |z - p| < R , where h=0 n R > -0 . I t i s clear that a Q 4s 0 . Let K be the f i r s t p o s i t i v e integer such that a f c + 0 . Then g(z) = aQ4c^. (z - p) k+..., f o r z c X and |z.- p| < R . Let h(z) = g(.z),..-. <*G , •!• . Then f o r each p' € X , p r+ P > there i s a ( z p ) power series S P„(z - p«) which converges f o r |z - p'| < R,, n=0 n x R, > 0 , and g(z) = ? B (z•- p » ) n f o r z e X and |z - p'| < 1 n=0 n 109. Thus' •, — • •• • / / •• • - • • _ . x i r ' v . .-• • •.. £ Q r V N T ( ( P < - P ) + ( Z - P ' ) ) * i(p' I -pJ+(*-p»)l k z e X and / z - p» | < R 1 where P Q = P Q ~ % and 6 ^ = , 1 = 1 , 2 , . . „ , w i l l he a power s e r i e s , say 2 Y _ ( z - p ' ) n which n = 0 n converges f o r [z = p'| < L , and evidently h ( z ) = 2 r / _,\n x n=0 q n ^ z " p ; f o r z € X and | z p « | < . Suppose p' = p . Then the power serie s E "Ct^^ ( z ° p ) n converges f o r | z - p| < R , n=0 and h ( z ) = E o u ( z - p ) n f o r z e X and |z, - p| < R . n=0 K + n Hence h e £%(X) . Now, g (z ) = a Q + (z - p) kh ( z ) , so that a Q = g (z ) - (z - p ) ^ 1 f o h ( z ) = g (z ) - (z - p ) * " 1 f ( z)oh ( z ) . Since f , g <s M which i s an i d e a l , aQ € M . However, i t i s impossible, as aQ 4s® i s a unit. Hence the assertion i s proved. (17 .7) Lemmas I f cp i s an isomorphism from (%(X) onto (%(Y) , then cp(Mp) i s a f i x e d maximal i d e a l . Proofs That ep(Mp) i s a maximal i d e a l i s clear. PrOm Lemma (17.6), there is'an f QVe such that Z ( f Q ) = [p| , and f Q belongs-to n© other maximal i d e a l . Consider Z C P(f 0)). If Z(cp(f Q)) = <j) , then cp(f 0) i s a unit so that cp(Mp) i s the whole ri n g , (%(X) . This i s impossible. Hence Z(q?(f )) 4 (j) . But i f Z ( ( j?(f 0)) contains more than one point, say and q 2 , then <p(f0) e M q and so that f Q would belong to at l e a s t two maximal ideals which i s again Impossible, for f 110. belongs t a only one maximal i d e a l . Hence - Z(<^(f ))' = {q}k, q € Y . Now, s i n c e «p(Mp) i s a maximal i d e a l , cp(f 0) € M q , and <p{f0) belongs t o only one maximal cp(Mp) = M q . Hence ep(Mp) I s a f i x e d maximal i d e a l . Q.E.D. (17.8) Theorem; Let X and. Y be two subsets of C, * and ep be an Isomorphism from (7[|Y) onto Q(JX) such, t h a t i t i s the i d e n t i t y .on the-constant f u n c t i o n s . Then cp Induces a mapping T t X -» Y , defined by ep(g) = g®T , and T i s a conformal mapping of X • ©**t®. Y . Proofs Befine T t@ be a mapping from X t o Y as f o l l o w s ; T ( P ) = nz[cp"1(Mp)3 . Since cp i s an Isomorphism, and onto, i t s Inverse mapping cp"1 i s an isomorphism of Qi{X) onto (%(Y) O By Lemma (17.7), c p " 1 ^ ) i s a f i x e d maximal i d e a l i n (%(Y) . Thus T i s a s i n g l e - v a l u e d mapping* E v i d e n t l y , ^ T ( P ) = c p" 1^ Mp ^ * L E T P A N D P T B E I N X 9 X 1 ( 1 P 4 S P ' • Then, by Lemma (17-7), cp~ 1(M p) = M q , and cp" 1(M p,) = M q l f o r some q,q ! ..