# Entropy plateaus, transitivity and bifurcation sets for the
$\beta$-transformation with a hole at $0$

Pieter Allaart  Mathematics Department, University of North Texas, 1155 Union
Cir #311430, Denton, TX 76203-5017, U.S.A<EMAIL_ADDRESS>and  Derong Kong
College of Mathematics and Statistics, Mathematical Research Center, Chongqing
University, Chongqing 401331, People’s Republic of China<EMAIL_ADDRESS>
###### Abstract.

Given $\beta\in(1,2]$, let $T_{\beta}$ be the $\beta$-transformation on the
unit circle $[0,1)$ such that $T_{\beta}(x)=\beta x\pmod{1}$. For each
$t\in[0,1)$ let $K_{\beta}(t)$ be the survivor set consisting of all
$x\in[0,1)$ whose orbit $\\{T^{n}_{\beta}(x):n\geq 0\\}$ never enters the
interval $[0,t)$. Letting $\mathscr{E}_{\beta}$ denote the bifurcation set of
the set-valued map $t\mapsto K_{\beta}(t)$, Kalle et al. [Ergodic Theory
Dynam. Systems, 40 (9): 2482–2514, 2020] conjectured that

$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}=\dim_{H}K_{\beta}(t)\qquad\forall\,t\in(0,1).$

The main purpose of this article is to prove this conjecture. We do so by
investigating dynamical properties of the symbolic equivalent of the survivor
set $K_{\beta}(t)$, in particular its entropy and topological transitivity. In
addition, we compare $\mathscr{E}_{\beta}$ with the bifurcation set
$\mathscr{B}_{\beta}$ of the map $t\mapsto\dim_{H}K_{\beta}(t)$ (which is a
decreasing devil’s staircase by a theorem of Kalle et al.), and show that, for
Lebesgue-almost every $\beta\in(1,2]$, the difference
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ has positive Hausdorff
dimension, but for every $k\in\\{0,1,2,\dots\\}\cup\\{\aleph_{0}\\}$, there
are infinitely many values of $\beta$ such that the cardinality of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ is exactly $k$. For a
countable but dense subset of $\beta$’s, we also determine the intervals of
constancy of the function $t\mapsto\dim_{H}K_{\beta}(t)$.

Some connections with other topics in dynamics, such as kneading invariants of
Lorenz maps and the doubling map with an arbitrary hole, are also discussed.

###### Key words and phrases:

$\beta$-transformation, survivor set, bifurcation set, Farey word, Lyndon
word, greedy expansion, renormalization, Hausdorff dimension, transitivity,
topological entropy, Lorenz map

###### 2020 Mathematics Subject Classification:

Primary: 37B10, 28A78; Secondary: 68R15, 26A30, 37E05, 37B40

###### Contents

  1. 1 Introduction
    1. 1.1 Connection with kneading sequences of Lorenz maps
    2. 1.2 Connection with the doubling map with a general hole
    3. 1.3 Organization of the paper
    4. 1.4 List of notation
  2. 2 Preliminaries
    1. 2.1 Farey words, Farey intervals and the set $E$
    2. 2.2 The substitution operator
  3. 3 Characterization of the bifurcation set $\mathscr{E}_{\beta}$
  4. 4 The key ideas
  5. 5 The case $\beta\in E_{L}$
    1. 5.1 Density of $\beta$-Lyndon intervals
  6. 6 Right endpoints of first-order basic intervals
  7. 7 The case of finitely renormalizable $\beta$
    1. 7.1 Density of $\beta$-Lyndon intervals
  8. 8 Comparison of the bifurcation sets
  9. 9 The case of infinitely renormalizable $\beta$
  10. 10 Interiors of basic intervals: Construction of non-transitivity windows
  11. 11 Proof of Theorem 10.3
  12. 12 Higher order basic intervals
  13. 13 Gaps between $\beta$-Lyndon intervals
  14. 14 Properties of extended $\beta$-Lyndon intervals
  15. 15 Proofs of the other main theorems

##  1\. Introduction

Let $1<\beta\leq 2$ and define the map $T_{\beta}:[0,1)\to[0,1)$ by
$T_{\beta}(x):=\beta x\\!\\!\mod 1$. Given a number $t\in(0,1)$, let

$K_{\beta}(t):=\\{x\in[0,1):T_{\beta}^{n}(x)\geq t\ \forall n\geq 0\\}.$

Up to countably many points, $K_{\beta}(t)$ is the survivor set of the open
dynamical system $(T_{\beta},O)$ with the hole $O=(0,t)$. Such open dynamical
systems were considered first by Urbański [33, 34], who proved, among many
other things, that the map $\eta_{2}:t\mapsto\dim_{H}K_{2}(t)$ is a decreasing
devil’s staircase. This result was extended more recently by Kalle et al. [22]
to all $\beta\in(1,2]$.

Other authors have studied the $\beta$-transformation with a hole. For
instance, Glendinning and Sidorov [19] considered the doubing map $T_{2}$ with
an arbitrary hole $(a,b)$ and proved several results about the size of the
survivor set. Their work was partially extended by Clark [12], who
characterized for the general $\beta$-transformation with $\beta\in(1,2]$ the
holes $(a,b)$ with an uncountable survivor set. Carminati and Tiozzo [10]
showed that the local Hölder exponent of $\eta_{2}$ at any bifurcation point
$t$ is precisely $\eta_{2}(t)$. In a recent paper [4], the present authors
determined for each $\beta\in(1,2]$ the critical value

(1.1) $\tau(\beta):=\min\\{t>0:\dim_{H}K_{\beta}(t)=0\\}$

and showed that the functon $\beta\mapsto\tau(\beta)$ has an infinite set of
discontinuities.

Note that the set-valued map $t\mapsto K_{\beta}(t)$ is non-increasing. We
define the bifurcation set

(1.2) $\mathscr{E}_{\beta}:=\\{t\in[0,1):K_{\beta}(t^{\prime})\neq
K_{\beta}(t)\ \forall\,t^{\prime}>t\\}.$

It was shown by Urbański [34] that

$\dim_{H}\big{(}\mathscr{E}_{2}\cap[t,1]\big{)}=\dim_{H}K_{2}(t)\qquad\forall\,t\in(0,1).$

Kalle et al. [22] investigated the topological structure of
$\mathscr{E}_{\beta}$, and conjectured that a similar identity should hold for
all $\beta\in(1,2]$, namely,

(1.3)
$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}=\dim_{H}K_{\beta}(t)\qquad\forall\,t\in(0,1).$

Baker and Kong [7] proved the conjecture for the special case when $\beta$ is
a multinacci number, i.e. the positive root of
$x^{n+1}=x^{n}+x^{n-1}+\dots+x+1$, where $n\in\mathbb{N}$. The main purpose of
this paper is to prove the conjecture for all $\beta\in(1,2]$.

######  Theorem 1.1.

The equation (1.3) holds for all $\beta\in(1,2]$.

The identity (1.3) is an instance of the interplay between the “parameter
space” (in this case, $\mathscr{E}_{\beta}$) and the “dynamical space” (in our
case $K_{\beta}(t)$) which was first observed by Douady [15] in the context of
dynamics of real quadratic polynomials. Theorems of this type have frequently
occurred in the literature in a variety of settings. For instance, Tiozzo [32]
considers for $c\in\mathbb{R}$ the set of angles of external rays which “land”
on the real slice of the Mandelbrot set to the right of $c$ (parameter space)
and the set of external angles which land on the real slice of the Julia set
of the map $z\mapsto z^{2}+c$ (dynamical space), showing that these two sets
have the same Hausdorff dimension. More recently, Carminati and Tiozzo [11]
proved an identity analogous to (1.3) in the setting of continued fractions,
and the present authors proved a similar result in the context of densities of
Cantor measure (see [5]).

Since the map $\eta_{\beta}:t\mapsto\dim_{H}K_{\beta}(t)$ is a decreasing
devil’s staircase, it is natural to consider also the dimension bifurcation
set

$\mathscr{B}_{\beta}:=\left\\{t\geq
0:\dim_{H}K_{\beta}(t^{\prime})<\dim_{H}K_{\beta}(t)~{}\forall
t^{\prime}>t\right\\},\qquad\beta\in(1,2].$

Clearly $\mathscr{B}_{\beta}\subseteq\mathscr{E}_{\beta}$. As noted in [22],
the inclusion is proper for most values of $\beta$, since
$\mathscr{E}_{\beta}$ typically has isolated points whereas
$\mathscr{B}_{\beta}$ does not. However, Baker and Kong [7] showed that
$\mathscr{B}_{\beta}=\mathscr{E}_{\beta}$ when $\beta$ is a multinacci number.
We extend this result here as follows:

######  Theorem 1.2.

The difference $\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ has positive
Hausdorff dimension for Lebesgue-almost every $\beta\in(1,2]$. On the other
hand, for each $k\in\\{0,1,2,\dots\\}\cup\\{\aleph_{0}\\}$, there are
infinitely many values of $\beta$ such that
$|\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}|=k$. However, there is no
$\beta\in(1,2]$ such that $\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$
is uncountable but of zero Hausdorff dimension.

Regarding the local dimension of $\mathscr{E}_{\beta}$ and
$\mathscr{B}_{\beta}$ we can prove the following.

######  Theorem 1.3.

Let $\beta\in(1,2]$.

  1. (i)

For each $t\in\mathscr{B}_{\beta}$, we have

$\lim_{\varepsilon\searrow
0}\dim_{H}\big{(}\mathscr{E}_{\beta}\cap(t-\varepsilon,t+\varepsilon)\big{)}=\lim_{\varepsilon\searrow
0}\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,t+\varepsilon)\big{)}=\dim_{H}K_{\beta}(t).$

  2. (ii)

If $\alpha(\beta)$ is eventually periodic, then for each
$t\in\mathscr{B}_{\beta}$, we also have

$\lim_{\varepsilon\searrow
0}\dim_{H}\big{(}\mathscr{B}_{\beta}\cap(t-\varepsilon,t+\varepsilon)\big{)}=\lim_{\varepsilon\searrow
0}\dim_{H}\big{(}\mathscr{B}_{\beta}\cap[t,t+\varepsilon)\big{)}=\dim_{H}K_{\beta}(t).$

Statement (i) is an easy consequence of Theorem 1.1, but (ii) will follow only
after we prove several other results. Therefore, we defer the proof until the
end of the paper. At present, we do not know how to compute the local
dimension of $\mathscr{E}_{\beta}$ at points of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$. Note that, since
$\mathscr{E}_{\beta}$ and $\mathscr{B}_{\beta}$ are defined as right
bifurcation sets, it is much less clear how to compute their left local
dimension functions.

