CONFORMAL ITERATED FUNCTION SYSTEMS WITH ... - CiteSeerX

CONFORMAL ITERATED FUNCTION SYSTEMS WITH APPLICATIONS TO THE GEOMETRY OF CONTINUED FRACTIONS by

R. Daniel Mauldin1 and

Mariusz Urbanski1 Department of Mathematics University of North Texas Denton, Texas 76203 May 18, 1998

ABSTRACT. In this paper we obtain some results about general conformal iterated function systems. We obtain a simple characterization of the packing dimension of the limit set of such systems and introduce some special systems which exhibit some interesting behavior. We then apply these results to the set of values of real continued fractions with restricted entries. We pay special attention to the Hausdor and packing measures of these sets. We also give direct interpretations of these measure theoretic results in terms of the arithmetic density properties of the set of allowed entries.

1 Research supported by NSF Grant DMS-9502952. AMS(MOS) subject classi cations(1980). Primary 28A80; Secondary 58F08, 58F11, 28A78 Key words and phrases. Iterated function systems, continued fractions, Hausdor dimension, Hausdor and packing measures, arithmetic densities. Typeset by

AMS-TEX

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Let I be a nonempty subset of N , the set of all positive integers such that I 6= N . Let JI be the set of all irrational numbers z whose standard continued fraction has the form 1 z= b1 + b + 1 b ... where each partial denominator bi is an element of I . We concern ourselves here with the geometric measure theoretic properties of the set J = JI . In particular, we are interested in the Hausdor, packing, and box dimension of J and corresponding measures. It is easy to see (comp. [MU], Section 6) that J is the limit set of the conformal iterated function system generated by the maps b (x) = 1=(b + x), b 2 I . Our investigations of J are based on this representation. We call the family S = fb : b 2 I g a continued fraction system and I the base for the continued fraction system. The paper is organized as follows. Later in this section we recall from [MU] some major features of general conformal iterated function systems. In section 2, we present some new results for general conformal iterated function systems. In particular, we introduce the absolutely regular systems which naturally occur among the continued fraction systems. For regular systems, we obtain some useful necessary and sucient conditions for the Hausdor measure of the limit set J to be positive and also necessary and sucient conditions for the packing measure to be nite where the dimension parameter for both of these is the Hausdor dimension of the limit set J . We also give a simple and useful characterization of the packing dimension of the limit set in terms of the Hausdor dimension of J and the box counting dimension of the set of rst iterates of a point in the limit set, J . In section 4, we apply these results to continued fraction systems. It turns out that when these characterizations are applied to a continued fraction system, these results have direct interpretations in terms of some arithmetic density properties of the set I . So, in section 3, we discuss some of these density notions. Some of these notions do not seem to have been discussed before. Again, in section 4, we give the relationship between these density properties and Hausdor measure and dimension. In section 5, we give the corresponding properties for packing measure and dimension. In section 6, we examine some particular continued fraction systems. The results of this paper include a detailed analysis of those continued fraction systems when the index set I is an arithmetic progression, the set of powers of a given integer, the set of all integers raised to a given exponent, and the set of prime numbers. Finally, we end the paper with some problems which remain unsolved. Many papers have been written on estimating or determining the Hausdor dimension of particular sets of continued fractions. The most detailed work has concerned the case where the index set I is nite. We mention here the papers of T.J. Cusick [Cu], I.J. Good [Go], and D. Hensley [He]. However, none of these papers have dealt with the ner geometry of these sets, e.g., whether the Hausdor measure in the dimension is positive or nite, but have mainly concentrated on other interesting aspects of these nite systems. 2

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Also, none of these papers have dealt with the corresponding properties of the packing measure. It is after all a relatively new concept introduced independently by D. Sullivan and C. Tricot in the 1980's. We shall be using several theorems concerning packing measures as presented in Mattila's book [Ma]. If the index set I is nite, then both the Hausdor and packing measures are positive and nite and each is up to a multiplicative constant the conformal measure corresponding to the system. Here we concentrate on new phenomena which occur when the index set I is in nite. In this paper, we demonstrate that there are many continued fraction systems where the Hausdor measure is trivial but the packing measure is, geometrically speaking, the correct measure or conversely. We also provide examples for which none of these measures is nontrivial. We now recall the setting and some of the results developed in [MU] which will be used in this paper. Let X be a nonempty compact subset of a Euclidean space R d . Let I be a countable index set with at least two elements and let S = fi : X ! X : i 2 I g be a collection of injective contractions from X into X for which there exists 0 < s < 1 such that (i (x); i(y))  s(x; y) for every i 2 I and for every pair of points x; y 2 X . Thus, the system S is uniformly contractive. Any such collection S of contractions is called an iterated function system. We are particularly interested in the properties of the limit set de ned by such a system. We canSde ne this set as the image of the coding space under a coding map as follows. Let I  = n1 I n , the space of nite words, and for ! 2 I n , n  1, let ! = !  ! 


 !n . If ! 2 I  [ I 1 and n  1 does not exceed the length of !, we denote by !jn the word !1!2 : : :!n . Since given ! 2 I 1 , the diameters of the compact sets !jn (X ), n  1, converge to zero and since they form a descending family, the set 1

2

1 \

n=0

!jn (X )

is a singleton and therefore, denoting its only element by (!), de nes the coding map  : I 1 ! X . The main object of our interest will be the limit set

J = (I 1) =

1 [ \

!2I 1 n=1

!jn (X );

S

Observe that J satis es the natural invariance equality, J = i2I i (J ). Notice that if I is nite, then J is compact. However, our main interest centers on systems S which are in nite. Some of the essential properties of J depend upon an object which appears only when I is in nite. Let X (1), the "asymptotic boundary," be the set of limit points of all sequences i (X ), i 2 I 0 , where I 0 ranges over all in nite subsets of I . The geometric behavior of the system at X (1) directly aects the geometric properties of the limit set J . For an in nite continued fraction system the only element of X (1) is 0. An iterated function system S = fi : X ! X : i 2 I g, is said to satisfy the Open Set Condition (abbreviated (OSC)) if there exists a nonempty open set U  X (in the topology

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of X) such that i (U )  U for every i 2 I and i (U ) \ j (U ) = ; for every pair i; j 2 I , i 6= j . An iterated function system S satisfying OSC, is said to be conformal (c.i.f.s.) if the following conditions are satis ed. (a) X is a compact connected subset of a Euclidean space R d and U = IntRd (X ). (b) There exist ; l > 0 such that for every x 2 @X  R d there exists an open cone Con(x; ux; ; l)  Int(X ) with vertex x, direction vector ux , central angle of Lebesgue measure , and altitude l. (c) There exists an open connected set X  V  R d such that all maps i , i 2 I , extend to C 1+" dieomorphisms on V and are conformal on V . (d) Bounded Distortion Property(BDP). There exists K  1 such that j0! (y)j  K j0! (x)j for every ! 2 I  and every pair of points x; y 2 V , where j0! (x)j means the norm of the derivative. Each continued fraction system S = fb (x) = 1=(b + x) : b 2 I g satis es properties (a) (c). We take X = [0; 1]. For V , we take an open interval such that X  V  (?1=4; 5=4): To check the bounded distortion property, we note that if ! = (b1; :::; bn); then 0! (x) = (?1)n =(qn + xqn?1 )2 : Thus, j0! (y)j  4j0! (x)j; for every pair of points x; y 2 X: So, we may take the distortion constant K as close to 4 as we like by adjusting the open interval V. There is one small point about these continued fraction systems. If 1 2 I , then the system is not uniformly contractive, since 01 (0) = ?1: However, this is not a real problem, since the system of second level maps, fb b : b1; b2 2 I g; has the same limit set and is uniformly contractive. 1 2

As was demonstrated in [MU], conformal iterated function systems naturally break into two main classes, irregular and regular. This dichotomy can be determined from either the existence of a zero of a natural pressure function or, equivalently, the existence of a conformal measure. The topological pressure function, P is de ned as follows. For every integer n  1 de ne X 0 t ( t ) = jj! jj : n and

!2I n

1 log (t): P (t) = nlim n !1 n

For a conformal system S , we sometimes set S = 1 = : The niteness parameter, S ; of the system S is de ned by inf ft : (t) < 1g = S . In [MU], it was shown that the topological pressure function P (t) is non-increasing on [0; 1), strictly decreasing, continuous and convex on [; 1) and P (d)  0. Of course, P (0) = 1 if and only if I is in nite. In [MU] (see Theorem 3.15) we have proved the following characterization of the Hausdor dimension of the limit set J , which will be denoted by dimH (J ) = hS .

Theorem 1.1. dimH (J ) = supfdimH (JF ) : F  I is niteg = inf ft : P (t)  0g. If P (t) = 0, then t = dimH (J ).

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We called the system S regular provided that there is some t such that P(t) = 0. It follows from [MU] that t is unique. Also, the system is regular if and only if there is a t-conformal measure. A Borel probability measure m is said to be t-conformal provided m(J ) = 1 and for every Borel set A  X and every i 2 I

m(i (A)) = and

Z

A

j0i jt dm

m(i (X ) \ j (X )) = 0;

for every pair i; j 2 I , i 6= j . A system S = fi gi2I is said to be strongly regular if 0 < P (t) < 1 for some t  0. As an immediate application of Theorem 1.1 we get the following

Theorem 1.2. A conformal system S is strongly regular if and only if h > . In [MU] we called a a system S = fi gi2I hereditarily regular or co nitely regular provided every nonempty subsystem S 0 = fi gi2I 0 , where I 0 is a co nite subset of I , is regular. A nite system is co nitely regular and for an in nite system, we showed in [MU] that whether a system is co nitely regular can be also determined from the pressure function:

Theorem 1.3. An in nite system S is co nitely or hereditarily regular if and only if P () = 1 , () = 1 , ft : P (t) < 1g = (; 1) , ft : (t) < 1g = (; 1). Theorem 1.4. Every co nitely regular system is strongly regular. We also need another characterization of the niteness parameter , Theorem 3.23 of [MU]:

Theorem 1.5. S = inf fdimH (JS0 ) : S 0is a co nite subsystem of Sg: 2. Results for general conformal systems. First, let us introduce another subclass of the regular conformal systems. De nition. A system S = fi gi2I is said to be absolutely regular if and only if every nonempty subsystem of S is regular. The following two statements are obvious. Theorem 2.1. A system S is absolutely regular if and only if every non-empty subsystem of S is co nitely regular. Corollary 2.2. Every absolutely regular system is co nitely regular.

