Badly approximable points on planar curves and winning

For any i,j>0 with i+j =1, let Bad(i,j) denote the set of points (x,y) \in R^2 such that max \{ ||qx||^{1/i}, \, ||qy||^{1/j} \}>c/q for some positive constant c = c(x,y) and all q in N. We show that \Bad(i,j) \cap C is winning in the sense of Schmidt games for a large class of planar curves C, namely, everywhere non-degenerate planar curves and straight lines satisfying a natural Diophantine condition. This strengthens recent results solving a problem of Davenport from the sixties. In short, within the context of Davenport's problem, the winning statement is best possible. Furthermore, we obtain the inhomogeneous generalizations of the winning results for planar curves and lines and also show that the inhomogeneous form of Bad(i,j) is winning for two dimensional Schmidt games.


Introduction
A real number x is said to be badly approximable if there exists a positive constant c(x) such that qx > c(x) q −1 ∀ q ∈ N .
Here and throughout · denotes the distance of a real number to the nearest integer. It is well known that the set Bad of badly approximable numbers is of Lebesgue measure zero but of maximal Hausdorff dimension; i.e. dim Bad = 1. In higher dimensions there are various natural generalizations of Bad. Restricting our attention to the plane R 2 , given a pair of real numbers i and j such that 0 < i, j < 1 and i + j = 1 , (1.1) Denote by Bad(i, j) the set of (i, j)-badly approximable points in R 2 . In the case i = j = 1/2, the set under consideration is the standard set of simultaneously badly approximable points. It easily follows from classical results in the theory of metric Diophantine approximation that Bad(i, j) is of (two-dimensional) Lebesgue measure zero. Regarding dimension, it was shown by Schmidt [14] in the vintage year of 1966 that dim Bad( 1 2 , 1 2 ) = 2. In fact, Schmidt proved the significantly stronger statement that Bad( 1 2 , 1 2 ) is winning in the sense of his now famous (α, β)-games -see §2.1. Almost fourty years later it was proved in [13] that dim Bad(i, j) = 2 and just recently the first author in [2] has shown that Bad(i, j) is in fact winning. The latter implies that any countable intersection of Bad(i, j) sets is of full dimension and thus provides a clean and direct proof of Schmidt's Conjecture -see also [1,3]. Now let C be a planar curve. Without loss of generality, we assume that C is given as a graph for some function f defined on an interval I ⊂ R. Throughout we will assume that f ∈ C (2) (I), a condition that conveniently allows us to define the curvature. Motivated by a problem of Davenport [11, p.52] from the sixties, the following statement regarding the intersection of Bad(i, j) sets with any curve C that is not a straight line segment has recently been established [5,6].
Theorem A Let (i t , j t ) be a countable number of pairs of real numbers satisfying (1.1) and suppose that lim inf t→∞ min{i t , j t } > 0 . (1.2) Let C := C f be a C (2) planar curve that is not a straight line segment. Then Bad(i t , j t ) ∩ C = 1 .
The theorem implies that there are continuum many points on the parabola V 2 := {(x, x 2 ) : x ∈ R} that are simultaneously badly approximable in the Bad( 1 2 , 1 2 ) sense and thus provides a solution to the specific problem raised by Davenport in [11]. It is worth mentioning that a consequence of [7,Theorem 1] is that the set Bad(i, j) ∩ C is of zero (induced) Lebesgue measure on C. Thus, the fact that it is a set of full dimension is not trivial.
The condition imposed on C is natural since the statement is not true for all lines. Indeed, let L a denote the vertical line parallel to the y-axis passing through the point (a, 0) in the (x, y)-plane. Then, it is easily verified, see [3, §1.3] for the details, that Bad(i, j) ∩ L a = ∅ for any a ∈ R satisfying lim inf q→∞ q 1/i qa = 0 . On the other hand, if the lim inf is strictly positive then dim(Bad(i, j) ∩ L a ) = 1. This is much harder to prove and is at the heart of the original proof of Schmidt's Conjecture established in [3]. Subsequently, it was shown in [1] that Bad(i, j) ∩ L a is winning. The following non-trivial extension of the full dimensional result to non-vertical lines has recently been established in [5]. Bad(i t , j t ) ∩ L a,b = 1 .
Both Theorem A and Theorem B should be true without the lim inf condition (1.2). Indeed, as pointed out in Remark 2 of [5, §1.2], it is very tempting and not at all outrageous to assert that Bad(i, j) ∩ C is winning at least on the part of the curve that is genuinely curved. If true it would imply Theorem A without assuming (1.2). In short, this is precisely what we show in this paper. We also obtain a winning statement for non-vertical lines that not only implies Theorem B without assuming (1.2) but replaces condition (1.3) by a weaker and essentially optimal Diophantine condition. Furthermore, by making use of a simple idea introduced in [8] that provides a natural mechanism for generalizing homogenous badly approximable statements to the inhomogeneous setting, we establish the inhomogeneous generalization of our winning results. The same idea is also exploited to prove that inhomogeneous Bad(i, j) is winning.