£ Y . I f q = q ! ', then cp°*1(Mp) = M q = ep'^Mjj,,) o r = ^p? • This I s Impossible f o r p 4 S P* • Thus, q 4= q v and so f ( p ) = q + f* = T ( P " ) . Hence :T-;.;JS -one-One. Let q^ be a r b i t r a r y i n Y * Then M ^ i s a maximal i d e a l i n (Jl{'f) , and ep(Ma ) = M f o r some "p € X . Thus ^© p o o q Q = nZlcp'=*1(Mp ) ] * T ( P 0 ) . This shows t h a t T I s onto. Now, f o r each g e 0C{Y) , and p <s X; , l e t cp(g)(p) = a » and a be the corresponding constant f u n c t i o n on X . Then, ep(g) -a € M p, g - cp" 1^) = cp°1(cp(g) - a) e M f ^ p j , g ( T ( p ) ) = cp" 1 ( a ) ( T ( p ) ) - S ( T ( P ) ) = a = cp(g)(p) . Hence cp(g) - g o t . S i m i l a r l y , 111. cp~ 1(f) « f . T " 1 ., where, T"1 : Y -» X i s defined hy T" 1(q) = nZ[cp(M )] . I f we Choose g( w) = w on Y , and f ( z ) = z on X , then T ( P ) = g»T(p) and T~ 1(q) = f » T ~ 1 ( q ) are analytic. Hence T i s a conformal mapping. Q. E.D. (17-9) Corollary; Let X and Y be two connected subsets of (£y , and cp be an isomorphism from onto (^ %(X) such that i t i s the i d e n t i t y on the r e a l constant functions. Then cp induces a mapping T : X >• Y . Either T or i t s conjugate mapping "T i s a conformal mapping according as cp(g) = g o r or cp(g) = g°T . Proof: By Theorem (17.8), the mapping T defined by r(p') = nZ[cp"'1(Mp) ] f: i s one-one and onto. Let a be the constant function of value a . We assume that cp(r) = r_ i f r i s r e a l and (cp(i.)) = cp(°l) = -1 , cp(i) = + i. as X i s connected. Hence, cp(a) = a or a . For each g e (/t(X) and p € X , l e t ep(g)(p) -.a . Then cp(g) - a Q e M , g - e p " 1 ^ ) = o • o -"p » 0 ^ vs©' «P° (cp(s) - OQ) € Mv^j , or g (T(p)) = cp" ( a 0 ) ( T ( p ) ) = a Q or a. according as cp(a_) ~ dn or a . . Thus cp(g) = g« T ' or goT according as cp(a) = a or a . S i m i l a r l y cp" 1 (f) = f o r " 1 or cp° 1(f) • f•T u- l according as cp(a) = a or a , where t " 1 i s defined by T™ 1(q) = nZ[cp(M ) ] . In p a r t i c u l a r , we choose g(w) = w on Y , and f ( z ) =» z on X , then r(p):«. g«t(p) and T" 1(q) = f » T " 1 ( q ) ; T(P) = g»T(p| and " ^ ( q ) = f«T**-L(q) are analytic. To show that T and T are one-one i s sim i l a r to what we have shown i n (17 .8). Hence T or T""' i s a con-formal mapping. Q. E.D. 112. Remarks: ( l ) In Theorem (17.8), the c o n d i t i o n t h a t cp i s the i d e n t i t y on the constant f u n c t i o n s can not he omitted. We know that an isomorphism of (%(Y) onto (%(X) always leaves the con-sta n t f u n c t i o n s w i t h r a t i o n a l values unchanged. Without t h i s r e s t r i c t i o n on cp , the r e s u l t of the Theorem w i l l not be t r u e . For example, consider X = [ p ] , Y=[<2}. Then#£x)={a :" ci '.€. (D and (%(Y) = {<x' : a* e (£} . In order t o e s t a b l i s h the r e q u i r e d r e s u l t , l e t us f i r s t prove the f o l l o w i n g Lemma: (17.10) Lemma: There e x i s t s a non-zero automorphism of (C onto i t s e l f which i s d i f f e r e n t from the mappings ep1 Z , the i d e n t i t y , and cp2 : z -• "E , the conjugate mapping. Proof: Let © and Ik be the r a t i o n a l and r e a l f i e l d s , r e s p e c t i v e l y * and so on. Let cp : (&£y2) - (&C/2) be defined by cp(,y2) » -J2 and «p(r) = r i f r e (bX . Then cp i s ah isomorphism of (Q,(»/2) onto i t s e l f . cp can be extended t o an Isomorphism, ep' of (Bl(,/2 , JZ>) onto i t s e l f (say a + \>J2 + c/3 --> a - h/2 + not the i d e n t i t y on the r e a l numbers. And so on. We have a l i n e a r l y ordered set of f i e l d s and a l i n e a r l y ordered set of isomorphisms. By Zorn's Lemma [10] , there e x i s t s a maximal f i e l d , t h a t i s , and a maximal isomorphism from (J^ onto i t s e l f . ^ i s then not the i d e n t i t y on the r e a l f i e l d . Hence i s n e i t h e r cp^ nor cp2 • Q. E. D. Now, def i n e cp : Ot{X) - Ot{Y) i n the obvious way. Then cp(a' ) I a 1 f o r some a 1 e (%(Y) . On the other hand ? cJ3 > or a - b^2 - c/5 » where a, b and c e (£L ) which i s 113. a ' o T s a S where T S I - * I i s defined by T ( p ) = q. Hence ep(a« ) 4 a 1 «T . However, L. Bers showed i n 1948 that i f X and Y are two domains i n (D and', i f &L(X) and £%(Y.) are isomorphic, then.there exists either a conformal or an anti-conformal mapping which maps X onto Y (Theorem 1,[3l). (2) In Corollary (17«9) , neither of the conditions "cp leaves a l l r e a l Constant functions unchanged" and "X i s connected" can be omitted. The Remark ( l ) above Shows that the r e s u l t i s not v a l i d i f X Is connected and cp does riot leave a l l the r e a l constant functions unchanged. Wow, consider X = {p-j^Pgl and Y = { q 1 5 q 2 } where p x + P 2 » and q x + q 2 . Then #(X) - C(oi/B) s a,p e. ' and 01{Y) = {(<x',p« ) : a',B« e £ } . Let cp s Ol{Y) - (%(X.) be defined as follows? cp((a,3)) = (a,"F) . That cp Is an isomorphism and leaves a l l the r e a l con-stant functions unchanged i s clear. Then T(P^ ) = n(Z[$~^(Mp )] = q A . Choose g = (a>p) e (X(Y), where £ i s a complex number. Then cp(g) = (ex,"^ ) 4 g where g»t = (a,P) . Hence «p(g),4-g»T . L. Bers has shown that l e t X and Y be two domains possessing boundary points. Then, every isomorphism of CPt{Y) onto induces a conformal or an -anti-conformal mapping of X onto Y (See Theorem 2 [3]). (17.11) Theorems Let X and Y be two subsets of CI/, and T be a conformal mapping of X onto Y . Then the induced mapping Tj defined by t'(g) = g»T Is an Isomorphism of 0\%Y) onto (X(X) which leaves the constant functions, unchanged. 114. Proof: Let *r'(g) = g»T for each g 6 , ^ ( Y ) ' . For each p e X , T ( P ) e Y . Since g € Oi(Y) and T ( P ) e Y , there exists a power series 2 •T'(p) ) n which converges f o r n=0 |w - T ( P ) | < R-, where R-, > 0 , and g(w) =*2 B_(w - T ( p ) ) n x x n=0 n f o r |w - T(P)| < R^ and w € Y . Since t i s ana l y t i c on X , there exists a power series 2 yn(z' - p ) n which converges ' n=0 n f o r |z - p| < Ro , where R 0 > G , and T ( Z ) = 2 Y„( z - P ) * * - * n=0 n f o r | z - H p| (,R9 and z € X . Therefore 2 B ( £ Y^ C z-p) 1 1 -m=0 m n=0 n T ( p ) ) m i s convergent and equal to g(T(z)) f o r each z e X such that |z - p| < R 2 , T(Z) e Y , and | T ( Z ) - T ( P ) | < R-^ . Moreover, there e x i s t s R > 0 , such that I B ( 2 Y n ( z - p ) n -m=0 n=0 T ( p ) ) m converges on | z - p | < R , and g(T(z)) = 2 B m=0 m ( S Y„(z -• P ) n - T ( p ) ) m f o r |z• - p | < R and z € X . By n=G n the Weierstrass's double-series theorem [ l l ] , 2 B m( 5 Y n m=0 n=0 (z - p ) n - T ( p ) ) m w i l l he a power series i n terms of (z - p), say, E ocn(z - p ) n which converges f o r each z, | z - p | < R , and • g ( T ( z ) ) = 2 a(z - p ) n f o r |z - p | n < R n=0 n and z e X . We know that p i s a r b i t r a r y i n X . Hence g»T e Ot(X) . S i m i l a r l y , f o r each f e Ql{X), f o r " 1 e (X(Y) . Also T ^ f ' T " ^ ( f o T " X ) T = f . Hence T ' i s on,t^- Now, i f T ' (g) = © , then g<>T(z) = G f o r each z € X or g ° T [ X ] = g [ Y ] = {0} as T i s onto. That i s , g = Q . This shows that T 1.15. i s one-one. F i n a l l y , f o r any constant function a ., T'(a)(z) -£L (T(Z)) = a f o r a l l z'e X . That i s , T« (a) * a . Q.E.D. Let Z ^ ( X ) = (Z(f) : f e (X(X)} , where Z(f) = {z e X : f ( z ) = G} as defined before. We know that not every closed set i s a zero-set of some ana l y t i c function. Hence Z ©o(X) i s not a base f o r the closed subset of X^ i n i t s r e l a t i v e topology i f X i s not a discrete subset of (£). How-ever, we have the following r e s u l t . (17.12) Theorem: Let X be a subset of (C. X i s compact i f f every maximal i d e a l i n (X(X) Is fixed. Proof: The necessity Is clear. Sufficiency: Suppose that f o r each maximal i d e a l M i n 0t(X) , M i s f i x e d , and X i s not compact. Then there exists a sequence of points* say {Z^ : n € N} c X which con-verges to a point ' Z X . Case I. I f Z Q = » , then {Z n : n € W] has no f i n i t e l i m i t point. By the Weierstrass factor Theorem [11], there e x i s t s an entire function, f n , which has zeros exactly at {Z^ : k J> n) f o r each n . Clea r l y f € OLW . Case I I . I f Z Q 4 0 0 > then consider the conformal mapping frt('Z) = — ^ — which Is ana l y t i c except at the ° : Z-.Zrt o point Z D . Then f y maps [ Z n : n e,N] Into a sequence {Wn : n € N] In the w-plane which has no f i n i t e l i m i t . By the Weierstrass f a c t o r Theorem [11], there e x i s t s an entire function g which has zeros exactly at {Wn : n;:e ,H} . Let h Q = (gof Q)|X . Then h Q e OL{X) which has zero exactly at {Z n : n e N] . S i m i l a r l y , we can f i n d an f e OliX) s u c n that 116. f has zeros exactly at {Zfc : k _> n] f o r each n . In either case we have { f n : n e N} with f has zeros exactly at £ Zk : k 2. n 3 • • L e t M 0 fee a m a x i m a l i d e a l which contains {f : n e N] . Then f|Z[M] e n Z(f ) = <}) . Thus, M 0 i s n=l n ° free. This' i s impossible. Hence X must be compact. Q.E.D. 117. BIBLIOGRAPHY [ l ] L. V. Ahlfors, Complex Analysis. McGraw-Hill Book Company, Inc., New York, 1953. [2] L. v . Ahlfors and L. Sario, Riemann Surfaces. Princeton Univ. Press," New Jersey, i960. [33 L. Bers, On Rings of Analytic Functions. B u l l . Amer. Math S o c , 54 (19^8), 311-315. [4 3 R. Bkoucke, Ideaux mous d'un Anneau Commutatif. Applications