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Mathematische Zeitschrift

, Volume 291, Issue 3–4, pp 1605–1619 | Cite as

Multiple expansions of real numbers with digits set \(\left\{ 0,1,q\right\} \)

  • Karma Dajani
  • Kan Jiang
  • Derong KongEmail author
  • Wenxia Li
Open Access
Article
  • 215 Downloads

Abstract

For \(q>1\) we consider expansions in base q with digits set \(\left\{ 0,1,q\right\} \). Let \({{\mathcal {U}}}_q\) be the set of points which have a unique q-expansion. For \(k=2, 3,\ldots ,\aleph _0\) let \(\mathcal {B}_k\) be the set of bases \(q>1\) for which there exists x having precisely k different q-expansions, and for \(q\in \mathcal {B}_k\) let \({{\mathcal {U}}}_q^{(k)}\) be the set of all such x’s which have exactly k different q-expansions. In this paper we show that
$$\begin{aligned} \mathcal {B}_{\aleph _0}=[2,\infty )\quad \text {and}\quad \mathcal {B}_k=(q_c,\infty )\quad \text {for any}\quad k\ge 2, \end{aligned}$$
where \(q_c\approx 2.32472\) is the appropriate root of \(x^3-3x^2+2x-1=0\). Moreover, we show that for any integer \(k\ge 2\) and any \(q\in \mathcal {B}_{k}\) the Hausdorff dimensions of \({{\mathcal {U}}}_q^{(k)}\) and \({{\mathcal {U}}}_q\) are the same, i.e.,
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q^{(k)}=\dim _H{{\mathcal {U}}}_q\quad \text {for any}\quad k\ge 2. \end{aligned}$$
Finally, we conclude that the set of points having a continuum of q-expansions has full Hausdorff dimension.

Keywords

Unique expansion Multiple expansion Countable expansion Hausdorff dimension 

Mathematics Subject Classification

Primary 11A63 Secondary 10K50 11K55 37B10 

1 Introduction

Expansions in non-integer bases were pioneered by Rényi [18] and Parry [16]. Unlike integer base expansions, for a given \(\beta \in (1, 2)\), it is well-known that typically a real number \(x\in I_\beta :=[0, 1/(\beta -1)]\) has a continuum of \(\beta \)-expansions with digits set \(\left\{ 0, 1\right\} \) (cf. [2, 19]), i.e., for Lebesuge almost every \(x\in I_\beta \) there exist a continuum of zero-one sequences \((x_i)\) such that \(x=\sum _{i=1}^\infty x_i/\beta ^i\). However, there still exist \(x\in I_\beta \) having a unique \(\beta \)-expansion (cf. [5, 10, 13]). Denote by \({{\mathcal {U}}}_\beta \) the set of all \(x\in I_\beta \) with a unique \(\beta \)-expansion. De Vries and Komornik [3] investigated the topological properties of \({{\mathcal {U}}}_\beta \). Komornik et al. [12] considered the Hausdorff dimension of \({{\mathcal {U}}}_\beta \), and concluded that the dimension function \(\beta \mapsto \dim _H{{\mathcal {U}}}_\beta \) behaves like a Devil’s staircase. Interestingly, for any \(k=2,3,\ldots \) or \(\aleph _0\) Erdős et al. [6, 7] showed that there exist \(\beta \in (1,2)\) and \(x\in I_\beta \) such that x has precisely k different \(\beta \)-expansions. For more information on expansions in non-integer bases we refer to [1, 21, 23], and the surveys [4, 11, 20].

In this paper we consider expansions with digits set \(\left\{ 0,1,q\right\} \). Given \(q>1\), the infinite sequence \((d_i)\) is called a q-expansion of x, if
$$\begin{aligned} x=((d_i))_q:=\sum _{i=1}^\infty \frac{d_i}{q^i},\quad d_i\in \left\{ 0,1,q\right\} \quad \text { for all } i\ge 1. \end{aligned}$$
We emphasize that the digits set\(\left\{ 0,1,q\right\} \) also depends on the base q.
For \(q>1\) let \(E_q\) be the set of points which have a q-expansion. Then \(E_q\) is the attractor of the iterated function system (IFS)
$$\begin{aligned} \phi _d(x)=\frac{x+d}{q},\quad d\in \left\{ 0,1,q\right\} . \end{aligned}$$
So, \(E_q\) is the non-empty compact set satisfying \(E_q=\bigcup _{d\in \left\{ 0,1,q\right\} }\phi _d(E_q)\) (cf. [8]). Observe that \(\phi _0(E_q)\cap \phi _1(E_q)\ne \emptyset \) for any \(q>1\). Then \(E_q\) is a self-similar set with overlaps. Ngai and Wang [15] gave the Hausdorff dimension of \(E_q\):
$$\begin{aligned} \dim _H E_q=\frac{\log q^*}{\log q}\quad \text {for any}\quad q>q^*, \end{aligned}$$
(1.1)
where \(q^*=(3+\sqrt{5})/2\). Yao and Li [22] considered all possible IFSs generating the set \(E_q\). Zou et al. [24] considered the set of points in \(E_q\) which have a unique q-expansion. In this paper, we investigate the set of points in \(E_q\) having multiple q-expansions.
For \(k= 1,2,\ldots , \aleph _0\) or \(2^{\aleph _0}\), let
$$\begin{aligned} \mathcal {B}_k:=\left\{ q\in (1,\infty ): \exists ~x\in E_q\text { with precisely }k\text { different }q\text {-expansions}\right\} . \end{aligned}$$
Accordingly, for \(q\in \mathcal {B}_k\) let
$$\begin{aligned} {{\mathcal {U}}}_q^{(k)}:=\left\{ x\in E_q: x~\text {has precisely}~k~\text {different}~q\text {-expansions}\right\} . \end{aligned}$$
For simplicity, we write \({{\mathcal {U}}}_q:={{\mathcal {U}}}_q^{(1)}\) for the set of \(x\in E_q\) having a unique q-expansion, and denote by \( {{\mathcal {U}}}'_q\) the set of all q-expansions corresponding to elements of \({{\mathcal {U}}}_q\).

In this paper we will describe the sizes of the sets \(\mathcal {B}_k\) and \({{\mathcal {U}}}_q^{(k)}\). Our first result is on the set \(\mathcal {B}_k\) for \(k= 1,2,\ldots ,\aleph _0\) or \(2^{\aleph _0}\). Clearly, when \(k=1\) we have \(\mathcal {B}_1=(1,\infty )\), since 0 always has a unique q-expansion for any \(q>1\). When \(k= 2,3,\ldots ,\aleph _0\) or \(2^{\aleph _0}\) we have the following

Theorem 1

Let \(q_c\approx 2.32472\) be the appropriate root of \(x^3-3x^2+2x-1=0.\) Then
$$\begin{aligned} \mathcal {B}_{2^{\aleph _0}}=(1,\infty ),\quad \mathcal {B}_{\aleph _0}=[2,\infty ),\quad \mathcal {B}_k=(q_c,\infty )\quad \text {for any}\quad k\ge 2. \end{aligned}$$

By Theorem 1 it follows that for \(q\in [2,q_c]\), any \(x\in E_q\) can only have a unique q-expansion, countably infinitely many q-expansions, or a continuum of q-expansions.

