2.1. Schmidt’s game

Given two real parameters $\alpha ,\beta \in (0,1)$ , Schmidt’s $(\alpha , \beta )$ game begins with Bob choosing an arbitrary closed ball $B_0\subseteq X$ . Alice and Bob then take turns to choose a sequence of nested closed balls (Alice chooses balls $A_i$ and Bob chooses balls $B_i$ ) satisfying both

$$ \begin{align*} B_0\supseteq A_1\supseteq B_1 \supseteq A_2\supseteq\cdots\supseteq B_i \supseteq A_{i+1} \supseteq B_{i+1}\supseteq \cdots \end{align*} $$

and

(2) $$ \begin{align} \mathrm{rad}(A_{i+1}) = \alpha \cdot \mathrm{rad}(B_i) \quad\textrm{and}\quad \mathrm{rad}(B_{i+1}) = \beta \cdot \mathrm{rad}(A_{i+1}) \quad\textrm{for all } i\geq 0. \end{align} $$

Since the radii of the balls tend to zero, at the end of the game, the intersection of Bob’s (or Alice’s) balls must consist of a single point. See Figure 2 for a pictorial representation of the game, in which the shaded area represents where Alice has specified Bob must play his next ball. A set $E\subseteq X$ is called $(\alpha ,\beta )$ winning if Alice has a strategy for placing her balls that ensures

(3) $$ \begin{align} \bigcap_{i =0}^\infty \, B_i \subseteq E. \end{align} $$

In this case, we say that Alice wins. The $\alpha $ game is defined by allowing Bob to choose the parameter $\beta $ and then continuing as in the $(\alpha ,\beta )$ game. We say that $E\subseteq X$ is $\alpha $ winning if Alice has a winning strategy for the $\alpha $ game. Note that $E\subseteq X$ is $\alpha $ winning if and only if E is $(\alpha ,\beta )$ winning for all $0<\beta <1$ . Finally, Schmidt’s game is defined by allowing Alice to choose $0<\alpha <1$ and then continuing as in the $\alpha $ game. We say $E\subseteq X$ is winning if Alice has a winning strategy for Schmidt’s game. Note that E is winning if and only if it is $\alpha $ winning for some $\alpha \in (0,1)$ .

Figure 2 Schmidt’s game.

It is easily observed that if a set E is $\alpha $ winning for some $\alpha \in (0,1)$ , then E is $\alpha '$ winning for every $\alpha ' \in (0, \alpha )$ , and that E is $\alpha $ winning for some $\alpha \in (1/2,1)$ if and only if $E=X$ (cf. [Reference Schmidt37], Lemmas 5 and 8, respectively); that is, the property of being $1/2$ winning is the strongest possible for a non-trivial subset.

We end the introduction of Schmidt games with the proof of the folklore Proposition 1.2.

Proof of Proposition 1.2

Fix $0 < \beta _0 < 1$ , let $\alpha _0$ be as in the statement. The case $\alpha _0=1$ is clear. For $\alpha _0<1$ , set

(4) $$ \begin{align} \alpha = \alpha_0\frac{1 - (\alpha_0 \beta_0)^2 - \beta_0 (1-\alpha_0)}{1 - (\alpha_0 \beta_0)^2 - \alpha_0 \beta_0 (1-\alpha_0)} < \alpha_0,\quad \alpha\beta = (\alpha_0\beta_0)^2. \end{align} $$

We describe the winning strategy for Alice for the $(\alpha _0,\beta _0)$ game (Game I) based on her strategy for the $(\alpha ,\beta )$ game (Game II). It can be given as follows.

1. (1) Whenever Bob plays a move $B_{2n} = B(b_{2n},\rho _{2n})$ in Game I, he correspondingly plays the move $B_n' = B(b_n',\rho _n')$ in Game II, where

   $$ \begin{align*} (1-\alpha)\rho_n' = (1-\alpha_0) \rho_{2n},\quad b_n' = b_{2n}. \end{align*} $$

2. (2) When Alice responds to this by playing a move $A_{n+1}' = B(a_{n+1}',\alpha \rho _n')$ in Game II (according to her winning strategy), she correspondingly plays the move $A_{2n+1} = B(a_{2n+1},\alpha _0 \rho _{2n})$ in Game I, where

   $$ \begin{align*} a_{2n+1} = a_{n+1}'. \end{align*} $$

3. (3) When Bob responds to this by playing a move $B_{2n+1} = B(b_{2n+1},\rho _{2n+1})$ in Game I, Alice responds by playing the move $A_{2n+2} = B(a_{2n+2},\alpha _0 \rho _{2n+1})$ in Game I, where

   $$ \begin{align*} a_{2n+2} = b_{2n+1}. \end{align*} $$
   This sets the stage for the next iteration.

To finish the proof, it suffices to check that all of these moves are legal. When $n=0$ , move 1 is legal because any ball can be Bob’s first move. Move 2 is legal because

$$ \begin{align*} \mathbf{d}(b_{2n},a_{2n+1}) = \mathbf{d}(b_n',a_{n+1}') \leq (1-\alpha)\rho_n' = (1-\alpha_0) \rho_{2n}, \end{align*} $$

and move 3 is legal because $\mathbf {d}(b_{2n+1},a_{2n+2}) = 0$ . When $n\geq 1$ , move 1 is legal because

$$ \begin{align*} \mathbf{d}(a_n',b_n') &= \mathbf{d}(a_{2n-1},b_{2n}) \leq \mathbf{d}(a_{2n-1},b_{2n-1}) + \mathbf{d}(a_{2n},b_{2n})\\ &\leq (1-\beta_0)\alpha_0 \rho_{2n-2} + (1-\beta_0)\alpha_0(\alpha_0\beta_0) \rho_{2n-2} = (1-\beta)\alpha\rho_{n-1}'.\\[-37.3pt] \end{align*} $$

For details and other properties of winning sets, see Schmidt’s book [Reference Schmidt38].

2.2. The absolute game

The absolute game was introduced by McMullen in [Reference McMullen33]. It is played similarly to Schmidt’s game but instead of choosing a region where Bob must play, Alice now chooses a region where Bob must not play. To be precise, given $0<\beta <1$ , the $\beta $ absolute game is played as follows: Bob first picks some initial ball $B_0 \subseteq X$ . Alice and Bob then take turns to place successive closed balls in such a way that

(5) $$ \begin{align} B_0 \: \supseteq\: B_0 \setminus A_1 \:\supseteq \:B_1 \supseteq \cdots \supseteq\: B_i \: \supseteq \: B_i \setminus A_{i+1} \:\supseteq \:B_{i+1} \:\supseteq \: \cdots, \end{align} $$

subject to the conditions

(6) $$ \begin{align} \mathrm{rad}(A_{i+1}) \leq \beta \cdot \mathrm{rad}(B_i) \quad \textrm{and} \quad \mathrm{rad}(B_{i+1}) \geq \beta \cdot \mathrm{rad}(B_i) \quad\textrm{for all } i \geq 0. \end{align} $$

See Figure 3 for a pictorial view of the game in $\mathbb R^2$ .

Figure 3 The absolute game.

If $r_i:= \mathrm {rad}(B_i) \not \rightarrow 0$ , then we say that Alice wins by default. We say a set $E \subseteq X$ is $\beta $ absolute winning if Alice has a strategy which guarantees that either she wins by default or equation (3) is satisfied. In this case, we say that Alice wins. The absolute game is defined by allowing Bob to choose $0<\beta <1$ on his first turn and then continuing as in the $\beta $ absolute game. 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 $\beta $ absolute winning for every $0<\beta <1$ .

A useful generalization of the absolute game was introduced in [Reference Fishman, Simmons and Urbański26, Appendix C]. Let $\mathcal {H}$ be any collection of non-empty closed subsets of X. The $\mathcal {H}$ absolute game is defined in a similar way to the absolute game, but Alice is now allowed to choose a neighborhood of an element $H\in \mathcal {H}$ , namely,

$$ \begin{align*} A_{i+1} = B(H,\beta\cdot \mathrm{rad}(B_i)). \end{align*} $$

McMullen’s absolute game is the same as the $\mathcal {H}$ absolute game when $\mathcal {H}$ is the collection of all singletons. The first generalization of the absolute game was the k-dimensional absolute game, which is the $\mathcal {H}$ absolute game where $X=\mathbb {R}^N$ and $\mathcal {H}$ is the collection of k-dimensional hyperplanes in $\mathbb {R}^N$ . When $k=N-1$ , the associated winning sets are often referred to by the acronym HAW, short for ‘hyperplane absolute winning’ [Reference Broderick, Bugeaud, Fishman, Kleinbock and Weiss10]. Some of the results in this note regarding the absolute game may be generalized to the context of $\mathcal {H}$ absolute game but we will not pursue this line further in the current note.

2.3. The potential game

Given two parameters $c, \beta>0$ , the $(c,\beta )$ potential game begins with Bob picking some initial ball $B_0 \subseteq X$ . On her $(i+1)$ st turn, Alice chooses a countable collection of closed balls $A_{i+1, k}$ , satisfying

(7) $$ \begin{align} \sum_{k} \mathrm{rad}(A_{i+1,k})^c \leq (\beta \cdot \mathrm{rad}(B_{i}))^c. \end{align} $$

By convention, we assume that a ball always has a positive radius; that is, Alice cannot choose $A_{i,k}$ to be singletons. Then, Bob on his $(i+1)$ st turn chooses a ball $B_{i+1}\subseteq B_i$ satisfying $\mathrm {rad}(B_{i+1}) \geq \beta \cdot \mathrm {rad}(B_i)$ . See Figure 4 for a demonstration of the playing of such a game in $\mathbb R^2$ (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).

Figure 4 The potential game.

If

(8) $$ \begin{align} \bigcap_{i} \, B_i \subseteq \bigcup_{i}\bigcup_{k} \, A_{i, k} \end{align} $$

or if $\mathrm {rad}(B_i) \not \rightarrow 0$ , we say that Alice wins by default.

A set $E \subseteq X$ is said to be $(c, \beta )$ potential winning if Alice has a strategy guaranteeing that either she wins by default or equation (3) holds. The $c_0$ potential game is defined by allowing Bob to choose in his first turn the parameters $c>c_0$ and $\beta>0$ , and then continue as in the $(c,\beta )$ potential game. A set E is $c_0$ potential winning if Alice has a winning strategy for E in the $c_0$ potential game. Note that E is $c_0$ potential winning if and only if E is $(c, \beta )$ potential winning for every $c> c_0$ and $\beta> 0$ . If X is Ahlfors $\delta $ regular for some $\delta>0$ , the potential game is defined by allowing Alice to choose $c_0<\delta $ and then continuing as in the $c_0$ potential game, and a set E is called potential winning if Alice has a winning strategy in the potential game. Note that E is potential winning if and only if it is $c_0$ potential winning for some $c_0<\delta $ .

