The range of dimensions of microsets
aa r X i v : . [ m a t h . C A ] F e b THE RANGE OF DIMENSIONS OF MICROSETS
RICH ´ARD BALKA, M ´ARTON ELEKES, AND VIKTOR KISS
Abstract.
We say that E is a microset of the compact set K ⊂ R d if thereexist sequences λ n ≥ u n ∈ R d such that ( λ n K + u n ) ∩ [0 , d convergesto E in the Hausdorff metric, and moreover, E ∩ (0 , d = ∅ .The main result of the paper is that for a non-empty set A ⊂ [0 , d ] thereis a compact set K ⊂ R d such that the set of Hausdorff dimensions attainedby the microsets of K equals A if and only if A is analytic and contains itsinfimum and supremum. This answers another question of Fraser, Howroyd,K¨aenm¨aki, and Yu.We show that for every compact set K ⊂ R d and non-empty analytic set A ⊂ [0 , dim H K ] there is a set C of compact subsets of K which is compactin the Hausdorff metric and { dim H C : C ∈ C} = A . The proof relies on thetechnique of stochastic co-dimension applied for a suitable coupling of fractalpercolations with generation dependent retention probabilities.We also examine the analogous problems for packing and box dimensions. Introduction
The notion of a microset is due to F¨urstenberg [6], it was introduced to under-stand the infinitesimal structure of a compact set K ⊂ R d . Microsets are obtainedby ‘zooming’ in on K and taking the Hausdorff limit of what we see. For thedefinition of Hausdorff metric and other notions see the Preliminaries section. Definition 1.1.
Let d ≥ K ⊂ R d be compact.We say that E is a microset of K if E ∩ (0 , d = ∅ and there exist homotheties S n : R d → R d defined as S n ( z ) = λ n z + u n with λ n ≥ u n ∈ R d such that S n ( K ) ∩ [0 , d converges to E in the Hausdorff metric. Then { λ n } n ≥ is called a scaling sequence of E .In order not to lose information about the ‘thinnest’ part of our set K , weadded the non-standard assumption E ∩ (0 , d = ∅ to the definition. Without thisproperty { } would be a microset of every Cantor set K ⊂ R . Besides this, thedefinition also varies across different sources. F¨urstenberg [6] allows to converge to E by subsets of S n ( K ) ∩ [0 , d , which approach only provides information about the‘thickest’ part of K . Bishop and Peres [1] assumes that λ n → ∞ . Fraser, Howroyd,K¨aenm¨aki, and Yu [5] define microsets without the restriction E ∩ (0 , d = ∅ , theyonly add this extra condition to the notion of gallery. Some authors also consider Mathematics Subject Classification.
Primary 28A78, 28A80; Secondary 28A05, 82B43.
Key words and phrases. microset, weak tangent, Hausdorff dimension, packing dimension, boxdimension, fractal percolation, Hawkes’ theorem.The first author was supported by the MTA Premium Postdoctoral Research Program andthe National Research, Development and Innovation Office – NKFIH, grant no. 124749. Thesecond author was supported by the National Research, Development and Innovation Office –NKFIH, grants no. 124749 and 129211. The third author was supported by the National Research,Development and Innovation Office – NKFIH, grants no. 124749, 129211, and 128273. weak tangents, which means that S n might be arbitrary expanding similarities inthe definition instead of homotheties. As the orthogonal group of R d is compact,this concept would not significantly differ from ours. Notation 1.2.
For a compact set K ⊂ R d define M K = { E : E is a microset of K } . Let dim H , dim B , dim B , and dim P denote the Hausdorff, lower box, upperbox, and packing dimensions, respectively. For the following inequalities see [10,Page 82]. Fact 1.3.
For any set E ⊂ R d we have dim H E ≤ dim B E ≤ dim B E and dim H E ≤ dim P E ≤ dim B E. The infimum part of the following theorem was proved by Fraser, Howroyd,K¨aenm¨aki, and Yu [5, Theorem 1.1], while the supremum part is basically due toFurstenberg [6, Theorem 5.1], see also [5, Theorem 2.4].
Theorem 1.4 (Fraser–Howroyd–K¨aenm¨aki–Yu and F¨urstenberg) . Let dim be oneof dim H , dim B , dim B , or dim P . Assume that d ≥ and K ⊂ R d is a non-emptycompact set. Then { dim E : E ∈ M K } contains its infimum and supremum. Fraser, Howroyd, K¨aenm¨aki, and Yu [5, Theorem 1.3] proved the following.Recall that a set is F σ if it is a countable union of closed sets. Theorem 1.5 (Fraser–Howroyd–K¨aenm¨aki–Yu) . Let d ≥ and let A ⊂ [0 , d ] bean F σ set which contains its infimum and supremum. Then there exists a compactset K ⊂ R d such that { dim H E : E ∈ M K } = A. They asked the following question [5, Question 7.4], see also [4, Question 17.3.2].
Question 1.6 (Fraser–Howroyd–K¨aenm¨aki–Yu) . If K ⊂ R d is compact, then is { dim H E : E ∈ M K } an F σ set? If not, does it belong to a finite Borel class? As the main result of this paper, in Section 3 we answer the above questions inthe negative. In fact, we prove a complete characterization as follows.
Theorem 3.12 (Main Theorem) . Let dim be one of dim H , dim B , or dim B . Let d ≥ and let A ⊂ [0 , d ] be a non-empty set. Then the following are equivalent:(1) There exists a compact set K ⊂ R d such that { dim E : E ∈ M K } = A ;(2) A is an analytic set which contains its infimum and supremum;(3) There exists a compact set K ⊂ R d such that dim H E = dim B E for all E ∈ M K and { dim H E : E ∈ M K } = A . For packing dimension we can only prove one direction due to measurabilityproblems, see Problem 5.3. The following corollary is immediate.
Corollary 1.7.
Let d ≥ and let A ⊂ [0 , d ] be a non-empty analytic set whichcontains its infimum and supremum. Then there exists a compact set K ⊂ R d with { dim P E : E ∈ M K } = A. HE RANGE OF DIMENSIONS OF MICROSETS 3
In Section 4 we consider the range of dimensions of compact families of compactsubsets of a given compact set K ⊂ R d . Let K ( K ) denote the set of non-empty com-pact subsets of K endowed with the Hausdorff metric. Subsection 4.1 is dedicatedto the proof of the following theorem, which relies on the stochastic co-dimensionmethod applied to suitable fractal percolations. Theorem 4.4.
Let K ⊂ R d be a non-empty compact set and let A ⊂ [0 , dim H K ] .Then the following are equivalent:(1) There is a compact set C ⊂ K ( K ) with { dim H C : C ∈ C} = A ;(2) A is an analytic set. In Subsection 4.2 we consider the analogous problems for box and packing di-mensions. In case of lower box dimension the analogue of Theorem 4.4 does nothold. The following theorem of Feng, Wen, and Wu [3, Theorem 3] demonstratesthat dim B even fails to satisfy the Darboux property. Theorem 1.8 (Feng–Wen–Wu) . There exists a compact set K ⊂ [0 , such that dim B K = 1 and any set E ⊂ K satisfies dim B E ∈ { , } . For upper box dimension an analogous version of Theorem 4.4 holds.
Theorem 4.7.
Let K be a non-empty compact metric space and A ⊂ [0 , dim B K ] .The following statements are equivalent:(1) There is a compact set C ⊂ K ( K ) with { dim B C : C ∈ C} = A ;(2) A is an analytic set. For packing dimension we will show the following.
Theorem 4.8.
Let K be a non-empty compact metric space and let A ⊂ [0 , dim P K ] be analytic. Then there is a compact set C ⊂ K ( K ) with { dim P C : C ∈ C} = A . Finally, we collect the open problems in Section 5.2.
