Schmidt's game on Hausdorff metric and function spaces: generic dimension of sets and images
Ábel Farkas, Jonathan M. Fraser, Erez Nesharim, David Simmons
aa r X i v : . [ m a t h . M G ] J u l Schmidt’s game on Hausdorff metric and functionspaces: generic dimension of sets and images
Ábel Farkas a , Jonathan M. Fraser b , Erez Nesharim c , & David Simmons d a Alfréd Rényi Institute of Mathematics, Hungarian Academy of Sciences, Budapest,Hungary. b Mathematical Institute, University of St Andrews, UK. c Faculty of Mathematics, Technion, Israel. d Department of Mathematics, York, UK.
Abstract
We consider Schmidt’s game on the space of compact subsets of a given metricspace equipped with the Hausdorff metric, and the space of continuous functionsequipped with the supremum norm. We are interested in determining the genericbehaviour of objects in a metric space, mostly in the context of fractal dimensions,and the notion of ‘generic’ we adopt is that of being winning for Schmidt’s game.We find properties whose corresponding sets are winning for Schmidt’s game thatare starkly different from previously established, and well-known, properties whichare generic in other contexts, such as being residual or of full measure.
Mathematics Subject Classification
Key words and phrases : Schmidt’s game, dimension.
We consider Schmidt’s game introduced by Schmidt in [S66]. The game is played in acomplete metric space ( X, d ) and it has some similarities to the Banach–Mazur game.The game is played by two players, Alice and Bob, and the rules are described below.Given < α, β < , Alice and Bob play the ( α, β ) -game as follows:1. Bob begins by choosing r > and x ∈ X . Denote B = B ( x , r ) and r n =( αβ ) n r for all n ∈ N .2. On Alice’s n th turn, she chooses y n such that d ( x n , y n ) + αr n ≤ r n .3. On Bob’s ( n + 1) st turn, he chooses x n +1 such that d ( y n , x n +1 ) + αβr n ≤ αr n .1. The inequalities above ensure that the closed balls { B n = B ( x n , r n ) } ∞ n =0 and (cid:8) A n = B (cid:0) y n , αr ( n − (cid:1)(cid:9) ∞ n =1 form a decreasing sequence that satisfies B ⊇ A ⊇ B ⊇ A ⊇ · · · ⊇ B n ⊇ A n +1 ⊇ B n +1 ⊇ · · · , (1)and hence intersect at a unique point which is called the outcome of the game.Given a set S ⊆ X , if Alice has a strategy guaranteeing that the outcome lies in S , then S is called ( α, β ) -winning . If for some fixed α , the set S is ( α, β ) -winning for all < β < ,then S is called α -winning . If S is α -winning for some < α < , then S is called winning .Sets which are winning should be thought of as being ‘big’ (cf. Theorem 1.1 below). Aproperty is generic with respect to Schmidt’s game if the set of points with that property iswinning. The Banach–Mazur game is played similarly, but without any restriction on thesize of the nested sequence of balls (1), that is, the Banach–Mazur game is a topologicalgame, whereas Schmidt’s game is metric. A fundamental fact which is important to keepin mind throughout this paper is that the winning sets for the Banach–Mazur game areprecisely the residual sets, that is, sets whose complement is a countable union of nowheredense sets.We list some well-known properties of Schmidt’s game. For reference see [S66]. Theorem 1.1.
Winning sets have the following properties:1. If S ⊆ X is winning, then S is dense. If S is α -winning for α > , then S = X .2. If S ⊆ X is α -winning and < α ′ ≤ α , then S is α ′ -winning.3. If S ⊆ R d is winning, then the Hausdorff dimension of S is d .4. If S k ⊆ X is α -winning for every k ∈ N , then S = T ∞ k =1 S k is α -winning.5. If S , . . . , S N ⊆ X are winning sets ( N finite) then S = T Nk =1 S k is winning.6. If S ⊆ X is winning, then X \ S is not winning. The property of being winning is subtle and how it relates to other notions of beinggeneric is of particular interest and a key theme of this paper. For example, the set ofbadly approximable numbers
BA = (cid:26) x ∈ R : inf q ∈ N , p ∈ Z q | qx − p | > (cid:27) is well-known to be of first category (co-residual) and have zero Lebesgue measure, butnevertheless be winning. Also, the set of numbers which are not normal to some fixedbase is well-known to have zero Lebesgue measure, and nevertheless to be residual andwinning, see [B09, S66, V59]. 2 Results
Let ( X, d ) be a complete separable metric space. Let K ( X ) be the set of nonemptycompact subsets of X equipped with the Hausdorff metric, denoted by d H . It is well-knownthat ( K ( X ) , d H ) is complete. Schmidt’s game will be played on K ( X ) and the winningproperties of subsets of K ( X ) defined according to dimension will be considered. Fivestandard notions of dimension will be used: the lower, Hausdorff, lower box, upper box,and Assouad dimensions, denoted by dim L , dim H , dim B , dim B , and dim A , respectively.We note that the lower dimension is sometimes referred to as the lower Assouad dimensionin the literature. The reader is referred to [F03, R11, KLV13] for the definitions of thesedimensions and their basic properties. It is of particular importance in the subsequentanalysis that for any compact set K in a metric space the following inequalities hold: dim L K ≤ dim H K ≤ dim B K ≤ dim B K ≤ dim A K. Although these inequalities are straightforward to establish, we refer the reader to [L67]for the first inequality, [F03] for the middle two, and [R11] for the final inequality.Recall that a metric space is doubling if and only if dim A X < ∞ , see [R11, Lemma9.4], and uniformly perfect if and only if dim L X > , see [KLV13, Lemma 2.1]. We saythat X is Assouad sharp if it is doubling and there exists