For the sake of completeness, we will verify properties (W1) and (W2) for potential winning sets. Property (W1) is verified for a class of Ahlfors regular spaces X in Theorem 4.7.

Proof. Fix arbitrary $c>\max \{c_1,c_2\}$ and $\beta>0$ . By assumption, $E_1$ and $E_2$ are both $(c, \beta ^2)$ potential winning. Now Alice uses the following strategy to win the $(c,\beta )$ game on the set $E_1\cap E_2$ . After Bob’s initial move $B_0$ , Alice chooses the collection $A_{1,k}$ from her $(c,\beta ^2)$ winning strategy for the set $E_1$ . After Bob’s first move $B_1$ , Alice chooses the collection $A^*_{1,k}$ from her $(c,\beta ^2)$ winning strategy for the set $E_2$ , assuming that Bob’s initial move was $B_1$ .

In general, after $B_{2m}$ , Alice chooses the collection $A_{m,k}$ following her $(c,\beta ^2)$ winning strategy for the set $E_1$ ; after $B_{2m+1}$ , she chooses the collection $A^*_{m,k}$ following her $(c,\beta ^2)$ winning strategy for the set $E_2$ .

It is easy to check that Alice’s moves are legal, because

$$ \begin{align*} \sum_{k} \mathrm{rad}(A_{i+1,k})^c \leq (\beta^2 \cdot \mathrm{rad}(B_{i}))^c\leq (\beta \cdot \mathrm{rad}(B_{i}))^c. \end{align*} $$

The same inequality is true for the collections $A^*_{i+1,k}$ . Also, by construction, either Alice wins by default or $\bigcap _i B_i \subseteq E_1\cap E_2$ .

2.4. The strong and the weak Schmidt games

One can define further variants of Schmidt’s original game. The strong game (introduced by McMullen in [Reference McMullen33]) coincides with Schmidt’s game except that the equalities in equation (2) are replaced by the inequalities

(9) $$ \begin{align} \mathrm{rad}(A_{i+1}) \geq \alpha \cdot \mathrm{rad}(B_i) \quad\textrm{and}\quad \mathrm{rad}(B_{i+1}) \geq \beta \cdot \mathrm{rad}(A_{i+1}) \quad\textrm{for all } i\geq 0. \end{align} $$

Similarly, we define the weak game so that Alice can choose her radii with an inequality but Bob must use equality. Obviously, every weak winning set is winning but the converse is not necessarily true (however, the weak game does not have the intersection property—see later).

One could also define a very strong game, where Alice must use equality but Bob may use inequality.

Later in the paper, we will need the following proposition which relates the very strong winning property of a set E and the weak winning property of its complement $X\setminus E$ .

2.5. Cantor winning

For the sake of clarity, we introduce a specialization of the framework presented in [Reference Badziahin and Harrap5], tailored to suit our needs. We first define the notion of a ‘splitting structure’ on X. This is in some sense the minimal amount of structure the metric space needs to allow the construction of generalized Cantor sets (cf. [Reference Badziahin and Harrap5]). Denote by $\mathcal {B}(X)$ the set of all closed balls. We assume that by definition, to specify a closed ball, it is necessary to specify its center and radius. Note that this means that in some cases, two distinct balls may be set-theoretically equal in the sense of containing the same points (see also Remark 2.1) in X and by $\mathcal {P}(\mathcal {B}(X))$ the set of all subsets of $\mathcal {B}(X)$ . A splitting structure is a quadruple $(X,\mathcal {S}, U,f)$ , where:

• • $U\subseteq \mathbb {N}$ is an infinite multiplicatively closed set such that if $u,v\in U$ and $u\mid v$ , then $v/u\in U$ ;

• • $f: U\to \mathbb {N}$ is a totally multiplicative function;

• • $\mathcal {S}\;:\mathcal {B}(X)\times U \to \mathcal P(\mathcal {B}(X))$ is a map defined in such a way that for every ball $B\in \mathcal {B}(X)$ and $u\in U$ , the set $\mathcal {S}(B,u)$ consists solely of balls $S\subseteq B$ of radius $\mathrm {rad}(B)/u$ .

Additionally, we require all these objects to be linked by the following properties:

1. (S1) $\#\mathcal {S}(B,u) = f(u)$ (in fact, one can make the definition of a splitting structure slightly more general by removing the function f out of it and replacing condition (S1) with

   1. (S1’) $\#\mathcal {S}(B,u)$ is finite.

   In other words, one can let $\#\mathcal {S}(B,u)$ vary with respect to B. In this case, instead of the function $f(u)$ , we define $f(B,u):=\#\mathcal {S}(B,u)$ . An enthusiastic reader can verify that in this more general setting, all the results of this paper are still satisfied after minor straightforward adjustments have been made. However, for convenience, we stick with the notation from [Reference Badziahin and Harrap5]);

2. (S2) if $S_1,S_2\in \mathcal {S}(B,u)$ and $S_1\neq S_2$ , then $S_1$ and $S_2$ may only intersect on the boundary, that is, the distance between their centers must be at least $\mathrm {rad}(S_1) + \mathrm {rad}(S_2) = 2\mathrm {rad}(B)/u$ ;

3. (S3) for all $u,v\in U$ ,

   $$ \begin{align*} \mathcal{S}(B,uv) = \bigcup_{S \,\in \,\mathcal{S}(B,u)} \mathcal{S}(S,v). \end{align*} $$

Broadly speaking, the set U determines the set of possible ratios of radii of balls in the successive levels of the upcoming Cantor set construction. For any fixed $B \in \mathcal {B}(X)$ , the function f tells us how many balls inside B of a fixed radius we may choose as candidates for the next level in the Cantor set construction, and the sets $\mathcal {S}(B, u)$ describe the position of these candidate balls.

If $f \equiv 1$ , we say the splitting structure $(X,\mathcal {S}, U,f)$ is trivial. Otherwise, fix a sequence $(u_i)_{i\in \mathbb {N}}$ with $u_i\in U$ such that $u_i\mid u_{i+1}$ and $u_i\underset {i\to \infty }{\longrightarrow } \infty $ . Then, one can show [Reference Badziahin and Harrap5, Theorem 2.1] that for any $B_0 \in \mathcal {B}(X)$ , the limit set

$$ \begin{align*} A_\infty(B_0):= \bigcap_{i=1}^\infty \: \bigcup_{B\, \in\, \mathcal{S}(B,u_i)} B \end{align*} $$

is compact and independent of the choice of sequence $(u_i)_{i\in \mathbb {N}}$ .

Example 2.7. The standard splitting structure on $X=\mathbb {R}^N$ is with the metric that comes from the sup norm: $\mathbf {d}(x,y):=\|x-y\|_\infty $ . The collection of splittings $\mathcal {S}(B,u)$ of any closed ball B according to the integer $u\in U = \mathbb {N}$ , into $f(u)=u^N$ parts, is defined as follows: cut B into $u^N$ equal square boxes whose edges have length u times less than the edges length of B. One can check that $(\mathbb {R}^N,\mathcal {S}, \mathbb {N}, f)$ satisfies properties (S1)–(S3) and is the unique splitting structure for $(\mathbb {R}^N, \mathbf {d})$ with $U=\mathbb {N}$ such that $A_\infty (B) = B$ for every $B\in \mathcal {B}(X)$ .

In a similar manner, the standard splitting structure on $\{0,1\}^{\mathbb {N}}$ is with the standard metric $\mathbf {d}(x,y):=2^{-\min \{i\geq 0: x_i\neq y_i\}}$ , $U=\{2^i: i\geq 0\}$ , $f(u)=u$ , and

$$ \begin{align*} \mathcal{S}([x_1,\ldots,x_n],2^i)=\{[x_1,\ldots,x_n,y_1,\ldots,y_i]: (y_1,\ldots,y_i)\in\{0,1\}^i \}, \end{align*} $$

where for any $x\in X$ and $n\geq 0$ , we define $[x_1,\ldots ,x_n]:=\{y: y_k=x_k \textrm { for all} 1\leq k\leq n\}= B(x,2^{-n})$ to be the cylinder with prequel $(x_1,\ldots ,x_n)$ , thought of as a ball of radius $2^{-n}$ (the choice of center is arbitrary).

For any collection $\mathcal {B} \subseteq \mathcal P(\mathcal {B}(X))$ of balls and for any $R\in U$ , we will write

$$ \begin{align*} \frac{1}{R} \mathcal{A}:= \bigcup_{A \in \mathcal{A}} \mathcal{S}(A, R). \end{align*} $$

Of course, for higher powers of R, we can write $1/{R^{n}} \mathcal {A} = 1/{R}(\cdots (1/{R}\mathcal {A})\cdots )$ .

We can now recall the definition of generalized Cantor sets. Fix some closed ball $B_0 \subseteq X$ and some $R \in U$ , and let

$$ \begin{align*} \mathbf{r}:= (r_{m,n}),\; m,n\in\mathbb{Z}_{\geqslant 0},\; m\leqslant n \end{align*} $$

be a two-parameter sequence of non-negative real numbers. Define $\mathcal {B}_0:=\{B_0\}$ . We start by considering the set $1/{R} \mathcal {B}_0$ . The first step in the construction of a generalized Cantor set involves the removal of a collection $\mathcal {A}_{0,0} $ of at most $r_{0,0}$ balls from $1/{R}\mathcal {B}_0$ . We label the surviving collection as $\mathcal {B}_1$ . Note that we do not specify here the removed balls, we just give an upper bound for their number. In general, for a fixed $n\geqslant 0$ , given the collection $\mathcal {B}_n$ , we construct a nested collection $\mathcal {B}_{n+1}$ as follows: let

(10) $$ \begin{align} \mathcal{B}_{n+1}: = \bigg(\frac{1}{R} \mathcal{B}_n\bigg) \setminus \bigcup_{m=0}^n \mathcal{A}_{m,n} \quad \text{where } \mathcal{A}_{m,n} \subseteq \frac{1}{R^{n-m+1}} \mathcal{B}_m \quad (0 \leq m \leq n) \end{align} $$

are collections of removed balls satisfying for every $B \in \mathcal {B}_m$ , the inequality

$$ \begin{align*}\# \mathcal{A}_{m,n}(B) \leq r_{m,n} \quad \text{where } \mathcal{A}_{m,n}(B):=\{ A \in \mathcal{A}_{m,n}: A \subseteq B \}.\end{align*} $$

For a pictorial demonstration in $\mathbb R^2$ of such a construction, see Figure 5, in which boxes removed in the current step are colored red.

Figure 5 Construction of a Cantor winning set.

