Schmidt games and Cantor winning sets
Dzmitry Badziahin, Stephen Harrap, Erez Nesharim, David Simmons
aa r X i v : . [ m a t h . N T ] A p r Schmidt games and Cantor winning sets
Dzmitry Badziahin, Stephen Harrap, Erez Nesharim, David Simmons
Abstract
Schmidt games and the Cantor winning property give alternative notions of largeness,similar to the more standard notions of measure and category. Being intuitive, flexible,and applicable to recent research made them an active object of study. We survey thedefinitions of the most common variants and connections between them. A new gamecalled the Cantor game is invented and helps with presenting a unifying framework. Weprove surprising new results such as the coincidence of absolute winning and 1 Cantorwinning in doubling metric spaces, and the fact that 1 / R , and we suggest a prototypical example of a Cantor winning set to showthe ubiquity of such sets in metric number theory and ergodic theory. When attacking various problems in Diophantine approximation, the application of certaingames has proven extremely fruitful. Many authors have appealed to Schmidt’s celebratedgame [37] (e.g. [10, 14, 17, 19, 22, 31, 34, 39]), McMullen’s absolute game [33] (e.g. [11,30, 25, 28, 32]), and the potential game [25] (e.g. [3, 4, 27, 35, 41, 40]). It is the amenableproperties of the winning sets associated with games of this type that make the games suchattractive tools. These properties most commonly include (here we use winning in a loosesense to mean winning for either the α game, the c potential game, or the absolute game):(W1) A winning set is dense within the ambient space. In R N winning sets are thick; i.e.,their intersection with any open set has Hausdorff dimension N .(W2) The intersection of a countable collection of winning sets is winning.(W3) The image of a winning set under a certain type of mapping (depending on the gamein question) is winning.At the same time, many related problems in Diophantine approximation have been solvedvia the method of constructing certain tree-like (or “Cantor-type”) structures inside a Dio-phantine set of interest [1, 2, 6, 7, 8, 15, 36, 42]. One of the key ingredients in the proofof a famous Schmidt conjecture in [6] was the construction of a certain generalised Cantorset in R . The main ideas were formalised in the subsequent paper [7] and then in [5] thetheory of generalised Cantor sets and Cantor winning sets was finalised in the general set-ting of complete metric spaces and a host of further applications to problems in Diophantineapproximation was given. The family of Cantor winning sets satisfies properties (W1)-(W3).It is therefore very natural to investigate how these two different approaches are related,and that is precisely the intention of this paper. In § winning α winningAbsolute winning 1 Cantor winning ε Cantor winning0 Potential winning c Potential winning R \ Figure 1: Main resultsThe main results of this paper can be summarised in Figure 1. In Figure 1 the solidarrows represent known implications, the dashed arrows represent our new results, and acrossed arrow stands for where an implication does not hold in general.The following connections are already known:
Proposition 1.1 ([33]) . If E ⊆ R N is absolute winning then E is α winning for every α ∈ (0 , / . One can actually slightly improve on this via the following folklore argument (see [27,Proposition 2.1] for a related observation):
Proposition 1.2. If E is winning and α := sup { α : E is α winning } , then E is α winning. For the sake of completeness we will prove this proposition in Section 2. Potential winningis related to absolute winning by:
Proposition 1.3 ([25, Theorem C.8]) . Let X be a complete doubling metric space. If E ⊆ X is c potential winning for every c > (i.e., potential winning), then E is absolute winning. Meanwhile, absolute winning is related to Cantor winning in the following way:
Proposition 1.4 ([5, Theorem 12]) . Absolute winning subsets of R N are Cantor winning.
In some sense, the above statements suggest that the construction of Cantor winning setsmimics the playing of a topological game. In § Theorem 1.5.
A set E ⊆ R N is ε Cantor winning if and only if it is N (1 − ε ) potentialwinning. In particular, E is Cantor winning if and only if E is potential winning. If E ⊆ R N is Cantor winning then it is absolute winning. See Definition 2.21 for the definition of doubling metric spaces. § Theorem 1.6.
There is a Cantor winning set in R that is not winning. Theorem 1.7.
There is a winning set in R that is not Cantor winning. However, a partial result connecting Cantor winning sets and winning sets when somerestrictions are placed on the parameters is possible:
Theorem 1.8.
Let E ⊆ R N be an ε Cantor winning set. Then for every c > N (1 − ε ) , thereexists γ > such that for all α, β > with α < γ ( αβ ) c/N , the set E is ( α, β ) winning. Even though Cantor winning does not imply winning, the intersection of the correspondingsets has full Hausdorff dimension:
Theorem 1.9.
The intersection in R N of a winning set for Schmidt’s game and a Cantorwinning set has Hausdorff dimension N . That being said, we prove a surprising result regarding subsets of the real line:
Theorem 1.10.
If a set E ⊆ R is / winning then it is Cantor winning.
In view of Theorem 1.10, Propositions 1.1 & 1.2, and Theorem 1.5, we deduce the followingremarkable equivalence:
Corollary 1.11.
The following statements about a set E ⊆ R are equivalent:1. E is / winning.2. E is Cantor winning.3. E is absolute winning. In [8] the notion of
Cantor rich subsets of R was introduced. Cantor rich sets were shownto satisfy some of the desirable properties associated with winning sets, and have been appliedsince elsewhere [42]. As an application of the results in this paper we can prove that Cantorrichness and Cantor winning coincide. Theorem 1.12.
A set E ⊆ R is Cantor rich in an interval B if and only if E is Cantorwinning in B . Another way of measuring the largeness of a set is by considering its intersections withother sets. Proving such intersection properties served as part of the motivation that led tothe invention of the absolute game. For example, it is well known that absolute winning setsare large in the following dual sense [11]:(W4) The intersection of an absolute winning set with any closed nonempty uniformly perfectset is nonempty.We demonstrate that a similar property holds for Cantor winning sets. In fact, Cantorwinning Borel sets are characterised by an analogue of property (W4):
Theorem 1.13.
Let E ⊆ R N . If E is ε Cantor winning then E ∩ K = (cid:31) for any K ∈ R N which is N (1 − ε ) Ahlfors regular. If E is Borel then the converse is also true. Theorem 1.14.
Let X = { , } N with the metric d ( x, y ) = 2 − v ( x,y ) , where v ( x, y ) denotesthe length of the longest common initial segment of x and y . Let T : X → X be the left shiftmap. If K ⊆ X satisfies dim H K < then the set E = n x ∈ X : { T i x : i ≥ } ∩ K = (cid:31) o (1) is − dim H K Cantor winning, or, equivalently, dim H K potential winning. Here, dim H stands for the Hausdorff dimension. The set E in (1) was first considered byDolgopyat [16], who proved that dim H E = 1. Unlike the property of having full Hausdorffdimension, the potential winning property passes automatically to subspaces. More precisely,if Y ⊆ X is a subspace of X and E ⊆ X is c potential winning, then E ∩ Y is c potentialwinning on Y . Hence, in particular, using the notation of Theorem 1.14, if Y ⊆ X is asubshift of finite type and K ⊆ X satisfies dim H ( K ∩ Y ) < dim H Y , then Theorems 1.14and 4.6 immediately imply that dim H ( E ∩ Y ) = dim H Y , which is exactly the content of [16,Theorem 1].The set of all points whose orbit closure misses a given subset is considered in variouscontexts, such as piecewise expanding maps of the interval and Anosov diffeomorphisms oncompact surfaces [16], certain partially hyperbolic maps [13, 41], rational maps on the Rie-mann sphere [12], and β shifts [20, 21]. We suggest that Theorem 1.14 might be generalisableto these contexts; however, we choose to present the proof in the above prototypical example.See also [9] for a related problem where the rate at which the orbit is allowed to approachthe subset K depends on the geometrical properties of K . We formally define the concepts introduced in §
1. Each of the games we describe below isplayed between two players, Alice and Bob, in a complete metric space ( X, d ). Given two real parameters α, β ∈ (0 , Schmidt’s ( α, β ) game begins with Bob choosing anarbitrary closed ball B ⊆ X . Alice and Bob then take turns to choose a sequence of nestedclosed balls (Alice chooses balls A i and Bob chooses balls B i ) satisfying both B ⊇ A ⊇ B ⊇ A ⊇ · · · ⊇ B i ⊇ A i +1 ⊇ B i +1 ⊇ · · · and rad( A i +1 ) = α · rad( B i ) and rad( B i +1 ) = β · rad( A i +1 ) for all i ≥ . (2)Since the radii of the balls tend to zero, at the end of the game the intersection of Bob’s (orAlice’s) balls must consist of a single point. See Figure 2 for a pictorial representation of thegame, in which the shaded area represents where Alice has specified Bob must play his nextball. A set E ⊆ X is called ( α, β ) winning if Alice has a strategy for placing her balls thatensures ∞ \ i =0 B i ⊆ E. (3)4 αr αβrB A B A rad( A i +1 ) = β rad( B i )rad( B i +1 ) = α rad( A i +1 )Figure 2: Schmidt’s gameIn this case we say that Alice wins. The α game is defined by allowing Bob to choose theparameter β and then continuing as in the ( α, β ) game. We say that E ⊆ X is α winning ifAlice has a winning strategy for the α game. Note that E ⊆ X is α winning if and only if E is ( α, β ) winning for all 0 < β <
1. Finally,
Schmidt’s game is defined by allowing Aliceto choose 0 < α < α game. We say E ⊆ X is winning ifAlice has a winning strategy for Schmidt’s game. Note that E is winning if and only if it is α winning for some α ∈ (0 , E is α winning for some α ∈ (0 ,
1) then E is α ′ winningfor every α ′ ∈ (0 , α ), and that E is α winning for some α ∈ (1 / ,
1) if and only if E = X ;i.e., the property of being 1 / Remark . Strictly speaking, since a ball in an abstract metric space X may not necessarilypossess a unique centre and radius, Alice and Bob should pick successive pairs of centres andradii for their balls, satisfying certain distance inequalities that formally guarantee that theappropriate subset relations hold between the corresponding balls. However, we will assumethat this nuance is accounted for in each player’s strategy.We end the introduction of Schmidt games with the proof of the folklore Proposition 1.2. Proof of Proposition 1.2.
Fix 0 < β <
1, let α be as in the statement, and set α = α − ( α β ) − β (1 − α )1 − ( α β ) − α β (1 − α ) < α , αβ = ( α β ) . (4)We describe the winning strategy for Alice for the ( α , β ) game (Game I) based on herstrategy for the ( α, β ) game (Game II). It can be given as follows:1. Whenever Bob plays a move B n = B ( b n , ρ n ) in Game I, he correspondingly plays5he move B ′ n = B ( b ′ n , ρ ′ n ) in Game II, where(1 − α ) ρ ′ n = (1 − α ) ρ n , b ′ n = b n .
2. When Alice responds to this by playing a move A ′ n +1 = B ( a ′ n +1 , αρ ′ n ) in Game II(according to her winning strategy), she correspondingly plays the move A n +1 = B ( a n +1 , α ρ n ) in Game I, where a n +1 = a ′ n +1 .
3. When Bob responds to this by playing a move B n +1 = B ( b n +1 , ρ n +1 ) in Game I,Alice responds by playing the move A n +2 = B ( a n +2 , α ρ n +1 ) in Game I, where a n +2 = b n +1 . This sets the stage for the next iteration.To finish the proof, it suffices to check that all of these moves are legal. When n = 0, move1 is legal because any ball can be Bob’s first move. Move 2 is legal because d ( b n , a n +1 ) = d ( b ′ n , a ′ n ) ≤ (1 − α ) ρ ′ n = (1 − α ) ρ n , and move 3 is legal because d ( b n +1 , a n +2 ) = 0. When n ≥
1, move 1 is legal because d ( a ′ n − , b ′ n ) = d ( a n − , b n ) ≤ d ( a n − , b n − ) + d ( a n , b n ) ≤ (1 − β ) α ρ n − + (1 − β ) α ( α β ) ρ n − = (1 − β ) αρ ′ n − . For details and other properties of winning sets see Schmidt’s book [38].