C > such that all x ∈ X and < r < R satisfy N r ( B ( x, R )) ≤ C (cid:18) Rr (cid:19) dim A X where N r ( E ) denotes the minimal number of open sets of diameter at most r requiredto cover a given set E ⊆ X . Examples of Assouad sharp spaces include Ahlfors regularmetric spaces, in particular R d . Note that for Ahlfors regular spaces all of the abovedimensions coincide, see [BG00]. Remark . If { B ( x i , r ) } Ni =1 is a maximal collection of disjoint balls of radius r withcenters in K then K ⊆ S Ni =1 B ( x i , r ) . On the other hand any cover of K with ballsof radius r must contain at least N balls. Hence if dim A ( K ) < ∞ then the minimumnumber of balls of radius r needed to cover K and the maximum number of disjoint ballsof radius r centered in K are comparable up to constant factor. This constant factor couldbe included in the constant C above. Throughout the paper we might switch back andforth between disjoint balls and covering balls, which is a common strategy in dimensiontheory. Theorem 2.2. (i) If X is uniformly perfect, then the set { K ∈ K ( X ) : dim L K > } is winning. (ii) For all ε > , the set { K ∈ K ( X ) : dim L K < dim L X − ε } is not winning. (iii) If X is Assouad sharp, then the set { K ∈ K ( X ) : dim A K < dim A X } is winning. If X is doubling, then for all ε > the set { K ∈ K ( X ) : dim A K > ε } is not win-ning.Remark . Assume that < a < b < c = dim L X = dim A X < ∞ . Without proof wenote that Bob, while choosing β small enough depending on α , can use a combination ofhis strategies in the proof of (ii) and (iv) to ensure that a < dim L K ≤ dim A K < b forthe outcome of the game. This implies that if for sets
A, B ⊆ (0 , c ) , the set { K ∈ K ( X ) : dim L K ∈ A, dim A K ∈ B, dim L K < dim A K } (2)is winning, then both A and B has uncountable intersection with the interval ( a, b ) . Wedon’t know if the converse holds, nor if there exists any other characterisation of thewinning property of the set in (2) in terms of A and B both being ‘big’ subsets of (0 , c ) ,where ‘big’ is chosen appropriately, for example ‘big’ might mean full Lebesgue measure,winning or co-meager. This is not clear even for X = R d . It is even less clear if X isuniformly perfect and doubling but satisfies dim L X < dim A X .Part (iii) has the weakness that we assumed X to be Assouad sharp which is oftennot the case. It is possible to drop the Assouad sharpness by assuming instead that X isuniformly perfect, though the proof becomes more complicated. Theorem 2.4.
Let X be a doubling uniformly perfect space. Then (i) { K ∈ K ( X ) : dim A K < dim A X } is winnig (ii) If furthermore dim B X < ∞ , then (cid:8) K ∈ K ( X ) : dim B K < dim B X (cid:9) is winning. Part (i) of Theorem 2.2 is sharp in the sense that the statement is not true if the‘uniformly perfect’ assumption is dropped. For example, if X has an isolated point, thenthe collection of sets with positive dimension cannot be winning because Bob can choosethe initial set to be the singleton consisting of the isolated point and the initial radiusto be so small that the outcome of the game is already forced to be the isolated point.However, one can construct a perfect set which also behaves similarly. Example 2.5.
Let X be the disjoint union of the unit ball in R d and a perfect compactset A of Hausdorff dimension (or even of Assouad dimension ). Then Bob can choosethe initial set to be contained in A and choose r to be small enough to make sure thatevery step stays inside A . Hence the outcome of the game is in A and so of dimension.On the other hand, having dimension is not winning because Bob can also make surethat the game is played only inside the unit ball after the initial step (where Theorem 2.2applies). The example shows that assuming that, for example, dim H X > is not enoughto conclude that dim H K > is a winning property.The converse of part (i) is not true, as can be seen from the following example of aset which is not uniformly perfect, but yet the conclusion of part (i) holds. Example 2.6.
Let X = { } ∪ ∞ [ n =1 (cid:2) /n − − n , /n + 2 − n (cid:3) dim L X = 0 and so X is not uniformly perfect. However, no matter which initial set K ⊆ X andradius r Bob picks, as long as α < / , eventually the radius r n = ( αβ ) n r will besmall enough such that Alice can choose a set K ′ n such that elements of B H ( K ′ n , αr n ) areuniformly bounded away from 0. Thus the problem reduces to the case when X = [0 , (more accurately, a finite collection of intervals) and it follows from Theorem 2.2(i) that { K ∈ K ( X ) : dim L K > } is winning.Part (i) of Theorem 2.2 is also sharp in another sense, since part (iv) shows that (inthe doubling case) we cannot replace dim L K > with dim L K > ε for any positive ε > .Part (iii) of Theorem 2.2 is sharp, since part (ii) shows that (provided dim L X =dim A X , e.g. if X is Ahlfors regular) the assumption dim A K < dim A X cannot bereplaced with dim A K < dim A X − ε for any positive ε > .To emphasise an important setting where our results are complete, we state the spe-cialisation of Theorem 2.2 to Euclidean space, noting that the same results also hold inany Ahlfors regular space, where d is replaced by the Hausdorff dimension of the space. Corollary 2.7.
The set (cid:8) K ∈ K ( R d ) : 0 < dim L K ≤ dim A K < d (cid:9) is winning, but for all ε > the sets (cid:8) K ∈ K ( R d ) : dim A K > ε (cid:9) and (cid:8) K ∈ K ( R d ) : dim L K < d − ε (cid:9) are not winning. This should be compared with the well-known result of Feng and Wu [FW97], whichconsiders the Banach–Mazur game instead of Schmidt’s game. One can see that theresults are rather different.
