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Main Results via Perron-Frobenius
Throughout this section let q 2 be a fixed integer. From the expansion (2.2) of a real number x 2 [0, 1] into an integer base q we see that we can redefine the set F(c, ), defined in (3.2), into a set of sequences of q symbols. We have F(c, q) = {x 2 S1(q) : n(x) c for all n 0}.
where clearly c is the sequence from the q-nary expansion of the real number c 2 [0, 1]. We define the intervals I(c, q) via the concept of minimal prefixes, that is, we let I(c, q) = x 2 S1(q) : c[1, nc] x c[1, nc]1 The next theorem gives that the definition of I(c, q) is independent of the choice of the representative c. Theorem 5.1 For any d 2 I(c, q) we have I(d, q) = I(c, q).
Proof: We may assume that c is a finite minimal sequence. Let d 2 I(c, q). Then Corollary 4.7 gives that d[1, nd] = c, and therefore we get I(d, q) = I(c, q).
We can now state a first result of the map q(c) = dimH F(c, q). On Numbers Badly Approximable by q-adic Rationals Theorem 5.2 The derivative of q is zero Lebesgue a.e. Proof: The sequences which give rise to one-point intervals I(c, q) = {c} are precisely the sequences c 2 IM (q). As IM (q) has Lebesgue measure 0, by Lemma 4.10 we must have that the complementary set, the set formed by the intervals, has full Lebesgue measure. Lemma 5.3 For any c 2 S(q) the dynamical system : F(c, q) ! F(c, q) is topologically mixing. Proof: Let U F(c, q) be the cylinder-set defined by [u1u2 . . . uk] and let similarly V F(c, q) bethe cylinder-set defined by [v1v2 . . . vj ]. By choosing the number N sufficiently large the cylinder-set [u1u2 . . . uk (q− 1)Nv1v2 . . . vj ] is a subset of U. This shows that the intersection n(U)\ V is non-empty for all n N.
For the next corollary recall from Example 2.4 how to associate a subshift with a transition matrix. Corollary 5.4 Let c 2 S(q) be such that F(c, q) is a subshift of finite type. Then the transition matrix Ac corresponding to F(c, q) is primitive. Theorem 5.5 The interval I(c, q) is the largest interval I on which q(d) = q(c) for d 2 I. Proof: We may assume that c is a finite minimal sequence. From Lemma 4.1 we have that F(c, q) = F(c1, q), and hence dimH F(c, q) = dimH F(c1, q). From Lemma 4.9 the sequences {ak(c, q)} can be assumed minimal. Let Aak be a transition matrix corresponding to F(ak, q) and let Ac be a transition matrix corresponding to F(c, q). We can scale them to beof the same size if needed. Since F(ak, q) \ F(c, q) is non-void, we have (Aak )ij (Ac)ij , entry by entry, where there is one pair of indices r, s such that the inequality is strict. As Aak is irreducible it follows from the Perron-Frobenius Theorem 2.2 that q(ak) > q(c). That we cannot go beyond the right endpoint of I(c, q) follows by the same arguments when considering the sequences {bk(c, q)}. Example 5.6 Let c = 0.25 and q = 2. Then c corresponds to the sequence c = 01, which is a minimal sequence, and we have that a transition matrix connected to F(c, q).
Main Results via the -shift
Recall the definition of the set F(c, ) from (3.2) for any > 1, F(c, ) = {x 2 S : nx c (mod 1) for all n 0} . We can clearly choose to study the symmetrically identical set F0(c, ) = {x 2 S : nx 1 − c (mod 1) for all n 0} By using the -expansion of real numbers we may turn into dealing with a set of sequences. That is, we let F(c) = {x 2 S : n(x) < d(1 − c, ) for all n 0} . (7.1) Note that we may change the inequality in (7.1) to be strict, as this only removes the periodic sequences, which are countable and hence does not affect the dimension. We emphasise also that dimH F(c, ) = dimH F(c). From Lemma 4.1 and Lemma 4.4 we have the following two corollaries
by, simply consider inverse sequences. Corollary 7.1 Let s 2 S(de − 1) be a finite sequence of length n not ending with a 0. Then n(x) (˜s) for all n 0 if and only if n(x) ((˜s))1 for all n 0. Corollary 7.2 A finite sequence s 2 S(de − 1) is minimal if and only if n((˜s)) < (˜s) for all 0 < n < |s|. An infinite sequence s 2 S1(de − 1) is minimal if and only if n(s) < s for all n > 0.
Fundamental Properties
From the definition of F(c, q) and by symmetry it is clear that we have the equaivalence x 2 F(c, q) if and only if x0 2 F(c, q). Lemma 10.1 Let c 2 S(q). Then F(c, q) = F(c1, q).
Proof: The lemma is a direct consequence from Lemma 4.1. Lemma 10.2 Let c 2 S(q) be of the form c = ˜u (u)ku0v for some k 0 and a finite sequence u. If x 2 F(c, q) contains the subsequence ˜u , (or symmetrically u0), then x must be of the form w˜u (u)k1 u0 uk2 ˜u (u)k3 u0 uk4 ˜u . . . , (10.1) with 0 ki k and where the sequence w does not contain the subsequence ˜u
Proof: Let n be the smallest integer such that n(x) = ˜u . . .. Let n(x) = ˜u a1 a2 . . ., with |ai| = |u|. Let m be the smallest integer such that am 6= u. From the inequality n(x) = ˜u a1 a2 . . . ˜u (u)ku0v we have that 1 m k + 1.
Table of contents :
Introduction
1. Diophantine Approximation and BAN
– q-adically Badly Approximable Numbers
2. Prerequisites
– General Symbolic Dynamics
– Dimension
I. One-sided q-adically BAN
3. Introduction
4. Minimal Sequences
5. Main Results via Perron-Frobenius
6. The -shift
7. Main Results via the -shift
8. Numerics
II. Two-sided q-adically BAN
9. Introduction
10. Fundamental Properties
11. Shift-Bounded Sequences
12. Minimal Sequences
13. The Set A
14. Dyadic Approximation
15. Triadic Approximation
16. q-adic Approximation
17. Numerics
18. Conclusion
List of Notation
References