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Like co nitely regular systems, whether an in nite conformal system is absolutely regular can also be completely determined by the behavior of the pressure function. Theorem 2.3. Let S = fi gi2I be in nite. The following conditions are equivalent: (1) S is absolutely regular. (2) S = 0. Proof. (2) ) (1). Assume (2). Then for every in nite subsystem S 0 , S0 = 0, and since, S 0 (0) = 1, S 0 is regular. Thus, S is absolutely regular. (1) ) (2). Suppose  = S > 0. Since S is in nite, we may assume that I = N = f1; 2; 3; : : : g. De ne inductively In , an increasing sequence of nite subsets of I , as follows. Set I0 = ; and suppose that In has been determined. Let Mn = max In . Since 0 k
2?n < ((1 ? 2?n )) = 1, there existsX a nite subset
of [ M + 1 ; 1 ) such that k  n n i 2?2n?1 for all i 2
n and 2n  k0i k(1?2?n )  2n+1 . Then by setting In+1 = In [
n , i2
n

we nish the recursion for the sequence fIn gn0 . Set F = in nite. Since

X

i2
n

k0i k  2?2n?1 F () =

X

i2
n

[

n0

In . By construction, F is

k0i k(1?2?n )  2?n , we conclude that

XX n0 i2
n

k0i k 

X n0

2?n = 2 < 1:

Also for every 0  t <  there exists k such that t  (1 ? 2?k ). Thus, F (t) 

XX

nk i2
n

k0i kt 

XX

nk i2
n

k0i k(1?2?n) 

X

nk

2n = 1:

Hence, F =  and F (F )  2 < 1. Therefore, the subsystem generated by F is not co nitely regular and the system S is not absolutely regular. For a regular conformal system with P (h) = 0, we know that Hh(J ) < 1, where Hh(J ) denotes the Hausdor h-dimensional measure. It is possible for the measure to be zero. In Lemma 4.11 of [MU] we gave a sucient condition for Hh(J ) > 0: In the next theorem we extend this result, by giving some necessary and sucient conditions for the Hausdor measure to be positive. Theorem 2.4. Let S = fi gi2I be regular. Then the following statements are equivalent. (1) Hh(J ) > 0. (10) 90 < L < 1 9  1 such that for all i 2 I and for all r > diami (X ); there is some x 2 i (X ) such that: m(B (x; r))  Lrh : (2) 90 < L < 1 9  1 and there exists a nite set F such that for all i 2 I n F and for all r > diami (X ); there is some x 2 i (X ) such that: m(B (x; r))  Lrh :

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(20) 90 < L < 1 8  1 there exists a nite set F such that for all i 2 I n F and for all r > diami (X ); there is some x 2 i (X ) such that:

m(B (x; r))  Lrh : (3) 90 < L < 1 9  1 and there exists a nite set F such that for all i 2 I n F and for all r > diami (X ), for all x 2 i (X ) :

m(B (x; r))  Lrh : (30) 90 < L < 1 8  1 there exists a nite set F such that for all i 2 I n F and for all r > diami (X ), for all x 2 i (X ) :

m(B (x; r))  Lrh :

Proof. Obviously (30 ) ) (3) ) (2) ) (10 ) and (30) ) (20 ) ) (2) ) (10). It is straightforward to show that (10) ) (2): Lemma 4.11 of [MU] shows that (10 ) ) (1): In order to

show that (1) implies (30) suppose that (30 ) fails. Then for every L > 1=disth (X; @V ) there exists j 2 I such that m(B (x; r)) > Lrh for some x 2 j (X ) and some r > diam(j (X )). Let J1 be the image under  of all words of I 1 such that each element of I occurs in nitely often. Consider z 2 J1 , z = (!) 2 I 1 such that !n+1 = j for some n  1. Then there exists zn 2 j (X ) such that z = !jn (zn ). Since r  1=L1=h  dist(X; @V ), all the geometric consequences of the bounded distortion property listed in Section 2 of [MU] are applicable to the ball B (x; r). In particular, we get j!jn (zn ) ? !jn (x)
j  Djj0!jn jjr and ?
B !jn (x); jj0!jn jjr  !jn (B (x; r)). Therefore, B (z; (D + 1)jj0!jn jjr  !jn (B (x; r)). By conformality and (BDP), this implies that

?





m B (z; (D + 1)jj0!jn jjr  K ?h jj0!jn jjhm(B (x; r))  K ?h Ljj0!jn jjhrh ?
= (D + 1)jj0!jn jjr h K h(DL+ 1)h : Using Theorem 2.8(1) of [MU], we get Hh(J1 )  C=L; for some constant C independent of L. Now, letting L ! 1 we conclude that Hh(J1) = 0. By Theorem 3.8 and Corollary 3.11 of [MU], m(J n J1 ) = 0. This in turn, in view of Lemma 4.2 of [MU], shows that Hh(J n J1) = 0. Thus, Hh(J ) = 0 and therefore (1) ) (30): The proof is nished. Similarly, we have some necessary and sucient conditions for the packing measure to be nite. We denote the h-dimensional packing measure by Ph: For its de nition and other informations about packing measures and dimensions the reader may look at the book [Ma] by P. Mattila for example.

Theorem 2.5. Let S = fi gi2I be regular. Then the following statements are equivalent (1) Ph (J ) < 1.

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(10)9L > 09 > 09  1 such that for all i 2 I and for all r with diami (X ) < r   there is some x 2 i (X ) such that

m(B (x; r))  Lrh : (2) 9L > 09 > 09  1 and there exists a nite set F such that for all i 2 I n F and for all r with diami (X ) < r   there is some x 2 i (X ) such that

m(B (x; r))  Lrh : (20) 9L > 09 > 08  1 there exists a nite set F such that for all i 2 I n F and for all r with diami (X ) < r   there is some x 2 i (X ) such that

m(B (x; r))  Lrh : (3) 9L > 09 > 09  1 and there exists a nite set F such that for all i 2 I n F and for all x 2 i (X ) and for all r with diami (X ) < r < 

m(B (x; r))  Lrh : (30) 9L > 09 > 08  1 there exists a nite set F such that for all i 2 I n F and for all x 2 i (X ) and for all r with diami (X ) < r < 

m(B (x; r))  Lrh :

Proof. It is straightforward to show that (2) ) (10 ): Lemma 4.10 of [MU] shows that (10 ) ) (1): Clearly, (30 ) ) (3) ) (2) and ((30) ) (20 ) ) (2). Finally, by way of contradiction, let us assume (1) holds and (30) fails. Fix L > 0;  > 0. Then there are  1, i 2 I and diami (X ) < r   such that for some x 2 i (X ), we have m(B (x; r))  Lrh : Since the system is regular, there is a Borel subset B of J with m(B ) = 1 and such that each point z of B has a unique code, !, and (n(!)) is in the ball B (x; r=2) for in nitely many n's. For such a point z and integer n  1, we have

m(!jn (B ((n(!)); r=2)))  k0!jnkh m((B ((n(!)); r=2)  k0!jn khLrh: But, by the bounded distortion property of the system,

!jn (B ((n(!)); r=2))  B (z; k0!jnkK ?1 r=2): So, m(B (z; k0!jnkr=2K ))  (k0!jnkr=2K )h(2K )hL: Using Theorem 2.9(1) of [MU], we get Ph (J )  Ph (J \ B )  (2K )?hL?1 . Now, letting L ! 0 we get Ph (J ) = 1. This contradiction nishes the proof.

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We close this section with a stronger form of Lemma 4.15 of [MU]. This improved version will be directly applied to continued fraction systems in Section 5. For completeness, and to avoid confusion, we have included a proof since part of the hypothesis was unfortunately left out of the statement of Lemma 4.15.

Theorem 2.6. Let fi : i 2 I g be a regular conformal iterated function system. Suppose that there exists a subset ; = 6 Z  X (1) such that for every z 2 Z there exist i(z) 2 I and a set R(z)  (0; dist(X; @V )) such that (a) i(z) (B (z; sup R(z)) \ J ) = i(z) (B (z; sup R(z))) \ J , (b) i(z) (B (z; sup R(z))  X , (c) inf f m(Br(hz;r)) : z 2 Z; r 2 R(z)g = 0. Then Ph(J ) = 1. Proof. First notice that since 'i(z) is one-to-one, i(z) (F \ J ) = i(z) (F ) \ J; for all z 2 Z and all F  B (z; sup R(z)). Let J1 = (1), where 1   is the set of all sequences

containing each nite word in nitely often. Of course, J1 has full measure. Fix " > 0 and take z 2 Z and r 2 R(z) such that m(B (z; r))  "rh . Fix x = (!), ! 2 1 : Then there exists q  1 such that !q (X ))  B (z; r=2) and !q = i(z). Now, x = !jq ((q !)); where  is the shift transformation on the coding space, I N : So, using (BDP.3) of [MU], we get m(B (x; K ?1jj0!jq jjr=2))  m(!jq (B ((q!); r=2))  m(!jq (B (z; r)): Using the facts that !q (B (z; r))  X; and condition (a) holds, we have +1

m(!jq (B (z; r))) = m(!jq? (!q (B (z; r))) = =

Z

1

Z

!q (B(z;r))\J

jj0!jq? jjh dm 1

jj0!jq? (x)jjh dm(x)

1 !q (B(z;r)\J ) Z = jj0!jq?1 (!q (y))jjhjj0!jq (y)jjh dm(y) ZB(z;r)\J = jj0!jq jjh dm  jj0!jq jjhm(B (z; r))  jj0!jq jjh"rh B(z;r))\J = (2K )?1(jj0!jq jjr)h"(2K )h:

Since we may require q to be as large as we wish and since r > 0 is bounded from above, the numbers (2K )?1jj!jq jjr converge to zero and we nish the proof applying Theorem 2.9(1) of [MU]. Recall that the Hausdor dimension of a probability measure m is de ned by dimH (m) = minfdimH (E ) : m(E ) = 1g: In Theorem 3.24 of [MU] we have shown that the Hausdor dimension of the conformal measure of every regular system for which the series

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P ? log(jj0 )jj)jj0 jjh converges is equal to the Hausdor dimension of the limit set. In i i i2I

the proof of Corollary 2.25 of [MU] we have demonstrated that this class of systems comprises all the strongly regular systems. Here we shall prove a complementary result for the packing dimension of the conformal measure m, dimP (m) = minfdimP (E ) : m(E ) = 1g:

Theorem 2.7. If S is a regular system andthe series Pi2I ? log(jj0i )jj)jj0ijjh converges ,

then dimP (m) = dimH (J ) = dimH (m). Proof. We shall rst show that

log(jj0!jn jj) lim = 1; n!1 log(jj0!jn? jj)

(2:1)

1

for almost every ! 2 I 1 . Indeed, applying Birkho's ergodic theorem similarly as it has been done in the beginning of the proof of Theorem 3.24 of [MU], we conclude that for almost every ! 2 I 1 thelimit 1 log?m( (J ))
lim ! jn n!1 n

exists and is independent of !. Since for all ! 2 I 1 and all n  1, K ?h jj0!jn jjh  m(!jn (J ))  jj0!jn jjh, formula (2.1) is therefore proved. Denote the set of points satisfying (2.1) by Z . Fix " > 0: Consider ! 2 Z . For n0 suciently large, we have log(jj0!jn jj)  (1+") log(jj0!jn? jj) for all n  n0 or equivalently jj0!jn jj=jj0!jn? jj  jj0!jn? jj" . For every r > 0, let n be the least number satisfying B ((!); r)  !jn (X ). Then m(B ((!); r))  K ?h jj0!jn jjh and r  Djj0!jn? jj, where D is given by (BDP.2) of [MU]. If r is small enough, then n  n0 and therefore 1

1

1

1

jj0!jn jjh ? h 0 h m(B ((!); r))  K jj!jn?1 jj
jj0 jjh !jn?1

 (DK )?h D?h"rh+h":

Thus, dimP ((Z ))  h + h" and since m(Z ) = 1, we conclude that dimP (m)  h. Since dimP (m)  dimH (m); the proof is nished. Finally, we close this section by characterizing the packing dimension of the limit set J of a conformal iterated function system. As recalled in Theorem 1.1, we showed in [MU] that the Hausdor dimension of J is given by dimH (J ) = inf ft : P (t)  0g. It turns out the packing dimension is determined by the box counting dimension of the \level one" portion of the orbits of points of J and the Hausdor dimension of J . For x 2 X; n 2 N , set Ln (x) = f! (x) : ! 2 I n g: We recall that Nr (E ) is the minimum number of balls of radius  r needed to cover a set E . We also make some notation. If F  X and R  I  ; we S denote the set !2R ! (F ) by O(F; R). If R = I n , n  1, we write O(F; n) for simplicity.

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In the sequel, we will need the following fact concerning conformal systems. Namely, from (BDP.2) of [MU], there is some number D > 1 such that diam(! (V ))  Djj0! jj; for all nite words !:

Lemma 2.8. Let fi : i 2 I g be a conformal iterated function system. Then dimP (J ) = M = supfdimH (J ); dimB (Ln (x)) : x 2 J; n 2 N g. Proof. Recall from Theorem 3.1 of [MU] that dimP (J ) = dimB (J ): Fix t > M: Since P (t) < 0, there is some Q such that if q  Q; then q (t) < 4?t and if j!j  Q; then jj0! jj  1=4: Fix q  Q and x 2 J . Choose A such that for all D  r > 0, Nr (Lq (x))  Ar?t : Now, choose B such that if 1  r  D, then Nr (J )  Br?t and such that B  4t A=(1 ? 4t q (t)). We will show by induction that for each n 2 N , if 1=n  r  D; then Nr (J )  Br?t: This inequality holds for 1. Suppose it holds for n?and 1=(n + 1)
 r?< 1=n: Let Cn+1
= f! 2 I q : diam(! (J ))  1=2(n + 1)g: Since J = [!2Cn ! (J ) [ [!2I q nCn ! (J ) ; +1

we have

+1

0 1 [ X ! (J )A + Nr (J )  N1=(n+1) (J )  N1=(n+1) @ N1=(n+1) (! (J )): !2Cn+1

!2I q nCn+1

For ! 2 I q n Cn+1 , we have N1=(n+1) (! (J ))  N1=((n+1)jj0! jj)(J )  N1=(2(n+1)jj0! jj) (J ): Since jj0! jj  1=4  (1=2)(n=n +1); we have 1=n  1=(2(n +1)jj0! jj): Since 1=(2(n +1)) < diam(! (J ))  Djj0! jj; we have 1=(2(n + 1)jj0! jj)  D: So, by the induction hypothesis, S t 0 N1=(n+1) (! (J ))  B (2(n + 1)jj! jj) : Next, we claim that N1=(n+1) ( !2Cn ! (J ))  N1=(2(n+1)) (Lq (x)): To see this, let B (yj ; 1=(2(n + 1))) be a collection of balls of radius 1=(2(n + 1)) covering Lq (x): Suppose z 2 ! (J ), where ! 2 Cn+1 : Then jz ? ! (x)j  diam(! (J ))  1=(2(nS+ 1)). For some j; j! (x) ? yj j  1=(2(n + 1)). So, the balls B (yj ; 1=(n + 1)) cover !2Cn ! (J ): Our claim follows from this. Since n + 1  2=r, +1

+1

Nr (t)  A2t (n + 1)t +

X

j!j=q

B 2t (n + 1)t jj0! jjt  4t [A + B q (t)] r?t  Br?t :

This completes the induction argument. It now follows that dimB (J )  t: From this we have, dimP (J ) = M: Our goal is to show that we can replace the supremum in Lemma 2.8 with a simple maximum. We use two propositions to accomplish this.

Proposition 2.9. If S is a c.i.f.s., then for all x; y 2 X , and all n  1 dimB (O(x; n)) = dimB (O(y; n)):

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Proof. First notice that it suces to prove this equality for n = 1 since for every n  1 the collection of maps f! : ! 2 I n g forms a conformal iterated function system again. With this setting notice that

(2:2) 9M  18r > 08z 2 Rd #fi 2 I : B (z; r) \ i(X ) 6= ; and diam(i (X ))  r=2g  M: To see this (cf., proof of Lemma 4.11, [MU]) denote the set of such i's by E and consider i 2 E with this property and x y 2 B (z; r) \ i (X ). We repeat here a crucial geometric condition from the de nition of a conformal system. The \cone condition" (2.10) of [MU] states: there exists 0 <   such that for all x 2 X and for all ! 2 I  (2:3)

?




?




! (Int(X ))  Con ! (x); ; D?1jj0! jj  Con ! (x); ; D?2diam(! (X )) ;

?




?




where Con ! (x); ; D?1jj0! jj and Con ! (x); ; D?2diam(X ) denote some cones with vertices at ! (x), angles , and altitudes D?1 jj0! jj and D?2 diam(X ) respectively. Thus, there exists a constant P > 0 such that

?

?

?

 i (X ) \ B (z; 2r))   i (X ) \ Con i (x); ; minfr; D?2 diam(i (X ))g ? ?   Con i (x); ; (2D2)?1 r))  Pr2;




where x 2 X is such that i (x) 2 B (z; r). Since all the sets i (Int(X )) \ B (z; 2r)),d i 2 E , d Vd  ( B ( z; 2 r )) 2 are mutually disjoint, #E  Pr  P . So, it suces to take M = 2 PVd . In order to prove the proposition, it is enough to show that dimB (O(x; 1))  dimB (O(y; 1)): Towards this goal, take 0 < r  diam(X ) and let Ir = fi 2 I : diam(i (X ))  r=2g. Then Nr (O(y; Ir ))  Nr=2 (O(x; Ir )). Clearly, Nr (O(z; I n Ir ))  #(I n Ir ); for all z 2 X . On the other hand by (2.2), Nr (O(z; I n Ir ))  #(I n Ir )=M . Hence, 2

Nr (O(y; 1))  Nr=2 (O(x; Ir )) + Nr (O(y; I n Ir ))  Nr=2 (O(x; I )) + MNr (O(z; I n Ir ))  (1 + M )Nr=2(O(x; 1)): Therefore, log Nr (O(y; 1))  lim log Nr (O(x; 1)) : lim r!0 r!0 log r log r

The proof is nished.

Lemma 2.10. If S is a c.i.f.s., then for all x 2 X and for all n  1, dimB (O(x; n)) = dimB (O(x; 1)): Proof. By Proposition 2.9, we may assume that x 2 Int(X ), so B (x; )  Int(X ) for some  > 0. First, we shall show that S  dimB (O(x; 1)): To see this, x t > t ? s > dimB (O(x; 1)). Then x " > 0 and consider the set I (") = fi 2 I : K"?1  jj0i (x)jj  2K"?1g:

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Since the balls B (i (x); ") with i 2 I (") are disjoint, N"(O(x; 1))  #I ("). Since for all " > 0 small enough "?t  "?s N" (O(x; 1)), we get for all k large enough, say k  k0 the following

X X

kk0 i2I (2?k )

jj0ijjt 

X

kk0

2t K t ?t2?kt #I (2?k )  2tK t ?t

 (2K?1)t

X kk0

X

kk0

2?kt N2?k (O(x; 1))

2?ks  (2K?1)t 1 ?12?s < 1:

S

Since limi2IN jj0i jj = 0, the set I n kk I (2?k ) is nite, and therefore t  S . Letting t ! dimB (O(x; 1)); we get S  dimB (O(x; 1)). Now, x t > dimB (O(x; 1)) again. We shall show by induction that for all n  1 there exists 0 < An < 1 such that Nr (O(x; n))  An r?t ; for all 0 < r  2D. Indeed, the existence of A1 is immediate as t > dimB (O(x; 1)). Suppose that 0 < An < 1 exists. To prove the existence of An+1 ; set I1 = f! 2 I n : diam(! (X )) < r=2g. Then Nr (O(x; I1  I ))  Nr=2 (O(x; I1))  Nr=2 (O(x; n))  2tAn r?t . If ! 2 I n n I1, then Nr (O(x; f!g  I ))  Nr=jj0! jj(O(x; 1))  A1 jj0! jjtr?t , where the second inequality sign holds since r=jj0! jj  2diam(! (X ))=jj0! jj  2D. Thus, since t > S ; 0

Nr (O(x; n + 1))  2t An r?t + A1 r?t

X

!2I n nI1 = (2tAn + A1 n (t))r?t:

jj0! jjt  2t An r?t + A1 n (t)r?t

The proof is completed by setting An+1 = 2tAn + A1 n (t). As a corollary of Lemma 2.8 and Propositions 2.9 and 2.10, we have a simple means of obtaining the packing dimension of the limit set.