Theorem 2.8 (Feng–Wu, 1997) . The set (cid:8) K ∈ K ( R d ) : dim L K = dim B K = 0 and dim B K = dim A K = d (cid:9) is winning for the Banach–Mazur game.Remark . McMullen [M10] introduced the concept of absolute winning in R n , whichcan be extended to any complete metric space (see, e.g., [BHNS13]). In the absolute gamein ( X, d ) , Bob chooses an initial closed ball B of radius r > and β ∈ (0 , . Alice thenchooses a closed ball A of radius at most βr , Bob chooses, if possible, a ball B of radiusat least βr inside B \ A and so on. If at some turn Bob has no legal move or if the radiiof Bob’s balls do not shrink to zero we say that Alice wins by default . Otherwise, we see B ⊇ B \ A ⊇ B ⊇ B \ A ⊇ B . . . and the outcome of the game is { x } = ∩ i ∈ N B i . A set S ⊆ X is called absolute winning if Alice has a strategy to make sure that either she wins by default or x ∈ S . McMullen[M10] showed that in R d absolute winning sets are winning. Let us consider the absolutegame in K ( R d ) . Assume β < / , r > and K ∈ K ( R d ) . Then Bob can play accordingto the strategy that at every step he chooses a translate of K to be the center of his5all. It is easy to see that since β < / no matter what Alice chooses Bob can alwaysmake a legal choice according to his strategy (worst thing Alice can do against Bob is tochoose the same center as Bob and maximal possible radius). Hence the outcome of thegame is a translate of K . This means that every absolute winning set in K ( R d ) containsa translate of every compact set. Therefore absolute winning is not the right notion toconsider when talking about generic properties of K ( R d ) . Let K be a compact metric space. Fix d ∈ N , and let F = F ( K, d ) be the space ofcontinuous functions from K to R d , endowed with the metric induced by the supremumnorm which is denoted by | · | . Write dim T ( K ) to denote the topological dimension of K . Theorem 2.10. (i) If dim T ( K ) ≥ d , then { f ∈ F : f ( K ) has nonempty interior } is winning. (ii) If dim T ( K ) < d , then the set { f ∈ F : dim A ( f ( K )) < d } is winning. (iii) If K is uncountable, then the set { f ∈ F : dim H ( f ( K )) > } is winning. (iv) If dim T ( K ) ≤ d , then the set { f ∈ F : H dim T ( K ) ( f ( K )) > } is winning. Inparticular, the set { f ∈ F : dim H ( f ( K )) ≥ dim T ( K ) } is winning.Remark . In the proof we show that the conclusion of part (iii) holds for every perfectset K . Every uncountable compact set K contains a perfect set F . In the more generalcase, Alice plays her strategy for the restriction of the functions to F . It follows fromthe general version of Tietze’s extension theorem that Alice can extend her choice to thewhole set K to be able to play a legal move. Remark . The conclusion of part (iv) can be deduced from part (i). Alice can play herstrategy in the first d coordinates completely ignoring the other coordinates, just leavingthem as Bob’s choice. Then they end up with a function whose projection onto the first d coordinates is of nonempty interior. Hence the conclusion of part (iv) follows.Again, this result can be compared with the analogous results in the Banach–Mazursetting, which are due to Balka–Farkas–Fraser–Hyde [BFFH13]. Theorem 2.13 (Balka–Farkas–Fraser–Hyde, 2013) . The set (cid:8) f ∈ F : dim H ( f ( K )) = dim B ( f ( K )) = min { d, dim T ( K ) } , dim B ( f ( K )) = dim A ( f ( K )) = d (cid:9) is winning for the Banach–Mazur game. In case dim T ( K ) ≥ d , these properties are similar for both Schmidt winning andBanach–Mazur winning; however, in the other case they are rather different.6 emark . We can again consider the absolute game in F . Assume β < / , r > and f ∈ F . Then Bob can play that at every step he chooses the center of his ball onthe line { f + λ : λ ∈ R d } . It is easy to see that Bob can always make a legal choice likethat since β < / (again the worst thing Alice can do against Bob is to choose the samecenter as Bob and maximal possible radius). The outcome of the game is f + λ for some λ ∈ R d . This means that an absolute winning set contains a constant translate of everyfunction in F . Hence, just like in the case of K ( R d ) , the absolute game does not seem tobe the right notion to consider in F . Finally, we consider analogous questions concerning the frequencies of digits in expan-sions of real numbers. Although this is somewhat incongruous with our other results,frequencies are inherently related to densities and therefore dimension. Moreover, ques-tions regarding the generic behaviour of digit frequencies are among the simplest andmost widely studied and therefore it is useful to see how our approach fits in here. Forsimplicity, we consider binary expansions of numbers x = x .x x · · · ∈ R where x is aninteger and x i ∈ { , } are the digits in the binary expansion of the fractional part of x .For definiteness take the lexicographically maximal expansion in the situations where x does not have a unique expansion and assume that all expansions are infinite, for example . . . . . For x ∈ R and j ∈ { , } write d + ( x, j ) = lim sup k →∞ { ≤ i ≤ k : x i = j } k and d − ( x, j ) for the same expression with lim sup replaced by lim inf , where if S is a finiteset then S stands for the number of elements in S . Theorem 2.15.
The set (cid:8) x ∈ R : 0 < d − ( x, j ) ≤ d + ( x, j ) < for j ∈ { , } (cid:9) is winning, but for all ε > and j ∈ { , } the sets { x ∈ R : ε < d + ( x, j ) } and { x ∈ R : d − ( x, j ) < − ε } are not winning. This theorem is sharp in the sense that we cannot replace by any ε > . Remark . Similarly to Remark 2.3, we can ask if for sets
A, B ⊆ (0 , , the set (cid:8) x ∈ R : d − ( x, j ) ∈ A, d + ( x, j ) ∈ B, d − ( x, j ) < d + ( x, j ) (cid:9) is winning if and only if A and B are both ‘big’ subsets of (0 , , where ‘big’ may standfor either full Lebesgue measure, winning or co-meager.There are obvious parallels between Theorem 2.15 and the results in the previoussections. Moreover, there is a stark comparison between this result and other resultsconcerning generic behaviour of digit frequencies.7 heorem 2.17 (Borel 1909, Volkmann 1959) . The set (cid:8) x ∈ R : d − ( x, j ) = d + ( x, j ) = 1 / for j ∈ { , } (cid:9) has full Lebesgue measure and the set (cid:8) x ∈ R : 0 = d − ( x, j ) and d + ( x, j ) = 1 for j ∈ { , } (cid:9) is winning for the Banach–Mazur game. The Lebesgue measure part of this theorem was proven by Borel in 1909 as an applic-ation of what became known as the Borel–Cantelli lemma [B09] and the Banach–Mazurpart was proven by Volkmann in 1959 [V59].