McMullen’s absolute game [33] is defined in such a way that instead of choosing a regionwhere Bob must play, Alice now chooses a region where Bob must not play. To be precise,given 0 < β <
1, the β absolute game is played as follows: Bob first picks some initial ball B ⊆ X . Alice and Bob then take turns to place successive closed balls in such a way that B ⊇ B \ A ⊇ B ⊇ · · · ⊇ B i ⊇ B i \ A i +1 ⊇ B i +1 ⊇ · · · , (5)subject to the conditionsrad( A i +1 ) ≤ β · rad( B i ) and rad( B i +1 ) ≥ β · rad( B i ) for all i ≥ . (6)See Figure 3 for a pictorial view of the game in R .If r i := rad( B i ) E ⊆ X is β absolute winning if Alice has a strategy which guarantees that either she wins by defaultor (3) is satisfied. In this case, we say that Alice wins. The absolute game is defined byallowing Bob to choose 0 < β < β absolutegame. A set is called absolute winning if Alice has a winning strategy for the absolute game.Note that a set is absolute winning if and only if it is β absolute winning for every 0 < β < H be any collection of nonempty closed subsets6 βr βrB A B A B rad( A i +1 ) ≤ β rad( B i )rad( B i +1 ) ≥ β rad( B i )Figure 3: McMullen’s absolute gameof X . The H absolute game is defined in a similar way to the absolute game, but Alice isnow allowed to choose a neighborhood of an element H ∈ H , namely, A i +1 = B ( H, β · rad ( B i )) . McMullen’s absolute game is the same as the H absolute game when H is the collection ofall singletons. The first generalisation of the absolute game was the k dimensional absolutegame , which is the H absolute game where X = R N and H is the collection of k dimensionalhyperplanes in R N . (When k = N − Remark . The definition we gave deserves some justification. We do not require 0 < β < as in McMullen’s original definition for X = R N . To cover those cases where Bob chooses ≤ β <
1, in which Bob might not have a legal move that satisfies (5) on one of his turns,we declare that Bob is losing if this happens. This gives the wanted effect when X = R N ,but puts us in a situation that a subset of an abstract metric space might be winning becausethe space X is not large enough to allow Bob not to lose by default. In this situation, eventhe empty set might be absolute winning (cf. Remark 4.8). This situation, however, cannotoccur when X is uniformly perfect (cf. Section 2.7).Also note that it is shown in [25, Proposition 4.5] that replacing the inequalities (6) withequalities does not change the class of absolute winning sets. The reason for introducingthese inequalities is to allow an easy proof of the invariance of absolute winning sets under“quasisymmetric maps” (cf. [33, Theorem 2.2]).7 .3 The potential game Given two parameters c, β >
0, the ( c, β ) potential game begins with Bob picking some initialball B ⊆ X . On her ( i + 1)st turn, Alice chooses a countable collection of closed balls A i +1 ,k ,satisfying X k rad ( A i +1 ,k ) c ≤ ( β · rad ( B i )) c . (7)By convention, we assume that a ball always has a positive radius; i.e. Alice can notchoose A i,k to be singletons. Then, Bob on his ( i + 1)st turn chooses a ball B i +1 satis-fying rad ( B i +1 ) ≥ β · rad ( B i ). See Figure 4 for a demonstration of the playing of such agame in R (in which, for simplicity, Bob’s balls do not intersect Alice’s preceding collections- however, it should be understood that this is not a necessary condition for Bob).If \ i B i ⊆ [ i [ k A i,k (8)or if rad ( B i ) E ⊆ X is said to be ( c, β ) potential winning if Alice has a strategy guaranteeingthat either she wins by default or (3) holds. The c potential game is defined by allowingBob to choose in his first turn the parameters c > c and β >
0, and then continue with asin the ( c, β ) potential game. A set E is c potential winning if Alice has a winning strategyfor E in the c potential game. Note that E is c potential winning if and only if E is ( c, β )potential winning for every c > c and β >
0. If X is Ahlfors δ regular for some δ >
0, the potential game is defined by allowing Alice to choose c < δ and then continuing as in the c potential game, and a set E is called potential winning if Alice has a winning strategy in thepotential game. Note that E is potential winning if and only if it is c potential winning forsome c < δ . rβrB A , A , A , A , A , B A , B X k (rad( A i +1 ,k )) c ≤ ( β rad( B i )) c rad( B i +1 ) ≥ β rad( B i )Figure 4: The potential game8 emark . The c potential game with c = 0 was introduced in [25, Appendix C] asthe potential game. In light of the connections between Cantor, absolute, and potentialwinning § §
4, it is natural to adjust the definition thatappears there to the one that appears here.For the sake of completeness we will verify Properties (W1) and (W2) for potential winninggames. Property (W1) is verified for a class of Ahlfors regular spaces X in Theorem 4.6. Proposition 2.4.
Let E and E from X be c and c potential winning respectively. Then E ∩ E is max { c , c } potential winning.Proof. Fix arbitrary c > max { c , c } and β >
0. By assumption, E and E are both ( c, β )potential winning. Now Alice uses the following strategy to win the ( c, β ) game on the set E ∩ E . After Bob’s initial move B Alice chooses the collection A ,k from her ( c, β ) winningstrategy for the set E . After Bob’s first move B Alice chooses the collection A ∗ ,k from her( c, β ) winning strategy for the set E , assuming that Bob’s initial move was B .In general, after B m Alice chooses the collection A m,k following her ( c, β ) winningstrategy for the set E ; after B m +1 she chooses the collection A ∗ m,k following her ( c, β )winning strategy for the set E .It is easy to check that Alice’s moves are legal, because X k rad ( A i +1 ,k ) c ≤ (cid:0) β · rad ( B i ) (cid:1) c ≤ ( β · rad ( B i )) c . The same inequality is true for the collections A ∗ i +1 ,k . Also, by construction, either Alice winsby default or S i B i ⊆ E ∩ E . One can define further variants of Schmidt’s original game. The strong game (introduced byMcMullen in [33]) coincides with Schmidt’s game except that the equalities (2) are replacedby the inequalitiesrad( A i +1 ) ≥ α · rad( B i ) and rad( B i +1 ) ≥ β · rad( A i +1 ) for all i ≥ . (9)Similarly, we define the weak game so that Alice can choose her radii with an inequalitybut Bob must use equality. Obviously, every weak winning set is winning but the converse isnot necessarily true (however, the weak game does not have the intersection property—seelater).One could also define a very strong game , where Alice must use equality but Bob mayuse inequality. Remark . It can be shown that a set is strong winning if and only if it is very strongwinning, but this fact will not be needed hence we skip its proof.Later in the paper we will need the following proposition which relates the very strongwinning property of a set E and the weak winning property of its complement X \ E . Proposition 2.6.
Let E ⊆ X satisfy the following property: for any β ∈ (0 , there exists α > such that E is ( α, β ) very strong winning. Then X \ E is not weak winning.Proof. We need to show that for any β > X \ E is not β weak winning. In orderto do that, we will provide a winning strategy for Bob that ensures that T ∞ i =0 B i ⊆ E . Onhis first turn Bob chooses α > E is ( α, β ) very strong winning, so that he is nowplaying the ( β, α ) weak game, and chooses an arbitrary ball B . In his subsequent movesBob follows Alice’s winning strategy for the ( α, β ) very strong game.9 .5 Cantor winning For the sake of clarity we introduce a specialisation of the framework presented in [5], tailoredto suit our needs. We first define the notion of a “splitting structure” on X . This is in somesense the minimal amount of structure the metric space needs to allow the construction ofgeneralised Cantor sets (cf. [5]). Denote by B ( X ) the set of all closed balls in X and by P ( B ( X )) the set of all subsets of B ( X ). A splitting structure is a quadruple ( X, S , U, f ),where • U ⊆ N is an infinite multiplicatively closed set such that if u, v ∈ U and u | v then v/u ∈ U ; • f : U → N is a totally multiplicative function; • S : B ( X ) × U → P ( B ( X )) is a map defined in such a way that for every ball B ∈ B ( X )and u ∈ U , the set S ( B, u ) consists solely of balls S ⊆ B of radius rad( B ) /u .Additionally, we require all these objects to be linked by the following properties:(S1) S ( B, u ) = f ( u );(S2) If S , S ∈ S ( B, u ) and S = S then S and S may only intersect on the boundary,i.e. the distance between their centers must be at least rad( S ) + rad( S ) = 2rad( B ) /u ;(S3) For all u, v ∈ U , S ( B, uv ) = [ S ∈ S ( B,u ) S ( S, v ) . Broadly speaking, the set U determines the set of possible ratios of radii of balls in thesuccessive levels of the upcoming Cantor set construction. For any fixed B ∈ B ( X ), thefunction f tells us how many balls inside B of a fixed radius we may choose as candidatesfor the next level in the Cantor set construction, and the sets S ( B, u ) describe the positionof these candidate balls.If f ≡ X, S , U, f ) is trivial. Otherwise, fix a sequence( u i ) i ∈ N with u i ∈ U such that u i | u i +1 and u i −→ i →∞ ∞ . Then, one can show [5, Theorem 2.1]that for any B ∈ B ( X ) the limit set A ∞ ( B ) := ∞ \ i =1 [ B ∈ S ( B,u i ) B is compact and independent of the choice of sequence ( u i ) i ∈ N . Example 2.7.
The standard splitting structure on X = R N is with the metric that comesfrom the sup norm: d ( x, y ) := k x − y k ∞ . The collection of splittings S ( B, u ) of any closedball B according to the integer u ∈ U = N , into f ( u ) = u N parts, is defined as follows:Cut B into u N equal square boxes which edges have length u times less than the edges lengthof B . One can easily check that ( R N , S , N , f ) satisfies properties (S1) – (S3) and is the uniquesplitting structure for (cid:0) R N , d (cid:1) with U = N such that A ∞ ( B ) = B for every B ∈ B ( X ). We assume that by definition, to specify a closed ball, it is necessary to specify its center and radius. Notethat this means that in some cases two distinct balls may be set-theoretically equal in the sense of containingthe same points.
10n a similar manner, the standard splitting structure on { , } N is with the standardmetric d ( x, y ) := 2 − min { i ≥ x i = y i } , U = { i : i ≥ } , f ( u ) = u , and S ([ x , . . . , x n ] , i ) = (cid:8) [ x , . . . , x n , y , . . . , y i ] : ( y , . . . , y i ) ∈ { , } i (cid:9) where for any x ∈ X and n ≥ x , . . . , x n ] := { y : y k = x k for all 1 ≤ k ≤ n } = B ( x, − n ) to be the cylinder with prequel ( x , . . . , x n ), thought of as a ball of radius 2 − n (thechoice of center is arbitrary).For any collection B ⊆ P ( B ( X )) of balls and for any R ∈ U we will write1 R A := [ A ∈A S ( A, R ) . Of course, for higher powers of R we can write R n A = R (cid:0) · · · (cid:0) R A (cid:1) · · · (cid:1) .We can now recall the definition of generalised Cantor sets. Fix some closed ball B ⊆ X and some R ∈ U , and let r := ( r m,n ) , m, n ∈ Z > , m n be a two-parameter sequence of nonnegative real numbers. Define B := { B } . We startby considering the set R B . The first step in the construction of a generalised Cantor setinvolves the removal of a collection A , of at most r , balls from R B . We label the survivingcollection as B . Note that we do not specify here the removed balls, we just give an upperbound for their number. In general, for a fixed n >
0, given the collection B n we construct anested collection B n +1 as follows: Let B n +1 := (cid:18) R B n (cid:19) \ n [ m =0 A m,n , where A m,n ⊆ R n − m +1 B m (0 ≤ m ≤ n )are collections of removed balls satisfying for every B ∈ B m the inequality A m,n ( B ) ≤ r m,n , where A m,n ( B ) := { A ∈ A m,n : A ⊆ B } . For a pictorial demonstration in R of such a construction, see Figure 5, in which boxesremoved in the current step are coloured red.We denote the limit set of such a construction by K ( B , R, r ). Note that the triple( B , R, r ) does not uniquely determine K ( B , R, r ) as there is a large degree of freedom inthe sets of balls A m,n removed in the construction procedure. We say a set K is a generalisedCantor set if it can be constructed by the procedure described above for some triple ( B , R, r ).In this case, we may refer to K as being ( B , R, r ) Cantor if we wish to make such a tripleexplicit and write K = K ( B , R, r ). Example 2.8.