When \(k=1\), the following theorem for the univoque set\({{\mathcal {U}}}_q={{\mathcal {U}}}_q^{(1)}\) was proven in [24].

Theorem 1.1

  1. (i)

    If \(q\in (1, q_c]\), then \({{\mathcal {U}}}_q=\left\{ 0,q/(q-1)\right\} \).

     
  2. (ii)

    If \(q\in (q_c, q^*)\), then \({{\mathcal {U}}}_q\) contains a continuum of points.

     
  3. (iii)

    If \(q\in [q^*,\infty )\), then \(\dim _H{{\mathcal {U}}}_q=\log q_c/\log q\).

     

Our second result complements Theorem 1.1, and shows that there is no difference between the Hausdorff dimensions of \({{\mathcal {U}}}_q^{(k)}\) and \({{\mathcal {U}}}_q\).

Theorem 2

  1. (i)

    \(\dim _H{{\mathcal {U}}}_{q}>0\) if and only if \(q>q_c\).

     
  2. (ii)

    For any integer \(k\ge 2\) and any \(q\in \mathcal {B}_{k}\) we have \( \dim _H{{\mathcal {U}}}_q^{(k)}=\dim _H{{\mathcal {U}}}_q. \)

     

As a result of Theorem 2 it follows that \(q_c\) is indeed the critical base, in the sense that \({{\mathcal {U}}}_q^{(k)}\) has positive Hausdorff dimension if \(q>q_c\), while \({{\mathcal {U}}}_q^{(k)}\) has zero Hausdorff dimension if \(q\le q_c\). In fact, by Theorems 1 and 1.1 (i) it follows that for \(q\le q_c\) the set \({{\mathcal {U}}}_q=\left\{ 0, q/(q-1)\right\} \) and \({{\mathcal {U}}}_q^{(k)}=\emptyset \) for any integer \(k\ge 2\).

Our final result focuses on the sizes of \({{\mathcal {U}}}_q^{(\aleph _0)}\) and \({{\mathcal {U}}}_q^{(2^{\aleph _0})}\).

Theorem 3

  1. (i)

    Let \(q\in \mathcal {B}_{\aleph _0}{\setminus }(q_c, q^*)\). Then \({{\mathcal {U}}}_q^{(\aleph _0)}\) is countably infinite.

     
  2. (ii)

    For any \(q>1\) we have \( \dim _H{{\mathcal {U}}}_q^{(2^{\aleph _0})}=\dim _H E_q. \)

     

Remark 1.2

In Lemma 5.5 we prove a stronger result of Theorem 3 (ii), and show that the Hausdorff measures of \({{\mathcal {U}}}_q^{(2^{\aleph _0})}\) and \(E_q\) are the same for any \(q>1\), i.e.,
$$\begin{aligned} \mathcal {H}^s\left( {{\mathcal {U}}}_q^{\left( 2^{\aleph _0}\right) }\right) =\mathcal {H}^s(E_q)\in (0, \infty ), \end{aligned}$$
where \(s=\dim _H E_q\).

The rest of the paper is arranged as follows. In Sect. 2 we recall some properties of unique q-expansions. The proof of Theorem 1 for the sets \(\mathcal {B}_k\) will be presented in Sect. 3, and the proofs of Theorems 2 and 3 for the sets \({{\mathcal {U}}}_q^{(k)}\) will be given in Sects. 4 and 5, respectively. Finally, in Sect. 6 we give some examples and end the paper with some questions.

2 Unique expansions

In this section we recall some properties of the univoque set \({{\mathcal {U}}}_q\) from [24]. Recall that
$$\begin{aligned} q_c\approx 2.32472\quad \text {and}\quad q^*=\frac{3+\sqrt{5}}{2}\approx 2.61803, \end{aligned}$$
(2.1)
where \(q_c\) is the appropriate root of the equation \(x^3-3 x^2+2x-1=0\). Note that for \(q\in (1,q^*]\) the attractor \(E_q=[0, q/(q-1)]\) is an interval. However, for \(q>q^*\) the attractor \(E_q\) is a Cantor set which contains neither interior nor isolated points.

Given \(q>1\), let \(\left\{ 0,1,q\right\} ^{\mathbb {N}}\) be the set of all infinite sequences \((d_i)\) over the alphabet \(\left\{ 0,1,q\right\} \). By a word \({\mathbf {c}}\) we mean a finite string of digits \({\mathbf {c}}=c_1\ldots c_n\) with each digit \(c_i\in \left\{ 0,1,q\right\} \). For two words \({\mathbf {c}}=c_1\ldots c_m\) and \(\mathbf d=d_1\ldots d_n\), we denote by \({\mathbf {c}}\mathbf d=c_1\ldots c_md_1\ldots d_n\) their concatenation. For a positive integer k we write \({\mathbf {c}}^k={\mathbf {c}}\cdots {\mathbf {c}}\) for the k-fold concatenation of \({\mathbf {c}}\) with itself. Furthermore, we write \({\mathbf {c}}^\infty ={\mathbf {c}}{\mathbf {c}}\cdots \) the infinite periodic sequence with periodic block \({\mathbf {c}}\). Throughout the paper we will use lexicographical ordering \(\prec , \preccurlyeq , \succ \) and \(\succcurlyeq \) between sequences. More precisely, for two sequences \((c_i), (d_i)\in \left\{ 0,1,q\right\} ^{{\mathbb {N}}}\) we say \((c_i)\prec (d_i)\) or \((d_i)\succ (c_i)\) if there exists an integer \(n\ge 1\) such that \(c_1\ldots c_{n-1}=d_1\ldots d_{n-1}\) and \(c_n<d_n\). Furthermore, we say \((c_i)\preccurlyeq (d_i)\) if \((c_i)\prec (d_i)\) or \((c_i)=(d_i)\).

Recall that \({{\mathcal {U}}}_q\) is the set of points in \(E_q\) with a unique q-expansion, and \({{\mathcal {U}}}_q'\) is the set of corresponding q-expansions. Then
$$\begin{aligned} {{\mathcal {U}}}_q'=\left\{ (d_i)\in \left\{ 0,1,q\right\} ^\mathbb {N}:((d_i))_q\in {{\mathcal {U}}}_q\right\} . \end{aligned}$$
The following lexicographical characterization of \({{\mathcal {U}}}'_q\) for \(q>q^*\) was established in [24, Lemma 3.1].

Lemma 2.1

Let \(q>q^*\). Then \((d_i)\in {{\mathcal {U}}}_q'\) if and only if
$$\begin{aligned} \left\{ \begin{array}{lll} (d_{n+i})\prec q 0^\infty &{}\quad \text {if}&{} \quad d_n=0,\\ (d_{n+i})\succ 1^\infty &{}\quad \text {if}&{} \quad d_n=1.\\ \end{array} \right. \end{aligned}$$
To describe \({{\mathcal {U}}}_q'\) for \(q\in (1, q^*]\) we need the following notation. Let
$$\begin{aligned} \alpha (q)=(\alpha _i(q)) \end{aligned}$$
be the quasi-greedyq-expansion of \(q-1\), i.e., the lexicographically largest q-expansion of \(q-1\) with infinitely many non-zero digits. We emphasize that \(\alpha (q)\) is well-defined for \(q\in (1, q^*]\). By (2.1) and a direct calculation one can verify that
$$\begin{aligned} \alpha (q_c)=q_c1^\infty ,\quad \alpha (q^*)=(q^*)^\infty . \end{aligned}$$
(2.2)
Note by Theorem 1.1 that for \(q\in (1, q_c]\) we have \({{\mathcal {U}}}_q=\left\{ 0, q/(q-1)\right\} \), and then \({{\mathcal {U}}}_q'=\left\{ 0^\infty , q^\infty \right\} \). So, it suffices to consider \({{\mathcal {U}}}'_q\) for \(q\in (q_c, q^*]\). The following lemma was obtained in [24, Lemmas 3.1 and 3.2].

