Badly approximable points on manifolds

This paper is motivated by two problems in the theory of Diophantine approximation, namely, Davenport’s problem regarding badly approximable points on submanifolds of a Euclidean space and Schmidt’s problem regarding the intersections of the sets of weighted badly approximable points. The problems have been recently settled in dimension two but remain open in higher dimensions. In this paper we develop new techniques that allow us to tackle them in full generality. The techniques rest on lattice points counting and a powerful quantitative result of Bernik, Kleinbock and Margulis. The main theorem of this paper implies that any finite intersection of the sets of weighted badly approximable points on any analytic nondegenerate submanifold of Rn\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}^n$$\end{document} has full dimension. One of the consequences of this result is the existence of transcendental real numbers badly approximable by algebraic numbers of any bounded degree.

fundamental result. It states that for every α ∈ R and any Q > 1 there exists q ∈ N and p ∈ Z such that |qα − p| < Q −1 and q ≤ Q. In particular, it implies that for every real irrational number α the inequality holds for infinitely many rational numbers p/q written as reduced fractions of integers p and q. A real number α is then called badly approximable if there exists a constant c = c(α) > 0 such that for all (q, p) ∈ N × Z. In what follows, the set of badly approximable real numbers will be denoted by Bad.
It is well known that a real irrational number α is badly approximable if and only if the partial quotients of its continued fraction expansion are uniformly bounded. For instance, any real quadratic irrational number is in Bad, since its continued fraction expansion is eventually periodic. 1 Using continued fractions one can easily produce continuum many examples of badly approximable real numbers. Beyond the cardinality, Jarník [27] established that dim Bad (the Hausdorff dimension of Bad) is 1. However, the Lebesgue measure of Bad is known to be zero. This is a trivial consequence of the divergence case of Khintchine's theorem [37], and can also be relatively easily proved using the Lebesgue density theorem, see [18] or [15, Corollary 2].

Higher dimensions: Schmidt's conjecture
Higher dimensions offer various ways of generalising the notion of badly approximable numbers. For now, we restrict ourselves to considering simultaneous Diophantine approximations by rationals. The point y = (y 1 , . . . , y n ) ∈ R n is called badly approximable if there exists a constant c = c(y) > 0 such that max 1≤i≤n qy i ≥ cq −1/n (2) for all q ∈ N, where x denotes the distance of x from the nearest integer. The quantities qy i are equal to |qy i − p i | for some p i ∈ Z and thus give rise to 'approximating' rationals p 1 /q, . . . , p n /q. Once again, the notion of badly approximable points is underpinned by Dirichlet's theorem, this time for R n , which implies that the inequality max 1≤i≤n qy i < q −1/n holds for infinitely many q ∈ N. The set of badly approximable points in R n will be denoted by Bad(n). Observe that Bad(1) = Bad.
The first examples of badly approximable points in R n were given by Perron [40] who used an algebraic construction and produced infinitely yet countably many elements of Bad(n). For instance, (α, . . . , α n ) ∈ Bad(n) whenever α is a real algebraic number of degree n + 1. However, it was not until 1954 when first Davenport [21] for n = 2 and then Cassels [19] for n ≥ 2 showed that Bad(n) was uncountable. The fact that Bad(n) has full Hausdorff dimension was proved by Schmidt [43] who introduced powerful ideas based on a specific type of games. The dimension result for Bad(n) comes about as a consequence of the fact that Bad(n) is winning for Schmidt's game. Furthermore, Schmidt proved that affine transformations of Bad(n) are winning and that the collection of winning sets in R n is closed under countable intersections.
In his 1983 paper [46] Schmidt formulated a conjecture that later became the catalysis for some remarkable developments. Schmidt's conjecture rests on the modified notion of badly approximable points in which approximations in each coordinate are given some weights, say r 1 , . . . , r n . In short, he conjectured that there exist points in R 2 that are simultaneously badly approximable with respect to two different collections of weights. The weights of approximation are required to satisfy the following conditions: r 1 + · · · + r n = 1 and r i ≥ 0 for all i = 1, . . . , n . ( Throughout this paper the set of all n-tuples r = (r 1 , . . . , r n ) subject to (3) will be denoted by R n . Formally, given r ∈ R n , the point y = (y 1 , . . . , y n ) ∈ R n will be called r-badly approximable if there exists c = c(y) > 0 such that max 1≤i≤n qy i 1/r i ≥ cq −1 (4) for all q ∈ N. Here, by definition, qy i 1/0 = 0. Again, a version of Dirichlet's theorem tells us that when c = 1 inequality (4) fails infinitely often.
The set of r-badly approximable points in R n will be denoted by Bad(r). As is readily seen, the classical set of badly approximable points Bad(n) is simply Bad( 1 n , . . . , 1 n ). Using this notation we can now specify the following concrete statement conjectured by Schmidt: It is worth mentioning that the sets Bad(r) have been studied at length in all dimensions and for arbitrary collections of weights, see [22,[34][35][36]41]. Partly the interest was fueled by natural links with homogeneous dynamics and Little-wood's conjecture in multiplicative Diophantine approximation, another long standing problem-see [13] for further details. Schmidt's conjecture withstood attacks for nearly 30 years. However, the recent progress has been dramatic.
In 2011 Badziahin et al. [13] made a breakthrough by proving that for any sequence r k = (i k , j k ) ∈ R 2 such that lim inf k→∞ min{i k , j k } > 0 ( 5 ) and any vertical line L θ = {(θ, y) : y ∈ R} ⊂ R 2 with θ ∈ Bad one has that dim k Bad(r k ) ∩ L θ = 1.
This readily gives that dim k Bad(r k ) = 2 and proves Schmidt's conjecture in a much stronger sense. Shortly thereafter, An [1] proves that for any r ∈ R 2 and any θ ∈ Bad the set Bad(r) ∩ L θ is winning for a Schmidt game in L θ . This immediately leads him to removing condition (5) from the theorem of Badziahin et al., since the collection of Schmidt's winning sets is closed under arbitrary countable intersections. In a related paper An [2] establishes that Bad(i, j) is winning for the 2-dimensional Schmidt game, thus giving another proof of Schmidt's conjecture. Generalising the techniques of [13] in yet another direction Nesharim [39], independently from An, proves that the set in the left hand side of (6) intersected with naturally occurring fractals embedded in L θ is uncountable for any sequence (r k ) k∈N . Subsequently, Nesharim jointly with Weiss establishes the winning property of these intersections-see Appendix B in [39]. As already mentioned, the sets Bad(r) and even their restrictions to naturally occurring fractals have been investigated in higher dimensions, see [26,[34][35][36]. In particular, the sets Bad(r) were shown to have full Hausdorff dimension for any r ∈ R n . However, the theory of their mutual intersections is a different story. In an apparent attempt to prove Schmidt's conjecture, Kleinbock and Weiss [36] introduced a modified version of Schmidt's games. As they have shown, winning sets for the same modified Schmidt game inherit the properties of classical winning sets. Namely, they have full Hausdorff dimension and their countable intersections are winning with respect to the same game. Also Kleinbock and Weiss have proved that Bad(r) is winning for a relevant modified Schmidt game. However, it was not possible to prove that the intersection Bad(r 1 ) ∩ Bad(r 2 ) was a winning set for some modified Schmidt game as, with very few exceptions, the corresponding modified Schmidt games were not 'compatible'. As a result the following key problem that generalises Schmidt's original conjecture has remained open in dimensions n ≥ 3: Problem 1 Let n ∈ N. Prove that for any finite or countable subset W of R n one has that dim r∈W Bad(r) = n .
The main result of this paper implies (7) in arbitrary dimensions n and for arbitrary countable subsets W of weights satisfying a condition similar to (5). For instance, the result is applicable to arbitrary finite collections of weights W . The proof will be given by restricting the sets of interest to a suitable family of curves in R n . Interestingly, this approach, which was innovated in [13] in the case n = 2, turns out to face another intricate problem that was first communicated by Davenport.