In the case that r m,n = 0 for every m < n , we refer to a ( B , R, r ) Cantor setas a local Cantor set and denote it simply by K ( B , R, { r n } ), where r n := r n,n . The simplestclass of local Cantor sets occur when the sequence { r n } is taken to be constant; i.e., r n := r .Such ( B , R, r ) Cantor sets are well studied and their measure theoretic properties are wellknown. Example 2.9.
The middle-third Cantor set K is an ( I, ,
1) Cantor set in R , and K is a (cid:0) I , , (cid:1) Cantor set in R . Definition 2.10.
Fix a ball B ⊆ X . Given a parameter ε ∈ (0 ,
1] we say a set E ⊆ X is ε Cantor winning on B (with respect to the splitting structure ( X, S , U, f ) ) if for every11 Step 1 B is subdivided into R boxes R B and thosein A , are removed,leaving the collection B Step 2( i )Each A ∈ B is subdividedinto R boxes R A and thosein A , ( A ) are removed. Step 2( ii )Then, any boxes in A , from R B are removed, leavingthe collection B Step 3( i )Each A ∈ B is subdividedinto R boxes R A and thosein A , ( A ) are removed Step 3( ii )Next, for each A ′ ∈ B anyboxes in A , ( A ′ )from R B are removed Step 3( iii )Finally, any boxes in A , from R B are removed, leavingthe collection B Figure 5: Construction of a Cantor winning set0 < ε < ε there exists R ε ∈ U such that for every R ε ≤ R ∈ U the set E contains a ( B , R, r )Cantor set satisfying r m,n ≤ f ( R ) ( n − m +1)(1 − ε ) for every m, n ∈ N , m n. If a set E ⊆ X is ε Cantor winning on B for every ball B ⊆ X then we say E is ε Cantor winning , and simply that E is Cantor winning if it is ε Cantor winning for some ε ∈ (0 , ε Cantor winning then it is also ε Cantor winning for any ε ∈ (0 , ε ). In particular, the property of being 1 Cantor winning is the strongest possibleCantor winning property. Remark . As for the generalised Cantor sets, whenever we mention Cantor winning setsnot in the context of a splitting structure we will refer to a standard splitting structure. Inparticular, this applies to all theorems in Section 1.2 where the standard splitting structurein R N is considered. Some of those results remain true in a more general context of completemetric spaces. The class of
Cantor rich subsets of R was introduced in [8]. As was noted in [5], this notion canbe generalised to the context of metric spaces endowed with a splitting structure. However,12e will show in § Definition 2.12.
Let M ≥ B be a ball in R . A set E ⊆ R is( B , M ) Cantor rich if for every
R > M (10)and 0 < y < E contains a ( B , R, r ) Cantor set, where r = ( r m,n ) ≤ m ≤ n is a two-parametersequence satisfying n X m =0 (cid:18) R (cid:19) n − m +1 r m,n ≤ y for every n ≥ .E is Cantor rich in B if it is ( B , M ) Cantor rich in B for some M ≥ E is Cantor rich if it is Cantor rich in some B .Note that, following [8], Cantor rich sets are Cantor rich in some ball, while Cantorwinning sets are Cantor winning in every ball. Also note that unlike in [8], the inequality(10) is strict. This enables a clearer correspondence with the parameter ε in Definition 2.10(cf. Corollary 3.11). Schmidt’s game was originally played on a complete metric space. However, in order toconclude that winning sets have properties such as large Hausdorff dimension, certain as-sumptions on the underlying space are required. The first time that large intersection withcertain subspaces of R n was proved using Schmidt’s game was in [23]. Later it was realised in[11] that more generally some spaces form a natural playground for the absolute game. Thefollowing notion generalises the definition of a diffuse set in R n as appeared in [11]. Definition 2.13.
For 0 < β <
1, a closed set K ⊆ X is β diffuse if there exists r > x ∈ K , y ∈ X and 0 < r ≤ r , we have (cid:0) B ( x, (1 − β ) r ) \ B ( y, βr ) (cid:1) ∩ K = (cid:31) . (11)If K is β diffuse for some β then we say it is diffuse .Condition (11) is satisfied if and only if there exists z ∈ K such that B ( z, βr ) ⊆ B ( x, r ) \ B ( y, βr ) . The definition of diffuse sets is designed so that the β absolute game on a β diffuse metricspace cannot end after finitely many steps (as long as the first ball of Bob is small enough).So, if Alice plays the game in such way that she is able to force the outcome to lie in thetarget set, then the reason for her victory is her strategy and not the limitations of the spacein which the game is played. Remark . In the definition that appears in [11], sets as in Definition 2.13 are called0-dimensionally diffuse. Moreover, equation (11) is replaced by (cid:0) B ( x, r ) \ B ( y, βr ) (cid:1) ∩ K = (cid:31) . (12)Note that B ( x, r ) \ B ( y, βr ) = B (cid:0) x, (cid:0) − β ′ (cid:1) ρ (cid:1) \ B (cid:0) y, β ′ ρ (cid:1) , for β ′ = β β and ρ = (cid:16) β (cid:17) r , so both (11) and (12) give rise to the same class of diffusesets. We will use (11) and (12) interchangeably without further notice.13 more standard definition that is equivalent to diffuseness (see [25, Lemma 4.3]) is thefollowing: Definition 2.15.
For 0 < c <
1, a closed set K ⊆ X is c uniformly perfect if there exists r > x ∈ K and 0 < r ≤ r we have (cid:0) B ( x, r ) \ B ( x, cr ) (cid:1) ∩ K = (cid:31) . (13)If K is c uniformly perfect for some c then we say it is uniformly perfect .Being diffuse is equivalent to having no singletons as microsets [11, Lemma 4.4]. Insteadof expanding the discussion on microsets, we refer the interested reader to [26], and mentionthat Ahlfors regular sets are diffuse. In the other direction, diffuse sets support Ahlforsregular measures (see Proposition 2.18 below). We will now give the relevant definitions andprovide the results which support the statements above. Definition 2.16.
A Borel measure µ on X is called Ahlfors regular if there exist
C, δ, r > x ∈ X and 0 < r ≤ r satisfy1 C r δ ≤ µ ( B ( x, r )) ≤ Cr δ . In this case we call δ the dimension of µ and say that µ is Ahlfors regular of dimension δ .We say that a closed set K ⊆ X is Ahlfors regular if it is the support of an Ahlfors regularmeasure. More exactly, K ⊆ X is δ Ahlfors regular if it is the support of an Ahlfors regularmeasure of dimension δ . This terminology gives rise to a dimensional property of K : Definition 2.17.
The
Ahlfors regularity dimension of K ⊆ X isdim R K = sup { δ : K supports an Ahlfors regular measure of dimension δ } . Note that it is a direct consequence of the Mass Distribution Principle (see [18]) thatdim R K ≤ dim H K , where dim H K stands for the Hausdorff dimension of K . Proposition 2.18.
A diffuse set K ⊆ X satisfies dim R K > . Conversely, if dim R K > then K contains a diffuse set.Proof. Suppose K is β diffuse; then every ball in K of radius ρ contains two disjoint balls ofradius βρ . Construct a Cantor set by recursively replacing every ball of radius ρ with twodisjoint subballs of radius βρ . Standard calculations show that this Cantor set is Ahlforsregular of dimension log(2) − log( β ) >
0, so dim R K > R K >
0; then K contains an Ahlfors regular subset of positive dimension,which is diffuse (cf. e.g. [11, Lemma 5.1]).The following property of Ahlfors regular sets will be used later in the paper. Proposition 2.19. If X is δ Ahlfors regular then there exists c > such that for any ball B ⊆ X and any α < there exist at least cα − δ balls B i ⊆ B ( i ∈ { , . . . , I } , I > cα − δ ) of radius α · rad( B ) , and separated in the following sense: d ( x i , x j ) > α · rad( B ) for any i = j I . roof. If α > / α / B ′ be a ball with the same center as B , but with radius (1 − α )rad( B ), and let D ⊆ X be an arbitrary ball of radius rad( D ) = 3 α · rad( B ). Then its δ Ahlfors regular measure canbe estimated as: µ ( D ) C · (3 α ) δ (rad( B )) δ C · (cid:18) α − α (cid:19) δ µ ( B ′ ) ≤ C α δ µ ( B ′ ) . Therefore, for any k balls D i of radius 3 α · rad( B ) with k < C − α − δ , there exists anotherball D whose center lies inside B ′ \ S ki =1 D i . Hence there are at least I = ⌈ C − α − δ ⌉ balls D i such that their centers lie inside B ′ and are distanced by at least 3 α · rad( B ) from eachother. For each 1 ≤ i ≤ I , let B i be a ball with the same center as of D i but with the radius α · rad( B ). Clearly the balls B i satisfy all the conditions of the proposition, where c > cα − δ ⌈ C − α − δ ⌉ for all α ∈ [0 , X being large enough. For example, absolutewinning sets may be empty if the Ahlfors regularity dimension is zero (cf. Remark 4.8).Various conditions are introduced in this context; for example, see [37, Section 11] and [23,Theorem 3.1]. More recently, the following condition was introduced in [5]:(S4) There exists an absolute constant C ( X ) such that any ball B ∈ B ( X ) cannot intersectmore than C ( X ) disjoint open balls of the same radius as B . Proposition 2.20 ([5, Corollary to Theorem 4.1]) . If X satisfies (S4) then for any B ∈ B ( X ) and any Cantor winning set E ⊆ X we have dim H ( E ∩ A ∞ ( B )) = dim H A ∞ ( B ) . Condition (S4) is similar to the following definition:
Definition 2.21.
A metric space X is doubling if there exists a constant N ∈ N such thatfor every r >
0, every ball of radius 2 r can be covered by a collection of at most N balls ofradius r .Doubling spaces satisfy condition (S4), but the converse does not hold. For example, inan ultrametric space any two non-disjoint balls of the same radius are set-theoretically equal.So an ultrametric space satisfies (S4) with C = 1, but is not necessarily doubling. As shownin [5], the space X needs to satisfy (S4) for Cantor winning sets in X to satisfy the largedimension (W1) and large intersection (W2) conditions, and if X is doubling then Cantorwinning sets satisfy (W3) with respect to bi-Lipschitz homeomorphisms of X . This makesdoubling spaces a natural setting for generalised Cantor sets, and it is under this conditionthat many of the results of the next section hold. To facilitate a more natural exposition, we introduce a new game. The “Cantor game”described below is similar to the games described earlier. The main difference is that this15ame is designed to be played on a metric space endowed with a splitting structure. Ournomenclature is so chosen because (as we shall demonstrate) under natural conditions thewinning sets for this game are precisely the Cantor winning sets defined in Definition 2.10.Let X be a complete metric space and let ( X, S , U, f ) be a splitting structure. Given ε ∈ (0 , ε Cantor game is defined as follows: Bob starts by choosing an integer 2 ≤ R ∈ U and a closed ball B ⊆ X . For any i ≥
0, given Bob’s i th choice B i , Alice chooses a collection A i +1 ⊆ R { B i } satisfying A i +1 ≤ f ( R ) − ε . Given A i +1 , Bob chooses a ball B i +1 ∈ R { B i } \ A i +1 , and the game continues. We say that E ⊆ X is winning for the ε Cantor game if Alice has a strategy which guarantees that (3)is satisfied. The
Cantor game is defined by allowing Alice first to choose ε ∈ (0 ,
1] and thencontinuing as in the ε Cantor game. We say that E is winning for the Cantor game if Alicehas a winning strategy for the Cantor game. Note that E is winning for the Cantor game ifand only if it is winning for the ε Cantor game for some ε ∈ (0 , E is winningfor the ε Cantor game on B if Alice has a winning strategy in the ε Cantor game given thatBob’s first ball is B . Remark . A version of the Cantor game may be played in metric spaces with no prescribedsplitting structure. Given c ≥ < β <
1, Bob will choose a sequence of decreasingballs with radii that shrink at rate β , while Alice will remove a collection of at most β − c closed balls of radius βr where r is the radius of Bob’s previous choice, instead of just onesuch ball as in the absolute game. When c = 0 this game coincides with the absolute game.It is clear that winning sets for the Cantor game contain Cantor winning sets. Indeed,by the construction, for any R ∈ U a winning set for the ε Cantor game on B containsa ( B , R, f ( R ) − ε ) Cantor set which is comprised of all Bob’s possible moves while playingagainst a fixed winning strategy of Alice. Therefore, it is ε Cantor winning on B . We showthat the converse is also true: Theorem 3.2.
