Intersections of thick compact sets in \mathbb{R}^d
IINTERSECTIONS OF THICK COMPACT SETS IN R d KENNETH FALCONER AND ALEXIA YAVICOLI
Abstract.
We introduce a definition of thickness in R d and obtain a lower boundfor the Hausdorff dimension of the intersection of finitely or countably many thickcompact sets using a variant of Schmidt’s game. As an application we prove thatgiven any compact set in R d with thickness τ , there is a number N ( τ ) such that theset contains a translate of all sufficiently small similar copies of every set in R d withat most N ( τ ) elements; indeed the set of such translations has positive Hausdorffdimension. We also prove a gap lemma and bounds relating Hausdorff dimension andthickness. Contents
1. Introduction. 12. A Gap Lemma in R d
63. Thickness and winning sets 104. A lower bound for the dimension of intersections of thick compact sets in R d R d Introduction.
The classical co-dimension formula states that if C , C are submanifolds of R d thatintersect transversally then(1) dim( C ∩ C ) = dim( C ) + dim( C ) − d provided the right hand side is non-negative, where dim denotes the dimension of themanifolds. There are various versions of (1) that are applicable in other settings, inparticular for more general sets using Hausdorff dimension dim H . For example, for Mathematics Subject Classification.
MSC 11B25, MSC 28A12, MSC 28A78, and MSC 28A80.
Key words and phrases.
Thickness, Intersections, Patterns, Dimension, Schmidt games, GapLemma. a r X i v : . [ m a t h . C A ] F e b KENNETH FALCONER AND ALEXIA YAVICOLI compact sets C , C ⊂ R d (2) dim H ( C ∩ ( C + x )) ≤ max { , dim H ( C × C ) − d } for Lebesgue almost-all x ∈ R d ; the right-hand side can be replaced by max { , dim H ( C )+dim H ( C ) − d } if, for example, either C or C has equal Hausdorff and upper box-counting dimension, see [17]. On the other hand, for all (cid:15) > H ( C ∩ σ ( C )) ≥ max { , dim H ( C ) + dim H ( C ) − d − (cid:15) } for a set of similarities σ of positive measure with respect to the natural measure onthe group of similarities σ on R d . The similarity group may be replaced by the groupof isometries if dim H C > ( d + 1) / d ≥ C of ‘limsup sets’ of Hausdorff dimension0 < s < d which are dense in R d with the property that the intersection of any countablecollection of similar copies of sets in C still has Hausdorff dimension s , see for example [8].It is natural to ask for specific conditions on compact sets that are ‘close enough’to each other that guarantee non-empty intersection, or even give a lower bound forthe dimension of their intersection. For subsets of the real line Newhouse [19] intro-duced a notion of thickness, see Definition 1, which depends on the relative sizes of thecomplementary open intervals of the set and showed that two Cantor-like sets, withneither contained in a gap of the other, must intersect if the product of their thicknessis greater than 1, see Theorem 2.In this paper we propose a definition of thickness for compact subsets of R d for all d ≥
1. We obtain a higher dimensional gap lemma, and show that given several compactsets in R d ( d ≥
1) that are not too far apart in a sense that will be made precise, if theirthicknesses are large enough then they have non-empty intersection, and we obtain alower bound for the Hausdorff dimension of this intersection.We first review the definition of thickness for subsets of the real line. Recall that everycompact set C on the real line can be constructed by starting with a closed interval I ≡ I (the convex hull of C ) and successively removing disjoint open complementaryintervals (they are the path-connected components of the complement of C ). Clearlythere are finitely or countably many disjoint open complementary intervals ( G n ) n , whichwe may assume are ordered so that their lengths | G n | are non-increasing; if severalintervals have the same length, we order them arbitrarily. The two unbounded path-connected components of R \ C are not included. For each n ∈ N the interval G n is asubset of some closed path-connected component I n of I \ ( G ∪ · · · ∪ G n − ). We saythat such a G n is removed from I n . Definition 1 (Thickness in R ) . Let C ⊂ R be compact with convex hull I , and let ( G n ) n be the ordered sequence of open intervals comprising I \ C . Each G n is removed froma closed interval I n , leaving behind two closed intervals L n and R n ; the left and right NTERSECTIONS OF THICK COMPACT SETS OF R d intervals of I n \ G n . The thickness of C ⊂ R is defined as τ ( C ) := inf n ∈ N min {| L n | , | R n |}| G n | . The sequence of complementary intervals ( G n ) n may be finite, in which case the infimumis taken over the finite set of indices.The thickness of a single point is taken to be , and that of a non-degenerate intervalto be + ∞ . If there are several complementary intervals of equal length, then the ordering of themdoes not affect the value of τ ( C ). See [1, 11, 20, 22] for more information on Newhousethickness and alternative definitions. Theorem 2 (Newhouse’s Gap Lemma) . Given two compact sets C , C ⊂ R , such thatneither set lies in a gap of the other, if τ ( C ) τ ( C ) > then C ∩ C (cid:54) = ∅ . Theorem 2 was proved only for subsets of R and it does not guarantee positiveHausdorff dimension of the intersection, nor does it generalise in any simple way tointersections of 3 or more sets.Here we give a definition of thickness for compact subsets of R d that enables usto generalize Theorem 2 to higher dimensions, and also obtain lower bounds for theHausdorff dimension of the intersection of several sets. For a different definition ofthickness for certain dynamically defined subsets of the complex plane see [3].Our setting throughout the paper is as follows. Given a compact subset C of R d ,we define ( G n ) ∞ n =1 to be the (at most) countably many open bounded path-connectedcomponents of C C and E to be the unbounded open path-connected component of C C (except when d = 1 when E consists of two unbounded intervals). We call E togetherwith G n ( n ∈ N ) the gaps of C . We may assume that the sequence of gaps ( G n ) ∞ n =1 is ordered by non-increasing diameter. Note that we make no assumption about theconnectedness or simply connectedness of C .We write dist for the usual distance in R d . Definition 3 (Thickness in R d ) . We define the thickness of C to be τ ( C ) := inf n ∈ N dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E )diam( G n ) , provided that E is not the only path-connected component of C .When the only complementary path-connected component is E , we define (4) τ ( C ) := + ∞ if C ◦ (cid:54) = ∅ if C ◦ = ∅ We say C is thick if τ ( C ) > . KENNETH FALCONER AND ALEXIA YAVICOLI
