Univoque bases and Hausdorff dimension

Abstract

Given a positive integer M and a real number $q >1$, a q -expansion of a real number x is a sequence $(c_i)=c_1c_2\ldots$ with $(c_i) \in \{0,\ldots ,M\}^\infty$ such that

$$\begin{aligned} x=\sum _{i=1}^{\infty } c_iq^{-i}. \end{aligned}$$

It is well known that if $q \in (1,M+1]$, then each $x \in I_q:=\left[ 0,M/(q-1)\right]$ has a q-expansion. Let $\mathcal {U}=\mathcal {U}(M)$ be the set of univoque bases $q>1$ for which 1 has a unique q-expansion. The main object of this paper is to provide new characterizations of $\mathcal {U}$ and to show that the Hausdorff dimension of the set of numbers $x \in I_q$ with a unique q-expansion changes the most if q “crosses” a univoque base. Denote by $\mathcal {B}_2=\mathcal {B}_2(M)$ the set of $q \in (1,M+1]$ such that there exist numbers having precisely two distinct q-expansions. As a by-product of our results, we obtain an answer to a question of Sidorov (J Number Theory 129:741–754, 2009) and prove that

$$\begin{aligned} \dim _H(\mathcal {B}_2\cap (q',q'+\delta ))>0\quad \text {for any}\quad \delta >0, \end{aligned}$$

where $q'=q'(M)$ is the Komornik–Loreti constant.

1 Introduction

Non-integer base expansions have received much attention since the pioneering works of Rényi [25] and Parry [24]. Given a positive integer M and a real number $q \in (1,M+1]$, a sequence $(d_i)=d_1d_2\ldots$ with digits $d_i\in \left\{ 0,1,\ldots ,M\right\}$ is called a q -expansion of x or an expansion of x in base q if

$$\begin{aligned} x=\pi _q((d_i)):=\sum _{i=1}^\infty \frac{d_i}{q^i}. \end{aligned}$$

It is well known that each $x \in I_q:=[0,M/(q-1)]$ has a q-expansion. One such expansion—the greedy q -expansion—can be obtained by performing the so called greedy algorithm of Rényi which is defined recursively as follows: if $d_1, \ldots ,d_{n-1}$ is already defined (no condition if $n=1$), then $d_n$ is the largest element of $\left\{ 0,\ldots ,M\right\}$ satisfying $\sum _{i=1}^n d_i q^{-i} \le x$. Equivalently, $(d_i)$ is the greedy q-expansion of $\sum _{i=1}^\infty d_i q^{-i}$ if and only if $\sum _{i=n+1}^\infty d_i q^{-i+n}<1$ whenever $d_n < M, n=1,2,\ldots$. Hence if $1<q < r \le M+1$, then the greedy q-expansion of a number $x \in I_q$ is also the greedy expansion in base r of a number in $I_r$.

Let ${{\mathcal {U}}}_q$ be the univoque set consisting of numbers $x\in I_q$ such that x has a unique q-expansion, and let ${{\mathcal {U}}}_q'$ be the set of corresponding expansions. Note that a sequence $(c_i)$ belongs to ${{\mathcal {U}}}_q'$ if and only if both the sequences $(c_i)$ and $(M-c_i):=(M-c_1)(M-c_2)\ldots$ are greedy q-expansions, hence ${{\mathcal {U}}}_q' \subseteq {{\mathcal {U}}}_r'$ whenever $1< q < r \le M+1$. Many works are devoted to the univoque sets ${{\mathcal {U}}}_q$ (see, e.g., [10, 11, 14]). Recently, de Vries and Komornik investigated their topological properties in [8]. Komornik et al. considered their Hausdorff dimension in [19], and showed that the dimension function $D: q\mapsto \dim _H{{\mathcal {U}}}_q$ behaves like a Devil’s staircase on $(1,M+1]$. For more information on the univoque set ${{\mathcal {U}}}_q$ we refer to the survey paper [15] and the references therein.

There is an intimate connection between the set ${{\mathcal {U}}}_q$ and the set of univoque bases ${{\mathcal {U}}}={{\mathcal {U}}}(M)$ consisting of numbers $q>1$ such that 1 has a unique q-expansion over the alphabet $\left\{ 0,1,\ldots ,M\right\}$. For instance, it was shown in [8] that ${{\mathcal {U}}}_q$ is closed if and only if q does not belong to the set $\overline{{{\mathcal {U}}}}$. It is well-known that ${{\mathcal {U}}}$ is a Lebesgue null set of full Hausdorff dimension (cf. [6, 12, 19]). Moreover, the smallest element of ${{\mathcal {U}}}$ is the Komornik–Loreti constant (cf. [16, 17])

$$\begin{aligned} q'=q'(M), \end{aligned}$$

while the largest element of ${{\mathcal {U}}}$ is (of course) $M+1$. Recently, Komornik and Loreti showed in [18] that its closure $\overline{{{\mathcal {U}}}}$ is a Cantor set (see also, [9]), i.e., a nonempty closed set having neither isolated nor interior points. Writing the open set $(1,M+1] {\setminus } \overline{{{\mathcal {U}}}}=(1,M+1){\setminus } \overline{{{\mathcal {U}}}}$ as the disjoint union of its connected components, i.e.,

$$\begin{aligned} (1,M+1]{\setminus }\overline{{{\mathcal {U}}}}= (1,q')\cup \bigcup \left( q_0,q_0^*\right) , \end{aligned}$$

(1)

the left endpoints $q_0$ in (1) run over the whole set $\overline{{{\mathcal {U}}}}{\setminus }{{\mathcal {U}}}$, and the right endpoints $q_0^*$ run through a subset of ${{\mathcal {U}}}$ (cf. [8]). Furthermore, each left endpoint $q_0$ is algebraic, while each right endpoint $q_0^*\in {{\mathcal {U}}}$ is transcendental (cf. [20]).

De Vries showed in [7], roughly speaking, that the sets ${{\mathcal {U}}}_q'$ change the most if we cross a univoque base. More precisely, it was shown that $q \in {{\mathcal {U}}}$ if and only if ${{\mathcal {U}}}_r' {\setminus } {{\mathcal {U}}}_q'$ is uncountable for each $r \in (q,M+1]$ and $r \in \overline{{{\mathcal {U}}}}$ if and only if ${{\mathcal {U}}}_r' {\setminus } {{\mathcal {U}}}_q'$ is uncountable for each $q \in (1,r)$.

The main object of this paper is to provide similar characterizations of ${{\mathcal {U}}}$ and $\overline{{{\mathcal {U}}}}$ in terms of the Hausdorff dimension of the sets ${{\mathcal {U}}}_r' {\setminus } {{\mathcal {U}}}_q'$ after a natural projection. Furthermore, we characterize the sets ${{\mathcal {U}}}$ and $\overline{{{\mathcal {U}}}}$ by looking at the Hausdorff dimensions of ${{\mathcal {U}}}$ and $\overline{{{\mathcal {U}}}}$ locally.