Lemma 2.2

Let \(q\in (q_c,q^*]\). Then
$$\begin{aligned} A_q\subseteq {{\mathcal {U}}}_q'\subseteq B_q, \end{aligned}$$
where \(A_q\) is the set of sequences \((d_i)\in \left\{ 0,1,q\right\} ^\mathbb {N}\) satisfying
$$\begin{aligned} \left\{ \begin{array}{lll} (d_{n+i})\prec 1\alpha (q)&{}\quad \text {if} &{} \quad d_n=0,\\ 1^\infty \prec (d_{n+i})\prec \alpha (q)&{}\quad \text {if} &{} \quad d_n=1,\\ (d_{n+i})\succ 0q^\infty &{}\quad \text {if} &{} \quad d_n=q,\\ \end{array} \right. \end{aligned}$$
(2.3)
and \(B_q\) is the set of sequences \((d_i)\in \left\{ 0,1,q\right\} ^\mathbb {N}\) satisfying the first two inequalities in (2.3).
For \(q>1\) let \(\Phi :\left\{ 0,1,q\right\} ^\mathbb {N}\rightarrow \left\{ 0,1,2\right\} ^\mathbb {N}\) be defined by
$$\begin{aligned} \Phi ((d_i))=(d_i'), \end{aligned}$$
where \(d_i'=d_i\) if \(d_i\in \left\{ 0,1\right\} \), and \(d_i'=2\) if \(d_i=q\). Clearly, \(\Phi \) is bijective and strictly increasing. The following lemma was given in [24, Lemma 3.2].

Lemma 2.3

The map \(q\rightarrow \Phi (\alpha (q))\) is strictly increasing in \((1,q^*]\).

By (2.2) and Lemma 2.3 it follows that for any \(q\in (q_c, q^*)\) we have \(q1^\infty \prec \alpha (q)\prec q^\infty \).

3 Proof of Theorem 1

In this section we will investigate the set \(\mathcal {B}_k\) of bases \(q>1\) in which there exists \(x\in E_q\) having k different q-expansions. Excluding the trivial case for \(k=1\) that \(\mathcal {B}_1=(1,\infty )\) we consider \(\mathcal {B}_k\) for \(k= 2,3,\ldots ,\aleph _0\) or \(2^{\aleph _0}\).

The following lemma was established in [24, Theorem 4.1] and [9, Theorem 1.1].

Lemma 3.1

Let \(q\in (1,2)\).
  1. (i)

    If \(q\in (1,2)\), then any \(x\in E_q\) has either a unique q-expansion, or a continuum of q-expansions.

     
  2. (ii)

    If \(q=2\), then any \(x\in E_q\) can only have a unique q-expansion, countably infinitely many q-expansions, or a continuum of q-expansions.

     
For \(q>1\) we recall that \(\phi _d(x)=(x+d)/q\) for \(d\in \left\{ 0,1,q\right\} \). Let
$$\begin{aligned} S_q:=\left( \phi _0(E_q)\cap \phi _1(E_q)\right) \cup \left( \phi _1(E_q)\cap \phi _q(E_q)\right) . \end{aligned}$$
(3.1)
Then \(S_q\) is associated with the switch region, since any \(x\in S_q\) has at least two q-expansions. More precisely, any \(x\in \phi _0(E_q)\cap \phi _1(E_q)\) has at least two q-expansions: one begins with the digit 0 and one begins with the digit 1. Accordingly, any \(x\in \phi _1(E_q)\cap \phi _q(E_q)\) also has at least two q-expansions: one starts with the digit 1 and one starts with the digit q. We point out that the union in (3.1) is disjoint if \(q>2\). In particular, for \(q>q^*\) the intersection \(\phi _1(E_q)\cap \phi _q(E_q)=\emptyset \).
For \(x\in E_q\) let \(\Sigma (x)\) be the set of all q-expansions of x, i.e.,
$$\begin{aligned} \Sigma (x):=\left\{ (d_i)\in \left\{ 0,1,q\right\} ^\mathbb {N}: ((d_i))_q=x\right\} , \end{aligned}$$
and denote its cardinality by \(|\Sigma (x)|\).
We recall from [1] that a point \(x\in S_q\) is called a q-null infinite point if x has an expansion \((d_i)\in \left\{ 0,1,q\right\} ^\mathbb {N}\) such that whenever
$$\begin{aligned} x_n:=(d_{n+1}d_{n+2}\ldots )_q\in S_q, \end{aligned}$$
one of the following quantities is infinity, and the other two are finite:
$$\begin{aligned} \left| \Sigma (\phi _0^{-1}(x_n))\right| ,\quad \left| \Sigma (\phi ^{-1}_{1}(x_n))\right| \quad \text {and}\quad \left| \Sigma (\phi ^{-1}_{q}(x_n))\right| . \end{aligned}$$
Then any q-null infinite point has countably infinitely many q-expansions.

First we consider the set \(\mathcal {B}_{\aleph _0}\), which is based on the following characterization (cf. [1, 23]).

Lemma 3.2

\(q\in \mathcal {B}_{\aleph _0}\) if and only if \(S_q\) contains a q-null infinite point.

Lemma 3.3

\(\mathcal {B}_{\aleph _0}=[2,\infty )\).

Proof

By Lemma 3.1 we have \(\mathcal {B}_{\aleph _0}\subseteq [2,\infty )\) and \(2\in \mathcal {B}_{\aleph _0}\). So, it suffices to prove \((2,\infty )\subseteq \mathcal {B}_{\aleph _0}\).

Take \(q\in (2,\infty )\). Note that \(0=(0^\infty )_q\) and \(q/(q-1)\in (q^\infty )_q\) belong to \({{\mathcal {U}}}_q\). We claim that
$$\begin{aligned} x=(0q^\infty )_q \end{aligned}$$
is a q-null infinite point. Note that \((10^\infty )_q=(0q0^\infty )_q\). Then by the words substitution \(10\sim 0q\) it follows that all expansions \(1^k0 q^\infty , k\ge 0,\) are q-expansions of x, i.e.,
$$\begin{aligned} \bigcup _{k=0}^\infty \left\{ 1^k0q^\infty \right\} \subseteq \Sigma (x). \end{aligned}$$
This implies that \(|\Sigma (x)|=\infty \). Furthermore, since \(q>2\), the union in (3.1) is disjoint. This implies
$$\begin{aligned} x=(0q^\infty )_q=(10q^\infty )_q\in \phi _0(E_q)\cap \phi _1(E_q){\setminus }\phi _q(E_q). \end{aligned}$$
Then \(\phi _0^{-1}(x)=(q^\infty )_q\in {{\mathcal {U}}}_q\), \(\phi _1^{-1}(x)=x\) and \(\phi _q^{-1}(x)\notin E_q\), i.e.,
$$\begin{aligned} |\Sigma (\phi _0^{-1}(x))|=1,\quad |\Sigma (\phi ^{-1}_1(x))|=\infty ,\quad |\Sigma (\phi _q^{-1}(x))|=0. \end{aligned}$$
By iteration it follows that x is a q-null infinite point. Hence, by Lemma 3.2 we have \(q\in \mathcal {B}_{\aleph _0}\), and therefore \((2,\infty )\subseteq \mathcal {B}_{\aleph _0}\). \(\square \)