We prove the main theorems above using tools from symbolic dynamics. A
critical role is played here by the greedy expansion $b(t,\beta)$, defined as
the lexicographically largest expansion of the number $t$ in base $\beta$, and
by the quasi-greedy expansion $\alpha(\beta)$ of 1 in base $\beta$, defined as
the lexicographically largest $\beta$-expansion of 1 not ending in
$0^{\infty}$.

We first define the symbolic equivalent of $K_{\beta}(t)$, that is,

$\mathcal{K}_{\beta}(t):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\prec\alpha(\beta)\
\forall n\geq 0\\},$

where $\sigma$ denotes the left shift map on $\\{0,1\\}^{\mathbb{N}}$. Note
that $K_{\beta}(t)=\pi_{\beta}(\mathcal{K}_{\beta}(t))$, where
$\pi_{\beta}:\\{0,1\\}^{\mathbb{N}}\to\mathbb{R}$ is the projection map given
by

$\pi_{\beta}((d_{i})):=\sum_{i=1}^{\infty}\frac{d_{i}}{\beta^{i}}.$

It is convenient to slightly enlarge the set $\mathcal{K}_{\beta}(t)$ and
consider the closed set

$\widetilde{\mathcal{K}}_{\beta}(t):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\
\forall n\geq 0\\}.$

Clearly $\mathcal{K}_{\beta}(t)\subseteq\widetilde{\mathcal{K}}_{\beta}(t)$,
and the difference
$\widetilde{\mathcal{K}}_{\beta}(t)\backslash\mathcal{K}_{\beta}(t)$ is at
most countable. The point is that $\widetilde{\mathcal{K}}_{\beta}(t)$ is
always a subshift, hence we can apply theorems from symbolic dynamics. It was
shown in [22] that

$\dim_{H}K_{\beta}(t)=\frac{h(\widetilde{\mathcal{K}}_{\beta}(t))}{\log\beta}\qquad\forall\beta\in(1,2],\quad\forall
t\in(0,1),$

where for a subshift $\mathcal{X}$ of $\\{0,1\\}^{\mathbb{N}}$,
$h(\mathcal{X})$ denotes the (topological) entropy of $\mathcal{X}$, that is,

$h(\mathcal{X}):=\lim_{n\to\infty}{\frac{\log\\#\mathcal{B}_{n}(\mathcal{X})}{n}}.$

Here $\mathcal{B}_{n}(\mathcal{X})$ is the set of all words of length $n$
occurring in some sequence in $\mathcal{X}$, and $\\#B$ denotes the number of
elements of the finite set $B$.

The reader may observe that $\widetilde{\mathcal{K}}_{\beta}(t)$ is a special
case of the subshifts

(1.4)
$\Sigma_{\mathbf{a},{\mathbf{b}}}:=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\mathbf{a}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq{\mathbf{b}}\
\forall\,n\geq 0\\},$

where $\mathbf{a}$ and ${\mathbf{b}}$ are sequences in
$\\{0,1\\}^{\mathbb{N}}$. These subshifts have many important applications in
dynamical systems and number theory. They arose in the 1990s in connection
with kneading sequences of Lorenz maps [17, 20, 21]; see the next subsection.
The symmetric case $\Sigma_{\mathbf{a},\tilde{\mathbf{a}}}$, where
$\tilde{\mathbf{a}}$ is the reflection of $\mathbf{a}$ obtained by
interchanging zeros and ones, plays a prominent role in the study of unique
expansions in non-integer bases (see [1, 2, 3, 13, 14, 16, 18, 23]). The
general subshifts $\Sigma_{\mathbf{a},{\mathbf{b}}}$ were studied in detail by
Labarca and Moreira [25] and more recently by Komornik, Steiner and Zou [24].
Several of our results here have new implications for these subshifts.

The complement of the bifurcation set $\mathscr{B}_{\beta}$ consists of
intervals on which $h(\widetilde{\mathcal{K}}_{\beta}(t))$ (and hence
$\dim_{H}K_{\beta}(t)$) is constant. For $\beta=2$ these intervals, called
entropy plateaus, were characterized by Nilsson [29]. Say a finite $0-1$ word
$\mathbf{w}$ is Lyndon if it is aperiodic and lexicographically smallest among
all its cyclic permutations. Nilsson proved that the plateaus of
$\eta_{2}:t\mapsto h(\widetilde{\mathcal{K}}_{2}(t))$ are precisely the
intervals $[\pi_{2}(\mathbf{w}0^{\infty}),\pi_{2}(\mathbf{w}^{\infty})]$, as
$\mathbf{w}$ ranges over all Lyndon words. It is easy to see that these
intervals are pairwise disjoint.

For $1<\beta<2$ the situation is more complex. First of all the sequences
$\mathbf{w}0^{\infty}$ and $\mathbf{w}^{\infty}$ may not be valid greedy
$\beta$-expansions. We say a Lyndon word $\mathbf{w}$ is $\beta$-Lyndon if
$\mathbf{w}^{\infty}=b(t,\beta)$ for some $t\in[0,1)$. By Lemma 2.2 below this
is the case if and only if $\sigma^{n}(\mathbf{w}^{\infty})\prec\alpha(\beta)$
for all $n\geq 0$. If $\mathbf{w}$ is a $\beta$-Lyndon word, we call the
interval
$[\pi_{\beta}(\mathbf{w}0^{\infty}),\pi_{\beta}(\mathbf{w}^{\infty})]$ a
$\beta$-Lyndon interval; see [7, Definition 1.2]. Furthermore, for each
$\beta$-Lyndon word $\mathbf{w}$ we define the extended $\beta$-Lyndon
interval (EBLI for short) by

$I_{\mathbf{w}}:=\begin{cases}[\pi_{\beta}(\mathbf{w}0^{\infty}),\pi_{\beta}(\mathbf{w}^{\infty})]&\mbox{if
$\sigma^{n}(\alpha(\beta))\succ\mathbf{w}^{\infty}$ for all $n\geq 0$},\\\
[\pi_{\beta}\big{(}\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\big{)},\pi_{\beta}(\mathbf{w}^{\infty})]&\mbox{otherwise},\end{cases}$

where in the second case, $m:=\min\\{n\geq
1:\sigma^{n}(\alpha(\beta))\preccurlyeq\mathbf{w}^{\infty}\\}$, and for any
word $\mathbf{w}=w_{1}\dots w_{n}$ ending in 1, $\mathbf{w}^{-}:=w_{1}\dots
w_{n-1}0$. In general, the EBLIs need not be disjoint. However, we will show
that if two EBLIs intersect then one contains the other. Our last main result
is the following generalization of Nilsson’s theorem:

######  Theorem 1.4.

Assume $\alpha(\beta)$ is eventually periodic. Then the plateaus of
$\eta_{\beta}:t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$ (hence of
$t\mapsto\dim_{H}K_{\beta}(t)$) are precisely $[\tau(\beta),1)$ and the
maximal (with respect to set inclusion) EBLIs in $[0,\tau(\beta)]$.

We will prove this theorem in Section 14, where we also indicate how to
determine whether a given EBLI is maximal.

Since every subshift $\Sigma_{\mathbf{a},{\mathbf{b}}}$ is essentially (up to
a countable set) of the form $\widetilde{\mathcal{K}}_{\beta}(t)$ for some
parameter pair $(\beta,t)$ (see Subsection 1.1 below), Theorem 1.4 also gives
the entropy plateaus for the map $\mathbf{a}\mapsto
h(\Sigma_{\mathbf{a},{\mathbf{b}}})$ for fixed ${\mathbf{b}}$, or, by
symmetry, for the map ${\mathbf{b}}\mapsto
h(\Sigma_{\mathbf{a},{\mathbf{b}}})$ for fixed $\mathbf{a}$. For completeness,
we state a precise result for the former:

######  Corollary 1.5.

Let ${\mathbf{b}}\succ 10^{\infty}$ be a sequence in $\\{0,1\\}^{\mathbb{N}}$,
and let ${\mathbf{b}}^{\prime}$ be the lexicographically largest sequence less
than or equal to ${\mathbf{b}}$ such that
$0^{\infty}\prec\sigma^{n}({\mathbf{b}}^{\prime})\preccurlyeq{\mathbf{b}}^{\prime}$
for all $n\geq 0$. Then ${\mathbf{b}}^{\prime}=\alpha(\beta)$ for some
$\beta\in(1,2]$, and if ${\mathbf{b}}^{\prime}$ is eventually periodic, the
entropy plateaus of $\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$
with strictly positive entropy value are the lexicographic intervals

$\tilde{I}_{\mathbf{w}}:=\begin{cases}[\mathbf{w}^{-}{\mathbf{b}}^{\prime},\mathbf{w}^{\infty}]&\mbox{if}\
I_{\mathbf{w}}=[\pi_{\beta}(\mathbf{w}0^{\infty}),\pi_{\beta}(\mathbf{w}^{\infty})],\\\
[\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty},\mathbf{w}^{\infty}]&\mbox{if}\
I_{\mathbf{w}}=[\pi_{\beta}\big{(}\mathbf{w}^{-}(\alpha_{1}\dots\alpha_{m}^{-})^{\infty}\big{)},\pi_{\beta}(\mathbf{w}^{\infty})],\end{cases}$

as $\mathbf{w}$ ranges over all $\beta$-Lyndon words such that
$I_{\mathbf{w}}\subseteq[0,\tau(\beta)]$.

Note that the symbolic plateaus in the first case extend a bit further to the
left, due to the fact that no sequence in
$[\mathbf{w}^{-}{\mathbf{b}}^{\prime},\mathbf{w}0^{\infty})$ is a greedy
$\beta$-expansion. Further, in the above corollary we did not describe the
$0$-plateau (that is, the plateau with entropy value $0$) of
$\mathbf{a}\mapsto h(\Sigma_{\mathbf{a},{\mathbf{b}}})$, as doing so requires
some notation to be developed later; see Remark 14.9. In any case, the
$0$-plateau was already determined previously in [25]; see also [24].