Define the limit set of such a construction to be

(11) $$ \begin{align} \bigcap_{n=0}\bigcup_{B\in\mathcal{B}_n}B \end{align} $$

and denote it by $\mathcal {K}(B_0, R,\mathbf {r})$ . Note that the triple $(B_0,R,\mathbf {r})$ does not uniquely determine $\mathcal {K}(B_0,R,\mathbf {r})$ as there is a large degree of freedom in the sets of balls $\mathcal {A}_{m,n}$ removed in the construction procedure. We say a set $\mathcal {K}$ is a generalized Cantor set if it can be constructed by the procedure described above for some triple $(B_0,R,\mathbf {r})$ . In this case, we may refer to $\mathcal {K}$ as being $(B_0,R,\mathbf {r})$ Cantor if we wish to make such a triple explicit and write $\mathcal {K}=\mathcal {K}(B_0,R,\mathbf {r})$ .

Definition 2.10. Fix a ball $B_0 \subseteq X$ . Given a parameter $\varepsilon _0 \in (0,1]$ , we say a set $E \subseteq X$ is $\varepsilon _0$ Cantor winning on $B_0$ (with respect to the splitting structure $(X, \mathcal {S}, U, f)$ ) if for every $0<\varepsilon <\varepsilon _0$ , there exists $R_\varepsilon \in U$ such that for every $R_\varepsilon \leq R\in U$ , the set E contains a $(B_0,R,\mathbf {r})$ Cantor set satisfying

$$ \begin{align*} r_{m,n} \:\leq \: f(R)^{(n-m+1)(1-\varepsilon)}\quad\text{for every }m,n\in\mathbb{N},\: m\leqslant n. \end{align*} $$

If a set $E \subseteq X$ is $\varepsilon _0$ Cantor winning on $B_0$ for every ball $B_0\subseteq X$ , then we say E is $\varepsilon _0$ Cantor winning, and simply that E is Cantor winning if it is $\varepsilon _0$ Cantor winning for some $\varepsilon _0 \in (0,1]$ .

It is obvious that if a set is $\varepsilon _0$ Cantor winning, then it is also $\varepsilon $ Cantor winning for any $\varepsilon \in (0, \varepsilon _0)$ . In particular, the property of being $1$ Cantor winning is the strongest possible Cantor winning property.

2.6. Cantor rich

The class of Cantor rich subsets of $\mathbb {R}$ was introduced in [Reference Beresnevich8]. As was noted in [Reference Badziahin and Harrap5], this notion can be generalized to the context of metric spaces endowed with a splitting structure.

Definition 2.12. Let $M\geq 4$ be a real number and $B_0$ be a ball in X. A set $E\subseteq X$ is $(B_{0},M)$ Cantor rich if for every R such that

(12) $$ \begin{align} f(R)> M \end{align} $$

and $0 < y < 1$ , E contains a $(B_{0},R,\mathbf {r})$ Cantor set, where $\mathbf {r} = (r_{m,n})_{0\leq m\leq n}$ is a two-parameter sequence satisfying

$$ \begin{align*} \sum_{m=0}^{n}\bigg(\frac{4}{f(R)}\bigg)^{n-m+1}r_{m,n}\leq y\quad\text{for every }n\geq0. \end{align*} $$

Here, E is Cantor rich in $B_{0}$ if it is $(B_{0},M)$ Cantor rich in $B_{0}$ for some $M\geq 4$ , and E is Cantor rich if it is Cantor rich in some $B_{0}$ .

Note that, following [Reference Beresnevich8], Cantor rich sets are Cantor rich in some ball, while Cantor winning sets are Cantor winning in every ball. Also note that unlike in [Reference Beresnevich8], the inequality in equation (12) is strict. This enables a clearer correspondence with the parameter $\varepsilon _0$ in Definition 2.10 (cf. Corollary 3.11).

2.7. Diffuse sets and doubling spaces

Schmidt’s game was originally played on a complete metric space. However, to conclude that winning sets have properties such as large Hausdorff dimension, certain assumptions on the underlying space are required. The first time that large intersection with certain subspaces of $\mathbb {R}^n$ was proved using Schmidt’s game was in [Reference Fishman24]. Later, it was realized in [Reference Broderick, Fishman, Kleinbock, Reich and Weiss11] that, more generally, some spaces form a natural playground for the absolute game. The following notion generalizes the definition of a diffuse set in $\mathbb {R}^n$ as appeared in [Reference Broderick, Fishman, Kleinbock, Reich and Weiss11].

Definition 2.13. For $0<\beta <1$ , a closed set $K\subseteq X$ is $\beta $ diffuse if there exists $r_{0}>0$ such that for every $x\in K$ , $y\in X$ and $0<r\leq r_{0}$ , we have

(13) $$ \begin{align} ( B(x,(1-\beta)r)\setminus B(y,2\beta r))\cap K\neq\emptyset. \end{align} $$

If K is $\beta $ diffuse for some $\beta $ , then we say it is diffuse.

The condition in equation (13) is satisfied if and only if there exists $z\in K$ such that

$$ \begin{align*} B(z,\beta r)\subseteq B(x,r)\setminus B(y,\beta r). \end{align*} $$

The definition of diffuse sets is designed so that the $\beta $ absolute game on a $\beta $ diffuse metric space 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 the target set, then the reason for her victory is her strategy and not the limitations of the space in which the game is played.

Remark 2.14. In the definition that appears in [Reference Broderick, Fishman, Kleinbock, Reich and Weiss11], sets as in Definition 2.13 are called $0$ -dimensionally diffuse. Moreover, equation (13) is replaced by

(14) $$ \begin{align} (B(x,r)\setminus B(y,\beta r))\cap K\neq\emptyset. \end{align} $$

Note that

$$ \begin{align*} B(x,r)\setminus B(y,\beta r)=B(x,(1-\beta')\rho)\setminus B(y,2\beta'\rho) \end{align*} $$

for $\beta '={\beta }/({2+\beta })$ and $\rho =(1+{\beta }/{2})r$ , so both equations (13) and (14) give rise to the same class of diffuse sets. We will use equations (13) and (14) interchangeably without further notice.

A more standard definition that is equivalent to diffuseness (see [Reference Fishman, Simmons and Urbański26, Lemma 4.3]) is the following.

Definition 2.15. For $0<c<1$ , a closed set $K\subseteq X$ is c uniformly perfect if there exists $r_{0}>0$ such that for every $x\in K$ and $0<r\leq r_{0}$ , we have

(15) $$ \begin{align} (B(x,r)\setminus B(x,cr))\cap K\neq\emptyset. \end{align} $$

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 [Reference Broderick, Fishman, Kleinbock, Reich and Weiss11, Lemma 4.4]. Instead of expanding the discussion on microsets, we refer the interested reader to [Reference Furstenberg27], and mention that Ahlfors regular sets are diffuse. In the other direction, diffuse sets support Ahlfors regular measures (see Proposition 2.18). We will now give the relevant definitions and provide the results which support the statements above.

Definition 2.16. A Borel measure $\mu $ on X is called Ahlfors regular if there exist $C,\delta ,r_{0}>0$ such that every x in the support of $\mu $ and $0<r\leq r_{0}$ satisfy

$$ \begin{align*} \frac{1}{C}r^\delta\leq\mu(B(x,r))\leq Cr^\delta. \end{align*} $$

In this case, we call $\delta $ the dimension of $\mu $ and say that $\mu $ is Ahlfors regular of dimension $\delta $ .

We say that a closed set $K\subseteq X$ is Ahlfors regular if it is the support of an Ahlfors regular measure. More exactly, $K\subseteq X$ is $\delta $ Ahlfors regular if it is the support of an Ahlfors regular measure of dimension $\delta $ . This terminology gives rise to a dimensional property of K.

Definition 2.17. The Ahlfors regularity dimension of $K\subseteq X$ is

$$ \begin{align*} \dim_R K=\sup \{\delta : K \text{contains the support of an Ahlfors regular measure of dimension }\delta\}. \end{align*} $$

Note that it is a direct consequence of the mass distribution principle (see [Reference Falconer19]) that $\dim _R K\leq \dim _H K$ , where $\dim _H K$ stands for the Hausdorff dimension of K.

The following property of Ahlfors regular sets will be used later in the paper.

Proof. It is enough to prove the proposition for all $\alpha \leq 1/2$ . Let $B'$ be a ball with the same center as B, but with radius $(1-\alpha )\mathrm {rad}(B)$ , and let $D\subseteq X$ be an arbitrary ball of radius $\mathrm {rad}(D) = 3\alpha \cdot \mathrm {rad}(B)$ . Then

$$ \begin{align*} \mu (D)\leqslant C\cdot (3\alpha)^\delta (\mathrm{rad}(B))^\delta \leqslant C^2\cdot \bigg(\frac{3\alpha}{1-\alpha}\bigg)^\delta \mu (B') \leq C_1 \alpha^\delta \mu (B'). \end{align*} $$

Therefore, for any k balls $D_i$ of radius $3\alpha \cdot \mathrm {rad}(B)$ with $k<C_1^{-1} \alpha ^{-\delta }$ , there exists another ball D whose center lies inside $B'\backslash \bigcup _{i=1}^k D_i$ . Hence, there are at least $I = \lceil C_1^{-1} \alpha ^{-\delta }\rceil $ balls $D_i$ such that their centers lie inside $B'$ and are distanced by at least $3 \alpha \cdot \mathrm {rad}(B)$ from each other. For each $1 \leq i \leq I$ , let $B_i$ be a ball with the same center as of $D_i$ but with the radius $\alpha \cdot \mathrm {rad}(B)$ . Clearly, the balls $B_i$ satisfy all the conditions of the proposition, where $c>0$ is chosen to ensure the inequality

$$ \begin{align*} c\alpha^{-\delta} \leqslant \lceil C_1^{-1} \alpha^{-\delta}\rceil \end{align*} $$

for all $\alpha \in [0,1]$ .

Although winning sets are usually dense, the large dimension property (W1) for some types of winning sets depends on the space X being large enough. For example, absolute winning sets may be empty if the Ahlfors regularity dimension is zero (cf. Remark 4.9). Various conditions are introduced in this context; for example, see [Reference Schmidt37, §11] and [Reference Fishman24, Theorem 3.1]. More recently, the following condition was introduced in [Reference Badziahin and Harrap5].

1. (S4) There exists an absolute constant $C(X)$ such that any ball $B \in \mathcal {B}(X)$ cannot intersect more than $C(X)$ disjoint open balls of the same radius as B.

Proposition 2.20. [Reference Badziahin and Harrap5, Corollary to Theorem 4.1]

Let X be a complete metric space and let $(X,\mathcal {S}, U,f)$ be a splitting structure. If X satisfies (S4), then for any $B \in \mathcal {B}(X)$ and any Cantor winning set $E\subseteq X$ , we have

$$ \begin{align*} \dim_H(E \cap A_\infty(B))=\dim_H A_\infty(B). \end{align*} $$

Condition (S4) is similar to the following definition.