Theorem 1.1

Let $q\in (1, M+1]$. The following statements are equivalent.

• (i) $q\in {{\mathcal {U}}}$.

• (ii) $\dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q')>0$ for any $r\in (q, M+1]$.

• (iii) $\dim _H{{\mathcal {U}}}\cap (q,r)>0$ for any $r\in (q, M+1]$.

Theorem 1.2

Let $q\in (1, M+1]$. The following statements are equivalent.

• (i) $q\in \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$.

• (ii) $\dim _H\pi _{M+1}({{\mathcal {U}}}_q'{\setminus }{{\mathcal {U}}}_p')>0$ for any $p\in (1, q)$.

• (iii) $\dim _H{{\mathcal {U}}}\cap (p, q)>0$ for any $p\in (1, q)$.

It follows at once from Theorems 1.1 and 1.2 that ${{\mathcal {U}}}$ (or, equivalently, $\overline{{{\mathcal {U}}}}$) does not contain isolated points.

We remark that the projection map $\pi _{M+1}$ in Theorem 1.1 (ii) can be replaced by $\pi _\rho$ for any $r\le \rho \le M+1$. Similarly, the projection map $\pi _{M+1}$ in Theorem 1.2 (ii) can also be replaced by $\pi _\rho$ with $q\le \rho \le M+1$. We also point out that Theorems 1.1 and  1.2 strengthen the main result of [7] where the cardinality of the sets ${{\mathcal {U}}}_q' {\setminus } {{\mathcal {U}}}_p'$ with $1< p < q \le M+1$ was determined.

Let $\mathcal {B}_2$ be the set of bases $q \in \left( 1,M+1\right]$ for which there exists a number $x\in [0, M/(q-1)]$ having exactly two q-expansions. It was asked by Sidorov [26] whether $\dim _H\mathcal {B}_2\cap (q', q'+\delta )>0$ for any $\delta >0$, where $q'$ is the Komornik–Loreti constant. Since ${{\mathcal {U}}}\subseteq \mathcal {B}_2$ (see [26, Lemma 3.1]¹), Theorem 1.1 answers this question in the affirmative.

Corollary 1

$\dim _H\mathcal {B}_2\cap (q', q'+\delta )>0$ for any $\delta >0.$

The rest of the paper is arranged as follows. In Sect. 2 we recall some properties of unique q-expansions. The proof of Theorems 1.1 and 1.2 will be given in Sect. 3.

2 Preliminaries

In this section we recall some properties of the univoque set ${{\mathcal {U}}}_q$. Throughout this paper, a sequence $(d_i)=d_1d_2\ldots$ is an element of $\left\{ 0,\ldots ,M\right\} ^\infty$ with each digit $d_i$ belonging to the alphabet $\left\{ 0,\ldots ,M\right\}$. Moreover, for a word ${\mathbf {c}}=c_1\ldots c_n$ we mean a finite string of digits with each digit $c_i$ from $\left\{ 0,\ldots ,M\right\}$. For two words ${\mathbf {c}}=c_1\ldots c_n$ and $\mathbf {d}=d_1\ldots d_m$ we denote by ${\mathbf {c}}\mathbf {d}=c_1\ldots c_nd_1\ldots d_m$ the concatenation of the two words. For an integer $k\ge 1$ we denote by ${\mathbf {c}}^k$ the k-times concatenation of ${\mathbf {c}}$ with itself, and by ${\mathbf {c}}^\infty$ the infinite repetition of ${\mathbf {c}}$.

For a sequence $(d_i)$ we denote its reflection by $\overline{(d_i)}:=(M-d_1)(M-d_2)\ldots$. Accordingly, for a word ${\mathbf {c}}=c_1\ldots c_n$ we denote its reflection by $\overline{{\mathbf {c}}}:=(M-c_1)\ldots (M-c_n)$. If $c_n<M$ we denote by ${\mathbf {c}}^+:=c_1\ldots c_{n-1}(c_n+1).$ If $c_n>0$ we write ${\mathbf {c}}^-:=c_1\ldots c_{n-1}(c_n-1)$.

We will use systematically the lexicographic ordering $<, \le ,>$ and $\ge$ between sequences and between words. For two sequences $(c_i), (d_i)\in \left\{ 0,1,\ldots , M\right\} ^\infty$ we say that $(c_i)<(d_i)$ 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 write $(c_i)\le (d_i)$ if $(c_i)<(d_i)$ or $(c_i)=(d_i)$. Similarly, we say $(c_i)>(d_i)$ if $(d_i)<(c_i)$, and $(c_i)\ge (d_i)$ if $(d_i)\le (c_i)$. We extend this definition to words in the obvious way. For example, for two words ${\mathbf {c}}$ and $\mathbf {d}$ we write ${\mathbf {c}}<\mathbf {d}$ if ${\mathbf {c}}0^\infty <\mathbf {d}0^\infty$.

A sequence is called finite if it has a last nonzero element. Otherwise it is called infinite. So $0^{\infty }:=00\ldots$ is considered to be infinite. For $q\in (1,M+1]$ we denote by

$$\begin{aligned} \alpha (q)=(\alpha _i(q)) \end{aligned}$$

the quasi-greedy q-expansion of 1 (cf. [5]), i.e., the lexicographically largest infinite q-expansion of 1. Let $\beta (q)=(\beta _i(q))$ be the greedy q-expansion of 1 (cf. [24]), i.e., the lexicographically largest q-expansion of 1. For convenience, we set $\alpha (1)=0^{\infty }$ and $\beta (1)=10^{\infty }$, even though $\alpha (1)$ is not a 1-expansion of 1.

Moreover, we endow the set $\left\{ 0,\ldots ,M\right\}$ with the discrete topology and the set of all possible sequences $\left\{ 0,1,\ldots ,M\right\} ^{\infty }$ with the Tychonoff product topology.

The following properties of $\alpha (q)$ and $\beta (q)$ were established in [24], see also [3].

Lemma 2.1

• (i) The map $q\mapsto \alpha (q)$ is an increasing bijection from $[1, M+1]$ onto the set of all infinite sequences $(\alpha _i)$ satisfying

  $$\begin{aligned} \alpha _{n+1}\alpha _{n+2}\ldots \le \alpha _1\alpha _2\ldots \quad \text {whenever}\quad \alpha _n<M. \end{aligned}$$

• (ii) The map $q\mapsto \beta (q)$ is an increasing bijection from $[1, M+1]$ onto the set of all sequences $(\beta _i)$ satisfying

  $$\begin{aligned} \beta _{n+1}\beta _{n+2}\ldots<\beta _1\beta _2\ldots \quad \text {whenever}\quad \beta _n<M. \end{aligned}$$

Lemma 2.2

• (i) $\beta (q)$ is infinite if and only if $\beta (q)=\alpha (q)$.

• (ii) If $\beta (q)=\beta _1\ldots \beta _m 0^\infty$ with $\beta _m>0$, then $\alpha (q)=(\beta _1\ldots \beta _m^-)^\infty .$

• (iii) The map $q\mapsto \alpha (q)$ is left-continuous, while the map $q\mapsto \beta (q)$ is right-continuous.