Inhomogeneous Bad(i, j) and our results
For θ = (γ, δ) ∈ R 2 , let Bad θ (i, j) denote the set of points (x, y) ∈ R 2 such that It is straight forward to deduce that Bad θ (i, j) is of measure zero from the inhomogeneous version of Khintchine's theorem with varying approximating functions in each co-ordinate. Surprisingly, the fact that dim Bad θ (i, j) = 2 is very much a recent development -see [10] for the symmetric i = j = 1/2 case and [8] for the general case.
One of the main goals of this paper is to prove the following full dimension statement which not only implies the inhomogeneous analogue of Theorem A but totally removes the lim inf condition (1.2). Theorem 1.1 Let (i, j) be a pair of real numbers satisfying (1.1) and let C := C f be a planar curve such that f ∈ C (2) (I) and that f ′′ (x) = 0 for all x ∈ I. Then, for any θ = (γ, δ) ∈ R 2 we have that Bad θ (i, j) ∩ C is a winning subset of C. Remark 1. The condition f ′′ (x) = 0 is often referred to as non-degeneracy at x. Note that if f ∈ C (2) (I) and f ′′ (x) = 0 for some point x ∈ I, then, by continuity, there exists an interval I * ⊂ I such that f ′′ (x) = 0 for all x ∈ I * . In other words, if the curve C f is not a straight line segment then it is always possible to find an arc of C f that is non-degenerate everywhere. Thus, Theorem 1.1 with I = I * and θ = (0, 0) implies Theorem A without assuming (1.2). Of course this makes use of the well know fact that any winning set is of full dimension and that any countable intersection of winning sets is again winning.
Remark 2. Given a subset X ⊂ R 2 , let π(X) denote the projection of X onto the x-axis. Regarding Theorem 1.1, we actually prove that π(Bad θ (i, j) ∩ C) is a 1/2-winning subset of I. In other words, for the projected set we prove α-winning with the best possible winning constant; i.e. α = 1/2. Now, since the projection map π is bi-Lipschitz on C and the image of a winning set under a bi-Lipschitz map is again winning [9,Proposition 5.3], it trivially follows that Bad θ (i, j) ∩ C is an α 0 -winning subset of C for some α 0 ∈ (0, 1/2]. The actual value of α 0 is dependent on the Lipschitz constant κ associated with (π| C ) −1 .
In fact, if we use the maximum norm on R 2 it is possible to obtain the winning statement with α 0 = 1/2. Essentially, if κ > 1 we consider the projection of Bad θ (i, j) ∩ C onto the y-axis rather then the x-axis.
For straight lines we prove the following counterpart statement. Then, for any θ = (γ, δ) ∈ R 2 we have that Bad θ (i, j) ∩ L a,b is a winning subset of L a,b . Moreover, if a ∈ Q the statement is true with ǫ = 0 in (1.4).
Remark 3. As with curves, the theorem is actually deduced on showing that the projected set π(Bad θ (i, j) ∩ L a,b ) is an 1/2-winning subset of R.
Remark 4. The Diophantine condition (1.4) imposed in the theorem is essentially optimal since To see that this is the case, assume for the moment that Bad(i, j) ∩ L a,b is nonempty. Then there exists some point x ∈ R such that (x, ax + b) ∈ Bad(i, j). In terms of the equivalent dual form representation of Bad(i, j) -see [3, §1.3], this means that there exists a constant c(x) > 0 such that for all A, B, C ∈ Z with (A, B) = (0, 0). Now, for any given B ∈ N we choose A, C ∈ Z such that |Ba + A| = Ba and |Bb + C| = Bb . Then Thus, on combining these estimates with (1.6), it follows that This establishes (1.5). Note than in view of (1.5) and the moreover part of the theorem, the Diophantine condition (1.4) with ǫ = 0 is optimal in the case a is rational.
Remark 5. The fact that a = 0 is excluded in the statement of the theorem is natural since as in Remark 4, on making use of the equivalent dual form representation of Bad(i, j) it is easily verified that On the other hand, if the above lim inf is strictly positive then it was shown in [1] that Bad(i, j) ∩ L 0,b is a winning subset of the horizontal line L 0,b . By making use of the mechanism developed in this paper, it is relatively straightforward to adapt the homogeneous proof given in [1] to show that if lim inf q→∞ q 1/j qb > 0, then for any θ ∈ R 2 the set Bad θ (i, j) ∩ L 0,b is a winning subset of the horizontal line L 0,b . Remark 6. Observe that when it comes to intersecting countably many (i t , j t ) pairs the Diophantine condition (1.4) imposes the condition that This is clearly weaker than condition (1.3) imposed in Theorem B and moreover in view of Remark 4 it is essentially optimal.
The proofs of Theorem 1.1 and Theorem 1.2 rely on first establishing the homogeneous cases and then making use of a natural mechanism that we develop for generalizing homogenous winning statements to the inhomogeneous setup. This mechanism is further exploited to prove that inhomogeneous Bad(i, j) is winning. Theorem 1.3 Let (i, j) be a pair of real numbers satisfying (1.1). Then, for any θ = (γ, δ) ∈ R 2 we have that Bad θ (i, j) is a (30 √ 2) −1 -winning subset of R 2 .
Remark 7. It is worth pointing out that the winning constant (30 √ 2) −1 is not optimal. Indeed, the ideas used to prove the above theorem and the argument given in [12] can be combined to show that Bad θ (i, j) is hyperplane absolute winning. This is a stronger version of winning and implies that Bad θ (i, j) is α-winning for any α < 1/2.

The main strategy
In this section we outline the key steps in establishing the homogeneous case (θ = (0, 0)) of Theorem 1.1. The general inhomogeneous statement is obtained by appropriately adapting the homogeneous argument and is carried out in §5. To begin with observe that for any planar curve C := C f and θ ∈ R 2 Recall, that π : R 2 → R is the projection map onto the x-axis. Also, for convenience and without loss of generality we will assume that j i. Thus, the homogeneous case of Theorem 1.1 is easily deduced from the following statement for Bad f (i, j) := Bad f (0,0) (i, j) -see Remark 2 above for the justification.
Theorem 2.1 Let (i, j) be a pair of real numbers satisfying 0 < j i < 1 and i + j = 1. Let I ⊂ R be a compact interval and f ∈ C (2) At this point it is useful to recall the definition of a winning set and the notion of rooted trees -a key 'structural' ingredient in establishing the above winning statement.