Preliminaries
Let d ≥ A ⊂ R d is analytic if there exist a Polishspace X and a continuous onto map f : X → A . Let ( X, ρ ) be a Polish space orcompact metric space. Let ( K ( X ) , d H ) be the set of non-empty compact subsets of X endowed with the Hausdorff metric , that is, for each K , K ∈ K ( X ) we have d H ( K , K ) = min { r : K ⊂ B ( K , r ) and K ⊂ B ( K , r ) } , where B ( A, r ) = { x ∈ X : ∃ y ∈ A such that ρ ( x, y ) | ≤ r } . Then ( K ( X ) , d H ) is aPolish space, and it is compact when X is compact, see [9, Theorems 4.25, 4.26].Let X be a metric space. For x ∈ X and r > B ( x, r ) be the closed ball ofradius r centered at x . For every s ≥ s -Hausdorff content of X is defined as H s ∞ ( X ) = inf ( ∞ X i =1 (diam E i ) s : X ⊂ ∞ [ i =1 E i ) , where diam E i denotes the diameter of E i . The Hausdorff dimension of X isdim H X = inf { s ≥ H s ∞ ( X ) = 0 } . RICH´ARD BALKA, M´ARTON ELEKES, AND VIKTOR KISS
Let N n ( X ) be the minimal number of closed balls of radius at most 2 − n needed tocover X . The lower box dimension and upper box dimension of E are defined asdim B X = lim inf n →∞ log N n ( X ) n log 2 and dim B X = lim sup n →∞ log N n ( X ) n log 2 , respectively. The packing dimension of X is defined bydim P X = inf ( sup i dim B E i : X ⊂ ∞ [ i =1 E i ) . For the following lemma see [1, Lemma 2.8.1].
Lemma 2.1.
Let X be a separable metric space.(i) If X is complete and dim B U ≥ α for each open set U ⊂ X , then dim P X ≥ α .(ii) If dim P X > α then there is a closed subset F ⊂ X such that dim P ( F ∩ U ) > α for each open set U intersecting F . The following fact easily follows from the finite stability of the dimensions inquestion.
Fact 2.2.
Let dim be one of dim H , dim B , or dim P . Let K be a non-empty compactmetric space. Then there exists y ∈ K such that dim B ( y, r ) = dim K for all r > . For the next claim see [2, Product formulas 7.2 and 7.5].
Claim 2.3.
Let A ⊂ R d and B ⊂ R m be compact sets. Then dim H A + dim H B ≤ dim H ( A × B ) ≤ dim B ( A × B ) ≤ dim B A + dim B B. Consult [2] or [10] for more on these concepts.For any x ∈ ω define a compact set K ( x ) ⊂ [0 ,
1] as K ( x ) = ( ∞ X i =0 a i − i − : a i = 0 if x ( i ) = 0 and a i ∈ { , } if x ( i ) = 1 ) . For n ∈ ω let x ↾ n be the restriction of x to its first n coordinates. We endow 2 ω with a metric d compatible with the product topology defined as d ( x, y ) = 2 − min { i : x ( i ) = y ( i ) } for all x, y ∈ ω . The following fact is straightforward.
Fact 2.4.
The map K : 2 ω → K ([0 , mapping x to K ( x ) is continuous, moreprecisely, d H ( K ( x ) , K ( y )) ≤ d ( x, y ) for all x, y ∈ ω . Define the left shift T : 2 ω → ω such that T ( x )( n ) = x ( n + 1) for all n ∈ ω. We define the lower density and upper density of x by ̺ ( x ) = lim inf n →∞ P n − i =0 x ( i ) n and ̺ ( x ) = lim sup n →∞ P n − i =0 x ( i ) n , respectively. If ̺ ( x ) = ̺ ( x ) then the common value ̺ ( x ) is called the density of x .For the following claim see e.g. [1, Example 3.2.3]. HE RANGE OF DIMENSIONS OF MICROSETS 5
Claim 2.5.
For any x ∈ ω we have dim H K ( x ) = dim B K ( x ) = ̺ ( x ) and dim P K ( x ) = dim B K ( x ) = ̺ ( x ) . Claims 2.3 and 2.5 allow us to calculate the dimensions of the Cartesian product K ( x ) d as follows. Fact 2.6.
Let dim be one of dim H , dim B , dim B or dim P . Assume that d ∈ N + and x ∈ ω are given such that ̺ ( x ) exists. Then dim K ( x ) d = d̺ ( x ) . For s, t ∈ <ω let s ⌢ t ∈ <ω denote the concatenation of s and t .3. Range of dimensions of microsets
The goal of this section is to prove Theorem 3.12 after some preparation.3.1.
A description of the microsets of K ( x ) d . The goal of this subsection is toprove Theorem 3.5, which provides us with a good tool to work with the microsetsof K ( x ) d . For technical reasons we generalize M K as follows. Definition 3.1.
For d ≥ F ⊂ K ( R d ) define M ( F ) as the set of compact sets K ⊂ [0 , d for which K ∩ (0 , d = ∅ and there exist K n ∈ F , λ n ≥
1, and u n ∈ R d such that ( λ n K n + u n ) ∩ [0 , d → K . Fact 3.2.
Let d ≥ and assume that C n ⊂ K n ⊂ R d are compact sets such that C n → C and K n → K . Then C ⊂ K . Definition 3.3.
For x ∈ ω and n ∈ ω define the finite set F n ( x ) = ( n − X i =0 a i − i − : a i = 0 if x ( i ) = 0 and a i ∈ { , } if x ( i ) = 1 ) and let D n ( x ) = { − n K ( T n ( x )) + u : u ∈ F n ( x ) } . Clearly, S D n ( x ) = K ( x ) and elements of D n ( x ) have pairwise non-overlappingconvex hulls. Lemma 3.4.
Let E ∈ M ( { K ( x ) : x ∈ ω } ) . Assume ( λ n K ( x n ) + u n ) ∩ [0 , → E for some λ n ≥ , x n ∈ ω , and u n ∈ R . Then there exist x ∈ ω , c ∈ R + , m n ∈ ω for all n , a subsequence of positive integers k n ↑ ∞ , a similar copy C ( x ) of K ( x ) , w = 0 and w , w , w ∈ R such that(1) T m n ( x k n ) → x ,(2) λ k n − m n → c ,(3) C ( x ) ⊂ E ⊂ S i =0 ( C ( x ) + w i ) .Proof. Define C in = 2 − i K ( T i ( x n )) for all i, n ∈ ω . Let ε ∈ (0 , /
2) such that E ∩ ( ε, − ε ) = ∅ . For all n let ϕ n : R → R be defined as ϕ n ( z ) = λ n z + u n and let p n ∈ ω be the minimal number such that there is a v n ∈ F p n ( x n ) such that(3.1) ϕ n ( C p n n + v n ) ⊂ [0 , . By the minimality of p n for all n there are at most 4 translations v ∈ F p n ( x n )satisfying ϕ n ( C p n n + v ) ∩ [0 , = ∅ , assume that they are v n , . . . , v ℓ n n for some ℓ n ∈ { , , , } . RICH´ARD BALKA, M´ARTON ELEKES, AND VIKTOR KISS
First we show λ n − p n ≥ ε/ n large enough. Indeed, if n is large enoughthen ϕ n ( K ( x n )) ∩ ( ε, − ε ) = ∅ . Thus for k = p n − D ∈ D k ( K ( x n ))with ϕ n ( D ) ∩ ( ε, − ε ) = ∅ . Assume to the contrary that λ n − p n < ε/
2. Thendiam ϕ n ( D ) ≤ λ n − k < ε , so ϕ n ( D ) ⊂ [0 , p n .Now let q n ≥ p n be the minimal integer for which λ n − q n ≤
2. We prove x n ( i ) = 0 for all i ∈ { p n , . . . , q n − } . Assume to the contrary that this is not thecase, and take the minimal i such that p n ≤ i < q n and x n ( i ) = 1. Then clearly C in = C p n n , so ϕ n ( C in + v n ) ⊂ [0 , x n ( i ) = 1, we have diam K ( T i ( x n )) ≥ / λ n − i − ≤ λ n − i diam K ( T i ( x n )) = diam ϕ n ( C in ) = diam ϕ n ( C in + v n ) ≤ . Hence λ n − i ≤
2, which contradicts the minimality of q n . As x n ( i ) = 0 for all i ∈ { p n , . . . , q n − } , we obtain C q n n = C p n n . Since λ n − q n ∈ [ ε/ ,
2] for all largeenough n , we can choose c ∈ [ ε/ ,
2] and a subsequence k n ↑ ∞ such that ℓ k n = ℓ ∈ { , , , } for all n and m n = q k n satisfies λ k n − m n → c as n → ∞ , so (2) holds. As 2 ω is compact, by choosing a subsequence we may assume thatthere exists x ∈ ω such that T m n ( x k n ) → x as n → ∞ , so (1) holds as well. As C q n n = C p n n for all n , by (3.1) and Fact 2.4 we may assumeby choosing a subsequence that ϕ k n ( C m n k n + v k n ) → cK ( x ) + v ⊂ [0 ,
1] as n → ∞ with some v ∈ R , and let C ( x ) = cK ( x ) + v . As ϕ k n ( C m n k n + v ik n ) ⊂ [ − ,
2] isisometric to ϕ k n ( C m n k n + v k n ) for every 1 ≤ i ≤ ℓ , we may suppose by choosing asubsequence that ϕ k n ( C m n k n + v ik n ) → C ( x ) + w i as n → ∞ for all 1 ≤ i ≤ ℓ with some w , . . . , w ℓ ∈ R . Let w = 0, by Fact 3.2 we obtain C ( x ) ⊂ E ⊂ ℓ [ i =0 ( C ( x ) + w i ) , so (3) holds no matter how we define w i for i > ℓ . The proof is complete. (cid:3) Theorem 3.5.