Since dim L X > , for all N ≥ there exists an α ∈ (0 , / (depending on X and N ) suchthat for every y ∈ X and < r ≤ there exist x , . . . , x N such that S Ni =1 B ( x i , αr ) ⊆ B ( y, r ) is a disjoint union. In what follows fixing N = 2 is sufficient.Let β ∈ (0 , be arbitrary. The game begins by Bob choosing a starting set K B ∈K ( X ) and a radius r > . Alice’s move is then as follows. Take x , . . . , x M ∈ K B , where M ≥ N , such that S Mi =1 B ( x i , αr ) is a disjoint union, and S Mi =1 B ( x i , αr ) covers K B .Alice then chooses the set K A = { x , . . . , x M } . This choice is legal since d H ( K A , K B ) < (1 − α ) r . Then Bob will have to choose a set K B which is contained in S Mi =1 B ( x i , (1 − β ) αr ) and contains at least one point in every B ( x i , (1 − β ) αr ) because the balls are αr -separated. Alice then repeats her strategy in each of the balls B ( x i , (1 − β ) αr ) simultaneously. At every step n in each of the balls of the previous step of the constuctionAlice finds at least N more new balls of radius (1 − β ) α ( αβ ) n − r . Let K be the outcomeof the game, x ∈ K and < R ≤ r . Let m ∈ N such that ( αβ ) m (1 − β ) αr < R ≤ ( αβ ) m − (1 − β ) αr . Then B ( x, R ) contains at least one of the balls of step m + 1 . Hence N (1 − β ) α ( αβ ) n r ( K ∩ B ( x, R )) ≥ N n − m ≥ ( αβ ) − ( n − m ) ε ≥ αβ − (cid:18) R (1 − β ) α ( αβ ) n r (cid:19) ε if < ε < is small enough that N ≥ ( αβ ) − ε . In particular, dim L K ≥ ε > , whichproves the result. Let ε > and fix α ∈ (0 , / . We need to show that there exists β ∈ (0 , such that { K ∈ K ( X ) : dim H K < dim L X − ε } is not ( α, β ) -winning. If dim L X = 0 then the statement is trivial, so assume that dim L X > . It follows that for dim L X > t > max { dim L X − ε, } C > and r such that for all β ∈ (0 , , for all < r < r and all x ∈ X thereare y , . . . , y N such that S Ni =1 B ( y i , βr ) ⊆ B ( x, r ) is a disjoint union, where N ≥ Cβ − t .Let t > s > dim L X − ε . Let β > be small enough that β < − β and Cα s ≥ β t − s ,i.e. Cβ − t ≥ ( αβ ) − s . Bob starts by choosing K B to be a single point and r was chosenabove. Then Bob adopts Alice’s strategy from part (i). Similarly it can be shown that N (1 − α )( αβ ) n r ( K ∩ B ( x, R )) ≥ ( Cβ − t ) n − m ≥ ( αβ ) − ( n − m ) s ≥ α s β s − s (cid:18) R (1 − α )( αβ ) n r (cid:19) s and therefore the outcome of the game satisfies dim L K ≥ s > dim L X − ε , proving theresult. Let α ∈ (0 , / and β ∈ (0 , and let d = dim A X < ∞ . Since by assumption X is Assouad sharp, it follows that there exists C > such that every bounded set B ofdiameter R can be covered by fewer than C ( R/r ) d balls of radius r ∈ (0 , R ) whichare centered in B , moreover we can assume that the balls with the same centers and ofradius r/ are disjoint. For now, assume that Bob starts the game by choosing K B tobe a subset of the ball B ( y, (1 − β ) r ) and a starting radius r = βr . Playing as Alice,choose x , . . . , x N ∈ K B such that K B ⊆ S Ni =1 B ( x i , βr ) where N ≤ C ((1 − β ) /β ) d and S Ni =1 B ( x i , αβr ) ⊆ S Ni =1 B ( x i , − βr ) is a disjoint union (since that α < / ). Moreover, if α < / then the distinct balls are αβr -separated. Alice then chooses K A = { x , . . . , x N } .For his next move, Bob will have to choose a set K B that is contained in S Ni =1 B ( x i , (1 − β ) αβr ) and the radius of the step is βαβr . Alice in her next move repeats the previousstrategy in all of the smaller balls B ( x i , (1 − β ) αβr ) simultaneously and proceeds similarlyat every step.Let us denote the outcome of the game by K . Let x ∈ K and < R ≤ r be fixed. Let m ∈ N be such that r ( αβ ) m +1 < R ≤ r ( αβ ) m . Let’s say Alice’s m th move was { y i } Mi =1 .Then it follows from the strategy that K is contained in S Mi =1 B ( x i , (1 − β )( αβ ) m r ) andthese balls are ( αβ ) m r -separated. Then the number M of these balls that intersects B ( x, R ) is at most the number of these balls that are contained in B ( x, α − β − R ) whichis at most C ( α − β − R ( αβ ) m r ) d ≤ C ( αβ ) d because these are disjoint balls centered in B ( x, R ) .Let n > m and observe that by the n th step of the game we have that each of these M balls of radius ( αβ ) m r contain at most (cid:16) C ( − ββ ) d (cid:17) n − m balls of radius (1 − β )( αβ ) n r such that K ∩ B ( x, R ) is contained in the union of these at most M (cid:16) C ( − ββ ) d (cid:17) n − m ≤ C ( αβ ) d (cid:0) Cβ − d (cid:1) n − m balls of radius (1 − β )( αβ ) n r ≤ ( αβ ) n r . Hence, noting that ( αβ ) m − n ≤ α − β − Rr ( αβ ) n , it follows that N ( αβ ) n r ( K ∩ B ( x, R )) ≤ C (cid:18) αβ (cid:19) d (cid:0) Cβ − d (cid:1) n − m ≤ C (cid:18) αβ (cid:19) d ( αβ ) − ( d − ε )( n − m ) ≤ C d α − d β − d (cid:18) Rr ( αβ ) n (cid:19) ( d − ε ) < α < / is small enough that C < α − d/ < α − ( d − ε ) / and < ε < d/ is small enough that α d/ < (cid:0) βα / (cid:1) ε . Therefore we can choose small enough α such thatfor every β we can choose a small enough ε that dim A K ≤ d − ε .Finally, suppose Bob does not begin by choosing a set inside B ( y (1 − βr )) , but ratheran arbitrary compact set K B , as he is of course permitted to do. In this case, on her firstmove, Alice plays her strategy with the only exception that on the first level we