Theorem 2.11. Let fi : i 2 I g be a conformal iterated function system. Then dimP (J ) = maxfdimH (J ); dimB (L1(x)) : x 2 J g = maxfdimH (J ); dimB (L1(x0))g; where x0 is any

given point in X .

x3. Arithmetic relations. In this section we collect some basic arithmetic de nitions and relations. We begin with the following notation. If I is a subset of N and 1  p  q  1 are two real numbers, then by #I (p; q) we denote the number of elements of the intersection I \ [p; q]. If p = 1, we frequently use the notation Sq (I ) for #I (p; q).

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Lemma 3.1. If I  N and 0 < k < l  2k, then for all s > 0 l 1 (n) #I (k; l) (kl)s X I s (l ? k) n=k n2s  (l ? k)s ;

where the comparability constant depends only on s. Proof. The proof follows immediately from the following computation. l 1 (n) l 1 (n) 2s X (kl)s X l I I  s 2 s s (l ? k) n=k n (l ? k) n=k n2s 2s 1 I (k; n) : l  (l ? k)s l2s #(I \ [k; n]) = #( (l ? k)s

Lemma 3.2. For each I  IN and for each 0 < t < 1 #I (n=2; n)  (1 ? 2?t ) lim #I (1; n) : lim n!1 n!1 nt nt

Proof. We may assume that limn!1 #In(1t;n) > 0. Let d = (1 ? 2?t )limn!1 #In(1t;n) .

#I (2k ; 2k+1) < d. Then there exists c < d such that for Suppose on the contrary that klim !1 2kt k k+1 every k  1 large enough, say k  S , we have #I (22kt; 2 )  c. So, for every k  S , k #I (2j ; 2j +1 )
2(j ?k)t X k #I (2S ; 2k+1)  X (j ?k)t  c 1 :  c 2 kt jt 2 2 1 ? 2?t j =S j =S

Thus,

limk!1 #I (12;kt2

k+1)

Now, since for every n  2,

S k+1 )

= limk!1 #I (22kt; 2

 1 ?c2?t :

#I (1; n)  #I (1; 2[log n]+1 ) ; nt (2[log n] )t 2

2

we get

; n)  c < lim #I (1; n) : limn!1 #I (1 n !1 t ? t n 1?2 nt

This contradiction nishes the proof.

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We now provide the reader with several de nitions of objects and properties associated with in nite subsets of N which are intended to measure the "size" of those sets. We rst de ne the lower density dimension of a set I  N . Given t  0; let



 #I (k; l)

%t (I ) = inf (l ? k)t : k < l and A( k2+kl l ) \ I 6= ; > 0; where given t  0; A(t) = N \ [t ? 1; t + 1]. Notice that

n

o

n

o

inf t : %t (I ) < 1 = sup t : %t (I ) > 0: This common value will be called the lower density dimension of I and will be denoted by %D(I ). Similarly, we de ne %D(I ), the upper density dimension of a set I  N , as follows. For each t  0; n 2 N , set

 #I (k; l)



 #I (k; l)



%t (I ) = sup (l ? k)t : k < l; k; l 2 I = sup (l ? k)t : k < l : Notice that

inf ft : %t (I ) < 1g = sup ft : %t (I ) > 0:g This common value will be called the upper density dimension of I and will be denoted by %D(I ). Clearly %D(I )  %D(I ), and if these two numbers are equal, the common value will be denoted by %D(I ). A subset I  N is said to have the strong arithmetic density 0 if for every t > 0, Sn (I ) = 0: lim n!1 nt We say that two subsets of N are strongly equivalent if their symmetric dierence is nite. Suppose that I  N . A subset A  I is said to be a cluster of I if and only if A \ [min(A); sup(A)] = A. By the length of A we mean the number sup(A) ? min(A). A subset A  I is said to be a punctured cluster of I if and only if there is x 2= A with min(A) < x < sup(A) such that [min(A); sup(A)]nfxg = A. Notice such an x is determined uniquely and by the lower length of A we mean the number minfx ? min(A); sup(A) ? xg. The following lemma, whose straightforward proof is left to the reader, provides some elementary properties of the notions introduced above.

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Lemma 3.3. Suppose that I; I 0  N . Then (a) 0  %D(I )  %D(I )  1.

(b) If I 0 is strongly equivalent with I , then %D(I 0 ) = %D(I ) and %D(I 0) = %D(I ). (c) If I contains arbitrarily long clusters, then %D(I ) = 1. (d) If N n I contains arbitrarily long punctured clusters, then %D(I ) = 0. (e) If p is a polynomial of degree d  1, then the set Ip = f[p(n)] : n 2 N g has density dimension 1=d. (f) If I is equivalent with a subset of a geometric sequence, then it has arithmetic density dimension and %D(I ) = 0. (g) If I is an in nite subset of N with upper density dimension zero, then I has strong density zero. If in turn I has strong density zero, then it is of lower density dimension zero. Let us relate the density dimensions of I to the niteness parameter of the continued fraction system with index set I .

Lemma 3.4. If I  N , then %D(I )  2(I )  %D(I ). Proof. Since I is in nite there exists an in nite subset F of I such that [n=2; 2n] \ [m=2; 2m] = ;; for all distinct elements m and n in F . Since I  0; in order to prove

the rst inequality we may assume that %D(I ) > 0. Fix then any 0 < s < %D(I ). By the de nition of the lower density dimension there exists a constant M > 0 such that #I (k; l)  M (l ? k)s; for all k < l with A( k2+kll ) \ I 6= ;. We then make the estimates 1 (s=2) 

2n 1 X1 XX X1 1 2  1 ( k )  I s s s
ns #I ( 3 n; 2n) n k 2 n2I n2F k= n n2F   s M 2s X 1 X 1 4

 M 2s

2 3

1 = 1: s 3 n = 3s n n2F n2F

Hence, s=2  (I ) and the rst inequality is proven. In order to prove the second inequality x t > %D(I ) and then auxilarily %D(I ) < s < t. By the de nition of the upper density dimension there exists a constant 0 < M < 1 such that #I (k; l)  M (l ? k)s for all k < l. Then n X 2X 1 1 (k)  X 1 #I (2n ; 2n+1)  M X 1 2ns = X 2(s?t)n < 1: ( t= 2)  1 kt I 2nt 2nt +1

n0 k=2n

n0

n0

n0

Hence, t=2  (I ) and we are done. We end this section with some basic results concerning sets of density zero.

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Theorem 3.5. Let I = fn1 < n2 < n3 < : : : g be an in nite subset of N . The following four statements are equivalent. (1) I has strong arithmetic density zero. P (2) For each t > 0, n2I 1t < 1.

n k = 0. (3) For each t > 0, klim !1 ntk (4) The continued fraction system S = fb gb2I is absolutely regular. Proof. For each t > 0, we have by summation by parts,

(3:1)

P



n X



n X 1 1I (k) kt = Sk (I ) k1t ? (k +1 1)t + (n +1 1)t Sn (I ): k=1 k=1

P

1 t Sn . Assume I has strong density zero. Then So, nk=1 1I (k) k1t  t nk=1 k1 Skkt + (n+1) Mt =Psupn1 Snn(tI ) < 1, limPn!1 Snn+1(It) = 0, and for all k, PSkk(tI )  Mkt=t= : It now follows that nk=1 1I (k) k1t  tMt=2 nk=1 k 1t= and consequently n2I n1t < 1: Now, assume statement (2) holds. Then from (3.1), it follows that for each t > 0; lim Sn(nt ) < 1: This in turn implies that that for each t > 0, limn!1 Sn(nt ) = 0: Hence (1) and (2) are equivalent. Now, given n  n1 take k such that nk  n < nk+1 . Then 2 2

1+

2

k  Sn (I )  k + 1 : nt ntk

ntk+1

Thus (3) and (1) are equivalent. Since condition (2) means that I = 0, the equivalence of (2) and (4) is established by Theorem 2.3 The proof is nished. One can use the summation by parts formula to obtain another characterization of the niteness parameter of a continued fraction system:

Theorem 3.6. Let I  N : Then the niteness parameter I of the continued fraction

system with index set I satis es:

Sn < 1g: I = inf ft : nlim !1 nt The following example completes the part (g) of Lemma 3.3.

Example 3.7. Consider I = f2n + i : n > 0; 0  i  n ? 1g: Then I has positive upper density dimension equal to one and also has strong density zero.

x4. Hausdor measures and dimensions.

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We begin this section with the following general theorem linking arithmetical properties of the set I and geometrical properties of the corresponding limit set.

Theorem 4.1. For a regular continued fraction system with index set I , the following conditions are equivalent: (a) Hh(JI ) = 0.

(

?

)

m(B i (X ); r)) (b) For some  1; sup : i 2 I; r  diam('i (X )) = 1: h r ( ? ) m(B i (X ); r)) (c) For each  1; sup : i 2 I; r  diam('i (X )) = 1: rh l 1 (n) h X ( kl ) I (d) sup (l ? k)h 2h = 1: k diam((X )) can be made arbitrarily large. Thus, (a) ) (b): Clearly,(c) ) (b): Next, let us assume 0  1 is such that the supremum in statement (b) is 1: Let L > 0: We will show that condition (30) of Theorem 2.4 fails with = 0 + 1. First, note that if j 2 I and r  diam('j(X )), then m(B (j (X ); r))=rh  ( diam(j (X ))?h : Let F  I be nite. Let T = max hL; ( diam'j (X ))?h : j 2 F : Choose i 2 I and r0  0diam('i (X )) such that m(B (i (X ); r0))  (T + 1)r0h: Thus, i 2 I n F . Let x 2 'i (X ) and r = (1 + 1= 0)r0. Then r  diam'i (X ) and B (x; r)  B ('i (X ); r0): So, m(B (x; r)) > Lrh: Thus, by Theorem 2.4, Hh(JI ) = 0: So, (b) ) (a): Now, assume statement (d) holds. We show that statement (b) holds with = 1: l 1 (n) h X I Choose k < l such that (l(?kl)k)h 2h > T > 0; where a positive lower bound on T n n=k will be speci ed later. Choose i 2 I such that the distance from 1=i to 1=2(1=k + 1=l) is minimum and let r = max fj1=(i + 1) ? 1=(l + 1)j; j1=k ? 1=ijg. So, r < 1=k ? 1=l and l 1 (n) m(B ('i(X ); r))  (K ?h) (kl)h X I ?h h h r (l ? k) n=k n2h  K T:

l 1 (n) h X ( kl ) I Let us note that the quantities (l ? k)h 2h with r < diam(i (X )) = 1=i(i+1); where n n=k r is chosen as above have a uniform upper bound. To see this note that in this setting k = i l 1 (n) h X ( kl ) (kl)h 1 2i : I and l < i(i + 1)=(i ? 1): So, (l ? k)h is bounded above by 2h (l ? k)h i2h (i ? 1) n=k n h If l  2k, then (l(?kl)k)h  (kl)h and using the bounds on l, we get the quantity uniformly

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h h ( kl ) i ( i + 1) bounded above. If 2k < l, then (l ? k)h  i ? 1 , and again the quantity is uniformly bounded above. Therefore, we also have r  diam(i (X )), for T suciently large. Thus, (d) ) (b): Finally, let us assume statement (c) holds and take = 1: Let m(B (i (X ); r))=rh > T; with r  diam(i (X )). Choose k < l such that [1=(l ? 1); 1=k + 1]  B (i (X ); r) and 1=k; 1=l 2= B (i (X ); r): Then 1=k ? 1=l  4r and l 1 (n) (kl)h X I ?h ) m(B (i (X ); r)) :  (4 h 2 h (l ? k) n=k n rh

From this, we see (c) ) (d).