Let d ≥ and let E ∈ M ( { K ( x ) d : x ∈ ω } ) . Assume that ( λ n K ( x n ) d + u n ) ∩ [0 , d → E for some λ n ≥ , x n ∈ ω , and u n ∈ R d . Then thereexist x ∈ ω , m n ∈ ω for all n , a subsequence of positive integers k n ↑ ∞ , a similarcopy C ( x ) of K ( x ) , and v , . . . , v ℓ ∈ R d such that(i) T m n ( x k n ) → x ,(ii) C ( x ) d + v ⊂ E ⊂ S ℓi =1 ( C ( x ) d + v i ) .Proof. Let u n = ( u n , . . . , u dn ) for all n . It easily follows that E = E × · · · × E d ,where E i ⊂ [0 ,
1] are compact sets such that E i ∩ (0 , = ∅ and( λ n K ( x n ) + u in ) ∩ [0 , → E i as n → ∞ for all 1 ≤ i ≤ d . Applying Lemma 3.4 for all 1 ≤ i ≤ d successively implies thatthere exist z i ∈ ω , c i ∈ R + , m i,n ∈ ω for all n , a subsequence of positive integers HE RANGE OF DIMENSIONS OF MICROSETS 7 k n ↑ ∞ (note that this does not depend on i ), a similar copy C ( z i ) of K ( z i ), and w i,j ∈ R for 0 ≤ j ≤ T m i,n ( x k n ) → z i ,(2) λ k n − m i,n → c i ,(3) C ( z i ) ⊂ E i ⊂ S j =0 ( C ( z i ) + w i,j ).By (2) for large enough n and for all 1 ≤ i ≤ j ≤ d we obtain that m i,n − m j,n isindependent of n . We may assume that m ,n − m i,n = ℓ i ∈ N for any 1 ≤ i ≤ d and any large enough n . Let m n = m ,n and x = z , then clearly (i) holds. By(1) we obtain x = T ℓ i ( z i ) for all 1 ≤ i ≤ d . Therefore, C ( z i ) is a union of at most2 ℓ i many translates of C ( x ) for all 1 ≤ i ≤ d , so taking the product of (3) for all i ∈ { , . . . , d } implies (ii) with ℓ = Q di =1 ℓ i +2 , which finishes the proof. (cid:3) Useful lemmas for analytic sets and balanced sequences.
The goal ofthis subsection is to prove Lemmas 3.8 and 3.10.
Fact 3.6.
Let A ⊂ [0 , ∞ ) be a non-empty analytic set. Then there exist a G δ set G ⊂ ω and a continuous map f : G → [0 , ∞ ) such that f ( G ) = A .Proof. Let g : 2 ω → [0 ,
1] be a continuous surjection, and let ψ : [0 , → [0 , ∞ ) be ahomeomorphism. Since g − ( ψ − ( A )) ⊂ ω is analytic, by [9, Exercise 14.3] we canfind a G δ set H ⊂ ω × ω such that π ( H ) = g − ( ψ − ( A )), where π denotes theprojection onto the first coordinate. Let h : 2 ω → ω × ω be a homeomorphism.As g ( g − ( ψ − ( A ))) = ψ − ( A ), taking G = h − ( H ) and f = ψ ◦ g ◦ π ◦ h | G finishesthe proof. (cid:3) Definition 3.7.
For s ∈ <ω we denote by length( s ) the number of coordinates of s , where length( ∅ ) = 0. Define[ s ] = { x ∈ ω : x ↾ length( s ) = s } . Lemma 3.8.
Let A ⊂ [0 , ∞ ) be a non-empty analytic set. Then there exists a map ϕ : 2 <ω → [0 , ∞ ) such that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n ) exists for each x ∈ ω , and the resulting function ϕ satisfies ϕ (2 ω ) = A .Proof. According to Fact 3.6 we can choose a G δ set G ⊂ ω and a continuous map f : G → [0 , ∞ ) such that f ( G ) = A . The set F = 2 ω \ G is F σ , so it can be writtenas F = S ∞ n =1 F n where F n ⊂ ω are closed and F n ⊂ F n +1 for all n ≥
1. We define ϕ on an element s ∈ <ω by induction on the length of s . Fix a ∈ A arbitrarilyand define ϕ ( ∅ ) = a . Now suppose that ϕ is already defined on s ∈ <ω , our taskis to define it on s ⌢ c where c ∈ { , } . If [ s ⌢ c ] ∩ F = ∅ then let ϕ ( s ⌢ c ) be anarbitrary element of f ([ s ⌢ c ]). If [ s ⌢ c ] ∩ G = ∅ then let ϕ ( s ⌢ c ) = ϕ ( s ).It remains to define ϕ ( s ⌢ c ) if [ s ⌢ c ] intersects both G and F . Let m ( s ⌢ c )and m ( s ) be the smallest indices with [ s ⌢ c ] ∩ F m ( s ⌢ c ) = ∅ and [ s ] ∩ F m ( s ) = ∅ ,respectively. If m ( s ⌢ c ) = m ( s ) then let ϕ ( s ⌢ c ) = ϕ ( s ). Otherwise, let ϕ ( s ⌢ c ) bean arbitrary element of f ([ s ⌢ c ] ∩ G ), concluding the definition of ϕ .It remains to check that ϕ satisfies the conditions of the lemma. First note that(3.2) ϕ ( s ) ∈ A for each s ∈ <ω , a fact that can be quickly checked by induction. Let x ∈ ω be fixed. It is enough toshow that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n ) exists, ϕ ( x ) ∈ A , and if x ∈ G then ϕ ( x ) = f ( x ). RICH´ARD BALKA, M´ARTON ELEKES, AND VIKTOR KISS
First assume that [ x ↾ m ] ∩ F = ∅ for some m . Then ϕ ( x ↾ n ) is an elementof f ([ x ↾ n ]) for each n ≥ m . Hence, using the continuity of f and the fact that x ∈ G , we obtain ϕ ( x ↾ n ) → f ( x ) ∈ A .Now suppose that [ x ↾ m ] ∩ G = ∅ for some m . The definition of ϕ and (3.2)imply that there exists a ∈ A such that ϕ ( x ↾ n ) = a for all n ≥ m . It follows that ϕ ( x ↾ n ) → a ∈ A . Note that in this case x ∈ F , so we do not have to check that ϕ ( x ) = f ( x ).Finally, assume that [ x ↾ n ] intersects both G and F for each n . Clearly m ( x ↾ n )increases as n → ∞ . Suppose first that m ( x ↾ n ) → ∞ . Then x F n for each n , hence x ∈ G . Also, for infinitely many n , the value of ϕ ( x ↾ n ) is chosen from f ([ x ↾ n ]), and when it is not, then ϕ ( x ↾ n ) = ϕ ( x ↾ ( n − ϕ ( x ↾ n ) → f ( x ) ∈ A . Finally, assume that m ( x ↾ n ) converges, that is, there exists m such that m ( x ↾ n ) = m for all large enough n . Then [ x ↾ n ] ∩ F m = ∅ for each n , hence x ∈ F m ⊂ F , so we do not need to check ϕ ( x ) = f ( x ). Using (3.2) it alsofollows that there exists a ∈ A such that ϕ ( x ↾ n ) = a if n is large enough. Theproof is complete. (cid:3) Definition 3.9.