In order to investigate the unique expansions we need the following lexicographic characterization of ${{\mathcal {U}}}_q'$ (cf. [3]).

Lemma 2.3

Let $q\in (1,M+1]$. Then $(d_i)\in {{\mathcal {U}}}_q'$ if and only if

$$\begin{aligned} \left\{ \begin{array}{lll} d_{n+1}d_{n+2}\ldots<\alpha _1(q)\alpha _2(q)\ldots &{}\text { whenever }&{}d_n<M,\\ d_{n+1}d_{n+2}\ldots>\overline{\alpha _1(q)\alpha _2(q)\ldots }&{}\text { whenever} &{}d_n>0. \end{array} \right. \end{aligned}$$

Note that $q\in {{\mathcal {U}}}$ if and only if $\alpha (q)$ is the unique q-expansion of 1. Then Lemma 2.3 yields a characterization of ${{\mathcal {U}}}$ (see also, [11, 17]).

Lemma 2.4

Let $q \in (1,M+1)$. Then $q\in {{\mathcal {U}}}$ if and only if $\alpha (q)=(\alpha _i(q))$ satisfies

$$\begin{aligned} \overline{\alpha (q)}<\alpha _{n+1}(q)\alpha _{n+2}(q)\ldots <\alpha (q)\quad \text {for all}\quad n\ge 1. \end{aligned}$$

Consider a connected component $(q_0, q_0^*)$ of $(q', M+1){\setminus }\overline{{{\mathcal {U}}}}$ as in (1). Then there exists a (unique) word ${{\mathbf {t}}}=t_1\ldots t_p$ such that (cf. [8, 20])

$$\begin{aligned} \alpha (q_0)={{\mathbf {t}}}^\infty \quad \text {and}\quad \alpha \left( q_0^*\right) =\lim _{n\rightarrow \infty }g^n({{\mathbf {t}}}), \end{aligned}$$

where $g^n=\underbrace{g\circ \cdots \circ g}_n$ denotes the n-fold composition of g with itself, and

$$\begin{aligned} g({\mathbf {c}}) :={\mathbf {c}}^+\overline{{\mathbf {c}}^+}\quad \text {for any word}\quad {\mathbf {c}}=c_1\ldots c_k\text { with }c_k<M. \end{aligned}$$

(2)

We point out that the word ${{\mathbf {t}}}=t_1\ldots t_p$ in the definitions of $\alpha (q_0)$ and $\alpha (q_0^*)$ is called an admissible block in [20, Definition 2.1] which satisfies the following lexicographical inequalities: $t_p<M$ and for any $1\le i\le p$ we have

$$\begin{aligned} \overline{t_1\ldots t_p}\le t_i\ldots t_pt_1\ldots t_{i-1}\quad \text {and}\quad t_i\ldots t_p\;\overline{t_1\ldots t_{i-1}}\le t_1\ldots t_p^+. \end{aligned}$$

We also mention that the limit $\lim _{n\rightarrow \infty }g^n({{\mathbf {t}}})$ stands for the infinite sequence beginning with ${{\mathbf {t}}}^+\overline{{{\mathbf {t}}}}\,\overline{{{\mathbf {t}}}^+}{{\mathbf {t}}}^+\;\overline{{{\mathbf {t}}}^+}{{\mathbf {t}}}\,{{\mathbf {t}}}^+\overline{{{\mathbf {t}}}}\ldots ,$ and the existence of this limit was shown by Allouche [2].

In this case $(q_0, q_0^*)$ is called the connected component generated by ${{\mathbf {t}}}$. The closed interval $[q_0, q_0^*]$ is the so called admissible interval generated by ${{\mathbf {t}}}$ (see [20, Definition 2.4]). Furthermore, the sequence

$$\begin{aligned} \alpha \left( q_0^*\right) =\lim _{n\rightarrow \infty }g^n({{\mathbf {t}}})={{\mathbf {t}}}^+\,\overline{{{\mathbf {t}}}}\;\overline{{{\mathbf {t}}}^+}\,{{\mathbf {t}}}^+~ \overline{{{\mathbf {t}}}^+}\, {{\mathbf {t}}}\, {{\mathbf {t}}}^+\,\overline{{{\mathbf {t}}}}\ldots \end{aligned}$$

is a generalized Thue–Morse sequence (cf. [20, Definition 2.2], see also [1]).

The following lemma for the generalized Thue–Morse sequence $\alpha (q_0^*)$ was established in [20, Lemma 4.2].

Lemma 2.5

Let $(q_0, q_0^*)\subset (q', M+1){\setminus }\overline{{{\mathcal {U}}}}$ be a connected component generated by $t_1\ldots t_p$. Then the sequence $(\theta _i)=\alpha (q_0^*)$ satisfies

$$\begin{aligned} \overline{\theta _1\ldots \theta _{2^n p-i}}<\theta _{i+1}\ldots \theta _{2^n p}\le \theta _1\ldots \theta _{2^n p-i} \end{aligned}$$

for any $n\ge 0$ and any $0\le i<2^n p$.

Finally, we recall some topological properties of ${{\mathcal {U}}}$ and $\overline{{{\mathcal {U}}}}$ which were essentially established in [8, 18] (see also, [9]).

Lemma 2.6

• (i) If $q\in {{\mathcal {U}}}$, then there exists a decreasing sequence $(r_n)$ of elements in $\bigcup \left\{ q_0^*\right\}$ that converges to q as $n\rightarrow \infty$;

• (ii) If $q\in \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$, then there exists an increasing sequence $(p_n)$ of elements in $\bigcup \left\{ q_0^*\right\}$ that converges to q as $n\rightarrow \infty$.

We remark here that the bases $q_0^*$ are called de Vries–Komornik numbers which were shown to be transcendental in [20]. By Lemma 2.6 it follows that the set of de Vries–Komornik numbers is dense in $\overline{{{\mathcal {U}}}}$.

3 Proofs of Theorems 1.1 and 1.2

3.1 Proof of Theorem 1.1 for (i) $\Leftrightarrow$ (ii).

For each connected component $(q_0, q_0^*)$ of $(q', M+1){\setminus }\overline{{{\mathcal {U}}}}$ we construct a sequence of bases $(r_n)$ in ${{\mathcal {U}}}$ strictly decreasing to $q_0^*$.

Lemma 3.1

Let $(q_0, q_0^*)\subset (q', M+1){\setminus }\overline{{{\mathcal {U}}}}$ be a connected component generated by $t_1\ldots t_p$, and let $(\theta _i)=\alpha (q_0^*)$. Then for each $n\ge 1$, the number $r_n\in {{\mathcal {U}}}$ determined by

$$\begin{aligned} \alpha (r_n)=\beta (r_n)=\theta _1\ldots \theta _{2^n p}\left( \theta _{2^n p+1}\ldots \theta _{2^{n+1}p}\right) ^\infty , \end{aligned}$$

belongs to ${{\mathcal {U}}}$. Furthermore, $(r_n)$ is a strictly decreasing sequence that converges to $q_0^*$.