Bad(r) on manifolds and Davenport's problem
In 1964 Davenport [22] established that, given a finite collection has the power of continuum. For instance, taking f 1 (x, y) = x, f 2 (x, y) = y and f 3 (x, y) = (x, y) shows that Bad(1, 0) ∩ Bad 1 2 , 1 2 ∩ Bad(0, 1) has the power of continuum. Another natural example obtained by taking f i (x) = x i for i = 1, . . . , k shows that there are continuum many α ∈ R such that α, α 2 , . . . , α k are all in Bad.
Clearly, the Jacobian condition above implies that m ≥ n i for every i. Commenting on this, Davenport writes [22, p. 52] "Problems of a much more difficult character arise when the number of independent parameters is less than the dimension of simultaneous approximation. I do not know whether there is a set of α with the cardinal of the continuum such that the pair (α, α 2 ) is badly approximable for simultaneous approximation". Essentially, if m < n i then f i (x) lies on a submanifold of R n i . Hence, Davenport's problem boils down to investigating badly approximable points restricted to submanifolds of Euclidean spaces.
In the theory of Diophantine approximation on manifolds, see for instance, [7][8][9]31,33], there are already well established classes of manifolds of interest. These include non-degenerate manifolds and affine subspaces and should likely be of primary interest when resolving Davenport's problem.
It is worth pointing out that the result of Perron [40] mentioned in § 1.1 implies the existence of algebraic badly approximable points on the Veronese curves V n = {(x, . . . , x n ) : x ∈ R}. However, there are only countably many of them. Khintchine [28] proved that Bad(n) ∩ V n had zero 1-dimensional Lebesgue measure. Baker [4] generalised this to arbitrary C 1 submanifold of R n . Apparently, Bad(n) can be relatively easily replaced with Bad(r) in Baker's result, though, to the best of author's knowledge, this has never been formally addressed. To make a long story short, until recently there has been no success in relation to Davenport's problem even for planar curves, let alone manifolds in higher dimension. The aforementioned work of Badziahin et al. [13] was the first step forward. Very recently, assuming (5), Badziahin and Velani [17] have proved (6) with L θ replaced by any C 2 planar curve which is not a straight line. In particular, this shows that there exist uncountably many real numbers α such that (α, α 2 ) is in Bad(2). Also they have dealt with a family of lines in R 2 satisfying a natural Diophantine condition. The most recent results established in [3] by An, Velani and the author of this paper remove condition (5) from the findings of [17] and at the same time settle Davenport's problem for a larger class of lines in R 2 defined by a near optimal condition. As a result, the following general version of Davenport's problem is essentially settled in the case n = 2: Problem 2 Let n, m ∈ N, B be a ball in R m , W be a finite or countable subset of R n and F n (B) be a finite or countable collection of maps f : B → R n . Determine sufficient (and possibly necessary) conditions on W and/or F n (B) so that Despite the success in resolving Problem 2 for planar curves, no progress has been made on Davenport's problem for n ≥ 3. The results of this paper imply (8) in arbitrary dimensions n and for arbitrary countable subsets W of weights satisfying a condition similar to (5) and arbitrary finite collection F n (B) of analytic non-degenerate maps. The proof introduces new ideas based on lattice points counting and a powerful quantitative result of Bernik, Kleinbock and Margulis. Indeed, the arguments presented should be of independent interest even for n = 2.

Main results and corollaries
In what follows, an analytic map f : B → R n defined on a ball B ⊂ R m will be called nondegenerate if the functions 1, f 1 , . . . , f n are linearly independent over R. The more general notion of nondegeneracy that does not require analyticity can be found in [33]. Given an integer n ≥ 2, F n (B) will denote a family of maps f : B → R n with a common domain B. To avoid ambiguity, let us agree from the beginning that all the intervals and balls mentioned in this paper are of positive and finite diameter. Recall that R n denotes the collection of weights of approximation and is defined by (3). Given r = (r 1 , . . . , r n ) ∈ R n , let that is τ (r) is the smallest strictly positive weight within r. The following result regarding Problem 2 represents the main finding of this paper.
Note that the corollary is applicable to M = R n , which is clearly analytic and nondegenerate. In this case Corollary 1 establishes an analogue of Schmidt's conjecture in arbitrary dimensions n ≥ 2 by settling Problem 1 subject to condition (10).