In order to prove the identity (1.3), it is necessary to better understand the
dynamics of the subshifts $\widetilde{\mathcal{K}}_{\beta}(t)$. We say a
subshift $\mathcal{X}$ of $\\{0,1\\}^{\mathbb{N}}$ is (topologically)
transitive111Our definition is slightly stronger than the usual one, which
requires that for any two finite words $\mathbf{u}$ and $\mathbf{w}$ in
$\mathcal{L}(\mathcal{X})$, there is a word $\mathbf{v}$ such that
$\mathbf{u}\mathbf{v}\mathbf{w}\in\mathcal{L}(\mathcal{X})$. However, it is
possible to show that for the subshifts $\widetilde{\mathcal{K}}_{\beta}(t)$,
the two definitions are equivalent. We adopt the stronger definition here for
convenience. if for any word $\mathbf{u}\in\mathcal{L}(\mathcal{X})$ and any
sequence $\mathbf{z}\in\mathcal{X}$ there is a word $\mathbf{v}$ such that
$\mathbf{u}\mathbf{v}\mathbf{z}\in\mathcal{X}$. Here
$\mathcal{L}(\mathcal{X})$ denotes the language of $\mathcal{X}$; that is, the
set of all finite words occurring in sequences in $\mathcal{X}$.
Unfortunately, $\widetilde{\mathcal{K}}_{\beta}(t)$ may fail to be transitive;
in that case, we try to identify a subshift of full entropy that contains the
sequence $b(t,\beta)$. Even this is not always possible; in Section 10 we will
identify so-called non-transitivity windows, i.e., intervals inside of which
no subshift of full entropy containing the sequence $b(t,\beta)$ exists. What
comes to the rescue in that situation is that the entropy
$h(\widetilde{\mathcal{K}}_{\beta}(t))$ is constant on these intervals. (The
closures of these non-transitivity windows are in fact EBLIs.)

The topological structure of the subshift $\widetilde{\mathcal{K}}_{\beta}(t)$
depends very heavily on $\beta$ (and, to a lesser extent, on $t$). We
decompose the interval $(1,2]$ into countably many disjoint subsets which are
of four essentially different types: basic intervals, the exceptional set, the
set of finitely renormalizable $\beta$, and the set of infinitely
renormalizable $\beta$. Different sections of the paper deal with different
parts of this decomposition. In the course of this work, we completely
characterize transitivity for all points $t$ whose greedy $\beta$-expansion is
purely periodic. Note that $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is a
subshift of the $\beta$-shift

(1.5)
$\Sigma_{\beta}:=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\
\forall n\geq 0\\},$

and therefore, this paper also characterizes transitivity for a large class of
subshifts of $\beta$-shifts. We are not aware of any earlier work of this
nature.

###  1.1. Connection with kneading sequences of Lorenz maps

The symbolic survivor set $\mathcal{K}_{\beta}(t)$ and the subshift
$\widetilde{\mathcal{K}}_{\beta}(t)$ are closely related to the set of
kneading sequences of Lorenz maps. Below we follow roughly the notion and
terminology of Hubbard and Sparrow [21]. A function $f:[0,1]\to[0,1]$ is
called a Lorenz map if there is a point $c\in(0,1)$ such that $f$ is
continuous and strictly increasing on $[0,c)$ and $(c,1]$, $f(c-)=1$ and
$f(c+)=0$. Such maps arise in the study of a geometric model of the Lorenz
differential equations. The function $T_{\beta}$ for $\beta\in(1,2]$ is a
Lorenz map, with $c=1/\beta$. For a point $x\in[0,1]$ that is not a pre-image
of $c$, the kneading sequence of $x$ is the sequence
$\mathbf{k}_{f}(x)=\varepsilon_{1}\varepsilon_{2}\dots\in\\{0,1\\}^{\mathbb{N}}$
given by

$\varepsilon_{n}=\begin{cases}0&\mbox{if $f^{n-1}(x)<c$},\\\ 1&\mbox{if
$f^{n-1}(x)>c$}.\end{cases}$

If $x$ is a pre-image of $c$, we define two kneading sequences
$\mathbf{k}_{f}^{+}(x)$ and $\mathbf{k}_{f}^{-}(x)$ by

$\mathbf{k}_{f}^{+}(x):=\lim_{y\searrow
x}\mathbf{k}_{f}(y),\qquad\mathbf{k}_{f}^{-}(x):=\lim_{y\nearrow
x}\mathbf{k}_{f}(y),$

where $y$ runs through points which are not pre-images of $c$, and the limits
are with respect to the product topology on $\\{0,1\\}^{\mathbb{N}}$. For
$f=T_{\beta}$, $\mathbf{k}_{f}^{+}(x)$ is precisely the greedy expansion of
$x$ in base $\beta$, whereas $\mathbf{k}_{f}^{-}(x)$ is the quasi-greedy
expansion. The kneading invariant of a Lorenz map $f$ is the pair
$(\mathbf{a}(f),{\mathbf{b}}(f))$ given by
$\mathbf{a}(f)=\mathbf{k}_{f}^{+}(0)$ and
${\mathbf{b}}(f)=\mathbf{k}_{f}^{-}(1)$. For $f=T_{\beta}$, note that
$\mathbf{a}(f)=0^{\infty}$ because $T_{\beta}(0)=0$, whereas
${\mathbf{b}}(f)=\alpha(\beta)$.

Hubbard and Sparrow [21] consider topologically expansive Lorenz maps, a
relatively weak notion of expansive which is equivalent to the set of pre-
images of $c$ being dense in $[0,1]$. The map $T_{\beta}$ is topologically
expansive, as is any Lorenz map with $C^{1}$ branches and derivative bounded
above 1. Hubbard and Sparrow prove that a pair $(\mathbf{a},{\mathbf{b}})$ of
sequences in $\\{0,1\\}^{\mathbb{N}}$ is the kneading invariant of some
topologically expansive Lorenz map if and only if

(1.6)
$\mathbf{a}\preccurlyeq\sigma^{n}(\mathbf{a})\prec{\mathbf{b}}\qquad\mbox{and}\qquad\mathbf{a}\prec\sigma^{n}({\mathbf{b}})\preccurlyeq{\mathbf{b}}\qquad\forall\,n\geq
0.$

Moreover, the corresponding map $f$ is unique up to conjugacy. Now observe
that, for any pair of sequences $(\mathbf{a},{\mathbf{b}})$ satisfying (1.6),
there is a parameter pair $(\beta,t)\in(1,2]\times[0,1)$ such that
$\mathbf{a}=b(t,\beta)$ and ${\mathbf{b}}=\alpha(\beta)$. (This follows from
Lemmas 2.1 and 2.2 below.) By the second main result of [21], the set of all
(upper or lower) kneading sequences of points in $[0,1]$ for the map $f$ is
then precisely $\widetilde{\mathcal{K}}_{\beta}(t)$.

Vice versa, suppose $\mathbf{a}$ and ${\mathbf{b}}$ are two $0-1$ sequences
beginning with $0$ and $1$, respectively, and consider the set
$\Sigma_{\mathbf{a},{\mathbf{b}}}$ from (1.4). If
$\Sigma_{\mathbf{a},{\mathbf{b}}}\neq\emptyset$, then there is a pair of
sequences $(\mathbf{a}^{\prime},{\mathbf{b}}^{\prime})$ such that
$\Sigma_{\mathbf{a},{\mathbf{b}}}=\Sigma_{\mathbf{a}^{\prime},{\mathbf{b}}^{\prime}}$
and

(1.7)
$\mathbf{a}^{\prime}\preccurlyeq\sigma^{n}(\mathbf{a}^{\prime})\preccurlyeq{\mathbf{b}}^{\prime}\qquad\mbox{and}\qquad\mathbf{a}^{\prime}\preccurlyeq\sigma^{n}({\mathbf{b}}^{\prime})\preccurlyeq{\mathbf{b}}^{\prime}\qquad\forall\,n\geq
0.$

(Simply take $\mathbf{a}^{\prime}:=\min\Sigma_{\mathbf{a},{\mathbf{b}}}$ and
${\mathbf{b}}^{\prime}:=\max\Sigma_{\mathbf{a},{\mathbf{b}}}$.) Labarca and
Moreira [25] call the set of pairs
$(\mathbf{a}^{\prime},{\mathbf{b}}^{\prime})$ satisfying (1.7) the
lexicographical world, and investigate it in detail. They show moreover that,
if $\Sigma_{\mathbf{a},{\mathbf{b}}}$ is uncountable, then there is a pair
$(\tilde{\mathbf{a}},\tilde{{\mathbf{b}}})$ of sequences satisfying the
kneading inequalities (1.6) such that
$\Sigma_{\tilde{\mathbf{a}},\tilde{{\mathbf{b}}}}\subseteq\Sigma_{\mathbf{a},{\mathbf{b}}}$
and
$\Sigma_{\mathbf{a},{\mathbf{b}}}\backslash\Sigma_{\tilde{\mathbf{a}},\tilde{{\mathbf{b}}}}$
is countable. Namely, let $\tilde{\mathbf{a}}$ be the smallest, and
$\tilde{{\mathbf{b}}}$ be the largest condensation point of
$\Sigma_{\mathbf{a},{\mathbf{b}}}$; see [25, p. 689]. (Recall that a point $x$
is a condensation point of a set $A$ in a topological space if every
neighborhood of $x$ contains uncountably many points of $A$.) The sequences
$\tilde{\mathbf{a}}$ and $\tilde{{\mathbf{b}}}$ are well defined and have the
required properties because the set of condensation points of
$\Sigma_{\mathbf{a},{\mathbf{b}}}$, like $\Sigma_{\mathbf{a},{\mathbf{b}}}$
itself, is closed and invariant under $\sigma$. While this characterization of
$\tilde{\mathbf{a}}$ and $\tilde{{\mathbf{b}}}$ is somewhat abstract, it is
also not too difficult to devise an algorithm for their construction.

