A Generalized Hausdorff Dimension for Functions and Sets Citation Buck, Robert Jay (1968) A Generalized Hausdorff Dimension for Functions and Sets. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/60CE-6945. https://resolver.caltech.edu/CaltechTHESIS:11232015-082345589 Abstract Let E be a compact subset of the n-dimensional unit cube, 1 n , and let C be a collection of convex bodies, all of positive n-dimensional Lebesgue measure, such that C contains bodies with arbitrarily small measure. The dimension of E with respect to the covering class C is defined to be the number d C (E) = sup(β:H β, C (E) > 0), where H β, C is the outer measure inf(Ʃm(C i ) β :UC i Ↄ E, C i ϵ C). Only the one and two-dimensional cases are studied. Moreover, the covering classes considered are those consisting of intervals and rectangles, parallel to the coordinate axes, and those closed under translations. A covering class is identified with a set of points in the left-open portion, 1’ n , of 1 n , whose closure intersects 1 n - 1’ n . For n = 2, the outer measure H β, C is adopted in place of the usual: Inf(Ʃ(diam. (C i )) β : UC i Ↄ E, C i ϵ C), for the purpose of studying the influence of the shape of the covering sets on the dimension d C (E). If E is a closed set in 1 1 , let M(E) be the class of all non-decreasing functions μ(x), supported on E with μ(x) = 0, x ≤ 0 and μ(x) = 1, x ≥ 1. Define for each μ ϵ M(E), d C (μ) = lim/c → inf/0 log ∆μ(c)/log c , (c ϵ C) where ∆μ(c) = v/x (μ(x+c) – μ(x)). It is shown that d C (E) = sup (d C (μ):μ ϵ M(E)). This notion of dimension is extended to a certain class Ӻ of sub-additive functions, and the problem of studying the behavior of d C (E) as a function of the covering class C is reduced to the study of d C (f) where f ϵ Ӻ. Specifically, the set of points in 1 1 , (*) {d B (F), d C (f)): f ϵ Ӻ} is characterized by a comparison of the relative positions of the points of B and C. A region of the form (*) is always closed and doubly-starred with respect to the points (0, 0) and (1, 1). Conversely, given any closed region in 1 2 , doubly-starred with respect to (0, 0) and (1, 1), there are covering classes B and C such that (*) is exactly that region. All of the results are shown to apply to the dimension of closed sets E. Similar results can be obtained when a finite number of covering classes are considered. In two dimensions, the notion of dimension is extended to the class M, of functions f(x, y), non-decreasing in x and y, supported on 1 2 with f(x, y) = 0 for x · y = 0 and f(1, 1) = 1, by the formula d C (f) = lim/s · t → inf/0 log ∆f(s, t)/log s · t , (s, t) ϵ C where ∆f(s, t) = V/x, y (f(x+s, y+t) – f(x+s, y) – f(x, y+t) + f(x, t)). A characterization of the equivalence d C 1 (f) = d C 2 (f) for all f ϵ M, is given by comparison of the gaps in the sets of products s · t and quotients s/t, (s, t) ϵ C i (I = 1, 2). Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Bohnenblust, Henri Frederic Thesis Committee: Unknown, Unknown Defense Date: 1 April 1968 Funders: Funding Agency Grant Number Caltech UNSPECIFIED Ford Foundation UNSPECIFIED Record Number: CaltechTHESIS:11232015-082345589 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:11232015-082345589 DOI: 10.7907/60CE-6945 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 9284 Collection: CaltechTHESIS Deposited By: INVALID USER Deposited On: 25 Nov 2015 17:08 Last Modified: 28 Mar 2024 22:05 Thesis Files Preview PDF - Final Version See Usage Policy. 12MB Repository Staff Only: item control page

A GENERALIZED HAUSDORFF DIMENSION FOR FUNCTIONS AND SETS Thesis by Robert Jay Buck In Partial Fulfillment of the Requirements For the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1968 {Submitted April 1, 1968) 11 ACKNOWLEDGMENTS I wish to thank Professor H. Bohnenblust for his guidance and constant encouragement during the preparation of this thesis. For financial support I am indebted to the California Institute of Technology and to the Ford Foundation, for various scholarships and summer research fellowships. iii ABSTRACT Let E be a compact subset of the n-dim.e nsional unit cube, 1 , n and l e t C be a collection of convex bodies, all of positive n- dimensional Lebesgue measure, such that C contains bodies with arbitrarily small measure. The dimension of E with respect to the covering class C is defined to be the number where Hf', C is the outer m e asure in£(6m(C .}13:uc. ::iE, C . EC}. 1 1 - 1 Only the one and two-dime nsional cases are ·studied. Moreover, the covering classes considered are those consisting of intervals and rectangles, parallel to the coordinate axes, and those closed under translations. A covering class is identified with a set of points in the left- open portion, l I , of 1 , whose closure intersects 1 - l' n n n n· For n = 2, the outer measure Hf', C is adopted in place of the usual: inf (6 (diam. (C.} }S:uc. ::i E, C . EC} • 1 1 - 1 for the purpose of studying the influence of the shape of the covering sets on the dimension dc(E}. If E is a closed set in 1 , let M(E) be the class of all non1 decreasing functions µ.(x}, µ.(x) = 1, x > 1. supported on E with µ.(x) = 0, x < 0 and Define for each µ. E M(E), iv = lim inf log 6µ(c) log c c ... 0 where 6µ.(c) = V (µ.(x+c) - µ(x) ). x (c EC) It is shown that This notion of dimension is extended to a certain class J of subadditive functions, and the problem of studying the behavior of dC(E) as a function of the covering class C is reduced to the study of dC(f) where f E J. Specifically, the set of points in 1 , 2 is characterized by a comparison of the relative positions of the points of B and C. A region of the form (':<) is always closed and doubly- starred with respect to the points (O , 0) and (1, 1). Conversely, given any closed region in 1 2 , doubly-starred with respect to (O,O) and (1, 1), there are covering classes B and C such that('!~) is exactly that region. All of the results are shown to apply to the dimension of closed sets E . Similar results can be obtained when a finite number of covering classes are considered. In two dimensions, the notion of dimension is extended to the class M, of functions f(x, y), non-decreasing in x and y, 1 2 supported on withf(x,y) = 0 for x•y= 0 andf(l,l):::: 1, by the formula = lim inf log M(s' t) log s. t where (s, t) E C v M(s, t) = (f(x+s, y+t) - f(x+s, y) - f(x, y+ t) + f(x, t)) . V x,y A characterization of the equivalence de (f) = de {f) for all f E M, 1 2 given by comparison of the gaps in the sets of products s· t and quotients s/t, {s,t) Ee . l {i = 1, 2). is vi TABLE OF CONTENTS Page ACKNOWLEDGMENTS ii ABSTRACT iii 1 INTRODUCTION CHAPTER I §1 §2 CHAPTER II §1 §2 §3 §4 CHAPTER III §1 §2 §3 CHAPTER IV §1 §2 §3 §4 §5 BIBLIOGRAPHY PRELIMINARIES Sub-additive Functions The Class M DIMENSION OF FUNCTIONS Covering Class e s , Dimension An Equivalence Relation on M Special Fune tions in ;J A Partial Ordering and Equivale nce Relation for Covering Classes COMPARISON OF COVERING CLASSES Doubly-Sta rred Sets Main Results A Characterization of R(S, T} 5 5 6 11 11 13 16 23 30 30 36 45 APPLICATIONS TO CLOSED SETS 53 Introduc tion and Pr e liminaries Frostman 1 s The or e m Generalized Cantor Sets Main Results Conclusions and Generalizations 53 56 60 79 91 94 1 INTRODUCTION Let E be a subset of the n-dimensional unit cube ln and let S be a collection of convex bodies, all of positive n-dimensional Lebesgue measure, such that S contains bodies with arbitrarily small measure. The dimension of E with respect to the covering class Sis defined to be the number where Hb is the outer measure inf(~ (m(C.) )b:uc . ::::>E, c. 1 1 - 1 Es). The dimension d (E) is always a number be.tween 0 and 1. 5 Moreover it is monotone with respect to the covering class in the sense that The purpose of this dissertation is to study the behavior of d (E) as a function ·of the covering class S . 5 dimensional cases are considered. Only the one and two- In the one-dimensional case con- vex sets are intervals but the question could arise as to the behavior of d (E) when covering classes other than those consisting of intervals 5 are considered. Although this ques ti.on is for the most part unan- swered, some· statement can be made. Theorem. If S is any covering class of open sets which contains the intervals and their finite unions, then d (E) = 0 ~ 5 d (E) = 1, !!;,! E has measure zero or positive outer measure. 5 2 If E has positive outer measure p, Proof: then any cover- ing (C .) by sets in S has the property that, 1 and so Hb(E) > 0, b ~ 1, which shows d (E) = 1. 5 On the other hand, if E has measure zero, there is a sequence of intervals ( 1.) such that J m(Ul.) < e, 6m(l.) <oo, J Ul.::>E. J J- For any r > 0, there is a seq. Nk such that Thus CX> 6 k=l and so let m be such that CX> r 6 € k=m+l ' then H r (E) < - ( m )~ + Nm U 1. ( j= 1 J r 00 6 k=m+ 1 ( m( )~ Nk U 1. j=Nk- l + l J Since e was arbitrary and r > 0, Hr(E) = 0, and d (E) = O. 5 r 3 Since part of the interest of a dimension such as ( 1) lies in the study of sets of Lebesgue measure zero, it appears that for this purpose the covering classes cannot be too large with respect to the intervals. The fact that the intervals themselves do not form too large a covering class was proved by Hausdorff [11]. The approach taken in this study is motivated by the se facts and so covering classes are considered here to be collections of intervals, closed under translation, which contain intervals of arbitrarily small length. Thus a covering class is completely determined by the lengths of its members. Of gre~t importance to this study is the fact that the study of dimension with respect to covering classes which consist of intervals can be reduced to the study of a dimension of a certain class of increasing functions. The notion of the dimension of a function considerably facilitates the investigations and carries a certain interest for its own sake. In two dimensions the situation is more interesting because the shape of the covering sets, along with the area , plays a fundamental role. Although the results are fragmentary, certain indications of the role of shape can be made. As a first attempt, covering classes which consist of rectangles are considered and an analysis made of the influence of shape on the dimension. When S is the collection of all convex bodies, the expression ( 1) is called the Hausdorff dimension of E. This dimension function on closed sets has been studied in connection with the theory of Trigonometric Series, for example, by Bari [l] andKahane and Salem [5 ] . For certain applications to Number Theory and some interesting 4 prope rties of the Hausdorff dimension see Be socovitch [8] and Eggleston [9 J. Finally, for the relationship of the Hausdorff· dimen- sion of a closed set to the topological dimension see Hurewitz and Wallman [4]. 