Let E ⊆ X . Let B ⊆ X be a closed ball and < ε ≤ . If E is ε Cantorwinning on B then E is winning for the ε Cantor game on B .Proof. We will define a winning strategy for Alice. Fix R ≥
2, and let δ ≥ f ( R ) = R δ . Choose 0 < η < ε so small that R δ (1 − ε + η ) ⌊ R δ (1 − ε ) ⌋ + 1 < . Set ε = ε − η , (14) ε = ε − η, (15)and let R ε be as in Definition 2.10. Then for any ℓ such that R ℓ ≥ R ε , since f (cid:0) R ℓ (cid:1) = f ( R ) ℓ = R δℓ , there exists a (cid:16) B , R ℓ , (cid:0) R ( n − m +1) ℓδ (1 − ε ) (cid:1) ≤ m ≤ n (cid:17) Cantor set in E . Let ℓ belarge enough so that R ℓ ≥ R ε , R δ (1 − ε ) ⌊ R δ (1 − ε ) ⌋ + 1 ! ℓ < , (16)and R − ℓδ η < · (17)16uch a value exists because of the choice of η .Choose a (cid:16) B , R ℓ , (cid:0) R ( n − m +1) ℓδ (1 − ε ) (cid:1) ≤ m ≤ n (cid:17) Cantor set in E . Let A = {A m,n } ≤ m ≤ n bethe removed balls, so that B n +1 = R ℓ B n \ S nm =0 A m,n for all n ≥
0, where B = { B } .Define a winning strategy for Alice in the ε Cantor game as follows: For each i ≥ ϕ i ( B ) = ⌊ i/ℓ ⌋ X m =0 ∞ X n = m { A ∈ A m,n : A ∩ B has nonempty interior } R − ( n +1) ℓδ (1 − ε ) . (18)Given Bob’s i th move B i , Alice’s strategy is to choose the subcollection A i +1 ⊆ R { B i } consisting of the ⌊ R δ (1 − ε ) ⌋ balls with the largest corresponding values of ϕ i +1 . We will showthat this is a winning strategy for Alice. Assume Bob chose the sequence of balls { B i } ∞ i =0 onhis turns, while Alice responded according to the above strategy. It is enough to prove thatfor every i ≥ ϕ iℓ ( B iℓ ) < R − iℓδ (1 − ε ) . (19)Indeed, this inequality would imply that B iℓ does not intersect any of the balls from A m,n with n < i except at the boundary, which implies that B iℓ ∈ B i .Fix i ≥ ≤ j ≤ ℓ −
1, note that j iℓ + jℓ k = i , so by theinduction hypothesis (19), we get ϕ iℓ + j ( B iℓ + j ) ϕ iℓ ( B iℓ ) < R − iℓδ (1 − ε ) . Therefore, for any n ≥ m ≥
0, if A ∈ A m,n is such that A ∩ B iℓ + j +1 has nonempty interior,then necessarily n + 1 > i , and thus by (S2), we have A ⊆ B iℓ + j +1 . So X A ∈ R { B iℓ + j } ϕ iℓ + j ( A ) = ϕ iℓ + j ( B iℓ + j ) , (20)and the definition of Alice’s strategy yields ϕ iℓ + j ( B iℓ + j +1 ) ≤ ⌊ R δ (1 − ε ) ⌋ + 1 ϕ iℓ + j ( B iℓ + j ) (21)for every 0 ≤ j ≤ l −
1. Indeed, there are at least ⌊ R δ (1 − ε ) ⌋ balls A , chosen by Alice in herturn, with ϕ iℓ + j ( A ) > ϕ iℓ + j ( B iℓ + j +1 ). Therefore, if (21) fails for for some ball B iℓ + j +1 , wehave ϕ iℓ + j ( B iℓ + j +1 ) + X A chosen by Alice ϕ iℓ + j ( A ) > ϕ iℓ + j ( B iℓ + j )which contradicts (20).In order to estimate ϕ ( i +1) ℓ (cid:0) B ( i +1) ℓ (cid:1) we need to additionally consider the potential comingfrom the sets ( A i +1 ,n ) n > i +1 . Clearly, for any A ∈ A i +1 ,n the condition that A ∩ B ( i +1) ℓ hasnonempty interior is equivalent to the inclusion A ⊆ B ( i +1) ℓ . Hence we have: ϕ ( i +1) ℓ (cid:0) B ( i +1) ℓ (cid:1) ≤ (cid:18) ⌊ R δ (1 − ε ) ⌋ + 1 (cid:19) ℓ ϕ iℓ ( B iℓ )+ ∞ X n = i +1 (cid:8) A ∈ A i +1 ,n : A ⊆ B ( i +1) ℓ (cid:9) R − ( n +1) ℓδ (1 − ε ) . { A ∈ A m,n : A ⊆ B mℓ } ≤ R ( n − m +1) ℓδ (1 − ε ) for any n ≥ m ≥
0, by (14), (15),and (19), we get: ϕ ( i +1) ℓ (cid:0) B ( i +1) ℓ (cid:1) ≤ (cid:18) ⌊ R δ (1 − ε ) ⌋ + 1 (cid:19) ℓ R − iℓδ (1 − ε ) + ∞ X n = i +1 R ( n − i ) ℓδ (1 − ε ) R − ( n +1) ℓδ (1 − ε ) = R δ (1 − ε ) ⌊ R δ (1 − ε ) ⌋ + 1 ! ℓ R − ( i +1) ℓδ (1 − ε ) + ∞ X n =1 R − nℓδ η R − ( i +1) ℓδ (1 − ε ) = R δ (1 − ε ) ⌊ R δ (1 − ε ) ⌋ + 1 ! ℓ + R − ℓδ η − R − ℓδ η R − ( i +1) ℓδ (1 − ε ) . By (16) and (17) we get: ϕ ( i +1) ℓ (cid:0) B ( i +1) ℓ (cid:1) < R − ( i +1) ℓδ (1 − ε ) , which is what we need to complete the induction.Theorem 3.2 in many cases allows us to simplify the proofs which involve ε Cantor winningsets. Indeed, given an ε Cantor winning set we may now assume much simpler and seeminglystronger conditions for it than those in Definition 2.10 of Cantor winning sets. We illustratethis point by the following corollary:
Corollary 3.3.
Let E ⊆ X . Let B ⊆ X be a closed ball and < ε . If E is ε Cantorwinning on B then for any R > it contains a ( B , R, R − ε ) Cantor set.
For the sake of completeness, let us define a version of the Cantor game called the “Cantorpotential game”. Its name is justified via the analogy with the potential game, and the keyrole that potential functions, like the one in (18), play in the proof above that relates it tothe Cantor game.Given ε ∈ (0 , ε Cantor potential game is defined as follows: Bob starts bychoosing 0 < ε < ε , and Alice replies by choosing R ε ≥
2. Then Bob chooses an integer R ≥ R ε , and a closed ball B ⊆ X . For any i ≥
0, given Bob’s i th choice B i , Alice choosescollections {A i +1 ,k } ∞ k =0 such that A i +1 ,k ⊆ R k { B i } and A i +1 ,k ≤ f ( R ) ( k +1)(1 − ε ) for all k ≥
0. Given {A i +1 ,k } ∞ k =0 , Bob chooses a ball B i +1 ∈ R { B i }\ S ik =0 A i +1 − k,k (if there’sno such ball, we say that Alice wins by default). We say that E ⊆ X is ε Cantor potentialwinning if Alice has a strategy which guarantees that either she wins by default or (3) issatisfied. The
Cantor potential game is defined by allowing Alice first to choose ε ∈ (0 , ε Cantor potential game. Note that, by definition, ε Cantorwinning is the same as ε Cantor potential winning, upon rewriting ( m, n ) as ( i + 1 , k + i + 1).Theorem 3.2 may then be reinterpreted as saying that the winning sets for the ε Cantorpotential game are the same as the winning sets for the ε Cantor game. Indeed, by definitionof the Cantor potential game, a set is ε Cantor potential winning if and only if it is ( B , ε )Cantor winning for every ball B ⊆ X . We prove Theorem 1.5 as a special case of a more general result connecting potential winningsets with Cantor winning sets, corresponding to the standard splitting structure on R N .18 heorem 3.4. Let ( X, S , U, f ) be a splitting structure on a doubling metric space X , and fix B ∈ B ( X ) . Let δ = dim H ( A ∞ ( B )) , so that δ = log f ( u )log( u ) for all u ∈ U . Fix ε ∈ (0 , , and let c = δ (1 − ε ) . Then a set E ⊆ A ∞ ( B ) is ε Cantor winning on B with respect to ( X, S , U, f ) if and only if E is c potential winning on A ∞ ( B ) . In particular, E is Cantor winning on B if and only if E ispotential winning on A ∞ ( B ) . Corollary 3.5.
The Cantor winning property in a doubling metric space X is independentof the choice of splitting structure, provided that the limit set A ∞ ( B ) is fixed. A combination of Proposition 1.3 and Theorem 3.4 immediately yields the following ad-ditional corollary:
Corollary 3.6.
Assume a complete metric space X is doubling and is endowed with a non-trivial splitting structure ( X, S , U, f ) , and let B ∈ B ( X ) . Then, a set E ⊆ X is Cantorwinning on B if and only if E is absolute winning on A ∞ ( B ) . In particular, for any split-ting structure such that B = A ∞ ( B ) for every ball B , a set E ⊆ X is Cantor winning ifand only if E is absolute winning.Remark . The statements in [5, Section 7] immediately provide us with applications ofCorollary 3.6. In particular, various sets from [5, Theorems 14 and 15] are in fact absolutelywinning.
Proof of Forward Direction.