From these observations, we conclude that

  1. (i)

For each pair $(\mathbf{a},{\mathbf{b}})$ such that $\mathbf{a}$ begins with
$0$ and ${\mathbf{b}}$ begins with $1$, there is a parameter pair $(\beta,t)$
such that
$\widetilde{\mathcal{K}}_{\beta}(t)\subseteq\Sigma_{\mathbf{a},{\mathbf{b}}}$
and
$\Sigma_{\mathbf{a},{\mathbf{b}}}\backslash\widetilde{\mathcal{K}}_{\beta}(t)$
is countable;

  2. (ii)

For each pair $(\beta,t)$ such that $\widetilde{\mathcal{K}}_{\beta}(t)$ is
uncountable, there is a parameter pair $(\beta^{\prime},t^{\prime})$ such that
$(\mathbf{a},{\mathbf{b}}):=(b(t,\beta),\alpha(\beta))$ satisfies (1.6),

$\widetilde{\mathcal{K}}_{\beta^{\prime}}(t^{\prime})\subseteq\widetilde{\mathcal{K}}_{\beta}(t),\qquad\mbox{and}\qquad\widetilde{\mathcal{K}}_{\beta}(t)\backslash\widetilde{\mathcal{K}}_{\beta^{\prime}}(t^{\prime})\
\mbox{is countable}.$

Thus, loosely speaking, each $\Sigma_{\mathbf{a},{\mathbf{b}}}$ is almost
realized as $\widetilde{\mathcal{K}}_{\beta}(t)$ for some pair $(\beta,t)$,
and each set $\widetilde{\mathcal{K}}_{\beta}(t)$ is almost the set of
kneading sequences for some topologically expansive Lorenz map, where “almost”
here means, up to a countable set.

Labarca and Moreira [25] give a further analysis of the sets
$\Sigma_{\mathbf{a},{\mathbf{b}}}$. For instance, they (implicitly) determine
for a given sequence $\mathbf{a}$ the critical ${\mathbf{b}}$ at which
$\Sigma_{\mathbf{a},{\mathbf{b}}}$ becomes uncountable, and show that when
$\Sigma_{\mathbf{a},{\mathbf{b}}}$ is uncountable, it has positive entropy and
hence positive Hausdorff dimension. More recently, their results have been
generalized by Komornik, Steiner and Zou [24]. Both of these works are also
very closely related to the recent paper [4] by the present authors.

###  1.2. Connection with the doubling map with a general hole

In 2015, Glendinning and Sidorov [19] considered the doubling map $T_{2}$ on
the circle $[0,1)$ with an arbitrary hole $(a,b)$, and determined for which
pairs $(a,b)$ with $0\leq a<b<1$ the survivor set

$K_{2}(a,b):=\\{x\in[0,1):T_{2}^{n}(x)\not\in(a,b)\ \forall\,n\geq 0\\}$

is uncountable. Replacing the numbers $a$ and $b$ with their greedy binary
expansions $\mathbf{a}$ and ${\mathbf{b}}$, respectively, the set $K_{2}(a,b)$
can be identified with

$\Omega_{\mathbf{a},{\mathbf{b}}}:=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\sigma^{n}(\mathbf{z})\preccurlyeq\mathbf{a}\
\mbox{or}\ \sigma^{n}(\mathbf{z})\succcurlyeq{\mathbf{b}}\ \forall\,n\geq
0\\}.$

Since it is easy to see that $\Omega_{\mathbf{a},{\mathbf{b}}}$ is uncountable
when $\mathbf{a}$ and ${\mathbf{b}}$ have the same first digit, Glendinning
and Sidorov focused on the case when $\mathbf{a}$ begins with $0$ and
${\mathbf{b}}$ begins with $1$, say $\mathbf{a}=0\mathbf{a}^{\prime}$ and
${\mathbf{b}}=1{\mathbf{b}}^{\prime}$. It can further be assumed that
$\mathbf{a}^{\prime}$ and ${\mathbf{b}}^{\prime}$ begin with $1$ and $0$,
respectively, since else $\Omega_{\mathbf{a},{\mathbf{b}}}$ consists of only
the two points $0^{\infty}$ and $1^{\infty}$. Now for such
$\mathbf{a}^{\prime}$ and ${\mathbf{b}}^{\prime}$, we have the relationship

(1.8)
$\Omega_{0\mathbf{a}^{\prime},1{\mathbf{b}}^{\prime}}=\bigcup_{n=0}^{\infty}0^{n}\Sigma_{{\mathbf{b}}^{\prime},\mathbf{a}^{\prime}}\cup\bigcup_{n=0}^{\infty}1^{n}\Sigma_{{\mathbf{b}}^{\prime},\mathbf{a}^{\prime}}\cup\\{0^{\infty},1^{\infty}\\}$

(see [24, Theorem 2.5 (vi)]). Hence $\Omega_{\mathbf{a},{\mathbf{b}}}$ has the
same cardinality and entropy as
$\Sigma_{{\mathbf{b}}^{\prime},\mathbf{a}^{\prime}}$, except in trivial cases
where the latter is finite and the former countably infinite. We can thus
relate the set $\Omega_{\mathbf{a},{\mathbf{b}}}$ to a symbolic survivor set
$\widetilde{\mathcal{K}}_{\beta}(t)$ for a suitable pair $(\beta,t)$, as
explained in the previous subsection.

###  1.3. Organization of the paper

After developing some notation and conventions, Section 2 introduces the main
building blocks of this article: Farey words and substitutions on such words.
Associated with these concepts are several important sets, not least of which
is the set $E$ of bases $\beta$ for which $\alpha(\beta)$ is Sturmian, i.e.
aperiodic and balanced. Here we also define three collections of intervals:
basic intervals, Farey intervals, and (higher order) Lyndon intervals arising
from the substitutions on Farey words. We see that the interval $(1,2]$ is
decomposed into several pieces, namely the set $E$; finitely renormalizable
$\beta$’s; infinitely renormalizable $\beta$’s; and the basic intervals.

In Section 3 we state and prove a useful characterization of the bifurcation
set $\mathscr{E}_{\beta}$. Section 4 then outlines the main ideas of the
proofs of our results; we state a more technical theorem whose proof takes up
much of the rest of the paper, and use it to prove Theorem 1.1.

Section 5 deals with the simplest case, when $\beta$ lies in the above-
mentioned set $E$. (In fact we include here also the left endpoints of first-
order basic intervals.) We show that here the subshift
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive for every
$\beta$-Lyndon interval $[t_{L},t_{R}]$, and that the $\beta$-Lyndon intervals
are dense in $[0,1-1/\beta]$. Section 6 then proves the same facts for the
right endpoints of first-order basic intervals, which require a somewhat
separate argument but behave much more like points in $E$ than like points in
the interiors of basic intervals.

Section 7 treats the case of finitely renormalizable $\beta$; that is, those
$\beta$’s that can be renormalized to a point of $E$ in a finite number of
steps. In this case we show inductively that
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive only for $t_{R}$ below
a certain threshold, but for $t_{R}$ above this threshold there still exists a
transitive subshift of full entropy containing the sequence $b(t_{R},\beta)$.
We show furthermore that the $\beta$-Lyndon intervals remain dense in this
case.

Section 8 considers the cardinality of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ when $\beta$ is an endpoint
of a basic interval or one of the special Lyndon intervals; the main result
here is a crucial element of the proof of Theorem 1.2.

Section 9 deals with the infinitely renormalizable case; the results are much
like those in Section 7, but the induction argument proceeds slightly
differently.

Sections 10, 11 and 12, the most complex of the paper, deal with the interiors
of basic intervals. Section 10 introduces a collection of intervals of
$t$-values called non-transitivity windows, inside of which
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is not transitive and does not
contain a transitive subshift of full entropy containing the sequence
$b(t_{R},\beta)$. These intervals are generated by a sequence
$(\mathbf{v}_{k})$ of Lyndon words which are extracted from the sequence
$\alpha(\beta)$. Most of this section is devoted to proving several necessary
facts about these words. Section 11 then proves the main result, which is that
(for $\beta$ in a first-order basic interval),
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive if and only if $t_{R}$
does not lie in any non-transitivity window, and the entropy of
$\widetilde{\mathcal{K}}_{\beta}(t)$ is constant in each non-transitivity
window. Section 12 then extends this result to the higher order basic
intervals, by combining the methods of the previous two sections with the
renormalization technique from Section 7.

Section 13 give a brief treatment of gaps between $\beta$-Lyndon intervals;
its main purpose is to show that for certain values of $\beta$, the difference
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ is countably infinite, thus
establishing part of Theorem 1.2.

Section 14 develops several properties of extended $\beta$-Lyndon intervals
(EBLIs), and proves Theorem 1.4 and Corollary 1.5. Finally, Section 15 gives
the proofs of the remaining theorems.

###  1.4. List of notation

  * •

$T_{\beta}(x):=\beta x\\!\\!\mod 1$ is the $\beta$-transformation on $[0,1)$

  * •

$K_{\beta}(t):=\\{x\in[0,1):T_{\beta}^{n}(x)\geq t\ \forall n\geq 0\\}$

  * •

$\pi_{\beta}((c_{i})):=\sum_{i=1}^{\infty}c_{i}\beta^{-i},\quad(c_{i})\in\\{0,1\\}^{\mathbb{N}}$

  * •

$\alpha(\beta)$ is the quasi-greedy expansion of $1$ in base $\beta$

  * •

$b(t,\beta)$ is the greedy expansion of $t$ in base $\beta$

  * •

$\sigma:\\{0,1\\}^{\mathbb{N}}\to\\{0,1\\}^{\mathbb{N}}$ is the left shift map

  * •

$\Sigma_{\beta}:=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\
\forall n\geq 0\\}$ is the $\beta$-shift

  * •

$\Sigma_{\mathbf{a},{\mathbf{b}}}:=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\mathbf{a}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq{\mathbf{b}}\
\forall n\geq 0\\}$

  * •

$\mathcal{K}_{\beta}(t):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\prec\alpha(\beta)\
\forall n\geq 0\\}$

  * •

$\widetilde{\mathcal{K}}_{\beta}(t):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:b(t,\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\
\forall n\geq 0\\}$

  * •

$h(\mathcal{X})$ is the topological entropy of the subshift $\mathcal{X}$

  * •

$\mathcal{L}(\mathcal{X})$ is the language of the subshift $\mathcal{X}$; i.e.
the set of all finite words that occur in some sequence in $\mathcal{X}$.