5 CHAPTER I PRELIMINARIES § 1. Sub - additive Fune tions A real valued function f(x). defined 0n x > 0, is said to be sub-additive provided that f(x+y) .:s, f(x) + f(y) , for all x, y ~ 0. Only non-negative functions are considered. Lemma 1. If f(x) is sub-additive and non-decreasing for x > 0, then for all t > 0 f(tx) .:s, 2(t v 1) f(x) . Proof: n < t < n+ 1. There is a non-negative integer n, such that Thus f(tx) .:s, f( (n+ l)x) ~ (n+ 1) f(x) .:s, (t+ 1) f(x) .:s, 2(t v 1) f(x) • 1 In particular, if t = -, x 2 f( 1) < - f(x) , -x or f( x ) ~ x f( 1) /2 • ( 1. 1) A second fact, due to Hille [ 3 J, us e ful in the following, is Lemma 2. is sub-additive. If f(t) /t is non-increasing for t > 0, the n f(t) 6 Proof: f(x+y) = x f(x+y) + y f(x+y) x+y x+y < x f~) + y f(;) = f(x) + f(y) A sub-additive function which will be important for later consider ations is the function ( 1. 2) t:if(s) = V {f(x+s) - f(x)} x where f is supposed non-decreasing. The fact that M is sub-additive can be seen by writing, f(x+s+t) - f(x) .s, {f(x+t+s) - f(x+t)} + {f(x+t) - f(x)} .s, t:if(s) + t:if(t) , and so, MC s+t) .s. M( s) + t.£( t) • Further, if f itself is sub-additive, then f(x+s) - f(x) .s, f(s) , and so if f(O) §2. = 0, M(s) = f(s). The Class M Let M denote the class of all real ........ alued, non-decreasing functions f(x), defined on x > 0 such that f(O) = 0, and M is bounded on [O, l]. Define a transformation T on M as follows: 7 x > 0 if T(f)(x) = ( 1. 3) x = 0 • 0 Some elementary properties of T are listed below. Theorem 1. Proof: a) T(f) < f , f E M. b) T(f)(x) /x is non- increasing for x > O. c) T(M) SM. d) T(f) is continuous at each x > 0. a) Since T(f)(x) when x > O. /\ f(y) /y < f(x) /x, it follows that - =x /\ f(y) < f(x) Y< ' - x y -. y~x For x = 0, the same is true. T(f)(x) /x = b) If x > 0, /\ Y<X f(y) /y , which is evidently non-increasing. (x+t) Let x, t > 0, c) T(f)(O) = 0 by definition. /\ = (x+t) ( /\ f(y) /\ /\ f(y)j !irL y y~x y x~y~x+t y} y~x+t > x /\ f(y) Y<X y /\, (x+t) f(x) x+t which shows that T(f) is non-decreasing. By Lemma 2, § 1 and the remarks following it, parts a) and b), above, imply ~T(f) then = T(f) < f 8 so that tiT(f)(x) < f(x):s_ f(l), d) Let x 0 > 0 and write g = T(f). If x > x , then 0 by c) and so, On the other hand, if x < x , 0 g(x) < x which implies Thus, which shows that g is continuous at x . 0 If f E M and if f(t) /t is non-increasing, it is clear that T(f) = £. In the case that f(x) is only sub-additive, there is the follow- ing estimation. Theorem 2. If f E M is sub -additive, then f(x) I 2 < T(f) (x) :s_ f(x) • 9 Proof: The fact that T(f) :s_ f was just proved above. By Lemma 1, = f(xy /y) :s_ 2( 1 v x/y) f(y) • f(x) If y ~ x, then f(x) :s_ 2x f(;) , f(x) :s_ 2x Denote by J, and so /\ Y:S, x f(y) = 2T(f)(x) y the class T(M), then J has the following properties: Theorem 3. a) If f E M, b) If f, then f E J if and only if f(t) /t is non- increasing. ra. gS E J, where o 0 c) g E J and O:s_ a. . e, a.+ e :s_ 1, then is defined to be zero. If f - a E J for all a E A, then cfEA fa E J .5!:.llil a¥A fa E J, provided these exist. Proof: a) is proved by Theorem 1, c) and the remark fol- lowing it. b} Let t > x. Then to:+e-1 ec:))a.. ( g~x)) e to:+s-1 1 = fX(x) ~gS(x) ( ~) a.+s- < 10 < c) are I\ a EA If f a (l(x) g f'(x) x is a non-decreasing for all a E A, fa and . V fa, and both are zero at x = O. a EA then so By a) it is suf- ficient to show (a~ A fa(t))/t and (a~ A fa(t))/t are non-increasing for t > 0. Indeed, since f ( I\ aEA a E ;;: , for x < t, fa(t1 · I\ = ( and f (x) a t a EA x V fa(x~ fa(t~ V aEA ( ·< t for the same reasons. I\ -< aEA t f ( t) a a EA x (a~A fa(x)) = x 11 CHAPTER II DIMENSION OF FUNCTIONS § 1. Covering Classes, Dimension A non-empty set S of points in (0, 1 J is called a covering class, provided that S has 0 as a limit point. Definition: Given a function f EM, f J, 0, and a covering class S, the dimension of f with respect to S is defined to be the number, ( 2. 1) ds(f) = lim inf s ... 0 log M(s) log s (s E S) , where Af. is defined by ( 1. 2). Remark 1: Observe that if f EM, and f '1:. 0, if x > O. then M(x) > 0, Indeed, if f(x ) > 0, for some x > 0 and M(x) 0 0 positive x, then let n > 1 be such that x 0 ~ nx. = 0 for some Then n f(x } < f(ruc} 0 = ~ f(kx} - f( (k- l)x} k=l < which is a contradiction. Remark 2: 1. n M(x) =0 , Thus dS(f) is well-defined. The dimension, dS(f), is always between 0 and Since f EM, 0 ~ M(s} ~ C, for some constant C, when s ES. Thus 12 . lo{c M(s) > log C og s - log s and so d (£) = O. 8 = o( l) s ... 0 ' On the other hand, since it was observed in § 1, Chapter 1, that t::J is sub-additive, log t::J(s) log s M( 1) /2 = 1 +log t::J.(1)/2 = 1 + o(l) log s < log s log s and so by ( 1. 1) it follows that d (f) .:s, 1. 8 Moreove r, x!1- is in M, for 0 < a..:S, 1 and d (x.9-) =a. . 5 0 The function x x 0 = defined by {l, 0, x = 0 ' 0 satisfies d (x ) 5 =O. Therefore d 5 (f) can take any value in [O, 1 ] . The following can be used as an alternate definition for d (f) . 8 Theorem. Proof: d 5 (£) = sup(13:s-13 t::J(s) = 0(1), s ES). Let r(S,f) suppose that d (£) > O. 5 = sup(13:s-1:3 t::J(s) = 0(1), s ES), and Ii 0 .:s, 13 < d (f), then for s E S and s small 8 enough, log M(s) , log s > 1:3' which implies tif(s) .:s, sl3. In this case', s -13 M(s) .:s, 1 when s E S, small enough so that 13 .:s, r(S, f). s Since r(S, f) 2: 0, On the other hand, suppose now that r(S, £) > 0, . and take 0.:::, 13 < r(S, £) . 13 s -13 M(s) ~ M , Then s ES , for some constant M, and so lot M(s) > log M og s log s + 13 ' which implies d (£) 2: 13. 5 § 2. Thus d (f) 2: r(S, £), and equality follows. 5 An Equivalence Relation on M For the study of the dimension d (f) when f E M, 5 there is a natural equivalence relation induced on M which conside rably simplifies the work. Namely, two functions f and g in M will be called equivalent, in symbols f,..., g, provided that d (£) = d (g) for all cover 5 5 ing classes S. The main result concerning this equivalence is the following: Given f EM, f I= 0, there is a function g E 3'., Theorem 1. such that f,...., g, that is, d (£) = d (g) for all covering classes S. 5 5 Proof: If f EM, define g mation defined by ( 1. 3). = T(6£) where T is the transfor- Since M is sub-additive, Theorem 2 ( § 2, I), gives T(6£) < M < 2T(M) . By Theorem 1, c) of that section and chapter, T( 6£) (x) /x is non- increasing and T(6£) is therefore sub-additive by Lemma 2 (§ 1, I). Thus the relation, ~ log s + log T(6f)(s) < log 6f(s} log s log s < log T(lf)(s) log s 14 implies that a (T(M)) = d (f), for all covering classes S, or 8 8 f...., T(M) E J. The advantage of being able to re place f E M by an equivalent g in 3 is that formula (2. 1) simplifies to: a (g) = lim inf log g ( s) 8 (s E S) • log s s -+ 0 For functions f, g in J which are not identically zero, the equivalence f...., g has an interesting interpretation. Recall that by Remark 1, if f E J and f i 0, then f(x) > 0 when x > O. Theorem 2. If f, g E J, and neither f nor g is ide ntically zero, then f,.., g, if and only if, f(x) lim x-+ 0 Proof: 1og i\if = 0 . log x Let a(x) = log f(x) , b(x) = log f(x) log x log x • For e v ery covering class S, lim sup (a(s)-b(s)) = -lim inf (b(s)-a(s)) s -+ 0 s -+ 0 > lim inf a(s) - lim inf b(s) s-+O · s-+0 > lim inf (a(s)-b(s)) , - s-+ (s E S) 0 since lim inf (A+ B) _;:: lim inf A+ lim inf B. If log f(x) / g (x) 1og x x -+ 0 lim it follows that d (f) 8 = lim inf a(s) = lim inf b(s) = d 8 (g) • = 0' 15 Conversely, if lim sup (a(x)-b(x)) > 0, then there .e xists S such that x ... 0 lim inf (a(s)-b(s)) > 0 , s ... o (s E S) which implies lim inf a(s) > lim inf b(s) s -+O s -o (s E S) • This would contradict f,...., g, sci that lim sup (a(s)-b(x)) < 0 • x-0 A similar argument shows that f,...., g implies lim sup (b(x)-a(x) .> < 0 , x-o and so lim I a(x) - b(x) I = 0, which completes the proof. x ... 0 For future use, the following lemma is established. Lemma 2. If f E 3' and 0 <a.< 1, then and Proof: Since both f and xa. are in :J, Theorem 3 (§ 2, I) implies that f /\ xa. and f V xa. are in 3'. Consequently log f(s) /\ = lim inf -""""-...,._...__ _sa. _ log s s - 0 (s E S) and d (f V xa.) = lim inf log f(s) v sa log s 8 s - 0 16 Since los f(s) A sa. log s = log f(s) log s v a. los f(s) v sa. log s = log s and log £( s) /\ a. ' it follows that ds(f /\ xa.) = a. v ds(f) ds(f v xa.) = a. " ds(f) . and §3. Special Functions in 3 Given a covering class S and any point s ES, define s* = sup(t:t< s, t ES) • If S is a covering class, denote by s*, the new covering class, Sn (O, l], where Sis the closure of S. For f E 3 define, V (f(s) Ax f(s):c) /s*) s ES* (x ~ 0) The next theorem lists some important properties of (f)S. Theorem 1. Proof: If f E 3 and S is a covering class, then: a) (f)s E 3. b) ds( (£) s> = ds(f). c) dT( (£) s> :::_ dT(f), for all covering classes T. a) For each s E S*, the function f(s) /\ x f(s*) /s* is in :J. By Theorem 3, c) (§2, I) it follows that (f)S E ;J. 17 Since s* < s and f E ;J, it follows that b) f ( s) .s_ s f ( s *) Is* and so, f(s) = f(s) /\ s f(s>:C) /s* • If t ES* and t > s, then t>:C > s, so that On the other hand, if t E S >:C and t ,S_ s, then f(t) ,S_ f(s), so that f(s) = v , {f(t) /\ s f(t*) /t*} = (f)S(s), tES':c holds for s E S* and in particular for s E S. c) This implies that Finally, observe that f(s) /\ x f(s>:<) /s* > f(x) . - w h enever s * _< x _< s, s ES*. , Since every x > 0, satisfies such a rela- ti on for some s ES*, it follows that (f) S ,;:: f, for x > 0, and thus for any covering class T . Remark: Observe that when s* ,S_ x ,S_ s, s E S*, it holds that (f) (x) = f(s) /\ x f(s>:C ) /s*. 