Corollary 4.2. If I  N generates a regular continued fraction system and if limn!1 #In(1h; n) = 1;

then Hh(JI ) = 0. Proof. For every k  1 we have 2k 1 (n) 2k (k
2k)h X I h kh X 1I (n)  2h kh #I (k; 2k) = 2?h #I (k; 2k) : = 2 2h (2k ? k)h n=k n2h (2k)2h kh n=k n

Thus an immediate application of Theorem 4.1 and Lemma 3.2 nishes the proof. As an immediate consequence of this result we get the following.

Corollary 4.3. Let I be a base for a regular continued fraction system and let h be the dimension of the system. If for some t > h, lim supn!1 #In(1t;n) > 0, then Hh(J ) = 0:

Proposition 4.4. If I is a base for a continued fraction system and Hh(JI ) > 0, then h  2. Proof. Since the Hausdor measure is a conformal measure, the system is regular. By Corollary 4.2, M = supn #In(1h;n) < 1. Suppose by way of contradiction that h < 2. Thus  > 0 and there exists 0 < t <  such that h < 2t. Then we can write 1 X 1 X 1 (t) = n2t =

X

1 #I [2n ; 2n+1 ) 1 X 2t 22tn n=0 k2I \[2n ;2n ) k n=0 n2I 1 h n ; 2n+1 ) X 2(h?2t)n  1 ? 22?M = #I [22nh (2t?h) < 1: n=0 +1

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Hence, 1(t) is nite which contradicts the de nition of  and nishes the proof. Since by Theorem 4.7 of [MU], h < 1 if I 6= N , as an immediate consequence of this proposition, we get the following

Corollary 4.5. If I 6= N is a base for a continued fraction system and Hh(JI ) > 0, then

 < 1=2. In particular, if I is the set of all prime numbers, then Hh(JI ) = 0: Proof. Since pn  nlogn, it easily follows that if I is the set primes, then I = 1=2, I (1=2) = 1 and the system is strongly regular.

Let us also note the following, in a sense stronger, consequence of Proposition 4.4.

Corollary 4.6. Let the in nite set I be the base for a continued fraction system. If

HhF (JF ) > 0, for every co nite subsystem F of I , then I has strong density 0.

Proof. Since the Hausdor measure is a conformal measure, the system is regular. Suppose

that I does not have strong density 0. Therefore by Theorem 2.3 and Theorem 3.5, (I ) > 0. Thus, applying Theorem 1.5 and Lemma 3.19 of [MU] we see that there is a co nite subsystem F of I such that hF < 2(F ). This contradicts Proposition 4.4 and nishes the proof.

Lemma 4.7. Suppose I  N , I 6= N and I contains arbitrarily long blocks, then Hh(JI ) = 0.

Proof. By way of contradiction, suppose Hh(JI ) > 0. By Theorem 4.1.7 of [MU], the

system is regular. If I has a block from k to l, then

l 1 (n) l 1 (kl)h X (kl)h X (kl)h (l ? k) = ( k )h (l ? k)1?h : I =  (l ? k)h n=k n2h (l ? k)h n=k n2h (l ? k)h l2h l

If additionally l  2k, then ( kl )h(l ? k)1?h  2?h (l ? k)1?h. Since h < 1 and since I has arbitrarily long blocks (l ? k ! 1) with the property that l  2k, we complete the proof by invoking Theorem 4.1.

Lemma 4.8. If I  N and h = dimH (JI ) > %D(I ), then Hh(JI ) > 0. Proof. Since h > %D(I ), there exists a constant M > 0 such that #I (k; l)  M (l ? k)h

for all 0 < k < l. Thus, it follows from Lemma 3.1 that

l 1 (h) h X I sup (l(?kl)k)h 2h < 1: n k 2k > 0, then (l(?klk))h  khlhlh = kh , and therefore, for l > 2k

l 1 (h) 1 1 (h) Xl 1I (h) h X (kl)h X I I h  k  k h 2 h 2 h 2h (l ? k) n=k n n=k n n=k n

1 #(I \ [2j k; 2j +1 k]) X h =k 2h  k (2j k)2h j =0 j =0 n=2j k n 1 1 (2j k)h j 2j +1 k]) M X X = k1h #(I \ [22jk;  2h kh j=0 22jh j =0 1 1 X = M 2jh = 1 ?M2?h < 1: j =0 1 2j X k?1 1 (h) X I h +1

By Lemma 3.4, h > %D(I )  2   and, therefore, by Theorem 1.2, I induces a regular system. Thus, an application of Theorem 4.1 nishes the proof.

Corollary 4.9 If I is strongly equivalent to fan : n 2 N g; for some a 2 N ; a  2, then

Hh(JI ) > 0:

Remark. As it follows from the proof of Lemma 4.8, it suces to require the existence of a constant M > 0 such that #I (k; l)  M (l ? k)h for all 0 < k < l which is weaker than the assumption h = dimH (JI ) > %D(I ).

Remark. Since there exist absolutely regular systems I with arbitrarily long blocks, it is possible to have Hh(JI ) = 0 for an absolutely regular system.

Theorem 4.10. If %D(I ) < 1, then the strong equivalence class of I contains an element F with HhF (JF ) > 0. More precisely, there exists a number q  1 such that if F is strongly equivalent with I and F  I [ [1; q], then HhF (JF ) > 0. Proof. In view of Theorem 1.1 there exists q  1 such that dimH (JF ) > %D(I ) provided F  [1; q]. So, we nish the proof applying Lemma 4.8. Remarks. Notice that combining Theorem 4.10 and Lemma 3.3(e) and (f) gives rise to

a method of producing a large class of sets I with HhI (JI ) > 0. We also note that the property of being co nitely regular is invariant under strong equivalency whereas regularity is not.

Lemma 4.11. Let I  N ; k  e2 ; and l  k: The function gk;l de ned by t 7! t P (kl) (l?k)t

l 1I (n) is non-increasing. n=k n2t

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Proof. One simply calculates

 Xl 1 (n)(1 ? ln n) t  ( kl ) kl I 0 gk;l (t) = (l ? k)t ln( l ? k ) ? 2 : 2t n n=k 0 (t)  0 and If k  e2 , then ln( l?klk ) ? 2  0 and 1 ? ln n < 0 for all n  k  e2 . Thus, gk;l we are done.

One may generalize Proposition 4.4:

Lemma 4.12. Let I  N : If ((l?klk))ss Pln=k 1nI (ns ) < M < 1; for some s  0, then s  2: Proof. Fix t > s=2: Then 2

n n 1 2X 1 2X 1 X X X 1 ( j ) 1 ( j ) I I 2 n ( s ? t ) ? s 22n(s?t)
M 2?ns 1 (t) = j 2s  2 j 2sj 2(t?s)  2 +1

+1

n=0

n=0 j =2n ?s  1 2? 2M s?2t

j =2n

n=0

< 1: Thus, t   and letting t & s=2, we get s=2  :

Theorem 4.13. If I  N and HhI (JI ) > 0, then HhE (JE ) > 0, for every system E  I such that E n I is nite. P Proof. AsPhE  hI , it follows from Theorems 4.1 and 4.11 that supk1 k?hE nk 1nI (hnE) < 1: Since nk 1nEh(nE) = Pnk 1nI (hnE) , for k large enough, supk1 k?hE Pnk 1nEh(nE) < 1: 2

2

2

2

Also, the system E is regular, since every system containing a regular system as a co nite subset, is regular. Invoking Theorem 4.1 again, we get HhE (JE ) > 0:

x5. Packing measures and dimensions. We begin this section by giving some necessary and sucient conditions for the packing measure to be nite for regular systems. Since the packing measure is more complex than Hausdor measure, we must analyze separately those index sets I which are co nite and those which are not. It is in the proof of Theorem 5.1 that the use of the harmonic mean of k and l, H (k; l), in connection with packing measure becomes essential.

Theorem 5.1. For a regular continued fraction system with index set I , the following

three conditions are equivalent: (a) Ph (J ) < 1 (b)

inf k 0: > 0 and inf k h 2h 1k n=k n2h ; (l ? k) n=k n

A( k2+kll )\I 6=

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(c)For some k0 ; n0, l 1 (n) 1 1 (n) h X X ( kl ) I I h inf n 0 and inf k h 2 h 2h > 0: k k + n0 . If the in mum of these quantities is 0, then there is some xed k  k0 and an in nite sequence of l's such that thePquantities converge to 0. But, the limit of this sequence of quantities is kh  ( 1 (h) ? kn=1 1nI (nh ) ) which is positive. Thus, (c) ) (b): Now, suppose that condition (a) is satis ed. We will show that the rst inequality in (c) holds with k0 = 6 and n0 large enough (we will indicate that n0  49 suces). Consider k + 49 < l, k > 6 and such that A( k2+kll ) \ I 6= ;. Let i 2 I be the point closest to 1 ; 1 ) \ JI . Set r = minf1=k ? x; x ? 1=lg. 2kl=(k + l). Then k < i < l and there exists x 2 ( i+1 i It can be shown under these conditions that the inequalities 1=(i + 1) ? 1=l > 1=i(i + 1) 1  x?r < and 1=k ? 1=i > 1=i(i + 1) hold. It now follows that r > diam('i (X )). Also, l+1 x + r  k1 and therefore 2