We call s ∈ <ω a subsequence of x ∈ ω if s = ( x ( k ) , x ( k + 1) , . . . , x ( k + n − k, n ∈ ω . Such a relation is denoted by s ∈ x . Let I ⊂ ω be a dis-crete interval if I = { k, . . . , k + n − } for some k, n ∈ ω , then define x ↾ I =( x ( k ) , . . . , x ( k + n − I, J ⊂ ω we write I < J ifmax { i : i ∈ I } < min { j : j ∈ J } . For s = ∅ let σ ( s ) = length( s ) − X i =0 s ( i ) and ̺ ( s ) = σ ( s )length( s ) . The sequence x ∈ ω is called balanced if for all n for any two subsequences s, t ∈ x of length n we have | σ ( s ) − σ ( t ) | ≤
1. It is known that for each a ∈ [0 ,
1] thereexists a balanced sequence x ∈ ω with ̺ ( x ) = a , see e.g. [13, Section 2.2] for thedetails. Lemma 3.10.
Let ≤ a ≤ b ≤ and assume that α, β ∈ ω are balanced with ̺ ( α ) = a and ̺ ( β ) = b . For each n let x n ∈ ω be defined as x n = s n⌢ s n⌢ s n . . . , where s kn ∈ <ω is either a subsequence of α or a subsequence of β for all k ∈ ω .Suppose that length( s n ) → ∞ and x n → x as n → ∞ . Then ̺ ( x ) ∈ { α, β } .Proof. Since α and β are balanced, it is easy to see that for any subsequences s ∈ α and t ∈ β of length n ≥ | ̺ ( α ) − ̺ ( s ) | ≤ n and | ̺ ( β ) − ̺ ( t ) | ≤ n . For N ∈ ω let I ( N ) denote the set of discrete intervals I satisfying I ⊂ [ N, ∞ ). Weclaim that it is enough to prove that there exists an N ∈ ω such that(3.4) either x ↾ I ∈ α for each I ∈ I ( N ), or x ↾ I ∈ β for each I ∈ I ( N ) . Suppose that (3.4) holds, we may assume by symmetry that x ↾ I ∈ α for each I ∈ I ( N ). We want to show that ̺ ( x ) = ̺ ( α ). Let y ∈ ω such that y ( n ) = x ( n + N ) HE RANGE OF DIMENSIONS OF MICROSETS 9 for each n ∈ ω . As ̺ ( y ) = ̺ ( x ), it is enough to prove that ̺ ( y ) = ̺ ( α ). By (3.4)for any n ≥ y ↾ n ∈ α , so (3.3) implies that | ̺ ( α ) − ̺ ( y ↾ n ) | ≤ n , thus ̺ ( y ) = ̺ ( α ) follows.Therefore it remains to show (3.4). Assume to the contrary that (3.4) failsfor all N . Then one can find non-empty discrete intervals I , I , I , I ⊂ ω with I < I < I < I such that x ↾ I α , x ↾ I β , x ↾ I α , and x ↾ I β .Fix k ∈ ω large enough so that I ∪ I ∪ I ∪ I ⊂ [0 , k ) and then fix n ∈ ω largeenough such that x n ↾ k = x ↾ k and length( s n ) > k . Then clearly x n ↾ I ∈ s n and x n ↾ I ∈ s n , or x n ↾ I ∈ s n and x n ↾ I ∈ s n . Using that both s n and s n aresubsequences of α or β , and the fact that x and x n coincide on these intervals, weobtain a contradiction. (cid:3) The proof of the Main Theorem.
Finally, in this subsection we are readyto prove Theorem 3.12. We need the following technical lemma which is implicitlycontained in [5].
Lemma 3.11.
Let K n ⊂ [0 , d be compact sets such that dim H E = dim B E forall E ∈ M ( { K n } n ≥ ) and let γ = sup { dim H K n : n ≥ } . Then there exists acompact set K ⊂ [0 , d such that dim H E = dim B E for all E ∈ M K and { dim H E : E ∈ M K } = { dim H E : E ∈ M ( { K n } n ≥ ) } ∪ { γ } . Proof.
Apply the proof of [5, Theorem 1.3] to the multiset Ω = { Q i } i ≥ , where Ω is an enumeration of { K n } n ≥ such that each set K n is repeated infinitely often. (cid:3) Theorem 3.12 (Main Theorem) . Let dim be one of dim H , dim B , or dim B . Let d ≥ and let A ⊂ [0 , d ] be a non-empty set. Then the following are equivalent:(1) There exists a compact set K ⊂ R d such that { dim E : E ∈ M K } = A ;(2) A is an analytic set which contains its infimum and supremum;(3) There exists a compact set K ⊂ R d such that dim H E = dim B E for all E ∈ M K and { dim H E : E ∈ M K } = A .Proof. The direction (3) ⇒ (1) is straightforward by Fact 1.3.Now we prove (1) ⇒ (2). Assume that A = { dim E : E ∈ M K } for somecompact set K ∈ K ( R d ). Theorem 1.4 yields that A contains its infimum andsupremum as well. To see that A is analytic, note that the set M K is F σ , sincethe set { E ∈ M K : E ∩ [ ε, − ε ] d = ∅} is closed for any ε ∈ (0 , / K ( R d ) → [0 , d ] is Borel measurable.Therefore, we obtain that A is the image of a Borel set under a Borel map, henceit is analytic by [9, Proposition 14.4].Finally, we show (2) ⇒ (3). Fix an analytic set A ⊂ [0 , d ] which contains itsinfimum and supremum. Define B = { z/d : z ∈ A } , then B ⊂ [0 ,
1] is analytic, andset a = min B and b = max B . If a = b = 0, then K can be a singleton. Hencewe may assume that b >
0. Applying Lemma 3.8 for the analytic set B yields amap ϕ : 2 <ω → [0 , ∞ ) such that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n ) exists for all x ∈ ω and ϕ (2 ω ) = B . We may assume that(3.5) a ≤ ϕ ( s ) ≤ b for each s ∈ <ω . Indeed, let us replace a value ϕ ( s ) by a if ϕ ( s ) < a , and replace ϕ ( s ) by b if ϕ ( s ) > b .As ϕ ( x ) ∈ B ⊂ [ a, b ] for each x ∈ ω , the values of ϕ do not change by modifying ϕ in this way.We now construct a continuous map ψ : 2 ω → ω and then use compact sets ofthe form ( K ( ψ ( x ))) d to construct K . Let α and β be the sequences provided byLemma 3.10 for a and b . To construct ψ , first we specify a mapping φ : 2 <ω → <ω such that φ ( s ) is a subsequence of either α or β for all s ∈ <ω . Let φ ( ∅ ) = ∅ . For s ∈ <ω with length( s ) = n ≥ φ ( s ) = ( α ↾ n ) ⌢ ( β ↾ k ), where k = k ( s ) is apositive integer such that √ n − < k < n √ n + 1 and(3.6) (cid:12)(cid:12)(cid:12)(cid:12) na + kbn + k − ϕ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ √ n . We show that such k exists. Indeed, (3.5) implies that ϕ ( s ) ∈ [ a, b ], and considerthe real function r : [ √ n − , n √ n + 1] → [ a, b ] , r ( z ) = an + bzn + z . The inequalities (cid:12)(cid:12) r (cid:0) √ n (cid:1) − a (cid:12)(cid:12) ≤ √ n , (cid:12)(cid:12) r (cid:0) n √ n (cid:1) − b (cid:12)(cid:12) ≤ √ n , and r ( z + 1) − r ( z ) ≤ n for all √ n − ≤ z ≤ n √ n easily imply the existence of k satisfying (3.6).For x ∈ ω define(3.7) ψ ( x ) = φ ( x ↾ ⌢ φ ( x ↾ ⌢ φ ( x ↾ . . . . As φ ( s ) = ∅ whenever s = ∅ , the map ψ is continuous. Since b > β isbalanced with ̺ ( β ) = b >