Reduction to curves
When m = 1 the nondegeneracy of an analytic map f = ( f 1 , . . . , f n ) is equivalent to the Wronskian of f 1 , . . . , f n being not identically zero. More generally, the map f (not necessarily analytic) defined on an interval I ⊂ R will be called nondegenerate at x 0 ∈ I if f is C n on a neighborhood of x 0 and the Wronskian of f 1 , . . . , f n does not vanish at x 0 . This definition of nondegeneracy at a single point is adopted within the following more general result for curves. Note that if f is nondegenerate at least at one point, then the functions 1, f 1 , . . . , f n are linearly independent over R. Theorem 2 Let n ∈ N, n ≥ 2, I ⊂ R be an open interval and F n (I ) be a finite family of maps defined on I nondegenerate at the same point x 0 ∈ I . Let W be a finite or countable subset of R n satisfying (10). Then Our immediate goal is to show that Theorem 1 is a consequence of Theorem 2. In metric Diophantine approximation the idea of reducing the case of manifolds to curves is not new. For instance, Badziahin et al. [13] use fibering of R 2 into vertical lines in their proof of Schmidt's conjecture. Underpinning our reduction of Theorem 1 to Theorem 2 is the following version of Marstrand's slicing lemma, see [24,Corollary 7.12] or [25,Theorem 10.11].
Marstrand's slicing Lemma Let m > 1 and S be a subset of R m . Let s > 0 and let U be a subset of We will also need the following formal statement which is a slightly modified extract from Sprindžuk's survey [48, pp. 9-10].
where E u ⊂ R is a neighbourhood of 0, are linearly independent over R.
Although the proof of the Fibering Lemma mostly follows the argument of [48, pp. 9-10], for completeness full details are given in Appendix C. Note that Sprindžuk's version of fibering involves the parametrisation φ u,i (t) = f i (u 1 t, u 2 t d , . . . , u m t d m−1 ).
Proof of Theorem 1 modulo Theorem 2 Let F n (B) be as in Theorem 1 and let f = ( f 1 , . . . , f n ) ∈ F n (B). Without loss of generality we will assume that B is centred at 0. Also assume that m ≥ 2 as otherwise there is nothing to prove. Let u 1 = 1, t 0 > 0 and δ 2 , . . . , δ m > 0 be sufficiently small numbers such that The existence of t 0 , δ 2 , . . . , δ m is guaranteed by the fact that 0 is an interior point of B. Let U be the set of u = (u 2 , . . . , u m ) satisfying the right hand side inequalities of (12) and D be the set of (t, u 2 , . . . , u m ) satisfying (12).
By the nondegeneracy of f, the functions 1, f 1 , . . . , f n are linearly independent. Since they are also analytic, by the Fibering Lemma, there exists d 0 (f) > 0 such that for every d > d 0 (f) and every u ∈ U the coordinate functions of the map defined on the interval I = 1 2 t 0 , t 0 together with 1 are linearly independent over R. Since F n (B) is finite, Then for every f ∈ F n (B) and every u ∈ U the coordinate functions of the map (13) together with 1 are linearly independent over R. By the well known criterion of linear independence, their Wronskian is not identically zero. Hence, the Wronskian of f u = d dt f u is not identically zero. As an analytic function, it has isolated zeros. Hence, for a fixed u, there are at most countably many points in I where the Wronskian of f u vanishes for some f ∈ F n (B). Hence, there exists a point x 0 ∈ I , which may depend on u, such that for every f ∈ F n (B) the Wronskian of f u is not zero, that is f u is non-degenerate at x 0 . Thus, Theorem 2 is applicable and we conclude that the following subset of I has Hausdorff dimension 1. Here, by definition, f −1 u (Bad(r)) is the set of Then, in view of the definitions of S, S u and f u , we have that Further, note that (14) maps D into B injectively and is bi-Lipschitz on D, since the map itself and its inverse (defined on the image of D) have continuous bounded derivatives. It is well known that bi-Lipschitz maps preserves Hausdorff dimension, see for example [24,Corollary 2.4]. Therefore, dim S = dim S ≥ m. By (15), and the fact that any subset of R m is of dimension ≤ m, we obtain (8) and thus complete the proof of Theorem 1 modulo Theorem 2.

The dual form of approximation
So far we have been dealing with simultaneous rational approximations. Here we introduce the dual definition of badly approximable points-see part (iii) of Lemma 1 below. This has two purposes. Firstly, it is the dual form that will be used in the proof of the results. Secondly, the dual form provides a natural environment for considering Diophantine approximation by algebraic numbers and will allow us to deduce further corollaries of our main results.
The equivalence of (i) and (ii) is a straightforward consequence of the definition of Bad(r). The equivalence of (ii) and (iii) is relatively well known, see Appendix in [13] for a similar statement. Indeed, this equivalence is essentially a special case of Mahler's version of Khintchine's transference Principle appearing in [38]. To make this paper self-contained we provide further details in Appendix A.

Approximation by algebraic numbers of bounded degree
There are two classical interrelated settings in the theory of approximation by algebraic numbers of bounded degree. One of them boils down to investigating small values of integral polynomials P with deg P ≤ n at a given number ξ . The other deals with the proximity of algebraic numbers α of degree ≤ n to a given number ξ , see [14] for further background. In particular, the long standing Wirsing-Schmidt conjecture [45, p. 258], which was motivated by Wirsing's theorem [49], states that for any n ∈ N and any real transcendental number ξ there is a constant C = C(ξ, n) > 0 such that holds for infinitely many algebraic numbers α of degree ≤ n, where H (α) denotes the height of α (to be recalled a few lines below). The n = 1 case of the conjecture is a trivial consequence of the theory of continued fractions. For n = 2 it was proved by Davenport and Schmidt [23]. However, there are only partial results for n > 2. Note, however, that using Dirichlet's theorem it is easily shown that for any ξ ∈ R there exists c 0 = c 0 (ξ, n) > 0 such that In this section we will deal with real numbers badly approximable by algebraic numbers. Given a polynomial P with integer coefficients, H (P) will denote the height of P, which, by definition, is the maximum of the absolute values of the coefficients of P. Given an algebraic number α ∈ C, H (α) will denote the (naive) height of α, which, by definition, is the height of the minimal defining polynomial P of α over Z. It is also convenient to introduce the following three sets: for infinitely many real algebraic α with deg α ≤ n , The sets B n and B * n are the natural generalisations of badly approximable numbers to the context of approximation by algebraic numbers. They are known to have Lebesgue measure zero, e.g., by a Khintchine type theorem proved in [10]. Within this paper we will deal with the following two conjectures that Bugeaud The proof of (18) is rather standard. Indeed, it rests on the Mean Value Theorem and Minkowski's theorem for convex bodies, see Appendix B for details. Here we establish the following Hausdorff dimension result that easily settles the above conjectures.