Now we turn to describe the set \(\mathcal {B}_k\). By Lemma 3.1 it follows that \(\mathcal {B}_k\subseteq (2,\infty )\) for any \(k\ge 2\). First we consider \(\mathcal {B}_2\) and need the following

Lemma 3.4

Let \(q>2\). Then \(q\in \mathcal {B}_2\) if and only if either
$$\begin{aligned} (0(a_i))_q=(1(b_i))_q\qquad \text {for some}\quad (a_i), (b_i)\in {{\mathcal {U}}}_q', \end{aligned}$$
or
$$\begin{aligned} (1(c_i))_q=(q(d_i))_q\qquad \text {for some}\quad (c_i), (d_i)\in {{\mathcal {U}}}_q'. \end{aligned}$$

Proof

First we prove the necessary condition. Take \(q\in \mathcal {B}_2\). Suppose \(x\in E_q\) has two different q-expansions, say
$$\begin{aligned} ((a_i))_q=x=((b_i))_q. \end{aligned}$$
Then there exists a least integer \(k\ge 1\) such that \(a_k\ne b_k\). Then
$$\begin{aligned} (a_k a_{k+1}\ldots )_q=(b_kb_{k+1}\ldots )_q\in S_q \quad \text {and}\quad (a_{k+i}), (b_{k+i})\in {{\mathcal {U}}}'_q. \end{aligned}$$
(3.2)
Since \(q>2\), it gives that the union in (3.1) is disjoint. Then the necessity follows by (3.2).
To prove the sufficiency, without loss of generality, we assume \((0(a_i))_q=(1(b_i))_q\) with \((a_i), (b_i)\in {{\mathcal {U}}}_q'\). Note by \(q>2\) that the union in (3.1) is disjoint. Then
$$\begin{aligned} (0(a_i))_q=(1(b_i))\in \phi _0(E_q)\cap \phi _1(E_q){\setminus }\phi _q(E_q). \end{aligned}$$
This implies that x has exactly two different q-expansions. So, \(q\in \mathcal {B}_2\). \(\square \)

Recall from (2.2) that \(q_c\approx 2.32472\) and \(q^*=(3+\sqrt{5})/2\) admit the quasi-greedy expansions \(\alpha (q_c)=q_c1^\infty \) and \(\alpha (q^*)=(q^*)^\infty .\) In the following lemma we describe the set \(\mathcal {B}_2\).

Lemma 3.5

\(\mathcal {B}_2=(q_c,\infty )\).

Proof

First we show that \(\mathcal {B}_2\subseteq (q_c,\infty )\). By Lemma 3.1 it suffices to prove that any \(q\in (2,q_c]\) is not contained in \(\mathcal {B}_2\). Take \(q\in (2,q_c]\). By Theorem 1.1 we have \({{\mathcal {U}}}'_q=\left\{ (0^\infty ), (q^\infty )\right\} \). Then by Lemma 3.4 it follows that if \(q\in \mathcal {B}_2\cap (2,q_c]\) then q must satisfy one of the following equations
$$\begin{aligned} (0q^\infty )_q=(10^\infty )_q \quad \text {or}\quad (1q^\infty )_q=(q0^\infty )_q. \end{aligned}$$
This is impossible since neither equation has a solution in \((2,q_c]\). Hence, \(\mathcal {B}_2\subseteq (q_c, \infty )\).
Now we turn to prove \((q_c,\infty )\subseteq \mathcal {B}_2\). By Lemmas 2.1 and 3.4, one can verify that for any \(q>q^*\) the number
$$\begin{aligned} x=(0q0^\infty )_q=(10^\infty )_q \end{aligned}$$
has precisely two different q-expansions. This implies that \((q^*,\infty )\subseteq \mathcal {B}_2\).
For \(q\in (q_c, q^*]\), one has by (2.2) that \(\alpha (q_c)=q_c1^\infty \) and \(\alpha (q^*)=(q^*)^\infty \). Then by Lemma 2.3 there exists an integer \(m\ge 0\) such that
$$\begin{aligned} \alpha (q)\succ q1^mq0^\infty . \end{aligned}$$
Hence, by Lemmas 2.2 and 3.4 one can verify that
$$\begin{aligned} y=(0q (1^{m+1}q)^\infty )_q=(10(1^{m+1}q)^\infty )_q \end{aligned}$$
has precisely two different q-expansions. So, \((q_c,q^*]\subseteq \mathcal {B}_2\), and the proof is complete. \(\square \)

Lemma 3.6

\(\mathcal {B}_k=(q_c,\infty )\) for any \(k\ge 3\).

Proof

First we prove \(\mathcal {B}_{k}\subseteq \mathcal {B}_2\) for any \(k\ge 3\). By Lemma 3.1 it follows that \(\mathcal {B}_k\subseteq (2,\infty )\). Take \(q\in \mathcal {B}_k\) with \(k\ge 3\). Suppose \(x\in E_q\) has exactly k different q-expansions. Since \(q>2\), the union in (3.1) is disjoint. This implies that there exists a word \(d_1\ldots d_n\) such that
$$\begin{aligned} \phi _{d_1}^{-1}\circ \cdots \circ \phi _{d_n}^{-1}(x) \end{aligned}$$
has exactly two different q-expansions. So, \(q\in \mathcal {B}_2\). Hence, \(\mathcal {B}_k\subseteq \mathcal {B}_2\) for any \(k\ge 3\).
Now we prove \(\mathcal {B}_2\subseteq \mathcal {B}_k\) for any \(k\ge 3\). Note by Lemma 3.5 that \(\mathcal {B}_2=(q_c, \infty )\). Then it suffices to prove \( (q_c,\infty )\subseteq \mathcal {B}_k.\) First we prove \((q^*,\infty )\subseteq \mathcal {B}_k\). Take \(q\in (q^*, \infty )\). We claim that for any \(k\ge 1\),
$$\begin{aligned} x_k=(0q^{k-1}(1q)^\infty )_q \end{aligned}$$
has precisely k different q-expansions. We will prove this by induction on k.
For \(k=1\) one can easily check by using Lemma 2.1 that \(x_1=(0(1q)^\infty )_q\in {{\mathcal {U}}}_q\). Suppose \(x_k\) has exactly k different q-expansions. Now we consider \(x_{k+1}\), which can be written as
$$\begin{aligned} x_{k+1}=(0q^k(1q)^\infty )_q=(10q^{k-1}(1q)^\infty )_q. \end{aligned}$$
By Lemma 2.1 we have \(q^k(1q)^\infty \in {{\mathcal {U}}}_{q}'\). Moreover, by the induction hypothesis \((0q^{k-1}(1q)^\infty )_q=x_k\) has exactly k different q-expansions. Then \(x_{k+1}\) has at least \(k+1\) different q-expansions. On the other hand, since \(q>q^*>2\), the union in (3.1) is disjoint. Then
$$\begin{aligned} x_{k+1}\in \phi _0(E_q)\cap \phi _1(E_q){\setminus }\phi _q(E_q). \end{aligned}$$
This implies that \(x_{k+1}\) indeed has \(k+1\) different q-expansions. By induction this proves the claim, and hence \((q^*, \infty )\subseteq \mathcal {B}_k\) for all \(k\ge 3\).
It remains to prove \((q_c,q^*]\subseteq \mathcal {B}_k\). Take \(q\in (q_c,q^*]\). By (2.2) and Lemma 2.3 there exists an integer \(m\ge 0\) such that
$$\begin{aligned} \alpha (q)\succ q1^mq0^\infty . \end{aligned}$$
(3.3)
We claim that
$$\begin{aligned} y_k=(0q^{k-1}(1^{m+1}q)^\infty )_q \end{aligned}$$
has exactly k different q-expansions. Again, this will be proven by induction on k.
If \(k=1\), then by using (3.3) in Lemma 2.2 it gives that \(y_1=(0(1^{m+1}q)^\infty )_q\) has a unique q-expansion. Suppose \(y_k\) has exactly k different q-expansions. Now we consider
$$\begin{aligned} y_{k+1}=(0q^k(1^{m+1}q)^\infty )_q=(10q^{k-1}(1^{m+1}q)^\infty )_q. \end{aligned}$$
By (3.3) and Lemma 2.2 it yields that \(q^{k}(1^{m+1}q)^\infty \in {{\mathcal {U}}}_{q}'\). Furthermore, by the induction hypothesis \((0q^{k-1}(1^{m+1}q)^\infty )_q=y_k\) has exactly k different q-expansions. This implies that \(y_{k+1}\) has at least \(k+1\) different q-expansions. On the other hand, note that \(q>q_c>2\), and therefore the union in (3.1) is disjoint. So, \( y_{k+1}\in \phi _0(E_q)\cap \phi _1(E_q){\setminus }\phi _q(E_q), \) which implies that \(y_{k+1}\) indeed has \(k+1\) different q-expansions. By induction this proves the claim, and then \((q_c, q^*]\subseteq \mathcal {B}_k\) for all \(k\ge 3\). This completes the proof. \(\square \)