8 Indeed, when t ES*, t > s, then t* > s so that x < s implies f(t*) ~ < f(t*) < f(t) . t"" - Moreover, since t* .2:. s*, x f(s*) /s* > x f(t*) It* . 18 Again x < s implies f(s) f(s) f(t>:C ) -X- -> -X- -> ~t)' ' , r so that when t > s, f( ''') f(t>'r) f(s) /\ x 2;. '"t • s >.- -> f(t) Ax -.... On the other hand, when t< s, t E S>le, then t.:: s >le , so that f(t) < f(t) f(t*) -t-<tr, x or f ( t >'.c ) f( t) /\ xt* = f(t) • Since f(t) .:: f(s) and f(t) .:: f(s >:C) .:: ~ f(s*) ' it follows that for all t E S>le, which verifys the statement. Of special interest are the £Unctions (xa.) , 0 <a. < 1. 8 In this case more can be said. Theorem 2. If f E 3' such that d (f) 8 = a., 0 _::: a. < 1, then dT(f) ~ dT( (xa.) ) for all covering classes T. 8 Proof: Lemma: The proof is based on the following. If g E ;J and if S is a cove ring class such that g(s) .:: s~ for s E S, ~ g.:: (x~)S for 0 < x < o for some positive o. 19 Proof of the Lemma: Let t* < x < t for t E s):c. By Theorem 1, d) ( § 2, 1.) and the definition of S*, g(x) < g(t) < ti3 and It follows that and so by the remark made following Theorem 1. This establishes the lemma. Returning to the proof of Theorem 2, suppose first that d (£) >ex.. 5 Then for all s E S, s sufficiently small, it follows that lofc f(s) > og s ex. ' or f(s) < sex. By the lemma, f(x) .:::, (x°') (x) when x is sufficiently small, so that 5 Now suppose that d {f) =a. and conside r the 5 for any covering class T. function f 1-t t •x , 0 < t < 1. This function is in J, by Theorem 3 {§ 2, I), and = lim inf log f 1-t( s) • s t s .... 0 log s 20 = (1-t) lim inf loglogf(s) + t s s .... 0 = (1 - t) <ls(£) . + t = (1 - t) a + t • Similarly, Since ( 1-t) a + t >a when 0 < t, it follows that The right-hand side of this last inequality approaches dT(f) as t approaches zero, and so dT( (xa)S) ~ dT(f), as desired. = 1, the situation is not so simple because while dT( (x)S) = 1, there can be f such that dS(f) = 1 but dT(f) < 1. The folIn the case a lowing theorem makes this clear. Theorem 3. If f E J and dS(f) = 1, then and there exists a function g E J such that dS(g) = 1 and dT(g) = sup dT( (xa) S) • a<l Proof: Suppose that f E J and dS(f) = 1. If dT(f) < sup dT( (xa.) S), then there exists y < 1 such that dT(f) < dT( (x y) s> • a.< 1 Consider now the function f(x) V xY. Then 21 and but these two statements are in contradiction to Theorem 2. Thus dT{f) .:::_ sup dT{ (xa.) S) . a.< 1 To prove the second part, suppose first that sup dT{ {xa.)S) = 1. a.< 1 In this case there is nothing to prove , since dT{ x ) = 1. Now l e t sup dT{ {xa.)S) = a.< 1 o< 1 • Select an increasing sequence {a. ) such that n and a.n > Since dT{ {xa.)S) .:::, o for all a. < 1, o {n= 1,2, .. . ). there is a d e creasing s e que nce {sk) of points in S which have the following properties : exists t For each k, th e re * tk.:::, sk and {xa.k.>s {~) .: :_ ~o+ l/k ; and = ~ E T such that sk.:::, the points sk satisfy: 1-a. k ):c) 1 -a. k+ 1 {sk ' {k = 1, 2 , .•• ) • Define It will now be shown that dS{g) = 1 and dT{g) = o. First, it is cle ar a.k >:C°'k-1 that g E 3, s;i.nce each o f the f unctions sk /\ x sk {k = 1, 2, .•. ), is in 3. Moreover, if s ES and sk+l.:::, s .:::, s~, the n 22 a. -1 ... k V s • s·.k g(s) Indeed, if t > k+ 1, then CL a. k+ 1 = k+l /\ 8 6 8 >:C .k+ 1 -1 k+l On the other hand when t < k, since a. -1 a. -1 a. a. -1 = s s~ t < s s~ k = skk /\ s · sk k * since s < s* and k l -a. k- l I 5 )I' l -a.k < s < s.. k* k k-1 l -a. t < s* - 1-a. k k-1 1 -a t < s* - 1-Clk t Thus log g(s) log s ,.. log sk+ 1 log sk = a.k+ l - - - - /\ l + (ak - 1) - - log s log s Since a.k approaches l as k approaches ... , it follows that dS(g) = 1. >:C Consider the sequence (tk) of points in T such that sk:::, tk:::, sk and 1 (xak)s (tk) Since, 2: t:+ k 23 it .follows that dT(g) :::_ o, and since g E ~ and by the first part of the theor e m, dT(g) = o. It is indeed possible that sup dT( (xcx') ) < 1. In fact, 8 ex.< 1 there exist covering classes S and T for which s.up dT( (xcx.) ) = O. Let 8 2 S be the sequence ( Z_zk ) (k 1, 2, •.. ) and T : 1 J. Then if o. < 1, Remark: ~O~ = 2 CX.2 k dT( (xcx.) S) < lim k-+oo k2 cx.2 Cl) lim = k-+ Cl) §4. 2 ex. = lim k-+ - cx.2(k+l)2 + 2 Ck+ 1) . ex. - ex. 22k+1 + 22k+l ex. =0 . ex. + 22k+ 1( 1 - ex.) A Partial Ordering and Equivalence Relation for Covering Classes As a first step in the study of d (f) as a function of the cover8 ing class, it is natural to consider the following partial ordering on the collection of covering classes: a covering class S is said to be less than, or equal to, the covering class T, in symbols S :::_ T, provided that d (f) :::_ dT(f) for all functions f in ~ • . It is clear that if T ~ S, 8 then S < T. Indeed , if for a given f in ~, s -!'f(s) is bounded for all the same is true for t E T. By the Theorem of § 1, II, s in S, then it follows that d (f) :::_ dT(f) and thus S :::_ T. 8 With the help of this re- mark, the following lemma can be proved. Lemma 1. If A, B are covering classes and f E ~. then 24 Proof: Since AU B contains both the sets A and B, the inequality, is immediate by the remarks made above. f E J, and suppose that dA(f).::. dB(f). To show the reverse, let If dA(f) = 0, then automatically. Thus assume dA(f) > 0, and let a. be such that 0 .::. o. < d A ( £) • Then a -a. f (a) .::. MA (a c A) • Since it is assumed that dA(f).::. dB(f), it follows that b -13f(b) < MB If M = max(MA' MB), then c -o.f(c) < M and so a..::. dAUB(f). (b EB) (c E AU B) , It then follows that which proves the lemma. The following theorem gives a characterization of the relation S < T in terms of a comparison of the gaps in the covering class e s S and T . 25 Theorem. A necessary and sufficient condition that S ::_ T, is that there exist a function g, g:T .... S, with the property lim log g(t) t .... o log t =1 (t E T) • Proof: Sufficiency: Let g:T ... S be such that lim log g(t) = 1 t .... 0 log t (t E T) • By Lemma 1, ( § 1, I), for f E 3', f(t) ::. 2(1 v gCt>J f(g(t) > • or {(1 /\ g~t)) f(t) < f(g(t)) . Thus it follows that ~ log log f(g(t)} c. log g(t) < log g(t) log + (1 /\ g(tt)) + log g(t) log f(t} log t log t log g(t) Since 1 - log g(t}' I = o(l} 1 log t (t .... 0) and log t 0 < log g(t) ::_ I l _ log ( t} log g(t) I+ 1 = 0 ( 1) + 1 ( t .... 0) it follows that log f(g(t}) < o(l) ( 1 + log f(t)'\ log g(t} log t ') + log f(t} log t ( t .... 0) • , 26 l + log f(t} log t By ( l. 1) and the fact that f E J, is bounded for t E T, so that logf(g(t}} _< o{l) log g(t) + logf(t} log t ( t ... O} , which implies that .nf log f(g(t)) < lim inf log f(t} = (£) = 1. i;: io log g(t) t ... o log t g(T) d Since g( T} CS, it follows that and thus S < T. Necessity: The following lemma is needed. Lemma 2, s < s):<. Proof of the Lemma: By what has been proved above, it is sufficient .to exhibit a mapping, h:S>:< ... S, such that lim log h(t) = 1 t ... 0 log t If t E s*, there is r E S, such that, either r < t and 1 < 1°g r < 1 + t, - 1og t - log r or t < r and l - t < t < l. - 1og h(t) = r. Choose any such r in S and write I Then it is clear that lirn log h( t) = 1 log t t ... 0 and the lemma is proved. Now define a function g: T ... s* as foUows: If s*.:::, t.:::, s, 27 s E s•:<, write s ' if log t < lo~ s >:C TOgS- log t g(t) = s*, lo~ s* if log t <~ og s Then g is defined on all of T , and since it is now assumed that d (f) :::_ dT(f) for all f E 3, it follows, in particular, that 8 for 0 <a. < 1. - - Fix a. such that 0 <a. , and consider 0 < e <a.. Then and so for t E T, t < i5, for some positive (S. From the definition of (x°') S' it follows that for s E s* and t < 5, and so, either s a. < r.rL- € (2. 2} or ( 2. 3) Thus given t E T, holds, then t < ~. choose s Es* such that s * < t < s. If (2. 2} 28 s a. a.-€ < t < s , and so By the definition of g(t), either log s = log g(t) log t log t or ~ > log g(t) IOgS - log t = Thus in any event, 9:..:.! < log g ( t) < a. log t a. - a. - € Simi liar 1y, if ( 2. 3) holds, then 1-a. s* < t < s*l-a.+e and so 1-a. < log t log 1-a.-e s* < 1 < log s* < 1-a.-e log t 1-a. Again, since either log s* log t = log g(t) log t or log t log s* < log g(t) = ~< 1 IOgt • log t it follows that 1-a. < log g(t) < 1-a.+ e 1-a.+e log t 1-a. 29 Since e: > 0 was arbitrary, these considerations show that lim log g (t} :a l t -+O log t (t E T} • Finally, consider the composition h{g}, where h is the function defined in Lemma 1. Then h(g} :T -+ S in such a way that lim log h(g(t}} log t t--0 =1 (t E T} , and this completes the proof of the theorem. Remark: The partial ordering S _s T, described above for covering classes, leads immediately to the equivalence S = T, when both S < T and T < S. In Lemma 2 it was shown that S < S):c . S 5= ~, s'f. < S, so that S =fi:C. it follows that Since For this reason it will be assumed henceforth that any covering class in question is closed in the left-open unit interval, unless specifically stated otherwis e . Further, since it is clear that the addition or deletion of a finite set of points to or from a covering class does not alte.r the dimension function, such additions or deletions will be assumed without specific mention whenever convenience dictates. Thus, for example, in later chapters the point, 1, will be assumed to belong to any covering class under consideration, whenever convenient. 30 CHAPTER III COMPARISON OF COVERING CLASSES § 1. Doubly-starred Sets Given any set A in 1 , 2 define the transpos e of A, tr A, by tr A = ((x, y):(y, x) EA) If (a, b) is a point of 1 (a,b) s 2 and a_;:: b, write = ((x,y):x_;::y, ay_;::bx, (t-b)(t-x)_;::(l-a)(l-y)) and if b > a , (a, b)s = tr (b, a) 8 •• Given any set A::; 1 , write 2 As = U((a, b)s : (a, b) EA) A set A will be called doubly-starre d with respe ct to the points (O, 0) and (1, 1), or more simply, doubly-starred in 1 , 2 provided that A = As. The next few lemmas describe some prop ~ rties of doubly-starred sets . Lemma 1. If A= A, then As = As • Proof: ' Suppose (x ,y) E As with x > y (n = 1, 2, ••. ) and n n n- n yn ... y. (a , b ) E A n n Then x _> y. Since A= A, Let a s > b n be such that (xn , y n ) E (an , b n ) , n- there is a subsequence (a . , b .) - (a, b) E A. J J 31 Since a.y . > b .x. , J J - J J (l-b.)