au anneaux de functions, C. R. Acad. Sc. P a r i s , t. 260 (21 j u i n , 1965), 6496-6498. [53 R. C. Buck, Advanced Calculus. McGraw-Hill Book Comp. Inc., New York, 1956. [6] R. M„ Orownover, Concerning Function Algebras. Studia Math., 25(1965), 353-365. [7] L. Gillman and M. Jerison, Rings of Continuous Functions. B. Van Nostrand Company,- Inc., New Jersey, 196O. [8] E. Hewitt, Rings of Real-Valued Continuous Functions I. Trans. Amer. Math. Soc., 64(1948), 45-99. [93 J. Go Hocking and G. S. Young, Topology. Addison-Wesley Publishing Company, Inc.? Massachusetts, 1961. [103 J. Lo Kelley,General Topology. D. Van Nostrand Company, Inc., New Jersey, 1955. [113 K. Khapp, Theory of Functions. Dover Publications, Inc., New York, 1945. [12 3 C. W0 Kohls, A Representation Theorem f o r Transformations of Rings of Continuous Functions. 111. J. Math., 6(1962), 674-681. [133 L. Ho Loomis, An Introduction to Abstract Harmonic Analysis. D. Van Nostrand Company, Inc., New Jersey, 1953. [14 3 K. E. M a g i l l , J r . , Some Embedding Theorems. Proc. Amer. MathoSoc, 16(1965), 126-130. , [153 J. Milnor, D i f f e r e n t i a l Topology. Princeton Univ. Press., 1958. [16] So Bo Myers, Algebras of Di f f e r e n t i a b l e Functions. Proc. Amer. Math. S o c , 5(1954), 917-922. [173 M. Nakai, Algebras of Some D i f f e r e n t i a b l e Functions on Riemannian Manifolds. Japan J. Math., 29(1959), 60-67. 118. [18] M. Ozawa and 1. Mlzum©t©, 'On Rings ©f Analytic. Functions. Japan J. Math., 29(1959)* 114-117. [ 1 9 3 Wo J * Pervin, Foundations.. -@f -General Topology.- Academic. Press, lew York,' 1964. [20] Lo. E . ' Fur s e l l , An Algebraic Characterization ©f Fixed. Ideals i n Certain Function•• Rings. P a c i f i c J. Math..,. 5(1955), 963-969. [21] c„ E„-':Riekart, ..'General Theory ©f Banaeh Algebras. D. Van Wog.trand Company,, Inc., Hew Jersey, i960. [22] Eo L„ Hoyden*, l i n g s , of Analytic and'Meramorphic Functions.. Trans. Amer. Math. S o c , 83(1956), 269-276. [231 W. M i n , Principles©f'Mathematical Analysis. McGraw-Hill Book Company, Inc.., lew 'fork, 1953.. [24'J W. Rudin, Some Theorems on Bounded Analytic Functions. Trans.' Amer. Math;- S o c , 78(1955), 333-3^2. [25] W. Rudin, An Algebraic Characterization, of Conformal Equivalence. B u l l . Amer. Math. Soc., 61(1955), 543. [26] ' 'Do' Eo Sherbert, Banaeh Algebras ©f LIpschltz Functions. P a c i f i c j ; Math., 13fl963), 1387-1399. [ 2 7 ] Mo H„ Stone, Applications ©f the Theory of Boolean .Rings to. General Topology. Trans. Arner^ Math. S o c , 41(1937)-, 375-481,, [28] B„ Lo van der Waerden, Modern Algebra. Frederick Ungar 'Publ. Co. .195©.- • • • -[29] Jo Yo- Whlttaker, Coincidence.sets and Transformations of Function:- spaces. Trans. Amer. Math. Soc., 101(1961), 459-^76. [30] H.- -Whitney, Analytic Extension of D i f f erentiable Functions Defined in.closed Setsj Trans. Amer. Math. S o c , 36(1934), 63-89. [31] H. Whitney, D i f f erentiable Functions Defined..in Arbitrary Subsets of Euclidean Space. Trans. Amer. Math. S o c , ^Q( 1936"), 309-317.