Proof of Theorem 1

By Lemmas 3.3, 3.5 and 3.6 it suffices to prove \(\mathcal {B}_{2^{\aleph _0}}=(1,\infty )\). This can be verified by observing that
$$\begin{aligned} x=((100)^\infty )_q\in {{\mathcal {U}}}_q^{(2^{\aleph _0})} \end{aligned}$$
for any \(q>1\), because by the word substitution \(10\sim 0q\) one can show that x indeed has a continuum of different q-expansions. \(\square \)

4 Proof of Theorem 2

For \(q>1\) and \(k\in \mathbb {N}\) we recall that \({{\mathcal {U}}}_q^{(k)}\) is the set of \(x\in [0, q/(q-1)]\) having precisely k different q-expansions. In this section we are going to investigate the Hausdorff dimension of \({{\mathcal {U}}}_q^{(k)}\). First we show that \(q_c\approx 2.32472\) is the critical base for \({{\mathcal {U}}}_q\).

Lemma 4.1

Let \(q>1\). Then \(\dim _H{{\mathcal {U}}}_q>0\) if and only if \(q>q_c\).

Proof

The necessity follows from Theorem 1.1 (i). For the sufficiency we take \(q\in (q_c,\infty )\). If \(q>q^*\), then by Theorem 1.1 (iii) we have
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q=\frac{\log q_c}{\log q}>0. \end{aligned}$$
So it remains to prove \(\dim _H{{\mathcal {U}}}_q>0\) for any \(q\in (q_c, q^*]\).
Take \(q\in (q_c, q^*]\). Recall from (2.2) that \(\alpha (q_c)=q_c1^\infty \) and \(\alpha (q^*)=(q^*)^\infty \). Then by Lemma 2.3 there exists an integer \(m\ge 0\) such that \( \alpha (q)\succ q 1^mq0^\infty . \) Whence, by Lemma 2.2 one can verify that all sequences in
$$\begin{aligned} \Delta '_m:=\prod _{i=1}^\infty \left\{ q1^{m+1}, 1^{m+2}\right\} \end{aligned}$$
excluding those ending with \(1^\infty \) belong to \({{\mathcal {U}}}_q'\). This implies that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q\ge \dim _H\Delta _m(q), \end{aligned}$$
(4.1)
where \(\Delta _m(q):=\left\{ ((d_i))_q: (d_i)\in \Delta _m'\right\} \). Note that \(\Delta _m(q)\) is a self-similar set generated by the IFS
$$\begin{aligned} f_1(x)=\frac{x}{q^{m+2}}+(q1^{m+1}0^\infty )_q,\quad f_2(x)=\frac{x}{q^{m+2}}+(1^{m+2}0^\infty )_q, \end{aligned}$$
which satisfies the open set condition (cf. [8]). Therefore, by (4.1) we conclude that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q\ge \dim _H\Delta _m(q)=\frac{\log 2}{(m+2)\log q}>0. \end{aligned}$$
\(\square \)

In the following we will consider the Hausdorff dimension of \({{\mathcal {U}}}_q^{(k)}\) for any \(k\ge 2\), and prove \(\dim _H{{\mathcal {U}}}_q^{(k)}=\dim _H{{\mathcal {U}}}_q\). The upper bound of \(\dim _H{{\mathcal {U}}}_q^{(k)}\) is easy.

Lemma 4.2

Let \(q>1\). Then \(\dim _H{{\mathcal {U}}}_q^{(k)}\le \dim _H{{\mathcal {U}}}_q\) for any \(k\ge 2\).

Proof

Recall that \(\phi _d(x)=(x+d)/q\) for \(d\in \left\{ 0,1,q\right\} \). Then the lemma follows by observing that for any \(k\ge 2\),
$$\begin{aligned} {{\mathcal {U}}}_q^{(k)}\subseteq \bigcup _{n=1}^\infty \bigcup _{d_1\cdots d_n\in \left\{ 0,1,q\right\} ^n}\phi _{d_1}\circ \cdots \circ \phi _{d_n}({{\mathcal {U}}}_q), \end{aligned}$$
and the countable stability of Hausdorff dimension. \(\square \)
For the lower bound of \(\dim _H{{\mathcal {U}}}_q^{(k)}\) we need more. By Lemmas 4.1 and 4.2 it follows that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q^{(k)}=0=\dim _H{{\mathcal {U}}}_q\quad \text {for any }q\le q_c. \end{aligned}$$
So, it suffices to consider \(q>q_c\). Let
$$\begin{aligned} F_q'(1):=\left\{ (d_i)\in {{\mathcal {U}}}_q': d_1=1\right\} \end{aligned}$$
be the follower set in \({{\mathcal {U}}}_q'\) generated by the word 1, and let \(F_q(1)\) be the set of \(x\in E_q\) which have a q-expansion in \(F_q'(1)\), i.e., \(F_q(1)=\{((d_i))_q: (d_i)\in F_q'(1)\}.\)

Lemma 4.3

Let \(q>q_c\). Then \(\dim _H{{\mathcal {U}}}_q^{(k)}\ge \dim _H F_q(1)\) for any \(k\ge 1\).