If the sequence of complementary intervals ( G n ) n is finite then the infimum is takenover the finite set of indices. Moreover, thickness is well-defined in the sense that if twogaps have the same diameter, interchanging their positions in the ordering does notchange the definition of thickness.Note that τ ∈ [0 , + ∞ ]. Also, τ is invariant under homothetic maps, and agrees withthe usual definition of thickness in the real line (recall Definition 1). Observation 4. If C ⊂ R d is a thick compact set, then either there are finitely manygaps ( G n ) n or lim n →∞ diam G n = 0 . To see this we can assume that E is not the onlycomplementary path-connected component. If diam G n ≥ c > for infinitely many n ,taking points x n ∈ G n with | x n − x i | ≥ cτ ( C ) for ≤ i < n contradicts the sequentialcompactness of E C . In Section 2, we obtain a higher dimensional gap lemma, Theorem 10. The gaplemma does not generalize in any simple way to intersections of three or more sets, sowe need to use other methods to study such intersections. To achieve this we obtainlower bounds for the Hausdorff dimension of the intersection of several thick compactsets in terms of their thicknessess, which is easy to estimate in many cases.Our main theorem, Theorem 6 which will follow from Theorem 18 which relatesthickness to ‘winning sets’.The following constants appear in many of our results:
Definition 5. In R d ( d ≥ , let (5) K := 2 d (24 √ d ) d log(16 √ d )1 − d and K := (cid:18) (24 √ d ) d (1 + 4 d − d (cid:19) . We now state our main theorem which will follow from applying Theorem 18 on ‘win-ning sets’ to thickness. We write E i for the unbounded open path-connected componentof C Ci (the union of two unbounded intervals when d = 1). Theorem 6 (Intersection of compact sets in R d ) . Let ( C i ) i be a family of countablymany compact sets in R d , where C i has thickness τ i > , such that:(i) sup i diam( C i ) < + ∞ ,(ii) there is a ball B such that B ∩ E i = ∅ for every i , where E i is the unboundedcomponent of C Ci ,(iii) there exists c ∈ (0 , d ) such that (cid:88) i τ − ci ≤ K β c (1 − β d − c ) where β := min (cid:110) , diam( B )sup i diam( C i ) (cid:111) . Then dim H (cid:16) B ∩ (cid:92) i C i (cid:17) ≥ d − K (cid:0)(cid:80) i τ − ci (cid:1) d/c β d | log( β ) | > . NTERSECTIONS OF THICK COMPACT SETS OF R d Note that condition (iii) comes from Theorem 18 and is needed both to obtain thelower bound for the dimension of the intersection and to ensure that this bound ispositive.The significance of Theorem 6 is that a condition on thicknesses can give a lowerbound for the dimension of intersection of a finite or countable collection of sets in R d so ensure that the intersection is non-empty. In practice, the thicknesses needed arerather large as a consequence of the large constants K and K .A very active research area involves finding conditions on a set that guarantees theset contains homothetic copies of a given finite set of points, called a pattern in thiscontext. It will follow from Theorem 6 that a set contains homothetic copies of anygiven pattern in R d provided it is sufficiently thick. Patterns and intersections arerelated: the set C contains a homothetic copy of A := { a , . . . , a n } if and only if thereexists λ (cid:54) = 0 such that (cid:84) ≤ i ≤ n ( C − λa i ) (cid:54) = ∅ .A consequence of the Lebesgue density theorem is that any set E ⊂ R d of positiveLebesgue measure contains a homothetic copy of every finite set at all sufficiently smallscales, so it is natural to seek conditions on sets of zero Lebesgue measure form whichthis remains true. Perhaps the most natural notion of size to consider is Hausdorff di-mension but there are constructions (see for example [6,13,14,16,18,21]) which indicatethat Hausdorff dimension cannot, in itself, detect the presence or absence of patternsin sets of Lebesgue measure zero, even in the most basic case of points in arithmeticprogressions.(cid:32)Laba and Pramanik [15] showed that if, in addition to having large Hausdorff dimen-sion, a subset of R supports a probability measure with appropriate Fourier decay, thenit contains arithmetic progressions of length 3. The hypotheses were relaxed and thefamily of patterns covered greatly enlarged in subsequent papers [5, 9, 10]. This workuses harmonic analysis, and such methods do not work easily for longer arithmetic pro-gressions. Moreover, the hypotheses may be difficult to check, and are not even knownto hold for natural classes of fractals such as central self-similar Cantor sets.Yavicoli [22] showed that Newhouse thickness, Definition 1, allows the detection ofhomothetic and more general copies of patterns inside fractal sets in the real line.Newhouse thickness is easy to compute or estimate for many classical fractal sets suchas self-similar sets or sets defined in terms of continued fraction coefficients. Our notionof thickness in higher dimensions, Definition 3, enables such results to be extended to R d . Theorem 7.
Let C ⊂ R d be a compact set with thickness τ := τ ( C ) , such that E C contains a ball B . Then C contains a homothetic copy of every set A with at most (6) N ( τ ) := (cid:22) β d | log( β ) | eK τ d log( τ ) (cid:23) elements, where β := min (cid:110) ,
15 diam( B )16 diam( C ) (cid:111) . KENNETH FALCONER AND ALEXIA YAVICOLI and K is as in (5) .Moreover, for all λ ∈ (cid:0) , diam( B )16 diam( A ) (cid:1) , there exists a set X of positive Hausdorff dimen-sion (depending on A , B , C and λ ) such that x + λA ⊆ C for all x ∈ X. We also discuss the relationship between Hausdorff dimension and thickness of a set.It is shown in [20, p.77] that for C ⊂ R ,(7) dim H ( C ) ≥ log 2log(2 + 1 /τ ( C )) , and in Section 6 we obtain analogous lower bounds for C ⊂ R d .2. A Gap Lemma in R d In this section we extend Theorem 2, Newhouse’s gap lemma on R , to R d for d ≥ U ⊂ R d by ∂U . Definition 8.
We say that U ⊆ R d and V ⊆ R d are linked gaps if U ∩ V (cid:54) = ∅ , ( ∂U ) \ V (cid:54) = ∅ and ( ∂V ) \ U (cid:54) = ∅ .We say that C and C are linked compact sets in R d if for every pair of gaps G and G of C and C respectively we have that either their intersection is empty or they arelinked gaps. We first obtain the conclusion when C and C are linked compact sets and thenin Theorem 10 we reduce to this case the weaker condition that neither C or C is contained in any gap of the other. Figure 1 illustrates how gaps may satisfy thehypotheses of Theorem 10 but not of Proposition 9. Proposition 9.
Let C and C be linked compact sets in R d , with τ ( C ) τ ( C ) > , then C ∩ C (cid:54) = ∅ .Proof. By definition of τ , τ := τ ( C ) := inf m dist (cid:0) G m , (cid:83) ≤ i ≤ m − G i ∪ E (cid:1) diam( G m )and τ := τ ( C ) := inf n dist (cid:0) G n , (cid:83) ≤ i ≤ n − G i ∪ E (cid:1) diam( G n )where C and C have gaps G n and G n and external path-connected components E and E respectively.We assume that C ∩ C = ∅ and will obtain a contradiction. Then, C ⊆ C C = (cid:91) i G i ∪ E and C ⊆ C C = (cid:91) i G i ∪ E . NTERSECTIONS OF THICK COMPACT SETS OF R d Figure 1.