  * •

$\mathscr{E}_{\beta}:=\\{t\in[0,1):K_{\beta}(t^{\prime})\neq K_{\beta}(t)\
\forall\,t^{\prime}>t\\}$

  * •

$\mathscr{E}_{\beta}^{+}:=\\{t\in[0,1):T_{\beta}^{n}(t)\geq t\ \forall n\geq
0\\}=\mathscr{E}_{\beta}$

  * •

$\mathscr{B}_{\beta}:=\left\\{t\geq
0:\dim_{H}K_{\beta}(t^{\prime})<\dim_{H}K_{\beta}(t)~{}\forall
t^{\prime}>t\right\\}$

  * •

$F^{*}$ is the set of all Farey words of length at least 2

  * •

$L^{*}$ is the set of all Lyndon words of length at least 2

  * •

$L^{*}(\beta)$ is the set of all $\beta$-Lyndon words of length at least 2

  * •

$\mathbb{L}({\mathbf{s}})$ is the lexicographically largest cyclic permutation
of ${\mathbf{s}}$

  * •

$w_{1}\dots w_{m}^{+}:=w_{1}\dots w_{m-1}(w_{m}+1)$ if $w_{m}=0$

  * •

$w_{1}\dots w_{m}^{-}:=w_{1}\dots w_{m-1}(w_{m}-1)$ if $w_{m}=1$

  * •

$\beta_{\ell}^{{\mathbf{s}}}$ is the base such that
$\alpha(\beta_{\ell}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{\infty}$

  * •

$\beta_{r}^{{\mathbf{s}}}$ is the base such that
$\alpha(\beta_{r}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty}$

  * •

$\beta_{*}^{{\mathbf{s}}}$ is the base such that
$\alpha(\beta_{*}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}$

  * •

$E:=(1,2]\backslash\bigcup_{s\in
F^{*}}[\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$

  * •

$E_{L}:=(1,2]\backslash\bigcup_{s\in
F^{*}}(\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$

  * •

$U_{0}$ is the substitution $0\mapsto 0$, $1\mapsto 01$

  * •

$U_{1}$ is the substitution $0\mapsto 01$, $1\mapsto 1$

  * •

$\tau(\beta):=\min\\{t>0:\dim_{H}K_{\beta}(t)=0\\}$

  * •

$\mathcal{T}_{R}(\beta)$ is the set of all right endpoints $t_{R}$ of
$\beta$-Lyndon intervals $[t_{L},t_{R}]$ in $[0,\tau(\beta)]$

  * •

${\mathbf{s}}\bullet{\mathbf{r}}:=\Phi_{\mathbf{s}}({\mathbf{r}})$ is the
substitution defined in (2.8)

  * •

$\Lambda_{k}:=\left\\{{\mathbf{S}}={\mathbf{s}}_{1}\bullet{\mathbf{s}}_{2}\bullet\cdots\bullet{\mathbf{s}}_{k}:\
{\mathbf{s}}_{i}\in F^{*}\textrm{ for any }1\leq i\leq k\right\\}$

  * •

$\Lambda:=\bigcup_{k=1}^{\infty}\Lambda_{k}$

  * •

$I^{\mathbf{S}}=[\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}]$ is the
basic interval generated by ${\mathbf{S}}\in\Lambda$

  * •

$J^{\mathbf{S}}=[\beta_{\ell}^{{\mathbf{S}}},\beta_{r}^{{\mathbf{S}}}]$ is the
Lyndon interval generated by ${\mathbf{S}}\in\Lambda$

  * •

$\Gamma({\mathbf{S}}):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:{\mathbf{S}}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{S}})^{\infty}\
\ \forall n\geq 0\\}$

##  2\. Preliminaries

First we introduce some notation and conventions. By a _word_ we mean a finite
string of zeros and ones. Let $\left\\{0,1\right\\}^{*}$ be the set of all
words over the alphabet $\left\\{0,1\right\\}$ together with the empty word
$\epsilon$. For a word ${\mathbf{c}}\in\left\\{0,1\right\\}^{*}$ we denote its
length by $|{\mathbf{c}}|$, and for a digit $a\in\left\\{0,1\right\\}$ we
denote by $|{\mathbf{c}}|_{a}$ the number of occurrences of $a$ in the word
${\mathbf{c}}$. The concatenation of two words $\mathbf{c}=c_{1}\ldots c_{m}$
and $\mathbf{d}=d_{1}\ldots d_{n}$ in $\left\\{0,1\right\\}^{*}$ is denoted by
$\mathbf{cd}=c_{1}\ldots c_{m}d_{1}\ldots d_{n}$. Similarly, $\mathbf{c}^{n}$
denotes the $n$-fold concatenation of $\mathbf{c}$ with itself, and
$\mathbf{c}^{\infty}$ denotes the periodic sequence with period block
$\mathbf{c}$. If $\mathbf{c}=c_{1}\ldots c_{m}$ with $c_{m}=0$, then we define
$\mathbf{c}^{+}:=c_{1}\ldots c_{m-1}1$; and if $\mathbf{c}=c_{1}\ldots c_{m}$
with $c_{m}=1$, we set $\mathbf{c}^{-}:=c_{1}\ldots c_{m-1}0$.

Throughout the paper we will use the lexicographical order
‘$\prec,\preccurlyeq,\succ$’ or ‘$\succcurlyeq$’ between sequences and words.
For example, for two sequences
$(c_{i}),(d_{i})\in\left\\{0,1\right\\}^{\mathbb{N}}$, we say
$(c_{i})\prec(d_{i})$ if $c_{1}<d_{1}$, or there exists $n\in\mathbb{N}$ such
that $c_{1}\ldots c_{n}=d_{1}\ldots d_{n}$ and $c_{n+1}<d_{n+1}$. For two
words $\mathbf{c},\mathbf{d}$, we say $\mathbf{c}\prec\mathbf{d}$ if
$\mathbf{c}1^{\infty}\prec\mathbf{d}0^{\infty}$. Finally, for a word
$\mathbf{c}$ and an infinite sequence $\mathbf{d}$, we say
$\mathbf{c}\prec\mathbf{d}$ if $\mathbf{c}1^{\infty}\prec\mathbf{d}$, and
define $\mathbf{c}\succ\mathbf{d}$ similarly. Thoughout this paper, when
discussing words and sequences, adjectives like smallest, greater, etc. will
always mean smallest, greater, etc. with respect to the lexicographical order.

Recall that $\alpha(\beta)$ is the quasi-greedy expansion of $1$ in base
$\beta$. The following useful result is well known (cf. [6]).

######  Lemma 2.1.

The map $\beta\mapsto\alpha(\beta)$ is strictly increasing and bijective from
$(1,2]$ to the set of sequences $(\alpha_{i})\in\\{0,1\\}^{\mathbb{N}}$
satisfying

$0^{\infty}\prec\sigma^{n}((\alpha_{i}))\preccurlyeq(\alpha_{i})\quad\forall
n\geq 1.$

We recall the following characterization of greedy $\beta$-expansions due to
Parry [30].

######  Lemma 2.2.

Let $\beta\in(1,2]$. The map $t\mapsto b(t,\beta)$ is an increasing bijection
from $[0,1)$ to

$\big{\\{}\mathbf{z}\in\left\\{0,1\right\\}^{\mathbb{N}}:\sigma^{n}(\mathbf{z})\prec\alpha(\beta)~{}\forall
n\geq 0\big{\\}}.$

Furthermore, the map $t\mapsto b(t,\beta)$ is right-continuous everywhere in
$[0,1)$ with respect to the order topology in
$\left\\{0,1\right\\}^{\mathbb{N}}$.

###  2.1. Farey words, Farey intervals and the set $E$

Farey words have attracted much attention in the literature due to their
intimate connection with rational rotations on the circle (see [28, Chapter
2]) and their one-to-one correspondence with the rational numbers in $[0,1]$
(see (2.1) below). In the following we adopt the definition from [9].

First we recursively define a sequence of ordered sets $F_{n},n=0,1,2,\ldots$.
Let $F_{0}=(0,1)$; and for $n\geq 0$ the ordered set
$F_{n+1}=(\mathbf{v}_{1},\ldots,\mathbf{v}_{2^{n+1}+1})$ is obtained from
$F_{n}=(\mathbf{w}_{1},\ldots,\mathbf{w}_{2^{n}+1})$ by

$\left\\{\begin{array}[]{lll}\mathbf{v}_{2i-1}=\mathbf{w}_{i}&\textrm{for}&1\leq
i\leq 2^{n}+1,\\\
\mathbf{v}_{2i}=\mathbf{w}_{i}\mathbf{w}_{i+1}&\textrm{for}&1\leq i\leq
2^{n}.\end{array}\right.$

In other words, $F_{n+1}$ is obtained from $F_{n}$ by inserting for each
$1\leq j\leq 2^{n}$ the new word $\mathbf{w}_{j}\mathbf{w}_{j+1}$ between the
two neighboring words $\mathbf{w}_{j}$ and $\mathbf{w}_{j+1}$. So,

$\begin{split}&F_{1}=(0,01,1),\qquad F_{2}=(0,001,01,011,1),\\\
F_{3}&=(0,0001,001,00101,01,01011,011,0111,1),\quad\dots\end{split}$

Note that for each $n\geq 0$ the ordered set $F_{n}$ consists of $2^{n}+1$
words which are listed from the left to the right in lexicographically
increasing order. We call $\mathbf{w}\in\left\\{0,1\right\\}^{*}$ a _Farey
word_ if $\mathbf{w}\in F_{n}$ for some $n\geq 0$. We denote by
$F:=\bigcup_{n=1}^{\infty}F_{n}$ the set of all Farey words, and by
$F^{*}:=F\backslash\\{0,1\\}$ the set of all Farey words of length at least
two. As shown in [9, Proposition 2.3], the set $F$ can be bijectively mapped
to $\mathbb{Q}\cap[0,1]$ via the map

(2.1)
$\xi:F\to\mathbb{Q}\cap[0,1];\quad{\mathbf{s}}\mapsto\frac{|{\mathbf{s}}|_{1}}{|{\mathbf{s}}|}.$

So, $\xi({\mathbf{s}})$ is the frequency of the digit $1$ in ${\mathbf{s}}$.