Schmidt games and rooted trees
Wolfgang M. Schmidt introduced the games which now bear his name in [14]. The simplified account which we are about to present is sufficient for the purposes of this paper. Suppose that 0 < α < 1 and 0 < β < 1. Consider the following game involving the two arch rivals Ayesha and Bhupen -often simply referred to as players A and B. First, B chooses a closed ball B 0 ⊂ R m . Next, A chooses a closed ball A 0 contained in B 0 of diameter α ρ(B 0 ) where ρ( . ) denotes the diameter of the ball under consideration. Then, B chooses at will a closed ball B 1 contained in A 0 of diameter β ρ(A 0 ). Alternating in this manner between the two players, generates a nested sequence of closed balls in R m : A subset X of R m is said to be (α, β)-winning if A can play in such a way that the unique point of intersection A n lies in X, regardless of how B plays. The set X is called α-winning if it is (α, β)-winning for all β ∈ (0, 1). Finally, X is simply called winning if it is α-winning for some α. Informally, player B tries to stay away from the 'target' set X whilst player A tries to land on X. As shown by Schmidt [14], the following are two key consequences of winning.
• If X ⊂ R m is a winning set, then dim X = m.
• The intersection of countably many α-winning sets is α-winning.
In the setting of Theorem 2.1, we have m = 1 and X = Bad f (i, j). Thus A n and B n are compact intervals. Note that more generally, we can replace R m in the above description of Schmidt games by a m-dimensional Riemannian manifold. It is this slightly more general form that is implicitly referred to within the context of Theorem 1.1.
We now turn our attention to rooted trees. Recall that a rooted tree is a connected graph T without cycles and with a distinguished vertex τ 0 , called the root of T . We identify T with the set of its vertices. Any vertex τ ∈ T is connected to τ 0 by a unique path. The length of the path is called the height of τ . The set of vertices of height n is called the n'th level of T and is denoted by T n . Thus T 0 = {τ 0 }. Next, given τ, τ ′ ∈ T we write τ ≺ τ ′ to indicate that the path between τ 0 and τ passes through τ ′ and in this case we call τ a descendant of τ ′ and τ ′ an ancestor of τ . By definition, every vertex is a descendant and an ancestor of itself. For V ⊂ T , we write τ ≺ V if V contains an ancestor of τ . If τ ≺ τ ′ and the height of τ is one greater than that of τ ′ , then τ is called a successor of τ ′ and τ ′ is called the predecessor of τ . Let T (τ ) denote the rooted tree formed by all descendants of τ . Thus the root of T (τ ) is τ and we denote by T suc (τ ) the set of all successors of τ . More generally, for V ⊂ T , we let T suc (V) := τ ∈V T suc (τ ). In this paper, we use the convention that a subtree of T has the same root as T . As a consequence, T (τ ) is not regarded as a subtree of T unless τ = τ 0 .
Let N ∈ N. We say that a rooted tree is N-regular if every vertex has exactly N successors. Note that an N -regular rooted tree is necessarily infinite. The following statement appears as Proposition 2.1 in [1].
Proposition 2.2 Let T be an N -regular rooted tree, S ⊂ T be a subtree, and 1 m N be an integer. Suppose that for every m-regular subtree R of T , we have that S ∩ R is infinite. Then S contains a (N − m + 1)-regular subtree.
This proposition will be required in establishing Proposition 2.4 below. As shown in §2.2.1, the latter is very much at the heart of the proof of Theorem 2.1.

The winning strategy for Theorem 2.1
Let β ∈ (0, 1). We want to prove that Bad f (i, j) is ( 1 2 , β)-winning. In the first round of the game, Bhupen chooses a closed interval B 0 ⊂ I. Now Ayesha chooses the closed interval A 0 ⊂ B 0 with diameter ρ(A 0 ) = 1 2 ρ(B 0 ) such that A 0 has the same center as B 0 . Let κ > 1 be sufficiently large so that for every x ∈ I we have that Clearly such a κ > 1 exists by the conditions imposed on f and I. Let Without loss of generality, we may assume that The point is that if this is not the case then Ayesha will choose her intervals A n−1 (n 1) arbitrarily until the interval B n chosen by Bhupen with diameter ρ(B n ) = (β/2) n ρ(B 0 ) satisfies 6κρ(B n )R 2 < 1. At this stage Ayesha chooses the closed interval A n with the same center as B n and half its diameter. We now simply relabel A n and B n as A 0 and B 0 respectively.
Choose µ > 0 such that and define λ 0 := 0 and λ k : In turn, let c := l 2 10 3 κ 5 R 4+λ 1 (2.6) and P := P = p q , r q : Note that R is large while l and c are small. Indeed, we have the inequalities R > 32, l < 10 −3 , c < 10 −10 l which we will use without further reference. Throughout, when a rational point in R 2 is expressed as P = ( p q , r q ), we assume that q > 0 and that the integers p, q, r are co-prime. Finally, for each P = ( p q , r q ) ∈ Q 2 we associate the interval (2.8) The following inclusion is a simple consequence of the manner in which the above quantities and objects have been defined. Lemma 2.3 Let A 0 , P and ∆(P ) be as above. Then

Proof.
Let x ∈ A 0 . Suppose x / ∈ Bad f (i, j). Then there exists P = ( p q , r q ) ∈ Q 2 such that In view of the fact that it follows that p q ∈ B 0 ⊂ I. Hence, using the Mean Value Theorem together with (2.1) we obtain the following estimate: The upshot is that x ∈ ∆(P ) with P ∈ P. This completes the proof of the lemma. ⊠ Now let T be an [R]-regular rooted tree with root τ 0 , where [ · ] denotes the integer part of a real number. We choose and fix an injective map I from T to the set of closed subintervals of A 0 satisfying the following conditions: • For any n 0 and τ ∈ T n , ρ(I(τ )) = lR −n . In particular, I(τ 0 ) = A 0 .
Note that for n 1 and τ ∈ T n−1 , any closed subinterval of I(τ ) of length 2lR −n must contain an I(τ ′ ) for some τ ′ ∈ T suc (τ ). Suppose that P is partitioned into a disjoint union We inductively define a subtree S of T as follows.
is a subtree of T and by construction Thus given a partition P n of P, the intervals {I(τ ) : τ ∈ S n } serve as possible candidates when it comes to Ayesha to turn to make a move. Moreover, we are able to choose the partition P n in such a way that S has the following key feature.

Proposition 2.4
There exists a partition P = ∞ n=1 P n such that the tree S has an ([R] − 10)-regular subtree.
Armed with this proposition we are able to describe the winning strategy that Ayesha will adopt.