Theorem 3 For any natural number n and any interval I in
Proof Without loss of generality we will assume that n ≥ 2. Let 1 ≤ k ≤ n be an integer and r k = 1 k , . . . , 1 k , 0, . . . , 0 ∈ R n , where the number of zeros is n − k. Let ξ ∈ R be such that f(ξ ) ∈ Bad(r k ). By Property (iii) of Lemma 1, there exists c(ξ, n, k) > 0 such that for any H ≥ 1 the only integer solution (a 0 , a 1 , . . . , a n ) to the system |a 0 + a 1 x + · · · + a n x n | < c(ξ, n, k)H −1 , is a 0 = · · · = a n = 0. Hence, for any non-zero polynomial P(x) = a k x k + · · · + a 0 ∈ Z[x] with H (P) < H 1/k we must have that |P(ξ )| ≥ c(ξ, n, k)H −1 > c(ξ, n, k)H (P) −k . By definition, this means that ξ ∈ B k . To sum up, we have just shown that By Theorem 2, for any interval I ⊂ R we have that dim n k=1 f −1 (Bad(r k )) ∩ I = 1. In view of the above inclusions the statement of Theorem 3 now readily follows.
Remark An interesting problem is to show that Theorem 3 holds when n = ∞.

Lattice points counting
The rest of the paper will be concerned with the proof of Theorem 2, which will rely heavily on efficient counting of lattice points in convex bodies. The lattices will arise upon reformulating Bad(r) in the spirit of Dani [20] and Kleinbock [30]. This will require the following notation. Given a subset of R n+1 , let where a ∞ = max{|a 0 |, . . . , |a n |} for a = (a 0 , . . . , a n ). Given 0 < κ < 1, let where y ∈ R n is regarded as a row and I n is the n × n identity matrix. Finally, given r ∈ R n , b > 1 and t ∈ R, define the (n + 1) Proof The necessity is straightforward as all one has to do is to take H = b t and divide each inequality in (17) by its right hand side. Then, assuming that y ∈ Bad(r), the non-existence of integer solutions to (17) would imply (22) with κ = c. The sufficiency is only slightly harder. Assume that for some κ and b inequality (22) holds for all t ∈ N, while y / ∈ Bad(r). Take c = κ/b. By definition, there is an H > 1 such that (17) has a non-zero integer solution (a 0 , . . . , a n ). Take t = [log H/ log b] + 1, where [·] denotes the integer part. Note that Hb −t < 1 and H −1 b t ≤ b. Then (17) implies that δ(g t r,b G(κ; y)Z n+1 ) < 1, contrary to (22). The proof is thus complete.
Remark Lemma 2 can be regarded as a variation of the Dani-Kleinbock correspondence between badly approximable points in R n and bounded orbits of certain lattices under the actions by the diagonal semigroup g t r,b : t > 0 , where b > 1. It is easily seen that this semigroup is independent of the choice of b > 1, which is usually taken to be e = exp(1). The correspondence was first established by Dani [20] in the case r = 1 n , . . . , 1 n and then extended by Kleinbock [30] to the case of arbitrary positive weights and can be stated as follows. The point y ∈ R n is r-badly approximable if and only if the orbit of the lattice G(1; y)Z n+1 under the action by g t r,e : t > 0 is bounded.
We proceed by recalling two classical results from the geometry of numbers. In what follows, vol (X ) denotes the -dimensional volume of X ⊂ R and # X denotes the cardinality of X . Also det will denote the determinant or covolume of a lattice .
Minkowski's Convex Body Theorem (see [45,Theorem 2B]) Let K ⊂ R be a convex body symmetric about the origin and let be a lattice in R . Suppose that vol (K ) > 2 det . Then K contains a non-zero point of .
Theorem (Blichfeldt [12]) Let K ⊂ R be a convex bounded body and let be a lattice in R such that rank (K ∩ ) = . Then The following lemma is a straightforward consequence of Blichfeldt's theorem.

Lemma 3 (cf. Lemma 4 in [34]) Let K be a convex bounded body in
Proof Assume the contrary, that is assume that rank (K ∩ Z ) = (note that the rank cannot be bigger than ). It means that K contains at least non-zero integer points. Since 0 ∈ K , we then have that # K ∩ Z ≥ + 1. However, since det Z = 1 and vol (K ) < 1/ !, by Blichfeldt's theorem, we conclude that contrary to the above lower bound.
The bodies K of interest will arise as the intersection of parallelepipeds with -dimensional subspaces of R n+1 , where θ = (θ 0 , . . . , θ n ) is an (n + 1)tuple of positive numbers. In view of this, we now obtain an estimate for the volume of the bodies that arise this way (Lemma 4 below) and then verify what Blichfeldt's theorem means for such bodies (Lemma 5 below).
Proof Since V is a linear subspace of R n+1 of dimension , it is given by n + 1 − linear equations. Using Gaussian elimination, we can rewrite these equations to parametrise V with a linear map f : Then note that vol ( θ ∩ V ) is bounded by the area of the intersection of V with the cylinder |x i j | ≤ θ i j for j = 1, . . . , . This area is equal to where · e is the Euclidean norm on R n+1 . Since |m i, j | ≤ 1, every coordinate of every partial derivative of f is bounded by 1 in absolute value. Hence ∂f/∂ x i j e ≤ √ n + 1 and the integrand in (24) is bounded above by ( √ n + 1) . This readily implies that the area given by (24) is bounded above by 2 (n + 1) /2 θ i 1 . . . θ i ≤ 2 (n + 1) /2 , whence the result follows.
We are now approaching the key counting result of this section. Let where u > 0, b > 1 and θ is given by (23). Given r ∈ R n , let Recall that τ (r), δ(·), g t r,b and (b, u) are given by (9), (19), (21) and (26) respectively, and [x] denotes the integer part of x.
Let be a discrete subgroup of R n+1 such that rank ≤ n − z and Then Proof Let x = (x 0 , . . . , x n ) ∈ be such that g t x ∈ (b, u). By the definitions of g t = g t r,b and (b, u), we have that b t |x 0 | < b u and b −r i t |x i | < 1 for i = 1, . . . , n. Equivalently, for s ∈ Z, 1 ≤ s ≤ u − 1, we have that This can be written as g t−s x ∈ θ , where θ = (b u−s , b r 1 s , . . . , b r n s ). Therefore, where = rank = rank ≤ n − z. Note that all the components of θ are ≥ 1 and exactly z of them equal 1. Then, since ≤ n − z and s = [λu], we get that Combining this estimate with (31) and the obvious fact that n +1 < c(n)b τ b λu gives (29).