The Farey words can also be obtained recursively via the substitutions

(2.2) $U_{0}:\left\\{\begin{array}[]{lll}0&\mapsto&0\\\
1&\mapsto&01,\end{array}\right.\quad\textrm{and}\quad
U_{1}:\left\\{\begin{array}[]{lll}0&\mapsto&01\\\
1&\mapsto&1.\end{array}\right.$

We extend the definitions of $U_{0}$ and $U_{1}$ to $\left\\{0,1\right\\}^{*}$
and $\left\\{0,1\right\\}^{\mathbb{N}}$ via homomorphism; that is,
$U_{0}(c_{1}\dots c_{m})=U_{0}(c_{1})\dots U_{0}(c_{m})$, etc. Observe that
the maps $U_{0}$ and $U_{1}$, when viewed as functions on
$\left\\{0,1\right\\}^{\mathbb{N}}$, are both strictly increasing with respect
to the lexicographical order.

The following lemma can be deduced from [9, Propositions 2.3 and 2.9]. (See
[4, Section 2] for more details.)

######  Lemma 2.3.

Let ${\mathbf{s}}\in F^{*}$; then one of the following holds:

  * (i)

${\mathbf{s}}=01$;

  * (ii)

${\mathbf{s}}=U_{0}(\hat{{\mathbf{s}}})$ for some Farey word
$\hat{{\mathbf{s}}}\in F^{*}$;

  * (iii)

${\mathbf{s}}=U_{1}(\hat{{\mathbf{s}}})$ for some Farey word
$\hat{{\mathbf{s}}}\in F^{*}$.

Vice versa, if ${\mathbf{s}}$ is a Farey word, then both $U_{0}({\mathbf{s}})$
and $U_{1}({\mathbf{s}})$ are Farey.

Observe that for any word $\mathbf{w}$ which begins with $0$ and ends with
$1$, both $U_{0}(\mathbf{w})$ and $U_{1}(\mathbf{w})$ begin with $0$ and end
with $1$. Furthermore, for any word $\mathbf{w}$, the word $U_{0}(\mathbf{w})$
does not contain $11$ as a subword, and $U_{1}(\mathbf{w})$ does not contain
$00$.

For a word $\mathbf{c}=c_{1}\ldots c_{m}\in\left\\{0,1\right\\}^{*}$, let
$\mathbb{S}(\mathbf{c})$ and $\mathbb{L}(\mathbf{c})$ be the smallest and
largest cyclic permutations of $\mathbf{c}$, respectively; that is, the
smallest and largest words among

$c_{1}c_{2}\ldots c_{m},\quad c_{2}\ldots c_{m}c_{1},\quad c_{3}\ldots
c_{m}c_{1}c_{2},\quad\cdots,\quad c_{m}c_{1}\ldots c_{m-1}.$

The following properties of Farey words are well known (see, e.g., [9,
Proposition 2.5]).

######  Lemma 2.4.

Let ${\mathbf{s}}=s_{1}\ldots s_{m}\in F^{*}$. Then

  1. (i)

$\mathbb{S}({\mathbf{s}})={\mathbf{s}}$ and
$\mathbb{L}({\mathbf{s}})=s_{m}s_{m-1}\ldots s_{1}$.

  2. (ii)

${{\mathbf{s}}^{-}}$ is a palindrome; that is, $s_{1}\ldots
s_{m-1}(s_{m}-1)=(s_{m}-1)s_{m-1}s_{m-2}\ldots s_{1}$.

We recall from [22] that a Farey interval is an interval
$J^{\mathbf{s}}=[\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$
defined by

$\alpha(\beta_{\ell}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{\infty},\qquad\alpha(\beta_{r}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty},$

where ${\mathbf{s}}\in F^{*}$. We define the sets

$E:=(1,2]\backslash\bigcup_{s\in
F^{*}}J^{\mathbf{s}}=(1,2]\backslash\bigcup_{{\mathbf{s}}\in
F^{*}}[\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$

and

(2.3) $E_{L}:=(1,2]\backslash\bigcup_{{\mathbf{s}}\in
F^{*}}(\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}].$

We let $\overline{E}$ denote the topological closure of $E$, and observe that

$\overline{E}=\overline{E_{L}}={[1,2]}\backslash\bigcup_{{\mathbf{s}}\in
F^{*}}(\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}).$

A word or sequence $\mathbf{w}$ is said to be balanced if for any two subwords
$\mathbf{u}$ and $\mathbf{v}$ of $\mathbf{w}$ with equal length,
$\big{|}|\mathbf{u}|_{1}-|\mathbf{v}|_{1}\big{|}\leq 1$. All Farey words are
balanced, and any subword of a balanced word or sequence is again balanced.

######  Lemma 2.5.

For each $\beta\in\overline{E}{\cap(1,2]}$, the sequence $\alpha(\beta)$ is
balanced.

###### Proof.

Write $\alpha(\beta)=\alpha_{1}\alpha_{2}\dots$. Since $\beta\in\overline{E}$,
$\alpha(\beta)$ is the component-wise limit of the sequence
$(\mathbb{L}({\mathbf{s}}_{n})^{\infty})$ for some sequence
$({\mathbf{s}}_{n})$ of Farey words. Without loss of generality we may assume
that the length of ${\mathbf{s}}_{n}$ strictly increases with $n$. Then for
each $k\in\mathbb{N}$ there is an integer $N_{k}$ such that
$\alpha_{1}\dots\alpha_{k}$ is a prefix of $\mathbb{L}({\mathbf{s}}_{n})$ for
all $n\geq N_{k}$. Since Farey words are balanced, it follows that
$\alpha(\beta)$ is balanced as well. ∎

######  Lemma 2.6.

Let ${\mathbf{s}}$ and $\hat{{\mathbf{s}}}$ be Farey words and suppose
${\mathbf{s}}=U_{0}(\hat{{\mathbf{s}}})$ or
${\mathbf{s}}=U_{1}(\hat{{\mathbf{s}}})$.

  1. (i)

If ${\mathbf{s}}=U_{0}(\hat{{\mathbf{s}}})$, then

$0\mathbb{L}({\mathbf{s}})=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}}))0,\qquad
0\mathbb{L}({\mathbf{s}})^{+}=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}})^{+}),\qquad{\mathbf{s}}^{-}=U_{0}(\hat{{\mathbf{s}}}^{-})0.$

  2. (ii)

If ${\mathbf{s}}=U_{1}(\hat{{\mathbf{s}}})$, then

$\mathbb{L}({\mathbf{s}})1=1U_{1}(\mathbb{L}(\hat{{\mathbf{s}}})),\qquad\mathbb{L}({\mathbf{s}})^{+}=1U_{1}(\mathbb{L}(\hat{{\mathbf{s}}})^{+}),\qquad{\mathbf{s}}^{-}1=U_{1}(\hat{{\mathbf{s}}}^{-}).$

###### Proof.

We prove (i); the proof of (ii) is similar.

Assume ${\mathbf{s}}=U_{0}(\hat{{\mathbf{s}}})$. By Lemma 2.4,
$\mathbb{L}(\hat{{\mathbf{s}}})$ is the word obtained from
$\hat{{\mathbf{s}}}$ by changing the first $0$ to a $1$ and the last $1$ to a
$0$. Hence $U_{0}(\mathbb{L}(\hat{{\mathbf{s}}}))$ is obtained from
$U_{0}(\hat{{\mathbf{s}}})$ by replacing the first block, $U_{0}(0)=0$, with
$U_{0}(1)=01$, and the last block, $U_{0}(1)=01$ with $U_{0}(0)=0$. On the
other hand, $\mathbb{L}({\mathbf{s}})=\mathbb{L}(U_{0}(\hat{{\mathbf{s}}}))$
is obtained from $U_{0}(\hat{\mathbf{s}})$ by replacing the first digit $0$
with $1$ and the last block $01$ with $00$. It follows that

(2.4) $0\mathbb{L}({\mathbf{s}})=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}}))0.$

Next, since $\mathbb{L}(\hat{{\mathbf{s}}})$ ends in a $0$ and
$U_{0}(1)=01=U_{0}(0)1$, by (2.4) we have

$U_{0}(\mathbb{L}(\hat{{\mathbf{s}}})^{+})=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}}))1=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}}))0^{+}=0\mathbb{L}({\mathbf{s}})^{+}.$

Finally, $\hat{{\mathbf{s}}}^{-}$ is obtained from $\hat{{\mathbf{s}}}$ by
changing the last digit 1 to a 0, so $U_{0}(\hat{{\mathbf{s}}}^{-})$ is
obtained from $U_{0}(\hat{{\mathbf{s}}})$ by removing the last 1, whereas
${\mathbf{s}}^{-}$ is obtained from $U_{0}(\hat{{\mathbf{s}}})$ by changing
the last 1 to a 0. Thus, ${\mathbf{s}}^{-}=U_{0}(\hat{{\mathbf{s}}}^{-})0$. ∎

######  Lemma 2.7.

If $\beta\in E_{L}$ and $\alpha(\beta)\neq(10)^{\infty}$, then $\alpha(\beta)$
does not end in $(01)^{\infty}$.

###### Proof.

Assume $\alpha(\beta)\neq(10)^{\infty}$. Note that $\alpha(\beta)$ is balanced
by Lemma 2.5. So if $\alpha(\beta)$ ends in $(01)^{\infty}$, then
$\alpha(\beta)$ cannot contain a block $11(01)^{k}1$ for any $k\geq 0$,
because this block has length $2k+3$ and contains $k$ zeros, whereas the block
$(01)^{k+1}0$ also has length $2k+3$ but contains $k+2$ zeros. This leaves
only the possibility $\alpha(\beta)=11(01)^{\infty}$, but then
$\beta=\beta_{r}^{{\mathbf{s}}}$ for ${\mathbf{s}}=01$, so $\beta\not\in
E_{L}$. Hence, $\alpha(\beta)$ cannot end in $(01)^{\infty}$. ∎

######  Lemma 2.8.

Let $\beta\in E_{L}$. Then there is a base $\hat{\beta}\in E_{L}$ such that

$0\alpha(\beta)=U_{0}(\alpha(\hat{\beta}))\qquad\mbox{or}\qquad
0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta})).$

Note that in the second case, it is possible that $\beta=\hat{\beta}$; this
happens when $\beta=2$, in which case
$\alpha(\beta)=\alpha(\hat{\beta})=1^{\infty}$.