We now prove Theorem 3.4. As in the proof of Theorem 3.2,since X is doubling, we have f ( u ) = u δ (cf. [5, Corollary 1]). Suppose that E is ε Cantorwinning, and fix β > c > c , and ρ >
0. Fix 0 < ε < ε such that δ (1 − ε ) ∈ ( c , c ). Fixlarge R ∈ U . Its precise value will be determined later. Potentially R may depend on ε , β , c , and ρ , and in particular it satisfies R ≥ max( R ε , /β ) and R δ (1 − ε ) − c < . Then by thedefinition of Cantor winning, E contains some ( B , R, r ) Cantor set K , where r m,n = f ( R ) ( n − m +1)(1 − ε ) = R δ ( n − m +1)(1 − ε ) for all m, n with m ≤ n .We now describe a strategy for Alice to win the ( c, β ) potential game on A ∞ ( B ), assumingthat Bob’s starting move has radius ρ . Recall that for any B ∈ B m , A m,n ( B ) := { A ∈ A m,n : A ⊆ B } . Let ρ denote the radius of B , let D k denote Bob’s k th move, and for each k ∈ N let m = m k ∈ N denote the unique integer such that β · rad ( D k ) < R − m ρ ≤ rad ( D k ), assumingsuch an integer exists. If such a number does not exist then we say that m k is undefined.Then Alice’s strategy is as follows: On turn k , if m k is defined, then remove all elements ofthe set [ n ≥ m A m,n = [ B ∈B m [ n ≥ m A m,n ( B )that intersect Bob’s current choice. If m k is undefined, then delete nothing. Obviously, thisstrategy, if executable, will make Alice win since the intersection of Bob’s balls will satisfy T k ∈ N D k ⊆ K ⊆ E . To show that it is legal, we need to show that X B ∈B m B ∩ D k = ∅ X n ≥ m r m,n (cid:16) R − ( n +1) ρ (cid:17) c ≤ ( β · rad( D k )) c . (22) We are denoting Bob’s k th move by D k instead of B m so as to reserve the letters B and m for the splittingstructure framework. A m,n ( B ) all have radius R − ( n +1) ρ .To estimate the sum in question, we use the fact that X is doubling. Since the elementsof B m have disjoint interiors and have radius R − m ρ ≍ β rad( D k ), the number of them thatintersect D k is bounded by a constant depending only on β . Call this constant C . Then theleft-hand side of (22) is less than C X n ≥ m R δ ( n − m +1)(1 − ε ) (cid:16) R − ( n +1) ρ (cid:17) c = C (cid:0) R − m ρ (cid:1) c ∞ X ℓ =1 R δ (1 − ε ) ℓ R − cℓ C (rad( D k )) c R δ (1 − ε ) − c . (cid:0) since R δ (1 − ε ) − c (cid:1) By choosing R large enough so that 2 C R δ (1 − ε ) − c < β c , we guarantee that the move is legal. Proof of Backward Direction.
Suppose that E ⊆ A ∞ ( B ) is c potential winning, and fix0 < ε < ε . Let R ε = 2, and fix R ∈ U such that R ≥ R ε . Fix a large integer q ∈ N to be determined, and let β = 1 /R q and c = δ (1 − ε ). Then E is ( c, β ) potential winning.Now for each k ∈ N and B ∈ B qk , let A ( B ) denote the collection of sets that Alice deletesin response to Bob’s k th move B k = B , assuming that it has been preceded by the moves B , B , . . . , B k − satisfying B i +1 ∈ S ( B i , R q ) for all i = 0 , . . . , k − Since Alice’s strategyis winning, we must have E ⊇ A ∞ ( B ) \ [ k ∈ N [ B ∈B qk [ A ∈A ( B ) A. Indeed, for every point on the right-hand side, there is some strategy that Bob can use toforce the outcome to equal that point (and also not lose by default). So to complete theproof, it suffices to show that the right-hand side contains a ( B , R, r ) Cantor set, where r m,n = f ( R ) ( n − m +1)(1 − ε ) = R δ ( n − m +1)(1 − ε ) = R c ( n − m +1) for all m, n with m ≤ n .For each m = qk ≤ n and B ∈ B m let C ( n ) be the collection of elements of A ( B ) whoseradius is between R − n ρ and R − ( n +1) ρ , where ρ is the radius of B . Define A m,n ( B ) := (cid:8) A ∈ S ( B, R n − m +1 ) : ∃ C ∈ C ( n ) such that A ∩ C = (cid:31) (cid:9) . By (7) and the definition of C ( n ), we have X C ∈C ( n ) (cid:16) R − ( n +1) ρ (cid:17) c ≤ (cid:0) βR − m ρ (cid:1) c ;i.e. C ( n ) ≤ β c (cid:0) R n − m +1 (cid:1) c . This implies that for m > n we get C ( n ) <
1, i.e. C ( n ) = (cid:31) and in turn A m,n ( B ) = (cid:31) . Nowwe construct a Cantor set K ∈ E by removing the collections A m,n ( B ) ( m ≤ n , B ∈ B m ).Then to complete the proof, we just need to show that A m,n ( B ) ≤ r m,n = R c ( n − m +1) . It is necessary to specify the history in order to uniquely identify Alice’s response because unlike the othervariants of Schmidt’s game, the potential game is not a positional game (see § B ∈ B m . By the doubling property each element of C ( n ) intersects a boundednumber of elements of S (cid:0) B, R n − m +1 (cid:1) . Thus A m,n ( B ) ≤ C C ( n ) ≤ C β c (cid:0) R n − m +1 (cid:1) c for some constant C > R . By letting β be sufficiently small (i.e. letting q be sufficiently large), we can get C β c ≤ C β c (cid:0) R n − m +1 (cid:1) c ≤ R ( n − m +1) c , which completes the proof.Finally, we briefly mention how the second statement of Theorem 1.5 follows from the firststatement. If E is 1 Cantor winning then, by the first part of the theorem, it is c potentialwinning for any c >
0. The statement follows as an immediate corollary of Proposition 1.3.
We now move our attention to the links between Schmidt’s original game and generalisedCantor set constructions. Thanks to Theorem 3.4, in δ Ahlfors regular spaces we can treatthe notions of Cantor winning sets and potential winning sets as interchangeable. The latternotion will be more convenient in this section. We prove Theorem 1.8 as a special case of amore general result:
Theorem 3.8.
Let X be a δ Ahlfors regular space and let E ⊆ X be a c potential winningset. Then for all c > c , there exists γ > such that for all α, β > with α < γ ( αβ ) c/δ , theset E is ( α, β ) very strong winning. To see that Theorem 1.8 follows from Theorem 3.4, simply notice that R N is obviously N Ahlfors regular and that a very strong ( α, β ) winning set is necessarily Schmidt ( α, β ) winning.When E is ε Cantor winning, the desired statement follows upon taking c = N (1 − ε ) in theabove. The reader may wish to compare the following proof with the proof of Theorem 3.2. Proof.
Consider a strategy for Alice to win the ( c, e β ) potential game for the set E , where e β = ( αβ ) q for some large q ∈ N . We want to show that Alice has a winning strategy inthe ( α, β ) very strong game. Given any sequence of Bob’s choices in the ( α, β ) game, say B , . . . , B qk , we can consider the corresponding sequence of choices in the ( c, e β ) potentialgame which correspond to Bob choosing the balls B , B q , · · · , B qk . Alice’s response to thissequence of moves is to delete a collection A ( B qk ). Now Alice’s corresponding response inthe ( α, β ) very strong game will be: if Bob has made the sequence of moves B , . . . , B m , thenshe will choose her ball A m +1 ⊆ B m so as to minimize the value at A m +1 of the potentialfunction ϕ m ( A ) := X k ∈ N qk ≤ m X C ∈A ( B qk ) C ∩ A = ∅ diam c ( C ) . We will now prove by induction that the inequality ϕ m ( A m +1 ) ≤ ( ε · rad ( B m )) c holds for all m , where ε > e β ). Supposethat it holds for some m , and let B m +1 be Bob’s next move. Define r = α · rad( B m +1 ) > α β · rad( B m ). By Proposition 2.19, there exists a set D := { B ( x i , r ) ⊆ B m +1 : i ∈ { , . . . , I }} , I = D ≫ α − δ , and for any distinct values 1 i = j I one has d ( x i , x j ) > r , i. e. all balls in D are separated by distances of at least r .By the induction hypothesis, for all C ∈ S qk ≤ m A ( B qk ) such that C ∩ A m +1 = (cid:31) , we havediam( C ) ≤ ε · rad( B m ). So by letting ε be small enough (i.e. ε ≤ α β ), we can guaranteethat all balls in D make disjoint contributions to ϕ m ( A m +1 ). In other words, X D ∈D ϕ m ( D ) ϕ m ( A m +1 ) . By choosing A m +2 to be the ball from D which minimises the function ϕ m , we get α − δ ϕ m ( A m +2 ) ≪ ϕ m ( A m +1 ) . By choosing γ small enough, the inequality α < γ ( αβ ) c/δ implies that ϕ m ( A m +2 ) ≤ ( αβ ) c ϕ m ( A m +1 ) / ε · rad( B m +1 )) c . Note that for the base of induction, m = −
1, the last estimate is straightforward, since ϕ m ( A ) = 0 for all balls A , and we can set A = B .When we consider ϕ m +1 ( A m +2 ) there may be a new term coming from a new collection A ( B m +1 ). By the definition of a ( c, e β ) potential game, that term is at most ( e β · rad( B m +1 )) c ,and thus by choosing q large enough we can guarantee that ϕ m +1 ( A m +2 ) ( ε ( αβ ) m +1 ρ ) c . This finishes the induction.Since Alice followed the winning strategy for ( c, e β ) potential game, we have either ∞ \ k =1 B kq ⊆ E or ∞ \ k =1 B kq ⊆ ∞ [ k =1 [ A ∈A ( B kq ) A. To finish the proof of the theorem, we need to show that the second inclusion is impossible.If we assume the contrary, then there exists A ∈ A ( B kq ) for some k such that every ball B m and in turn every ball A m has nonempty intersection with A . But in that case we have φ m ( A m +1 ) ≥ diam c ( A ), which contradicts the fact that φ m ( A m +1 ) → m → ∞ .As a corollary of Theorem 3.8, we get a general form of Theorem 1.9: Corollary 3.9.
In an Ahlfors regular space, the intersection of a weak winning set and apotential winning set has full Hausdorff dimension.Proof.
Let E ⊆ X be a c potential winning set for some c < δ . Fix c < c < δ . ByTheorem 3.8, there exists γ > α, β > α < γ ( αβ ) c/δ , E is ( α, β )very strong winning. In particular, for all β > α > E is ( α, β )very strong winning. Hence we can apply Proposition 2.6 which says that X \ E is not weakwinning (and thus does not contain a weak winning set). So every potential winning setintersects every weak winning set nontrivially.Now by contradiction suppose that T is a weak winning set such that dim H ( E ∩ T ) < δ .This implies that for any c ∈ (dim H ( E ∩ T ) , δ ) and for any ε > E ∩ T by a countable collection ( A i ) i ∈ N of balls such that ∞ X i =1 diam c ( A i ) < ε.
22n other words, this cover can be chosen by Alice in a single move of the c potential game, so X \ ( E ∩ T ) is potential winning. So by the intersection property of potential winning sets, E \ T = E ∩ ( X \ ( E ∩ T )) is potential winning. But then X \ T is a potential winning setwhose complement is weak winning, contradicting what we have just shown. To prove Theorem 1.10 we employ the same notation, terminology, and observations as in [37].Denote by B ( R ) the set of all closed intervals in R .In Schmidt’s ( α, β ) game, the method Alice uses to determine where to play her in-tervals A i is called a strategy . Formally, a strategy F := { f , f , . . . } is a sequence offunctions f i : B ( R ) i → B ( R ) satisfying rad( f i ( B , B , . . . , B i − )) = α · rad( B i − ) and f i ( B , B , . . . , B i − ) ⊆ B i − . A set E is a winning set for the ( α, β ) game if and only if thereexists a strategy determining where Alice should place her intervals A i := f i ( B , B , . . . , B i − )so that, however Bob chooses his intervals B i ⊆ A i , the intersection point T i B i lies in E . Sucha strategy is referred to as a winning strategy (for E ) . Schmidt [37, Theorem 7] observed thatthe existence of a winning strategy guarantees the existence of a positional winning strategy;that is, a winning strategy for which the placement of a given interval by Alice needs only todepend upon the position of Bob’s immediately preceding interval, not on the entirety of thegame so far. To be precise, a winning strategy F := ( f , f , . . . ) is positional if there is a func-tion f : B ( X ) → B ( R ) such that each function f i ∈ F satisfies f i ( B , . . . , B i − ) = f ( B i − ).Employing a positional strategy will not give us any real advantage in the proof that follows;however, it will allow us to simplify our notation somewhat. Without confusion we will write A i = F ( B i − ) = f ( B i − ) where F is a given positional winning strategy for Alice and f isthe function witnessing F ’s positionality.We refer to a sequence { B , B , . . . } of intervals as an F -chain if it consists of the movesBob has made during the playing out of an ( α, β ) game with target set E in which Alice hasfollowed the winning strategy F . By definition we must have T i B i ⊆ E . Furthermore, wesay a finite sequence { B , . . . B n − } is an F n -chain if there exist B n , B n +1 , . . . for which theinfinite sequence { B , . . . B n − , B n , B n +1 . . . } is an F -chain. One can readily verify (see [37,Lemma 1]) that if { B , B , . . . } is a sequence of intervals such that for every n ∈ N the finitesequence { B , B , . . . B n − } is an F n -chain, then { B , B , . . . } is an F -chain. Outline of the strategy.