 1 1 [  1 1 B (x; r) \ JI  l + 1 ; k \ JI = ; j \ JI : j + 1 j 2[k;l]\I

It also follows from the conditions on k and l that the following inequalities hold 1=k ? 1=l < 4(1=k ? 1=i) and 1=k ? 1=l < 4(1=i + 1 ? 1=l). From this we get 1=k ? 1=l < 4r: Since condition (20) of Theorem 2.5 with = 1 holds, we nd there is a positive number L such that l 1 (n) (kl)h X I ?h m(B (x; r))  L:  4 h 2 h (l ? k) n rh n=k

So, the rst in mum in (c) is positive. To see that the second in mum is positive, note P l m ( B (0 ; 1 =k )) h that for each k, 1=kh  k n=k 1nI (nh) . Consider two cases. First, suppose that there exists 2  j 2 I such that j ? 1 2= I . If the in mum is zero, then the assumptions of Theorem 2.6 are satis ed with Z = f0g, i(0) = j , and R(0) = f1=k : k  4g. Hence, Ph(JI ) = 1 and we have a contradiction. Otherwise, I = N or I = N n f1g. Then  = 1=2 and for all k  2 1 1 1 1 (n) X X I h h 1?2h = k1?h : h k n2h = k n2h  k k 2

n=k

n=k

Since limk!1 k1?h = 1, our in mum is also positive in this case. Finally, suppose that condition (b) is satis ed. We will show condition (2) of Theorem 2.5 holds. Let M > 0 be the rst in mum appearing in (b). Fix  1 which will be

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speci ed later on. Consider i 2 I and =i(i + 1) < r < 1=i. Set k = [1=( 1i + r)] + 1 1 + 1 < 2 < 1 + 1 which equivalently means that and l = [1=( 1i ? r)] ? 1. Then l+2 k i k?1 l+1 H (k ? 1; l + 1)  i  H (k; l + 2); where H(a,b) = 2ab/a+b, the harmonic mean of a and b. Since H (k ? 1; l + 1)  H (k; l + 1)  H (k; l + 2) and H (k; l + 2) < H (k; l + 1) + 1 < H (k ? 1; l + 1) + 2, there exists (a; b) 2 f(k ? 1; l + 1); (k; l + 1); (k; l + 2)g such that 1 )  4( 1 ? 1 ). jH (a; b) ? ij  1 which means that i 2 A(a; b) \ I . Moreover, r  21 ( k?1 1 ? l+2 a b S Choose x 2 J \ (1=i + 1; 1=i) so close to 1=i that B (x; r)  lj?=1k [1=j + 1; 1=j ]. We get (ab)h

l?1 1 (n) ! X I

m(B (x; r))  4h rh (b ? a)h n=k n2h ! b 1 (n) h X 1 1 1 1 ( ab ) I  4h (b ? a)h 2h ? (k ? 1)2h ? (k)2h ? (l + 1)2h ? (l + 2)2h n n=a ! b h h X ( ab ) ( ab ) 1 ( n ) I h 1+ h  4 (b ? a)h 2h ? 4 (b ? a)h (k ? 1)2h : n=a n Now,

(ab)h (ab)h a )h
bh M;   (  (b ? a)h(k ? 1)2h (b ? a)h(a ? 1)2h a ? 1 (b ? a)hah 8 provided that k and l?k are large enough (depending only on M ) and then m(B (x; r))=rh  4?h M=2. But k will be as large as we wish by taking i suciently large and since l ? k 

i  =2, l ? k will be as large as we wish choosing large enough. Applying now ri2  i+1 Theorem 2.5(2) we nish the case when r < 1=i. In case r  1=i, set k = [1=( 1i + r)] + 1 as before. Then 1=i + r  1=k and therefore takingSx 2 J \ (1=(i + 1); 1=i) suciently close to 1=i; we get B (x; r) \ J  (0; 1=(k + 1)) \ J = nk+1 [1=(n + 1); 1=n] \ J . Thus 1 1 (n) m(B (x; r))  K ?h X I : h h r r n=k+1 n2h

Since 1=r  i +1 r  k ? 1; 1

 k ? 1 h 1 1 (n) 1 1 (n) X m(B (x; r))  K ?h (k ? 1)h X I : I h ? h ( k + 1) = K h 2 h 2h r k+1 n=k+1 n n=k+1 n

Taking i large enough, we get k  2 and kk?+11  1=3. Since the second in mum in (b) is positive, the proof is completed.

Lemma 5.2. If Ph (J ) < 1, then h  2.

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Proof. Let A > 0 be the rst in mum appearing in Theorem 5.1(b). If l 2 I is large (2l) ? lj < 1: So, enough and k = [2l=3], then j 2kk+2 l 2l 1 (n) A < (k)h(2l)h X I h #I (2l=3; 2l) = #I (2l=3; 2l) :  l h 2 h 2 (k + 2l) n=k n l2h lh

Since I is in nite, there exists an in nite sequence F  I such that [2l=3; 2l] \ [2s=3; 2s] = ; for all distinct elements l and s of F . Hence, 2l 1 X 1 X1 X X A X 2?h = 1:   # I (2 l= 3 ; 2 l )  1 (h=2)  lh nh (2l)h 2 l2I

l2F n=2l=3

l2F

l2F

Thus, h=2   and we are done.

Combining this lemma and Proposition 4.4 we get the following

Proposition 5.3. If Hh(J ) > 0 and Ph(J ) < 1, then h = 2. The proof of the following consequence of Lemma 5.2 is similar to the proof of Theorem 4.9.

Theorem 5.4. Let I  N induce a regular continued fraction system and suppose  < 1=2. Then there exists a number q  1 such that if F is strongly equivalent with I and F  [1; q], then PhF (JF ) = 1. Proof. In view of Theorem 1.1, there exists q  1 such that dimH (JF ) > 2I if F  [1; q]. Now applying Lemma 5.2 along with the fact that strongly equivalent sets have the same niteness parameter, , nishes the proof.

We shall now prove the following.

Lemma 5.5. If N n I contains punctured clusters of arbitrarily large lower lengths, then Ph(JI ) = 1. Proof. By assumption, I contains an in nite sequence of triples (a; n; b) and a < n < b, I \ [a; b] = fng such that min(b ? n; n ? a) ! 1. For each such triple, let r = inf fs : 3=4  s  1; a  [sn]; [rn=2r ? 1]  b ? 1g, let k = [rn] and l = [rn=2r ? 1]. Then k ! 1 and 6 ;, and [k; l] \ I is a singleton l ? k ! 1. Also, if n is large enough, then A( k2+kll ) \ I = contained in A( k2+kll ). Therefore,

l 1 (n) h lh X h lh  k + l 2h k (k + l)2h k I Mk;l = (l ? k)h   2h (l ? k)h kl (l ? k)h(kl)h n=k n h 2h  4h (l ? kl)h (kl)h = 4h kh(l l? k)h :

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Since maxfk; l ? kg  l=2 and since both numbers k and (l ? k) diverge to 1, it follows from this estimate that Mk;l ! 0 over such pairs of k and l. Thus an application of Theorem 5.1 nishes the proof.

Corollary 5.6. Let I be the set of prime numbers. Then Ph (JI ) = 1: Proof. It is known that the primes have arbitrarily large two sided gaps (see [E],[M]). The corollary follows.

Theorem 5.7. Let I  N be a proper in nite subset of N . If #I (k; l)  (l ? k)h for all k < l with A( k2+kll ) \ I 6= ;, then the rst in mum in Theorem 5.1(b) is positive. Proof. Consider k < l such that A( k2+kll ) \ I =6 ;. Suppose rst that l  2k. Then l 1 (n) k2h #I (k; l) 1  (l ? k)h = 1: (kl)h X I  (l ? k)h n=k n2h (l ? k)h k2h (l ? k)h

j 2j ) 6= ;. Since the If l  2k, then we can nd j such that [k; l]  [j; 2j ] and A( k2+kll ) \ A( j2+2 j 2 j 2 j 2 kl 4 points of A( k+l ) are of order k and j+2j = 3 j , we see that the numbers k and j are of the same order. Hence 2j 1 (n) l 1 (n) Xl 1I (n) h X (kl)h X I I h  k  k h 2 h 2 h 2h (l ? k) n=k n n=j n n=k n

 kh #I (j; 2j )
j12h

 khj h j ?2h =

 1:

 k h j

This implies the rst in mum in Theorem 5.1(b) is nite.

Lemma 5.8. If lim inf n!1 Snn > 1 ? 21h , then the second in mum in Theorem 5.1(b) is positive and  = 1=2 . Proof. It is straightforward to check that (1=2) = 1. By summation by parts and the fact that h > 1=2, we have

"X 1

1 h X 1I (n)

#

2hSn ? Sk?1 : h k  k 2 h 2h+1 k2h n=k n n=k n Now, there exists c > 1 ? 21h such for all suciently large k, 1 h X 1I (n)

"X 1

#

1 ? Sk?1 h k  (2 h ) ck 2 h n2h kh n=k n n = k  2h    2h S k ? 1 1 ? h 1 ? h k 2h ? 1 c ? k  k 2h ? 1 c ? 1 :

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Since 2h2?h 1 c ? 1 > 0, this implies the second in mum in Theorem 5.1(b) is positive.

Corollary 5.9. If I  N has bounded gaps, in particular, if I contains an in nite arithmetic progression, then Ph (J ) < 1. Proof. If I has gaps bounded by d, then the lower arithmetic density of I is  1=d: Also, we have #I (k; l)  d1 (l ? k) for k and l ? k large enough. So, by Theorems 5.7 and 5.8, both in ma in Theorem 5.1(c) are positive and Ph(J ) < 1. Remark. There are subsets I with bounded gaps and which do not contain an in nite arithmetic progression. We shall now formulate a sucient condition for the rst in mum in Theorem 5.1(c) to be positive.