###### Proof.

If $\alpha(\beta)=(10)^{\infty}$, then
$0\alpha(\beta)=U_{0}(1^{\infty})=U_{0}(\alpha(2))$, and $2\in E_{L}$. So
assume $\alpha(\beta)\neq(10)^{\infty}$. Then $\alpha(\beta)$ does not end in
$(01)^{\infty}$ by Lemma 2.7. Furthermore, since $\alpha(\beta)$ is balanced
by Lemma 2.5, it either does not contain the word $00$ or does not contain the
word $11$.

Case 1: $\alpha(\beta)$ does not contain $11$. Then
$0\alpha(\beta))=U_{0}(\mathbf{d})$ for some sequence
$\mathbf{d}=d_{1}d_{2}\dots\in\\{0,1\\}^{\mathbb{N}}$ with $d_{1}=1$. Since
$\alpha(\beta)$ does not end in $0^{\infty}=U_{0}(0^{\infty})$, $\mathbf{d}$
does not end in $0^{\infty}$ either. We claim that
$\sigma^{k}(\mathbf{d})\preccurlyeq\mathbf{d}$ for all $k\geq 1$. Fix
$k\in\mathbb{N}$. We can choose $n\geq 1$ so that

$U_{0}(\sigma^{k}(\mathbf{d}))=\sigma^{n}(U_{0}(\mathbf{d}))=\sigma^{n}(0\alpha(\beta)),$

and moreover, $\sigma^{n}(0\alpha(\beta))$ begins with a $0$. Hence,

$U_{0}(\sigma^{k}(\mathbf{d}))=\sigma^{n}(0\alpha(\beta))=0\sigma^{n}(\alpha(\beta))\preccurlyeq
0\alpha(\beta)=U_{0}(\mathbf{d}),$

and since $U_{0}$ is strictly increasing, it follows that
$\sigma^{k}(\mathbf{d})\preccurlyeq\mathbf{d}$. Thus, by Lemma 2.1,
$\mathbf{d}=\alpha(\hat{\beta})$ for some base $\hat{\beta}$. In other words,

(2.5) $0\alpha(\beta)=U_{0}(\alpha(\hat{\beta})).$

We next verify that $\hat{\beta}\in E_{L}$. Suppose, by way of contradiction,
that
$\hat{\beta}\in(\beta_{\ell}^{\hat{{\mathbf{s}}}},\beta_{r}^{\hat{{\mathbf{s}}}}]$
for some Farey word $\hat{{\mathbf{s}}}$. That is,

(2.6)
$\mathbb{L}(\hat{{\mathbf{s}}})^{\infty}\prec\alpha(\hat{\beta})\preccurlyeq\mathbb{L}(\hat{{\mathbf{s}}})^{+}\hat{{\mathbf{s}}}^{\infty}.$

Set ${\mathbf{s}}:=U_{0}(\hat{{\mathbf{s}}})$. Then ${\mathbf{s}}$ is Farey,
and since $U_{0}$ is strictly increasing, by (2.5) and Lemma 2.6 (i) we obtain

$0\mathbb{L}({\mathbf{s}})^{\infty}=U_{0}(\mathbb{L}(\hat{{\mathbf{s}}})^{\infty})\prec
U_{0}(\alpha(\hat{\beta}))=0\alpha(\beta)\preccurlyeq
U_{0}\big{(}\mathbb{L}(\hat{{\mathbf{s}}})^{+}\hat{{\mathbf{s}}}^{\infty}\big{)}=0\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty},$

Thus, $\beta\in(\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$,
contradicting that $\beta\in E_{L}$. We conclude that $\hat{\beta}\in E_{L}$.

Case 2: $\alpha(\beta)$ does not contain $00$. Then
$0\alpha(\beta)=U_{1}(0\mathbf{d})$ for some sequence
$\mathbf{d}=d_{1}d_{2}\dots\in\\{0,1\\}^{\mathbb{N}}$ with $d_{1}=1$. Since
$\alpha(\beta)$ does not end in $(01)^{\infty}=U_{1}(0^{\infty})$,
$\mathbf{d}$ does not end in $0^{\infty}$.

We next show that $\sigma^{k}(\mathbf{d})\preccurlyeq\mathbf{d}$ for all
$k\geq 1$. Fix $k\in\mathbb{N}$. Since $\mathbf{c}\preccurlyeq 1\mathbf{c}$
for any sequence $\mathbf{c}$, we may assume without loss of generality that
$d_{k}=0$. We can choose $n\geq 1$ so that
$U_{1}(0\sigma^{k}(\mathbf{d}))=U_{1}(\sigma^{k-1}(\mathbf{d}))=\sigma^{n}(0\alpha(\beta))$,
and this expression begins with a $U_{1}(0)=01$. Hence,

$U_{1}(0\sigma^{k}(\mathbf{d}))=\sigma^{n}(0\alpha(\beta))=0\sigma^{n}(\alpha(\beta))\preccurlyeq
0\alpha(\beta)=U_{1}(0\mathbf{d}).$

Since $U_{1}$ is strictly increasing, it follows that
$\sigma^{k}(\mathbf{d})\preccurlyeq\mathbf{d}$, as desired. Thus,
$\mathbf{d}=\alpha(\hat{\beta})$ for some $\hat{\beta}\in(1,2]$, and then
$0\alpha(\beta)=U_{1}(0\mathbf{d})=U_{1}(0\alpha(\hat{\beta}))$.

We next verify that $\hat{\beta}\in E_{L}$. Suppose, by way of contradiction,
that
$\hat{\beta}\in(\beta_{\ell}^{\hat{{\mathbf{s}}}},\beta_{r}^{\hat{{\mathbf{s}}}}]$
for some Farey word $\hat{{\mathbf{s}}}$. That is, (2.6) holds. Set
${\mathbf{s}}:=U_{1}(\hat{{\mathbf{s}}})$. Then ${\mathbf{s}}$ is Farey, and
since $U_{1}$ is strictly increasing, we obtain from (2.6) and Lemma 2.6 (ii)
that

$\displaystyle 0\mathbb{L}({\mathbf{s}})^{\infty}$
$\displaystyle=01U_{1}(\mathbb{L}(\hat{{\mathbf{s}}})^{\infty})=U_{1}(0\mathbb{L}(\hat{{\mathbf{s}}})^{\infty})$
$\displaystyle\prec U_{1}(0\alpha(\hat{\beta}))=0\alpha(\beta)\preccurlyeq
U_{1}(0\mathbb{L}(\hat{\mathbf{s}})^{+}\hat{\mathbf{s}}^{\infty})$
$\displaystyle=01U_{1}(\mathbb{L}(\hat{\mathbf{s}})^{+}\hat{\mathbf{s}}^{\infty})=0\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty}.$

Hence
$\mathbb{L}({\mathbf{s}})^{\infty}\prec\alpha(\beta)\preccurlyeq\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty}$.
But then $\beta\in(\beta_{\ell}^{{\mathbf{s}}},\beta_{r}^{{\mathbf{s}}}]$,
contradicting that $\beta\in E_{L}$. Thus, $\hat{\beta}\in E_{L}$. ∎

The last result in this subsection may be of independent interest. Recall that
a $0-1$ sequence is Sturmian if it is balanced and not eventually periodic
(cf. [28]).

######  Proposition 2.9.

Let $\beta\in(1,2)$.

  1. (i)

$\alpha(\beta)$ is balanced if and only if $\beta\in\overline{E}$.

  2. (ii)

$\alpha(\beta)$ is Sturmian if and only if $\beta\in E$.

###### Proof.

(i) We have already seen from Lemma 2.5 that, if $\beta\in\overline{E}$, then
$\alpha(\beta)$ is balanced. Conversely, suppose $\beta\not\in\overline{E}$.
Then $\beta_{\ell}^{{\mathbf{s}}}<\beta<\beta_{r}^{{\mathbf{s}}}$ for some
Farey word ${\mathbf{s}}=s_{1}\dots s_{m}$, and so
$\mathbb{L}({\mathbf{s}})^{\infty}\prec\alpha(\beta)\prec\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{\infty}$.
By Lemma 2.1 this implies that $\alpha(\beta)$ begins with
$\mathbb{L}({\mathbf{s}})^{+}$ and there is an $n\in\mathbb{N}$ such that
$\alpha_{n+1}\dots\alpha_{n+m}\prec{\mathbf{s}}$, where
$(\alpha_{i}):=\alpha(\beta)$. Let $k$ be the smallest integer such that
$\alpha_{n+k}\neq s_{k}$. Then $\alpha_{n+1}\dots\alpha_{n+k}=s_{1}\dots
s_{k}^{-}=0s_{2}\dots s_{k}^{-}$. By Lemma 2.4, $\mathbb{L}({\mathbf{s}})^{+}$
is obtained from ${\mathbf{s}}$ by changing the first 0 to a 1, so
$\mathbb{L}({\mathbf{s}})^{+}$ begins with $1s_{2}\dots s_{k}$. Thus,
$\alpha_{n+1}\dots\alpha_{n+k}$ has two fewer 1’s than
$\alpha_{1}\dots\alpha_{k}=1s_{2}\ldots s_{k}$, and we conclude that
$\alpha(\beta)$ is not balanced.

(ii) If $\alpha(\beta)$ is Sturmian, then it is balanced so
$\beta\in\overline{E}$ by part (i). But $\beta\neq\beta_{\ell}^{{\mathbf{s}}}$
or $\beta_{r}^{{\mathbf{s}}}$ for any Farey word ${\mathbf{s}}$, because
$\alpha(\beta_{\ell}^{{\mathbf{s}}})$ and $\alpha(\beta_{r}^{{\mathbf{s}}})$
are (eventually) periodic. Thus, $\beta\in E$.