0, we obtain that ψ ( x ) = for all x ∈ ω , where ∈ ω is the zero sequence. We now claim that ψ satisfies(3.8) ̺ ( ψ ( x )) = ϕ ( x ) for each x ∈ ω .Let us fix x ∈ ω , and let ψ n ( x ) = φ ( x ↾ ⌢ φ ( x ↾ ⌢ . . . ⌢ φ ( x ↾ n ) . An elementary calculation using k ( s ) = o ( n ) as length( s ) = n → ∞ shows thatlength( φ ( x ↾ n ))length( ψ n ( x )) → n → ∞ , hence it is enough to show that ̺ ( ψ n ( x )) → ϕ ( x ). Therefore, it is enough to showthat ̺ ( φ ( x ↾ n )) → ϕ ( x ). Since α and β are balanced, for k = k ( x ↾ n ) we obtain(3.9) | ̺ ( α ↾ n ) − a | ≤ n and | ̺ ( β ↾ k ) − b | ≤ k . Then (3.6) and (3.9) imply that | ̺ ( φ ( x ↾ n )) − ϕ ( x ↾ n ) | = (cid:12)(cid:12)(cid:12)(cid:12) n̺ ( α ↾ n ) + k̺ ( β ↾ k ) n + k − ϕ ( x ↾ n ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ n + k + 2 √ n , which tends to 0. This implies that ̺ ( φ ( x ↾ n )) → ϕ ( x ), so the proof of (3.8) iscomplete. HE RANGE OF DIMENSIONS OF MICROSETS 11
Finally, we can construct K . As ψ is continuous, Fact 2.4 implies that the map(3.10) x K ( ψ ( x )) is also continuous.Let { x n } n ≥ be a dense subset of 2 ω with ϕ ( x ) = b . Let K n = K ( ψ ( x n )) d for all n ≥
1. First we show that(3.11) A ⊂ { dim H E : E ∈ M ( { K n } n ≥ ) } . Let z ∈ A be arbitrary. By the definition of ϕ there exists x ∈ ω such that ϕ ( x ) = z/d ∈ B . Fact 2.6 and (3.8) implydim H ( K ( ψ ( x )) d ) = d̺ ( ψ ( x )) = z. As { x n } n ≥ is dense in 2 ω , we can find a sequence k n such that x k n → x . By (3.10)we obtain that K k n = K ( ψ ( x k n )) d → K ( ψ ( x )) d as n → ∞ . As ψ ( x ) = , we obtain K ( ψ ( x )) ∩ (0 , = ∅ , hence K ( ψ ( x )) d ∩ (0 , d = ∅ . Thus K ( ψ ( x )) d ∈ M ( { K n } n ≥ ), so z ∈ { dim H E : E ∈ M ( { K n } n ≥ ) } proving (3.11).Now it is enough to prove that(3.12) dim H E = dim B E ∈ A for all E ∈ M ( { K n } n ≥ ) . Indeed, note that sup { dim H K n : n ≥ } = dim H K and K ∈ M ( { K n } n ≥ ).Then (3.11), (3.12), and Lemma 3.11 will immediately imply (3).Finally, we prove (3.12). Let E ∈ M ( { K n } n ≥ ) and let dim be one of dim H ordim B , we will calculate dim E independently of the choice of the dimension. Sincewe are only interested in dim E , by Theorem 3.5 we may suppose that E = K ( y ) d for some y ∈ ω for which there exists a subsequence of positive integers k n ↑ ∞ and m n ∈ ω such that T m n ( ψ ( x k n )) → y . We may assume by choosing a subsequencethat x k n → x for some x ∈ ω .First suppose that { m n } n ≥ is bounded. By choosing a subsequence we mayassume that m n = m for all n . The continuity of ψ implies y = T m ( ψ ( x )). As ̺ ( T m ( ψ ( x ))) = ̺ ( ψ ( x )), using (3.8) and ϕ ( x ) ∈ B we obtaindim E = dim( K ( y ) d ) = dim( K ( ψ ( x )) d ) = d̺ ( ψ ( x )) = dϕ ( x ) ∈ A, and we are done.Finally, assume that { m n } n ≥ is not bounded. We may suppose by choosing asubsequence that m n → ∞ . Applying Lemma 3.10 for T m n ( ψ ( x k n )) ∈ ω impliesthat ̺ ( y ) = a or ̺ ( y ) = b , sodim E = dim( K ( y ) d ) = d̺ ( y ) ∈ { min A, max A } , which finishes the proof of (3.12). The proof of the theorem is complete. (cid:3) Compact families of compact sets
The case of Hausdorff dimension.
The goal of this subsection is to proveTheorem 4.4 after some preparation. We define parametrized fractal percolationsin axis parallel cubes Q ⊂ R d as follows. Definition 4.1.
Let Q ⊂ R d be an axis parallel cube with side length r . For all n ∈ N + we denote by D n the collection of compact dyadic subcubes of Q of sidelength r − n . Given α n ∈ [0 , d ] for all n ≥
1, we construct a random compact setΓ( { α n } n ≥ ) ⊂ Q as follows. We keep each of the 2 d cubes in D with probability2 − α . Let ∆ ⊂ D be the collection of kept cubes and let S = S ∆ be their union. If ∆ n − ⊂ D n − and S n − = S ∆ n − are already defined, then we keep eachcube in D ∈ D n for which D ⊂ S n − independently with probability 2 − α n . Denoteby ∆ n ⊂ D n the collection of kept cubes and by S n = S ∆ n their union. Defineour percolation limit set with generation dependent retention probabilities − α n asΓ( { α n } n ≥ ) = ∞ \ n =1 S n . If α n = α for all n ≥ α ) instead of Γ( { α n } n ≥ ).We say that a random set X stochastically dominates Y if they can be definedon a common probability space (coupled) such that Y ⊂ X almost surely. For twosets A, B we write A ⊂ ⋆ B if A \ B is countable.The following theorem is due to Hawkes [7, Theorem 6] in the context of trees,see also [12, Theorem 9.5]. Theorem 4.2 (Hawkes) . For every β ∈ [0 , d ] and every compact set K ⊂ Q thefollowing properties hold(1) if dim H K < β , then almost surely, K ∩ Γ( β ) = ∅ ,(2) if dim H K > β , then K ∩ Γ( β ) = ∅ with positive probability,(3) if dim H K > β , then almost surely, dim H ( K ∩ Γ( β )) ≤ dim H K − β . Lemma 4.3.
Let K ⊂ Q be compact with dim H K = γ and let < β < γ .Then there exists a constant c = c ( Q, K, β ) > such that the following holds. If α n ∈ [0 , β ] for all n ≥ and α n → α , then (4.1) P (dim H ( K ∩ Γ( { α n } n ≥ )) ≥ β − α ) ≥ c. Proof.