have nocontrol on the number of balls N . However, this is not an issue because it is just somefixed number and on the latter levels we have exactly the same control over the balls. Itfollows that for some α > the set { K ∈ K ( X ) : dim A K < dim A X } is ( α, β ) -winning for all β ∈ (0 , and therefore winning. Let ε > and fix α ∈ (0 , / . We need to show that there exists β ∈ (0 , such that { K ∈ K ( X ) : dim A K > ε } is not ( α, β ) -winning. To achieve this, Bob will adopt a similar strategy to Alice’s strategyin the proof of Part (iii). Since X is doubling, it follows that there exist C, d > such thatevery ball of radius R can be covered by fewer than C ( R/r ) d balls of radius r ∈ (0 , R ) .Note that d may be strictly greater then dim A X in case X is not Assouad sharp, butthis is not an issue. Bob begins by choosing K B = { y } , where y ∈ X is arbitrary, and r = 1 . Alice is then forced to choose a set K A ⊆ B ( y, − α ) . Bob covers K A byballs of radius αr with centers in y , . . . , y N ∈ K where N ≤ C ((1 − α ) /α ) d is such that S Ni =1 B ( y i , βαr ) ⊆ S Ni =1 B ( y i , − αr ) is a disjoint union, provided since β < / . Moreoverif β < / then the balls are βα -separated. Alice’s next choice must be contained in S Ni =1 B ( y i , (1 − α ) βαr ) . Bob repeats his strategy in each of these balls. Let K be theoutcome of the game. Via a similar argument to that in the proof of Part (iii) one candeduce that for ( αβ ) m +1 < R ≤ ( αβ ) m and n > mN ( αβ ) n ( K ∩ B ( x, R )) ≤ C (cid:18) αβ (cid:19) d (cid:0) Cα − d (cid:1) n − m ≤ C (cid:18) αβ (cid:19) d ( αβ ) − ε ( n − m ) ≤ C d α − d β − d (cid:18) R ( αβ ) n (cid:19) ε if < β < / is small enough such that Cα − d < β − ε < ( αβ ) − ε . Therefore dim A K ≤ ε , as required. Since X is uniformly perfect, i.e. dim L X > , we can find a small enough < α < / such that inside every ball B of radius r/ we can find two disjoint balls of radius αr such10hat they are αr -separated from each other and from the boundary of B . Let β ∈ (0 , be arbitrary and choose ε > small enough that ( αβ ) − ε ≤ / . (3)Write s = dim A X < ∞ , and therefore there exists C ε > such that N r ( B ( x, R )) ≤ C ε ( R/r ) s + ε (4)for every x ∈ X and < r < R . Fix k ∈ N such that C ε ≤ k/ (5)and let ρ n = ( αβ ) n r where r = βr .First assume that Bob starts by choosing K B ⊆ B ( y, (1 − β ) r ) ⊆ B ( y, r ) for some y ∈ X . Then Alice can find x , . . . , x N ∈ K B such that K B ⊆ S Ni =1 B ( x i , βr/ and S Ni =1 B ( x i , βr/ is a disjoint union. Then the balls B ( x i , βr/ are disjoint and βr/ -separated. By the choice of α , inside every ball B ( x i , βr/ there exists two balls B ( y i , αβr ) and B ( y i , αβr ) which are disjoint and αβr -separated from each other andfrom the boundary of B ( x i , βr/ . Without loss of generality we can assume that g N ρ ( B ( y i , αβr )) ≤ g N ρ ( B ( y i , αβr )) where f N ρ ( B ) denotes the maximum number ofdisjoint balls of radius ρ inside B for a ball B . Alice’s move then is K A = S Ni =1 { y i } .Bob’s next move K B must be contained in S Ni =1 B (cid:0) K A , (1 − β ) αβr (cid:1) ⊆ S Ni =1 B ( y i , αβr ) and the radius of the step is βα ( βr ) , noting that K B also contains at least point fromeach of these balls. Assume that on the m th round of the game Bob plays K Bm which,by Alice’s strategy, is contained in S N m i =1 B (cid:0) y i,m , (1 − β )( αβ ) m r (cid:1) ⊆ S N m i =1 B (cid:0) y i,m , ( αβ ) m r (cid:1) where these balls are ( αβ ) m r -separated. Let m = kn + l where n ≥ , l =0 , . . . , k − . Then in each of these balls similarly to the first step of the game Alicecan find x m,i , . . . , x m,iN im ∈ K B ∩ B (cid:0) y i,m , ( αβ ) m r (cid:1) such that K B ∩ B (cid:0) y i,m , ( αβ ) m r (cid:1) ⊆ S Ni =1 B (cid:0) x m,ij , β ( αβ ) m r/ (cid:1) and S N im i =1 B (cid:0) x m,ij , β ( αβ ) m r/ (cid:1) is a disjoint union. Then theballs B (cid:0)(cid:0) x m,ij , β ( αβ ) m r/ (cid:1)(cid:1) are disjoint and β ( αβ ) m r/ -separated. By the choice of α inside every ball B (cid:0)(cid:0) x m,ij , β ( αβ ) m r/ (cid:1)(cid:1) there exist two balls B (cid:0) y ,m,ij , ( αβ ) m +1 r (cid:1) and B (cid:0) y ,m,ij , ( αβ ) m +1 r (cid:1) which are disjoint and ( αβ ) m +1 r -separated from each other and fromthe boundary of B ( x i , βr/ . Without loss of generality we can assume that ^ N ρ n +1 (cid:0) B (cid:0) y ,m,ij , ( αβ ) m +1 r (cid:1)(cid:1) ≤ ^ N ρ n +1 (cid:0) B (cid:0) y ,m,ij , ( αβ ) m +1 r (cid:1)(cid:1) . (6)Alice’s move is then K Am +1 = S N m i =1 S N im j =1 { y ,m,ij } = S N m +1 i =1 { y i,m +1 } and Bob’s next movemust be contained in S N m +1 i =1 B (cid:0) y i,m +1 , (1 − β )( αβ ) m +1 r (cid:1) .Now let us consider what happens inside one ball at the nk th level of the con-struction B = B (cid:0) y i,nk , ( αβ ) nk r (cid:1) at the ( n + 1) k th level of the construction. The ball B (cid:0) y i,nk , ( αβ ) nk r (cid:1) contains some M balls from the collection B (cid:16) y i, ( n +1) k , ( αβ ) ( n +1) k r (cid:17) .On the (( n + 1) k − th level we kept the balls B (cid:16) y i, ( n +1) k − , ( αβ ) ( n +1) k − r (cid:17) and elimin-ated the balls B (cid:16) y i, ( n +1) k − , ( αβ ) ( n +1) k − r (cid:17) and by (6) it follows that inside the balls at11he (( n + 1) k − th level that