Preliminaries for Proposition 2.4
The following simple but important lemma was established in [3].
Proof. Since the proof is only a few lines we reproduce it here. By Minkowski's theorem for systems of linear forms there is (A, B, C) ∈ Z 3 \ {0} such that |Ap + Br + Cq| < 1, |A| q i and |B| q j .
Since Ap + Br + Cq is an integer it must be zero. If (A, B) = (0, 0) then qC = 0 and, since q = 0 we also have that C = 0, a contradiction. Hence (A, B) = (0, 0) and the proof is complete. ⊠ In view of Lemma 3.1, to each point P = ( p q , r q ) ∈ P, we can assign a rational line passing through P where A P , B P , C P ∈ Z are co-prime with (A P , B P ) = (0, 0) and such that |A P | q i and |B P | q j .
If there is more than one line satisfying the above conditions, we choose any one. Further, for P ∈ P we define the function F P : I → R by In this section we gather basic information regarding the sets Θ(P ) and associated quantities.
Lemma 3.2 Let P = ( p q , r q ) ∈ P and x ∈ ∆(P ). Then In particular, we have that ∆(P ) ⊂ Θ(P ). Furthermore, if x ∈ Θ(P ) then Proof. On using the fact that P lies on the line L P and (2.7) we obtain the following For the second inequality, by the Mean Value Theorem there exists a point ξ ∈ I such that In particular, it follows that for x ∈ ∆(P ), we have that and so by definition x ∈ Θ(P ).
Finally, if x ∈ Θ(P ), then ⊠ Our next goal is to describe the structure of Θ(P ), in particular, to estimate its size. To this end, we introduce the following quantities. Let and let Then we define the height of P by Proof. In view of (2.1) and the fact that j i, it follows that and hence In the above, to each P ∈ P we have attached a rational line L P with coefficients A P , B P , C P and the function F P and the quantity E P . For ease of notation and the sake of clarity, we shall drop the subscript P if there is no ambiguity or confusion caused.
Lemma 3.4 If P ∈ P * (resp. P ∈ P \ P * ), then Θ(P ) is contained in one (resp. at most two) open interval(s) of length at most . Proof.
Case (1). Suppose P ∈ P * . Then B = 0. For any x ∈ Θ(P ), it follows from Lemma 3.2 and (3.5) that where ξ ∈ I. This implies that The upshot of this is that Θ(P ) is contained in an open interval of length 42κ 3 c/H(P ).
Now suppose B = 0. Then, by (3.5), we have that E = 0. Consider the closed interval We first prove that Θ(P ) ∩ Ω is contained in an open interval of length 42κ 3 c/H(P ). If x ∈ Θ(P ) ∩ Ω, then it follows from Lemma 3.2 that This implies that Hence Θ(P ) ∩ Ω is contained in an open interval of length 12κc/H(P ) < 42κ 3 c/H(P ). In particular, this implies the desired statement if Θ(P ) ⊂ Ω.
Suppose that Θ(P ) ⊂ Ω. Then the following three observations imply that Θ(P ) \ Ω is a connected interval.
• If x 0 is an end point of Ω and is contained in the interior of I, then |F (x 0 )| 2κc/q. To see this, note that the assumption on x 0 and (3.5) imply that which together with (3.9) implies that x 0 / ∈ Θ(P ). Hence the desired inequality follows from (3.3).
• The function F : I → R is either convex or concave.
We next claim that Assuming this claim for the moment, it follows that for any where ξ ∈ Θ(P ) \ Ω. Thus Hence Θ(P ) \ Ω is contained in an open interval of length 8κ 2 c/H(P ) < 42κ 3 c/H(P ) and thereby completes the proof of the lemma modulo (3.10).
We now prove (3.10) in the instance that f ′′ > 0 and B > 0. The other cases are similar and left to the reader. Let ξ ∈ Θ(P ) \ Ω, and consider the functions G : R → R and H : R → R given by Thus, for x ∈ I we have that G(x) F (x) H(x). It follows that for any x ∈ I, Note that F ′′ = Bf ′′ > 0. So the end point of Ω that is contained in the interval with end points p q and ξ is equal to p q − E κB . In particular, we have that p q − E κB ∈ I. Thus (3.11) implies that On the other hand, we have that On combining the previous two inequalities, we find that and (3.10) follows. This completes the proof of Lemma 3.4. ⊠

Proof of Proposition 2.4
For n 1, let H n := 42κ 3 cl −1 R n and P n : Note that if P ∈ P n , then, by (3.8), we have that Next let P n,0 := P n ∩ P * (4.3) and where λ k are defined by (2.5).
Lemma 4.1 With P n and P n,k as above, we have that P = ∞ n=1 P n and P n = n k=0 P n,k .

Proof.
It is easily verified via (2.6) that H(P ) 3κc 1 2 for any P ∈ P, and that This together with (4.2) implies that P n = n k=0 P n,k . ⊠ We claim that the partition of P given by Lemma 4.1 satisfies the requirement of Proposition 2.4. In other words, P gives rise to a tree S as described in §2.2 that contains an ([R] − 10)-regular subtree. Recall that S is itself a subtree of an [R]-regular rooted tree T . The key towards establishing the claim is the following lemma and its corollary. For k 0, we let k + := max{k, 1} .
The following lemma contains a crucial property of the lines L P defined by (3.1).
Lemma 4.2 For any n 1, 0 k n and τ ∈ S n−k + , the map P → L P is constant on We postpone the proof for the moment and continue by stating an important consequence of the lemma. Corollary 4.3 For any n 1, 0 k n and τ ∈ S n−k + , we have

Proof.
We may assume that P n,k (τ ) = ∅. Let P 0 = ( p 0 q 0 , r 0 q 0 ) ∈ P n,k (τ ) be such that q 0 q for any ( p q , r q ) ∈ P n,k (τ ). By Lemma 4.2, for any P ∈ P n,k (τ ) we have L P = L P 0 and so Θ(P ) ⊂ Θ(P 0 ). Thus it follows from Lemma 3.2 that By Lemma 3.4, if k = 0 (resp. k 1), then Θ(P 0 ) is contained in one (resp. at most two) open interval(s) of length at most Since the intervals {I(τ ′ ) : τ ′ ∈ T n } are of length lR −n and have mutually disjoint interiors, there can be at most 2 (resp. 4) of them that intersect Θ(P 0 ). This proves the corollary. ⊠ We are now in the position to prove Proposition 2.4. In view of Proposition 2.2, it suffices to prove that the intersection of S with every 11-regular subtree of T is infinite. Let R ⊂ T be an 11-regular subtree and let R ′ := R ∩ S and a n := #R ′ n (n 0) , where R ′ n is the n'th level of the tree R ′ . Then a 0 = 1. We prove that R ′ is infinite by showing that a n > 3a n−1 (n 1) . (4.5) We use induction. For n 1, let It follows that a n 11a n−1 − #U n .
This completes the induction step and thus establishes (4.5). In turn this completes the proof of Proposition 2.4 modulo the truth of Lemma 4.2. ⊠