- Library Home /
- Search Collections /
- Open Collections /
- Browse Collections /
- UBC Theses and Dissertations /
- Algebraic properties of certain rings of continuous...
Open Collections
UBC Theses and Dissertations
Featured Collection
UBC Theses and Dissertations
Algebraic properties of certain rings of continuous functions Su, Li Pi 1966
pdf
Page Metadata
Item Metadata
Title | Algebraic properties of certain rings of continuous functions |
Creator |
Su, Li Pi |
Publisher | University of British Columbia |
Date Issued | 1966 |
Description | We study the relations between algebraic properties of certain rings of functions and topological properties of the spaces on which the functions are defined. We begin by considering the relation between ideals of rings of functions and z-filters. Let [fomula omitted] be the ring of all m-times differentiable functions on a [formula omitted] differentiable n-manifold X , [formula omitted] the ring of all Lc-functions on a metric space X , and [formula omitted] the ring of all analytic functions on a subset X of the complex plane. It is proved that two m-(resp. Lc-) realcompact spaces X and Y are [formula omitted] diffeomorphic (resp. Lc-homeomorphic) iff [formula omitted] are ring isomorphic. Again if X and Y are m-(resp. Lc-) realcompact spaces, then X can be [formula omitted] (resp.Lc-) embedded as an open [resp. closed] subset in Y iff [formula omitted] homomorphic image of [formula omitted]. The subrings of [formula omitted] which determine the [formula omitted] diffeomorphism (resp. Lc-homeomorphism) of the spaces are studied. We also establish a representation for a transformation, more general than homomorphism, from a ring of [formula omitted] differentiable functions to another ring of [formula omitted] differentiable functions. Finally, we show that, for arbitrary subsets X and Y of the complex plane, if there is a ring isomorphism from [formula omitted] which is the identity on the constant functions, then X and Y are conformally equivalent. |
Subject |
Functions, Continuous Rings (Algebra) |
Genre |
Thesis/Dissertation |
Type |
Text |
Language | eng |
Date Available | 2011-09-09 |
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. |
IsShownAt | 10.14288/1.0080603 |
URI | http://hdl.handle.net/2429/37206 |
Degree |
Doctor of Philosophy - PhD |
Program |
Mathematics |
Affiliation |
Science, Faculty of Mathematics, Department of |
Degree Grantor | University of British Columbia |
Campus |
UBCV |
Scholarly Level | Graduate |
AggregatedSourceRepository | DSpace |
Download
- Media
- 831-UBC_1966_A1 S8.pdf [ 7.26MB ]
- Metadata
- JSON: 831-1.0080603.json
- JSON-LD: 831-1.0080603-ld.json
- RDF/XML (Pretty): 831-1.0080603-rdf.xml
- RDF/JSON: 831-1.0080603-rdf.json
- Turtle: 831-1.0080603-turtle.txt
- N-Triples: 831-1.0080603-rdf-ntriples.txt
- Original Record: 831-1.0080603-source.json
- Full Text
- 831-1.0080603-fulltext.txt
- Citation
- 831-1.0080603.ris
Full Text
Cite
Citation Scheme:
Usage Statistics
Share
Embed
Customize your widget with the following options, then copy and paste the code below into the HTML
of your page to embed this item in your website.
<div id="ubcOpenCollectionsWidgetDisplay">
<script id="ubcOpenCollectionsWidget"
src="{[{embed.src}]}"
data-item="{[{embed.item}]}"
data-collection="{[{embed.collection}]}"
data-metadata="{[{embed.showMetadata}]}"
data-width="{[{embed.width}]}"
async >
</script>
</div>
Our image viewer uses the IIIF 2.0 standard.
To load this item in other compatible viewers, use this url:
http://iiif.library.ubc.ca/presentation/dsp.831.1-0080603/manifest