'Dangerous' intervals
In view of Lemma 2, when proving Theorem 2 we will aim to avoid the solutions of the inequalities δ g t r,b G x Z n+1 < 1, where G x = G(κ; y) with y = f(x) and κ is a sufficiently small constant. For fixed r, b, t, f and κ the above inequality is equivalent to the existence of (a 0 , a) ∈ Z n+1 with a = 0 satisfying Here the dot means the usual inner product. That is a.b = a 1 b 1 + · · · + a n b n for any given a = (a 1 , . . . , a n ) and b = (b 1 , . . . , b n ). In this section we study intervals arising from (32) that, for obvious reasons, are referred to as dangerous (see [45] for similar terminology). We will consider several cases that are tied up with the magnitude of a.f (x); i.e., the derivative of a 0 + a.f(x)-see Propositions 1 and 2 below. Throughout F n (I ) and x 0 are as in Theorem 2. First we discuss some conditions that arise from the nondegeneracy assumption on maps in F n (I ). Let f = ( f 1 , . . . , f n ) ∈ F n (I ). Since f is nondegenerate at x 0 ∈ I , there is a sufficiently small neighborhood I f of x 0 such that the Wronskian of f 1 , . . . , f n , which, by definition, is the determinant det f (i) j 1≤i, j≤n , is non zero everywhere in I f . Then every coordinate function f j is non-vanishing at all but countably many points of I f ⊂ I -see, e.g., [5,Lemma 3]. Since f ∈ C n and F n (I ) is finite, we can choose a compact interval I 0 ⊂ f∈F n (I ) I f ⊂ I satisfying Property F There are constants 0 < c 0 < 1 < c 1 such that for every map f = ( f 1 , . . . , f n ) ∈ F n (I ), for all x ∈ I 0 , 1 ≤ i ≤ n and 0 ≤ j ≤ n one has that det f Next, we prove two auxiliary lemmas that are well known in a related context. Lemma 7 (cf. Lemma 5 in [5]) Let I 0 ⊂ I be a compact interval satisfying Property F. Let 2c 2 = c 0 c −n+1 1 n! −1 , where c 0 and c 1 arise from (33). Then for any f ∈ F n (I ), any a = (a 1 , . . . , a n ) ∈ Z n \{0} and any x ∈ I 0 there exists i ∈ {1, . . . , n} such that |a.f (i) (x)| ≥ 2c 2 max 1≤ j≤n |a j |. Proof Solving the system a 1 f by Cramer's rule with respect to a i and using (33) to estimate the determinants involved in the rule we obtain for each j = 1, . . . , n, whence the statement of lemma readily follows.
Lemma 8 (cf. Lemma 6 in [5]) Let I 0 ⊂ I and c 2 be as in Lemma 7. Then there is δ 0 > 0 such that for any interval J ⊂ I 0 of length |J | ≤ δ 0 , any f ∈ F n (I ) and a = (a 1 , · · · , a n ) ∈ Z n \{0}, there is an i ∈ {1, . . . , n} satisfying Proof Since I 0 is compact, for each f ∈ F n (I ) and 1 ≤ i ≤ n, the map f (i) is uniformly continuous on I 0 . Hence, there is a δ i,f > 0 such that for any x, y ∈ I 0 with |x − y| ≤ δ i,f we have f (i) (x) − f (i) (y) < c 2 /n. Let J ⊂ I 0 be an interval of length |J | ≤ δ i,f and x, y ∈ J . By Lemma 7, there is Finally, let t ∈ N, ∈ Z ≥0 , b > 1, a ∈ Z n \{0}, a 0 ∈ Z, 0 < κ < 1 and Then, there is a constant c 3 > 0 depending on n, |I 0 |, c 1 , c 2 and δ 0 only such that the set D 1 t, ,r,b,κ,f (a 0 , a) can be covered by a collection D 1  t, ,r,b,κ,f (a 0 , a)
It remains to estimate the length of each q . To this end, take any x 1 , x 2 ∈ q . By the construction of q , the numbers f (x 1 ) and f (x 2 ) have the same sign and satisfy the inequality | f (x i )| < κb −t . Hence, . This estimate together with the obvious equality | q | = sup x 1 ,x 2 ∈ q |x 1 − x 2 | implies that | q | ≤ κb −(1+γ )(t− ) . Thus, if i > 1, the set D 1 can be covered by at most n(n + 1)/2 + 1 intervals of length κb −(1+γ )(t− ) . Now consider the case i = 1. Recall that f (x) = a 0 + a.f(x). Then, by the definition of D 1 and (33), for x ∈ D 1 we get Further, (34) Therefore, f is monotonic on I 0 and D 1 is covered by a single interval defined by the inequality | f (x)| < κb −t . Arguing as above and using (37) we get Thus, by splitting into smaller intervals if necessary, D 1 can be covered by at most 2c 1 n c 2 Proposition 2 Let I 0 ⊂ I be a compact interval satisfying Property F and γ = γ (r) be given by (36). Then there are constants K 0 > 0 and 0 < κ 0 < 1 such that for any f ∈ F n (I ), any r ∈ R n , t ∈ N, 0 ≤ ε < γ , b > 1 and 0 < κ < κ 0 the set and where δ t = κ b −t (1+γ −ε) and α = 1 (n+1)(2n−1) . Proposition 2 will be derived from a theorem due to Bernik, Kleinbock and Margulis using the ideas of [6]. In what follows |X | denotes the Lebesgue measure of a set X ⊂ R. The following is a simplified version of Theorem 1.4 from [11] that refines the results of [33]. x ∈ I : ∃a ∈ Z n \{0} We will also use the following elementary consequence of Taylor's formula.
Proof of Proposition 2 Fix any f ∈ F n (I ). We will abbreviate D 2 t,ε,r,b,κ,f as D 2 and naturally assume that it is non-empty as otherwise there is nothing to prove. By (33), f is nondegenerate at any x ∈ I 0 and therefore Theorem 4 is applicable. Let J = J (x) be the interval centred at x that arises from Theorem 4. Since I 0 is compact there is a finite cover of I 0 by inter- Let 0 < κ < κ 0 , r ∈ R n , t ∈ N, 0 ≤ ε < γ , b > 1 and let Note that since ε < γ and c 1 > 1 we have that K > 2. Also note that ω < 2κ.
Next, by Theorem 4, condition r 1 + · · · + r n = 1 and (44) we conclude that where E f = s max 1≤i≤s E J (x i ) . By (43),D 2 can be written as a union of disjoint intervals of length By splitting some of these intervals if necessary, we get a collectionD 2 of disjoint intervals such that 1 Then, by (46) and the above inequality, we get Let D 2 be the collection of all the intervals inD 2 together with the 2s intervals It is easily seen that 2s is less than or equal to the right hand side of (47). Then, by (47) and the definition of D 2 , we get (38) and (39). Also, by construction, we see that D 2 is a cover of D 2 . The proof is thus complete.