Proof

Using (2) one may verify that the sequence $(\theta _i)$ satisfies

$$\begin{aligned} \theta _{2^n p+k} =\overline{\theta _k }\quad \text {for all}\quad 1\le k<2^n p;\quad \theta _{2^{n+1}p} =\overline{\theta _{2^n p} }\,^+ \end{aligned}$$

for all $n\ge 0$. Now fix $n\ge 1$. We claim that

$$\begin{aligned} \sigma ^i\left( \theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty \right) <\theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty \end{aligned}$$

(3)

for all $i\ge 1$, where $\sigma$ is the left shift on $\left\{ 0,\ldots , M\right\} ^\infty$ defined by $\sigma ((c_i))=(c_{i+1})$. By periodicity it suffices to prove (3) for $0<i<2^{n+1}p$. We distinguish between the following three cases: (I) $0< i<2^np$; (II) $i=2^n p$; (III) $2^n p<i<2^{n+1}p$.

Case (I). $0< i<2^n p$. Then by Lemma 2.5 it follows that

$$\begin{aligned} \theta _{i+1}\ldots \theta _{2^n p}\le \theta _1\ldots \theta _{2^n p-i} \end{aligned}$$

and

$$\begin{aligned} \theta _{2^n p+1}\ldots \theta _{2^np +i}=\overline{\theta _1\ldots \theta _i}<\theta _{2^n p-i+1}\ldots \theta _{2^n p}. \end{aligned}$$

This implies (3) for $0<i<2^n p$.

Case (II). $i=2^n p$. Note by [17] that $\alpha _1(q')=[M/2]+1$ (see also, [4]), where [y] denotes the integer part of a real number y. Then by using $q_0^*>q'$ in Lemma 2.1 we have

$$\begin{aligned} \theta _1=\alpha _1\left( q_0^*\right) \ge \alpha _1(q')>\overline{\alpha _1(q')}\ge \overline{\theta _1}. \end{aligned}$$

This, together with $n\ge 1$, implies

$$\begin{aligned} \theta _{2^n p+1}\ldots \theta _{2^{n+1} p}=\overline{\theta _1\ldots \theta _{2^np}}\,^+<\theta _1\ldots \theta _{2^n p}. \end{aligned}$$

So, (3) holds true for $i=2^n p$.

Case (III). $2^n p< i<2^{n+1}p$. Write $j=i-2^n p$. Then $0< j<2^n p$. Once again, we infer from Lemma 2.5 that

$$\begin{aligned} \theta _{i+1}\ldots \theta _{2^{n+1}p}=\overline{\theta _{j+1}\ldots \theta _{2^n p}}\,^+\le \theta _1\ldots \theta _{2^n p-j} \end{aligned}$$

and

$$\begin{aligned} \theta _{2^n p+1}\ldots \theta _{2^n p+j}=\overline{\theta _1\ldots \theta _j}<\theta _{2^np-j+1}\ldots \theta _{2^n p}. \end{aligned}$$

This yields (3) for $2^n p<i<2^{n+1}p$.

Note by Lemma 2.5 that

$$\begin{aligned} \sigma ^i\left( \theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty \right) >\overline{\theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty } \end{aligned}$$

for any $i\ge 0$. Then by (3) and Lemma 2.4 it follows that there exists $r_n\in {{\mathcal {U}}}$ such that

$$\begin{aligned} \alpha (r_n)=\beta (r_n)=\theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty . \end{aligned}$$

In the following we prove $r_n\searrow q_0^*$ as $n\rightarrow \infty$. For $n\ge 1$ we observe that

$$\begin{aligned} \beta (r_{n+1})&=\theta _1\ldots \theta _{2^{n+1}p}(\theta _{2^{n+1}p+1}\ldots \theta _{2^{n+2}p})^\infty \\&=\theta _1\ldots \theta _{2^{n}p}\overline{\theta _1\ldots \theta _{2^n p}}\,^+\overline{\theta _1\ldots \theta _{2^n p}}\ldots \\&<\theta _1\ldots \theta _{2^n p}\,\left( \overline{\theta _1\ldots \theta _{2^n p}}^+\right) ^\infty =\beta (r_n). \end{aligned}$$

Then by Lemma 2.1 (ii) we have $r_{n+1}<r_{n}$. Note that $\beta (q_0^*)=\alpha (q_0^*)=(\theta _i)$, and

$$\begin{aligned} \beta (r_n)\rightarrow (\theta _i)=\beta \left( q_0^*\right) \quad \text {as}\quad n\rightarrow \infty . \end{aligned}$$

Hence, we conclude from Lemma 2.2 (iii) that $r_n\searrow q_0^*$ as $n\rightarrow \infty$. $\square$

Lemma 3.2

Let $(q_0, q_0^*)\subset (q',M+1){\setminus }\overline{{{\mathcal {U}}}}$ be a connected component generated by $t_1\ldots t_p$, and let $(\theta _i)=\alpha (q_0^*)$. Then for any $n\ge 1$ and any $0\le i< 2^{n}p$ we have

$$\begin{aligned}&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}<\sigma ^i(\xi _n\overline{\xi _n})<\theta _1\ldots \theta _{2^{n+1}p-i},\nonumber \\&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}< \sigma ^i(\xi _n\overline{\xi _n^-}) \le \theta _1\ldots \theta _{2^{n+1}p-i},\\&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}< \sigma ^i(\xi _n^-\xi _n) <\theta _1\ldots \theta _{2^{n+1}p-i},\nonumber \end{aligned}$$

(4)

and thus (by symmetry),

$$\begin{aligned}&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}<\sigma ^i(\overline{\xi _n}\xi _n)<\theta _1\ldots \theta _{2^{n+1}p-i},\\&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}\le \sigma ^i(\overline{\xi _n} {\xi _n^-})<\theta _1\ldots \theta _{2^{n+1}p-i},\\&\overline{\theta _1\ldots \theta _{2^{n+1}p-i}}< \sigma ^i(\overline{\xi _n^-}\,\overline{\xi _n}) <\theta _1\ldots \theta _{2^{n+1}p-i}, \end{aligned}$$

where $\xi _n:=\theta _1\ldots \theta _{2^n p}$.

Proof

By symmetry it suffices to prove (4).

Note that $\xi _n\overline{\xi _n}=\theta _1\ldots \theta _{2^{n+1}p}^-$ and $\xi _n\overline{\xi _n^-}=\theta _1\ldots \theta _{2^{n+1}p}.$ Then by Lemma 2.5 it follows that

$$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n+1}p-i}}<\sigma ^i(\xi _n\overline{\xi _n})<\theta _1\ldots \theta _{2^{n+1}p-i} \end{aligned}$$

and

$$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n+1}p-i}}<\sigma ^i(\xi _n\overline{\xi _n^-})\le \theta _1\ldots \theta _{2^{n+1}p-i} \end{aligned}$$

for any $0\le i<2^{n}p$.