Proof of Lemma 4.2
To begin with we prove the following result.
(1) If k = 0 and q 1 q 2 , then (4.10) (2) If k = 1, then (4.14) Proof. We first prove that for any 0 k n, To establish (4.15), note that for s = 1, 2 there exists x s ∈ I(τ ) ∩ ∆(P s ) = ∅ and that |x 1 − x 2 | ρ(I(τ )) = lR −n+k + . Then, it follows from (2.8) that Regarding (4.16), we expand F 2 ( p 1 q 1 ) using Taylor's formula at the point p 2 q 2 and estimate as follows The above argument can be trivially modified to show that the same upper bound is valid for q 1 |F 1 ( p 2 q 2 )|. Turning our attention to (4.17), using (2.7) we estimate as follows The same argument with obvious modifications yields (4.18). Finally, regarding (4.19), using the definition (3.4) for E 1 and E 2 , we have that Having established (4.15)-(4.19), we are now in the position to prove the lemma.
Part (1). Suppose k = 0 and q 1 q 2 . It follows from the definition (4.3) for P n,0 that Since the left hand side of the above inequality is an integer, it follows that A 2 p 1 + B 2 r 1 + C 2 q 1 = 0. Regarding (4.10), note that for s = 1, 2 These inequalities together with (2.3) imply that Part (2). Suppose k = 1. It follows from (4.4) that This implies that The left hand side is an integer and so must be zero. The same argument involving (4.18) rather than (4.17) shows that |A 1 p 2 + B 1 r 2 + C 1 q 2 | = 0.
Part (3). Suppose k 2. We first prove that (4.21) It follows from (4.4) that In view of the fact that This establishes (4.20) and (4.21). It now follows that Finally, since P s ∈ P * , we have that q s E 2 s 9κ 2 c|B s | and H(P s ) = q s |E s | for s = 1, 2. Then |B 2 | H(P 2 ) 2 9κ 2 cq 2 and using (3.2) and the fact that H n+1 = RH n , we get that This thereby completes the proof of Part (3) and thus the lemma. ⊠ We now proceed with the proof of Lemma 4.2. Let P 1 = ( p 1 q 1 , r 1 q 1 ) and P 2 = ( p 2 q 2 , r 2 q 2 ) be distinct points in P n,k (τ ). We need to prove that L P 1 = L P 2 . We consider three separate cases.
Case (1). Suppose k = 0. Without loss of generality, we assume that q 1 q 2 . Then in view of (4.9) we have that A 2 p 1 + B 2 r 1 + C 2 q 1 = 0. Hence L P 2 passes through P 1 . We prove that L P 1 = L P 2 by contradiction. Thus, assume that L P 1 = L P 2 . Since the two lines intersect at P 1 , it follows that In particular, since p 1 , q 1 , r 1 are co-prime, the non-zero integer A 1 B 2 − A 2 B 1 is divisible by q 1 . Thus which is of course impossible.
Case (2). Suppose k = 1. Then in view of (4.11) we have that A 2 p 1 + B 2 r 1 + C 2 q 1 = 0 = A 1 p 2 + B 1 r 2 + C 1 q 2 . Hence, L P 2 passes through P 1 and L P 1 passes through P 2 . By definition, L P 2 passes through P 2 and L P 1 passes through P 1 . The upshot is that both lines pass through the points P 1 and P 2 , and so we must have that L P 1 = L P 2 .
Case (3). Suppose k 2. We prove that L P 1 = L P 2 by contradiction. Thus, assume that L P 1 = L P 2 . We first consider the case where L P 1 is parallel to L P 2 . Then, it is easily verifed that Since H n c, it follows via (4.12) and (4.13) that which is of course impossible. Now suppose L P 1 is not parallel to L P 2 . Let P 0 = ( p 0 q 0 , r 0 q 0 ) ∈ Q 2 be the point of intersection of L P 1 and L P 2 . Then it follows that the non-zero integer A 1 B 2 − A 2 B 1 is divisible by q 0 and so q 0 We first prove that ∆(P 1 ) ⊂ ∆(P 0 ) and that P 0 ∈ P. It is easily verified that .
Hence, on using Cramer's rule, we obtain that and (4.24) If x ∈ ∆(P 1 ), then (4.14) and (4.23) imply that Thus x ∈ ∆(P 0 ) and the upshot is that ∆(P 1 ) ⊂ ∆(P 0 ). In particular, (4.25) This implies that p 0 q 0 ∈ B 0 ⊂ I. Also note that by (4.14) and (4.24), we have that Thus P 0 ∈ P and so there exists a unique integer n 0 1 such that P 0 ∈ P n 0 . Suppose for the moment that n 0 n − k. Then there exists τ ′ ∈ S n 0 such that τ ≺ τ ′ , and hence This contradicts the fact that P 1 ∈ P n,k (τ ). Thus On the other hand, we have that This contradicts the above lower bound for H(P 0 ) and so completes the proof of Case (3) and indeed the lemma. ⊠ 5 The inhomogeneous case: establishing Theorem 1.1 Theorem 1.1 is easily deduced from the following statement.
Theorem 5.1 Let (i, j) be a pair of real numbers satisfying 0 < j i < 1 and i + j = 1.
Let I ⊂ R be a compact interval and f ∈ C (2) (I) such that f ′′ (x) = 0 for all x ∈ I. Then, for any θ = (γ, δ) ∈ R 2 we have that Bad f θ (i, j) is a 1/2-winning subset of I. To prove Theorem 5.1, the idea is to merge the inhomogeneous constraints into the homogeneous construction. More precisely, we show that S ′ has an ([R] − 12)-regular subtree Q ′ that incorporates the inhomogeneous constraints. With this in mind, let where c is defined in (2.6), and let V := (p, r, q) ∈ Z 2 × N : Furthermore, for each v = (p, r, q) ∈ Z 2 × N, we associate the interval Then, with A 0 ⊂ B 0 as in §2.2, the following is the inhomogeneous analogue of Lemma 2.3.
Lemma 5.2 Let A 0 , V and ∆ θ (v) be as above. Then

Proof.
The proof is similar to the homogeneous proof but is included for the sake of completeness. Let x ∈ A 0 . Suppose x / ∈ Bad f θ (i, j). Then there exists v = (p, r, q) ∈ Z 2 × N such that In view of the fact that it follows that p+γ q ∈ B 0 ⊂ I. Hence Thus x ∈ ∆ θ (v) and v ∈ V . This completes the proof of the lemma. ⊠ Observe that H ′ 1 = 2c ′ l −1 R 1 and so it follows that V = ∞ n=1 V n . We inductively define a subtree Q of S ′ as follows.
is a subtree of S ′ and, by construction, we have that Armed with the following result, the same arguments as in §2.2.1 with the most obvious modifications enables us to prove Theorem 5.1. In view of this the details of the proof of Theorem 5.1 modulo Proposition 5.3 are omitted. In order to establish the proposition, it suffices to prove the following statement.
Lemma 5.4 For any n 1 and τ ∈ Q n−1 , there is at most one v ∈ V n such that I(τ ) ∩ ∆ θ (v) = ∅. Moreover, ρ(∆ θ (v)) lR −n . Therefore, Proof. Suppose v s = (p s , r s , q s ) ∈ V n and I(τ ) ∩ ∆ θ (v s ) = ∅, s = 1, 2. We need to prove that v 1 = v 2 . Without loss of generality, assume that q 1 q 2 . Let x s ∈ I(τ ) ∩ ∆ θ (v s ). Then It follows that Suppose for the moment that q 1 > q 2 and let We show that and that P 0 ∈ P, where P and ∆(P 0 ) are defined by (2.7) and (2.8), respectively. In view of (5.1), it follows that So x 1 ∈ ∆(P 0 ) and (5.3) follows. Also, the above inequality implies that and since x 1 ∈ A 0 , it follows that p 1 −p 2 q 1 −q 2 ∈ B 0 ⊂ I. Moreover, by making use of (5.2) we have that Thus P 0 ∈ P and so there exists a unique integer n 0 1 such that P 0 ∈ P n 0 . Suppose for the moment that n 0 n − 1. Then there exists τ ′ ∈ S n 0 such that τ ≺ τ ′ and it follows that I(τ ) ∩ ∆(P 0 ) ⊂ I(τ ′ ) ∩ ∆(P 0 ) = ∅ contrary to (5.3). Thus, n 0 n and so On the other hand, we have that This contradicts the above lower bound for H(P 0 ) and we conclude that q 1 = q 2 . Since q 2 H ′ n+1 cl −1 R n−1 , it now follows from (5.1) and (5.2) that Thus, p 1 = p 2 and r 1 = r 2 . In other words, v 1 = v 2 and this proves the main substance of the lemma. To prove the 'moreover' part, it is easily verified that for any v ∈ V n we have that The 'therefore' part of the lemma is a direct consequence of this and the fact that there is at most one v ∈ V n such that I(τ ) ∩ ∆ θ (v) is non-empty. ⊠ As already mentioned, given Proposition 5.3, the proof of Theorem 5.1 follows on adapting the arguments of §2.2.1.