Proof

For \(k\ge 1\) and \(q>q_c\) let
$$\begin{aligned} \Lambda ^k_q:=\left\{ ((d_i))_q: d_1\ldots d_k=0q^{k-1}, (d_{k+i})\in F'_q(1)\right\} . \end{aligned}$$
Then \(\Lambda ^k_q=\phi _0\circ \phi _q^{k-1}(F_q(1))\), and therefore \(\dim _H\Lambda ^k_q=\dim _H F_q(1).\) So it suffices to prove \(\Lambda ^k_q\subseteq {{\mathcal {U}}}_q^{(k)}\). Arbitrarily take
$$\begin{aligned} x_k=\left( 0q^{k-1}(c_i)\right) _q\in \Lambda ^k_q\quad \text { with}\quad (c_i)\in F'_q(1). \end{aligned}$$
We will prove by induction on k that \(x_k\) has exactly k different q-expansions.
For \(k=1\), by Lemmas 2.1 and 2.2 it follows that \( x_1=(0(c_i))_q\in {{\mathcal {U}}}_q. \) Suppose \(x_k=(0q^{k-1}(c_i))_q\) has precisely k different q-expansions. Now we consider \(x_{k+1}\), which can be expanded as
$$\begin{aligned} x_{k+1}=\left( 0q^k(c_i)\right) _q=(10q^{k-1}(c_i))_q. \end{aligned}$$
By Lemmas 2.1 and 2.2 we have \(q^k(c_i)\in {{\mathcal {U}}}'_q\), and by the induction hypothesis it yields that \((0q^{k-1}(c_i))_q=x_k\) has k different q-expansions. This implies that \(x_{k+1}\) has at least \(k+1\) different q-expansions. On the other hand, since \(q>q_c>2\), it gives that the union in (3.1) is disjoint. So, \(x_{k+1}\in \phi _0(E_q)\cap \phi _1(E_q){\setminus }\phi _q(E_q),\) which implies that \(x_{k+1}\) indeed has \(k+1\) different q-expansions.

By induction this proves \(x_k\in {{\mathcal {U}}}_q^{(k)}\) for all \(k\ge 1\). Since \(x_k\) was taken arbitrarily from \(\Lambda _q^k\), we conclude that \(\Lambda _q^k\subseteq {{\mathcal {U}}}_q^{(k)}\) for any \(k\ge 1\). The proof is complete. \(\square \)

Lemma 4.4

Let \(q>q_c\). Then \(\dim _H F_q(1)\ge \dim _H{{\mathcal {U}}}_q\).

Proof

First we consider \(q>q^*\). By Lemma 2.1 one can show that \({{\mathcal {U}}}'_q\) is contained in an irreducible sub-shift of finite type \(X'_A\) over the states \(\left\{ 0,1,q\right\} \) with adjacency matrix
$$\begin{aligned} A=\left( \begin{array}{c@{\qquad }c@{\qquad }c} 1&{}1&{}0\\ 0&{}1&{}1\\ 1&{}1&{}1 \end{array} \right) . \end{aligned}$$
(4.2)
Moreover, the complement set \(X'_A{\setminus }{{\mathcal {U}}}_q'\) contains all sequences ending with \(1^\infty \). This implies that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q=\dim _H X_A(q), \end{aligned}$$
(4.3)
where \(X_A(q):=\left\{ ((d_i))_q: (d_i)\in X_A'\right\} \). Note that \(X_A(q)\) is a graph-directed set satisfying the open set condition (cf. [24, Theorem 3.4]), and the sub-shift of finite type \(X_A'\) is irreducible. Then by (4.3) it follows that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q=\dim _H X_A(q)=\dim _H F_q(1). \end{aligned}$$
Now we consider \(q\in (q_c,q^*]\). By Lemma 2.2 it follows that
$$\begin{aligned} \begin{aligned} {{\mathcal {U}}}_q'\subseteq \left\{ q^\infty \right\} \cup \bigcup _{k=0}^\infty \left\{ q^k 0^\infty \right\} \cup \bigcup _{k=0}^\infty \bigcup _{m=0}^\infty \left\{ q^k 0^m F_q'(1)\right\} , \end{aligned} \end{aligned}$$
where
$$\begin{aligned} q^k 0^m F_q'(1):=\left\{ (d_i): d_1\ldots d_{k+m}=q^k0^m, (d_{k+m+i})\in F_q'(1)\right\} . \end{aligned}$$
This implies that \( \dim _H{{\mathcal {U}}}_q\le \dim _H F_q(1). \)\(\square \)

Proof of Theorem 2

The theorem follows directly by Lemmas 4.14.4. \(\square \)

5 Proof of Theorem 3

In this section we will consider the set \({{\mathcal {U}}}_q^{(\aleph _0)}\) which consists of all \(x\in E_q\) having countably infinitely many q-expansions.

Lemma 5.1

For any \(q\in \mathcal {B}_{\aleph _0}\) the set \({{\mathcal {U}}}_q^{(\aleph _0)}\) contains infinitely many points.

Proof

Let \(q\in \mathcal {B}_{\aleph _0}\). By Theorem 1 we have \(q\in [2,\infty )\). Then it suffices to show that for any \(k\ge 1\),
$$\begin{aligned} z_k:=(0^kq^\infty )_q \end{aligned}$$
is a q-null infinite points, and thus \(z_k\in {{\mathcal {U}}}_q^{(\aleph _0)}\).

If \(q>2\), then by the proof of Lemma 3.3 it yields that \(z_1=(0q^\infty )_q\) is a q-null infinite point. Moreover, note that \( z_k=\phi _0^{k-1}(z_1)\notin S_q \) for any \(k\ge 2\). This implies that all of these points \(z_k, k\ge 1\), are q-null infinite points. So, \( \left\{ z_k: k\ge 1\right\} \subseteq {{\mathcal {U}}}_q^{(\aleph _0)}. \)

If \(q=2\), then by using the substitutions
$$\begin{aligned} 0q\sim 10,\quad 0q^\infty =1^\infty =q0^\infty , \end{aligned}$$
one can also show that \(z_k\) is a q-null infinite point. In fact, all of the q-expansions of \(z_k=(0^kq^\infty )_q\) are of the form
$$\begin{aligned} 0^kq^\infty ,\quad 0^{k-1}1^\infty ,\quad 0^{k-1}1^m0q^\infty \quad \text {and}\quad 0^{k-1}1^{m-1}q0^\infty , \end{aligned}$$
where \(m\ge 1\). Therefore, \(z_k\in {{\mathcal {U}}}_q^{(\aleph _0)}\) for any \(k\ge 1\). \(\square \)

By Lemma 5.1 it follows that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is at least countably infinite for any \(q\in \mathcal {B}_{\aleph _0}=[2,\infty )\). In the following lemma we show that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is indeed countably infinite if \(q\ge q^*\).

Lemma 5.2

Let \(q\ge q^*\). Then \({{\mathcal {U}}}_q^{(\aleph _0)}\) is at most countable.

Proof

Let \(x\in {{\mathcal {U}}}_q^{(\aleph _0)}\). Then x has a q-expansion \((d_i)\) such that
$$\begin{aligned} \left| \Sigma (x_n)\right| =\infty \quad \text {for infinitely many }n\in \mathbb {N}, \end{aligned}$$
where \(x_n:=((d_{n+i}))_q\). This implies that \((d_i)\) can not end in \({{\mathcal {U}}}_q'\).