An example of gaps G and G which intersect but are notlinked and which might be parts of the complements of compact sets C and C which satisfy the hypotheses of Theorem 10 but not of Proposition9.We will construct inductively a sequence ( U i , V i ) i ∈ N of pairs of linked gaps that occur inthe construction of C and C respectively, such that either diam U i → V i → To start the induction:
Since E ∩ E (cid:54) = ∅ and C and C are linked, E and E arelinked gaps, so we take ( U , V ) := ( E , E ). Inductive step:
Given that we have defined a pair of linked gaps ( U k , V k ) of C and C defined, we now define ( U k +1 , V k +1 ).Since ( U k , V k ) is a pair of linked gaps, there is a k ∈ ∂U k \ V k . Since a k ∈ ∂U k , we have a k ∈ C , hence by assumption a k / ∈ C , so there is a gap G n k of C such that a k ∈ G n k .Note that ( U k , G n k ) are linked because they intersect and C and C are linked.In the same way, since ( U k , V k ) is a pair of linked gaps there is b k ∈ ∂V k \ U k . Since b k ∈ ∂V k , then b k ∈ C , hence b k / ∈ C , so there is G m k a gap of C such that b k ∈ G m k Again ( G m k , V k ) are linked.We will show that we can choose ( U k +1 , V k +1 ) to be either ( U k , G n k ) or ( G m k , V k ) insuch a way the diameters of either U k or V k tends to 0.We observe that for a fixed pair n, m ∈ N the following two inequalities cannot holdsimultaneously: • dist( G m , (cid:83) ≤ i ≤ m − G i ∪ E ) ≤ diam( G n ) • dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E ) ≤ diam( G m ) . For if both hold, then by definition of thickness,diam( G n ) ≥ τ diam( G m ) and diam( G m ) ≥ τ diam( G n ) . Using the hypothesis that τ τ > G m ) ≥ τ diam( G n ) ≥ τ τ diam( G m ) > diam( G m ) , KENNETH FALCONER AND ALEXIA YAVICOLI a contradiction.The gaps U k and V k can be identified as U k := G m and V k := G n for some n, m ∈ N .In the case dist( G m , (cid:83) ≤ i ≤ m − G i ∪ E ) > diam( G n ),we also know that ( U k , V k ) arelinked, so V k does not intersect G i for every 1 ≤ i ≤ m −
1. Also b k ∈ ∂V k \ U k . Then b k ∈ G m k with m k > m , and we take ( U k +1 , V k +1 ) := ( G m k , V k ).In the case dist( G m , (cid:83) ≤ i ≤ m − G i ∪ E ) ≤ diam( G n ), by the previous observationwe have dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E ) > diam( G m ). Analogously to the previous case a k ∈ G n k with n k > n , and we take ( U k +1 , V k +1 ) := ( U k , G n k ).Since one or other of these cases occurs infinitely many times, we get a sequence( U k , V k ) of linked gaps of C and C , where at least one of the diameter sequencestends to 0. Assume, by symmetry, that diam( U k ) →
0. Take x k ∈ ∂U k ⊆ C , and y k ∈ U k ∩ ∂V k ⊆ C . Then, dist( x k , y k ) ≤ diam( U k ) → . Since ( x k ) k ⊆ C there exists ( x k j ) j a subsequence ( x k j ) j convergent to x ∈ C . Since( y k j ) j ⊆ C , we also get ( y k j ) j → x ∈ C . So x ∈ C ∩ C contradicting the assumptionthat C ∩ C = ∅ . (cid:3) We can now relax the hypotheses of Proposition 9.
Theorem 10 (Gap Lemma in R d ) . Let C and C be compact sets in R d such thatneither of them is contained in a gap of the other and τ ( C ) τ ( C ) > . Then C ∩ C (cid:54) = ∅ .Proof. We write τ := τ ( C ) and τ := τ ( C ). By hypothesis, C and C are thickcompact sets.We will show that if the theorem is not trivially true then there are sets (cid:101) C and (cid:101) C with thicknesses ˜ τ ≥ τ and ˜ τ ≥ τ that satisfy the conditions of Proposition 9 suchthat (cid:101) C ∩ (cid:101) C = C ∩ C , from which the theorem follows immediately. We do this usinga sequence of steps to modify the sets so that we can assume that the sets satisfy suchstronger conditions.Note that in these steps G will always be a gap of C and G will be a gap of C ;such gaps may be unbounded unless stated otherwise.(0) We may assume that there is at least one bounded gap in the construc-tion of C , and similarly for C . Otherwise C = E C . But by hypothesis C is not contained in E , so C ∩ C (cid:54) = ∅ and the theorem is trivially true.(1) We may assume that ∂G ∩ ∂G = ∅ for all gaps G and G of C and C respectively . Otherwise there exist gaps G and G of C and C such that ∂G ∩ ∂G (cid:54) = ∅ , so C ∩ C (cid:54) = ∅ and the theorem is trivially true.(2) We may assume that ∂E (cid:42) E and ∂E (cid:42) E . Otherwise ∂E ⊆ E (or vice-versa). Since C and C are compact, there exists a closed ball B R ( x )such that C ∪ C ⊆ B R ( x ). We define r := R/ (2 τ + 1) ∈ (0 , R ), ˜ x ∈ R d such that dist( x, ˜ x ) > R and (cid:101) C := C ∪ B R (˜ x ) \ B r (˜ x ). Thus the externalpath-connected component of (cid:101) C is (cid:101) E = E \ B R (˜ x ), and there is a new gap NTERSECTIONS OF THICK COMPACT SETS OF R d G := B r (˜ x ) that was not in the construction of C . Then τ = ˜ τ by definitionof r .We take ˜ r ∈ (0 , r ) and define ˜ C := C ∪ B ˜ r (˜ x ). Then the external path-connected component of ˜ C is ˜ E := E \ B ˜ r (˜ x ) and ˜ τ = τ .By construction (cid:101) C and (cid:101) C are compact sets, with the same thicknesses as C and C , such that (cid:101) C ∩ (cid:101) C = C ∩ C , and ∂ (cid:101) E (cid:42) (cid:101) E and ∂ (cid:101) E (cid:42) (cid:101) E .(3) We may assume that no bounded gap of C is contained in a gap of C , and vice-versa . If there are bounded gaps G i of C contained in boundedgaps G j of C , we set (cid:101) C := C ∪ (cid:91) j (cid:91) G i ⊆ G j G i ;thus (cid:101) C is obtained from C by ‘filling in’ the gaps that are contained in a gapof C . Then (cid:101) C is compact with ˜ τ ≥ τ and (cid:101) C ∩ C = C ∩ C and no gaps of (cid:101) C are contained in gaps of C .Now we can apply the same argument to C and (cid:101) C (filling in certain gaps of C ) to obtain a set (cid:101) C . Hence, (cid:101) C and (cid:101) C are compact sets such that (cid:101) τ (cid:101) τ > (cid:101) C ∩ (cid:101) C = C ∩ C .(4) We may assume that there are no bounded gaps G of C such that ∂G ⊆ G and G (cid:42) G for any gap G of C , and vice-versa . If thisis not the case, we can inductively replace each gap G of C by (cid:101) G := G ∪ (cid:83) ∂G j ⊆ G G j (since ∂G ⊆ G this is intuitively G with some holes filled in).Then diam( G ) = diam( (cid:101) G ), possibly infinity. Note that a priori in this newsequence of gaps we could have gaps contained in another gap, but that can beeasily fixed by removing (in order of the sequence) the gaps that are contained ina previous gap. In this way we obtained a compact set (cid:101) C that satisfies ˜ τ ≥ τ and C ∩ C = C ∩ (cid:101) C .In a symmetric manner, we may repeat this procedure with C and (cid:101) C toobtain (cid:101) C and (cid:101) C with ˜ τ ≥ τ and ˜ τ ≥ τ and C ∩ C = (cid:101) C ∩ (cid:101) C , and such thatall gaps satisfy condition (4). (There may remain gaps of (cid:101) C fully contained ingaps of (cid:101) C or vive-versa, and these may be removed by Step (3).)(5) We may assume that ∂G (cid:42) G for every bounded gap G of C andevery gap G of C , and vice-versa. This combines Steps 3 and 4.(6)
We may assume that ∂G (cid:42) G and ∂G (cid:42) G for all gaps G of C andall gaps G of C . This means that we can assume that C and C satisfy thehypothesis of Theorem 9.We consider in turn the cases when G and G are unbounded and boundedgaps. • Case G = E and G = E : was proved in Step 2. • Case G bounded and G = E : By Step 5 we have that ∂G (cid:42) E . Tocheck that ∂E (cid:42) G , note that if ∂E ⊆ G with G a bounded gap of C , then C ⊆ G , contradicting the gap containment hypothesis of thisTheorem. • Case G bounded and G = E : as in the previous case. • Case G and G bounded: was proved in Step 5.Thus we can replace C and C by a pair of sets with the same intersection and atleast the same thicknesses which satisfy the hypotheses of Proposition 9, and applyingit completes this proof. (cid:3) Thickness and winning sets
Schmidt’s game and its variants are a powerful tool for investigating properties ofintersections of sequences of sets, see [2] for a survey. We will define a game andprove that every set with positive thickness can be seen as a winning set with certainparameters for the game. We will show that game has good properties, for examplemonotonicity in its parameters, invariance under similarities, and that the intersectionof winning sets is again a winning set with different parameters. Theorem 18, provedin the Appendix, gives a lower bound for the Hausdorff dimension of winning sets forthis game and this leads to Theorem 6 on the dimension of intersections.