So, it suffices to prove the inequalities

$$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n+1}p-i}}<\sigma ^i(\theta _1\ldots \theta _{2^n p}^-\theta _1\ldots \theta _{2^n p})<\theta _1\ldots \theta _{2^{n+1}p-i} \end{aligned}$$

(5)

for any $0\le i<2^{n}p$. By Lemma 2.5 it follows that for any $0\le i<2^n p$ we have

$$\begin{aligned} \overline{\theta _1\ldots \theta _{2^n p-i}}\le \theta _{i+1}\ldots \theta _{2^n p}^-<\theta _1\ldots \theta _{2^n p-i} \end{aligned}$$

and

$$\begin{aligned} \theta _1\ldots \theta _i>\overline{\theta _{2^n p-i+1}\ldots \theta _{2^n p}}. \end{aligned}$$

This proves (5). $\square$

Lemma 3.3

Let $(q_0, q_0^*)\subset (q',M+1){\setminus }\overline{{{\mathcal {U}}}}$ be a connected component generated by $t_1\ldots t_p$. Then $\dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_{q_0^*}')>0$ for any $r\in (q_0^*, M+1]$.

Proof

Take $r\in (q_0^*, M+1]$. By Lemma 3.1 there exists $n\ge 1$ such that

$$\begin{aligned} r_{n}\in \left( q_0^*,{ r}\right) \cap {{\mathcal {U}}}. \end{aligned}$$

Write $(\theta _i)=\alpha (q_0^*)$ and let $\xi _n=\theta _1\ldots \theta _{2^n p}$. Denote by $X_A^{(n)}$ the subshift of finite type over the states $\left\{ \xi _n, \xi _n^-, \overline{\xi _n}, \overline{\xi _n^-}\right\}$ with adjacency matrix

$$\begin{aligned} A=\left( \begin{array}{c@{\quad }c@{\quad }c@{\quad }c} 0&{}0 &{}1 &{}1 \\ 1&{}0&{}0 &{}0 \\ 1&{}1&{}0 &{}0\\ 0&{}0&{}1&{}0 \end{array} \right) . \end{aligned}$$

Note that $\alpha (r_n)=\theta _1\ldots \theta _{2^n p}(\theta _{2^n p+1}\ldots \theta _{2^{n+1}p})^\infty$. Then by Lemmas 3.2 and 2.3 it follows that

$$\begin{aligned} X_A^{(n)}\subseteq {{\mathcal {U}}}_{r_{n}}'\subseteq {{\mathcal {U}}}_r'. \end{aligned}$$

(6)

Furthermore, note that

$$\begin{aligned} \xi _n\overline{\xi _n^-}(\overline{\xi _n} \xi _n)^3&=\theta _1\ldots \theta _{2^{n+1}p}\left( \overline{\theta _1\ldots \theta _{2^{n+1}p}}\,^+\right) ^3\\&=\theta _1\ldots \theta _{2^{n+2}p}\left( \overline{\theta _1\ldots \theta _{2^{n+1}p}}\,^+\right) ^2\\&>\theta _1\ldots \theta _{2^{n+2}p}\overline{\theta _1\ldots \theta _{2^{n+1}p}\theta _{2^{n+1}p+1}\ldots \theta _{2^{n+2}p}}\,^+\\&=\theta _1\ldots \theta _{2^{n+2}p}\theta _{2^{n+2}p+1}\ldots \theta _{2^{n+3}p}. \end{aligned}$$

Then by Lemmas 2.3 and 3.1 it follows that any sequence starting at

$$\begin{aligned} {\mathbf {c}}:=\xi _n^- \xi _n\overline{\xi _n^-}(\overline{\xi _n} \xi _n)^3 \end{aligned}$$

can not belong to ${{\mathcal {U}}}_{r_{n+2}}'$. Therefore, by (6) we obtain

$$\begin{aligned} X_A^{(n)}({\mathbf {c}}) :=\left\{ (d_i)\in X_A^{(n)}: d_1\ldots d_{(2^{n+3}+2^n)p}={\mathbf {c}}\right\} \subseteq X_A^{(n)}{\setminus }{{\mathcal {U}}}_{r_{n+2}}' \subset {{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_{q_0^*}'. \end{aligned}$$

(7)

Note that the subshift of finite type $X_A^{(n)}$ is irreducible (cf. [22]), and the image $\pi _{M+1}(X_A^{(n)})$ is a graph-directed set satisfying the open set condition (cf. [23]). Then by (7) it follows that

$$\begin{aligned} \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_{q_0^*}'\right)&\ge \dim _H\pi _{M+1}(X_A^{(n)}({\mathbf {c}}))\\&=\dim _H\pi _{M+1}(X_A^{(n)}) =\frac{\log \left( (1+\sqrt{5})/2\right) }{2^n p\log (M+1)}>0. \end{aligned}$$

$\square$

The following lemma can be shown in a way which resembles closely the analysis in [21, pp. 2829–2830]. For the sake of completeness we include a sketch of its proof.

Lemma 3.4

Let $(q_0, q_0^*)\subset (q',M+1){\setminus }\overline{{{\mathcal {U}}}}$ be a connected component. Then $\dim _H\pi _{M+1}({{\mathcal {U}}}_{q_0^*}'{\setminus }{{\mathcal {U}}}_{q_0}')=0$.

Proof

(Sketch of the proof) Suppose that $(q_0, q_0^*)$ is a connected component generated by ${{\mathbf {t}}}=t_1\ldots t_p$. Then

$$\begin{aligned} \alpha (q_0)={{\mathbf {t}}}^\infty \quad \text {and}\quad \alpha \left( q_0^*\right) =\lim _{n\rightarrow \infty }g^n({{\mathbf {t}}})={{\mathbf {t}}}^+\,\overline{{{\mathbf {t}}}}\;\overline{{{\mathbf {t}}}^+}\,{{\mathbf {t}}}^+ \ldots , \end{aligned}$$

(8)

where $g(\cdot )$ is defined in (2).

For $n\ge 0$ let ${\omega }_n:=g^n({{\mathbf {t}}})^+$. Take $(d_i)\in {{\mathcal {U}}}_{q_0^*}'{\setminus }{{\mathcal {U}}}_{q_0}'$. Then by using (8) and Lemma 2.3 it follows that there exists $m\ge 1$ such that

$$\begin{aligned} {{\mathbf {t}}}^\infty =\alpha (q_0)\le d_{m+1}d_{m+2}\ldots <\alpha \left( q_0^*\right) ={{\mathbf {t}}}^+\overline{{{\mathbf {t}}}} \ldots , \end{aligned}$$

(9)

or symmetrically,

$$\begin{aligned} {{\mathbf {t}}}^\infty =\alpha (q_0)\le \overline{d_{m+1}d_{m+2}\ldots }<\alpha \left( q_0^*\right) ={{\mathbf {t}}}^+\overline{{{\mathbf {t}}}} \ldots . \end{aligned}$$

(10)