Proposition 5.10. Let I = fan : n  1g be a subsequence of positive integers such that if an ? 1 2= I , then I \ [an; 1)  [an ; 2an]. Then with h = 1 the rst in mum in

Theorem 5.1(c) is positive. Proof. Let us x k; l 2 IN such that A( k2+kll )\I 6= ; and l ?k  10. Choose c 2 A( k2+kll )\I 6= ; and then x a cluster of the form [a; 2a] containing c. Let us explore several cases: Case 1: [k; l]  [a; 2a]. Then

Xl 1I (n)

kl l ? k = k  1 :  l ? k n=k n2 l ? k l2 l 2 Case 2: [a; 2a]  [k; l]. Since a  c  2k, we get kl

2a 1 X kl kl a = 1 kl  1 k  1 :   2 2 l ? k n=k n l ? k n=a n l ? k 4a2 4 (l ? k)a 4 a 8

kl

Xl 1I (n)

Case 3: k 2= [a; 2a] and l 2 [a; 2a]. Since c 2 A( k2+kll ) we get l ? c  l(ll+?kk) ? 1. Since also 2k  c  a  l=2, we get

Xl 1 kl l ? a kl l ? k n=k n2  l ? k n=a n2  l ? k l2  kl((ll ?? kc))  l(l kl?(kl )(?l k+) k) ? l(l ?k k) = l +k k ? l(l ?k k)  51 ? 101 1: = 10 kl

Xl 1I (n)

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(c+1) , we get Case 4: k 2 [a; 2a] and l 2= [a; 2a]. Since l?klk  2ck+1 ?k

2a 1 2k(c + 1) 2a ? k Xl 1I (n) 2k(c + 1) X  l ? k n=k n2 c + 1 ? k n=k n2  c + 1 ? k
4a2

kl

2  4a22a(c(2+a 1??k)k) = 21 c 2+a 1??kk  21 c +c ?1 ?k k

The proof is nished.

 41 :

Remark. By Lemma 4.11, the in mum considered in Proposition 5.10 is positive for all 0  h  1. We shall now construct two examples showing that in general the two inequalities in Theorem 5.1(c) are mutually independent.

Example 5.11. Here we construct an example of a regular system showing that the second

in mum in Theorem 5.1(c) may happen to be zero although the rst one is positive. It goes as follows. Fix 1=2S< s < 1. We will de ne by induction an in nite sequence fan : n  1g such that for I1 = n1 [an; 2an], the second in mum, taken with this s, in Theorem 5.1(c) fails to be positive. Indeed, set a1 = 1 and suppose that an is already de ned. The rst restriction on an+1 is that an+1 > 2an . Then (2an + 1)s

1 1 1 X 1  3sas X s as 1 a1?2s  3 n n 2s ? 1 n+1 2 s 2 s j j

j =2an +1

j =an+1 s 1?2s = 32s2 ? 1 asn an1?+12s

Since 1 ? 2s < 0, we can nd an+1 > 2an so large that s 21?2s 3 s a1?2s  1 : a (5:1) 2s ? 1 n n+1 n + 1 The construction of I1 is nished. The rst in mum in Theorem 5.1(c) is positive by Proposition 5.10 and the second one is zero by (5.1). We now de ne the set I adding to I1 an initial segment of the form [1; 2p] so long that dimH (JI ) > maxfs; I g. Then I induces a regular system (see Theorem 1.2), I continuous to satisfy the assumptions of Proposition 5.10 and the second in mum in Theorem 5.1(c) is zero by (5.1), Lemma 4.11 and since I \ [2p + 1; 1) = I1 \ [2p + 1; 1).

Example 5.12. We shall now describe a regular system for whichSthe rst in mum in ? n Theorem 5.1(c) is zero but the second one is positive. Indeed, let I = n0 [4 ; 2
4n] [f3


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2n g . Then the complement of I contains arbitrarily long punctured clusters and therefore the rst in mum in Theorem 5.1(c) is zero by Theorem 5.5. In order to check that the second in mum is positive, given k  1, consider n  0 such that 4n  k  4n+1 . Then

k

1 1 (j ) X I j =k

j2

 4n

2
X 4n+1 1 1 = 1: n
4n+1
 4 2 4
42n+2 16 j =4n+1 j

Since I = 1=2 and I (1=2) = 1, the system generated by I is co nitely regular. We are done.

?




Example 5.13. Consider I = Sn0 [4n ; 2
4n ] [ f3
2n g . Then I has positive arithmetic density, unbounded gaps and Ph(JI ) < 1. This is so, since the assumptions of Proposition 5.10 are satis ed and since the second in mum in Theorem 5.1(c) is positive which we check in exactly the same way as in Example 5.12. Notice that the sets I considered here and in Example 5.12 dier only by a rather thin set f3
2n g but the limit sets they generate have substantially dierent geometrical properties.

Theorem 5.14. If Ph(JI ) < 1; then lim supn!1 Snnh > 0: Proof. By summation by parts,

 Xl 1I (n) Xl  1 Sl ? Sk?1 : 1 + = S ? n 2 h 2 h 2 h n (n + 1) (l + 1)2h (k)2h n=k n n=k

If lim supn!1 Snnh = 0; then, for each k

"X 1 1 1 1 (n) X I h h Sn n2h k 2h = k n=k n=k n



#

? (n +11)2h ? Skk2?h1  sup Snnh ? Skk?h 1 : nk But, the right-hand side converges to 0 as k ! 1. Thus, the second in mum in Theorem 5.1(b) is not positive and we have a contradiction.

Corollary 5.15. If I is the base for an absolutely regular system, then Ph(JI ) = 1: Question. Does 0 < Ph(JI ) < 1 imply lim inf n!1 Snnh > 0?

Lemma 5.16. Let I  N : If for some s  0; we have ks Pln=k 1nI (ns ) > L > 0, for all k  1, then s  2: Proof. For all k  1, X 1I (n) X 1I (n)ks s X 1I (n) ns  n2s = k n2s  L: 2

nk

nk

P 1I (n) = 1. So,   s=2. Therefore, 1 (s=2) = 1 n=1 ns

nk

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Theorem 5.17. If I  N and PhI (JI ) < 1, then PhE (JE ) < 1, for every co nite regular

subsystem E of I. Proof. Let k0 be such that E \ [k0; 1) = I \ [k0 ; 1). Using Lemma 4.11 we see that condition (c) of Theorem 5.1 is satis ed for the system with base E. Thus, PhE (JE ) < 1.

We nish this section with its most constructive theorem whose proof shows how to produce sets of arithmetic density zero, but whose limit sets have nite packing measure.

Theorem 5.18. There exist in nite sets I  IN such that the induced continued fraction systems are strongly regular, Hh(JI ) = 0, Ph (JI ) < 1, and both numbers I and hI are arbitrarily close to zero. Proof. Fix an integer p > 4 and 3 <  < p ? 1. We will show that if the integer w is large enough, then the systems generated by the sets of the form

I = I (p; ; w) =

[

nw

[np + n ]

satisfy the requirements of our theorem. We shall show rst that the system generated by the set of entries I = I (p; ; w) is strongly regular and I (p;;w) = 1+2p . Indeed, this follows from the following computation +n 1 X 1 X npX X 1  ( t ) =  n = 1 j 2t n2pt n2pt? nw j =np

nw

nw

Let, as usual, h denote the Hausdor dimension of JI , the limit set generated by the set I (p; ; w). By Lemma 4.7, Hh(JI ) = 0. In order to prove that Ph(JI ) < 1, we shall demonstrate that the assumptions of Theorem 5.1(c) are satis ed. Indeed, in order to verify that the second in mum inPTheorem 5.1(c) is positive it suces to check that p +n ?2h P n ph lim inf k!1 Mk > 0, where Mk = k nk j =np j . We do it as follows. +n X X X npX ph j ?2h  kph n?2phn = kph n?2ph Mk = k nk nk nk j =np Z1 ph  kph x?2ph dx =  ? k2ph + 1 [x?2ph+1]1 k: k

Notice now that since h > I = 1+2p , we have  ? 2ph +1 < 0, and therefore Mk  k?ph+1 . Hence

a:1

lim inf M > 0 ,   ph + 1: k!1 k

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In order to check that the rst condition of Theorem 5.1(c) is satis ed, set l h X ( kl ) Mk;l = (l ? k)h n?2h11I (p;w;)(n): n=k

We want to show that Mk;l with A( k2+kll ) \ I (p; w; ) 6= ; are bounded away from zero. Let n be the only integer with (n ? 1)p  k < np , and let m be the only integer with mp  l < (m + 1)p. We shall consider several cases: Case 1. m  n + 2. Then we may estimate the number Mk;l from below as follows. +j 1 ?1 j X ?1 1 h mX h mX ( kl ) ( kl )  Mk;l  (l ? k)h  j 2 h h 2 ph (l ? k) j=n j j =n s=j p s p 

?1 h mX h Z m?1 ( kl ) ( kl )  ? 2 ph j  (l ? k)h = (l ? k)h x?2ph dx n j =n h ?
 (l(?kl)k)h  ? 21ph + 1 (m ? 1)?2ph+1 ? n?2ph+1 ? )ph ?2ph+1 ? n?2ph+1
 ((m + 1)(nm p ? (n ? 1)p )h (m ? 1) ph ? ph ( nm ) = ((m + 1)p ? (n ? 1)p )h ((m ? 1)p ? np ) x p ?1 : for some x, where np  x  (m ? 1)p . We continue the above estimates as follows. ? ph?p Mk;l  12 (nm)phx p (mp ? np)1?h : Since 1 +  ? 2ph ? p < 0, we get ? ph?p Mk;l  (nm)ph m p (mp ? np )1?h  nph (mp ? np )1?hm ? php p h?p : Thus, if a:2 1 +  ? 2ph + p2 h ? p  0; then the quantities Mk;l with m  n + 2 are bounded away from zero. Case 2a. We now assume that (n ? 1)p  k  np and (n + 1)p + (n + 1)  l < (n + 2)p. Then  kl h (n + 1) n2ph n Mk;l  l ? k n2ph  (l ? k)h
n2ph  n : n = (l ?n k)h  ? 
(n + 2)p ? (n ? 1)p h n(p?1)h 1+

1+

1+

2

2

1+

2

+ 2

2

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So, if

a:3

 ? (p ? 1)h  0;

then Mk;l is bounded away from 0. Case 2b. (n ? 1)p  k < np and (n + 1)p  l < (n + 1)p + (n + 1). We shall show that in this case if n is large enough, then A( k2+kll ) \ I = ;. And indeed, for this intersection to be empty it suces to know that np + n < k2+kl l ? 1 and k2+kl l + 1 < (n + 1)p : But the harmonic mean k2+kll takes on its minimum if k and l are minimal and it takes on its maximum if k and l are maximal. Thus our task reduces to check that ?(n + 1)p + (n + 1)
p p p 2 n 2( n ) ( n + 1) np + n < (n ? 1)1 p + (n + 1)p ? 1 and np + (n + 1)p + (n + 1) < (n + 1)p : provided n is large enough. This can be veri ed by a straightforward computation. Case 3. (n ? 1)p  k < np and np  l < (n + 1)p. We shall consider three subcases. Case 3a. np + n  l < (n + 1)p . Then

 kl h 2ph Mk;l  l ? k
n
n21ph  (ln? k)h
n
n21ph  = (l ?n k)h  ?