Define c = P ( K ∩ Γ( β ) = ∅ ), we have c > n ≥ δ n = β − α n ≥ K ∩ Γ( { α n } n ≥ ) ∩ Γ( { δ n } n ≥ ), where Γ( { α n } n ≥ )and Γ( { δ n } n ≥ ) are independent. Since Γ( { α n } n ≥ ) ∩ Γ( { δ n } n ≥ ) stochasticallydominates Γ( β ), we obtain(4.2) P ( K ∩ Γ( { α n } n ≥ ) ∩ Γ( { δ n } n ≥ ) = ∅ ) ≥ c. As Γ( { δ n } n ≥ ) is stochastically dominated by the union of finitely many affinecopies of Γ( δ ) for any given δ < β − α , Theorem 4.2 (1), the independence ofΓ( { α n } n ≥ ) and Γ( { δ n } n ≥ ), and Fubini’s theorem imply P ( K ∩ Γ( { α n } n ≥ ) ∩ Γ( { δ n } n ≥ ) = ∅ and dim H ( K ∩ Γ( { α n } n ≥ )) < β − α ) = 0 . This and (4.2) imply (4.1), and the proof is complete. (cid:3)
Theorem 4.4.
Let K ⊂ R d be a non-empty compact set and let A ⊂ [0 , dim H K ] .The following statements are equivalent:(1) There is a compact set C ⊂ K ( K ) with { dim H C : C ∈ C} = A ;(2) A is an analytic set.Proof. First we prove (1) ⇒ (2). Since dim H : K ( R d ) → [0 , d ] is Borel measurableby [11, Theorem 2.1], if { dim H C : C ∈ C} = A then A must be analytic as theimage of a compact set under a Borel map, see e.g. [9, Proposition 14.4].Now we show (2) ⇒ (1). If A = ∅ then C = ∅ works, so we may suppose that A ⊂ [0 , dim H K ] is non-empty and analytic. Let γ = dim H K . We may assumethat A ⊂ (0 , γ ], since if an appropriate family C for A ∩ (0 , dim H K ] is constructed,then C ∪ works for A for any y ∈ K . By Fact 2.2 we can fix y ∈ K such thatdim H ( K ∩ U ) = dim H K for any open neighborhood U of y . HE RANGE OF DIMENSIONS OF MICROSETS 13
Fix a sequence of positive numbers β k ↑ dim H K , and for all k ≥ Q k bea cube around y of side length 1 /k . Fix k ≥ c k > K ∩ Q k ⊂ Q k and β k . Choose i k ∈ N + large enough so that(4.3) (1 − c k ) i k < . We will run i k independent, parameterized family of percolations inside Q k . Forall n ≥ D kn denote the collection of dyadic subcubes of Q k with side length(1 /k )2 − n and let D k = S ∞ n =1 D kn . For any D ∈ D k let u ki ( D ) be a random variableuniformly distributed in [0 ,
1] such that the family { u ki ( D ) : k ≥ , i ≤ i k , D ∈ D k } is independent. Assume that a sequence { α n } n ≥ is given such that α n ∈ [0 , γ ) forall n ≥ α n → α ∈ [0 , γ ). We define Γ ki ( { α n } n ≥ ) as follows. Let S = Q k and D k = { Q k } , and for each 1 ≤ i ≤ i k let∆ = ∆ ki, ( { α n } n ≥ ) = { D ∈ D k : u ki ( D ) ≤ − α } ,S = S ki, ( { α n } n ≥ ) = [ ∆ . If ∆ m − and S m − are already defined, let∆ m = ∆ ki,m ( { α n } n ≥ ) = { D ∈ D km : D ⊂ S m − and u ki ( D ) ≤ − α m } ,S m = S ki,m ( { α n } n ≥ ) = [ ∆ m . Finally, we define Γ ki ( { α n } n ≥ ) = ∞ \ m =0 S m . Since for any δ < α the percolation Γ ki ( { α n } n ≥ ) is stochastically dominated by theunion of finitely many affine copies of Γ( δ ), by Theorem 4.2 (3) for all 1 ≤ i ≤ i k we obtain(4.4) P (cid:0) dim H (cid:0) K ∩ Γ ki ( { α n } n ≥ ) (cid:1) ≤ γ − α (cid:1) = 1 . Moreover, if α n ∈ [0 , β k ] for all n ≥
1, then Lemma 4.3, (4.3), and the independenceof the processes Γ ki ( { α n } n ≥ ) for 1 ≤ i ≤ i k imply(4.5) P dim H K ∩ i k [ i =1 Γ ki ( { α n } n ≥ ) !! ≥ β k − α ! ≥ . For each k ≥ D ∈ D k satisfying D ∩ K = ∅ we choose a point z D ∈ D ∩ K . For all n ≥ C kn = { D ∈ D kn : D ∩ K = ∅} , and define the countable random set F ki ( { α n } n ≥ ) = ∞ [ n =0 (cid:8) z D : D ∈ C kn , D ⊂ S n , and ∄ C ∈ C kn +1 with C ⊂ S n +1 ∩ D (cid:9) . We now claim that(4.6) F ki ( { α n } n ≥ ) ∪ Γ ki ( { α n } n ≥ ) is compact for each k ≥ ≤ i ≤ i k . Indeed, F ki ( { α n } n ≥ ) \ S m is finite for all m ≥
1, hence S ∗ m = S m ∪ F ki ( { α n } n ≥ ) iscompact. Therefore F ki ( { α n } n ≥ ) ∪ Γ ki ( { α n } n ≥ ) = T ∞ m =1 S ∗ m is compact as well,which completes the proof of (4.6). For all 1 ≤ i ≤ i k letΓ ∗ k,i ( { α n } n ≥ ) = F ki ( { α n } n ≥ ) ∪ Γ ki ( { α n } n ≥ ) , and define Γ ∗ ( { α n } n ≥ ) = { x } ∪ ∞ [ k =1 i k [ i =1 Γ ∗ k,i ( { α n } n ≥ ) . Using (4.6) and the fact that Γ ∗ k,i ( { α n } n ≥ ) ⊂ Q k and Q k → { x } as k → ∞ , it isclear that Γ ∗ ( { α n } n ≥ ) is compact. As α < γ and α n < γ for all n ≥
1, we havesup { α n : n ≥ } ≤ β k for all large enough k , so we can apply (4.5) for large valuesof k . Therefore (4.4), (4.5), and the independence of the processes defining eachΓ ki ( { α n } n ≥ ) yield P (dim H ( K ∩ Γ ∗ ( { α n } n ≥ )) = γ − α ) = 1 . Our coupling of percolations clearly implies the following monotonicity: Almostsurely, for all sequences { α n } n ≥ and { α ′ n } n ≥ we have(4.7) Γ ∗ ( { α n } n ≥ ) ⊂ ⋆ Γ ∗ ( { α ′ n } n ≥ ) if α n ≥ α ′ n for each n. Let Q = Q ∩ [0 , γ ), and define the set Q ∗ = {{ α n } n ≥ : α n ∈ Q for all n and α n is eventually constant } . Clearly, Q ∗ is countable, therefore, with probability 1 we have(4.8) dim H ( K ∩ Γ ∗ ( { α n } n ≥ )) = γ − α for all { α n } n ≥ ∈ Q ∗ with α n → α. Now we are ready to define our family of compact sets C . By Lemma 3.8 thereexists a map ϕ : 2 <ω → [0 , ∞ ) such that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n ) exists for all x ∈ ω and ϕ (2 ω ) = A . Since A ⊂ (0 , γ ], we may assume that(4.9) 0 < ϕ ( s ) ≤ γ for all s ∈ <ω . Indeed, otherwise for all s ∈ n we can replace ϕ ( s ) by γ − n if ϕ ( s ) ≤ γ if ϕ ( s ) > γ , which modifications do not change the limit ϕ . For each x ∈ ω let α ( x ) = { α ( x ) n : n ≥ } such that α ( x ) n = γ − ϕ ( x ↾ n ) for all n ≥ . Let us define C as C = { K ∩ Γ ∗ ( α ( x )) : x ∈ ω } . It is clear that C is a random family of compact sets. Now we prove that, almostsurely, { dim H C : C ∈ C} = A . Assume that the event of (4.8) holds, it is enoughto show that dim H ( K ∩ Γ ∗ ( α ( x ))) = ϕ ( x ) for all x ∈ ω . Let x ∈ ω be fixed.The definition of ϕ and (4.9) imply that α ( x ) n ∈ [0 , γ ) for all n , and α ( x ) n con-verges to α x = γ − ϕ ( x ) ∈ [0 , γ ). Hence for any ε >