Suppose $(d_{m+i})\ne {{\mathbf {t}}}^\infty$ and $(d_{m+i})\ne \overline{{{\mathbf {t}}}^\infty }$. Then there exists $u\ge m$ such that

$$\begin{aligned} d_{u+1}\ldots d_{u+p}={{\mathbf {t}}}^+={\omega }_0\quad \text {or}\quad d_{u+1}\ldots d_{u+p}=\overline{{{\mathbf {t}}}^+}=\overline{{\omega }_0}. \end{aligned}$$

• If $d_{u+1}\ldots d_{u+p}={\omega }_0={{\mathbf {t}}}^+$, then by (9) and Lemma 2.3 it follows that

  $$\begin{aligned} d_{u+p+1}\ldots d_{u+2p}=\overline{{{\mathbf {t}}}^+} \quad \text {or}\quad d_{u+p+1}\ldots d_{u+2p}=\overline{{{\mathbf {t}}}}. \end{aligned}$$

  This implies $d_{u+1}\ldots d_{u+2p}={{\mathbf {t}}}^+\overline{{{\mathbf {t}}}^+} ={\omega }_0\,\overline{{\omega }_0}$ or $d_{u+1}\ldots d_{u+2p}={{\mathbf {t}}}^+\overline{{{\mathbf {t}}}}={\omega }_1.$

• If $d_{u+1}\ldots d_{u+p}=\overline{{\omega }_0}=\overline{{{\mathbf {t}}}^+}$, then by (10) and Lemma 2.3 it follows that

  $$\begin{aligned} d_{u+p+1}\ldots d_{u+2p}={{{\mathbf {t}}}^+} \quad \text {or}\quad d_{u+p+1}\ldots d_{u+2p}={{\mathbf {t}}}. \end{aligned}$$

  This yields that $d_{u+1}\ldots d_{u+2p}=\overline{{\omega }_0}\,{\omega }_0$ or $d_{u+1}\ldots d_{u+2p}=\overline{{\omega }_1}$.

Note that for each $n\ge 0$ the word $g^n({{\mathbf {t}}})^+\,\overline{g^n({{\mathbf {t}}})}$ is a prefix of $\alpha (q_0^*)$. By iteration of the above arguments, one can show that if $d_{v+1}\ldots d_{v+2^n p}={\omega }_n$, then $d_{v+1}\ldots d_{v+2^{n+1}p}={\omega }_n\overline{{\omega }_n}$ or ${\omega }_{n+1}$. Symmetrically, if $d_{v+1}\ldots d_{v+2^n p}=\overline{{\omega }_n}$, then $d_{v+1}\ldots d_{v+2^{n+1}p}=\overline{{\omega }_n}{\omega }_n$ or $\overline{{\omega }_{n+1}}$.

Hence, we conclude that $(d_{i})$ must end with

$$\begin{aligned} {{\mathbf {t}}}^* ({\omega }_{i_0}\overline{{\omega }_{i_0}})^{*}({\omega }_{i_0}\overline{{\omega }_{j_0}})^{s_0}({\omega }_{i_1}\overline{{\omega }_{i_1}})^{*} ({\omega }_{i_1}\overline{{\omega }_{j_1}})^{s_1}\ldots ({\omega }_{i_n}\overline{{\omega }_{i_n}})^{*}({\omega }_{i_n}\overline{{\omega }_{j_n}})^{s_n}\ldots \end{aligned}$$

or its reflections, where $s_n\in \left\{ 0,1\right\}$ and

$$\begin{aligned} 0=i_0<j_0\le i_1<j_1\le i_2<\cdots \le i_n<j_n\le i_{n+1}<\cdots . \end{aligned}$$

Here $*$ is an element of the set $\left\{ 0,1,2,\ldots \right\} \cup \left\{ \infty \right\}$.

Since the length of ${\omega }_n=g^n({{\mathbf {t}}})^+$ grows exponentially fast as $n\rightarrow \infty$, we conclude that $\dim _H\pi _{M+1}({{\mathcal {U}}}_{q_0^*}'{\setminus }{{\mathcal {U}}}_{q_0}')=0$. $\square$

Proof of Theorem 1.1 for (i) $\Leftrightarrow$ (ii) First we prove (i) $\Rightarrow$ (ii). If $q=q_0^*$ is the right endpoint of a connected component of $(q', M+1){\setminus }\overline{{{\mathcal {U}}}}$, then by Lemma 3.3 we have

$$\begin{aligned} \dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q')>0\quad \text {for any }\quad r\in (q, M+1]. \end{aligned}$$

Clearly, it is trivial when $q=M+1$. Now we take $q\in ({{\mathcal {U}}}{\setminus }\left\{ M+1\right\} ){\setminus }\bigcup \left\{ q_0^*\right\}$ and take $r\in (q, M+1]$. By Lemma 2.6 (i) one can find $q_0^*\in (q,r)$, and therefore by Lemma 3.3 we obtain

$$\begin{aligned} \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q'\right) \ge \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_{q_0^*}'\right) >0. \end{aligned}$$

Now we prove (ii) $\Rightarrow$ (i). Take $q\in (1, M+1]{\setminus }{{\mathcal {U}}}$. We will show that $\dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q')=0$ for some $r\in (q, M+1]$. Note that $\bigcup \left\{ q_0\right\} =\overline{{{\mathcal {U}}}}{\setminus }{{\mathcal {U}}}$. Then by (1) it follows that

$$\begin{aligned} q\in (1,q')\cup \bigcup \left[ q_0, q_0^*\right) . \end{aligned}$$

Therefore, it suffices to prove $\dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q')=0$ for some $r\in (q, M+1]$. We distinct the following two cases.

Case (I). $q\in (1,q')$. Then for any $r\in (q,q')$ we have

$$\begin{aligned} \dim _H\pi _{M+1}({{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q')\le \dim _H\pi _{M+1}({{\mathcal {U}}}_r')=0, \end{aligned}$$

where the last equality follows by [21, Theorem 4.6] (see also [4, 14]).

Case (II). $q\in [q_0, q_0^*)$. Then for any $r\in (q,q_0^*)$ we have by Lemma 3.4 that

$$\begin{aligned} \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_r'{\setminus }{{\mathcal {U}}}_q'\right) \le \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_{q_0^*}'{\setminus }{{\mathcal {U}}}_{q_0}'\right) =0. \end{aligned}$$

$\square$

3.2 Proof of Theorem 1.1 for (i) $\Leftrightarrow$ (iii)

The following property for the Hausdorff dimension is well-known (cf. [13, Proposition 2.3]).

Lemma 3.5

Let $f: (X, d_1)\rightarrow (Y, d_2)$ be a map between two metric spaces . If there exist constants $C>0$ and $\lambda >0$ such that

$$\begin{aligned} d_2(f(x), f(y))\le C d_1(x, y)^\lambda \end{aligned}$$

for any $x, y\in X$, then $\dim _H X\ge \lambda \dim _H f(X)$.