We begin with some interval b ⊆ R of length 2 · rad( b ), some ε ∈ (0 , R sufficiently large, and a 1 / E ⊆ R . Fix α = 1 / β = 2 /R ,so that αβ = 1 /R . We will construct for every R ≥ / (1 − ε ) a local ( b , R,
10) Cantor set K lying inside E . This is sufficient to prove that E is 1 Cantor winning.By assumption the set E is (1 / , /R ) winning; thus, there exists a positional winningstrategy F such that however we choose to place Bob’s intervals B i in the game, the placementof Alice’s subsequent intervals A i = F ( B i − ) guarantees that the unique element of T i B i will fall inside E . We will in a sense play as Bob in many simultaneous (1 / , /R ) games,each of which Alice will win by following her prescribed strategy F .Our procedure is as follows. The construction of the local Cantor set K comprises theconstruction of subcollections B n of intervals in R n B for each n ∈ N , where B := { b } .Construction will be carried out iteratively in such a way that for any interval b ∈ B n thereexists an F n -chain { B , B , . . . , B n − } satisfying b ⊆ B n − . Moreover, for each ancestor b ′ of b (that is, the unique interval b ′ ∈ B k with b ⊆ b ′ , for some given 0 ≤ k < n ) the subsequence { B , B , . . . , B k − } will coincide with the F k -chain constructed for b ′ satisfying b ′ ⊆ B k − from the earlier inductive steps. We write B n ( b ) if the dependence of the interval B n on aninterval b is not clear from context. In view of the previous discussion, upon completion of23he iterative procedure it follows that for any sequence of intervals { b i } i ∈ N with b i ∈ B i theassociated sequence { B , B , . . . } must be an F -chain satisfying b i ⊆ B i − . In this way, forevery point x in the Cantor set K we will establish the existence of an F -chain { B , B , . . . } for which { x } = T i B i ; namely, we may simply choose the F -chain associated with a sequenceof intervals { b i } i ∈ N with b i ∈ B i satisfying T i ∈ N b i = { x } . It follows that x must fall within E and, since x ∈ K was arbitrary, that K ⊆ E as required.We begin the first step of the inductive process. Given B = { b } , the idea is to constructthe collection B ⊆ R B in such a way that for every b ∈ B there exists a place Bob mayplay his first interval B in a (1 / , /R ) game so that Alice’s first interval A := F ( B ) (asspecified by the winning strategy F ) contains b ; that is, we construct B so that for every b ∈ B there exists an F -chain { B } satisfying b ⊆ A := F ( B ). For technical reasons, inpractice we will ensure the stronger condition that b ⊆ (1 − /R ) A , where κB denotes theinterval with the same center as B but with radius multiplied by κ .Recall that for his opening move in a (1 / , /R ) game Bob may choose any interval in R .In particular, he may choose any interval of radius 2 · rad( b ) with centre lying in b , thereforecovering b . Let W ( b ) ⊆ B ( R ) be the set of all “winning” intervals for Alice (as prescribedby the winning strategy F ) corresponding to any opening interval of this kind that Bob mightchoose to begin the game with; that is, let W ( b ) : = { F ( B ) : B ∈ B ( R ) , rad( B ) = 2 · rad( b ) , cent( B ) ∈ b } . Note that this set could either be uncountable or, since for each A ∈ W ( b ) we have rad( A ) = α · (2 · rad( b )) = rad( b ), it could simply coincide with the singleton B . Moreover, since α = 1 / B that cent( B ) is contained in F ( B ), and so W ( b )is a cover of b . This property is unique to the case α = 1 / Lemma 3.10.
Let b ∈ B ( R ) be any closed interval and let W ⊆ B ( R ) be a cover of b byclosed intervals of radius rad( b ) . Then, for any strictly positive ε < rad( b ) there exists asubset W ∗ ( ε ) ⊆ W of cardinality at most two that covers b except for an open interval oflength at most ε .Proof of Lemma 3.10. Fix n such that n − rad( b ) ≤ ε , and let cent( b ) − rad( b ) = x , . . . , x n =cent( b ) + rad( b ) be an evenly spaced sequence of points starting and ending at the twoendpoints of b . For each i = 0 , . . . , n let W i be an interval in W containing x i . Now let j bethe smallest integer such that x n ∈ W j . If j = 0, then W = b and we are done. Otherwise,since x n / ∈ W j − , we have x ∈ W j − and thus [ x , x j − ] ⊆ W j − , whereas [ x j , x n ] ⊆ W j .Thus the intervals W j − and W j cover b except for the interval ( x j − , x j ), which is of lengthat most ε .We continue the proof of Theorem 1.10. Take b = b , ε = rad( b ) /R , and W = W in Lemma 3.10. Then there exists a subcollection W ∗ := { A L , A R } of W of cardinalityat most two covering b except for an open interval I ( b ) of length at most rad( b ) /R . Itis possible that I ( b ) is empty. If W ∗ contains only one interval, say W ∗ = { b ′ } , then set A L = A R = b ′ . Let B L and B R be two choices for Bob’s opening move with centres in b that satisfy A L = F ( B L ) and A R = F ( B R ); such choices exist by the definition of W . If wehad A L = A R then assume B L = B R .We now discard the “bad” intervals that will not appear in the generalised Cantor set K .Let Bad := (cid:26) b ′ ∈ R B : b ′ ∩ I ( b ) = (cid:31) or b ′ (cid:0) (1 − /R ) A L ∪ (1 − /R ) A R (cid:1)(cid:27) . b ) /R may intersect at most two intervals from R B , and thereare at most five intervals (of length at most rad( b ) /R ) that need to be intersected by b ′ for itto be in Bad , it is immediate that Bad ≤
10. It follows that B := R B \ Bad satisfiesthe required conditions for our generalised Cantor set. For each b ∈ B we associate one ofthe F -chains (cid:8) B L (cid:9) or (cid:8) B R (cid:9) , depending on whether b is a subset of A L or A R respectively.If b lies in both A L and A R (which is possible if I ( b ) was empty) then we will assign as aprecedent whichever of the intervals B L or B R has the ‘leftmost’ centre. This completes thefirst step of the inductive construction process.Assume now that we have constructed the collection B i so that for each b ∈ B i we have an F i -chain { B , . . . , B i − } with rad( B i − ) = 2 R · rad( b ) and b ⊆ (1 − /R ) F ( B i − ). We outlinethe procedure to construct the subsequent collection B i +1 and the corresponding F i +1 -chains:Fix b ∈ B i with associated F i -chain { B , . . . , B i − } . Assume Alice has just played her i th move A i := F ( B i − ), an interval of radius R · rad( b ), in a (1 / , /R ) game correspondingto this F i -chain. Bob may then choose any interval of radius 2 · rad( b ) inside A i . Since b ⊆ (1 − /R ) A i , Bob is free to choose as B i any interval of radius 2 · rad( b ) centred in b ;every such choice b ′ will constitute a legal move since rad( b ′ ) = R rad( A i ) and b ⊆ (1 − /R ) A i together imply b ′ ⊆ A i . In particular, the collection W i +1 ( b ) : = (cid:8) F ( b ′ ) : b ′ ∈ B ( R ) , rad( b ′ ) = 2 · rad( b ) /R i , cent( b ′ ) ∈ b (cid:9) is a cover of b and consists of candidate “winning” moves that could be played by Aliceafter any legal move B i (with centre in b ) played by Bob. As before, we apply Lemma 3.10to find a subset of W i +1 ( b ) of cardinality at most two that covers b except for an opensubinterval I ( b ) of length at most rad( b ) /R . If the subcover contains two intervals denotethem by A Li +1 ( b ) and A Ri +1 ( b ); otherwise, if the subcover consists of a single ball b ′ then let A Li +1 ( b ) = A Ri +1 ( b ) = b ′ . Denote by B Li ( b ) and B Ri ( b ) some choice of intervals centred in b forwhich A Li +1 ( b ) = F ( B Li ( b )) and A Ri +1 ( b ) = F ( B Ri ( b )) respectively. Define Bad i +1 ( b ) := (cid:26) b ′ ∈ R { b } : b ′ ∩ I ( b ) = (cid:31) or b ′ (cid:0) (1 − /R ) A Li +1 ( b ) ∪ (1 − /R ) A Ri +1 ( b ) (cid:1)(cid:27) and let B i +1 ( b ) := R { b } \ Bad i +1 ( b ). By the same arguments as before it is clear that Bad i +1 ( b ) ≤
10 for each b ∈ B i .Finally, for each b ′ ∈ B i +1 ( b ) we assign the F i +1 -chain (cid:8) B , . . . , B i − , B Li ( b ) (cid:9) or (cid:8) B , . . . , B i − , B Ri ( b i ) (cid:9) depending on whether b ′ lies in A Li +1 ( b i ) or A Ri +1 ( b i ) respectively.Again, if I ( b ) was empty and b was contained in both A Li +1 ( b ) and A Ri +1 ( b ) then we choosewhichever of B Li ( b ) and B Ri ( b ) has the leftmost centre as a convention.Upon setting B i +1 := [ b ∈B i B i +1 ( b )we see that the conditions of our generalised Cantor set K ( b , R,
10) are satisfied. Thiscompletes the inductive procedure.
In this section we prove the following statement which implies Theorem 1.12:
Theorem 3.11.
Let B ⊆ R be any ball, and let ε > be a real number. If E ⊆ R is ε Cantor winning in B then it is ( B , M ) Cantor rich with M = 4 ε . onversely, let M > be a real number. If E is ( B , M ) Cantor rich then it is ε Cantorwinning in B , for ε which is the solution of the equation ⌊ M ⌋ + 1 = 4 ε . (23) Proof.