The proof of Theorem 1.2
The basic strategy towards establishing the winning result for lines is the same as when considering curves. To begin with observe that for any line L a,b given by As in the case of curves, without loss of generality we will assume that j i. Thus, the homogeneous case of Theorem 1.2 is easily deduced from the following statement.
Theorem 6.1 Let (i, j) be a pair of real numbers satisfying 0 < j i < 1 and i + j = 1. Given a, b ∈ R, suppose there exists ǫ > 0 such that Then Bad f (i, j) is a 1/2-winning subset of R. Moreover, if a ∈ Q the statement is true with ǫ = 0 in (6.1).
Note that in view of Remark 5 after the statement of Theorem 1.2, we do not require that a = 0 in Theorem 6.1 since j i.

The winning strategy for Theorem 6.1
Let β ∈ (0, 1). We want to prove that Bad f (i, j) is ( 1 2 , β)-winning. In the first round of the game, Bhupen chooses a closed interval B 0 ⊂ R. Now Ayesha chooses any closed The fact that c 0 > 0 follows from the Diophantine condition (6.1). Recall, that by hypothesis ǫ = 0 if a is rational and ǫ > 0 otherwise. We denote κ := |a| + 1, and choose µ 1 such that If a ∈ Q , we also require that where d ∈ N is the smallest positive integer such that da ∈ Z. Next, if a / ∈ Q, so that ǫ > 0, we let λ > 0 be such that If a ∈ Q, we simply let λ = 0. In turn, let λ 0 := 0 and λ k := λ j −k + k i + µ for k 1, and let c := min For each rational point P = ( p q , r q ) ∈ R 2 we associate the interval ∆(P ) := x ∈ R : x − p q < c q 1+i and we let The following inclusion is a simple consequence of the manner in which the above quantities and objects have been defined. Lemma 6.2 Let A 0 , P and ∆(P ) be as above. Then

Proof.
Let x ∈ A 0 . Suppose x / ∈ Bad f (i, j). Then there exists P = ( p q , r q ) ∈ Q 2 such that The upshot is that x ∈ ∆(P ) with P ∈ P. This completes the proof of the lemma. ⊠ Next, just as in §2.2, but with A 0 and P as above, let • T be an [R]-regular rooted tree with root τ 0 , • I be an injective map from T to the set of closed subintervals of A 0 , • S = ∞ n=0 S n be a subtree of T associated with a partition P n of P.
The following proposition is the lines analogue of Proposition 2.4. It enables us to deduce Theorem 6.1 (and thus the homogeneous case of Theorem 1.2) by adapting the arguments of §2.2.1 in the most obvious manner. In view of this the details of the proof of Theorem 6.1 modulo Proposition 6.3 are omitted. Proposition 6.3 There exists a partition P = ∞ n=1 P n such that the tree S has an ([R] − 5)-regular subtree.

Proof of Proposition 6.3
As in the 'curves' proof, to each point P = ( p q , r q ) ∈ P, we attach a rational line passing through P where A, B, C ∈ Z are co-prime with (A, B) = (0, 0) and such that |A| q i and |B| q j .
Associated with each point P ∈ P, we also consider the quantity Then Note that if x ∈ ∆(P ), then The following statement enables us to construct the desired partition in Proposition 6.3.
Lemma 6.4 For any P = ( p q , r q ) ∈ P, we have Proof. If B = 0, then q 1−jǫ |E| = q 1−jǫ |A| 1 c 1 and we are done. If B = 0, then it is easily verified that If |E| is the maximum in the above then again we are done. So suppose q 1−jǫ |Bb+C| c 0 . Since P ∈ P, there exists x ∈ A 0 ∩ ∆(P ) and it follows that This proves the lemma. ⊠ A particular consequence of (6.8) is that E = 0. Thus every line L P intersects at the line L a,b given by y = f (x) = ax + b at a single point.
For n 1, let H n := 4κcl −1 R n and P n := P = p q , r q ∈ P : H n q|E| < H n+1 .
Lemma 6.5 With P n and P n,k as above, we have that P = ∞ n=1 P n and P n = n k=1 P n,k .
Proof. Note that by (6.8), for any P ∈ P we have that q|E| c 1 and by definition Thus, P = ∞ n=1 P n . To prove the second conclusion, we first show that κq 1+i < H n R λn for P ∈ P n . (6.11) We consider two separate cases. If a / ∈ Q, on combining the fact that q|E| < H n+1 with (6.8) implies that q jǫ < c −1 1 H n+1 . It then follows that 1.
If a ∈ Q, then |E| 1/d, and hence the fact q|E| < H n+1 implies that q < dH n+1 . It follows that κq 1+i H n R λn < This proves (6.11). Now (6.11) together with (6.9) implies that P n = n k=1 P n,k . ⊠ We claim that the partition of P given by Lemma 6.5 satisfies the requirement of Proposition 6.3. The key towards establishing the claim is the following lemma. It is the lines analogue of Lemma 4.2.