Doubling spaces satisfy condition (S4), but the converse does not hold. For example, in an ultrametric space, any two non-disjoint balls of the same radius are set-theoretically equal. So an ultrametric space satisfies condition (S4) with $C=1$ , but is not necessarily doubling. As shown in [Reference Badziahin and Harrap5], the space X needs to satisfy condition (S4) for Cantor winning sets in X to satisfy the large dimension condition (W1) and large intersection condition (W2), and if X is doubling, then Cantor winning sets satisfy condition (W3) with respect to bi-Lipschitz homeomorphisms of X. This makes doubling spaces a natural setting for generalized Cantor sets, and it is under this condition that many of the results of the next section hold.

3.1. The Cantor game

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 this game is designed to be played on a metric space endowed with a splitting structure. Our nomenclature is so chosen because (as we shall demonstrate) under natural conditions, the winning sets for this game are precisely the Cantor winning sets defined in Definition 2.10.

Let X be a complete metric space and let $(X,\mathcal {S}, U,f)$ be a splitting structure. Given $\varepsilon \in (0,1]$ , the $\varepsilon $ Cantor game is defined as follows: Bob starts by choosing an integer $2\leq R\in U$ and a closed ball $B_0\subseteq X$ . For any $i\geq 0$ , given Bob’s ith choice $B_i$ , Alice chooses a collection $\mathcal {A}_{i+1}\subseteq {1}/{R}\{B_i\}$ satisfying

$$ \begin{align*} \#\mathcal{A}_{i+1}\leq f(R)^{1-\varepsilon}. \end{align*} $$

Given $\mathcal {A}_{i+1}$ , Bob chooses a ball $B_{i+1}\in {1}/{R}\{B_i\}\setminus \mathcal {A}_{i+1}$ , and the game continues. We say that $E\subseteq X$ is winning for the $\varepsilon $ Cantor game if Alice has a strategy which guarantees that equation (3) is satisfied. The Cantor game is defined by allowing Alice first to choose $\varepsilon \in (0,1]$ and then continuing as in the $\varepsilon $ Cantor game. We say that E is winning for the Cantor game if Alice has a winning strategy for the Cantor game. Note that E is winning for the Cantor game if and only if it is winning for the $\varepsilon $ Cantor game for some $\varepsilon \in (0,1]$ . We say that E is winning for the $\varepsilon $ Cantor game on $B_0$ if Alice has a winning strategy in the $\varepsilon $ Cantor game given that Bob’s first ball is $B_0$ .

It is clear that winning sets for the Cantor game contain Cantor winning sets. Indeed, by the construction, for any $R\in U$ , a winning set for the $\varepsilon $ Cantor game on $B_0$ contains a $(B_0, R, f(R)^{1-\varepsilon })$ Cantor set which comprises all Bob’s possible moves while playing against a fixed winning strategy of Alice. Therefore, it is $\varepsilon $ Cantor winning on $B_0$ . We show that the converse is also true.

Proof. We will define a winning strategy for Alice. Fix $R\geq 2$ and let $\delta \geq 0$ be chosen so that $f(R) = R^\delta $ . Choose $0<\eta <\varepsilon _0$ so small that

$$ \begin{align*} \frac{R^{\delta(1-\varepsilon_0 + \eta)}}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1} < 1. \end{align*} $$

Set

(16) $$ \begin{align} \varepsilon_1 &=\varepsilon_0 - \frac{\eta}{2}, \end{align} $$

(17) $$ \begin{align} \varepsilon_2 &=\varepsilon_0 - \eta, \end{align} $$

and let $R_{\varepsilon _1}$ be as in Definition 2.10. Then, for any $\ell $ such that $R^\ell \geq R_{\varepsilon _1}$ , since $f(R^\ell )=f(R)^\ell =R^{\delta \ell }$ , there exists a $(B_{0},R^\ell ,(R^{(n-m+1)\ell \delta (1-\varepsilon _1)})_{0\leq m\leq n})$ Cantor set in E. Let $\ell $ be large enough so that $R^{\ell }\geq R_{\varepsilon _1}$ ,

(18) $$ \begin{align} \bigg(\frac{R^{\delta(1-\varepsilon_2)}}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\bigg)^\ell<\frac{1}{2}, \end{align} $$

and

(19) $$ \begin{align} R^{-\ell\delta({\eta}/{2})}<\frac{1}{3} \cdot \end{align} $$

Such a value exists because of the choice of $\eta $ .

Choose a $(B_{0},R^{\ell },(R^{(n-m+1)\ell \delta (1-\varepsilon _1)})_{0\leq m\leq n})$ Cantor set in E. Let $\mathcal {A}=\{ \mathcal {A}_{m,n}\} _{0\leq m\leq n}$ be the removed balls, so that $\mathcal {B}_{n+1}=({1}/{R^{\ell }})\mathcal {B}_{n}\setminus \bigcup _{m=0}^{n}\mathcal {A}_{m,n}$ for all $n\geq 0$ , where $\mathcal {B}_{0}=\{ B_{0}\} $ .

Define a winning strategy for Alice in the $\varepsilon _0$ Cantor game as follows. For each $i\geq 0$ , define a potential function on balls:

(20) $$ \begin{align} \varphi_{i}(B)=\sum_{m=0}^{\lfloor i/\ell\rfloor }\sum_{n=m}^{\infty}\#\{ A\in\mathcal{A}_{m,n}: A\cap B \text{ has non-empty interior}\} R^{-(n+1)\ell\delta(1 - \varepsilon_2)}. \end{align} $$

Given Bob’s ith move $B_i$ , Alice’s strategy is to choose the subcollection $\mathcal {A}_{i+1} \subseteq {1}/{R}\{B_i\}$ consisting of the $\lfloor R^{\delta (1-\varepsilon _0)} \rfloor $ balls with the largest corresponding values of $\varphi _{i+1}$ . We will show that this is a winning strategy for Alice. Assume Bob chose the sequence of balls $\{B_i\}_{i=0}^\infty $ on his turns, while Alice responded according to the above strategy. It is enough to prove that for every $i\geq 0$ ,

(21) $$ \begin{align} \varphi_{i\ell}(B_{i\ell}) < R^{-i\ell\delta(1- \varepsilon_2)}. \end{align} $$

Indeed, this inequality would imply that $B_{i\ell }$ does not intersect any of the balls from $\mathcal {A}_{m,n}$ with $n<i$ except at the boundary, which implies that $B_{i\ell }\in \mathcal {B}_i$ .

Fix $i\geq 0$ such that equation (21) holds. For every $0\leq j\leq \ell -1$ , note that $\lfloor ({i\ell +j})/({\ell })\rfloor = i$ , so by the induction hypothesis in equation (21), we get

$$ \begin{align*} \varphi_{i\ell+j}(B_{i\ell+j})\leqslant \varphi_{i\ell}(B_{i\ell})<R^{-i\ell\delta(1- \varepsilon_2)}. \end{align*} $$

Therefore, for any $n\geq m \geq 0$ , if $A\in \mathcal {A}_{m,n}$ is such that $A\cap B_{i\ell +j+1}$ has non-empty interior, then necessarily $n+1> i$ , and thus by condition (S2), we have $A \subseteq B_{i\ell +j+1}$ . So

(22) $$ \begin{align} \sum_{A\in\frac{1}{R}\{B_{i\ell+j}\}}\varphi_{i\ell+j}(A)=\varphi_{i\ell+j}(B_{i\ell+ j}), \end{align} $$

and the definition of Alice’s strategy yields

(23) $$ \begin{align} \varphi_{i\ell+j}(B_{i\ell+j+1})\leq\frac{1}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\varphi_{i\ell+j}(B_{i\ell+j}) \end{align} $$

for every $0\leq j\leq l-1$ . Indeed, there are at least $\lfloor R^{\delta (1-\varepsilon _0)}\rfloor $ balls A, chosen by Alice in her turn, with $\varphi _{i\ell +j}(A)\geqslant \varphi _{i\ell +j}(B_{i\ell +j+1})$ . Therefore, if equation (23) fails for some ball $B_{i\ell +j+1}$ , we have

$$ \begin{align*} \varphi_{i\ell+j}(B_{i\ell+j+1}) + \sum_{A\text{ chosen by Alice}} \varphi_{i\ell+j}(A)> \varphi_{i\ell+j}(B_{i\ell+j}) \end{align*} $$

which contradicts equation (22).

To estimate $\varphi _{(i+1)\ell }(B_{(i+1)\ell })$ , we need to additionally consider the potential coming from the sets $(\mathcal {A}_{i+1,n})_{n\geqslant i+1}$ . Clearly, for any $A\in \mathcal {A}_{i+1,n}$ , the condition that $A\cap B_{(i+1)\ell }$ has non-empty interior is equivalent to the inclusion $A\subseteq B_{(i+1)\ell }$ . Hence, we have

$$ \begin{align*} \varphi_{(i+1)\ell}(B_{(i+1)\ell}) \leq & \bigg(\frac{1}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\bigg)^{\ell}\varphi_{i\ell}(B_{i\ell}) \\ & + \sum_{n=i+1}^{\infty}\#\{ A\in\mathcal{A}_{i+1,n}: A\subseteq B_{(i+1)\ell}\} R^{-(n+1)\ell\delta(1-\varepsilon_2)}. \end{align*} $$

Since $\#\{ A\in \mathcal {A}_{m,n}: A\subseteq B_{m\ell }\} \leq R^{(n-m+1)\ell \delta (1-\varepsilon _1)}$ for any $n\geq m\geq 0$ , by equations (16), (17), and (21), we get

$$ \begin{align*} \varphi_{(i+1)\ell}(B_{(i+1)\ell}) & \leq \bigg(\frac{1}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\bigg)^{\ell}R^{-i\ell\delta(1-\varepsilon_2)}+\sum_{n=i+1}^{\infty}R^{(n-i)\ell\delta(1-\varepsilon_1)}R^{-(n+1)\ell\delta(1-\varepsilon_2)}\\ & = \bigg(\frac{R^{\delta(1-\varepsilon_2)}}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\bigg)^{\ell}R^{-(i+1)\ell\delta(1-\varepsilon_2)}+\sum_{n=1}^{\infty}R^{-n\ell\delta({\eta}/{2})}R^{-(i+1)\ell\delta(1-\varepsilon_2)}\\ & = \bigg(\bigg(\frac{R^{\delta(1-\varepsilon_2)}}{\lfloor R^{\delta(1-\varepsilon_0)}\rfloor+1}\bigg)^{\ell}+\frac{R^{-\ell\delta({\eta}/{2})}}{1-R^{-\ell\delta({\eta}/{2})}}\bigg)R^{-(i+1)\ell\delta(1-\varepsilon_2)}. \end{align*} $$

By equations (18) and (19), we get

$$ \begin{align*} \varphi_{(i+1)\ell}(B_{(i+1)\ell})<R^{-(i+1)\ell\delta(1-\varepsilon_2)}, \end{align*} $$

which is what we need to complete the induction.

Theorem 3.2 can be rephrased in terms of local Cantor sets (cf. Example 2.8).

Corollary 3.3 means that in Definition 2.10, it is enough to use local Cantor sets. This will be useful in the proof of Theorem 1.13.