A Cantor sets framework
Let R ≥ 2 be an integer. Given a collection I of compact intervals in R, let The intervals lying in I q will be called to be of level q. Thus, the intervals of level q are obtained from intervals of level q − 1 by, firstly, splitting the intervals of I q−1 into R equal parts to form 1 R I q−1 , and, secondly, removing some of the intervals from 1 R I q−1 to form I q . Given q ∈ N, the intervals that are being removed in this procedure will be denoted by Naturally, I q will denote any interval from the collection I q , that is any interval of level q. Observe that By definition, given I q ∈ I q with q ≥ 1, there is a unique interval I q−1 ∈ I q−1 such that I q ⊂ I q−1 ; this interval I q−1 will be called the precursor of I q . Obviously it is independent of the choice of the R-sequence (I q ) q≥0 with I q ∈ I q .
We also define the limit set of (I q ) q≥0 as This is a Cantor type set. The classical middle third Cantor set can be constructed this way in an obvious manner with R = 3 and I 0 = [0, 1]. Theorem 2 will be proved by finding suitable Cantor type sets K((I q ) q≥0 ). The construction of the corresponding R-sequences will be based on removing the intervals that intersect dangerous intervals-see Sect. 4. Note that if I q = ∅ for all q so that (I q ) q≥0 is genuinely an infinite sequence, then K((I q ) q≥0 ) = ∅. However, ensuring that K((I q ) q≥0 ) is large requires better understanding of the sets I q . There are various techniques in fractal geometry that are geared towards this task-see [24]. We shall use a recent powerful result of Badziahin and Velani [16] restated below using our notation. Naturally, if we expect that the Cantor set K((I q ) q≥0 ) is large, then the number of removed intervals at level q, that is the cardinality of I q , should be relatively small. In what follows, given q ∈ N and an interval J , let This denotes the subcollection of removed intervals (when going from level q − 1 to level q) that lie over a given interval J . The key characteristic that is 'assessing' the proportion of removed intervals at a particular level is given by where the minimum is taken over all partitions { I q, p } q−1 p=0 of I q , that is I q = q−1 p=0 I q, p . Also define the corresponding global characteristic as The goal is to ensure that d((I q ) q≥0 ) is small. Then as we shall shortly see the corresponding Cantor set is large. Note that when estimating d q (I q ) the key is to arrange the removed intervals into a partition q−1 p=0 I q, p which makes the sum on the right of (51) small. [16]) Let R ≥ 4 be an integer, I 0 be a compact interval in R and (I q ) q≥0 be an R-sequence in I 0 . If d((I q ) q≥0 ) ≤ 1 then

Theorem 5 (Theorem 4 in
In order to facilitate the comparison of Theorem 5 to [16,Theorem 4] we summarise the correspondence between the notation and objects used in this paper and in [16]: Our notation/object Corresponding notation/object in [16] q n + 1 R R n (allowed to vary with n) Given the above correspondence table, it is readily verified that our condition d((I q ) q≥0 ) ≤ 1 corresponds to condition (16) within [16,Theorem 4]. Hence Theorem 5 above is an immediate consequence of Theorem 4 from [16].
Let M > 1, X ⊂ R and I 0 be a compact interval. We will say that X is M-Cantor rich in I 0 if for any ε > 0 and any integer R ≥ M there exists an R-sequence (I q ) q≥0 in I 0 such that K((I q ) q≥0 ) ⊂ X and d((I q ) q≥0 ) ≤ ε. We will say that X is Cantor rich in I 0 if it is M-Cantor rich in I 0 for some M. We will say that X is Cantor rich if it is Cantor rich in I 0 for some compact interval I 0 . The following statement readily follows from Theorem 5 and our definitions.
Theorem 6 Any Cantor rich set X satisfies dim X = 1.
We now proceed with a discussion of the intersections of Cantor rich sets. To some extent this already appears in [16,Theorem 5] and in [13]. First we prove the following auxiliary statement.

Lemma 10 Let
be a family of R-sequences in I 0 indexed by j. Given q ∈ Z ≥0 , let J q = j I j q . Then (J q ) q≥0 is an R-sequence in I 0 such that and Proof The validity of (48) for (J q ) q≥0 follows from the uniqueness of the precursor of an interval in any R-sequence from that sequence and the fact that I j 0 = {I 0 } for all j, which means that J 0 = j I j 0 = {I 0 }. Thus, (J q ) q≥0 is truly an R-sequence. The inclusion (53) is obvious for q = 0 for both sides of the inclusion are empty sets in this case. To see (53) for q > 0, observe that J q−1 ⊂ I j q−1 and this implies that 1 Finally, by the inclusion J q ⊂ I j q , we have that J q ⊂ I j q for each pair of j and q, where the union is taken over J q ∈ J q and I j q ∈ I j q respectively. Hence, by (50), we have that K((J q ) q≥0 ) ⊂ K((I For q ∈ Z ≥0 define J q = j∈N I j q and J q, p = J q ∩ j∈N I j q, p . Since I j q = q−1 p=0 I j q, p for each j, by (53), we have that J q = q−1 p=0 J q, p , where q > 0. Then, for each q > 0 we get that This inequality together with (55) and the definition of d((J q ) q≥0 ) implies that d((J q ) q≥0 ) ≤ ε. By (54) and the fact that K I j q q≥0 ⊂ X j for each j, we have that K((J q ) q≥0 ) ⊂ j X j . Thus the intersection j X j meets the definition of M-Cantor rich sets and the proof is complete.
The winning sets in the sense of Schmidt have been used a lot to investigate various sets of badly approximable points. Hence we suggest the following Problem 3 Verify if an α-winning set in R as defined by Schmidt [45] is M-Cantor rich for some M and, if this so, find an explicit relation between M and α.