0, we can find sequences { α ′ n } n ≥ , { α ′′ n } n ≥ ∈ Q ∗ such that α ′ n ≤ α ( x ) n ≤ α ′′ n and α ′′ n − α ′ n ≤ ε for each n. Then α ′ n → α ′ and α ′′ n → α ′′ such that(4.10) α ′ ≤ α x ≤ α ′′ and α ′′ − α ′ ≤ ε. By (4.8) we have(4.11) dim H ( K ∩ Γ ∗ ( { α ′ n } n ≥ )) = γ − α ′ and dim H ( K ∩ Γ ∗ ( { α ′′ n } n ≥ )) = γ − α ′′ . HE RANGE OF DIMENSIONS OF MICROSETS 15
Monotonicity (4.7) yields(4.12) Γ ∗ ( { α ′′ n } n ≥ ) ⊂ ⋆ Γ ∗ ( α ( x )) ⊂ ⋆ Γ ∗ ( { α ′ n } n ≥ ) . As ε > H ( K ∩ Γ ∗ ( α ( x ))) = γ − α x = ϕ ( x ) , which proves that { dim H C : C ∈ C} = A almost surely.Finally, we check that C is compact with probability 1. Since 2 ω is compact and Q k → { y } as k → ∞ , it is enough to prove that for an arbitrarily given k ≥ ≤ i ≤ i k the map x C x def = K ∩ Γ ∗ k,i ( α ( x )) is continuous. Assume that x and y are two sequences with x ↾ n = y ↾ n for some n ≥
1, it is enough to show that(4.13) d H ( C x , C y ) ≤ − n diam Q k . The construction implies that S ki,n ( α ( x )) = S ki,n ( α ( y )) def = S n coincide, and also C x \ S n = C y \ S n . Therefore,(4.14) d H ( C x , C y ) ≤ d H ( C x ∩ S n , C y ∩ S n ) . Let D ∈ D kn be arbitrary such that D ⊂ S n . By the construction we obtain that C x ∩ D = ∅ iff D ∈ C kn iff C y ∩ D = ∅ , so diam D = 2 − n diam Q k yields(4.15) d H ( C x ∩ S n , C y ∩ S n ) ≤ − n diam Q k . Since (4.14) and (4.15) imply (4.13), the proof is complete. (cid:3)
Packing and box dimensions.
We prove Theorems 4.7 and 4.8 in this sub-section. For the following equivalent version of the upper box dimension and formore alternative definitions see [2, Chapter 3].
Definition 4.5.
In a metric space (
X, ρ ) we say that S ⊂ X is a δ -packing if ρ ( x, y ) > δ for all distinct x, y ∈ S . Let P n ( X ) be the cardinality of a maximal2 − n -packing in X . Fact 4.6.
For any metric space X we have N n ( X ) ≤ P n ( X ) ≤ N n +1 ( X ) , so dim B X = lim sup n →∞ log P n ( X ) n log 2 . Theorem 4.7.
Let K be a non-empty compact metric space and A ⊂ [0 , dim B K ] .The following statements are equivalent:(1) There is a compact set C ⊂ K ( K ) with { dim B C : C ∈ C} = A ;(2) A is an analytic set.Proof. The direction (1) ⇒ (2) is analogous to the one in Theorem 4.4.Now we prove (2) ⇒ (1). We may assume that A = ∅ , otherwise C = ∅ works.Let α = dim B K . Choose a sequence α n ↑ α . By Fact 2.2 we can fix y ∈ K suchthat dim B B ( y , r ) = α for all r > . Let g ( n ) = max { n + 1 , P n ( K ) } for all n ∈ N . We can choose a sequence k n ↑ ∞ with k = 0 and a positive integer j = j ( n ) such that g ( k n ) ≤ j ≤ k n +1 − P j ( B ( y , − g ( k n ) )) ≥ α n j . By Lemma 3.8 there is a map ϕ : 2 <ω → [0 , ∞ ) such that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n ) exists for each x ∈ ω , and the resulting function ϕ satisfies ϕ (2 ω ) = A . We mayassume that ϕ ( s ) ≤ α n for all s ∈ n and n ∈ N , otherwise we can replace ϕ ( s ) by α n without changing ϕ .Let m ( ∅ ) = 1, y ∅ = y , and C ( ∅ ) = B ( y , s ∈ n , a positiveinteger m ( s ), and points y i ( s ) ∈ K are given such that y ( s ) = y and their pairwisedistance is more than 2 − k n , and C ( s ) = m ( s ) [ i =1 B ( y i ( s ) , − k n ) , so the distance between distinct balls of the form B ( y i ( s ) , − k n ) is bigger than 2 − k n .Let c ∈ { , } and t = s ⌢ c . Let ℓ ( t ) be the minimal positive integer such that g ( k n ) ≤ ℓ ( t ) ≤ k n +1 − P ℓ ( t ) ( B ( y , − g ( k n ) )) ≥ ϕ ( t ) ℓ ( t ) , by (4.16) the number ℓ ( t ) is well-defined. Then we can choose a 2 − ℓ i -packing S ofsize exactly ⌊ ϕ ( t ) ℓ ( t ) ⌋ in B ( y i ( s ) , − g ( k n ) ), where ⌊·⌋ denotes the integer part. Byreplacing a suitable element of S with y we can obtain a 2 − ℓ ( t ) − -packing T in B ( y i ( s ) , − g ( k n ) ) such that y ∈ T and T = S . Indeed, this is straightforwardif S ∪ { y } is a 2 − ℓ i − -packing. Otherwise y ∈ B ( y, − ℓ i − ) for some y ∈ S , andreplacing y with y provides a suitable T . Let m ( t ) = m ( s ) + ( T ) − y i ( t ) = y i ( s ) if 1 ≤ i ≤ m ( s ) and { y i ( t ) } m ( s )
For the other direction we need(4.20) dim B C ( x ) ≤ ϕ ( x ) . Let γ > ϕ ( x ) be arbitrary, we use that { y i ( x ↾ n ) } ≤ i ≤ m ( x ↾ n ) forms a 2 − k n -packingin K , so m ( x ↾ n ) ≤ g ( k n ). Then for all large enough n and g ( k n ) ≤ ℓ ≤ k n +1 − N ℓ ( C ( x )) ≤ m ( x ↾ n ) + N ℓ (cid:16) C ( x ) ∩ B (cid:16) y , − g ( k n ) (cid:17)(cid:17) ≤ g ( k n ) + P ℓ (cid:16) C ( x ) ∩ B (cid:16) y , − g ( k n ) (cid:17)(cid:17) ≤ ℓ + 2 ϕ ( x ↾ ( n +1)) ℓ ≤ γℓ . For all large enough n and k n − < ℓ < g ( k n ) the inequality N ℓ ( C ( x )) ≤ γℓ clearlyholds. As γ > ϕ ( x ) was arbitrary, we obtain (4.20). Then (4.19) and (4.20) imply(4.18), so the proof is of the theorem is complete. (cid:3) For the packing dimension we prove the following theorem.
Theorem 4.8.