Proof.
Let P 1 = ( p 1 q 1 , r 1 q 1 ) and P 2 = ( p 2 q 2 , r 2 q 2 ) be distinct points in P n,k (τ ). We need to prove that L P 1 = L P 2 . We let A s , B s , C s and E s be the respective quantities associated with P s , s = 1, 2, and consider two separate cases.
Since the left hand side of the above inequality is an integer, it follows that A 1 p 2 + B 1 r 2 + C 1 q 2 = 0. Similarly, we obtain that A 2 p 1 + B 2 r 1 + C 2 q 1 = 0. The upshot is that both the lines L P 1 and L P 2 pass through both the points P 1 and P 2 , and so we must have that L P 1 = L P 2 .
Case (2). Suppose k 2. We prove that L P 1 = L P 2 by contradiction. Thus, assume that L P 1 = L P 2 . We first establish various preliminary estimates. Let With s = 1 or 2, let (x s , ax s + b) denote the intersection point of L Ps with the line L a,b . Then and so Hence This completes the preliminaries. Recall that we are assuming that L P 1 = L P 2 and the name of the game is to obtain a contradiction. We first consider the case that L P 1 is parallel to L P 2 . Then there exist (A, B) ∈ Z 2 \ {(0, 0)} and nonzero integers t 1 , t 2 such that .
and it follows that This together with (6.13) implies that and this contradicts (6.14). If B 1 = 0, then |t 1 | |B 1 | and This together with (6.14) implies that However, this contradicts the fact that c l −i . Hence, if L P 1 is parallel to L P 2 then we must have that L P 1 = L P 2 .
Now suppose L P 1 is not parallel to L P 2 . Let P 0 = ( p 0 q 0 , r 0 q 0 ) ∈ Q 2 be the intersection of L P 1 and L P 2 . Then it follows that A 1 B 2 − A 2 B 1 is a nonzero integer and is divisible by q 0 and so q 0 We first prove that ∆(P 1 ) ⊂ ∆(P 0 ) and that P 0 ∈ P. In view of the fact that It then follows that So if x ∈ ∆(P 1 ), then Thus x ∈ ∆(P 0 ) and the upshot is that ∆(P 1 ) ⊂ ∆(P 0 ). In turn, since A 0 ∩ ∆(P 1 ) = ∅, it follows that A 0 ∩ ∆(P 0 ) = ∅. In view of this, in order to prove that P 0 ∈ P we need to we have that Hence < κc and this established (6.16). Thus P 0 ∈ P and so there exists a unique integer n 0 1 such that P 0 ∈ P n 0 . Suppose for the moment that n 0 n − k. Then there exists τ ′ ∈ S n 0 such that τ ≺ τ ′ , and hence This contradicts the fact that P 1 ∈ P n,k (τ ). Thus and so q 1+i 0 (6.9) On the other hand, we have that This contradicts the above lower bound for q 1+i 0 and so completes the proof of Case (2) and indeed the lemma. ⊠ An important consequence of Lemma 6.6 is the following lines analogue of Corollary 4.3.
This implies that P ∈P n,k (τ ) ∆(P ) is contained in the open interval which has length lR −n . Since the intervals {I(τ ′ ) : τ ′ ∈ T n } are of length lR −n and have mutually disjoint interiors, there can be at most 2 of them that intersect the interval (6.17). This proves the corollary. ⊠ We are now in the position to prove Proposition 6.3. In view of Proposition 2.2, it suffices to prove that the intersection of S with every 6-regular subtree of T is infinite. Let R ⊂ T be a 6-regular subtree and let R ′ := R ∩ S and a n := #R ′ n (n 0) .
Then a 0 = 1. We prove that R ′ is infinite by showing that a n > 2a n−1 (n 1) . (6.18) We use induction. As in §4, for n 1, let Then and it follows that a n 6a n−1 − #U n .

(6.19)
On the other hand, as in §4, we have that Thus, Corollary 6.7 implies that With n = 1 in (6.21), we find that a 1 4. Hence, (6.18) holds for n = 1. Now assume n 2 and that (6.18) holds with n replaced by 1, . . . , n − 1. Then for any 1 k n, we have that a n−k 2 −k+1 a n−1 .
This completes the induction step and thus establishes (6.18). In turn this completes the proof of Proposition 6.3. ⊠ 6.2 The inhomogeneous case: establishing Theorem 1.2 Theorem 1.2 is easily deduced from the following statement.
Theorem 6.8 Let (i, j) be a pair of real numbers satisfying 0 < j i < 1 and i + j = 1. Given a, b ∈ R, suppose there exists ǫ > 0 such that (6.1) is satisfied. Then Bad f θ (i, j) is a 1/2-winning subset of R.
We have already established the homogeneous case (γ = δ = 0) of the Theorem 6.8; namely Theorem 6.1. With reference to §6.1, the crux of the 'homogeneous' proof involved constructing a partition P n (n 1) of P (given by Lemma 6.5) such that the subtree S of an [R]-regular rooted tree T has an ([R] − 5)-regular subtree S ′ -the substance of Proposition 6.3. To prove Theorem 6.8, the idea is to merge the inhomogeneous constraints into the homogeneous construction as in the case of curves in §5. More precisely, we show that S ′ has an ([R] − 7)-regular subtree Q ′ that incorporates the inhomogeneous constraints. With this in mind, let Furthermore, for each v = (p, r, q) ∈ Z 2 × N, we associate the interval Then, with A 0 ⊂ B 0 as in §6.1, the following is the inhomogeneous analogue of Lemma 6.2. Lemma 6.9 Let A 0 , V and ∆ θ (v) be as above. Then

Proof.
The proof is similar to the homogeneous proof but is included for sake of completeness. Let x ∈ A 0 . Suppose x / ∈ Bad f θ (i, j). Then there exists v = (p, r, q) ∈ Z 2 × N such that It follows that Thus x ∈ ∆ θ (v) and v ∈ V . This completes the proof of the lemma.