For the sake of completeness, let us define a version of the Cantor game called the ‘Cantor potential game’. Its name is justified via the analogy with the potential game and the key role that potential functions, like the one in equation (20), play in the proof above that relates it to the Cantor game.

Given $\varepsilon _0\in (0,1]$ , the $\varepsilon _0$ Cantor potential game is defined as follows: Bob starts by choosing $0<\varepsilon <\varepsilon _0$ and Alice replies by choosing $R_\varepsilon \geq 2$ . Then Bob chooses an integer $R\geq R_\varepsilon $ and a closed ball $B_0\subseteq X$ . For any $i\geq 0$ , given Bob’s ith choice $B_i$ , Alice chooses collections $\{\mathcal {A}_{i+1,k}\}_{k=0}^\infty $ such that $\mathcal {A}_{i+1,k}\subseteq {1}/{R}^k\{B_i\}$ and

$$ \begin{align*} \#\mathcal{A}_{i+1,k}\leq f(R)^{(k+1)(1-\varepsilon)} \end{align*} $$

for all $k\geq 0$ . Given $\{\mathcal {A}_{i+1,k}\}_{k=0}^\infty $ , Bob chooses a ball $B_{i+1}\in {1}/{R}\{B_i\}\setminus \bigcup _{k=0}^{i}\mathcal {A}_{i+1-k,k}$ (if there is no such ball, we say that Alice wins by default). We say that $E\subseteq X$ is $\varepsilon $ Cantor potential winning if Alice has a strategy which guarantees that either she wins by default or equation (3) is satisfied. The Cantor potential game is defined by allowing Alice first to choose $\varepsilon _0\in (0,1]$ and then continuing as in the $\varepsilon _0$ Cantor potential game.

By definition, $\varepsilon _0$ Cantor winning is the same as $\varepsilon _0$ Cantor potential winning. Moreover, there is a natural way to pass between the collections that define a Cantor winning set to the winning strategy in the Cantor potential game. Indeed, if $E\subseteq X$ is $\varepsilon _0$ Cantor winning in $B_0$ given by the collections $\mathcal {B}_n$ and $\mathcal {A}_{m,n}$ for any $n\geq 0$ and any $0\leq m\leq n$ , then a winning strategy for Alice for the $\varepsilon _0$ Cantor potential game may be given by $\mathcal {A}^{\prime }_{i+1,k} = \mathcal {A}_{i+1,k+i+1}(B_i)$ where $B_i$ is Bob’s ith move. In the other direction, let $\mathcal {A}'$ denote a winning strategy for Alice for the $\varepsilon _0$ Cantor potential game on $B_0$ for E, that is, if B is the ith move of Bob, denote by $\mathcal {A}^{\prime }_{i+1,k}(B_i)$ Alice’s $i+1$ st move according to this winning strategy. Define the collections $\mathcal {B}_n$ for $n\geq 0$ and $\{\mathcal {A}_{m,n}\}_{0\leq m \leq n}$ for $n\geq 1$ by recursion as follows: $\mathcal {B}_0=\{B_0\}$ ,

$$ \begin{align*} \mathcal{A}_{m,n} = \bigcup_{B\in\mathcal{B}_m} \mathcal{A}^{\prime}_{m+1,n-m}(B), \end{align*} $$

for every $n\geq 1$ and $0\leq m\leq n$ , and define $\mathcal {B}_n$ as usual via equation (10). Since $\mathcal {A}'$ is a winning strategy, it implies that the limit set as in equation (11) is contained in E. Theorem 3.2 may then be reinterpreted as saying that the winning sets for the $\varepsilon $ Cantor potential game are the same as the winning sets for the $\varepsilon $ Cantor game.

3.2. Cantor, absolute, and potential winning

We prove Theorem 1.5 as a special case of a more general result connecting potential winning sets with Cantor winning sets, corresponding to the standard splitting structure on $\mathbb {R}^N$ .

Theorem 3.4. Let $(X,\mathcal {S},U,f)$ be a splitting structure on a doubling metric space X, fix $B_0\in \mathcal {B}(X)$ , and denote

$$ \begin{align*} \delta = {\dim_H}(A_\infty(B_0)). \end{align*} $$

Fix $\varepsilon _0 \in (0,1] $ and let $c_0 = \delta (1 - \varepsilon _0)$ . Then a set $E\subseteq A_\infty (B_0)$ is $\varepsilon _0$ Cantor winning on $B_0$ with respect to $(X,\mathcal {S},U,f)$ if and only if E is $c_0$ potential winning on $A_\infty (B_0)$ .

In particular, E is Cantor winning on $B_0$ if and only if E is potential winning on $A_\infty (B_0)$ .

Since potential winning is defined on metric spaces without using a splitting structure, we get the following observation.

A combination of Proposition 1.3 and Theorem 3.4 yields the following corollary.

We now prove Theorem 3.4.

Proof of forward direction

Since X is doubling, we may assume that $f(u) = u^{\delta }$ for every $u\in U$ (cf. [Reference Badziahin and Harrap5, Corollary 1]). Suppose that E is $\varepsilon _0$ Cantor winning, and fix $\beta> 0$ , $c> c_0$ , and $\rho _0> 0$ . Fix $0 < \varepsilon < \varepsilon _0$ such that $\delta (1 - \varepsilon ) \in (c_0,c)$ . Fix large $R\in U$ . Its precise value will be determined later. Potentially, R may depend on $\varepsilon $ , $\beta $ , c, and $\rho _0$ , and, in particular, it satisfies $R \geq \max (R_\varepsilon ,1/\beta )$ and $R^{\delta (1-\varepsilon )-c}<\tfrac 12$ . Then by the definition of Cantor winning, E contains some $(B_0,R,\mathbf {r})$ Cantor set $\mathcal {K}$ , where

$$ \begin{align*} r_{m,n} = f(R)^{(n - m + 1)(1 - \varepsilon)} = R^{\delta(n - m + 1)(1 - \varepsilon)} \quad\text{for all} m,n \text{with} m\leq n. \end{align*} $$

We now describe a strategy for Alice to win the $(c,\beta )$ potential game on $A_\infty (B_0)$ , assuming that Bob’s starting move has radius $\rho _0$ . Recall that for any $B\in \mathcal {B}_m$ ,

$$ \begin{align*}\mathcal{A}_{m,n}(B):=\{ A \in \mathcal{A}_{m,n}:\, A \subseteq B \}.\end{align*} $$

Let $\rho $ denote the radius of $B_0$ , let $D_k$ denote Bob’s kth move (we are denoting Bob’s kth move by $D_k$ instead of $B_m$ so as to reserve the letters B and m for the splitting structure framework), and for each $k\in \mathbb {N}$ , let $m = m_k\in \mathbb {N}$ denote the unique integer such that $\beta \cdot \mathrm {rad}(D_k) < R^{-m}\rho \leq \mathrm {rad}(D_k)$ , assuming such 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 of the set

$$ \begin{align*} \bigcup_{n\geq m} \mathcal{A}_{m,n} = \bigcup_{B\in \mathcal{B}_m} \bigcup_{n\geq m} \mathcal{A}_{m,n}(B) \end{align*} $$

that intersect Bob’s current choice. If $m_k$ is undefined, then delete nothing. Obviously, this strategy, if executable, will make Alice win since the intersection of Bob’s balls will satisfy $\bigcap _{k \in \mathbb {N}}D_k \subseteq \mathcal {K} \subseteq E$ . To show that it is legal, we need to show that

(24) $$ \begin{align} \sum_{\substack{B\in \mathcal{B}_m \\ B\cap D_k \neq \varnothing}} \sum_{n \geq m} r_{m,n} (R^{-(n + 1)}\rho)^c \leq (\beta\cdot \mathrm{rad}(D_k))^c. \end{align} $$

This is because elements of $\mathcal {A}_{m,n}(B)$ all have radius $R^{-(n + 1)}\rho $ .

To estimate the sum in question, we use the fact that X is doubling. Since the elements of $\mathcal {B}_m$ have disjoint interiors and have radius $R^{-m} \rho \asymp _\beta \mathrm {rad}(D_k)$ , the number of them that intersect $D_k$ is bounded by a constant depending only on $\beta $ . Call this constant $C_1$ . Then the left-hand side of equation (24) is less than

$$ \begin{align*} &C_1 \sum_{n\geq m} R^{\delta(n - m + 1)(1 - \varepsilon)} (R^{-(n + 1)}\rho)^c\\ &\quad= C_1 (R^{-m} \rho)^c \sum_{\ell = 1}^\infty R^{\delta(1 - \varepsilon)\ell} R^{-c\ell}\\ &\quad\leqslant 2 C_1 (\mathrm{rad}(D_k))^c R^{\delta(1 - \varepsilon) - c} \quad\big(\text{since }R^{\delta(1 - \varepsilon) - c}\leqslant \tfrac12\big). \end{align*} $$

By choosing R large enough so that

$$ \begin{align*} 2C_1 R^{\delta(1 - \varepsilon) - c} < \beta^c, \end{align*} $$

we guarantee that the move is legal.

Proof of backward direction

Suppose that $E \subseteq A_\infty (B_0)$ is $c_0$ potential winning and fix $0 < \varepsilon < \varepsilon _0$ . Let $R_\varepsilon = 2$ and fix $R\in U$ such that $R\geq R_\varepsilon $ . Fix a large integer $q\in \mathbb {N}$ to be determined, and let $\beta = 1/R^q$ and $c = \delta (1 - \varepsilon )$ . Then E is $(c,\beta )$ potential winning. Now for each $k\in \mathbb {N}$ and $B\in \mathcal {B}_{qk}$ , let $\mathcal {A}(B)$ denote the collection of sets that Alice deletes in response to Bob’s kth move $B_k=B$ , assuming that it has been preceded by the moves $B_0,B_1,\ldots ,B_{k-1}$ satisfying $B_{i + 1}\in \mathcal {S}(B_i,R^q)$ for all $i = 0,\ldots ,k - 1$ . (It is necessary to specify the history to uniquely identify Alice’s response because unlike the other variants of Schmidt’s game, the potential game is not a positional game (see §3.4 and [Reference Schmidt37, Theorem 7]). This is because the regions that were deleted at previous stages of the game affect the overall situation and therefore may affect Alice’s strategy.) Since Alice’s strategy is winning, we must have

$$ \begin{align*} E\supseteq A_\infty(B_0) \setminus \bigcup_{k\in\mathbb{N}} \bigcup_{B\in \mathcal{B}_{qk}} \bigcup_{A\in \mathcal{A}(B)} A. \end{align*} $$