(1-x .) > (l-a .)(1-y .) , J J J J it follows that ay .=::bx and (1-b)(l-x) .=:: (1-a)(l-y). CA 8 , whichshows A 5 Thus (x,y) E (a, b)s 5 =A • If A is a set in 1 , 2 then the sets tA and A+ s for real s and t are defined by tA = ((ta, tb):(a, b) EA} , A+ s = ((a+s, b+s) : (a, b) EA} Lemma 2. A = A 6 , if and only if tA U ( tA + 1- t) ~ A ; for all 0 < t < 1. Proof: s If A= A , 0 _:: t _:: 1, (a, b) E A with a.=:: b, ta· > tb , a(tb) _=:: b(ta) , (1-b)(l-ta) = (1-b)(l-a) + a(l-b)(l-t) > (1-b)(l-a) + b(l-a)(l-t) = (1-a)(l-tb) . This shows (ta, tb) E (a, b)s. Moreover, since ta+ 1-t _=:: tb + 1-t , a( tb+ 1-t) .=:: b( ta+ 1-t) , (1-b)(l-(ta+l-t)) = (1-b)t(l-a) = (1-a)(l-(tb+l-t)) , then 32 s it follows that (ta+l-t, tb+l-t) E (a, b) . Thus tA U( tA+ 1- t) .'.:: As = A • On the other hand, let (x, y) E (a, b) If x = y, generality, suppose a> b. c A, by hypothesis. s with (a, b) E A. Without loss of then ( 1, 1) EA implies (x, y) E xA Thus suppose x > y, and so a> b. t = Write (x-y) (x-y} + (by-ax} and s = (x-y) + (by-ax) a-b Since by.?: ax, it follows that 0 ~ t ~ 1 and s > O. Writing (x-y) +(by-ax)= (1-a)(l-y) - (1-b)(l-x) + a - b , shows that s < 1. Since and s(tb+ 1-t) = s( y(a-b) ~ = y , x-y +by - axJ it follows that (x, y) E s(tA+ 1-t). (x, y) EA. Thus (a, b) Lemma 3. Proof: s (x, y) E (a, b) • s ~A (As ) 6 Since s(tA+ 1. t) ~ A, it follows that which shows that A c: A s . = A6 • Suppose (a, b) E A with a> b and assume Since, for 0 ~ t ~ 1, 33 tx ~ ty ' aty > btx , (1-b)(l-tx) = (1-b)(l-x) + x(l-t)(l-b) > ( 1 - a)( 1 -y) + y( 1- t)( 1 - a) = ( 1 - a)( 1- ty) it follows that (tx, ty) E (a, b) s. Moreover, tx+ 1-t > ty+ 1-t' a(ty+l-t) = aty + a(l-t) ~ btx + b(l-t) = b(tx+l-t) , (1-b)(l-(tx+l-t)) = (1-b)(l-x)t > ( 1 - a)( 1-y) t = ( 1 - a)( 1 -( ty+ 1 - t) ) , . s and so (tx+l-t, ty+l-t) E (a, b) • t( a, b) s U ( t( a, b) s + 1-t) for all t, 0 < t < 1. It follows that =. (a, b) 5 ((a, b) E A) This implies tAs U ( tAs + 1 - t) =- As , for 0 < t < l; and s·o by Lemma 2, (As}s = As. The next lemma asserts that the transpose of a doublystarred set is again doubly-starred. Lemma 4. Proof: s If A= A , then trA =(tr A) s tr(As) = tr(U ((a, b)s:(a, b) EA}) = U (tr(a, b)s:(a, b) EA} = U((b,a)s:(a,b) EA} 34 = U ((b, a) s: (b, a) E tr A} = (tr A) s • s . s Since A= A, it follows that trA= (trA). Under the same assumption made in Lemma 4, it can be - s shown that A = (A) • Lemma 5. Proof: tA c A. A= A s -s implies A= (A) • tA c A. By Lemma 2, Since tA = tA, it follows that Similarly tA + 1-t = (tA+l-t) c A - s so that by Lemma 2, A = (A) . Given any set A~ 1 , and t, 0 < t < 1, define a function 2 A(t) by A(t) = inf(y:(t, y) E A) where A(T) is defined to be equal 1, if (y:(t, y) E A) = ¢. Lemma 6 . s H A= A , then the functions A(t) and trA(t) ~ continuous on (0, 1), non-decreasing on [O, l] , and ( 3. 1) A( t) It , tr A( t) It , ( l -A( t)) I ( l - t) , ( 1 - tr A( t)) I ( 1-t) , are all non-decreasing on (O, l). Proof: - Given 0 < t < 1, it follows that (t, A(t)) E A. -s Lemma 5, A= (A) , so that for 0 <a.:::_ 1, (a.t, a.A(t)) EA and By 35 . (at+l-a, aA(t)+l-a.) EA. Then A(a. t) _:: a A( t) and A(a t+ 1-a.) _:: a.A( t) + 1-a. • These inequalities imply A(a. t) < A( t) ctt t and 1 - A(a.t+l-a. ) < 1 - A(t) 1 - (at+ 1-a) - 1 - t ' which shows that A(t) It and ( 1-A(t)) I( 1-t) are non-decreasing on (O, 1). Since tr A = (tr A) s, it follows that tr A(t) It and ( 1-tr A(t)) I( 1-t} are nondecreasing on (0, 1). If 0 < x < y, then A(x) < ~ A(y) < A(y) , - y so that A(t) is non-decreasing. The same is true for trA(t). Since A= As implies (O, O) EA and (x, x) EA, it follows that A(O) = 0 and A(x) _:: x, which shows A(x) is continuous at x = O. If t > 0 and 0 < x < 1, 1 - A(x+t) l - (x+ t) > 1 - A(x) l - x so that A(x} .::_ A(x+t) _:: l :x + ( 1 -: l :x) A(x) . By the same reasoning, A(x) > A(x-t) ~ ( 1 + l:x) A(~) - l:x · 36 The s e relations show that A(x) is continuous at each 0 < x < 1, and the lemma is proved. § 2. Main Results In Chapter II, the partial ordering on cove ring classes: d (£) .:S. dT(f) for all f E J, was introduced and studie d. 8 The question arises as to what can be said in general about the behavior of d (£) and 8 dT(f) when the covering classes S and T are not necessarily comparable in this ordering. For this purpose the set is now studied. The main results concerning the dimension of functions state that the set R(S, T) is closed and doubly-st'3;rred in 1 ; and conversely, 2 any closed, doubly..: starr e d set in 1 covering classes S and T. results, 2 is of the form R(S, T) for some Apart from the inherent interest in these they take on significance when, in Chapter IV it is shown that they have direct application to the study of the dimension of closed sets. Theorem 1. Proof: s R(S, T) = (R(S,T)). Suppose (a., 13) E R(S, T). If 0 < t < 1, in J by Theorem 3 (§2,I). and Since Then the r e is f E J with then both f(x) t t 1-t and f(x) · x are 37 for every coyering class A, it follows that (ta., t13) E R(S, T) and (ta+ 1-t, tl3 + 1-t) E R(S, T). By Lemma 2, § 1, it follows that R(S, T) s =R(S,T). The set R(S, T) can be described in terms of the special functions introduced in §3, Chapter II. Lemma. where (a., 13) E 1 R(S, T) = ((a, 13):a;:: 13;:: f(a) or 13 >a;:: g(a)} , and, 2 (3. 2) f( 1) = sup f(a.) a.< 1 and ( 3. 3) g( 1) = sup g(a.) a.< 1 If f E J and d (£) =a. , a. < 1, then 8 Proof: by Theorem 2 ( § 3, II). Further, by Theorem 3 ( § 3, II), f E J and d (f) = 1 implies dT{f) ;:: sup dT( (xa.) ) , for any covering classes S 8 8 a.< 1 and T. Consequently R(S, T) ,:: ((a., 13) :a.;:: 13 ;:: f(a.) or On the other hand, suppose Then put f = (xa.) 5 /\ xl3. By Lemma 2 (§2,II) 13 ;:: a;:: g(13)} • 38 and Now, if ex. = 1 and 1 _?: 13 _?: sup dT( (xcx.)S) , let g be the function defined ex.< 1 in the proof of Theorem 3, ( § 3, II) such that d (g) = 1 and dT(g) 5 = sup dT( (xcx.) ). 5 Write h = g /\xi' and, as before, it follows that d (h) 5 =1 dT(h) = 13 This shows that [(ex., 13) :ex._?: 13 _?: f(cx.) or 13 _?:ex. > g(cx.)} ~ R(S, T) and the proof is complete. Theorem 2. Proof: R(S, T) is closed in 1 . 2 Writing R = R(S, T), R(t) the lemma above implies that = f(t) trR ( t) = g ( t) , where f and g are defined by (3. 2) and (3. 3). are continuous on [O, 1 J so that R(S, T) By .Lemma 6, §1, f and g is closed. The values that (d (f), dT(F)) may take, in general, are 5 quite unrestricted as the following theorem demonstrates. Theorem 3. ..!!_ R is any closed, doubly-starred set in 1 2 , there are covering classes S and T such that R . = R(S, T). 39 Proof: For each positive integer k, define and = Ti< V trR(x) . gk(x) Then the functions fk and gk decrease monotonically to R(t) and trR(t) respectively; and they are continuous on [0, l], since R(t) and trR(t) are continuous. Further, since gk(x) = 1 V trR(x) x "Zk x 1- R(x) 1 - x/2k ---/\ l - x 1 - x = and 1 - trR(x) 1 - x/2k = 1- x /\ 1 - x 1 - x/Zk b · · · x, it · is · c 1ear th at the functi.ons l _x eing non- d ecreasing in fk(x) and gk(x) satisfy condition (3. 1) for ·each k. Define functions as follows: vk(x) = wk(x) x f"0X' x = gk(x) c- fk(x)) 1 - x ~l - gk(x)) 1 - x 40 x =~ x = gk(x) . Let ... ' r n ' ... be an enumeration of the rational numbers in (O, 1) and write Now define sequences pk and qk as follows: (k = 1, 2, ••• ) • Since vk 2: uk 2: 1 and wk 2: zk 2: 1 , it follows that and .:: ~2k • If for some k > 1, both vk(x) = 1 and wk(x) _ 1, then it fol- lows that x = or which implies fk(x) = x. Similarly, wk(x) = 1 implies gk( x ) = x. 41 These relations in turn imply that R(x) x = x and trR(x) = x since {k= 1,2, ••• ) . 7k < x Thus taking any covering classes S, T with S - T, would give R(S, T) =R . Therefore, without loss of generality assume that for some k _;:: 1, there is a point x , 0 < x < l, such that, vk(x );:: s > 1. There is then a 0 0 0 s-1 'n eighborhood, U, of x on which vk(y) > 1 + - - > 1. Since j _;:: k 0 2 implies f.(x) < fk(x), J v.(y) J it follows that = y 1jlY1 (1-f/y)) 1- y = _Y_(l -f/y)) 1- y f /Y) y (1 - fk(y)\ ;:: 1 - y l f;k(y) ) = and so, s-1 v .(y) > 1 + - - > 1 ' J 2 for y in U and j _;:: k. Since r . is in U for infinitely many indices J. J ' it follows that there are infinitely many j for which and thus the product v fini ty. 1 ••• v. approaches infj.nity as j approaches inJ Since and since the sequence (pk) is non-increasing, it follows that pk tends to zero as k tends to infinity. Thus covering classes S and T can 42 be defined by writing: s = (pk) u (q2k) (k=l,2, .•• ) T = (pk) U (q2k- l) (k=l,2, ••. ) The next step is to show that for 0 <a. < l and and this will complete the proof. Fir st consider the function (xa.) . 8 the points pk' (k = 1, 2, .•• ) since pk E s. (k It remains to determine the V?-lues (xa.) s< q2k- l) = 1, 2, ••• ) • Sinc·e p 2 k < q 2 k- l .::_ p 2 k- l' it follows that by the remarkmade following Theorem l (§3,II). By definition, and so that vk 1-( 1-a) Uk q2k-l At 43 Using the definitions of vk and uk and substituting, yields a.fk(rk) V (.1 - (l-a.)(1-fk(rk) = rk qZk-1 >) 1-rk Thus, for t ET, if tsp a. ' log (xa.) log t k (k= 1, 2, .•• ) s< t) = Since and Since the rational numbers are everywhere dense in (0, 1), it follows that which in turn implies or 44 and thus Similarly, if rk~a., then ( 1-a.)( 1-fk(r k) ) 1 - - - - -1-r ----k Hence It is now easy to verify that and when 0 <a.< 1. so that is immediate. Indeed, when rk >a., 45 On the other hand, given e > 0, there is o > 0, such that a. :::_ r k :::_ a. + o implies Since a:::_ rk:::_a. + o, for arbitrarily large k and since fk(a.) tends to R(a.) as k tends to infinity, it follows that there is k, with a:::_ rk and, and this shows that = R(a.) • Similarly, when rk <a, so again > R(a.) • An analogous argument shows that dS( (xa.)