⊠
For n 1, let H ′ n := 2c ′ l −1 R n and V n := {(p, r, q) ∈ V : H ′ n q 1+i < H ′ n+1 }. Observe that H ′ 1 = 2c ′ l −1 R 1 and so it follows that V = ∞ n=1 V n . We inductively define a subtree Q of S ′ as follows. Let Q 0 = {τ 0 }. If Q n−1 (n 1) is defined, we let is a subtree of S ′ and by construction Armed with the following result, the same arguments as in §2.2.1 with the most obvious modifications enables us to prove Theorem 6.8. In view of this the details of the proof of Theorem 6.8 modulo Proposition 6.10 are omitted. In order to establish the proposition, it suffices to prove the following statement. Lemma 6.11 For any n 1 and τ ∈ Q n−1 , there is at most one v ∈ V n such that I(τ ) ∩ ∆ θ (v) = ∅. Moreover, ρ(∆ θ (v)) lR −n . Therefore, Proof. Suppose v s = (p s , r s , q s ) ∈ V n and I(τ ) ∩ ∆ θ (v s ) = ∅, s = 1, 2. We need to prove that v 1 = v 2 . Without loss of generality, assume that q 1 q 2 . Let x 1 ∈ I(τ ) ∩ ∆ θ (v 1 ). The same arguments as in the proofs of (5.1) and (5.2) show that and (6.23) Suppose for the moment that q 1 > q 2 and let We show that I(τ ) ∩ ∆(P 0 ) = ∅ and that P 0 ∈ P. Similar to the proof of Lemma 5.4, it follows from (6.22) that So x 1 ∈ ∆(P 0 ) and it follows that I(τ ) ∩ ∆(P 0 ) = ∅. Moreover, by making use of (6.23) we have that Thus P 0 ∈ P and so there exists a unique integer n 0 1 such that P 0 ∈ P n 0 . The same argument as in the proof of Lemma 5.4 shows that n 0 n, and so On the other hand, we have that This contradicts the above lower bound for (q 1 − q 2 )|E P 0 | and we conclude that q 1 = q 2 . It now follows from (6.22) and (6.23) that |p 1 − p 2 | < 1 and |r 1 − r 2 | < 1.
Thus, p 1 = p 2 and r 1 = r 2 . In other words, v 1 = v 2 and this proves the main substance of the proposition. The proofs of the remaining parts are the same as those for Lemma 5.4. ⊠ As already mentioned, given Proposition 6.10, the proof of Theorem 6.8 follows on adapting the arguments of §2.2.1.

The proof of Theorem 1.3
We need the notion of regular colorings of rooted trees. Let D ∈ N. A D-coloring of a rooted tree T is a map γ : T → {1, . . . , D}. For V ⊂ T and 1 i D, we denote V (i) = V ∩ γ −1 (i). Let N ∈ N be an integer multiple of D, and suppose that T is Nregular. We say that a D-coloring of T is regular if for any τ ∈ T and 1 i D, we have #T suc (τ ) (i) = N/D. The following two types of subtrees are of interest to us.
• The subtree S is of type (I) if for any τ ∈ S and 1 i D, we have #S suc (τ ) (i) = 1.
Roughly speaking, in the proof of Theorem 1.3, the two types of subtrees correspond to strategies of the two players in Schmidt's game. We will make use of the following criterion for the existence of subtree of type (I). It appears as Proposition 2.2 in [2].
Proposition 7.1 Let T be an N -regular rooted tree with a regular D-coloring, and let S ⊂ T be a subtree. Suppose that for every subtree R ⊂ T of type (II), S ∩ R is infinite. Then S contains a subtree of type (I).
7.1 The winning strategy for Theorem 1.3 Let α 0 := (30 √ 2) −1 , β ∈ (0, 1). We want to prove that Bad θ (i, j) is (α 0 , β)-winning. In the first round of the game, Bhupen chooses a closed disc B 0 ⊂ R 2 . Now Ayesha chooses any closed disc A 0 ⊂ B 0 with diameter ρ(A 0 ) = α 0 ρ(B 0 ). Let By a square we mean a set of the form where ℓ(Σ) > 0 is the side length of Σ. Let Σ 0 be the circumscribed square of A 0 . Then ℓ(Σ 0 ) = l. Let T be an m 2 [R/m] 2 -regular rooted tree with a regular [R/m] 2 -coloring. We choose and fix an injective map Φ from T to the set of subsquares of Σ 0 satisfying the following conditions: • For any n 0 and τ ∈ T n , we have ℓ(Φ(τ )) = lR −n . In particular, the root τ 0 of T is mapped to Σ 0 .
For n 1, let H n := 6cl −1 R n , H ′ n := 3c ′ l −1 R n and define P n := P = p q , r q ∈ Q 2 : H n q max{|A P |, |B P |} < H n+1 (7.4) and V n := {v = (p, r, q) ∈ Z 2 × N : H ′ n q 1+max{i,j} < H ′ n+1 }, (7.5) where A P and B P are as in §3. In view of (7.1), we have that H ′ 1 H 1 = 6cl −1 R 1. Thus We inductively define a subtree S of T as follows. Let S 0 = {τ 0 }. If n 1 and S n−1 is defined, we let The following proposition is the key to proving Theorem 1.3.

Proposition 7.2
The tree S contains a subtree of type (I).

Proof of Theorem 1.3 modulo Proposition 7.2
So Ayesha can choose A n to be the inscribed closed disc of Φ(τ n ). This proves (7.8).
In view of (7.8), (7.7) and (7.3), we have This proves the theorem assuming the truth of Proposition 7.2. ⊠

Proof of Proposition 7.2
Let w > 0. By a strip of width w, we mean a subset of R 2 of the form L := {x ∈ R 2 : |x · u − a| w/2}, where the dot denotes the standard inner product, u ∈ R 2 is a unit vector, and a ∈ R.
The following result is proved in [2, Corollary 4.2].
Lemma 7.3 For any n 1, there exists a partition P n = n k=1 P n,k such that for any 1 k n and τ ∈ S n−k , there is a strip of width 2 3 lR −n which contains all the rectangles {∆(P ) : P ∈ P n,k , Φ(τ ) ∩ ∆(P ) = ∅}.
We now prove a corresponding result which takes into consideration the inhomogeneous approximation aspect. Lemma 7.4 For any n 1 and τ ∈ S n−1 , there is at most one v ∈ V n such that Φ(τ ) ∩ ∆ θ (v) = ∅. Moreover, ∆ θ (v) is contained in a strip of width 2 3 lR −n .

⊠
The following result proved in [2,Lemma 4.3] gives an upper bound for the number of certain squares which intersect a thin strip.
Lemma 7.5 Let R ⊂ T be a subtree of type (II), let n 1, and let L be a strip of width 2 3 lR −n . Then for any 1 k n and τ ∈ R n−k , we have that On combining Lemmas 7.3, 7.4 and 7.5, we obtain the following statement.
We are now in the position to prove Proposition 7.2. The proof is essentially the same as the proof of Proposition 3.3 in [2]. However for the sake of completeness we have included the argument. In view of Proposition 7.1, it suffices to prove that the intersection of S with every subtree of type (II) is infinite. Let R ⊂ T be a subtree of type (II), and let R ′ := R ∩ S, and a n := #R ′ n (n 0) .
This completes the induction step and thus establishes (7.12). In turn this completes the proof of Proposition 7.2. ⊠