Proof of Theorem 2
The following proposition is a key step to establishing Theorem 2. We will use the Vinogradov symbol to simplify the calculations. The expression X Y will mean that X ≤ CY for some C > 0, which only depends on n, the family of maps F n (I ) from Theorem 2 and the interval I 0 occurring in Property F. and let

Proposition 3 Let
and Then for any f ∈ F n (I ), r ∈ R n and any integers m ≥ m 0 and R ≥ R 0 , there exists an R-sequence (I q ) q≥0 in I 0 such that (i) for any t ∈ N and any I t+m ∈ I t+m we have that where g t = g t r,b is given by (21) with b 1+γ = R, γ = γ (r) and G x = G(κ; f(x)) is given by (20) with κ = R −m ; (ii) if q ≤ m then # I q = 0; (iii) if q = t + m for some t ∈ N then I q can be written as the union I q = q−1 p=0 I q, p such that for integers p = t + 3 − 2 with 0 ≤ ≤ t = [t/2n] + 1 and I p ∈ I p we have that and I q, p = ∅ for all other p < q, where λ = λ(r) is given by (27).
Proof Note that since 0 , 1 > 1 and c 0 < 1, we have that R 0 > 4. Let m ≥ m 0 and R ≥ R 0 be any integers. Define I 0 = {I 0 } and then for q = 1, . . . , m let I q = 1 R I q−1 . In this case conditions (i) and (iii) are irrelevant, while (ii) is obvious. Continuing by induction, let q = t + m with t ≥ 1 and let us assume that I q with q < q are given and satisfy conditions (i)-(iii). Define I q to be the collection of intervals from 1 R I q−1 that satisfy (59). By construction, (i) holds, (ii) is irrelevant and we only need to verify condition (iii). We shall assume that I q = ∅ as otherwise (iii) is obvious. By construction, I q consists of intervals I q such that δ g t G x Z n+1 < 1 for some x ∈ I q . Recall that this is equivalent to the existence of (a 0 , a) ∈ Z n+1 with a = 0 satisfying the system (32). We shall use Propositions 1 and 2 and Lemma 6 to estimate the number of these intervals I q . Before we proceed with the estimates note that, by (33) and (36), the validity of (32) implies that |a.f (x)| ≤ nc 1 max 1≤ j≤n |a j | ≤ nc 1 max 1≤ j≤n b r j t = nc 1 b γ t . Thus, The arguments split into two cases depending on the size of t as follows. Note that in view of our choice of m 0 we have that and so Proposition 2 is applicable as appropriate.
Case 1: t ≤ 2nm. In this case let I q,0 = I q and I q, p = ∅ for 0 < p < q.
Then, the only thing we need to verify is (61). Let ε = 0. Then, by (62), we have that where D 2 0 = D 2 t,ε,r,b,κ,f (with ε = 0) as defined in Proposition 2. Hence, # I q,0 is bounded by the number of intervals in 1 R I q−1 that intersect an interval from the corresponding collection D 2 0 of intervals arising from Proposition 2. By (49), the intervals in 1 R I q−1 are of length R −q |I 0 |. By (38), the intervals from Then, by (39), we get Using q = t + m, κ = R −m and t ≤ 2nm we obtain from (64) that Recall from Proposition 2 that α = Case 2: t > 2nm. Let ε = (2n) −1 . Since i r i = 1 and γ = max{r 1 , . . . , r n }, we have that γ ≥ 1/n. Hence ε < γ . Recall that R > nc 1 . Then, by (62) and the choice of ε, for any x ∈ I 0 such that δ(g t G x Z n+1 ) < 1 we have that either Then, once again using the equivalence of δ g t G x Z n+1 < 1 to the existence of (a 0 , a) ∈ Z n+1 with a = 0 satisfying (32), we write that By definition, intervals in I q are characterised by having a non-empty intersection with the left hand side of (66). We now use the right hand side of (66) to define the subcollections I q, p of I q . More precisely, for p = t +3−2 with 0 ≤ ≤ t let I q, p consist of the intervals I q ∈ I q that intersect D 1 (a 0 , a) for some a ∈ Z n \{0} and a 0 ∈ Z. Next, let I q,0 consist of the intervals I q ∈ I q that intersect D 2 . Finally, define I q, p = ∅ for all other p < q. By (66), it is easily seen that I q = q−1 p=0 I q, p . It remains to verify (60) and (61).
• Verifying (61) This is very much in line with Case 1. The goal is to count the number intervals in 1 R I q−1 that intersect some interval from the collection D 2 arising from Proposition 2. By (49), the intervals in 1 R I q−1 are of length R −q |I 0 |. By (38), the intervals from D 2 have length ≤ δ t = κ b −t (1+γ −ε) . Hence, each interval from D 2 can intersect at most δ t /(R −q |I 0 |) + 2 δ t R q intervals from 1 R I q−1 . Then, by (39), we get Using κ = R −m , b 1+γ = R, q = t + m and 0 < γ ≤ 1, we obtain that Once again using the value of α from Proposition 2 we verify that σ ≥ 1− 1 2 εα and so (67) implies (61) as required.
• Verifying (60). Let p = t + 3 − 2 with 0 ≤ ≤ t and I p ∈ I p . Let S(I p ) be the set of points (a 0 , a) ∈ Z n+1 with a = 0 such that D 1 (a 0 , a) ∩ I p = ∅. By Proposition 1, for every (a 0 , a) ∈ S(I p ) any interval in D 1 (a 0 , a) is of length as κ = R −m , b 1+γ = R and q = t + m. Then, by (49), any interval from D 1 (a 0 , a) intersects R intervals from 1 R I q−1 . By Proposition 1, #D 1 (a 0 , a) 1. Hence, and our main concern becomes to obtain a bound for #S(I p ). We shall prove that #S(I p ) R τ 1+γ +λ(m+ −1) .
Step 1: We show that for any (a 0 , a) ∈ S(I p ) and any x ∈ I p we have where 0 and 1 are given by (56).
First we prove the right hand side of (70). To this end, fix any (a 0 , a) ∈ S(I p ) and let x 0 ∈ D 1 (a 0 , a) ∩ I p . To simplify notation define f (x) = a 0 + a.f(x). By the Mean Value Theorem, for any x ∈ I p we have wherex 0 is a point between x and x 0 . By the definition of D 1 (a 0 , a) implied by (49), we get Since ≤ t < t/4 + 1, one easily verifies that (t + 3 − 2 ) > ( − 1). Therefore (73) implies the right hand side of (70). Now we prove the left hand side of (70). Again fix any (a 0 , a) ∈ S(I p ), x 0 ∈ D 1 (a 0 , a) ∩ I p and let f (x) = a 0 + a.f(x). By the Mean Value Theorem, for any x ∈ I p we have that where x 0 is a point between x and x 0 . In particular, x 0 ∈ I p and therefore, by the right hand side of (70), which we have already established, . By the definition of D 1 (a 0 , a), . Hence, using these estimates together with inequality (72) and equation b 1+γ = R, we get from (74) that Since m ≥ m 0 ≥ 4 we have that −m(1 + γ ) ≤ (1 + γ )( − 2) for all ≥ 0. Therefore (75) implies the left hand side of (70).
Step 2: Now we utilize (70) to show that rank S(I p ) ≤ n − z. First of all, observe that if (a 0 , a) ∈ S(I p ), where a = (a 1 , . . . , a n ), then |a j | < b r j t = 1 whenever r j = 0. Since a j ∈ Z in this case, we have that ∀ (a 0 , a) ∈ S(I p ) a j = 0 whenever r j = 0.
Let J = { j : r j = 0} and J = {1, . . . , n}\ J . Note that J contains exactly n − z > 0 elements, where z = z(r) is the number of zeros in r. Let J 0 be the subset of J obtained by removing the smallest index j 0 such that r j 0 = γ (r).
Note that if r has only one non-zero component then J 0 = ∅. Let x ∈ I p . Then, using (76) and (70) we obtain that every (a 0 , a) ∈ S(I p ) satisfies the system where 0 and 1 ar given by (56). Let B p,x denote the set of (a 0 , a 1 , . . . , a n ) ∈ R n+1 satisfying (77). Then, S(I p ) ⊂ B p,x . Clearly, B p,x is a convex body lying over the n − z + 1 dimensional linear subspace of R n+1 given by the equations a j = 0 for j ∈ J . As is well known the n − z + 1-dimensional volume of B p,x is equal to where is the determinant of the system of linear forms in the variables a j , j ∈ J ∪ {0}, staying in the first three lines of (77). Note that | = | f j 0 (x) .
Hence, using (33) and the fact that b 1+γ = R ≥ R 0 , we conclude that the volume of B p,x is In this case, Lemma 3 is applicable and we have that rank S(I p ) ≤ n − z as claimed at the start of Step 2.
Step 3 : Finally, we obtain (69). To this end, let denote the Z-span of S(I p ).
The following key statement is essentially a corollary of Proposition 3.