Definition of the Game.
We define a game in R d similar to the potential gamefrom [4] but adapted to our purposes: Definition 11.
Given α, β, ρ > and c ≥ , Alice and Bob play the ( α, β, c, ρ ) -gamein R d under the following rules: • For each m ∈ N Bob plays first, and then Alice plays. • On the m -th turn, Bob plays a closed ball B m := B [ x m , ρ m ] , satisfying ρ ≥ ρ ,and ρ m ≥ βρ m − and B m ⊆ B m − for every m ∈ N . • On the m -th turn Alice responds by choosing and erasing a finite or countably in-finite collection A m of open sets. Alice’s collection must satisfy (cid:80) i (diam A i,m ) c ≤ ( αρ m ) c if c > , or diam A ,m ≤ αρ m if c = 0 (in the case c = 0 Alice can erasejust one set). • lim m →∞ ρ m = 0 (Note that this is a non-local rule for Bob. One can define thegame without this rule, adding that Alice wins if lim m →∞ ρ m (cid:54) = 0 . But to makethe definitions simpler we added this condition as a rule for Bob.) Alice is allowed not to erase any set, or equivalently to pass her turn.There exists a single point x ∞ = (cid:84) m ∈ N B m called the outcome of the game . We saya set S ⊂ R d is an ( α, β, c, ρ )- winning set , or just a winning set when the game is clear,if Alice has a strategy guaranteeing that if x ∞ / ∈ (cid:83) m ∈ N (cid:83) i A i,m , then x ∞ ∈ S .Note that the conditions B ⊇ B ⊇ · · · and lim m →∞ ρ m = 0 imply β < NTERSECTIONS OF THICK COMPACT SETS OF R d Good properties of the game.Proposition 12 (Countable intersection property) . Let J be a countable index set,and for each j ∈ J let S j be an ( α j , β, c, ρ ) -winning set, where c > . Then, theset S := (cid:84) j ∈ J S j is ( α, β, c, ρ ) -winning where α c = (cid:80) j ∈ J α cj (assuming that the seriesconverges). To see this, it is enough to consider the following strategy for Alice: in the turn k she plays the union over j of all the strategies of turn k . Proposition 13 (Monotonicity) . If S is ( α, β, c, ρ ) -winning and ˜ α ≥ α , ˜ β ≥ β , ˜ c ≥ c and ˜ ρ ≥ ρ , then S is ( ˜ α, ˜ β, ˜ c, ˜ ρ ) -winning. This holds because (cid:16) (cid:88) i α ˜ ci (cid:17) / ˜ c ≤ (cid:16) (cid:88) i α ci (cid:17) /c when c ≤ ˜ c, so Alice can answer in the ( ˜ α, ˜ β, ˜ c, ˜ ρ )-game using her strategy to answer from the( α, β, c, ρ )-game. Proposition 14 (Invariance under similarities) . Let f : R d → R d be a similarity satis-fying dist( f ( x ) , f ( y )) = λ dist( x, y ) for all x, y ∈ R d . Then a set S is ( α, β, c, ρ ) -winning if and only if the set f ( S ) is ( α, β, c, λρ ) -winning. This holds by “translating” the strategies being played through f . Remark 15 (Relationship with the potential game in [4]) . Let P be the set of singletonsin R d . Since every set A is contained in a ball of radius diam( A ) , if S ⊆ R d is an ( α, β, c, ρ ) -winning set, then it is an ( α, β, c, ρ, P ) -potential winning set in the gamedefined in [4]). Relationship between thickness and winning sets.
We now establish the keyproperty that relates winning sets to thickness.
Proposition 16.
Let C ⊂ R d be compact with unbounded complement E and write S := C ∪ E . If τ := τ ( C ) > , then S is (cid:0) τβ , β, , β diam( C )2 (cid:1) -winning for every β ∈ (0 , .Proof. We first describe a strategy for Alice. Given a move B by Bob, how doesAlice respond? If there exists n ∈ N such that B intersects G n and diam( B ) < dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E ), then B ∩ G n (cid:54) = ∅ and B ∩ G k = ∅ for all 1 ≤ k < n and B ∩ E = ∅ . Alice erases G n if it is a legal move, otherwise Alice does not eraseanything.To show that this strategy is winning, suppose that x ∞ / ∈ (cid:83) m A m . We want to showthat x ∞ ∈ S . Otherwise x ∞ / ∈ S so there exists n such that x ∞ ∈ G n . We will showthat Alice erases G n at some stage of the game. By definition x ∞ ∈ B m for all m ∈ N ,and we assumed x ∞ ∈ G n , so x ∞ ∈ B m ∩ G n for all m ∈ N . Since τ >
0, then dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E ) >
0. Also lim m →∞ diam( B m ) = 0, so taking m n ∈ N to bethe smallest integer such that dist( G n , (cid:83) ≤ i ≤ n − G i ∪ E ) > diam( B m n ), we know that B m n ∩ G n (cid:54) = ∅ and B m n ∩ G k = ∅ for all 1 ≤ k < n . If m n = 0, thendiam( B ) = 2 ρ ≥ ρ = β diam( C ) ≥ β dist (cid:16) G n , (cid:91) ≤ i ≤ n − G i ∪ E (cid:17) . If m n >
0, thendiam( B m n ) ≥ β diam( B m n − ) ≥ β dist (cid:16) G n , (cid:91) ≤ i ≤ n − G i ∪ E (cid:17) . So diam( B m n ) ≥ β dist (cid:0) G n , (cid:83) ≤ i ≤ n − G i (cid:83) E (cid:1) . Hence,diam( G n ) ≤ τ dist (cid:16) G n , (cid:91) ≤ i ≤ n − G i ∪ E (cid:17) ≤ τ β diam( B m n ) = α diam( B m n ) . This means that it is legal for Alice to erase G n in the m n -th turn, and her strategyspecifies that she does so. Finally, if m i = m j then the first gap intersecting B m i = B m j is G j and also G i , so i = j ; thus the elements of { m n : n ∈ N } are all different. (cid:3) Observation 17.