~) = trR(a.) w hen a< 1. Since R is closed, the assertion of the theorem fo llows by the Lemma above. §3. A Characte rization of R(S, T) Let S and T be given covering classes. For each point t E T, there is associated a unique point (a(t), b(t)) in the unit square 1 • 2 This association is made as follows: Given · t, the re are points 46 SO' sl in s, defined by so = sup (s: s :::_ t, s ES) sl = inf ( s: s 2:. t;', s E S) where s s 0 1 is defined to be 1, if there are no s E S with s > t. = t°' and s 1 = tl3, where a. 2:_ 12:_ 13· Write Observe that if either a.= l or 13 = l, then both are equal 1. In this case define a(t) = b(t) = 1/2, when a.> l > 13, the equations in a(t), b(t), b(t) 1 - b( t) = 13 a(t) , = a ( 1-a( t) ) have the solution a(t) = a-1 a-13 ' b(t) = 13(a- l) t) a-1 a-13 Altogether then, a(t) = l 2' a. = l ; a( =a.-13' a> 1 (3. 4) b ( t) = ~ • 13 .= 1 ; b ( t) = 13~~-al) • 13 < 1 • Note that a(t) > 0 for all t E T and that b(t) > 0 for t E T sufficiently small. Also, if 13 < 1 then both 13(a- l) < 1 a-13 and a-1 -- < 1 • a.-13 47 Thus, in any event a(t) < l and b(t) < 1, with a(t) ~ b(t), for all t E T. It follows that = tb ( t) I a( t) sl and = t( 1-b(t)) /( 1-a(t)) so Define An(S:T) = {ca(t). b(t) ):t E T. t:::. ~} • and finally write, A(S:T) = (3. 5) n n=l A (S:T) , n where A denotes the closure of A. In a similar manner, there is .associated with each point s ES, a point (c(s), d(s)) in 1 , such that 2 t 0 =s 1-c(s) c(s) l-d(s) s d\51 ' .. where t , t are the analogues of s , s above. 1 0 0 1 A(T:S) = (3. 6) Then the set II A (T:S) n n=l can be defined, where An(T:S) = {(c(s), d{s) ):s E S, 's,::: Since An(S:T) ~} • ? An+l(S:T) and each of these sets is non- empty, it follows that the sets A(T:S) and A(S: T) are non-empty. The interest in the sets A(S:T) and A(T:S) is that while they are defined solely in terms of the points of S and T, the smallest doubly-starred 48 set containing A(S: T) U A( T:S) is precisely R(S, T), as the following results make clear. . (A(S: T) U A(T:S) f Theorem 1. Proof: = R(S, T) • Since R(S, T) is doubly-starred, it is sufficient to show that = A(S: T) U A( T:S) R(S, T) • The proof demonstrates only that A(S: T) ::; R(S, T) , the proof that A( T:S) :=; R(S, T) being similar. Hence let co n A (S:T), and 0 < a< 1. Then there is a decreasing sen=l n quence (t ) c T, such that t -+ 0 and (a, b) E n - n a n = a(t ) ... a n (n -+co) b Write s x 0 ::::. tn::::. s <t <s n- n- 1 1 n n a 1-a and put x n = s 1 s , then for tn satisfying 0 , ab /a = tn ' n and for s • = b(t ) ... b n < t < x , 0 - n- n a-1 (x~S(tn) = to t n 1-b n (a-1) 1-a + l n = tn 49 Thus in any case ab ( 1-b ) n 1 + (a-1) (l-a) n /\ t ab n a = tn n n n ( 1-bn) V 1 + (a-1) {1-a) n and so ab log t a n 1-b n V 1 + (a-1) _ _ n 1-a n n Since b(t ) ... b and a(t ) ... a as n ... co, it follows that n n which shows (a, b) E R(S, T). Now, if a= 0 .and (a, b) E A(S, T), b = 0, and (0, 0) is clearly in R(S, T). is clearly in R(S, T). large, then (a, b) The only remaining case is a = 1, b < 1. let a(t ) ... a, b{t ) ... b n If a= 1 and b = 1, n t n E T, t a(tn) < 1 for the same n. n -+ O. n Let a.< 1 and consider (xa.)S(tn). and s< tn) a. b n 1-b ) ------< V 1 - ( 1-a.) ( l-a: a log t · n n Again Since b(t ) < 1 for n sufficiently above log {xa.) then As 50 Since a(t ) ..... 1, it follows that n log (xa.) s< tn) a bn - - - - - - < - - .... ab, log t a n n and the re fore dT( (xa.) ) ~ab, for all a < 1. 5 This implies By the lemma of §2 it follows that (a, b) E R(S, T}. Theorem 2. Proof: R(S, T) ~ (A(S: T) U A( T:S)) s Let f E :J with d (f) ~ dT(f). 5 If actually d (f) 5 dT(f) then (d (£), dT(f)) E (A(S: T) U A(T:S) )s automatically. 5 = Thus assume d (f) > dT(f). and let a, 13 be such that 8 By Lemma 1, (§1,I) for any t ET, s ES, f(t) ~ (1 V !) f(s) , and so, Since a< d (£), it follows by the Theorem of §1, Chapter II, that 8 ( t E T) ( s E S) • Since 13 > dT(f) there is a set T 0 ~ T, T 0 a covering class , such that t E T 0, implies 51 t-13f(t) > M a. • Thus for s E S and t E T , 0 Choosing t E T , and s (t) _::: t_::: s (t). (si(t) ES) 1 0 0 (i = 0, 1) , it follows that and Since Sl(t) -- tb{t)/a(t) and so(t) -- tl-b(t)/1-a(t). t h ese equa 1ities imply that -13 +a. b( t) affj l<t and 1-13 + (a.-1) (1-b(t)j 1-a(ijj . 1 < t Since t _::: 1, it follows that b(t) < ]. anr - a. 1- 13 and 1-b( t) T=a _::: 1- a( t) Let (a , b) E A(S, T ) ~ A(S, _T ) and let tn be a 'sequence in T such that 0 0 a(t ) ... a, n b(t ) ... b • n Then it follows that ~ < ]. a a. and 1-13 1-b < 1-a. - 1-a 52 which implies that (a., 13) E A(S, T ) 0 s ~ A(S:T) s s ::; (A(S:T) U A(T:S)) . Since (a., 13) can be chosen arbitrarily close to (d (£), dT(f)) it follows 5 that (d (f), dT(f)) E (A(S: T) U A(T:S)) s, this latter set being closed by 8 Lemma 1, § 1. The case d (£) < dT(f) is treated similarly, and so the 8 proof is complete. Theorems l and 2 combine to give Theorem 3. R(S, T) = (A(S: T) U A(T:S)) s • 53 CHAPTER IV APPLICATIONS TO CLOSED SETS § 1. Introduction and Preliminaries Hausdorff [11] introduced the following outer measure on the subsets of the real line. H (E) p If 0 ~ p ~ 1, = sup (inf e>O (6 t( lj) P: t( 1.) < J ~) J where ( 1.) is any countable callee tion of open intervals containing E, J and t( 1.) denotes the length of 1.. The set function H (E) is called the J p J p-th dimensional Hausdorff measure of E. For an elementary discus- sion of the properties of H p (E) see Halmos . [ 2] or Munroe [ 7] . The interest in the outer measure H the fact that given any closed set E, p for this work lies in there is precisely one value p, 0 < p < 1 for which r < p implies H (E) = CD, and r > p implies - - H (E) = O. r r The value of H (E) may be any non-negative real number p or CD (for a proof of these facts, see Hurewitz [4]). p is called the Hausdorff dimension of E This unique value and is denoted d 8 (E). A slightly different, but more useful description of the Hausdorff dimension of a closed set E is given by the following: Lemma 1. For each 0 ~ p ~ 1, define :X. (E) = inf p {6 ~ E} • t ( 1.) p: U 1. J J - 54 Proof: H (E) > O. p !£ A. (E) > 0, p then since A. (E) < H (E) it follows that p p On the other hand, if H (E) > 0, then there are . 6> 0, b > 0, p such that if (J.} is any sequence of intervals containing E 1 such that .t(J.) < o, then 1 - 6 .t(J.)p >b > 0. J - For any other cov:ering ( lk) of E by intervals, Thus A. (E) > 0. p Hence the dimension ~(E) can be defined by ~(E) = sup (p:A.p(E) > 0) • This notion, of course, depends upon the particular kind of covering class used to cover the set E. A study is now made of this dependence when the intervals belong to a given class C. Further, this class C is assumed to be closed under translations, that is, all real t. 1 E C => l+t E C for Thus whether or not a given interval 1 belongs to C de- pends only on .t( 1). Moreover it will be assumed that C contains intervals of arbitrarily small length, and that .l( 1) ~ 1 if 1 E C. class C will be called a covering class, S( C) = S = ( .l( 1): 1 E C} is a covering class in the sense of § 1, II. functions, Observe that the set Then given C new set Such a 55 A.p,C(E) can be defined. H p, = in£(6 £,(lj)P: u lj;;) E, lj E c) , If Hp, C (E) is defined by C(E) = sup (in£(6 (.t(l.) )p:U 1. :? E, .t(l .) < e, 1. E e> 0 ~ J J J J c\)l , then the proof of Lemma 1, shows that: Lemma 2. Since A. p, H p, C(E) = 0 if and only if A. p, C(E) = O. C(E) is clearly a non-increasing function of p it follows that there is precis:ely one 0 :::_ q :::_ 1 such that r > q implies H r, C(e) = 0 and r < q implies H r, C(E) = oo. This value q will be called the Hausdorff dimension of E with respect to the covering class C (or S(C)) and is denoted dC(E). Remark: Thus it follows that To avoid confusion, the notation d (E), 5 A.p, 5 (E) will be used where S refers to a covering class in the sense of § 1, Chapter II. A final lemma is needed for later considerations. L e mma 3. If E 1 and E 2 are any closed sets in [O, 1 J and S any covering class, then Proof: so that For each p, the set function A. p, 5 (E) is monotone, 56 (i = 1, 2) A p, s<E 1 u E 2) > A p, s<E.) l • It follows then that On the other hand, assume 0 < d (E U E 2), 1 8 and take 0 < p < d (E U E ). 2 8 1 Thus Since A.p,S is sub-additive, either A.p,S(E 1) > 0 or A.p,S(E 2) > O. In any event, which shows If d (E 1 U E 2 ) = 0, 8 §2. there is nothing to prove. Frostman's Theorem Frostman [io] proved that the Hausdorff dimension of a closed set E.::, [O, l], has the property that if 0 < a.< ~(E), there exists a function µ(x) defined on (-co, co) which satisfies the following conditions: µ.(x) .is non-decr e asing, µ(x) = 0 when x < 0, µ(x) = 1 whe n x > 1, and if (a, b) is any open interval not intersecting E, µ.(a) = µ(b). then Further µ(x) satisfies a Lipschitz condition of order a. at each point of [O, 1 ]. In this section, a generalization of this result to dimension with respect to a covering class is given. If µis a non-decreasing function defined on (-co, co), support of µ. is defined to be the closed set, the 57 g µ. = (U((a, b):µ.(a) = µ.(b)}) c , where xc denotes the compliment of x with respect to (-co, co). Let M(E) denote the collection of all non-decreasing functions µ. defined on (-co,co) with g < E and µ.(x) = 0 for x < 0, µ.(x) = l for x > 1. µ.- - - Obs erve that if µ. E M(E) for some closed set E, in § 2 of Chapter I. then µ. E M, Indeed, the requirements µ.(O) = 0, and 6µ. bounded on [O, l] are all satisfied. defined µ. non-decreasing With this observation the following fundamental theorem can be stated. Theorem. Proof: Given any closed set · E, and any covering class S, First suppose µ. E M(E), and consider dS(µ.). By the Theorem of § 1, Chapter II, ds(µ.) =sup (13:s- 13 6µ.(s) = 0(1), s ES). Let 13 < dS(µ.). Then there is a finite constant M, :::_ sl'M, for all s ES. t( lk) E S, such that 6µ. (s) If ( lk} is a covering of E by intervals such that then Since E is compact, a finite number of the intervals lk, (k = 1, 2, ..