Assume E is ε Cantor winning in B . Then, by Corollary 3.3, for every R ≥ (cid:0) B , R, R − ε (cid:1) Cantor set contained in E . Fix an integer R > M and a real number y >
0. Since R ε <
1, we can choose ℓ > R · (cid:0) R ε (cid:1) ℓ < y . Consider a (cid:0) B , R ℓ , R ℓ (1 − ε ) (cid:1) Cantor set contained in E and denote its Cantor sequence by {B n } ∞ n =0 . Define another Cantorsequence {B ′ n } ∞ n =0 by setting B ′ ℓk = B k for all k ≥ B ′ n = R B n − whenever n = ℓk . Thenobviously the limit set ∞ \ n =1 [ B ∈B ′ n B is a (cid:16) B , R, ( r m,n ) ≤ m ≤ n (cid:17) generalised Cantor set, with r ℓ ( k − ,ℓk = R ℓ (1 − ε ) for all k ≥
0, and r m,n = 0 for all other pairs. Then, for any k > n = ℓk , we get: ℓk X m =0 (cid:18) R (cid:19) ℓk − m +1 r m,ℓk = (cid:18) R (cid:19) ℓ +1 R ℓ (1 − ε ) = 4 R · (cid:18) R ε (cid:19) ℓ < y. For every n = ℓk , we have n X m =0 (cid:18) R (cid:19) n − m +1 r m,n = 0 < y. For the other direction, suppose that E is ( B , M ) Cantor rich, and let c = 1 − ε . ByTheorem 3.4, it is enough to show that E is c potential winning in B . We argue as in theproof of Theorem 3.4. Set R = ⌊ M ⌋ + 1 = 4 /ε and fix any 0 < β ≤ R , c > c , and ρ > y > E contains some ( B , R, r ) Cantor set K , where r satisfies n X m =0 (cid:18) R (cid:19) n − m +1 r m,n < y for every n ∈ N . In particular, for every m, n ∈ N we have r m,n < y (cid:18) R (cid:19) n − m +1 = yR ( n − m +1) c . We now describe a strategy for Alice to win the ( c, β ) potential game on A ∞ ( B ), assumingthat Bob’s starting move has radius ρ . As in the proof of Theorem 3.4, let ρ denote theradius of B , let D k denote Bob’s k th move, and for each k ∈ N let m = m k ∈ N denote aninteger such that β · rad( D k ) < R − m ρ ≤ rad( D k ). The inequality β /R ensures that such m exists. Then Alice’s strategy is as follows: on turn k remove all elements of the set [ B ∈B m [ n ≥ m A m,n ( B )that intersect Bob’s current choice. Obviously, this strategy, if executable, will make Alicewin since the intersection of Bob’s balls will satisfy T k ∈ N D k ⊆ K ⊆ E . To show that it islegal, we need to show that X B ∈B m B ∩ D k = ∅ X n ≥ m r m,n (cid:16) R − ( n +1) ρ (cid:17) c ≤ ( β · rad( D k )) c . (24)26his is because elements of A m,n ( B ) all have radius R − ( n +1) ρ . Since the elements of B m aredisjoint and have radius R − m ρ ≍ β rad( D k ), the number of them that intersect D k is boundedby a constant depending only on β . Call this constant C . Then the left-hand side of (24) isless than C X n ≥ m yR ( n − m +1) c (cid:16) R − ( n +1) ρ (cid:17) c = C (cid:0) R − m ρ (cid:1) c y ∞ X ℓ =1 R ( c − c ) ℓ C (rad( D k )) c R ( c − c ) − R ( c − c ) y. By choosing y so that y < β c (cid:0) − R ( c − c ) (cid:1) C R ( c − c ) , we guarantee that the move is legal. The concept of winning sets gives a notion of largeness which is orthogonal to both categoryand measure. It was first noticed in [11] that absolute winning subsets of R N are also largein a dual sense: they intersect every nonempty diffuse set. The Cantor and the potentialgames, together with the definition of the Ahlfors regularity dimension, allow us to quantifythis observation. In fact, based on the Borel determinacy property for Schmidt games [24]we prove a converse result for Borel sets: Theorem 4.1. If E ⊆ X is c potential winning then E ∩ K = (cid:31) for every closed set K ⊆ X with dim R K > c . If E is Borel, then the converse is also true. Combining this with Propositions 2.18 and 1.3 yields the following corollary:
Corollary 4.2. If E ⊆ X is Borel and E ∩ K = (cid:31) for any diffuse set K , then E is absolutewinning.Remark . One can easily check that Theorem 1.13 follows from Theorem 4.1 together withTheorem 3.4. Indeed, by Theorem 3.4 the set E ∈ R N if ε Cantor winning if and only if it is c potential winning for c = N (1 − ε ). Then, Theorem 1.13 is a strightforward corollary ofTheorem 4.1. Proof.
Assume that E ⊆ X is c potential winning, and let K ⊆ X be a c Ahlfors regular setwith c > c . Then E ∩ K is also winning for the c potential game restricted to K . Therefore,by Theorem 3.8, E ∩ K contains an ( α, β ) winning set for certain pairs ( α, β ) and hence it isnonempty.For the converse, assume E is Borel, and that E is not c potential winning. By the Boreldeterminacy theorem [24, Theorem 3.1], Bob has a winning strategy to make sure the c potential game ends in X \ E . Fix this strategy and let β > c > c be Bob’s choicesaccording to this strategy. We will show that X \ E contains Ahlfors regular sets of dimensionarbitrarily close to c . In fact, Theorem 3.1 in [24] does not deal with potential games. However its proof can be easily modifiedto cover them. In the notation of that paper, it is clear from the definition of who wins the potential gamethat there exists a Borel set A ⊆ Z × X such that if ω ∈ E Γ is a play of the game, then ω is a win for Alice ifand only if either ω / ∈ Z or ( ω, ι ( ω )) ∈ B . Combining with [24, Lemma 3.3] shows that the set of plays of thegame that result in a win for Alice is Borel. γ >
0. Denote by A i +1 the collection of balls chosen by Alice inresponse to Bob’s i th move B i . We will consider strategies for Alice where she only choosesnonempty collections on some turns, namely A i +1 = (cid:31) only ifrad( B i ) ≤ γ n rad( B ) < rad( B i − ) for some n ∈ N . Such turns i = i ( n ) will be called good turns . If B , . . . , B i ( n ) vs. A , . . . , A i ( n )+1 is a historyof the game, then we let f ( A i ( n )+1 ) denote the next ball that Bob plays on a good turn if thegame continues with Bob playing his winning strategy and Alice playing dummy moves (i.e.choosing the empty collection), i.e. f ( A i ( n )+1 ) := B i ( n +1) . Next, we let N = (cid:22)(cid:18) β γ (cid:19) c (cid:23) . If B , . . . , B i ( n ) vs. A , . . . , A i ( n ) is a history of the game, then we define a sequence of balls (cid:0) g k ( B i ( n ) ) (cid:1) Nk =1 by recursion: we let g ( B i ( n ) ) := f ( (cid:31) ), and if g , . . . , g k − are defined for some k ≥
1, then we let g k ( B i ( n ) ) = f (cid:0)(cid:8) N ( g ( B i ( n ) ) , γ n +1 diam( B )) , . . . , N ( g k − ( B i ( n ) ) , γ n +1 diam( B )) (cid:9)(cid:1) . Here N ( A, d ) denotes the d -neighborhood of a set A , i.e. N ( A, d ) := { x ∈ X : d ( A, x ) d } .We will show later (using the definition of N ) that it is legal for Alice to play the collectionappearing in the right-hand side, hence the right-hand side is well-defined. Note that since B i ( n ) is the n th good turn in the history B , . . . , B i ( n ) vs. A , . . . , A i ( n ) , for each k = 1 , . . . , N , g k ( B i ( n ) ) is the ( n +1)st good turn in its corresponding history. We call these balls the children of B i ( n ) . The children of B i ( n ) are disjoint, because if g j ( B i ( n ) ) ∩ g k ( B i ( n ) ) = (cid:31) for some j, k with j < k , then g k ( B i ( n ) ) ⊆ N ( g j ( B i ( n ) ) , γ n +1 diam( B )), and thus Alice wins by default inthe game in which Bob played his winning strategy ending in move g k ( B i ( n ) ), a contradiction.Now construct a Cantor set K as follows: let B = { B } , and if B n is defined, then let B n +1 be the collection of children of elements of B n according to the above construction.Note that B n consists of moves for Bob that are the n th good turns with respect to theircorresponding histories, and in particular βγ n rad( B ) < rad( B ) ≤ γ n rad( B ) for all B ∈ B n .Finally, let K = T n ∈ N S B ∈B n B . Any element of K is the outcome of some game in whichBob played his winning strategy, so K ⊆ X \ E . Moreover, from the construction of K it isclear that K is Ahlfors regular of dimension log( N ) − log( γ ) . From the definition of N , we see thatthis dimension tends to c as γ → (cid:8) N ( g ( B i ( n ) ) , γ n +1 diam( B )) , . . . , N ( g k − ( B i ( n ) ) , γ n +1 diam( B )) (cid:9) for any k ≤ N . Indeed, the cost of this collection is k − X j =1 (cid:0) rad( g j ( B i ( n ) )) + γ n +1 diam( B j ) (cid:1) c ≤ N (3 γ n +1 rad( B )) c ≤ ( β γ n rad( B )) c ≤ ( β · rad( B i ( n ) )) c . Remark . A similar characterisation is possible in the generality of H potential games.In fact, potential winning sets not only have nonempty intersection with Ahlfors regularsets, but the intersection has full Hausdorff dimension. To prove this, let us first prove thefollowing: 28 roposition 4.5. If dim H ( E ) ≤ c , then X \ E is c potential winning.Proof. Fix β > c > c , and r >
0. Since the c dimensional Hausdorff measure of E iszero, there exists a cover C of E such that X C ∈C diam c ( C ) ≤ ( βr ) c . Thus if Bob’s first ball has radius r , then Alice can legally remove the collection C , thusdeleting the entire set E on her first turn. Theorem 4.6. If c < δ = dim R X , then every c potential winning set in X has Hausdorffdimension δ .Proof. By contradiction, suppose that E ⊆ X is c potential winning but c := dim H ( E ) < δ .Then by Proposition 4.5, X \ E is c potential winning, so by the intersection property (W2)(see Proposition 2.4), the empty set (cid:31) is c potential winning, where c = max( c , c ) < δ .But then by Theorem 3.8, the empty set (cid:31) is ( α, β ) winning for some α, β >
0, which is acontradiction.
Remark . In view of Theorem 3.4, if a complete doubling metric space admits a splittingstructure ( X, S , U, f ) such that B = A ∞ ( B ) for every ball B , then the notions of Cantorwinning and potential winning sets are equivalent. However, not every metric space X admitssuch a splitting structure. Indeed, spaces with splitting structures satisfying (S4) must havedimensions of the form log( n )log( m ) ( m, n ∈ N ) (cf. [5, Theorem 3]), whereas one can construct acomplete doubling metric space of arbitrary Hausdorff dimension. In that sense, the notionof potential winning sets is strictly more general than the notion of Cantor winning sets. Remark . In the above proof, we needed to be careful due to the fact that c potentialwinning sets are not automatically nonempty (as the winning sets for Schmidt’s game are);indeed, by Proposition 4.5, the empty set is dim H ( X ) potential winning.Lastly, the same observation that is used to prove Proposition 4.5 may be used to proveTheorem 1.14: Proof of Theorem 1.14.
Recall that every closed ball in X of radius ≤ B ( x, r ) = { ( y k ) ∞ k =1 : y k = x k for any 1 ≤ k ≤ n } whenever x ∈ X and 2 − n ≤ r < − n +1 . Note that X is 1 Ahlfors regular so it is enough toprove that the exceptional set defined in (1), i.e. E = n x ∈ X : { T i x : i ≥ } ∩ K = (cid:31) o , is dim H K potential winning. We now define a winning strategy for Alice. Assume Bobchoose 0 < β < c > dim H K . Without loss of generality assume that B = X and r = 1. Then there exists a unique positive integer ℓ such that2 − ℓ ≤ β < − ℓ +1 . Since dim H K < c , we may choose a collection of open balls C such that K ⊆ S C ∈C C and X C ∈C rad( C ) c < ℓ − (2 − ℓ β ) c . (25)29et { B i } ∞ i =0 denote any sequence of balls that are chosen by Bob which satisfy 2 − ( i +2) ℓ ≤ rad( B i ) < − ( i +1) ℓ . If this sequence is not infinite then Alice wins by default. By abuse ofnotation, we will think of B i as Bob’s i th choice, and it will be sufficient to define Alice’sreaction to these moves. On her ( i + 1)st move, Alice will remove the collection A i +1 = ℓ − [ j =0 n C : C ∈ T − ( iℓ + j ) C and C ∩ B i = (cid:31) o . (26)Here, T − j C denotes the set of all images of elements of C under the inverse branches of T j .Note that if Alice still doesn’t win by default, i.e. if (8) is not satisfied, then necessarily ∞ \ i =0 B i ⊆ ( x ∈ X : { T i x : i ≥ } ⊆ X \ [ C ∈C C ) ⊆ E. Therefore, we are done once we show that the collection (26) is a legal move for Alice, i.e.,that it satisfies (7). Firstly, for any C ∈ C and j = 0 , . . . , ℓ −
1, we have that T − ( iℓ + j ) ( { C } )is a collection of balls of radius 2 − ( iℓ + j ) rad( C ), pairwise separated by distances of at least2 − ( iℓ + j ) ≥ − iℓ − ℓ . Since rad( B i ) < − iℓ − ℓ , at most one of them intersects B i . Therefore, X C ∈A i +1 rad ( C ) c ≤ ℓ − X j =0 X C ∈C (cid:16) − ( iℓ + j ) · rad( C ) (cid:17) c ≤ ℓ cℓ X C ∈C rad( C ) c · rad( B i ) c , which is smaller than ( β · rad( B i )) c by (25). The original papers of Schmidt [37, Theorem 5] and McMullen [33] provide examples of setsthat are winning (respectively, strong winning) but not absolute winning. We provide twoadditional counterexamples.