Let C be a compact set in R d and τ := τ ( C ) > . Then, by Propo-sition 16 and monotonicity, S := C ∪ E is a (cid:0) τβ , β, c, β diam( C ) (cid:1) -winning set for all β ∈ (0 , and all c ≥ . A lower bound for the dimension of intersections of thick compactsets in R d Whilst the gap lemma, Theorem 10, concerns the intersection of just two sets, itis of interest to obtain conditions that ensure that finitely many, or even countablymany compact subsets of R d have non-empty intersection. Using the game introducedin Definition 11 we not only obtain conditions involving thickness that ensure that suchcollection of sets in has non-empty intersection, but also get a lower bound for theHausdorff dimension of this intersection, as stated in Theorem 6.To achieve this we use the following technical theorem that gives a lower bound forthe dimension of winning sets, based on [4, Theorem 5.5] and [22, Theorem 4] andproved in the Appendix A. The parameters of a winning set provide a measure of itssize and we translate this in terms of thickness which is a single number that is easy tocompute and work with. Theorem 18.
Let S ⊆ R d be an ( α, β, c, ρ ) -winning set with c < d and β ≤ . Thenfor every ball B of radius larger than ρ , dim H ( S ∩ B ) ≥ d − K α d | log( β ) | > if α c ≤ K (1 − β d − c ) , where K and K are as in (5) . NTERSECTIONS OF THICK COMPACT SETS OF R d We now prove Theorem 6 by combining Theorem 18 with the fact that sets of positivethickness can be regarded as winning sets.
Proof of Theorem 6.
By Observation 17, for each iS i := E i ∪ C i is a (cid:18) τ i β , β, c, β C i ) (cid:19) -winning setfor all β ∈ (0 ,
1) and all c ≥
0. We fix c ∈ (0 , d ) and β ∈ (0 , ] from the hypothesis(iii). We define ρ := β sup i diam( C i ) which is a finite number by hypothesis (i). Bymonotonicity, Proposition 13, S i a (cid:0) τ i β , β, c, ρ (cid:1) -winning set. Hence, by Proposition 12, S := (cid:92) i S i is a ( α, β, c, ρ )-winning set , where(8) α := (cid:16) (cid:88) i ( τ i β ) − c (cid:17) /c = 1 β (cid:16) (cid:88) i τ − ci (cid:17) /c . By hypothesis (ii) there exists a ball B such that B ∩ E i = ∅ for all i and we take r to be the radius of B . By definition of ρ and β , we have r ≥ ρ . By hypothesis (iii) andequation (8) we have α c ≤ K (1 − β d − c ), hence we can apply Theorem 18 to getdim H ( S ∩ B ) ≥ d − K α d | log( β ) | > , and we know by definition of α that d − K α d | log( β ) | = d − K ( (cid:80) i τ − ci ) d/c β d | log( β ) | .Since B does not intersect any E i , S i ∩ B ⊆ S i ∩ E Ci ∩ B = C i ∩ B for every i, so S ∩ B ⊆ B ∩ (cid:84) i C i . The conclusion follows. (cid:3) Application: Patterns in thick compact sets of R d In this section we deduce Theorem 7 on the existence of small copies of pattensin sufficiently thick sets from Theorem 6 and illustrate this in the case of Sierpi´nskicarpets.
Proof of Theorem 7.
We write B := B for the ball with the same centre as B butwith radius rad( B ). Given a finite set A and λ ∈ (cid:0) , diam( B )16 diam( A ) (cid:1) we seek translates of λA := { b , · · · , b n } with b i ∈ R d where we can assume b = 0. As diam( λA ) < diam( B )16 then λA ⊆ B (0 , diam( B )16 ).We define C i := C − b i which is a compact set with thickness τ for every 1 ≤ i ≤ n . Byhypothesis there is a ball B ⊆ E C , so there is a ball (cid:101) B ⊆ (cid:84) ≤ i ≤ n ( B − b i ) ⊆ (cid:84) ≤ i ≤ n E Ci of diameter diam( B )(1 − ) = diam( B ). We take β := min { , diam( ˜ B )diam( C ) } , α := 1 /τ β and c := d − / log( τ β ). Then α c = eα d and d − c = 1 / log( τ β ).By Theorem 6, if(9) nα c ≤ K (1 − β d − c ) or equivalently n ≤ K α − c (1 − β d − c )then dim H ( (cid:101) B ∩ (cid:84) ≤ i ≤ n C i ) > α , β and c , and using that f ( τ ) := log( τ )(1 − β / log( τβ ) ) is a decreasingfunction with lim τ →∞ f ( τ ) = | log( β ) | ,1 K α − c (1 − β d − c ) = 1 eK ( τ β ) d (1 − β / log( τβ ) )= 1 eK τ d log( τ ) β d log( τ )(1 − β / log( τβ ) ) ≥ eK β d | log( β ) | τ d log( τ )Setting N ( τ ) := (cid:22) β d | log( β ) | eK τ d log( τ ) (cid:23) it follows from if (9) that if n ≤ N ( τ ) then dim H ( (cid:101) B ∩ (cid:84) ≤ i ≤ n C i ) >
0. If x ∈ X := (cid:101) B ∩ (cid:84) ≤ i ≤ n C i , then x + b i ∈ C i + b i = C for every 1 ≤ i ≤ n , so C ⊇ x + { b , · · · , b n } = x + λA as required. (cid:3) Sierpi´nski carpets and sponges.
Sierpi´nski carpets and sponges provide examplesof sets for which thickness is easily found and which satisfy Theorem 7 .Let n , · · · , n d ∈ N ≥ be odd natural numbers. Let D = (cid:8) i := ( i , . . . , i d ) : 1 ≤ i k ≤ n k , with ( i , . . . , i d ) (cid:54) = (cid:0) ( n + 1) , . . . , ( n d + 1) (cid:1)(cid:9) . The family of affine maps (cid:8) f i : R d → R d : i ∈ D (cid:9) , where f i ( x , . . . , x d ) = (cid:16) x + i − n , . . . , x d + i d − n d (cid:17) , forms an iterated function system, which defines a unique non-empty compact set C ⊂ R d such that C = (cid:83) i ∈ D f i ( C ), see [7]. Then C is a self-affine Sierpi´nski sponge (carpet if d = 2) which can also be realised iteratively by repeatedly substituting the coordinateparallelepipeds obtained by dividing the unit cube [0 , d into n × · · · × n d smallerparallelepipeds, with the central one removed, into themselves. In other words C = ∞ (cid:92) k =0 (cid:91) i ,..., i k ∈ D f i ◦ · · · ◦ f i k ([0 , d ) . NTERSECTIONS OF THICK COMPACT SETS OF R d We will find the thickness of C . Each parallelepiped at the k th step of the iterativeconstruction has side-lengths 1 /n ki (1 ≤ i ≤ d ). Thus the central parallelepipeds thatare removed and which form gaps at the k th step have diameterdiam k := (cid:115) (cid:88) ≤ k ≤ d n ki . The minimal distance of a gap removed at the k th step from the previous gaps and theexternal complementary component E isdist k := min ≤ i ≤ d n ki n i − . Hence, the thickness of C is τ := τ ( C ) = inf k ∈ N dist k diam k = inf k ∈ N min ≤ i ≤ d n ki n i − (cid:113)(cid:80) ≤ k ≤ d n ki . Thus, with β := min { , √ d } , Theorem 7 gives that C contains homothetic copiesof every pattern with at most N ( τ ) points where N ( τ ) is given by (6).For example, the self-similar Sierpi´nski carpet C n in R , taking n = n = n above,has thickness τ = ( n − (cid:14) √
2, so there is a homothetic copy in C n of every configurationof up to N ( τ ) points. Because K is large, n needs to be large to guarantee even thatsimilar copies of all triangles can be found in C n . On the other hand, for C n to containcopies of all k -point configurations, n = O (( k log k ) / ) which does not increase toorapidly for large k . 6. Thickness and Hausdorff dimension
In this section we obtain two different lower bounds for the Hausdorff dimension ofsets in R d in terms of their thickness.Firstly, Theorem 6 yields a lower bound by taking a single set C . Corollary 19.