Note by the proof of Lemma 4.4 that \({{\mathcal {U}}}_q'\subseteq X_A'\), where \(X_A'\) is a sub-shift of finite type over the state \(\left\{ 0,1,q\right\} \) with adjacency matrix A defined in (4.2). Moreover, \(X_A'{\setminus } {{\mathcal {U}}}_q'\) is at most countable (cf. [24, Theorem 3.4]). Note that the expansion \((d_i)\) of \(x\in {{\mathcal {U}}}_q^{(\aleph _0)}\) does not end in \({{\mathcal {U}}}_q'\). Then it suffices to prove that the sequence \((d_i)\) must end in \(X_A'\).

Suppose on the contrary that \((d_i)\) does not end in \(X_A'\). Then by (4.2) the word 0q or 10 occurs infinitely many times in \((d_i)\). Using the word substitution \(0q\sim 10\) this implies that \(x=((d_i))_q\) has a continuum of q-expansions, leading to a contradiction with \(x\in {{\mathcal {U}}}_q^{(\aleph _0)}\). \(\square \)

Furthermore, we can prove that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is also countably infinite for \(q\in [2,q_c]\).

Lemma 5.3

Let \(q\in [2,q_c]\). Then \({{\mathcal {U}}}_q^{(\aleph _0)}\) is at most countable.

Proof

Take \(q\in [2, q_c]\). By Theorems 1 and 1.1 it follows that any \(x\in E_q\) with \(|\Sigma (x)|<\infty \) must belong to \({{\mathcal {U}}}_q=\left\{ 0,q/(q-1)\right\} \). Suppose \(x\in {{\mathcal {U}}}_q^{(\aleph _0)}\). Then there exists a word \(d_1\ldots d_n\) such that
$$\begin{aligned} \phi _{d_1}^{-1}\circ \cdots \circ \phi _{d_n}^{-1}(x)\in {{\mathcal {U}}}_q. \end{aligned}$$
This implies that the set \({{\mathcal {U}}}_q^{(\aleph _0)}\) is at most countable, since
$$\begin{aligned} {{\mathcal {U}}}_q^{(\aleph _0)}\subseteq \bigcup _{n=1}^\infty \bigcup _{d_1\ldots d_n\in \left\{ 0,1,q\right\} ^n}\phi _{d_1}\circ \cdots \circ \phi _{d_n}\left( {{\mathcal {U}}}_q\right) . \end{aligned}$$
\(\square \)

When \(q\in (q_c, q^*)\), one might expect that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is also countably infinite. Unfortunately, we are not able to prove this. Instead, we show that the Hausdorff dimension of \({{\mathcal {U}}}_q^{(\aleph _0)}\) is strictly smaller than \(\dim _H E_q=1\).

Lemma 5.4

For \(q\in (q_c, q^*)\) we have \( \dim _H{{\mathcal {U}}}_q^{(\aleph _0)}\le \dim _H{{\mathcal {U}}}_q<1\).

Proof

Take \(q\in (q_c, q^*)\). Note that
$$\begin{aligned} {{\mathcal {U}}}_q^{(\aleph _0)}\subseteq \bigcup _{n=1}^\infty \bigcup _{d_1\ldots d_n\in \left\{ 0,1,q\right\} ^n}\phi _{d_1}\circ \cdots \circ \phi _{d_n}({{\mathcal {U}}}_q). \end{aligned}$$
By using the countable stability of Hausdorff dimension this implies that \(\dim _H{{\mathcal {U}}}_q^{(\aleph _0)}\le \dim _H{{\mathcal {U}}}_q\). In the following it suffices to prove \(\dim _H{{\mathcal {U}}}_q<1\).
Note that \({{\mathcal {U}}}_q'\subseteq X_A'\), where \(X_A'\) is the sub-shift of finite type over the state \(\left\{ 0,1, q\right\} \) with adjacency matrix A defined in (4.2). Then
$$\begin{aligned} {{\mathcal {U}}}_q\subseteq X_A(q)=\left\{ ((d_i))_q: (d_i)\in X_A'\right\} . \end{aligned}$$
Note that \(X_A(q)\) is a graph-directed set (cf. [14]). This implies that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q\le \dim _H X_A(q)\le \frac{\log q_c}{\log q}<1. \end{aligned}$$
\(\square \)

At the end of this section we investigate the set \({{\mathcal {U}}}_q^{(2^{\aleph _0})}\) which consists of all points having a continuum of q-expansions, and show that \({{\mathcal {U}}}_q^{(2^{\aleph _0})}\) has full Hausdorff measure.

Lemma 5.5

For any \(q>1\) we have
$$\begin{aligned} \mathcal {H}^{\dim _H E_q} \left( {{\mathcal {U}}}_q^{\left( 2^{\aleph _0}\right) }\right) =\mathcal {H}^{\dim _H E_q}(E_q)\in (0,\infty ). \end{aligned}$$

Proof

Clearly, for \(q\in (1,q^*]\) we have \(E_q=[0, q/(q-1)]\), and then \(\mathcal H^{\dim _H E_q} (E_q)\in (0,\infty )\). Moreover, for \(q>q^*\) we have by (1.1) that \(\dim _H E_q=\log q^*/\log q\), and the set \(E_q\) has positive and finite Hausdorff measure (cf. [15]). Therefore,
$$\begin{aligned} 0< \mathcal {H}^{\dim _H E_q}(E_q)<\infty \quad \text {for any}\quad q>1. \end{aligned}$$
(5.1)
First we prove the lemma for \(q\le q^*\). By Theorems 1 and 1.1 it follows that for any \(q\in (1,q^*]\),
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q^{(k)}=\dim _H{{\mathcal {U}}}_q<1=\dim _H E_q \quad \text {for any}\quad k\ge 2. \end{aligned}$$
Moreover, by Lemmas 5.25.4 we have \(\dim _H{{\mathcal {U}}}_q^{(\aleph _0)}<1.\) Observe that
$$\begin{aligned} E_q={{\mathcal {U}}}_q^{(2^{\aleph _0})}\cup {{\mathcal {U}}}_q^{(\aleph _0)}\cup \bigcup _{k=1}^\infty {{\mathcal {U}}}_q^{(k)}\quad \text {for any }q>1. \end{aligned}$$
(5.2)
Therefore, by (5.1) and (5.2) we have \(\mathcal {H}^{\dim _H E_q}({{\mathcal {U}}}_q^{(2^{\aleph _0})})=\mathcal {H}^{\dim _H E_q}(E_q)\in (0,\infty ). \)
Now we consider \(q> q^*\). By Theorems 1.1 (iii), 2 and (1.1) it follows that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q^{(k)}=\frac{\log q_c}{\log q}<\frac{\log q^*}{\log q}=\dim _H E_q \end{aligned}$$
for any \(k\ge 1\). Moreover, by Lemma 5.2 we have \(\dim _H{{\mathcal {U}}}_q^{(\aleph _0)}=0\). Again, by (5.1) and (5.2) it follows that \(\mathcal {H}^{\dim _H E_q}({{\mathcal {U}}}_q^{(2^{\aleph _0})})=\mathcal {H}^{\dim _H E_q}(E_q)\in (0,\infty ). \) This completes the proof. \(\square \)

Proof of Theorem 3

The theorem follows by Lemmas 5.15.3 and 5.5. \(\square \)

6 Examples and final remarks

In this section we consider some examples. The first example is an application of Theorems 13 to expansions with deleted digits set.