are contained in B there are at least M balls of radius ( αβ ) ( n +1) k r . Similarly inside the balls at the (( n + 1) k − the level of the constructionthat are contained in B there are at least M balls of radius ( αβ ) ( n +1) k r . Continuing thisthrough k steps we get that inside B there are k M balls of radius ( αβ ) ( n +1) k r . Hence M ≤ − k N ( αβ ) ( n +1) k r ( B ) ≤ − k C ε ( R/r ) s + ε = 2 − k C ε ( αβ ) − εk ( αβ ) − ( s − ε ) k ≤ ( αβ ) − ( s − ε ) k (7)by (3), (4) and (5). So inside each ball at the nk th level of the construction there are atmost ( αβ ) − ( s − ε ) k balls from the ( n + 1) k level of the construction while the ratio betweenthe radii are ( αβ ) k . Hence for the outcome of the game K dim A K ≤ s − ε. Now let us consider the situation when Bob does not start by choosing K B ⊆ B ( y, (1 − β ) r ) . Then Alice plays her strategy exactly the same way. The only difference is thatnow on the first level, and so on the first k levels, we have no control over the numberof balls of the construction. However, from then on inside each ball we have exactly thesame control over the number of the balls. Hence this has no effect on dim A K and theproof is finished. Note that N ρ n ( K ) ≤ N ( αβ ) nk r ( B ( K Ank , ρ n )) ≤ ( αβ ) − ( s − ε ) nk N βr/ ( K B ) = ρ − ( s − ε ) n N βr/ ( K B ) . (8) We prove (ii) similarly to (i). We apply a trick to replace dim A X with dim B X , but theprice is that we cannot keep k fixed and therefore only get estimates for the lower boxdimension of the outcome. The Assouad spectrum, introduced by Fraser and Yu [ FY18b ] ,of a compact set X is the function θ dim θA X , where θ varies in (0 , and dim θA X := inf (cid:26) α : there exists C > such that all < R < and x ∈ X satisfy N R /θ (cid:0) B ( x, R ) (cid:1) ≤ C (cid:18) RR /θ (cid:19) α (cid:27) . It was shown in [ FY18b ] that dim θA X is continuous in θ and satisfies dim θA X ≤ dim B X − θ and therefore dim θA X → dim B X as θ → . Therefore, in the above proof we can replace(4) with the statement that there exist a small θ ∈ (0 , and a constant C ε,θ > suchthat N R /θ ( B ( z, R )) ≤ C ε,θ ( R/R /θ ) s + ε (9)for all z ∈ X and R ∈ (0 , where s is now the upper box dimension of X , that is, s = dim B X (note that we choose θ to be less than / which ensures that /θ − ≥ ).12he only place this estimate is needed in the proof of (i) is (7) but to make (9) work herewe would ideally choose ρ n +1 = ρ /θn which means that k needs to change at every step of the strategy. That is, at step n wegroup the turns in Schmidt’s game into groups of length k n defined inductively as follows.Let k ∈ N be such that C ε,θ ≤ k/ . We need to make sure that k n ≥ k for every n . Choose k = k and, given k , . . . , k n ≥ k ,choose k n +1 to be the largest integer satisfying ρ n +1 := ( αβ ) P n +1 i =1 k i r ≥ (cid:16) ( αβ ) P ni =1 k i r (cid:17) /θ = ρ /θn . It is easy to see that if r ≤ (which can be assumed because otherwise Alice just playsrandomly until r m /β ≤ and starts playing her strategy then), then k n +1 ≥ (1 /θ − n X i =1 k i ≥ nk ≥ k. Clearly this forces k n to grow very quickly and, moreover, ρ /θn ( αβ ) − ≥ ρ n +1 ≥ ρ /θn . This is enough to apply (9) to make an estimation like in (7) for all n up to a constantfactor which can be absorbed into the constant C ε,θ .The proof then works as before leading to the estimate (similar to (8)) N ( αβ ) S ( n ) r ( K ) ≤ ( αβ ) − S ( n )( s − ε ) N βr/ ( K B ) for all n ∈ N where S ( n ) = P ni =1 k i . Therefore we conclude that dim B K ≤ s − ε < dim B X completing the proof. The following notation is used in this section: vectors in R d and functions to R d aredenoted by boldface letters and their coordinates by a normal font with a subscript, e.g.,for f : K → R d , the functions f , . . . , f d : K → R are such that f = ( f , . . . , f d ) .Fix any α ∈ (0 , / and β ∈ (0 , , and let f ∈ F and r > be Bob’s first move andstarting radius. Write r = (1 − α ) r , and for each m ∈ Z d let K m = d \ i =1 f − i ([ m i r, ( m i + 1) r ]) . Then K = S m ∈ Z d K m , so by the sum theorem [E89, Theorem 1.5.3], there exists m ∈ Z d for which dim T ( K m ) ≥ d. F , . . . , F d ⊆ K m and relatively open sets U , . . . , U d ⊆ K m such that F i ⊆ U i for each ≤ i ≤ d , and suchthat if F i ⊆ W i ⊆ U i for some relatively open sets W , . . . , W d ⊆ K n , then T d ∂W i = ∅ .Now Alice’s strategy is as follows: on the first move, choose g ∈ B ( f , (1 − α ) r ) so that g i = m i r on F i and g i = ( m i + 1) r on K m \ U i for some m ∈ Z d for every ≤ i ≤ d , andplay arbitrarily on later moves. The outcome function h must lie in B ( g , αr ) , hence itsatisfies h i ≤ (cid:18) m i + α − α (cid:19) r on F i ,h i ≥ (cid:18) m i + 1 − α − α (cid:19) r on K m \ U i , for every ≤ i ≤ d . It follows that h ( K m ) has a nonempty interior as it contains thecube P := d Y i =1 (cid:18)(cid:18) m i + α − α (cid:19) r, (cid:18) m i + 1 − α − α (cid:19) r (cid:19) . Indeed, if p ∈ P , then for each i the set W i = { h i < p i } is open and satisfies F i ⊆ W i ⊆ U i .Thus by our hypothesis there exists x ∈ T d ∂W i , and it follows that h ( x ) = p . Fix any α ∈ (0 , / and β ∈ (0 , , and let r > be Bob’s initial radius. Let S ⊆ R d bethe set of all points with at least one integer coordinate, i.e. S = d [ i =1 (cid:8) x ∈ R d : x i ∈ Z (cid:9) . Alice’s strategy is as follows: On turn n , choose a ball with center f such that f ( K ) ⊆ r n S .Such a choice is possible because dim T ( K ) < d . Indeed, dim T ( K ) < d implies that the set { f ∈ F : f ( K ) is nowhere dense } is dense in F . This follows from [BFFH13, Theorem2.4] since if f ( K ) is not nowhere dense, then dim B f ( K ) = d . If Bob’s n th move is a ballwith center g , let h ∈ F be such that h ( K ) is nowhere dense and | h − g | < (1 / − α ) r n .Let P be the collection of cubes whose boundaries form r n S . For each P ∈ P , choose x ∈ P \ h ( K ) , and let π P : P \ { x } → ∂P be the radial projection from x . Finally, let f = π ◦ h , where π ( y ) = π P ( y ) for all y ∈ P . Alice can choose a ball with center f as hernext move since | f − g | ≤ | f − h | + | h − g | ≤ (1 − α ) r n . Using this strategy Alice guarantees that the outcome of the game is a function whoseimage is a subset of F d = ∞ \ n =1 B (cid:16) r n S, αr n (cid:17) . (10)14he set F d is porous , i.e., there exists ε > such that for all x ∈ R d and r > , thereexists y ∈ R d such that B ( y, εr ) ⊆ B ( x, r ) and B ( y, εr ) ∩ F d = ∅ . First we show that F is porous. Set ε = αβ (1 / − α ) and assume r > . Fix n ≥ to be the unique integer that satisfies r n ≤ r < r n − .By (10), F is contained in B (cid:0) r n Z , αr n (cid:1) which is a union of intervals of length αr n forwhich the centers are at a distance of r n / apart. Hence, the length of the gap betweenconsecutive intervals is (1 / − α ) r n . Since r ≥ r n , any ball of radius r contains such agap. Since r < r n − we get that (1 / − α ) r n r ≥ (1 / − α ) r n r n − = ε, which finishes the proof of the one-dimensional case. So we showed that for the scale r there is a hole of size εr in F ∩ B ( x, r ) . If we take the product of d holes of F then itforms a hole in F d . Hence in every ball we can find a hole of comparable size (the constant ε depends on α, β and d ). This shows the porosity of F d .It follows from [FY18a, Theorem 2.4] that dim A ( F d ) < d , so the image of the outcomehas Assouad dimension strictly smaller than d . Specifically, it follows from [FY18a, The-orem 2.4] that if the Assouad dimension of a set E ⊂ R d is d , then the unit ball is a weaktangent to E . Having the unit ball as a weak tangent clearly prevents the set E frombeing porous. Fix α < / small enough and let S ⊆ R d be a α -separated subset of the Euclidean ballof radius / around zero in R d . Set N = S . We can assume that N ≥ by choosing α small enough. Alice’s strategy will involve constructing a sequence of sets E n ⊆ K ,such that E n = N n . The set E is any singleton. Suppose that the set E n has beenconstructed, and that the game has been played up until turn n , with g ∈ F being thecenter of the current ball, i.e. Bob’s last move. Also suppose that | g ( y ) − g ( x ) | > αr n − for all x, y ∈ E n distinct. (11)For each x ∈ E n , let F n,x ⊆ K be a set of cardinality N such that for all y ∈ F n,x , | g ( y ) − g ( x ) | ≤ r n / . Such a set exists because K is perfect and g is continuous. By Tietze’s extension theoremthere exists g ∈ F such that g ( y ) = g ( x ) for every y ∈ F n,x and x ∈ E n and | g ( y ) − g ( y ) | ≤ r n / for every y ∈ K . Now let E n +1 = S x ∈ E n F n,x . By (11), the sets { F n,x } x ∈ E n are disjoint,so E k +1 = N k +1 . Alice will choose the center f ∈ F of her next ball so as to guaranteethat { f ( y ) : y ∈ F n,x } = g ( x ) + r n S (12)15or all x ∈ E n . Namely, for each x ∈ E n , let σ x : F n,x → S be any bijection, let { φ y } y ∈ E n +1 be any family of continuous real valued functions with disjoint supports such that | φ y | ≤ and φ y ( y ) = 1 . Then the function f := g + r n X x ∈ E n X y ∈ F n,x φ y σ x ( y ) (13)satisfies (12). Now (13) implies that | f − g | ≤ | f − g | + | g − g | ≤ r n r n ≤ (1 − α ) r n so f is a legal move. If h is the center of Bob’s next ball, then h must satisfy | h − g | < (1 − β ) αr n < αr n . Since S is α -separated it follows that | h ( y ) − h ( x ) | ≥ | f ( y ) − f ( x ) | − αr n > αr n for all x, y ∈ F n,z distinct and z ∈ E n . If x ∈ F n,z and y ∈ F n,z for distinct z , z ∈ E n then | h ( y ) − h ( x ) | ≥ | f ( y ) − f ( x ) | − αr n ≥ αr n − − r n / − αr n ≥ (1 − / − α ) αr n − ≥ αr n − / > αr n by (11), (12) and that α < / . Thus (11) is satisfied for n + 1 , allowing Alice to completethe game by induction.Alice’s strategy results in a branching construction. Namely, if g n is Bob’s n thmove and E n has been constructed then the collection of balls in the branching is S x ∈ E n B ( g ( x ) , r n ) . The balls in the construction are clearly disjoint by (11). Then Alicechooses f such that f ( F n,x ) ⊆ B ( g ( x ) , r n / and for his next move Bob can only changethe values by (1 − β ) αr n hence the values stay inside B ( g ( x ) , r n / . Thus the ballsof radius r n +1 around these modified values are contained in B ( g ( x ) , r n ) . So indeed S x ∈ E n B ( g ( x ) , r n ) is a branching construction with limit set contained in the image of theoutcome of the game. Hence by the mass distribution principle the image of the outcomeof the game has Hausdorff dimension at least log N − log αβ > . We first prove that the set X := (cid:8) x ∈ R : 0 < d − ( x, (cid:9) is winning. Fix α ∈ (0 , / , β ∈ (0 , . Bob starts the game by choosing x ∈ R and r > . As usual, we can assume r < / since if it was not, Alice could play arbitrarilyuntil r n < / . Let I be a dyadic interval at level k := ⌈− log r / log 2 ⌉ which iscompletely contained inside B ( x , r ) . Let l = max { k ∈ N : 2 − k − k > αr } y be the centre of the leftmost level ( k + l ) dyadic interval inside I . In particular, B ( y , αr ) ⊆ I ⊆ B ( x , r ) and for all numbers in B ( y , αr ) all of the binary digits fromthe ( k + 1) th position to the ( k + l ) th position are equal to 0.This argument is then repeated. Assuming Bob has chosen x n as the center of his n th ball, let I n be a dyadic interval at level k n := ⌈− log r n / log 2 ⌉ which is completelycontained inside B ( x n , r n ) . Let l n = max { k ∈ N : 2 − k n − k > αr n } and let y n be the centre of the leftmost level ( k n + l n ) dyadic interval inside I n . Inparticular, B ( y n , αr n ) ⊆ I n ⊆ B ( x n , r n ) and for all numbers in B ( y n , αr n ) all of thebinary digits from the ( k n + 1) th position to the ( k n + l n ) th position are equal to 0.Let x be the outcome of the game and note that l n ≥ − log(4 α ) / log 2 > . It followsthat d − ( x,
0) = lim inf k →∞ { ≤ i ≤ k : x i = 0 } k ≥ lim inf n →∞ { ≤ i ≤ k n : x i = 0 } k n +1 ≥ lim inf n →∞ P n − i =0 l i k n +1 ≥ lim inf n →∞ − n log(4 α ) / log 21 − log(( αβ ) n +1 r ) / log 2= log(4 α )log( αβ ) > which proves X is ( α, β ) -winning and therefore winning. Proving that the other sets fromthe theorem are winning is very similar and so we omit the proofs.Let ε > . We now prove that the set Y := (cid:8) x ∈ R : ε < d + ( x, (cid:9) is not winning. However, this is also very similar to the above proof, but with Bobadopting Alice’s strategy and therefore we only point out the differences.Fix α ∈ (0 , / and choose β ∈ (0 , such that log 4 − log( αβ ) ≤ ε. Bob starts the game by choosing x = 1 ∈ R and r = 1 . Alice then chooses y ∈ R suchthat B ( y , α ) ⊆ B ( x , . Let I ′ be a dyadic interval at level k ′ := ⌈− log α/ log 2 ⌉ whichis completely contained inside B ( y , α ) . Let l ′ = max { k ∈ N : 2 − k ′ − k > αβ } and Bob then chooses x to be the centre of the right most level ( k ′ + l ′ ) dyadic intervalinside I ′ . In particular, B ( x , αβ ) ⊆ I ′ ⊆ B ( y , α ) and for all numbers in B ( x , αβ ) , allof the binary digits from the ( k ′ + 1) th position to the ( k ′ + l ′ ) th position are equal to 1.17his argument is then repeated. When Alice chooses y n as the center of her n thball, let I ′ n be a dyadic interval of length k ′ n := ⌈− log( αr n ) / log 2 ⌉ which is completelycontained inside B ( y n , αr n ) . Let l ′ n = max { k ∈ N : 2 − k ′ n − k > αβr n } and let x n be the centre of the rightmost level ( k ′ n + l ′ n ) dyadic interval inside I ′ n . Inparticular, B ( x n , αβr n ) ⊆ I ′ n ⊆ B ( y n , αr n ) and for all numbers in B ( x n , αβr n ) all of thebinary digits from the ( k ′ n + 1) th position to the ( k ′ n + l ′ n ) th position are equal to 1.Let x be the outcome of the game and note that l ′ n ≥ − log(4 βα ) / log 2 > . It followsthat d + ( x,
0) = lim sup k →∞ { ≤ i ≤ k : x i = 0 } k ≤ lim sup n →∞ { ≤ i ≤ k ′ n : x i = 0 } k ′ n − ≤ lim sup n →∞ k ′ n − P n − i =0 l ′ n k ′ n − ≤ − lim inf n →∞ − n log(4 αβ ) / log 2 − log( α ( αβ ) n − ) / log 2= log 4 − log( αβ ) ≤ ε which proves Y is not ( α, β ) -winning and therefore not winning.Proving that the other sets from the theorem are not winning is very similar and sowe omit the proofs.We note that Alice’s and Bob’s strategy in the above proofs ensure that the outcome isas desired for at least one valid binary expansion. The proof so far did not pay attentionto the fact that a number may have two valid binary expansions: if eventually thereare only s in the binary expansion then it has another expansion where it has only seventually. However, it is easy to modify the proofs above so that no outcome has twodifferent valid expansions. For example, every now and then (but rarely) the players playa round with the opposite digit than they usually play for. We exclude the exact details. Acknowledgements
This work began during the semester programme on
Fractal Geometry, HyperbolicDynamics and Thermodynamical Formalism hosted by ICERM in Spring 2016. Itcontinued at the semester programme on
Fractal Geometry and Dynamics hosted bythe Institut Mittag-Leffler in Fall 2017. AF was financially supported by an
ERCConsolidator Grant (772466) and by
The MTA Momentum Project (LP2016-5). JMFwas financially supported by a
Leverhulme Trust Research Fellowship (RF-2016-500) andan
EPSRC Standard Grant (EP/R015104/1). EN and DS were supported by an EPSRCProgramme Grant (EP/J018260/1). 18 eferences [BHNS13] D. Badziahin, S. Harrap, E. Nesharim, D.S. Simmons. Schmidt games andCantor winning sets, arxiv. org/ abs/ 1804. 06499 , (2018), 1–36.[BFFH13] R. Balka, Á. Farkas, J. M. Fraser, J. T. Hyde. Dimension and measure forgeneric continuous images,
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Arch. Math. , , (1959), 235–240. Ábel Farkas, E-mail: [email protected]
Jonathan M. Fraser, E-mail: [email protected]
Erez Nesharim, E-mail: [email protected]
David Simmons, E-mail: [email protected]@gmail.com