Lemma 3.6

Let $q\in {{{\mathcal {U}}}}{\setminus }\left\{ M+1\right\}$. Then for any $r\in (q, M+1)$ we have

$$\begin{aligned} \dim _H{{\mathcal {U}}}\cap (q, r)\ge \dim _H\pi _{M+1}\left( \left\{ \alpha (p): p\in {{\mathcal {U}}}\cap (q, r)\right\} \right) . \end{aligned}$$

Proof

Fix $q\in {{{\mathcal {U}}}}{\setminus }\left\{ M+1\right\}$ and $r\in (q, M+1)$. Then Lemma 2.6 yields that ${{\mathcal {U}}}\cap (q, r)$ contains infinitely many elements. Take $p_1, p_2\in {{\mathcal {U}}}\cap (q, r)$ with $p_1<p_2$. Then by Lemma 2.1 we have $\alpha (p_1)<\alpha (p_2)$. So, there exists $n\ge 1$ such that

$$\begin{aligned} \alpha _1(p_1)\ldots \alpha _{n-1}(p_1)=\alpha _1(p_2)\ldots \alpha _{n-1}(p_2)\quad \text {and}\quad \alpha _n(p_1)<\alpha _n(p_2). \end{aligned}$$

(11)

This implies

$$\begin{aligned} \pi _{M+1}(\alpha (p_2))-\pi _{M+1}(\alpha (p_1))= & {} \sum _{i=1}^\infty \frac{\alpha _i(p_2)-\alpha _i(p_1)}{(M+1)^i}\nonumber \\\le & {} \sum _{i=n}^\infty \frac{M}{(M+1)^i}=(M+1)^{1-n}. \end{aligned}$$

(12)

Note that $r<M+1$. By Lemma 2.1 we have $\alpha (r)<\alpha (M+1)=M^\infty$. Then there exists $N\ge 1$ such that

$$\begin{aligned} \alpha _1(r)\ldots \alpha _N(r)<\underbrace{M\ldots M}_N. \end{aligned}$$

Therefore, by (11) and Lemma 2.3 we obtain

$$\begin{aligned} \sum _{i=1}^n\frac{\alpha _i(p_2)}{p_1^i}\ge \sum _{i=1}^\infty \frac{\alpha _i(p_1)}{p_1^i}=1=\sum _{i=1}^\infty \frac{\alpha _i(p_2)}{p_2^i}>\sum _{i=1}^n\frac{\alpha _i(p_2)}{p_2^i}+\frac{1}{p_2^{n+N}}. \end{aligned}$$

Note that $p_1, p_2$ are elements of ${{\mathcal {U}}}$. Then $p_2>p_1\ge q'$. This implies

$$\begin{aligned} \frac{1}{(M+1)^{n+N}}&<\frac{1}{p_2^{n+N}}<\sum _{i=1}^n\left( \frac{\alpha _i(p_2)}{p_1^i}-\frac{\alpha _i(p_2)}{p_2^i}\right) \\&\le \sum _{i=1}^\infty \left( \frac{M}{p_1^i}-\frac{M}{p_2^i}\right) =\frac{M(p_2-p_1)}{(p_1-1)(p_2-1)}\le \frac{M(p_2-p_1)}{(q'-1)^2}. \end{aligned}$$

Therefore, by (12) it follows that

$$\begin{aligned} \pi _{M+1}(\alpha (p_2))-\pi _{M+1}(\alpha (p_1))\le (M+1)^{1-n}\le \frac{(M+1)^{2+N}}{(q'-1)^2}(p_2-p_1). \end{aligned}$$

Furthermore, by Lemma 2.1 it follows that $\pi _{M+1}(\alpha (p_2))-\pi _{M+1}(\alpha (p_1))\ge 0$. Hence, by using

$$\begin{aligned} f=\pi _{M+1}\circ \alpha : {{\mathcal {U}}}\cap (q, r)\rightarrow \pi _{M+1}(\left\{ \alpha (p): p\in {{\mathcal {U}}}\cap (q, r)\right\} ) \end{aligned}$$

in Lemma 3.5 we establish the lemma. $\square$

Lemma 3.7

Let $(q_0, q_0^*)$ be a connected component of $(q', M+1){\setminus }\overline{{{\mathcal {U}}}}$. Then $\dim _H{{\mathcal {U}}}\cap (q_0^*, r)>0$ for any $r\in (q_0^*, M+1]$.

Proof

Suppose that $(q_0, q_0^*)$ is a connected component generated by $t_1\ldots t_p$. Let $(\theta _i)=\alpha (q_0^*)$. For $n\ge 2$ we write $\xi _n=\theta _1\ldots \theta _{2^n p}$, and denote by

$$\begin{aligned} \varGamma _n':= \left\{ (d_i): d_1\ldots d_{2^{n+1}p}=\xi _{n-1}\left( \overline{\xi _{n-1}}\,^+\right) ^3, \quad (d_{2^{n+1}p+i}) \in X_A^{(n)}(\overline{\xi _n}) \right\} . \end{aligned}$$

Here $X_A^{(n)}(\overline{\xi _n})$ is the follower set of $\overline{\xi _n}$ in the subshift of finite type $X_A^{(n)}$ defined in (7). Now we claim that any sequence $(d_i)\in \varGamma _n'$ satisfies

$$\begin{aligned} \overline{(d_i)}<\sigma ^j((d_i))<(d_i)\quad \text {for all} \quad j\ge 1. \end{aligned}$$

(13)

Take $(d_i)\in \varGamma _n'$. Then we deduce by the definition of $\varGamma _n'$ that

$$\begin{aligned} d_1\ldots d_{2^{n+1}p+2^{n-1} p}=\theta _1\ldots \theta _{2^{n-1}p}\left( \overline{\theta _1\ldots \theta _{2^{n-1}p}}\,^+\right) ^3\, \overline{\theta _1\ldots \theta _{2^{n}p}}. \end{aligned}$$

(14)

We will split the proof of (13) into the following five cases.

• (a) $1\le j<2^{n-1}p$. By (14) and Lemma 2.5 it follows that

  $$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n-1}p-j}}<d_{j+1}\ldots d_{2^{n-1}p}=\theta _{j+1}\ldots \theta _{2^{n-1}p}\le \theta _1\ldots \theta _{2^{n-1}p-j}, \end{aligned}$$

  and

  $$\begin{aligned} d_{2^{n-1}p+1}\ldots d_{2^{n-1}p+j}=\overline{\theta _1\ldots \theta _j}<\theta _{2^{n-1}p-j+1}\ldots \theta _{2^{n-1}p}. \end{aligned}$$

  This implies that (13) holds for all $1\le j<2^{n-1}p$.