We give an explicit example of a set which is potential winning in R (or equivalently, byTheorem 1.5, is Cantor winning), but is not winning. In fact, we will prove more by showingthat the set in question is not weak winning (see § Proof of Theorem 1.6.
By an iterated function system on R , or IFS , we mean a finite collec-tion { f i : R → R } , i = 1 , , . . . , N, of contracting similarities. By the limit set of the IFSwe mean the unique compact set S which is equal to the union of its images f i ( S ) underthe elements of the IFS. We call an IFS on R rational if its elements all preserve the set ofrationals. Note that there are only countably many rational IFSes.Let E ⊆ R be the complement of the union of the limit sets of all rational IFSes whoselimit sets have Hausdorff dimension ≤ /
2. We will show that E is not winning for Schmidt’sgame. On the other hand, since dim( R ∩ E ) = 1 / <
1, it follows from Proposition 4.5 that E is 1 / α >
0, and let β > α, β ) game. Let I be a rational interval contained in Alice’sfirst move, whose length is at least half of the length of Alice’s first move. Without loss of30enerality, suppose that I = [ − , λ < λ + λα ≥
1, and let γ > λ / + 2 γ / ≤ . (27)Consider the IFS on the interval I consisting of the following three contractions: u ( x ) = λx, u ( x ) = γ ( x −
1) + 1 , u − ( x ) = γ ( x + 1) − , and note that condition (27) guarantees that the dimension of the limit set of this IFS is atmost 1 / k , if Alice just made the move A k , then find the largest interval J k ⊆ A k which can be written as the image of I under a composition of elements of the IFS, and selectthe unique ball B k of radius β · rad( A k ) whose centre is the midpoint of this interval. Wewill show by induction that B k ⊆ J k , thus proving that this choice is legal. Indeed, supposethat this holds for k . By “blowing up the picture” using inverse images of the elements ofthe IFS, we may without loss of generality suppose that J k = I . Then we have B k = [ − r, r ],where r = rad( B k ) ≤
1. By “blowing up the picture” further using an iterate of the inverseimage of the contraction u , we may assume without loss of generality that r > λ . ThenAlice’s next choice A k +1 is an interval of length at least 2 λα contained in [ − , {± λ q : q = 1 , , . . . , N } intersects every interval of length 1 − λ contained in [ − , N isa large constant. Since λα ≥ − λ by assumption, there exist q ∈ { , . . . , N } and ε ∈ {− , } such that ελ q ∈ A k +1 . But then J = u q u ε ( I ) is an interval coming from a cylinder in the IFSconstruction which intersects A k +1 , and whose length is 2 γλ q ∈ [2 λ N γ, λγ ]. By choosing γ ≤ α , we can guarantee that the length of J satisfies 2rad( J ) ≤ λα ≤ rad( A k +1 ).Combining with the fact that J ∩ A k +1 = (cid:31) shows that J ⊆ A k +1 . So then by the definitionof J k +1 , we have 2rad( J k +1 ) ≥ J ) ≥ λ N γ , and thus by letting β = λ N γ we get B k +1 ⊆ J k +1 , completing the induction step. Finally, we will show that there exists a set in R which is winning, but is not potential winning(or equivalently, by Theorem 1.5, Cantor winning). Remark . The above statement remains true if “winning” is replaced by “weak winning”,but the resulting statement is weaker and so we prove the original (stronger) statementinstead. We will however need the weak ( α, β ) game in the proof. It appears that this is anontrivial application of the weak game.By Corollary 3.9, it suffices to show that there exists a winning set E whose complementis weak winning. Note that this also shows that the weak game does not have the intersectionproperty (W2), since E and its complement are both weak winning, but their intersection isempty, and the empty set is not weak winning. We will use the result below as a substitutefor the intersection property. Definition 5.2.
A set E is finitely weak winning if Alice can play the weak game so as toguarantee that after finitely many moves her ball A n is a subset of E . Lemma 5.3.
The intersection of countably many finitely weak α winning sets is weak α winning.Proof. Let ( E n ) n ∈ N be a countable collection of finitely weak α winning sets. Alice can usethe following strategy to ensure that T ∞ m =0 B m lies inside T ∞ n =0 E n . She starts by following31he strategy to ensure that A m ⊆ E . After that she starts playing the weak winning gamefor the set E , assuming that Bob’s initial move is B m . Therefore she can ensure that A m + m ⊆ E ∩ E . Then she switches her strategy to E and so on.Finally, we have that for any N ∞ \ m =1 A m ⊆ A P Nk =1 m k ⊆ N \ n =1 E n . By letting N tend to infinity we prove the lemma.In what follows, the intersection of countably many finitely weak α -winning sets will becalled σ finitely weak α -winning. Proof of Theorem 1.7.
Throughout this proof we let Z = { , } , and we let π denote thecoding map for the binary expansion, so that π : Z N → [0 , Definition 5.4. A Bohr set is a set of the form A ( γ, δ ) := { n ∈ N : nγ ∈ ( − δ, δ ) mod 1 } , where γ, δ > ω = ( ω n ) n ∈ N ∈ Z N , we let E ( ω ) = { n ∈ N : ω n = 1 } .We now define the set E ⊆ R : E : = Z + { π ( ω ) : ω ∈ Z N , E ( ω ) contains a Bohr set } E : = [ r ∈ Q rE . We claim that E is 1 / / / < α < / β, ρ >
0. We show that Alice has a strategy to win the ( α, β )Schmidt game assuming that Bob has chosen a ball of radius ρ with his first move. Choose0 < r ∈ Q so that 3 < ρ /r < α − , and consider the set A := { n ∈ Z : ∃ m = m n ∈ N such that ( αβ ) m ρ /r ∈ (3 · − ( n +1) , α − · − ( n +1) ) } . The bounds on α guarantee that the values m n for all n ∈ A are distinct. By taking logarithmsand rearranging the expression for the set A , we get n ∈ A ⇔ ∃ m log 3 < ( n + 1) log 2 + m log( αβ ) + log( ρ /r ) < log( α − ) ⇔ n log 2 − log( αβ ) ∈ log (cid:16) ρ /r (cid:17) − log( αβ ) , log (cid:16) α − ρ /r (cid:17) − log( αβ ) mod 1 . Since the interval on the right-hand side contains 0, this implies that A contains a Bohr set.Now fix n ∈ A , and let m = m n ∈ N be the corresponding value. On the m th turn,Bob’s interval B m is of length 2( αβ ) m ρ ≥ · − n r , so if we subdivide R into intervals oflength 2 − n r , then B m must contain at least two of them. Since Alice’s next interval is oflength 2 α ( αβ ) m ρ ≤ − n r , this means that Alice can choose a subinterval of either one ofthese intervals on her turn. So Alice can control the n th binary digit of x/r , where x is theoutcome of the game, and in particular she can set the n th digit to 1. Since every value m corresponds to only one value n ∈ A , Alice can set all of the digits of x/r whose indices are32n A to 1. This means that x/r ∈ E and thus x ∈ E , so Alice can force the outcome of thegame to lie in E . So by Proposition 1.2, E is 1 / R \ E is σ finitely weak 1 / R \ rE is σ finitely weak 1 / r >
0, and since the intersection of countably many σ finitely weak 1 / σ finitely weak 1 / R \ E is σ finitelyweak 1 / r = 1. To prove that R \ E is indeed σ finitely weak 1 / Lemma 5.5.
Fix < α < / and β > . Then there exists N ∈ N such that for any ρ > there exists a ( ρ ) so that for any sequence of integers ( a k ) k M with a k +1 − a k ≥ N ∀ k, α > α ( ρ ) , (28) Alice has a strategy in the finite weak ( α, β ) game to control the digits indexed by the integers ( a k ) k M .Proof of Lemma 5.5. Suppose that Bob has just played a move of radius ρ . Then Alice canplay any radius between αρ and ρ , forcing Bob to play a predetermined radius between αβρ and βρ . After iterating n times, Alice can force Bob to play a ball of any given radius between α n β n ρ and β n ρ . Since α n < β for sufficiently large n , there exists γ > ≤ γρ (after a sufficient number of turnsdepending on the radius). Now Alice uses the following strategy: for each k = 1 , . . . , M , firstforce Bob to play a ball whose radius is 3 · − a k , and then respond to Bob’s choice in a waysuch that the a k th digit of the outcome is guaranteed (this is possible by the argument fromthe previous proof). Forcing Bob to play these radii is possible as long as 3 · − a ≤ γρ and3 · − a k +1 ≤ γα · − a k for all k . By (28), the latter statement is true as long as N and a ( ρ )are sufficiently large.To continue the proof of Theorem 1.7, fix 0 < α < / β >
0, and let N ∈ N be asin Lemma 5.5. Then: • For all k ∈ N , the set A k := Z + { π ( ω ) : T ( ω ) contains an arithmetic progression of length k and gap size N } is finitely weak winning, where T ( ω ) := N \ E ( ω ) = { n ∈ N : ω n = 0 } . • For all q, i ∈ N , the set A q,i := Z + { π ( ω ) : T ( ω ) ∩ ( q N + i ) = (cid:31) } is finitely weak winning.So the intersection A = \ k A k ∩ \ q,i A q,i is σ finitely weak winning. To finish the proof, we need to show that R \ E ⊇ A . Indeed,fix n + π ( ω ) ∈ A . Then T ( ω ) contains arbitrarily long arithmetic progressions of gap size N ,and intersects every infinite arithmetic progression nontrivially.Now let A ( γ, δ ) be a Bohr set, and we will show that A ( γ, δ ) intersects T ( ω ) nontrivially.First suppose that γ is irrational. Then so is N γ , so by the minimality of irrational rotations,there exists M such that the finite sequence 0 , N γ, . . . , M N γ is δ -dense in R / Z . Now let n, n + N, . . . , n + M N be an arithmetic progression of length M and gap size N contained33n T ( ω ). Then nγ, nγ + N γ, . . . , nγ + M N γ is also δ -dense in R / Z , and in particular mustcontain an element of ( − δ, δ ). So ( n + iN ) γ ∈ ( − δ, δ ) mod 1 for some i = 0 , . . . , M . But then n + iN ∈ A ( γ, δ ) ∩ T ( ω ).On the other hand, suppose that γ is rational, say γ = p/q . Then A ( γ, δ ) contains theinfinite arithmetic progression q N , which by assumption intersects T ( ω ) nontrivially.Thus every Bohr set intersects T ( ω ) nontrivially, so E ( ω ) does not contain any Bohr set,i.e. n + π ( ω ) / ∈ E . So A ⊆ R \ E , completing the proof. References [1] Jinpeng An,
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