Indeed, for every point on the right-hand side, there is some strategy that Bob can use to force the outcome to equal that point (and also not lose by default). So to complete the proof, it suffices to show that the right-hand side contains a $(B_0,R,\mathbf {r})$ Cantor set, where

$$ \begin{align*} r_{m,n} = f(R)^{(n - m + 1)(1 - \varepsilon)} = R^{\delta(n - m + 1)(1 - \varepsilon)} = R^{c(n - m + 1)} \quad\text{for all } m,\ n \text{ with } m\leq n. \end{align*} $$

For each $m = qk \leq n$ and $B\in \mathcal {B}_m$ , let $\mathcal {C}(n)$ be the collection of elements of $\mathcal {A}(B)$ whose radius is between $R^{-n}\rho $ and $R^{-(n + 1)}\rho $ , where $\rho $ is the radius of $B_0$ . Define

$$ \begin{align*} \mathcal{A}_{m,n} (B):= \{ A\in \mathcal{S}(B,R^{n - m + 1}): \text{ there exists } C\in\mathcal{C}(n) \text{ such that } A\cap C\neq\emptyset\}. \end{align*} $$

By equation (7) and the definition of $\mathcal {C}(n)$ , we have

$$ \begin{align*} \sum_{C\in\mathcal{C}(n)} (R^{-(n+1)}\rho)^c \leq (\beta R^{-m}\rho)^c; \end{align*} $$

that is,

$$ \begin{align*} \#\mathcal{C}(n) \leq \beta^c (R^{n - m+1})^c. \end{align*} $$

This implies that for $m>n$ , we get $\#\mathcal {C}(n)<1$ , that is, $\mathcal {C}(n) = \emptyset $ and, in turn, $\mathcal {A}_{m,n} (B) = \emptyset $ . Now we construct a Cantor set $\mathcal {K}\in E$ by removing the collections $\mathcal {A}_{m,n}(B)$ ( $m\leq n$ , $B\in \mathcal {B}_m$ ). Then to complete the proof, we just need to show that

$$ \begin{align*} \#\mathcal{A}_{m,n}(B) \leq r_{m,n} = R^{c(n - m + 1)} \end{align*} $$

for every $B\in \mathcal {B}_m$ . By the doubling property, each element of $\mathcal {C}(n)$ intersects a bounded number of elements of $\mathcal {S}(B,R^{n - m + 1})$ . Thus,

$$ \begin{align*} \#\mathcal{A}_{m,n}(B) \leq C_3 \#\mathcal{C}(n)\leq C_3 \beta^c (R^{n - m+1})^c \end{align*} $$

for some constant $C_3> 0$ depending on R. By letting $\beta $ be sufficiently small (that is, letting q be sufficiently large), we can get $C_3 \beta ^c \leq 1$ and therefore

$$ \begin{align*} C_3 \beta^c (R^{n - m+1})^c \leq R^{(n - m + 1)c}, \end{align*} $$

which completes the proof.

Finally, we briefly mention how the second statement of Theorem 1.5 follows from the first statement. If E is 1 Cantor winning then, by the first part of the theorem, it is c potential winning for any $c>0$ . The statement follows as an immediate corollary of Proposition 1.3.

3.3. Schmidt’s winning and Cantor winning

We now move our attention to the links between Schmidt’s original game and generalized Cantor set constructions. Thanks to Theorem 3.4, in $\delta $ Ahlfors regular spaces, we can treat the notions of Cantor winning sets and potential winning sets as interchangeable. The latter notion will be more convenient in this section. We prove Theorem 1.9 as a special case of a more general result.

To see that Theorem 1.9 follows from Theorem 3.8, notice that $\mathbb {R}^N$ is N Ahlfors regular, and that very strong $(\alpha ,\beta )$ winning sets are $(\alpha ,\beta )$ winning. When E is $\varepsilon $ Cantor winning, the desired statement follows upon taking $c_0=N(1-\varepsilon )$ in the above. 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 $(c,\widetilde \beta )$ potential game for the set E, where $\widetilde \beta = (\alpha \beta )^q$ for some large $q\in \mathbb {N}$ . We want to show that Alice has a winning strategy in the $(\alpha ,\beta )$ very strong game. Given any sequence of Bob’s choices in the $(\alpha ,\beta )$ game, say $B_0,\ldots ,B_{qk}$ , we can consider the corresponding sequence of choices in the $(c,\widetilde \beta )$ potential game which correspond to Bob choosing the balls $B_0,B_q,\ldots ,B_{qk}$ . Alice’s response to this sequence of moves is to delete a collection $\mathcal {A}(B_{qk})$ . Now Alice’s corresponding response in the $(\alpha ,\beta )$ very strong game will be: if Bob has made the sequence of moves $B_0,\ldots ,B_m$ , then she will choose her ball $A_{m+1} \subseteq B_m$ so as to minimize the value at $A_{m+1}$ of the potential function

$$ \begin{align*} \varphi_m(A) := \sum_{\substack{k\in\mathbb{N} \\ qk \leq m}} \sum_{\substack{C\in\mathcal{A}(B_{qk}) \\ C\cap A \neq \varnothing}} \mathrm{diam}^c(C). \end{align*} $$

We will now prove by induction that the inequality

$$ \begin{align*} \varphi_m(A_{m+1}) \leq (\varepsilon \cdot \mathrm{rad}(B_m))^c \end{align*} $$

holds for all m, where $\varepsilon> 0$ will be chosen later (small but independent of $\widetilde \beta $ ). Suppose that it holds for some m and let $B_{m + 1}$ be Bob’s next move. Define $r = \alpha \cdot \mathrm {rad}(B_{m+1})\geqslant \alpha ^2\beta \cdot \mathrm {rad}(B_m)$ . By Proposition 2.19, there exists a set

$$ \begin{align*} \mathcal{D}:= \{B(x_i,r)\subseteq B_{m+1}: i\in \{1,\ldots,I\}\}, \end{align*} $$

which satisfies the following conditions: $I = \#\mathcal {D} \gg \alpha ^{-\delta }$ , and for any distinct values $1\leqslant i\neq j\leqslant I$ , one has $d(x_i,x_j)\geqslant 3r$ , i.e. all balls in $\mathcal {D}$ are separated by distances of at least r.

By the induction hypothesis, for all $C\in \bigcup _{qk\leq m} \mathcal {A}(B_{qk})$ such that $C\cap A_{m+1} \neq \emptyset $ , we have $\mathrm {diam}(C) \leq \varepsilon \cdot \mathrm {rad}(B_m)$ . So by letting $\varepsilon $ be small enough (that is, $\varepsilon \leq \alpha ^2\beta $ ), we can guarantee that all balls in $\mathcal {D}$ make disjoint contributions to $\varphi _m(A_{m+1})$ . In other words,

$$ \begin{align*} \sum_{D\in \mathcal{D}} \varphi_m(D) \leqslant \varphi_m(A_{m+1}). \end{align*} $$

By choosing $A_{m + 2}$ to be the ball from $\mathcal {D}$ which minimizes the function $\varphi _m$ , we get

$$ \begin{align*} \alpha^{-\delta} \varphi_m(A_{m + 2}) \ll \varphi_m(A_{m+1}). \end{align*} $$

By choosing $\gamma $ small enough, the inequality $\alpha < \gamma (\alpha \beta )^{c/\delta }$ implies that

$$ \begin{align*} \varphi_m(A_{m + 2}) \leq (\alpha\beta)^c \varphi_m(A_{m+1})/2 = \tfrac12 (\varepsilon \cdot\mathrm{rad}(B_{m+1}))^c. \end{align*} $$

Note that for the base of induction, $m=-1$ , the last estimate is straightforward, since $\varphi _m(A) = 0$ for all balls A, and we can set $A_0=B_0$ .

When we consider $\varphi _{m+1}(A_{m+2})$ , there may be a new term coming from a new collection $\mathcal {A}(B_{m + 1})$ . By the definition of a $(c,\widetilde \beta )$ potential game, that term is at most $(\widetilde \beta \cdot \mathrm {rad}(B_{m+1}))^c$ , and thus by choosing q large enough, we can guarantee that

$$ \begin{align*} \varphi_{m+1}(A_{m+2}) \leqslant (\varepsilon (\alpha\beta)^{m+1}\rho)^c. \end{align*} $$

This finishes the induction.

Since Alice followed the winning strategy for $(c,\widetilde \beta )$ potential game, we have either

$$ \begin{align*} \bigcap_{k=1}^\infty B_{kq} \subseteq E\quad\text{or}\quad \bigcap_{k=1}^\infty B_{kq} \subseteq \bigcup_{k=1}^\infty \bigcup_{A\in\mathcal{A}(B_{kq})} A. \end{align*} $$

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 $A\in \mathcal {A}(B_{kq})$ for some k such that every ball $B_m$ and, in turn, every ball $A_m$ has non-empty intersection with A. However, in that case, we have $\varphi _m(A_{m+1}) \geq \mathrm {diam}^c(A)$ , which contradicts the fact that $\varphi _m(A_{m+1}) \to 0$ as $m\to \infty $ .

As a corollary of Theorem 3.8, we get a general form of Theorem 1.10.

Proof. Let $E\subseteq X$ be a $c_0$ potential winning set for some $c_0<\delta $ . Fix $c_0 < c < \delta $ . By Theorem 3.8, there exists $\gamma> 0$ such that for all $\alpha ,\beta>0$ with $\alpha < \gamma (\alpha \beta )^{c/\delta }$ , E is $(\alpha ,\beta )$ very strong winning. In particular, for all $\beta> 0$ , there exists $\alpha> 0$ such that E is $(\alpha ,\beta )$ very strong winning. Hence, we can apply Proposition 2.6 which says that $X\setminus E$ is not weak winning (and thus does not contain a weak winning set). So every potential winning set intersects every weak winning set non-trivially.

Now by contradiction, suppose that T is a weak winning set such that ${\dim _H}(E\cap T) < \delta $ . This implies that for any $c \in ({\dim _H}(E\cap T), \delta )$ and for any $\varepsilon>0$ , one can construct a cover of $E\cap T$ by a countable collection $(A_i)_{i\in \mathbb {N}}$ of balls such that

$$ \begin{align*} \sum_{i=1}^\infty \mathrm{diam}^c(A_i) <\varepsilon. \end{align*} $$

In other words, this cover can be chosen by Alice in a single move of the c potential game, so $X\setminus (E\cap T)$ is potential winning. So by the intersection property of potential winning sets, $E\setminus T = E\cap (X\setminus (E\cap T))$ is potential winning. However, then $X\setminus T$ is a potential winning set whose complement is weak winning, contradicting what we have just shown.

3.4. Proof of Theorem 1.11

In this section, we prove that $1/2$ winning implies absolute winning for subsets of the real line. The same notation, terminology, and observations as in [Reference Schmidt37] are employed. Recall that $\mathcal {B}(\mathbb {R})$ stands for the set of all closed intervals in $\mathbb {R}$ .