• (b) $2^{n-1}p\le j<2^n p$. Let $k=j-2^{n-1}p$. Then $0\le k<2^{n-1}p$. Clearly, if $k=0$, then by using $\theta _1>\overline{\theta _1}$ and $n\ge 2$ it yields that

  $$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n-1}p}}<d_{j+1}\ldots d_{2^n p}=\overline{\theta _1\ldots \theta _{2^{n-1}p}}\,^+<\theta _1\ldots \theta _{2^{n-1}p}. \end{aligned}$$

  Now we assume $1\le k<2^{n-1}p$. Then by (14) and Lemma 2.5 it follows that

  $$\begin{aligned} \overline{\theta _1\ldots \theta _{2^{n-1}p-k}}<d_{j+1}\ldots d_{2^n p}=\overline{\theta _{k+1}\ldots \theta _{2^{n-1}p}}\,^+\le \theta _1\ldots \theta _{2^{n-1}p-k}, \end{aligned}$$

  and

  $$\begin{aligned} d_{2^{n}p+1}\ldots d_{2^n p+k}=\overline{\theta _1\ldots \theta _k}<\theta _{2^{n-1}p-k+1}\ldots \theta _{2^{n-1}p}. \end{aligned}$$

  Therefore, (13) holds for all $2^{n-1}p\le j<2^n p$.

• (c) $2^n p\le j<2^n p+2^{n-1}p$. Let $k=j-2^n p$. Then in a similar way as in Case (b) one can prove (13).

• (d) $2^np+2^{n-1}p\le j<2^{n+1}p$. Let $k=j-2^np-2^{n-1}p$. Again by the same arguments as in Case (b) we obtain (13).

• (e) $j\ge 2^{n+1}p$. Note that

  $$\begin{aligned} d_1\ldots d_{2^{n+1}p}=\theta _1\ldots \theta _{2^{n-1}p}\left( \overline{\theta _1\ldots \theta _{2^{n-1}p}}\,^+\right) ^3>\theta _1\ldots \theta _{2^{n+1}p}. \end{aligned}$$

  Then (13) follows by Lemma 3.2.

Therefore, by (13) and Lemma 2.4 it follows that any sequence in $\varGamma _n'$ corresponds to a unique base $q\in {{\mathcal {U}}}$. Furthermore, by (14) and Lemma 3.1 each sequence $(d_i)\in \varGamma _n'$ satisfies

$$\begin{aligned} \alpha \left( q_0^*\right) =(\theta _i)<(d_i)<\theta _1\ldots \theta _{2^{n-1}p}\left( \overline{\theta _1\ldots \theta _{2^{n-1}p}}\,^+\right) ^\infty =\alpha (r_{n-1}). \end{aligned}$$

Then by Lemma  2.1 it follows that

$$\begin{aligned} \alpha (q)\in \varGamma _n'\Longrightarrow q\in {{\mathcal {U}}}\cap \left( q_0^*, r_{n-1}\right) . \end{aligned}$$

Fix $r>q_0^*$. So by Lemma 3.1 there exists a sufficiently large integer $n\ge 2$ such that

$$\begin{aligned} \varGamma _n'\subset \left\{ \alpha (q): q\in {{\mathcal {U}}}\cap \left( q_0^*, r\right) \right\} . \end{aligned}$$

(15)

Note by the proof of Lemma 3.3 that $X_A^{(n)}$ is an irreducible subshift of finite type over the states $\left\{ \xi _n, \xi _n^-, \overline{\xi _n}, \overline{\xi _n^-}\right\}$. Hence, by (15) and Lemma 3.6 it follows that

$$\begin{aligned} \dim _H{{\mathcal {U}}}\cap \left( q_0^*, r\right) \ge \dim _H\pi _{M+1}(\varGamma _n')&=\dim _H\pi _{M+1}(X_A^{(n)})\\&=\frac{\log \left( (1+\sqrt{5})/2\right) }{2^n p\log (M+1)} \;>0. \end{aligned}$$

$\square$

Proof of Theorem  1.1 for (i) $\Leftrightarrow$ (iii) First we prove (i) $\Rightarrow$ (iii). Excluding the trivial case $q=M+1$ we take $q\in {{\mathcal {U}}}{\setminus }\left\{ M+1\right\}$. Suppose that $r \in (q,M+1]$. If $q=q_0^*$, then by Lemma 3.7 we have $\dim _H{{\mathcal {U}}}\cap (q, r)>0.$

If $q\in ({{\mathcal {U}}}{\setminus }\left\{ M+1\right\} ) {\setminus }\bigcup \left\{ q_0^*\right\}$, then by Lemma 2.6 (i) there exists $q_0^*\in (q, r)$. So, by Lemma 3.7 we have

$$\begin{aligned} \dim _H{{\mathcal {U}}}\cap (q,r)\ge \dim _H{{\mathcal {U}}}\cap \left( q_0^*, r\right) >0. \end{aligned}$$

Now we prove (iii) $\Rightarrow$ (i). Suppose on the contrary that $q\in (1,M+1]{\setminus }{{\mathcal {U}}}$. We will show that ${{\mathcal {U}}}\cap (q,r)=\emptyset$ for some $r\in (q, M+1]$. Take $q\in (1, M+1]{\setminus }{{\mathcal {U}}}$. By (1) it follows that

$$\begin{aligned} q\in (1, q')\cap \bigcup [q_0, q_0^*). \end{aligned}$$

This implies that ${{\mathcal {U}}}\cap (q,r)=\emptyset$ for $r \in (q, M+1]$ sufficiently close to q. $\square$

3.3 Proof of Theorem 1.2

Proof of Theorem 1.2 (i) $\Rightarrow$ (ii) Take $q\in \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$ and $p\in (1,q)$. By Lemma 2.6 (ii) there exists $q_0^*\in (p,q)$. Hence, by Lemma 3.3 it follows that

$$\begin{aligned} \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_q'{\setminus }{{\mathcal {U}}}_p'\right) \ge \dim _H\pi _{M+1}\left( {{\mathcal {U}}}_q'{\setminus }{{\mathcal {U}}}_{q_0^*}'\right) >0. \end{aligned}$$

(ii) $\Rightarrow$ (i). Suppose on the contrary that $q\notin \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$. Then by (1) we have

$$\begin{aligned} q\in (1,q']\cup \bigcup \left( q_0, q_0^*\right] . \end{aligned}$$

By using Lemma 3.4 it follows that for $p\in (1,q)$ sufficiently close to q we have $\dim _H\pi _{M+1}({{\mathcal {U}}}_q'{\setminus }{{\mathcal {U}}}_p')=0$.

(i) $\Rightarrow$ (iii). Take $q\in \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$ and $p\in (1,q)$. By Lemma 2.6 (ii) there exists $q_0^*\in (p,q)$. Hence, by Lemma 3.7 it follows that

$$\begin{aligned} \dim _H{{\mathcal {U}}}\cap (p,q)\ge \dim _H{{\mathcal {U}}}\cap \left( q_0^*, q\right) >0. \end{aligned}$$

(iii) $\Rightarrow$ (i). Suppose $q\notin \overline{{{\mathcal {U}}}}{\setminus }(\bigcup \left\{ q_0^*\right\} \cup \left\{ q'\right\} )$. Then by (1) we have $q\in (1,q']\cup \bigcup (q_0, q_0^*].$ So, for $p\in (1, q)$ sufficiently close to q we have ${{\mathcal {U}}}\cap (p,q)=\emptyset$. $\square$