• ) say (a.,b.) (j = 1,2, ••• ,n). If l = (a,b), then 6µ.(.t(l)) J J > µ.(b) - µ.(a). Since g c E and µ.( 1) = 1, it follows that µ.cover E, 58 1 n ~ .t(lk)l3 > M ~ µ.(b .) - µ.(a.) > 1 • - Consequently A. 13, J j=l J - S(E) > 0, and it follows that Since 13 was arbitrary < <ls(µ.), this shows that <ls(µ.) ~ dS(E) or that To show the reverse inequality, suppose that d (E) > 0. 8 there would be nothing to prove. Let 0 < 13 < d (E). 8 If d (E) 8 = 0, Then O < A. , (E) < oo, and it can be assumed that A. , (E) = 1, multiplying by 13 8 13 8 an appropriate constant, if necessary. Put µ.(x) = A.l3,S(E n [O,x]) .if x>O µ.(x) = 0 if x < 0 Then µ.(x) is non-:decreasing since A. , S is monotone. 13 Since A. l3, S is · a sub-additive set function and since ds(A) = 0, if A is a finite set, it follows that µ.( 0) = 0 and for x > 1 µ.(x) = A.l3,S(E n [O,x] < A.l3,S(E) + A.l3,S(E n [l,x] 1 = A. l3, 8 (E) = 1 • = 1 when x > 1. More- over, if (a, b) is any open interval not intersecting E, then E n [a, b] The fact that µ is non-decr e asing implies µ.(x) consists of at most two points so that µ.(b) = A. 13 ,s(E n [O,b] < A. 13 , 5 (E n [O,a]) + A.13 , 5 (E n [a,b]) = µ.(a) • 59 This implies that g c E and hence µ. E M(E). µ. - µ.(x+s) = A.S, S (E n [O, x+s] Now, r' J ~ A- 13 ,s(E n [o,x]J + A.s,s(<E n [x,x+s]) Since A.!3, s(E n [x, x+s ~ ~ sl', it follows that for every x, µ.(x+s) - µ.(x) and so t.µ.(s) < s 13. ~ s a, This implies that and so sup (d (µ.):µ. E M(E)) > 13. 8 Since S was arbitrary <a., it follows that and the proof of the theorem is complete. Remark 1: To obtain Frostrnan's result from the above theorem, take S = (O, 1 ]. 0 < 13 < dS(E), Then given E with d (E) > 0 and 8 there is µ. E M(E) such that d (µ.) > 5 of § 1, Chapter II, s -13 f:lµ.(s) = 0( 1) for s E (0, 1 ]. 13. By the Theorem It follows that µ. satisfies a Lipschitz condition of order 13 at each point x, Remark 2: It follows immediately from this theorem that S < T implies d (E) ~ dT(E) for every closed s~t, where 8 11 ~11 is the partial ordering introduced in §4 of Chapter II. · Indeed, since S < T 60 implies d (f) < dT(f) for every f E J, 5 the same is true for every g E M It follows that d (E) .S, dT(E). 5 and thus for every µ. E M(E). The con- verse is also true, that is, if d (E) < dT(E) for every closed set E in 5 1 , then S < T. 1 This fact will follow from the results in §4 of the present chapter, §3. Generalized Cantor Sets It will be useful to distinguish a certain class of closed sets in 1 . 1 Let (~) be a given sequence of positive integers, whose k-th k -1 < oo, partial products Nk = n 1 , .•• , nk satisfy, "'"' Nk O Define = X( 0, 1, ... , nk -1} , k where the set (0,1, ... ,~_ } has the discrete topology and o the topo- 1 logy of pointwise convergence, Kelley [ 6].) (For definiti_ons and notations see In this topology O is compact and satisfies the first axiom of countability. The space O is totally ordered by the lexico- graphic ordering: If a= (a.) and b = (b.) are distinct points in o, 1 1 a< b provided that if p is the smallest integer for which a p =!. b , then p Define a mapping cp:O ..... [O, 1 J by "<;"""' - cp{a) = L.J a.N. ( 4. 1) i 1 1 1 • Then cp has the following properties: Theorem 1, a) If a, b, E O, and a< b or a= b, cp(a).::, cp(b). b) For all a E O, 0 .S, cp(a) .S, 1, then then 61 Proof: a) c) cp is onto. d) cp is continuous. Let a, b E o. If a= b, hence suppose that a< b. i= 1, •.. , k-1, and ~ < bk. then clearly cp(a) = cp(b), Then there exists k such that a. = b., l l In this case, ~ -1 cp (b) - cp(a) = LJ b.l N.l ~ -1 - LJ a. N. l -1 > Nk - 6 -1 = Nk - 6 l -1 n.N. j>k J J -1 N. j>k J + 6 -1 -1 N. + 6 N J. = 0 , J j >k j~k and so cp(a) 2 cp (b) . b) Clearly cp(a) ~ 0 for all a E O. Further, since a E O , implies a< w = (n. -1), it follows from part a) that l ~a) and Let a l l Now let 0 2 x 2 1, and define a = (ai ) in 0 as fol- c) lows. ~ -1 _< cp(w) = LJ (n. - l)N. 1 be the greatest integer n, satisfying : 62 If a 1 ,a , ••• ,ak-l have already been defined, choose ak to be the 2 greatest integer n satisfying: and k-1 1 6 a Nr=l r r -1 + n Nk < x • The point a = (a .) so defined clearly satisfies 1 cp(a) ~ x • Suppose that cp(a) < x, then two cases arise. integers j, a . < n.-1, then choose j large enough so that J J I and put a. = a.+ l. J J If for infinitely many Then, j-1 I -1 -1 -1 a N + a.N . < cp(a) + N . r r J J J r=l 6 < cp(a) + < x x-~(a) 2 ' which would contradict the choice of a . made above. J suppose there exists an integer r ak Moreover a. J- 1 <n. J- 1 such that for all k ~ r, =~ - 1 • Let j be the least such integer r. j > l. On the other hand, -1. Since cp(a) <x< 1, it follows that Thus 63 j-2 <P (a) = 6 r=l -1 a N r r j-2 = 6 r=l -1 a N r r j-2 = 6 r=l -1 a N r r -1 -1 + 6 . N. + a J1 J r ~j -1 + a.J- 1N J. (n -1) N r r -1 + N.J + (a.J- l + 1) N J. < x ' which contradicts the choice of a. • J- 1 Thus the only possibility is <P (a) = x which was to be proved. d) to b E o. Suppose a(n) is a sequence in O such that a(n) converges k > M . implies a . (k) = b.. - L; r>k there exists M. such that Then for each integer i= 1, 2, ... , 1 1 1 Let N be such that k > N implies 1 (n -1) N-l < e and put r r M = If n _> M, then a . (n) = b . , 1 1 max (M.) 1 i<i<N i=l, ••. , N and so ""' -1 = <P (a(n)) = LJ a.1 {n) N.1 N 6 b.N~ i= 1 1 1 + 1 ""' LJ a.(n) N.-1 i>N i i Thus N 1 ~ b.N~ < cp(a(n}} i=l and since it follows that 1 1 N < ~ b.N.-1 i= 1 1 1 + e, 64 cp(b) - e < cp(a(n)) _:: cp(b) + e . Since o is first countable, this shows that cp is continuous on o. Assume that the sequence of positive integers (~) is fixed. Given positive sequences (ak) and (sk), the triple (nk, O'k' sk) is said to be admissi .ble, provided that (k = 1 , 2, .•. ) , and (4. 2) (k= 2,3, ..• ) . Given an admissible triple (nk' O'k' sk) and a point a E O, sum n ~ a .s . . j= 1 J J By (4. 2) and the fact that a. < n. 1 - n n n ~ a .s. < ~ (n. -1) s . = ~ n .s. j= 1 J J j= 1 J J j= 1 J J n ~ < O' . 1 j=l J- 1- 1 consider the , n ~ - j= 1 S·J n - ~ O' . j= 1 J =1-a<l. nThus the series ~ a.i;. is convergent, and d e fines a number v(a). 1 1 function v:o _, [O, 1] so defined is called the derive d mapping of (~, O'k' sk). Such a function ~ satisfies: Theorem 2. ~(a) < ~(b). a) a< b, if and only if, b) 0 _:: ~(a) _:: 1, for all a E O. c) ~ is continuous. The 65 Proof: i=l, 2, ••• , k-1, a) If a, b E O and a< b, and ak <bk. let k be such that a . = b., 1 1 Then ijr(b) -v(a) =L)b.t;. -L)a. i;. 1 = 1 1 ~ {b . -a .) s. j~k J J J > sk > sk - ~ j~k+l .6 j~ k+ 1 = sk - ak > 0 Therefore w(b) > w(a). b) {n. -1) s. J J a k-1 .6 + j~k+l a. J . It follows that ijr(b) > ~(a) implies b It is clear that ijr(a) ~ 0. the Theorem established that c) 1 > a. The remarks made preceeding ~(a) ::_ 1. Since .6 (n. -1) s . is convergent, the same argume nt as 1 1 that used to show the continuity of <P suffic e s. A set E is called the G e n e ralized Cantor Se t of type (~,a k' sk) , or more briefly, the GCS of type (nk' ak' sk), provide d that (~, ak' sk) is admissible and that E = ijr( O), where ~ i s the de rived mapping of (nk' ak' sk). Since ijr is continuous and O compact, the following theorem is immediate. Theore m 3. If E is a GCF of type (~, ak' sk), then E is a closed subset of [O, 1 J. Given an admissible triple (~, ak' sk), µ.(x) on (-00,00) by writing: define a function 66 µ.{x) =0 , if x < 0 = sup (cp (b): v(b) :s x) , µ.{x) if x > 0 where cp is defined by ( 4. 1) and * is the de rived mapping of (~, O'k' sk). The function µ. is called the Generalized Cantor Function of type (~, O'k' sk), or the GCF of type (nk, O'k' sk). The following lemma will be useful for the study of Generalized Cantor Functions. Lemma 1. Proof: If x = *(a), then· µ. (x) = cp (a) . If x = w(a), then by definition µ.(x) 2::. cp (a). Since cp(b) > cp{a) only if b >a, and µ.(x) > cp{a) implies the existence of b E O such that cp (b) > cp (a) and v(b) :S x, it follows from Theorem 2, a) that µ.(x) :S cp (a). The first general statement about Generalized Cantor functions is: Theore m 4. If µ. is a GCF of type (nk, O'k' sk), µ. E M( Proof: Suppose x :Sy. it follows that µ.(x) :S µ.(y). then *( O) ) • Then v(b) :S x implies *(b) < y, and Since a< w for a E O, µ.(x) :S cp(w) = 1. The fact that v(w) = :E (n.1 - 1) s.1 -< 1 , and cp (w) = 1, implies that µ.(x) = 1 when x 2::, 1. Now write E suppose that the open interval (x, y) is disjoint from E. If y = w(O) and f- E, then 67 it is clear that (x,y) n E = ¢ implies = sup(cp(c):~(c).:::_x) = sup(cp(c):~(c).:::_y) = µ.(y). µ.(x) Suppose then that y E E. such that x If also x EE, = V(a), y = iV(b). then there exist points a, b E O Then by Lemma 1, µ.(x) If cp(a) < cp(b) then there would be c E O , = cp(a), µ.(y) = cp(b). such that cp (a) < cp( c) < cp(b) , since cp is onto. This would mean a< c < b, and so iV (a) < ~(c) < ~ (b) which would contradict En (x,y) = ¢. µ.(x) = µ.(y). Thus in the case x,y E E, Finally, if x r/. E and x .:::_ 0, µ.(x) since 0 EE. If x > 0, then y .:::_ 0 and = µ.( O) = µ.( y) , then x ~ sup (t:t EE, t,:::. x) since E is closed. Further (t , y) n E 1 µ.(y) and so µ.(x) = t 1 EE = ¢ and so = µ.(t 1) ,:::. µ.(x) , = µ. (y), which shows that µ. E M( i¥(0)). The next theorem gives an alternate definition of Theorem 5. (4.3) µ.(x)~ For all x, µ.(x) = inf(cp(c):v(c)~x), where the infimum is d efine d to be 1 in the case that there is no c E O such that w(c) ~ x. 68 Proof: µ.(x) If x is such that x > *(c) for all c in o, = 1, which agrees with (4. 3). If x< 0, then then inf(cp(c):*(c) _2: x) = 0 , so that µ.(x) = inf (cp .(c) : ~(c) > x) in this case. and that there is c E o for which > x. ~(c) Thus suppose that 0 < x Again write E = *(O) and let t 0 = sup (t: t E E, t ~ x) .t l = inf ( t: t E E , t ,2: x) • Since E is closed, t , t belong to E. 0 1 Then there are points a, b E O such that If µ.(x) < cp(b), ~(a) = t0 < x ~(b) = tl > x • then cp(a) .:S, µ.(x) < cp(b), and so there would be c E o with c.o(a) < cp(c) < cp(b) or a< c < b. This would imply w(a) < *(c) < *(b) which contradicts the choice of t , t • 0 1 µ(x) .2: inf(cp(c):w(c) .2: x). Thus µ.