Theorem 8
Let F n (I ) be as in Theorem 2, I 0 ⊂ I be a compact interval satisfying Property F. Then there is a constant M 0 ≥ 4 such that for any r ∈ R n and any f ∈ F n (I ) the set f −1 (Bad(r)) is M-Cantor rich in I 0 for any M > max M 0 , 16 1+1/τ , where τ = τ (r) is defined by (9).
Let M > max M 0 , 16 1+1/τ , R ≥ M and m ≥ m 0 . Take any f ∈ F n (I ) and r ∈ R n . Let (I q ) q≥0 denote the R-sequence in I 0 that arises from Proposition 3. By (59) and Lemma 2, we have that K((I q ) q≥0 ) ⊂ f −1 (Bad(r)). Thus, by definition, the fact that f −1 (Bad(r)) is M-Cantor rich in I 0 will follow on showing that d((I q ) q≥0 ) can be made ≤ ε for any ε > 0.
Once again, by conditions (ii) and (iii) of Proposition 3, for q ≤ m we have that # I q,0 I 0 = 0, while for q > m (81) as m → ∞. By (51), combining (80) and (81) gives d q (I q ) ≤ ε for all q > 0 provided that m is sufficiently large. This completes the proof.

Final remarks
In this section we discuss possible generalisations of our main results and further problems. First of all, the analyticity assumption within Theorem 1 can be relaxed by making use of more general fibering techniques such as that of [42]. This however leaves the question of whether Theorem 1 holds for arbitrary nondegenerate submanifold of R n as defined in [33] open. Beyond nondegenerate manifolds, it would be interesting to obtain generalisations of Theorems 1 and 2 for friendly measures as defined in [32] as well as for affine subspaces of R n and their submanifolds-see [31] for a related context. In another direction, it would be interesting to develop the theory of badly approximable systems of linear forms. Removing condition (10) is another appealing problem that would be settled if the sets of interest were shown to be winning in the sense of Schmidt (see [1], [3], [17, §1.3] and [45]). However, the techniques of this paper could also help accomplishing this task: the key is to make the lower bound on M appearing in Theorem 8 independent of τ (r). Finally, all of the above questions make sense and are of course interesting in the case of Diophantine approximation over Q p and in positive characteristic.
where λ = δ n . This implies (17) with H = Q and c = nδ. Note that c → 0 as c → 0. Thus if there is c > 0 such that the only integer solution to (17) is a 0 = a 1 = · · · = a n = 0, then there must exist a c > 0 such that the only integer solution to (16) is q = p 1 = · · · = p n = 0. The converse is proved in exactly the same way by swapping the roles of L i and L i and taking T 0 = c H −1 , T i = H r i and Q = (n + 1)H .
The case when r contains a zero is treated by induction. The case n = 1 meaning r = (r 1 ) with r 1 = 0 has already been done. Assume that n > 1 and our desired statement holds for smaller dimensions. Assume that r contains a zero component. Without loss of generality assume that r n = 0. Since x 1/0 = 0, we have that max 1≤i≤n qy i 1/r i = max 1≤i≤n−1 qy i 1/r i . Therefore, y ∈ Bad(r) if and only if y = (y 1 , . . . , y n−1 ) ∈ Bad(r ), where r = (r 1 , . . . , r n−1 ). By induction, this is equivalent to the existence of c > 0 such that for any H ≥ 1 the only integer solution (a 0 , a 1 , . . . , a n−1 ) to the system |a 0 + a 1 y 1 + · · · + a n−1 y n−1 | < cH −1 , |a i | < H r i (1 ≤ i ≤ n − 1) is a 0 = · · · = a n−1 = 0. In turn, the latter statement is equivalent to (iii), since, by r n = 0, the inequality |a n | < H r n implies that a n = 0 whenever a n ∈ Z.