Example 6.1

Let \(q=3\). We consider q-expansions with digits set \(\left\{ 0,1,3\right\} \). This is a special case of expansions with deleted digits (cf. [17]). Then
$$\begin{aligned} E_3=\left\{ \sum _{i=1}^\infty \frac{d_i}{3^i}: d_i\in \left\{ 0,1,3\right\} \right\} . \end{aligned}$$
By Theorems 1.1 and 2 we have
$$\begin{aligned} \dim _H{{\mathcal {U}}}_3^{(k)}=\dim _H{{\mathcal {U}}}_3=\frac{\log q_c}{\log 3}\approx 0.767877 \end{aligned}$$
for any \(k\ge 2\). This means that the set \({{\mathcal {U}}}_3^{(k)}\) consisting of all points in \(E_3\) with precisely k different triadic expansions has the same Hausdorff dimension \(\log q_c/\log 3\) for any integer \(k\ge 1\). Moreover, by Theorem 3 it follows that \({{\mathcal {U}}}_3^{(\aleph _0)}\) is countably infinite, and
$$\begin{aligned} \dim _H{{\mathcal {U}}}_3^{(2^{\aleph _0})}=\dim _H E_3=\frac{\log q^*}{\log 3}\approx 0.876036. \end{aligned}$$

Theorem 1.1 gives a uniform formula for the Hausdorff dimension of \({{\mathcal {U}}}_q\) for \(q\in [q^*, \infty )\). Excluding the trivial case for \(q\in (1, q_c]\) that \({{\mathcal {U}}}_q=\left\{ 0, q/(q-1)\right\} \), it would be interesting to ask whether the Hausdorff dimension of \({{\mathcal {U}}}_q\) can be determined for \(q\in (q_c, q^*)\). In the following we give an example for which the Hausdorff dimension of \({{\mathcal {U}}}_q\) can be explicitly calculated.

Example 6.2

Let \(q=1+\sqrt{2}\in (q_c, q^*)\). Then
$$\begin{aligned} (q0^\infty )_q=(1qq0^\infty )_q\quad \text {and}\quad \alpha (q)=(q1)^\infty . \end{aligned}$$
Moreover, the quasi-greedy q-expansion of \(q-1\) with alphabet \(\left\{ 0, q-1, q\right\} \) is \(q(q-1)^\infty \). Therefore, by Lemmas 3.1 and 3.2 of [24] it follows that \({{\mathcal {U}}}_q'\) is the set of sequences \((d_i)\in \left\{ 0,1,q\right\} ^\infty \) satisfying
$$\begin{aligned} \left\{ \begin{array}{lll} d_{n+1}d_{n+2}\cdots \prec (1q)^\infty &{} \text {if}&{} \quad d_n=0,\\ 1^\infty <d_{n+1}d_{n+2}\cdots \prec (q1)^\infty &{}\text {if}&{} \quad d_n=1,\\ d_{n+1}d_{n+2}\cdots \succ 01^\infty &{}\text {if}&{} \quad d_n=q. \end{array} \right. \end{aligned}$$
Let \(X_A'\) be the sub-shift of finite type over the states
$$\begin{aligned} \left\{ 00, 01, 11, 1q, q0, q1,qq\right\} \end{aligned}$$
with adjacency matrix
$$\begin{aligned} A=\left( \begin{array}{c@{\qquad }c@{\qquad }c@{\qquad }c@{\qquad }c@{\qquad }c@{\qquad }c} 1&{}1&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}1&{}1&{}0&{}0&{}0\\ 0&{}0&{}1&{}1&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}1&{}1&{}0\\ 0&{}1&{}0&{}0&{}0&{}0&{}0\\ 0&{}0&{}1&{}1&{}0&{}0&{}0\\ 0&{}0&{}0&{}0&{}1&{}1&{}1 \end{array} \right) . \end{aligned}$$
Then one can verify that \({{\mathcal {U}}}_q'\subseteq X_A'\), and \(X_A'{\setminus }{{\mathcal {U}}}_q'\) contains all sequences ending with \(1^\infty \) or \((1q)^\infty \). This implies that
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q=\dim _H X_A(q), \end{aligned}$$
where \(X_A(q)=\left\{ ((d_i))_q: (d_i)\in X_A'\right\} \). Note that \(X_A(q)\) is a graph-directed set satisfying the open set condition (cf. [14]). Then by Theorem 2 we have
$$\begin{aligned} \dim _H{{\mathcal {U}}}_q^{(k)}=\dim _H{{\mathcal {U}}}_q=\frac{h(X_A')}{\log q}\approx 0.691404. \end{aligned}$$
Furthermore, by the word substitution \(q00\sim 1qq\) and in a similar way as in the proof of Lemma 5.2 one can show that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is countably infinite. Finally, by Theorem 3 we have \(\dim _H{{\mathcal {U}}}_q^{(2^{\aleph _0})}=\dim _H E_q=1\).

Question 1. Can we give a uniform formula for the Hausdorff dimension of \({{\mathcal {U}}}_q\) for \(q\in (q_c, q^*)\)?

In beta expansions we know that the dimension function of the univoque set has a Devil’s staircase behavior (cf. [12]).

Question 2. Does the dimension function \(D(q):=\dim _H{{\mathcal {U}}}_q\) have a Devil’s staircase behavior in the interval \((q_c, q^*)\)?

By Theorem 3 one has that \({{\mathcal {U}}}_q^{(\aleph _0)}\) is countable for any \(q\in \mathcal {B}_2{\setminus }(q_c, q^*)\). Moreover, in Lemma 5.4 we show that \(\dim _H{{\mathcal {U}}}_q^{(\aleph _0)}\le \dim _H{{\mathcal {U}}}_q<1\) for any \(q\in (q_c, q^*)\). In view of Example 6.2 we ask the following

Question 3. Does there exist a \(q\in (q_c, q^*)\) such that \({{\mathcal {U}}}_q^{(\aleph _0)}\) has positive Hausdorff dimension?

Notes

Acknowledgements

The second author was supported by NSFC no. 11701302 and K. C. Wong Magna Fund at Ningbo University. The third author was supported by NSFC no. 11401516 and Jiangsu Province Natural Science Foundation for the Youth no. BK20130433. The forth author was supported by NSFC nos. 11271137, 11571144, 11671147 and in part by Science and Technology Commission of Shanghai Municipality (no. 18dz2271000)

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© The Author(s) 2018

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors and Affiliations

  • Karma Dajani
    • 1
  • Kan Jiang
    • 1
    • 2
  • Derong Kong
    • 3
    • 4
    Email author
  • Wenxia Li
    • 5
  1. 1.Department of MathematicsUtrecht UniversityUtrechtThe Netherlands
  2. 2.Department of MathematicsNingbo UniversityNingboPeople’s Republic of China
  3. 3.School of Mathematical ScienceYangzhou UniversityYangzhouPeople’s Republic of China
  4. 4.College of Mathematics and StatisticsChongqing UniversityShapingbaChina
  5. 5.School of Mathematical Sciences, Shanghai Key Laboratory of PMMPEast China Normal UniversityShanghaiPeople’s Republic of China

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