(x) = ~b) which implies The fact that µ.(x) .:s, inf(cp(c):*(c) .2: x) follows from the definition of µ.. The foregoing proof also establishes: Lemma 2. that ~(b) ~ x Given 0 < x and µ.(x) < 1, there is b E o such and µ.(x) = cp(b). The following lemma will be instrumental in the proof of Theor em 6 below. 69 Given a and b in ,....,, ~• Lemma 3. there is a sequence ( E:.) -K where ek = 0 2!. ek = 1 (k= 1, 2, ••• ), such that the point c = (ci) defined (i = 1, 2' ..• ) ' belongs too. Proof: a, b E o. proved as follows. The existance of the required (E:k) is Define CXl A= X (0,1}, k=l where ( 0, 1} has the discrete topology and A, the topology of pointwise convergence relative to the compact spaces ( 0, 1} . For each positive integer k define A. = ((e .):O <a . + b . + e.+l - E: .n . < n . -1; -l<. J J J J J JJ Thus ~ j = 1, ... 'k} . ~ is closed in A for each k, since if (ekn)) is a sequence in converting to (ok) then there exists M such that n > M implies ( n) = ok+ , and so ( uk I'. ). • • A k. e (1n) = o l' •.. , ek+ is in 1 1 F ur ther i· t is · o b v1ous · that ~ 2~+ 1 for all k, and that ~ is non-empty for each k. Sinc e A is compact and the sets ~ have the finite intersection property, it CXl follows that n ~ is non-empty. Hence there is a sequence ( ek) which k=l satisfies the required conditions. Remark: by the lemma above. Let a, b be points of O and c a point determined Then 70 cp(a) + cp(b) = cp(c) + e 1 • Moreover, if v is the derived mapping of the admissible triple v(c) .2: *(a) + *(b) - €1 • Indeed, if there exists (ek) such that c = (ci) defined by (i=l,2, ••• ) is in O, then "" -1 cp( c) = L.J N.1 (a.1 + b 1. + e 1. + 1 - e 1. n.) 1 = cp(a) + cp(b) - e 1 • Moreover, s. s. s. s. - 6 e. n. s. 1f (c) = 6 c 1 . = 6 a. + 6 b 1. 1 + 6 e 1. + 1 1 1 11 = *(a) + *(b) - Theor e m 6. 111 e1 . If µ(x) is a GCF of type (~, O'k' sk), then µ.(x) is sub-additive. Proof: If either x < 0 or y < 0, µ.(x+y) :::_ µ.(xVy) :::_ µ.(x) + µ(y). then x+y :::_ xVy, so that If either µ(x) = l or µ(y) = 1, clearly µ.(x + y) < µ(x) + µ.(y) • then 71 y .:::_ 0, µ.(x) < 1 and µ.(y) < 1. Thus suppose that x .:::_ 0, By Lemma 2, there are points a, b E O such that v(a) .:::. x' and µ.(x) = <P(a) , µ.(y) = cp (b) • According to Lemma 3, and the remark following it, there is c E O such that cp(c) = <P(a) + cp(b) - e 1 and If e (e 1 = 0, then v(c) .:::_ x+y. 1 = 0 or 1) • By Theorem 5, this implies µ.(x+y) .S, cp( c) = cp( a) + cp(b) = µ.(x) + µ.( y) • On the other hand, if e 1 = 1, then µ.(x+y) < µ.('¥(a) + ~(b)) < µ.( v(c) + 1) < 1 + µ. ( v(c)) . Since µ.( ~(c)) = cp(c) = cp(a) + cp(b) - 1 , by Lemma 1, it follows that µ.(x+y) < cp(a) + cp(b) Hence in all cases µ.(x+y) .:::, µ.(x) + µ.(y), = µ.(x) + µ.(y) . which proves the theor em. It will now be shown that a GCS, E, of type (nk' O'k' sk) and its GCF, µ. have the same dimension with respect to any covering class, but fir st some useful lemmas. 72 For each positive integer k, define a set Bk by Bk = (b:b E O, bj = 0, j > k} • Then Bk has exactly Nk points and has the following property: Lemma 4. .!f {nk' O'k' sk) is admissible and ~ its derived mapping, then {k= 1,2, ••• ). Proof: Let c E O. b{c) = Since Bk is finite, let max {b:b < c) bEBk - this maximum being taken with respect to th.e lexicographic ordering on o. Then \f{c) ~ ~{b{c)) . k ~ {c) > I; b . {c) s . + ~+ 1sk+ 1 j= 1 > since J k ~ b.{c)i;. + ~ {n.-1) S · , j= 1 J J j >k J J I; { n. - 1) s . .:S, n.+ 1s l + 1 . j>k J would follow that J If b . {c) J = n. - 1 £or 1 < j < k, then it J K ~{c) > 1, which is false. <: integer r < k such that b {c) < n -1, r r J - - Thus there is a largest and so 73 ~(c) > r-1 6 j= 1 b.(c)s . + b (c)s J J r r + 6 j >r (n.-l)s . J J r-1 > 6 j= 1 b . (c)s. + (b (c) + l)s . · J J r r Then there would be a point b' in Bk with b'. = b . (c) (i=l, ..• ,r-1), l b I r l = b ( c) + 1, and r b(c)<b'<c, which contradicts the choice of b(c). ~{b(c)) < Consequently ~(c) ~ ~(b(c)) + ~sk and the lemma is established. With the help of Lemma 4, it is easy to verify L e mma 5. Proof: and sk = ~( (o jk) ). by Lemma 4. Let ojk = 0, if j -J. k, = 1 if j = k. Then (ojk) E O By Lemma 1, Since µ. is non-decreasing and µ.( 1) and so for some t E Bk, µ.(t + O'k) -1 µ.(t) ~ Nk = 1, it follows that Since µ. is sub-additive, 74 Lemma 6. Proof: For all s > 0, s ) Nk -1 • ~(s) .:::_ 2 ( l V Sk By Lemma l (§ l, I), Theorem 6, and Lemma 5, The next theorem is a partial answer to the question: what sets E is it true that there is µ E M(E) such that d (µ) 8 For = d 8 (E) for all S? If µ is the GCF of type (~, O'k' sk) and E the Theorem 7. GCS of the same type, then for every covering class S. Proof: E If w is the derived mapping of (nk' O'k' sk) I = *(O) by definition and µ E M(E) by Theorem 4. of §2, it follows that d (µ.) :::_ d (E). 5 s E S with O'k .:::_ s .:::_ O'k- l' 5 then From the Theorem On the other hand, if A. E M(E) and then c U [t,t+s], - tE ~(Bk) 75 by Lemma 4. Then 1 < I; A. ( t + s) - A. ( t) , t E v(Bk) and it follows that it follows that On the othe r hand, if O'k ::_ s ::_ nksk' then (4. 4) and Lemma 1 ( § 1, I) imply from which it follows that Altogether then, for O'k S. s S. O'k- l s Nk - 1 V Nk - 1) /\ ( µ.(s) V Nk - 1) > 21 ( sk ~ 1 4 µ.(s) /\ µ.(s) = 1 4 µ.(s), . 76 by Lemma 6 and the fact that µ.(s) ,:::. µ.( O'k) It follows that for all s E S, log tiA.(s) < log 1/4 + log µ. (s) log s log s log s The next result is concerned with the computation of d (µ.), 5 where µ. is a GCF of type (~, O'kSk), in terms of some of the concepts introduced in §3 of Chapter III. Given a covering class S and a pos- itive sequence (ak) decreasing to 0, recall the definition of a(ak) and b{ak) relative to S, given by ( 3. 4). If µ. is a GCF of type (nk, O'k' sk) where ~sk Theorem 8. and if S is any covering class, then ( 4. 5) d ( µ,) 5 log µ(ak)~ ( 1 log µ.(crk) 1 ) I\ lim inf + 1 - uk 1og O'k k-+ro vk log O'k vk ' = lim inf( k-+ro Proof: 1 Since ( O':k) and ( O':k) are contained in S and are covering classes, it follows from Theorem 1 (§2, II) and Lemma 1 (§4 , II), that 77 . and thus . . ( 1 log µ.(ak)) d(au.k ~. (µ.) < hm inf l<.J - k -+ ex> Uk 1og (J k vk Moreover, the fact that (]k .::;_ (]k and the sub-additivity of µ. imply that by Lemma 1, (§ 1, I). Thus which implies that It follows that, By Lemma 6 and the fact that ~gk = O'k- l' (4. 6) Suppose now that s ES and ak .::;_ s .::;_ (]k- l" Then necessarily 78 Uk Vk-1 O'k < s .:::_ O'k- l , and this with ( 4. 6) implies log µ.(s) > log 2 + (log log s log s > + log O'k 1 log s log 2 ( 4. 7) N~ /\ log \ s N~: 1 /O'k- 1)) log s 1 17 log Nk-1 k-1)) (~log µ(cr k) /\ ( 1 --log O'k Uk ( 1 log µ (crk) = o( 1) + uk log ak vk-1 ( 1 /\-vk-l log Ok-1 log µ(crk_ 1) log a k-l l )) +1--vk-l If s E S and k{s) denotes the integer since k such that O'k < s .:::_ O'k- l' then k{s) ... CXl as s ... O. Thus ( 4. 7) implies that Since lim inf (Ak /\ Bk) = lim inf ~ /\ lim inf Bk' equality in ( 4. 5) is proved. The last theorem of this section asserts essentially that if -1 log Nk <\r(E) = lim inf - - - k ... CXl Theorem 9. log O'k If µ.' is a GCF of type (~, aksk), with nksk = O'k-l' ~ d 8 (µ.) ~ d(a )(µ.) for every covering class S. k 79 Proof: By Lemma 6, s ) Nk -1 µ.(s) :::_ 2 1 V i;k ( then for every 0 < 13 < d(a ) (µ.) If ak ::_ s ::_ ak- l' , k < 2C l3 , where c is a constant inde pendent of k. By the Theorem of§ 1, 13 Chapter II, it follows that 13 ::_ d (µ.), and so d(a )(µ.) :::_ d (µ.). Note that 8 8 if d(ak)(µ.) = 0, the result is trivial. Remark: k Theorem 9 can also be proved as a consequence of Theorem 8. § 4. Main Results The purpose of this section is to show that the r e sults of § 2 , Chapte r III apply to the dimension of closed sets with respect to a covering class. More precisely, let C be the collection of closed sub- s e ts of [ 0, 1 J and where S and T are any covering classes. is identical with R(S, T). It will be shown that M(S, T) To show M(S, T) ~ R(S, T) is relatively e asy: 80 The orem 1. Proof: M{S, T) ~ R{S, T) Let E be a closed set such that There exists µ. E M{E) such that by the Theorem of § 2. Since µ. E M, there is £ E 3' such that £,...., µ., that is d A{f) = d A{µ.) for all A, by Theorem 1 {§ 2, II). Now dT(E) f A x E 3', and the lemma of § 2, Chapte r II implie s that and since µ. E M(E). It follows that and, since d (µ.) can b e chosen arbitrarily close to d {E), and since 5 5 R(S, T) is closed , this implie s The case d {E) < dT{E) is treated similarly. 8 If d {E) 8 = dT{E) , then the fact that R{S, T) is doubly-starred shows that {d {E), dT(E)) 5 E R(S, T), and this completes the proof. 81 Theorem 2. class S, Given f3 such that 0 < f3 < 1, and any covering there is a GCF, µ. of type (~, ak' sk), with nksk = ak- l, (a k) :;: S and Proof: Let (ak) :;: S be a decreasing sequence satisfying: (k Write n1 = = =0 1 2 J = ak- l /~. ~ak < ak- l' For J ••• ) • '[a i 13J+ 1 ' -~~13 /Nk-1]_ + 1 J (k = 2, 3, ... ) J where [x] denotes the greatest integer less than or equal x. put sk J Further, In order to show that ak < sk' it suffices to show Indeed, for k = 1, k~ 2, Nk-l = Nk_ 2 ~-l > Nk_ 2 a~~ 1 /Nk-Z = a~~l' and it follows that ~ak 5_ ( a~l3 /Nk-1 + i) ak 5_ ( a~l3 la~~ l + i) ak 5. (2 a~-13 la~=~) Dk-1 < ak-1 . -13 and_"LJ ak 13 is a convergent series, it follows that "LJ Nk -1 Since Nk ~ ak is convergent. It follows that there exists a function µ. which is a GCF of type (~, ak' sk). Since 82 -1 log Nk = lim inf - - - k ..... m log O'k and Nk ~ O'~~ , it follows that On the other hand, given k sufficiently large, there are integers j ::_ k for which n. > 2. J- Let jk be the largest of these integers. Then or since jk :S, k. Thus log 2- 1 <~+--- log O'k which implies d(O' ) ( µ.) :S, ~, or k d (µ.) ::_ ~. 5 But by Theorem 9, the theorem. Given covering classes S and T, recall the definitions of A(S:T) and A(T:S) given by (3. 5) and (3. 6) of §3, Chapter III. Theorem 3. . 