Let C be a compact set in R d with positive thickness τ (so diam( C ) < + ∞ and there is a ball B such that B ∩ E = ∅ ). If there exists c ∈ (0 , d ) such that τ − c ≤ K β c (1 − β d − c ) for β := min { , diam( B )diam( C ) } then dim H ( C ) ≥ dim H ( B ∩ C ) ≥ d − K τ − d β d | log( β ) | > . Secondly, we can get a lower bound in the case of convex sets with convex gaps byconsidering 1-dimensional sections.
Proposition 20.
Let C be a proper compact convex set in R d where d ≥ , and let C = C \ (cid:83) ∞ k =1 G k , where { G k } k are open convex gaps ordered by decreasing diameters.Then τ ( C ∩ L ) ≥ τ ( C ) for every straight line L that properly intersects C . Proof.
Let L be a straight line that properly intersects C . Let { I i } ∞ i =1 be the (countableor finite) set of open intervals I i := G k ( i ) ∩ L in L ordered so that | I i | ≤ | I j | if i ≥ j ,where | | denotes the length of an interval. Let 1 ≤ i ≤ j −
1. There are two cases:(a) if k ( i ) < k ( j ) thendist( I i , I j ) ≥ dist( G k ( i ) , G k ( j ) ) ≥ τ ( C ) diam( G k ( j ) ) ≥ τ ( C ) | I j | ;(b) if k ( j ) < k ( i ) thendist( I i , I j ) ≥ dist( G k ( i ) , G k ( j ) ) ≥ τ ( C ) diam( G k ( i ) ) ≥ τ ( C ) | I i | ≥ τ ( C ) | I j | . In both cases dist( I i , I j ) ≥ | I j | for all 1 ≤ i ≤ j − τ ( C ∩ L ) ≥ τ ( C ) from thedefinition of thickness. (cid:3) We can now obtain a lower bound for the Hausdorff dimension for these sets in termsof thickness using the bound (7) fore sets in R . Proposition 21.
Let C ⊆ R d be a proper compact convex set, and let C = C \ (cid:83) ∞ i =1 G i where { G i } i are open convex gaps. Then (10) dim H ( C ) ≥ d − /τ ( C )) where τ ( C ) is the thickness of C .Proof. Let L be a straight line that properly intersects C . Combining the relation-ship between thickness and Hausdorff dimension for subsets of R stated in (7) withProposition 20,dim H ( C ∩ L ) ≥ log 2log(2 + 1 /τ ( C ∩ L )) ≥ log 2log(2 + 1 /τ ( C )) . This is true for all lines L in a given direction that properly intersect C , so by astandard result relating the Hausdorff dimension of a set to the Hausdorff dimensionsof parallel sections, see for example, [7, Corollary 7.10], inequality (10) follows. (cid:3) Observation 22.
When d = 1 Proposition 21 is better than Corollary 19. For d ≥ ,Corollary 19 gives a better bound than Proposition 21 when τ is large but when τ issmall Proposition 21 is better. Appendix A. Proof of Theorem 18
The proof of Theorem 18 is based on [4, Theorem 5.5] and [22, Theorem 4] andadapted to our particular setting.
Proof.
We can assume without loss of generality that the radius of B is ρ . We let x be the center of B , ρ n := β n ρ radii of balls, and E n := ρ n Z d + x centers of balls of the family E n := (cid:110) B ( ρ n z + x , ρ n ) : z ∈ Z d (cid:111) . NTERSECTIONS OF THICK COMPACT SETS OF R d We will take Bob’s move of the n -turn from E n .We also define D n := 3 ρ n Z d + x ⊂ E n , D n := { B (3 ρ n z + x , ρ n ) : z ∈ Z d } ⊂ E n . Note that the elements of D n are disjoint (moreover they are at distance ρ n ).We fix γ ∈ (0 , α , β , c and ρ ). Let N := (cid:98) γ d α d (cid:99) .We define the function π n : E n +1 → E n , B (cid:55)→ π n ( B ) in the following way: • When n (cid:54) = jN for all j : we define π n ( B ) as the element of E n that contains B such that B is as centered as possible inside that element. • When n = jN for some j : If there exists B (cid:48) ∈ D jN containing B , we define π n ( B ) := B (cid:48) (it is well defined because in that case there is only one elementbelonging to D jN ). If not, we define the function as before.Intuitively the function π n carries the elements of level n + 1 to its ancestor of level n . We use the following notation: for m < n and B ∈ E n , π m ( B ) := π m ◦ π m +1 ◦· · · ◦ π n − ( B ) ∈ E m . This is to say, we carry B to its ancestor of level m via thefunctions π . If Bob plays B ∈ E n in the turn n , we consider that in the previous turns m ∈ { , · · · , n − } Bob has played π m ( B ). Then, we have the following inclusions ofmovements from the turn n to the turn 0: B ⊂ π n − ( B ) ⊂ · · · ⊂ π ( B ) . We defined the function in this way to guarantee that Bob’s moves are legal. Aliceresponds under her winning strategy. If in the turn n Bob plays B ∈ E n , we define A ( B ) as Alice’s answer (each A ∈ A ( B ) is a countable collection of sets A := { A i,n } i ,and a legal movement as an answer for B , i.e.: (cid:80) i diam( A i,n ) c ≤ ( αρ n ) c ). Let A ∗ m ( B ) := { A ∈ A ( π m ( B )) : B ∩ A (cid:54) = ∅} be Alice’s answer (this is a list of sets) to the ancestor of B of level m < n .Given any ball B , we denote by B the ball with the same center as B and the halfof the radius.Note that as β ≤ , if B ∈ D jN and B (cid:48) ∈ E jN +1 satisfying that B (cid:48) ∩ B (cid:54) = ∅ , then B (cid:48) ⊂ B , so π jN ( B (cid:48) ) = B . It follows that(11) if n > jN, B (cid:48) ⊂ B with B (cid:48) ∈ E n and B ∈ D jN , then π jN ( B (cid:48) ) = B. This is true because if we look at the ancestor of B (cid:48) of level jN + 1, since π n choosesthe element belonging to E n that contains B such that B is as centered as possible, thatelement must intersect B .We define for every B ∈ D j φ j ( B ) := (cid:88) n Observation 23. If B ∈ D (cid:48) jN , we have that rad ( B ) = β jN ρ , and rad ( B (cid:48) ) = β ( j +1) N ρ for every B (cid:48) ∈ D ( j +1) N . We can cover B with enlarged balls from D ( j +1) N ( B ) (withradii ρ ( j +1) N √ d ). This gives us a lower bound for D ( j +1) N ( B ) : L d ( 12 B ) ≤ D ( j +1) N ( B ) L d ( B ρ ( j +1) N √ d ) , so β − Nd d d √ d d ≤ D ( j +1) N ( B ) . Proposition 24. If α c ≤ K (1 − β d − c ) where K := max { γ − d , γ − d log( γ − d ) } , we havethat D ( j +1) N ( B ) ∩ D (cid:48) ( j +1) N ) ≥ β − Nd (cid:32) d d √ d d − d γ d (1 + 4 d (cid:33) for all B ∈ D (cid:48) jN . We denote by rad( B ) the radius of the ball B . We start by proving two preliminarylemmas: Lemma 25. a) For all n ∈ N and B (cid:48) ∈ E n we have that (cid:88) A ∈A ( B (cid:48) ) min (cid:26) , diam( A i,n ) c ( γρ ( j +1) N ) c (cid:27) (cid:18) diam( A i,n ) + 2 ρ ( j +1) N rad( B (cid:48) ) (cid:19) d ≤ d α c max (cid:40) α d − c , γ − c (cid:18) ρ ( j +1) N rad( B (cid:48) ) (cid:19) d − c (cid:41) . b) If B ∈ D (cid:48) jN then (cid:88) n Firstly, splitting into the cases x ≤ y and y ≤ x , it is easy to seethat(12) min (cid:26) , x c ( γy ) c (cid:27) ( x + 2 y ) d ≤ d x c max (cid:26) x d − c , y d − c γ c (cid:27) for all x, y > . Secondly, we will prove that if n ∈ N and B (cid:48) ∈ E n then the claim a) holds. By applyingthe inequality (12) to x := diam( A i,n )rad( B (cid:48) ) and y := ρ ( j +1) N rad( B (cid:48) ) , summing over all A i,n ∈ A ( B (cid:48) ) NTERSECTIONS OF THICK COMPACT SETS OF R d and using that Alice is playing legally, we have that (cid:88) A i,n ∈A ( B (cid:48) ) min (cid:26) , diam( A i,n ) c ( γρ ( j +1) N ) c (cid:27) (cid:18) diam( A i,n ) + 2 ρ ( j +1) N rad( B (cid:48) ) (cid:19) d ≤ d (cid:88) A i,n ∈A ( B (cid:48) ) (cid:18) diam( A i,n )rad( B (cid:48) ) (cid:19) c max (cid:40)(cid:18) diam( A i,n )rad( B (cid:48) ) (cid:19) d − c , γ − c (cid:18) ρ ( j +1) N rad( B (cid:48) ) (cid:19) d − c (cid:41) ≤ d max (cid:40)(cid:18) max A i,n ∈A ( B (cid:48) ) diam( A i,n )rad( B (cid:48) ) (cid:19) d − c , γ − c (cid:18) ρ ( j +1) N rad( B (cid:48) ) (cid:19) d − c (cid:41) (cid:88) A i,n ∈A ( B (cid:48) ) (cid:18) diam( A i,n )rad( B (cid:48) ) (cid:19) c ≤ d α c max (cid:40) α d − c , γ − c (cid:18) ρ ( j +1) N rad( B (cid:48) ) (cid:19) d − c (cid:41) . Finally, we prove the claim b). By applying the inequality (12) to x := diam( A i,n )rad( B ) and y := ρ ( j +1) N rad( B ) , summing over all elements of (cid:83) n Proof of Proposition 24. D ( j +1) N ( B ) \ D (cid:48) ( j +1) N ) ≤ (cid:26) B (cid:48) ∈ D ( j +1) N ( B ) : φ ( j +1) N ( B (cid:48) )( γρ ( j +1) N ) c > (cid:27) ≤ (cid:88) B (cid:48) ∈D ( j +1) N ( B ) min (cid:26) , φ ( j +1) N ( B (cid:48) )( γρ ( j +1) N ) c (cid:27) ≤ (cid:88) B (cid:48) ∈D ( j +1) N ( B ) min , (cid:88) n< ( j +1) N (cid:88) A i,n ∈A ∗ n ( B (cid:48) ) diam( A i,n ) c ( γρ ( j +1) N ) c ≤ (cid:88) B (cid:48) ∈D ( j +1) N ( B ) (cid:88) n< ( j +1) N (cid:88) A i,n ∈A ∗ n ( B (cid:48) ) min (cid:26) , diam( A i,n ) c ( γρ ( j +1) N ) c (cid:27) ≤ (cid:88) B (cid:48) ∈D ( j +1) N ( B ) (cid:88) n 1, and thus(17) N ≥ γ d α − d . On the other hand, using the hypotheses, c ∈ (0 , d ) and α ∈ (0 , α d ≤ α c ≤ K (1 − β d − c ) ≤ K | log( β d − c ) | = 1 K ( d − c ) | log( β ) | , where in the last inequality we have used that d − c ∈ (0 , β ∈ (0 , ], z := β d − c ∈ (0 , f ( z ) := log( z ) + z + 1 is a positive function on (0 , − z ≤ log( z ). Then,(19) N α d K ≤ N ( d − c ) | log( β ) | . By inequalities (17) and (19) and the definition of K , N ( d − c ) | log( β ) | ≥ N α d K ≥ γ d K ≥ | log( γ d ) | which is equivalent to claim (3).This concludes the proof of Proposition 24. (cid:3) Conclusion of the proof. For each γ ∈ (0 , 1) we proceed as follows:By definition, B ∈ D . Moreover, φ ( B ) := 0 < ( γρ ) c , so B ∈ D (cid:48) . We willconstruct a Cantor set F as the intersection of a sequence of unions of closed sets: • B := { B } ⊂ D (cid:48) . • Given a collection B j ⊂ D (cid:48) jN we construct the next level of sets B j +1 ⊂ D (cid:48) ( j +1) N by replacing each element of B ∈ B j by M := (cid:108) β − Nd (cid:16) d d √ d d − d γ d (1 + 4 d (cid:17) (cid:109) elements of D ( j +1) N ( B ) ∩ D (cid:48) ( j +1) N ; this is possible by Proposition 24.We define F := (cid:92) j ∈ N (cid:91) B ∈B j B. NTERSECTIONS OF THICK COMPACT SETS OF R d By a standard argument (see e.g. [7, Example 4.6]),dim H ( F ) ≥ log( M ) | log( β N ) | ≥ log( β − Nd ) + log (cid:16) d d √ d d − d γ d (1 + 4 d (cid:17) N | log( β ) | = d + log (cid:16) d d √ d d − d γ d (1 + 4 d (cid:17) N | log( β ) |≥ d − α d log (cid:18)(cid:16) d d √ d d − d γ d (1 + 4 d (cid:17) − (cid:19) γ d | log( β ) | , where we have used (17).This last inequality holds for every γ ∈ (0 , γ ∈ (0 , d γ d (1 + 4 d 2) = (cid:0) − d (cid:1) √ d ) d (this is not sharp, but it is close enough). Forthis γ we get dim H ( F ) ≥ d − K α d | log( β ) | , where K := 2 d (24 √ d ) d (cid:16) log(2) + log(8 √ d ) (cid:17) − d making K := (cid:32) (24 √ d ) d (1 + 4 d − d (cid:33) . It remains to prove that F ⊂ S ∩ B , since thendim H ( S ) ≥ dim H ( S ∩ B ) ≥ dim H ( F ) ≥ d − K α d | log( β ) | . Clearly F ⊂ B , by definition of F . We need to show that F ⊂ S . Let x ∈ F . Forevery j ∈ N there exists a unique B