Let K be a non-empty compact metric space and let A ⊂ [0 , dim P K ] be analytic. Then there is a compact set C ⊂ K ( K ) with { dim P C : C ∈ C} = A .Proof. We may assume that A = ∅ , otherwise C = ∅ works. Let dim P K = α . Firstsuppose that A ⊂ [0 , β ] for some β < α . By Lemma 2.1 (ii) we may assume byshrinking K if necessary that dim B U > β for any non-empty open set U ⊂ K . ByLemma 3.8 there exists a map ϕ : 2 <ω → [0 , ∞ ) such that ϕ ( x ) = lim n →∞ ϕ ( x ↾ n )exists for each s ∈ ω , and the resulting function ϕ satisfies ϕ (2 ω ) = A . We mayassume that ϕ ( s ) ≤ β for all s ∈ <ω . Let g ( n ) = max { n + 1 , P n ( K ) } for all n ∈ N . By compactness we can choose a sequence k n ↑ ∞ with k = 0 suchthat for all y ∈ K and n ∈ N there exists a positive integer j = j ( n, y ) such that g ( k n ) ≤ j ≤ k n +1 − P j ( B ( y, − g ( k n ) )) ≥ βj . Let m ( ∅ ) = 1, y ∅ ∈ K be arbitrary, and C ( ∅ ) = B ( y ∅ , s ∈ n anda positive integer m ( s ), points y i ( s ) ∈ K with pairwise distance more than 2 − k n are given, and C ( s ) = m ( s ) [ i =1 B ( y i ( s ) , − k n ) , so the distance between distinct balls of the form B ( y i ( s ) , − k n ) is bigger than 2 − k n .Let c ∈ { , } and t = s ⌢ c . For all 1 ≤ i ≤ m ( s ) let ℓ i = ℓ i ( t ) be the minimalnumber such that g ( k n ) ≤ ℓ i ≤ k n +1 − P ℓ i ( B ( y i ( s ) , − g ( k n ) )) ≥ ϕ ( t ) ℓ i , by (4.21) the number ℓ i are well-defined. For all 1 ≤ i ≤ m ( s ) let us choose a2 − ℓ i -packing S i of size exactly ⌊ ϕ ( t ) ℓ i ⌋ in B ( y i ( s ) , − g ( k n ) ), where ⌊·⌋ denotes the integer part. Let S = S m ( s ) i =1 S i and m ( t ) = S , and let us define y i ( t ) ∈ K suchthat S = { y i ( t ) } ≤ i ≤ m ( t ) . Let C ( t ) = m ( t ) [ i =1 B ( y i ( t ) , − k n +1 ) , so the distance between distinct balls B ( y i ( t ) , − k n +1 ) is bigger than 2 − k n +1 , andclearly C ( t ) ⊂ C ( s ). Thus we defined C ( s ) for all s ∈ <ω . For x ∈ ω let C ( x ) = ∞ \ n =1 C ( x ↾ n ) . The construction clearly implies that C ( x ) ⊂ K is compact. If x ↾ n = y ↾ n ,then C ( x ) and C ( y ) are covered by the same balls of radius 2 − k n which they bothintersect, so the map x C ( x ) is continuous. Therefore, the definition C = { C ( x ) : x ∈ ω } yields a compact set C ⊂ K ( K ). In order to prove { dim P C : C ∈ C} = A let x ∈ ω be arbitrarily fixed, it is enough to show that(4.23) dim P C ( x ) = ϕ ( x ) . First we prove(4.24) dim P C ( x ) ≥ ϕ ( x ) . Let U ⊂ K be an arbitrary open set intersecting C ( x ), by Lemma 2.1 (i) it isenough to show that dim B ( C ( x ) ∩ U ) ≥ ϕ ( x ). For all large enough n we can fix1 ≤ i ≤ m ( s ) such that B = B ( y i ( x ↾ n ) , − k n ) ⊂ U . Let ℓ ( n ) = ℓ i ( x ↾ ( n + 1)), bythe construction and (4.22) the ball B contains ⌊ ϕ ( x ↾ ( n +1)) ℓ ( n ) ⌋ balls with pairwisedistance greater than 2 − ℓ ( n ) such that all of them intersect C ( x ) ∩ U . Thus P ℓ ( n ) ( C ( x ) ∩ U ) ≥ ⌊ ϕ ( x ↾ ( n +1)) ℓ ( n ) ⌋ , implying dim B ( C ( x ) ∩ U ) ≥ ϕ ( x ) . Hence (4.24) holds. For the other direction it is enough to prove that(4.25) dim B C ( x ) ≤ ϕ ( x ) . Let γ > ϕ ( x ) be arbitrary, we use that { y i ( x ↾ n ) } ≤ i ≤ m ( x ↾ n ) form a 2 − k n -packingin K , so m ( x ↾ n ) ≤ g ( k n ). Then for all large enough n and g ( k n ) ≤ ℓ ≤ k n +1 − N ℓ ( C ( x )) ≤ m ( x ↾ n ) X i =1 N ℓ (cid:16) C ( x ) ∩ B (cid:16) y i ( s ) , − g ( k n ) (cid:17)(cid:17) ≤ m ( x ↾ n ) X i =1 P ℓ (cid:16) C ( x ) ∩ B (cid:16) y i ( s ) , − g ( k n ) (cid:17)(cid:17) ≤ g ( k n )2 ϕ ( x ↾ ( n +1)) ℓ ≤ γℓ . For all large enough n and k n − < ℓ < g ( k n ) the inequality N ℓ ( C ( x )) ≤ γℓ clearlyholds. As γ > ϕ ( x ) was arbitrary, we obtain (4.25). Inequalities (4.24) and (4.25)imply (4.23), and the proof is complete if A ⊂ [0 , β ] for some β < α . HE RANGE OF DIMENSIONS OF MICROSETS 19
Finally, we prove the theorem in case of A ⊂ [0 , α ]. If A = { α } then C = { K } works, so we may assume that there exists β ∈ A ∩ [0 , α ). Take a sequence β n ↑ α ,and define A n = A ∩ [0 , β n ] for all n ≥
0. By Fact 2.2 we can choose y ∈ K suchthat dim P B ( y , r ) = α for all r >
0. By the first part we can choose a compact set K ⊂ K with dim P K = β and we may assume that y ∈ K . By the first partwe can also choose compact sets C n ⊂ K ( B ( y , /n )) such that { dim P C : C ∈ C n } = A n for all n ≥ . Define C = C ∪ { K } ∪ { K ∪ C : C ∈ C n , n ≥ } . It is easy to see that
C ⊂ K ( K ) is compact and satisfies { dim P C : C ∈ C} = A ,and the proof of the theorem is complete. (cid:3) Open problems
Our first problem is the most ambitious one. For the sake of simplicity we onlyformalize it in case of the Hausdorff dimension.
Problem 5.1.
Let K ⊂ R d be compact. Characterize the sets A ⊂ [0 , dim H K ] forwhich there exists a compact set C ⊂ K such that { dim H E : E ∈ M C } = A . J¨arvenp¨a¨a, J¨arvenp¨a¨a, Koivusalo, Li, Suomala, and Xiao [8, Lemma 2.3] provedan analogue of Hawkes’ theorem in complete metric spaces satisfying a mild dou-bling condition. Therefore, the proof of Theorem 4.4 possibly works in compactmetric spaces with a suitable doubling condition as well. As the case of arbitrarycompact metric spaces can be out of reach with this method, we ask the following.
Problem 5.2.
Let K be a non-empty compact metric space and let A ⊂ [0 , dim H K ] be an analytic set. Is there a compact set C ⊂ K ( K ) with { dim H C : C ∈ C} = A ? As Mattila and Mauldin [11, Theorem 7.5] proved that the packing dimensiondim P : K ( R d ) → [0 , d ] is not Borel measurable, we do not know whether analogueversions of Theorems 3.12 and 4.4 hold for the packing dimension. We state themeasurability problems as follows. Problem 5.3. If K ⊂ R d is a compact set, is the set { dim P E : E ∈ M K } analytic? If C ⊂ K ( R d ) is compact, is the set { dim P E : E ∈ C} analytic? If the answer to the above problem is negative, we can ask for a characterization.
Problem 5.4.
Let d ≥ . Characterize the sets A ⊂ [0 , d ] for which there exists acompact set K ⊂ R d with { dim P E : E ∈ M K } = A . Problem 5.5.
Let K ⊂ R d be compact. Characterize the sets A ⊂ [0 , dim P K ] forwhich there exists a compact set C ⊂ K ( K ) with { dim P C : C ∈ C} = A . Acknowledgments.
We are indebted to Jonathan M. Fraser for some illuminatingconversations and for providing us with the reference [3]. We thank Ville Suomalafor pointing our attention to the paper [8]. We also thank Ignacio Garc´ıa forpointing out that a result stated in an earlier version of the manuscript was alreadyknown.
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