0 ::_ t :S, 1, Given any (a, b) E A(S: T) U A(T:S), and any t, there are closed sets E and F, such that 83 and (ds(F), dT(F)) = (ta+ 1 - t, tb + 1 - t) . Proof: Since any finite set E has d (E) = 0, and any closed 5 interval F of positive length has d (F) = 1, for all S, only the values 5 t, 0 < t < 1 need to be considered. Assume (a, b) E A(S: T), is a sequence (t ) , t = 1, (t ) c T, 0 r r lim a(t ) = a and lim b(t ) = b. r r r-+oo r-oo Case 1. If a> b > 0, (t ) decreasing to zero such that r Four cases are considered. b < 1, and 0 < t < 1, then there is a by Theorem 9, §3, and Theorem 2 above. that ak - a, bk ... b, then the re Since (O'k) c (t ) , it follows - r (k-+ oo), where ~ = a(ak), bk = (O'k), and thus Theorem 8, §3 implies that ds(µ.) = ta /\ (tb ta+ 1. - tb) . 1-a Since ta< 1 and T:'b,:: 1, it follows that 1-a 1-a T=b ta + 1 - 1-b ~ ta , so that By the same reasoning, since 0 < t < 1 implies 0 < tb + 1 - t < 1, is a GCF, A., of type (~,a~, s~) with dT(A.} = tb + l - t. ~S~ = a~_ 1, (a~} ::; (tr), and Applying Theorem 8, §3, and reducing, implies there 84 d (A.) = ~ (tb+ 1- t) "(ta+ 1-.t) 5 = ta+l-t, since a> b. Thus taking E to be the GCS of type (nk, ak' sk) and F the GCS of type (~,a~, s~) and applying Theorem 7, §3, proves the theorem in this case. If a= b = 1, and 0 < t< 1, then there is a GCF, µ; Case II. of type (~,ak,i;k), ~sk = ak-l' and (ak)::: (tr) with dT(µ) = .d (ak)(µ) = t. Then by Theorem 9, §3, d (µ) 2: t. 5 On the other hand, by Theorem 8, §3, = d(CT ) ( µ) = t , k and so d (µ) = t. 5 In this case take E to be the GCS of type (~, ak' sk) and F any closed interval of positive length •. Case III. If a = b = 0 and 0 < t < 1, then let µ be the GCF described in Case II ,, with t replaced by 1-t. (1-ak log µ(CTk) . . d ( µ) < hm inf - S k ... 00 1-bk log CT k =d C1k (µ) + 1 1-a k - 1-bk and in this case take E to be a the GCS of type (nk' O'k, sk). Case IV. The only remaining case is a > b = 0. then tb + 1 - t > 0, and there is a GCF, µ, such that ) = 1-t . It follows that d (µ.) = 1-t c dT( µ.), 5 finite set and F Then If 0 < t < 1, of type (~, ak' sk), ~sk =C1k- l 85 dT( µ.) = d(O' ) ( µ.) = tb + l - t = l - t > 0 . k Since bk ..... o, ak ..... a>O and d(O'k)(µ.) >O, it f ollowsfromThe orem 8 , §3, that ds(µ.) c-a = liminf k ..... 00 + log µ(erk) l-bk k 1_1-akj ~ log O'k = ( 1-a) d(ak) ( µ.) + l - ( 1-a) = ta + l - t . Thus let F be the GCS of type (~, O'k' sk). set E The existance of a closed such that (t > 0) • requires special consideration. Without los.s of g e nerali t y, it can b e assume d that the s e quence (t ) is chosen so that b(t ) d e cr e ases to r b = 0, and that between t r and t r r- 1 there is a point of S, for every r. This implies that b(t ) /a(t ) t r r r < t , r- l or -tb(t ) t r r > t ta(t ) r -1 r Since a(t ) ..... a> 0 and t > 0, it follows that r -tb(t ) t ( 4. 8) r r ..... 00 r-.co • Further, a(t ) ..... a > 0 and b(t ) ..... 0 implies that r r 86 (r -+ ro) • ( 4. 9) On the basis of statements (4. 8) and (4. 9), there is a subsequence, (ak) of (tr) which has the following properties: ( 4. 10) 1 a k (bk - - t) ~ 2 Given such a sequence (ak) ::; T, define nl = ~~tbl] + 1 = ~~tbk /Nk-1] + 1 where N r . (k= 2,3, ... ) = n 1 , ••. , n r and [x] denotes the greatest integer in x, Further, put In order to find a GCF of type (~, O'k' sk), the sequences (~), (ak), (sk) must satisfy: (k=l,2, ... ) 87 Since it follows that this latter series being convergent by (4. 10). follows from the relation sk = O'k- l /1\:. that ak < sk' whenever nk"k < O'k- l" The fact that 1\:Sk::, O'k- l This same relation implies Now r (, -tb ) \_°'1 1 + 1 O'l .:::. \_°'1-112 + 1' ) 0'1 3 = O' 1/2 + O' 1 -< -4 -< 1 -- O' 0 1 by (4. 10). To verify ~ak < ak-l fork> 1, observe that Indeed, N = n -tbl 1 1 .:s_ a 1 -tb2 + 1 .:s_ a 2 by (4. 10). - tb ' k-1 Nk- l < ""' LI r=l Suppose that ar r + 1 . Then ~kk-1k + 1 Nk-1 -tb = 1\:~k-1 .:S, ( ) -tbk = O'k + Nk-1 < k 6 r=l -tb O' r r + 1 . 88 . -tbk+ 1 The choice of O'k+ 1 described by (4. 10) implies that Nk.:::, O'k+ 1 . This fact implies that by (4. 10). Thus let µ. be the GCF of type (nk' O'k' sk). As has already -tbk -tbk been observed, Nk 2: O'k . It is also true that Nk.:::, 2 O'k . Indeed, -th since Nk_ 1 .:::,ak k, it follows that -th ak k ~ .:::, 1 + N=k---1- < -th 2crk k Nk-1 and multiplying both sides of this inequality by Nk- l gives It then follows, since log µ.(O'k) ( 4. 11) th < k - log O'k log 1 /2 < log O' + tbk k 89 and, bk ... 0 implies Since (O'k).:;: T, dT(µ.) = 0 as well, By Theorem 8, §3, The estimation (4. 11), implies Since (k-+ o:i) it follows that d (µ.) 5 = ta /\a = ta. (~, O'k' sk), and in this case d 5 Hence take E (E) = ta, ' to be the GCS of type dT(E) = O. Thus cases I-N show the theorem true for (a, b) in A(S: T}. A symmetric argument establishes the result when (a, b} E A(T:S). Therefore the theorem is proved. The principal theorem concerning the dimension of closed sets is the following: Theorem 4. Proof: M(S, T) = R(S, T) . By Theorem l and Theorem 3 ( § 3, III}, it suffices to show that (A(S:T) U A(T:S)) s (x, y) E (a, b) , closed set with s .:;: M(S,T). and (c, d) E A(T:S). If · x Let (a,b) E A(S:T), = ta, 0 .:S:. t .:S:. 1, let E 1 be the 90 There exists 0.:::;, s.:::;, 1 such that either y = sd or y = sd + 1-s, and a closed set E 2 with dT{E ) = y. 2 In either case, d {E 2 ).:::, y. 5 ay ~bx, it follows that y ~ tb. By Lemma 3, § 1, d A(E = d A(E ) V d A(E ) for any A, 2 1 so that 1 Since UE ) 2 and If x =ta+ 1 - t for 0.:::;, t.:::. 1, then let E s If (x,y) E (a,b) , (1-y).:::. t(l-b). E 2 3 be the closed set such that then (1-a) (1-y).:::;, (1-b) (1-x) which implies that It follows that y _::: tb+ 1- t. defined above. Consider the set E 3 UE 2 , Then and It follows that (a, b) s:;: M{S, T) if (a, b) :;: A(S: T). ment shows that (a, b) s A symmetric argu- :;: M(S, T) when (a, b) E A(T:S). prove that (A(S:T) U A(T:S) )s:;: M(S, T) , and hence R(S, T).::; M(S, T), which proves the the orem. These f acts 91 § 5. Conclusions and G e n e rali zations Theorem 4 of § 3 shows that the results obtained in Chapter III apply to the dimension of closed sets. Thus, for example, a knowledge of the relative positions of the covering classes S and T, that is a knowledge of the sets A(S:T) and A(T:S) introduced in § 3 of Chapter III, gives a complete description of the values of {d (E), dT(E)) as E varies 5 over the collection of closed sets in [O, 1 J. Mor e ove r, it is clear from the material of § 4, that questions of this sort may b e dealt with entirely by considering finite unions of Generalized Cantor Sets. In general, when a finite number of covering classes are considered, a similar analysis can be made using essentially the same methods. The question arises as to what can be said for closed sets in the unit square. If C is any collection of open r ectangle s, whose sides are parallel to the coordinate axe s, and if C is clo s ed under translations and contains rectangle s of arbitrarily small ar e a, the dimension of a closed set E with respect to C is defined to b e the number where The covering class C can be associated with a s e t left-ope n portion, 1 I 2 of points in the , of the unit square 1 , whose closure intersects 2 I 12 - 12. If M denotes the class of functions f(x, y), d e fined on x 2:: 0, y 2:: 0, non-decreasing in x and in y such that f (x, y) = 0 if 92 x· y = 0, and such that M(a, b) = \/ fi(x+a, y+b) - f(x+a, y) - f(x, y+b) + f(x, y)) (x, y) ER+ ( is bounded in 1 , then for f-$. 0, the dimension of f with respect to the 2 covering class S is defined to be the number = lim inf log M(s' t) log st st .... 0 (s, t) E S • Since M(a, b) is · sub-additive in each variable, it follows that M(ta, rb) ::, 4(t Y 1) (r Y i) M(a, b) for r,t~O. In particular, for (x,y) E 1 , this implies 2 C > M(x, y) > x · y M( 1, 1) I 4 , where C is some positive constant. f E M. It follows that 0::, d (f) .::::, 1 for 8 On the other hand, the functions (xy)a,, O _:::a, < 1 (with the con- vention 0 ° = 0) are sub-additive in each variable and so for a, > 0 M(a, b} = Y ( (x+ a)a, - xa,) (y+b}a, - ya,} x,y < (ab}a, = f(a, b} , while for a, = 0, M(a, b) = 1. This implies that d ( (xy}a,} =a,, 8 the dimension can take any value between 0 and 1. so that The interest in studying the dimension of functions in M with respect to various covering classes lies in the study of the influence of the shape, along with the area, of the members of the covering class. In the case of rec- tangles the shape of the members of a covering class, S, is given by 93 the set of quotients s It where (s, t) E S. The following theorem is a first indication of the role of shape on the dimension of functions in M. Theorem. A necessary and sufficient condition that d (f) .:::_ dT(f) for all f EM, is that there exist a function g, g:T - S with 8 the properties: log g ( t ) g (t ) 1 2 log t t 1 2 =1 and =0 • The proof is carried out in a similar manner to that of the Theorem of § 4, Chapter IL 94 BIBLIOGRAPHY [l] Bari, N. K. A Treatise on Trigonometric Series. (translation by M. F. Mullins) Permagon Press, New York, 1964, Vol. 2, pp. 301,302, 400-404, 417. [2] Halmos, P. R. Measure Theory, D. Van Nostrand Co., Inc. Princeton, New Jersey, 1950, p. 53. [3] Hille, E. Functional Analysis and Semi-Groups, Am. Math. Soc. Coll. Pub. , Vol. XXXI, 1948, p . 132. [4] Hurewitz, W. and Wallman, H. Dimension Theory. Princeton University Press, Princeton, New Jersey, 1948, p . 10 2. [5] Kahane, J. and Salem, R. Ensembles parfaits et series trigo nometriques. Hermann, Paris. 1963, pp. 23-51. [6] Kelley, J. L. General Topology, D. Van Nostrand Co., Inc . Princeton, New Jersey. 1955. [7] Munroe, M. E. Introduction to Measure and Integration. Addison-Wesley, Cambridge, M_ass. 1953, 133. [8] Besocovltch, A. S. , "On linear sets of fractional dimensions." Math. Annalen. 101 (19 29) pp. 161-193. [9] Eggleston, H. C. , "Se ts of fractional dimensions which occur in some problems of number theory. 11 Proc. London Math. Soc. 54 ( 1952) pp. 42-93. [10] Frostman, O., "Potential d'equilbre et capacite des ensembles avecquelques applications a la theorie des functions. 11 Lund. Universitet. Medd. 1935, Band 3 , pp. 56, 57, 85-91. [ 11] Hausdorff, F. "Dimension und ~us seres Mass. 11 Math. Annalen. 79 (1919) pp. 157-179.