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

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.

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

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

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

• (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

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

• (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

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

• (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 $q\in (1,2)$.

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

• (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 4.1–4.4. $\square$

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 5.2–5.4 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 5.1–5.3 and 5.5. $\square$

6 Examples and final remarks

In this section we consider some examples. The first example is an application of Theorems 1–3 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?