jN ∈ B j containing x . By definition of B j +1 wehave that B ( j +1) ⊂ B jN . By (11), π jN ( B ( j +1) N ) = B jN . The sequence ( B jN ) j can beextended in a unique way to a sequence ( B n ) n satisfying B n ∈ E n and B n := π n ( B n +1 )for all n . We interpret this sequence as Bob’s moves, to which Alice responds accordingto her winning strategy.Thus, for each x ∈ F we construct a sequence ( B n ) n as before, where x is the onlyelement of (cid:84) n B n (so x = x ∞ is the outcome of the game). We will show that x ∈ S by contradiction. Otherwise, suppose that x / ∈ S where S is an ( α, β, c, ρ )-winning set.Then, x ∈ (cid:83) m ∈ N (cid:83) i A i,m , where (cid:80) i (diam A i,m ) c ≤ ( αρ m ) c = ( αβ m ρ ) c (since it is alegal move for Alice we know that (cid:83) i A i,m ∈ A ( B m )). So x ∈ A ∈ A ( B m ) for some m ,and as x ∈ B m we have x ∈ A ∩ B m . Since A ∗ m ( B n ) = A ( B m ) for every n > m (because π m ( B n ) = B m ), then φ j ( B jN ) ≥ (diam A ) c for every j such that jN > m (because(diam A ) c is just one term in the sum of the definition of φ j ( B jN ) when A ∈ A ∗ m ( B n )).On the other hand, since B jN ∈ D (cid:48) j , then φ j ( B jN ) ≤ ( γρ jN ) c . Putting everythingtogether, diam A ≤ γρ jN for all j such that jN > m . Letting j → ∞ , we get diam A =0, a contradiction. So x ∈ S , that is F ⊂ S . Finally, using (18), and that K /K < K α d | log( β ) | ≤ ( d − c ) K K < d, so d − K α d | log( β ) | > α c ≤ K (1 − β d − c ) . This concludes the proof. (cid:3) Acknowledgements Alexia Yavicoli was financially supported by the Swiss National Science Foundation,grant n ◦ P2SKP2 184047. References [1] Stephen Astels. Cantor sets and numbers with restricted partial quotients. Trans. Amer. Math.Soc. , 352(1):133–170, 2000.[2] Dzmitry Badziahin, Stephen Harrap, Erez Nesharim, and David Simmons. Schmidt games andCantor winning sets. arXiv:1804.06499, 2020.[3] S´ebastien Biebler. A complex gap lemma. Proc. Amer. Math. Soc. , 148(1):351–364, 2020.[4] Ryan Broderick, Lior Fishman, and David Simmons. Quantitative results using variants ofSchmidt’s game: dimension bounds, arithmetic progressions, and more. Acta Arith. , 188(3):289–316, 2019.[5] Vincent Chan, Izabella (cid:32)Laba, and Malabika Pramanik. Finite configurations in sparse sets. J.Anal. Math. , 128:289–335, 2016.[6] Roy O. Davies, John M. Marstrand, and S. J. Taylor. On the intersections of transforms of linearsets. Colloq. Math. , 7:237–243, 1959/60.[7] Kenneth Falconer. Fractal Geometry: Mathematical Foundations and Applications . John Wiley &Sons, Ltd., Chichester, third edition, 2014.[8] Kenneth J. Falconer. Sets with large intersection properties. J. London Math. Soc. (2) , 49(2):267–280, 1994.[9] Robert Fraser, Shaoming Guo, and Malabika Pramanik. Polynomial Roth theorems on sets offractional dimensions. arXiv:1904.11123, 2019.[10] Kevin Henriot, Izabella (cid:32)Laba, and Malabika Pramanik. On polynomial configurations in fractalsets. Anal. PDE , 9(5):1153–1184, 2016.[11] Brian R. Hunt, Ittai Kan, and James A. Yorke. When Cantor sets intersect thickly. Trans. Amer.Math. Soc. , 339(2):869–888, 1993.[12] Jean-Pierre Kahane. Sur la dimension des intersections. In Aspects of mathematics and its appli-cations , volume 34 of North-Holland Math. Library , pages 419–430. North-Holland, Amsterdam,1986.[13] Tam´as Keleti. A 1-dimensional subset of the reals that intersects each of its translates in at mosta single point. Real Anal. Exchange , 24(2):843–844, 1998/99.[14] Tam´as Keleti. Construction of one-dimensional subsets of the reals not containing similar copiesof given patterns. Anal. PDE , 1(1):29–33, 2008.[15] Izabella (cid:32)Laba and Malabika Pramanik. Arithmetic progressions in sets of fractional dimension. Geom. Funct. Anal. , 19(2):429–456, 2009.[16] Andr´as M´ath´e. Sets of large dimension not containing polynomial configurations. Adv. Math. ,316:691–709, 2017. NTERSECTIONS OF THICK COMPACT SETS OF R d [17] Pertti Mattila. Geometry of sets and measures in Euclidean spaces , volume 44 of CambridgeStudies in Advanced Mathematics . Cambridge University Press, Cambridge, 1995. Fractals andrectifiability.[18] Ursula Molter and Alexia Yavicoli. Small sets containing any pattern. Math. Proc. CambridgePhilos. Soc. , 168(1):57–73, 2020.[19] Sheldon E. Newhouse. Nondensity of axiom A(a) on S . In Global Analysis (Proc. Sympos. PureMath., Vol. XIV, Berkeley, Calif., 1968) , pages 191–202. Amer. Math. Soc., Providence, R.I.,1970.[20] Jacob Palis and Floris Takens. Hyperbolicity and sensitive chaotic dynamics at homoclinic bifur-cations , volume 35 of Cambridge Studies in Advanced Mathematics . Cambridge University Press,Cambridge, 1993. Fractal dimensions and infinitely many attractors.[21] Alexia Yavicoli. Large sets avoiding linear patterns. Proceedings of AMS , in press, 2017.arXiv:1706.08118.[22] Alexia Yavicoli. Patterns in thick compact sets. Israel Journal of mathematics , in press, 2019.arXiv:1910.10057. School of Mathematics and Statistics, University of St. Andrews Email address : [email protected] School of Mathematics and Statistics, University of St. Andrews, UK. Currentaddress: Department of Mathematics, the University of British Columbia. 1984 Math-ematics Road, Vancouver BC V6T 1Z2, Canada Email address ::