In Schmidt’s $(\alpha , \beta )$ game, the method Alice uses to determine where to play her intervals $A_i$ is called a strategy. Formally, a strategy $F:=\{f_1, f_2, \ldots \}$ is a sequence of functions $f_i: \mathcal {B}(\mathbb {R})^i \rightarrow \mathcal {B}(\mathbb {R})$ satisfying $\mathrm {rad}(f_i(B_0, B_1, \ldots , B_{i-1})) = \alpha \cdot \mathrm {rad}(B_{i-1})$ and $f_i(B_0, B_1, \ldots , B_{i-1}) \subseteq B_{i-1}$ . A set E is a winning set for the $(\alpha , \beta )$ game if and only if there exists a strategy determining where Alice should place her intervals $A_{i}:=f_i(B_0, B_1, \ldots , B_{i-1})$ , so that however Bob chooses his intervals $B_{i} \subseteq A_{i}$ , the intersection point $\bigcap _{i} B_i$ lies in E. Such a strategy is referred to as a winning strategy (for E). Schmidt [Reference Schmidt37, Theorem $7$ ] observed that the 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 to depend upon the position of Bob’s immediately preceding interval, not on the entirety of the game so far. To be precise, a winning strategy $F:=(f_1, f_2, \ldots )$ is positional if there is a function $f_0: \mathcal {B}(X) \rightarrow \mathcal {B}(\mathbb {R})$ such that each function $f_i \in F$ satisfies $f_i(B_0, \ldots , B_{i-1})=f_0(B_{i-1})$ . 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 $A_i=F(B_{i-1}) = f_0(B_{i-1})$ where F is a given positional winning strategy for Alice and $f_0$ is the function witnessing F’s positionality.

We refer to a sequence $\{B_0, B_1, \ldots \}$ of intervals as an F-chain if it consists of the moves Bob has made during the playing out of an $(\alpha , \beta )$ game with target set E in which Alice has followed the winning strategy F. By definition, we must have $\bigcap _{i} B_i \, \subseteq E$ . Furthermore, we say a finite sequence $\{B_0, \ldots B_{n-1}\}$ is an $F_n$ -chain if there exist $B_{n}, B_{n+1}, \ldots $ for which the infinite sequence $\{B_0, \ldots B_{n-1}, B_{n}, B_{n+1} \ldots \}$ is an F-chain. One can readily verify (see [Reference Schmidt37, Lemma $1$ ]) that if $\{B_0, B_1, \ldots \}$ is a sequence of intervals such that for every $n \in \mathbb {N}$ , the finite sequence $\{B_0, B_1, \ldots B_{n-1}\}$ is an $F_n$ -chain, then $\{B_0, B_1, \ldots \}$ is an F-chain.

Outline of the strategy. We begin with some interval $b_0 \subseteq \mathbb {R}$ of length $2\cdot \mathrm {rad}(b_0)$ , some $\varepsilon \in (0, 1)$ , some R sufficiently large, and a $1/2$ winning set $E \subseteq \mathbb {R}$ . Fix $\alpha =1/2$ and $\beta =2/R$ , so that $\alpha \beta = 1/R$ . We will construct for every $R \geq 10^{1/(1-\varepsilon )}$ , a local $(b_0, R, 10)$ Cantor set $\mathcal {K}$ lying inside E. This is sufficient to prove that E is $1$ Cantor winning.

By assumption, the set E is $(1/2, 2/R)$ winning; thus, there exists a positional winning strategy F such that however we choose to place Bob’s intervals $B_i$ in the game, the placement of Alice’s subsequent intervals $A_i=F(B_{i-1})$ guarantees that the unique element of $\bigcap _i B_i$ will fall inside E. We will, in a sense, play as Bob in many simultaneous $(1/2, 2/R)$ 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 $\mathcal {K}$ comprises the construction of subcollections $\mathcal {B}_n$ of intervals in $1/{R}^n \mathcal {B}_0$ for each $n \in \mathbb {N}$ , where $\mathcal {B}_0:=\{ b_0\}$ . Construction will be carried out iteratively in such a way that for any interval $b \in \mathcal {B}_n$ , there exists an $F_n$ -chain $\{B_0, B_1, \ldots , B_{n-1}\}$ satisfying $b \subseteq B_{n-1}$ . Moreover, for each ancestor $b'$ of b (that is, the unique interval $b' \in \mathcal {B}_k$ with $b \subseteq b'$ , for some given $0 \leq k < n$ ), the subsequence $\{B_0, B_1, \ldots , B_{k-1}\} $ will coincide with the $F_k$ -chain constructed for $b'$ satisfying $b' \subseteq B_{k-1}$ from the earlier inductive steps. We write $B_n(b)$ if the dependence of the interval $B_n$ on an interval b is not clear from context. In view of the previous discussion, upon completion of the iterative procedure, it follows that for any sequence of intervals $\{b_i\}_{i \in \mathbb {N}}$ with $b_i \in \mathcal {B}_i$ , the associated sequence $\{B_0, B_1, \ldots \}$ must be an F-chain satisfying $b_i \subseteq B_{i-1}$ . In this way, for every point $\mathbf {x}$ in the Cantor set $\mathcal {K}$ , we will establish the existence of an F-chain $\{B_0, B_1, \ldots \}$ for which $\{\mathbf {x}\}= \bigcap _{i}B_i$ ; namely, we may simply choose the F-chain associated with a sequence of intervals $\{b_i\}_{i \in \mathbb {N}}$ with $b_i \in \mathcal {B}_i$ satisfying $\bigcap _{i \in \mathbb {N}}b_i=\{\mathbf {x}\}$ . It follows that $\mathbf x$ must fall within E and, since $\mathbf {x}\in \mathcal {K}$ was arbitrary, that $\mathcal {K} \subseteq E$ as required.

We begin the first step of the inductive process. Given $\mathcal {B}_0=\{b_0\}$ , the idea is to construct the collection $\mathcal {B}_1 \subseteq 1/R \mathcal {B}_0$ in such a way that for every $b \in \mathcal {B}_1$ , there exists a place Bob may play his first interval $B_0$ in a $(1/2, 2/R)$ game so that Alice’s first interval $A_1:=F(B_0)$ (as specified by the winning strategy F) contains b; that is, we construct $\mathcal {B}_1$ so that for every $b \in \mathcal {B}_1$ , there exists an $F_1$ -chain $\{ B_0 \}$ satisfying $b \subseteq A_1 := F(B_0)$ . For technical reasons, in practice, we will ensure the stronger condition that $b \subseteq (1-2/R)A_1$ , where $\kappa B$ denotes the interval with the same center as B but with radius multiplied by $\kappa $ .

Recall that for his opening move in a $(1/2, 2/R)$ game, Bob may choose any interval in $\mathbb {R}$ . In particular, he may choose any interval of radius $2 \cdot \mathrm {rad}(b_0)$ with center lying in $b_0$ , therefore covering $b_0$ . Let $\mathcal {W}_1(b_0) \subseteq \mathcal {B}(\mathbb {R})$ be the set of all ‘winning’ intervals for Alice (as prescribed by the winning strategy F) corresponding to any opening interval of this kind that Bob might choose to begin the game with; that is, let

$$ \begin{align*} \mathcal{W}_1(b_0): \: = \: \{F(B): \: B \in \mathcal{B}(\mathbb{R}),\: \mathrm{rad}(B) = 2\cdot\mathrm{rad}(b_0), \, \mathrm{cent}(B) \in b_0 \}. \end{align*} $$

Note that this set could either be uncountable or, since for each $A \in \mathcal {W}_1(b_0)$ , we have $\mathrm {rad} (A)=\alpha \cdot (2\cdot \mathrm {rad}(b_0)) = \mathrm {rad}(b_0)$ , it could simply coincide with the singleton $\mathcal {B}_0$ . Moreover, since $\alpha =1/2$ , we must have for every interval B that $\mathrm {cent}(B)$ is contained in $F(B)$ , and so $\mathcal {W}_1(b_0)$ is a cover of $b_0$ . This property is unique to the case $\alpha =1/2$ .

We continue the proof of Theorem 1.11. Take $b=b_0$ , $\varepsilon =\mathrm {rad}(b_0)/R$ , and $\mathcal {W}= \mathcal {W}_1$ in Lemma 3.10. Then there exists a subcollection $\mathcal {W}_1^\ast :=\{A_1^L, A_1^R\}$ of $\mathcal {W}_1$ of cardinality at most two covering $b_0$ except for an open interval $I(b_0)$ of length at most $\mathrm {rad}(b)/R$ . It is possible that $I(b_0)$ is empty. If $\mathcal {W}_1^\ast $ contains only one interval, say $\mathcal {W}_1^\ast = \{ b'\}$ , then set $A_1^L=A_1^R=b'$ . Let $B_0^L$ and $B_0^R$ be two choices for Bob’s opening move with centers in $b_0$ that satisfy $A_1^L=F(B_0^L)$ and $A_1^R=F(B_0^R)$ ; such choices exist by the definition of $\mathcal {W}_1$ . If we had $A_1^L=A_1^R$ , then assume $B_0^L=B_0^R$ .

We now discard the ‘bad’ intervals that will not appear in the generalized Cantor set $\mathcal {K}$ . Let

$$ \begin{align*}\mathbf{Bad}_1:= \bigg\{b' \in \frac{1}{R}\mathcal{B}_0: b'\cap I(b_0) \neq \emptyset\;\text{or}\; b' \not\subseteq ( (1-2/R)A_1^L \cup (1-2/R)A_1^R ) \bigg\}. \end{align*} $$

Since any interval of length $\mathrm {rad}(b_0)/R$ may intersect at most two intervals from ${1}/{R}\mathcal {B}_0$ , and there are at most five intervals (of length at most $\mathrm {rad}(b_0)/R$ ) that need to be intersected by $b'$ for it to be in $\mathbf {Bad}_1$ , it is immediate that $\#\mathbf {Bad}_1 \leq 10$ . It follows that $\mathcal {B}_1:= {1}/{R}\mathcal {B}_0 \setminus \mathbf {Bad}_1$ satisfies the required conditions for our generalized Cantor set. For each $b \in \mathcal {B}_1$ , we associate one of the $F_1$ -chains $\{ B_0^L \}$ or $\{ B_0^R \}$ , depending on whether b is a subset of $A_1^L$ or $A_1^R$ , respectively. If b lies in both $A_1^L$ and $A_1^R$ (which is possible if $I(b_0)$ was empty), then we will assign as a precedent whichever of the intervals $B_0^L$ or $B_0^R$ has the ‘leftmost’ center. This completes the first step of the inductive construction process.

Assume now that we have constructed the collection $\mathcal {B}_i$ so that for each $b \in \mathcal {B}_i$ , we have an $F_i$ -chain $\{B_0, \ldots , B_{i-1} \}$ with $\mathrm {rad}(B_{i-1})=2R \cdot \mathrm {rad}(b)$ and $b \subseteq (1-2/R)F(B_{i-1})$ . We outline the procedure to construct the subsequent collection $\mathcal {B}_{i+1}$ and the corresponding $F_{i+1}$ -chains.