= n?(p?1)h :

n n 
(n + 1)p ? (n ? 1)p h n(p?1)h

So, again if  ? (p ? 1)h  0, then we are done. Case 3b. np  l < np + n and (n ? 1)p + (n ? 1)  k. Then

 kl h 1 ? np : Mk;l  l ? k l2h (l ? np + 1)  (ll ? k )h

But since A( k2+kll ) \ I 6= ;, we conclude that k2+kll  np ? 1. So, l ? np  l ? q ? 1. But since l ? k2+kll  (l ? k)=3, we get 1 (l ? k) ? 1 1?h  1: 3 Mk;l  h  (l ? k)

(l ? k) Case 3c. np  l < np + n and (n ? 1)p  k < (n ? 1)p + (n ? 1). Then for all n large enough ?
?
2np (n ? 1)6p > np + (n ? 1)p (n ? 1)p + (n ? 1) + 1

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2np (n ? 1)p > (n ? 1)p + (n ? 1) + 1: np + (n ? 1)p

2kl  2np (n ? 1)p > (n ? 1)p + (n ? 1) + 1 k + l np + (n ? 1)p ?
and consequently A k2+kll  np . But then

 kl h 1  l ? q h  1 h h Mk;l  l ? k
kh (l ? q)  l ? k  3 (l ? k)h  1:

and we are done in this case. Case 4. (n?1)p  k < l < np. In order for A( k2+kll \I 6= ;, (n?1)p  k  (n?1)p +(n?1) . Case 4a. l  (n ? 1)p + (n ? 1). Then

 kl h 1 Mk;l  l ? k
l2h (l ? k)  (l ? k)1?h  1:

and we are done. Case 4b. (n ? 1)p + (n ? 1) < l. Let q = k2+kll . We then get

 kl h Mk;l  l ? k
?

?(n ? 1)p + (n ? 1) ? k
1
(n ? 1)p + (n ? 1) 2h p   (n ? 1) (+l ?(nk)?h 1) ? k  (lq??kk)h

Now, q ? k  (l ? k)=3, so Mk;l  (l ? k)1?h  1 and we have nished this last case. Since  > p + 1 and h !  as w ! 1, if w is large enough  > ph + 1. The proof is completed.

x6. Dimension relations. In this section, we give some examples of strongly regular systems with dimH (J ) < dimP (J ) and some examples with equality of these two dimensions. Theorem 6.1. Let Ip = fnp : p  1g: If p  2, then dimP (JIp ) = hIp > 1=p; Hh(JIp ) > 0 and Ph (JIp ) = 1: Proof. Since N n Ip has punctured clusters of arbitrarily long lower lengths, Ph (JIp ) = 1; follows from Lemma 5.5. We will show that h > 1=p by showing that P (1=p) > 0: Since it is easy to calculate that dimB (O(0; Ip)) = 1=(p + 1), it would then follow from Theorem 2.11

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that the box and Hausdor dimensions of JIp are equal. First, we estimate n+1 (t) from below as follows: X 0 t X XX 1 1 ( t ) = jj  jj = = n+1 ! 2t 2t !2I n b2I (bqn + qn?1 ) !2I n !2I n (qn+1 (! )) X 1 X X 1 1 = [ ]  ( t ) n 2t 2t 2t : !2I n qn b2I (b + qn?1 =qn ) b2I (b + 1) +1

+1

Therefore, by induction we have

n (t) 

P

X 1! X b2I

b2t

1 2t b2I (b + 1)

!n?1

:

P 

2=p

1 1 t > 1, then P (t) > 0: It can be checked that 1 > 1, for all So, if b2I (b+1) k=1 kp +1 p  2: Finally, since %D(I ) = 1=p, it follows from Lemma 4.8 that Hh(JIp ) > 0: 2

Theorem 6.2. For every p  2, there exists q  1 such that if l  q and Il = fnp : n  lg, then dimH (Jl ) < dimB(Jl )  dimB (Jl ) = dimP (Jl ): Proof. First notice that for every l  1, Il = 1=2p and Il is a regular subset of N .

According to Theorem 1.5, liml!1 dimH (Jl ) = l = 1=2p and since 1=2p < 1=(p +1), there exists q  1 so large that dimH (Jl ) < dimB (Jl ) for all l  q. The last two equality signs in Theorem 6.2 are consequences of Theorem 3.1 in [MU] and Theorem 2.11.

Remark. Notice that in contrast to the case p  2, for p = 1 and every system strongly equivalent with I1, we have dimH (Jl ) = dimB (Jl ) = dimP (Jl ): This follows from Corollary 5.9 and Theorem 3.1 in [MU].

Theorem 6.3. If S = fi : i 2 I g is a conformal iterated function system and the index set I is in nite, then for every 0 < t <  there exists a set It  I such that dimH (JIt ) = t. Proof. Without loosing generality we may assume that I = N . First we shall show that for every set E  N such that N n E is in nite and for every " > 0 there exists k 2 N n E such that dimH (JE[fkg )  dimH (JE ) + ". Indeed, let h = dimH (JE ). By Theorem 1.2, PE (h + ") < 0 and by the de nition of pressure there exists 0 < a < 1 and j0  1 such that E;j (h + ") < aj ; if j > j0 . But, for every k 2 N n E; we have n n X

E;j (h + ")jj0k jj(n?j )(h+") K (n?j )(h+") j 3 2 j  j=0 X n 4 E;j (h + ")K (n?j )(h+") 5jj0k jj(n?j )(h+") + (a + (K jj0k jj)(h+"))n j j =0

E [fkg;n (h + ")  0

0

 j0 sup f 0j 

E;j (h + ")gnj0 K n(h+") jj0k jj(n?j0 )(h+") + (a + (K jj0k jj)(h+") )n :

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Since jj0k jj is suciently small for k suciently large, we have E[fkg;n(h + ") < 1 for all n large enough. This implies PE[fkg (h + ") < 0 and consequently dimH (JE[fkg )  h + ". The claim is proved. Passing to the actual proof, x 0 < t < N : We shall build the set It by constructing inductively an increasing sequence In of nite subsets of I satisfying dimH (JIn ) < t for S all n  1. We then will show that setting It = n1 In we have dimH (JIt ) = t. Indeed, let I1 = f1g and suppose that In is constructed and dimH (JIn ) < t. By the claim proved above there exists k > maxfIn g such that dimH (JIn [fkg) < t. Let kn+1 be such Sminimal k and let In+1 = In [ fkn+1 g. The inductive construction is nished. Let It = n1 In . This set is in nite. By Theorem 1.2 dimH (JIt )  t. If the set N n It were nite, then because of Theorem 1.3 dimH (JIt )  IN  t, and we would have a contradiction. Thus, N n It is in nite. If dimH (JIt ) = t; we are done. Otherwise, due to our claim we can nd an element q 2 N n It such that kn+1 > q > kn and dimH (JIn [fqg) < t . But then dimH (JIn [fqg)  dimH (JIt [fqg) < t which contradicts the choice of kn+1 and nishes the proof of our theorem. In general Theorem 6.3 fails to be true for t > . Indeed, below we provide an example.

Example 6.4. Consider a system of similarity maps on the interval [0,1] given by two

generators  and with contraction coecients 1=4 and the maps n with with contraction coecients cn , where c is so small that the sets ([0; 1]), ([0; 1]), and n ([0; 1]), n  1 are mutually disjoint. Then dimH (Jf; g) = 1=2 but the Hausdor dimension of any subsystem missing either  or is bounded from above by the solution to the equation (1=4)t + ct =(1 ? ct ) = 0. But t = t(c), the solution to this equation converges to 0 if c ! 0. Therefore if c is taken so small that t = t(c) < 1=4, we have a gap of Hausdor dimension between t(c) and 1=4.

Example 6.5. We give an example of an irregular continued fraction system. First notice that if I  N is an index set, we may obtain upper bounds on the the functions n (t) by a similar method to that given in Example 6.1 for obtaining lower bounds. Thus, using the Bounded Distortion Property with K = 4 and using the facts that b < b + qn?1 =qn < b + 1 and b + 1  2b, we have !n X 1 !n X 1 : ?(n?1)t  ( t )  4 n 2 t 2t b2I b b2I b From this we have

X 1!

X 1!

b2t  PI (t)  ?t log 4 + log b2I b2t : P In particular, if p > 1=P 2 and we set I = f[n(log n)p ] : n  n0g; then b2I 1b < 1; provided n0 is large enough and b2I b1s = 1; if s < 1: Thus, PI (1=2) < 0 and PI (t) = 1 if t < 1=2. So, this system is irregular. log

b2I

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x7. Some Problems. 1. Is there a nontrivial subset I of N such that 0 < Hh(JI ) and Ph(JI ) < 1? If there is such an I, we know that 0 < limsupn!1 Snnh < 1: 2. Is there a Hausdor gauge function g of the form g(t) = thL(t), where L is a slowly varying function such that 0 < Hh(JI ) or Ph(JI ) < 1, where I is the set of prime numbers? Since some detailed information is known about the distribution of the two sided gaps in the primes one can at least determine a class of g for which these measures are either 0 or 1. 3. By Theorem 6.3 we know that for every 0 < t < 1=2 there exists a continued fraction system whose limit set has dimension t. We conjecture that this remains true for all t 2 (0; 1].

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REFERENCES [Cu] T.W. Cusick, Hausdor dimension of sets of continued fractions, Quart. J. Math. Oxford (2), 41 (1990), 277-286. [E] P. Erdos, On the dierence of consecutive primes, Quart. J. Oxford 6 (1935), 124-128. [Go] I.J. Good, The fractional dimension theory of continued fractions, Proc. Camb. Philos. Soc. 37(1941), 199-228; Corrigenda, Math. Proc. Camb. Phil. Soc. 105(1989), 607; Addenda, J. Statist. Comput. and Simul. 32, Nos. 1-2 (1989), 64-68. [He] D. Hensley, Continued fraction Cantor sets, Hausdor dimension, and functional analysis, J. Number Th. 40(1992), 336-358. [M] H. Maier, Chains of large gaps between consecutive primes, Advances Math. 39 (1981), 257-269. [Ma] P. Mattila, Geometry of sets and measures in euclidean spaces, Cambridge Studies in Advanced Mathematics 44(Cambridge University Press, 1995). [MU] R.D. Mauldin, M. Urbanski, Dimensions and measures in in nite iterated function systems, Proceedings London Math. Soc., (3) 73(1996), 105-154. [Ra] G. Ramharter, Some metrical properties of continued fractions, Mathematika 30 (1983), 117-132.