Fix $b \in \mathcal {B}_i$ with associated $F_i$ -chain $\{B_0, \ldots , B_{i-1} \}$ . Assume Alice has just played her ith move $A_i:=F(B_{i-1})$ , an interval of radius $R \cdot \mathrm {rad}(b)$ , in a $(1/2, 2/R)$ game corresponding to this $F_i$ -chain. Bob may then choose any interval of radius $2 \cdot \mathrm {rad}(b)$ inside $A_i$ . Since $b \subseteq (1-2/R)A_i$ , Bob is free to choose as $B_i$ any interval of radius $2 \cdot \mathrm {rad}(b)$ centered in b; every such choice $b'$ will constitute a legal move since $\mathrm {rad}(b')=2/R \mathrm {rad}(A_i)$ and $b \subseteq (1-2/R)A_i$ together imply $b' \subseteq A_i$ . In particular, the collection

$$ \begin{align*} \mathcal{W}_{i+1}(b): = \{F(b'): b' \in \mathcal{B}(\mathbb{R}), \mathrm{rad}(b') = 2\cdot\mathrm{rad}(b_0)/R^i, \mathrm{cent}(b') \in b \} \end{align*} $$

is a cover of b and consists of candidate ‘winning’ moves that could be played by Alice after any legal move $B_i$ (with center in b) played by Bob. As before, we apply Lemma 3.10 to find a subset of $\mathcal {W}_{i+1}(b)$ of cardinality at most two that covers b except for an open subinterval $I(b)$ of length at most $\mathrm {rad}(b)/R$ . If the subcover contains two intervals, denote them by $A_{i+1}^L(b)$ and $A_{i+1}^R(b)$ ; otherwise, if the subcover consists of a single ball $b'$ , then let $A_{i+1}^L(b)=A_{i+1}^R(b)=b'$ . Denote by $B_{i}^L(b)$ and $B_{i}^R(b)$ some choice of intervals centered in b for which $A_{i+1}^L(b)=F(B_{i}^L(b))$ and $A_{i+1}^R(b)=F(B_{i}^R(b))$ , respectively. Define

$$ \begin{align*}\mathbf{Bad}_{i+1}(b):= \bigg\{&b' \in \frac{1}{R}\{b\}: b'\cap I(b) \neq \emptyset\;\text{or}\; b'\\ &\not\subseteq ( (1-2/R)A_{i+1}^L(b) \cup (1-2/R)A_{i+1}^R(b) ) \bigg\} \end{align*} $$

and let $\mathcal {B}_{i+1}(b):= {1}/{R}\{b\} \setminus \mathbf {Bad}_{i+1}(b)$ . By the same arguments as before, it is clear that $\#\mathbf {Bad}_{i+1}(b) \leq 10$ for each $b \in \mathcal {B}_i$ .

Finally, for each $b' \in \mathcal {B}_{i+1}(b)$ , we assign the $F_{i+1}$ -chain $\{B_0, \ldots , B_{i-1}, B_{i}^L(b) \}$ or $\{B_0, \ldots , B_{i-1}, B_{i}^R(b_i) \}$ depending on whether $b'$ lies in $A_{i+1}^L(b_i)$ or $A_{i+1}^R(b_i)$ , respectively. Again, if $I(b)$ was empty and b was contained in both $A_{i+1}^L(b)$ and $A_{i+1}^R(b)$ , then we choose whichever of $B_{i}^L(b)$ and $B_{i}^R(b)$ has the leftmost center as a convention.

Upon setting

$$ \begin{align*}\mathcal{B}_{i+1}:= \bigcup_{b \in \mathcal{B}_i}\mathcal{B}_{i+1}(b),\end{align*} $$

we see that the conditions of our generalized Cantor set $\mathcal {K}(b_0,R,10)$ are satisfied. This completes the inductive procedure.

3.5. Cantor rich and Cantor winning

In this section, we prove the following statement which implies Theorem 1.13.

Theorem 3.11. Let $(X,\mathcal {S},U,f)$ be a splitting structure, let $B_0\subseteq X$ be any ball, and let $\varepsilon>0$ be a real number. If $E\subseteq X$ is $\varepsilon $ Cantor winning in $B_0$ , then it is $(B_0,M)$ Cantor rich with

$$ \begin{align*} M = 4^{{1}/{\varepsilon}}. \end{align*} $$

Conversely, assume that X is doubling and let $M\geqslant 4$ be a real number. If E is $(B_0,M)$ Cantor rich, then it is $\varepsilon $ Cantor winning in $B_0$ , for any $\varepsilon $ for which there exists $R\in U$ such that

(25) $$ \begin{align} M < f(R) \leq 4^{{1}/{\delta\varepsilon}}. \end{align} $$

In particular, if $X=\mathbb {R}$ with the standard splitting structure, then E is M Cantor rich if and only if it is $\varepsilon $ Cantor winning with

$$ \begin{align*} \varepsilon = \log_4 \bigg(\frac{1}{\lfloor M\rfloor +1}\bigg). \end{align*} $$

Proof. Assume E is $\varepsilon $ Cantor winning in $B_0$ . Then, by Corollary 3.3, for every $2\leq R\in U$ , there exists a $(B_{0},R,f(R)^{1-\varepsilon })$ Cantor set contained in E. Fix an integer R such that $f(R)>M$ and a real number $y>0$ . Since ${4}/{f(R)}^{\varepsilon }<1$ , we can choose $\ell>0$ such that ${4}/{f(R)}\cdot ({4}/{f(R)}^{\varepsilon })^{\ell }<y$ . Consider a $(B_{0},R^{\ell },f(R)^{\ell (1-\varepsilon )})$ Cantor set contained in E and denote its Cantor sequence by $\{ \mathcal {B}_{n}\} _{n=0}^{\infty }$ . Define another Cantor sequence $\{ \mathcal {B}_{n}'\} _{n=0}^{\infty }$ by setting $\mathcal {B}^{\prime }_{\ell k}=\mathcal {B}_{k}$ for all $k\geq 0$ and $\mathcal {B}_{n}'={1}/{R}\mathcal {B}_{n-1}$ whenever $n\neq \ell k$ . Then the limit set

$$ \begin{align*} \bigcap_{n=1}^\infty \bigcup_{B\in \mathcal{B}_{n}'} B \end{align*} $$

is a $(B_{0},R,(r_{m,n})_{0\leq m\leq n})$ generalized Cantor set, with $r_{\ell (k-1), \ell k}=f(R)^{\ell (1-\varepsilon )}$ for all $k\geq 0$ , and $r_{m,n}=0$ for all other pairs. Then, for any $k>0$ and $n = \ell k$ , we get

$$ \begin{align*} \sum_{m=0}^{\ell k}\bigg(\frac{4}{f(R)}\bigg)^{\ell k-m +1}r_{m,\ell k}=\bigg(\frac{4}{f(R)}\bigg)^{\ell+1}f(R)^{\ell(1-\varepsilon)}=\frac{4}{f(R)}\cdot \bigg(\frac{4}{f(R)^{\varepsilon}}\bigg)^{\ell}<y. \end{align*} $$

For every $n\neq \ell k$ , we have

$$ \begin{align*} \sum_{m=0}^{n}\bigg(\frac{4}{f(R)}\bigg)^{n-m+1}r_{m,n}=0<y. \end{align*} $$

For the other direction, suppose that E is $(B_0,M)$ Cantor rich. Let $c_0 = \delta (1-\varepsilon )$ . By Theorem 3.4, it is enough to show that E is $c_0$ potential winning in $B_0$ . We argue as in the proof of Theorem 3.4. Choose R such that $M < f(R) \leq 4^{1/\varepsilon }$ and fix any $0 < \beta \leq {1}/{R}$ , $c> c_0$ . Fix a small enough $y>0$ whose precise value will be determined later. By the definition of Cantor rich, E contains some $(B_0,R,\mathbf {r})$ Cantor set $\mathcal {K}$ , where $\mathbf {r}$ satisfies

$$ \begin{align*} \sum_{m=0}^{n} \bigg(\frac{4}{f(R)}\bigg)^{n-m+1}r_{m,n} < y \end{align*} $$

for every $n\in \mathbb {N}$ . In particular, for every $m,n\in \mathbb {N}$ , we have

$$ \begin{align*} r_{m,n} < y\bigg(\frac{f(R)}{4}\bigg)^{n-m+1} \leq yR^{(n-m+1)c_0}. \end{align*} $$

We now describe a strategy for Alice to win the $(c,\beta )$ potential game on $A_\infty (B_0)$ . As in the proof of Theorem 3.4, let $\rho $ denote the radius of $B_0$ , let $D_k$ denote Bob’s kth move, and for each $k\in \mathbb {N}$ , let $m = m_k\in \mathbb {N}$ denote an integer such that $\beta \cdot \mathrm {rad}(D_k) < R^{-m}\rho \leq \mathrm {rad}(D_k)$ . The inequality $\beta \leqslant 1/R$ ensures that such m exists. Then Alice’s strategy is as follows: on turn k, remove all elements of the set

$$ \begin{align*} \bigcup_{B\in \mathcal{B}_m} \bigcup_{n\geq m} \mathcal{A}_{m,n}(B) \end{align*} $$

that intersect Bob’s current choice. This strategy, if executable, will make Alice win since the intersection of Bob’s balls will satisfy $\bigcap _{k \in \mathbb {N}}D_k \subseteq \mathcal {K} \subseteq E$ . To show that it is legal, we need to show that

(26) $$ \begin{align} \sum_{\substack{B\in \mathcal{B}_m \\ B\cap D_k \neq \varnothing}} \sum_{n \geq m} r_{m,n} (R^{-(n + 1)}\rho)^c \leq (\beta\cdot \mathrm{rad}(D_k))^c. \end{align} $$

This is enough because elements of $\mathcal {A}_{m,n}(B)$ all have radius $R^{-(n + 1)}\rho $ . Since the elements of $\mathcal {B}_m$ are disjoint and have radius $R^{-m} \rho \asymp _\beta \mathrm {rad}(D_k)$ , the number of them that intersect $D_k$ is bounded by a constant depending only on $\beta $ . Call this constant $C_1$ . Then the left-hand side of equation (26) is less than

$$ \begin{align*} C_1 \sum_{n\geq m} yR^{(n - m + 1)c_0} \bigg(R^{-(n + 1)}\rho\bigg)^c &= C_1 (R^{-m} \rho)^c y\sum_{\ell = 1}^\infty R^{(c_0 - c)\ell}\\ &\leqslant C_1(\mathrm{rad}(D_k))^c\frac{R^{(c_0 - c)}}{1 - R^{(c_0 - c)}} y. \end{align*} $$

By choosing y so that

$$ \begin{align*} y < \frac{\beta^c (1 - R^{(c_0 - c)})}{C_1R^{(c_0 - c)}}, \end{align*} $$

we guarantee that the move is legal.