Vice versa, take $\beta\in E$. Then $\alpha(\beta)$ is balanced, but we still
need to show that it is not eventually periodic. Suppose $\alpha(\beta)$ is
eventually periodic. Since $\beta<2$, $\alpha(\beta)\neq 1^{\infty}$. By
cyclical shifting, we may assume that the minimal period block of
$\alpha(\beta)$ is a Lyndon word $\mathbf{w}$. Note that
$\alpha(\beta)\neq(10)^{\infty}$, since otherwise
$\beta=\beta_{\ell}^{01}\notin E$. By Lemma 2.7, $\mathbf{w}\neq 01$. Since
$\alpha(\beta)$ is balanced, so is $\mathbf{w}$, and therefore either
$\mathbf{w}=U_{0}(\hat{\mathbf{w}})$ or $\mathbf{w}=U_{1}(\hat{\mathbf{w}})$
for some Lyndon word $\hat{\mathbf{w}}$, as $\mathbf{w}$ cannot contain both
of the words $00$ and $11$. (Note $\hat{\mathbf{w}}$ is also Lyndon because
$U_{0}$ and $U_{1}$ are strictly increasing.) If
$\mathbf{w}=U_{0}(\hat{\mathbf{w}})$, then
$0\alpha(\beta)=0U_{0}(\alpha(\hat{\beta}))$ for some $\hat{\beta}\in E$,
since Lemma 2.8 obviously holds also for $E$ instead of $E_{L}$. But then
$\alpha(\hat{\beta})$ is eventually periodic with the shorter period block
$\hat{\mathbf{w}}$. A similar argument holds when
$\mathbf{w}=U_{1}(\hat{\mathbf{w}})$. We can now repeat the same argument over
and over again, to obtain an infinite sequence of Lyndon words
$\mathbf{w}_{1},\mathbf{w}_{2},\dots$, each one shorter than the last. But
this is absurd. Therefore, $\alpha(\beta)$ is not eventually periodic. ∎

###  2.2. The substitution operator

######  Definition 2.10.

For a Lyndon word ${\mathbf{s}}\in L^{*}$, we define the substitution map
$\Phi_{\mathbf{s}}:\\{0,1\\}^{\mathbb{N}}\to\\{0,1\\}^{\mathbb{N}}$ by

(2.7)
$\displaystyle\begin{split}\Phi_{\mathbf{s}}(0^{k_{1}}1^{l_{1}}0^{k_{2}}1^{l_{2}}\dots)={\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{k_{1}-1}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{l_{1}-1}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{k_{2}-1}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{l_{2}-1}\dots,\\\
\Phi_{s}(1^{k_{1}}0^{l_{1}}1^{k_{2}}0^{l_{2}}\dots)=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{k_{1}-1}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{l_{1}-1}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{k_{2}-1}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{l_{2}-1}\dots,\end{split}$

where $1\leq k_{i},l_{i}\leq\infty$ for all $i$. We allow one of the exponents
$k_{i}$ or $l_{i}$ to take the value $+\infty$, in which case we ignore the
remainder of the sequence. We denote the range of $\Phi_{\mathbf{s}}$ by
$X({\mathbf{s}})$. We define $\Phi_{\mathbf{s}}({\mathbf{r}})$ for a finite
word ${\mathbf{r}}$ in the same way, and set
$X^{*}({\mathbf{s}}):=\Phi_{\mathbf{s}}\big{(}\\{0,1\\}^{*}\big{)}$.

The order in which the four blocks
${\mathbf{s}},{\mathbf{s}}^{-},\mathbb{L}({\mathbf{s}})$ and
$\mathbb{L}({\mathbf{s}})^{+}$ can appear in $\Phi_{\mathbf{s}}({\mathbf{r}})$
is illustrated in Figure 1.

${\mathbf{s}}$${\mathbf{s}}^{-}$
$\mathbb{L}({\mathbf{s}})^{+}$Start-$1$$\mathbb{L}({\mathbf{s}})$Start-$0$$1$$1$$0$$0$$1$$1$$0$$1$$0$$0$
Figure 1. The directed graph illustrating the map $\Phi_{\mathbf{s}}$.

For example,

$\Phi_{\mathbf{s}}(0110^{\infty})={\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty},\qquad\Phi_{\mathbf{s}}(11100)=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{2}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}}).$

Now for any two words ${\mathbf{s}}\in L^{*}$ and
${\mathbf{r}}\in\left\\{0,1\right\\}^{*}$ we define the substitution operation

(2.8) ${\mathbf{s}}\bullet{\mathbf{r}}:=\Phi_{\mathbf{s}}({\mathbf{r}}).$

######  Example 2.11.

Let ${\mathbf{s}}=01$ and ${\mathbf{r}}=011$. Then

${\mathbf{s}}\bullet{\mathbf{r}}=\Phi_{\mathbf{s}}({\mathbf{r}})=\Phi_{\mathbf{s}}(011)={{\mathbf{s}}^{-}}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}=001101.$

The following lemma collects properties of the map $\Phi_{\mathbf{s}}$ and the
substitution operator $\bullet$. The proofs can be found in [4, Section 3]. To
summarize, $\Phi_{\mathbf{s}}$ is increasing and commutes with the operators
$\cdot^{+},\cdot^{-}$ and $\mathbb{L}$; the set $L^{*}$ of Lyndon words is
closed under $\bullet$, and $\bullet$ is associative.

######  Lemma 2.12.

Let ${\mathbf{s}}\in L^{*}$.

  1. (i)

The map $\Phi_{\mathbf{s}}$ is strictly increasing on
$\left\\{0,1\right\\}^{\mathbb{N}}$.

  2. (ii)

For any word ${\mathbf{d}}=d_{1}\ldots d_{k}\in\\{0,1\\}^{*}$ with $k\geq 2$,
we have

$\left\\{\begin{array}[]{lll}\Phi_{\mathbf{s}}({\mathbf{d}}^{-})=\Phi_{\mathbf{s}}({\mathbf{d}})^{-}&\textrm{if}&d_{k}=1,\\\
\Phi_{\mathbf{s}}({\mathbf{d}}^{+})=\Phi_{\mathbf{s}}({\mathbf{d}})^{+}&\textrm{if}&d_{k}=0.\end{array}\right.$

  3. (iii)

For any two sequences
${\mathbf{c}},{\mathbf{d}}\in\left\\{0,1\right\\}^{\mathbb{N}}$, we have the
implications

$\sigma^{n}({\mathbf{c}})\prec{\mathbf{d}}~{}\forall n\geq
0\quad\Longrightarrow\quad\sigma^{n}(\Phi_{\mathbf{s}}({\mathbf{c}}))\prec\Phi_{\mathbf{s}}({\mathbf{d}})~{}\forall
n\geq 0$

and

$\sigma^{n}({\mathbf{c}})\succ{\mathbf{d}}~{}\forall n\geq
0\quad\Longrightarrow\quad\sigma^{n}(\Phi_{\mathbf{s}}({\mathbf{c}}))\succ\Phi_{\mathbf{s}}({\mathbf{d}})~{}\forall
n\geq 0.$

  4. (iv)

For any ${\mathbf{r}}\in L^{*}$, we have ${\mathbf{s}}\bullet{\mathbf{r}}\in
L^{*}$ and
$\mathbb{L}({\mathbf{s}}\bullet{\mathbf{r}})={\mathbf{s}}\bullet\mathbb{L}({\mathbf{r}})$.

  5. (v)

For any ${\mathbf{r}}\in L^{*}$ and
${\mathbf{t}}\in\\{0,1\\}^{\mathbb{N}}\cup\\{0,1\\}^{*}$, we have
$({\mathbf{r}}\bullet{\mathbf{s}})\bullet{\mathbf{t}}={\mathbf{r}}\bullet({\mathbf{s}}\bullet{\mathbf{t}})$.

Observe that the map $\Phi_{\mathbf{s}}$ is not a homomorphism on
$\\{0,1\\}^{*}$: For instance,
$\Phi_{\mathbf{s}}(00)={\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})\neq{\mathbf{s}}^{-}{\mathbf{s}}^{-}=\Phi_{\mathbf{s}}(0)\Phi_{\mathbf{s}}(0)$.
Nonetheless, $\Phi_{\mathbf{s}}$ is a partial homomorphism in the sense
described below.

Say a finite or infinite sequence of words
${\mathbf{b}}_{1},\dots,{\mathbf{b}}_{n}$ or
${\mathbf{b}}_{1},{\mathbf{b}}_{2},\dots$ is connectible if for each $i$, the
last digit of ${\mathbf{b}}_{i}$ differs from the first digit of
${\mathbf{b}}_{i+1}$. Thus, for instance, the sequence $1101,00111$ is
connectible whereas the sequence $11010,0111$ is not.

######  Lemma 2.13.

([4, Section 3])

  1. (i)

Let ${\mathbf{b}}_{1},{\mathbf{b}}_{2},\dots$ be a (finite or infinite)
connectible sequence of words. Then for any ${\mathbf{s}}\in L^{*}$,

$\Phi_{\mathbf{s}}({\mathbf{b}}_{1}{\mathbf{b}}_{2}\dots)=\Phi_{\mathbf{s}}({\mathbf{b}}_{1})\Phi_{\mathbf{s}}({\mathbf{b}}_{2})\dots.$

  2. (ii)

Let ${\mathbf{s}},{\mathbf{r}}\in L^{*}$. Then
$\Phi_{\mathbf{s}}({\mathbf{r}}^{\infty})=\Phi_{\mathbf{s}}({\mathbf{r}})^{\infty}$
and
$\Phi_{\mathbf{s}}(\mathbb{L}({\mathbf{r}})^{\infty})=\Phi_{\mathbf{s}}(\mathbb{L}({\mathbf{r}}))^{\infty}$.

Now for $k\in\mathbb{N}$ define

(2.9)
$\Lambda_{k}:=\left\\{{\mathbf{S}}={\mathbf{s}}_{1}\bullet{\mathbf{s}}_{2}\bullet\cdots\bullet{\mathbf{s}}_{k}:\
{\mathbf{s}}_{i}\in F^{*}\textrm{ for any }1\leq i\leq k\right\\},$

and set

$\Lambda:=\bigcup_{k=1}^{\infty}\Lambda_{k}.$

If ${\mathbf{S}}\in\Lambda_{k}$, we call $k$ the degree of ${\mathbf{S}}$.
Note that $F^{*}\subseteq\Lambda\subseteq L^{*}$ by Lemma 2.12 (iv). Both
inclusions are proper, e.g. $001011=01\bullet 001\in\Lambda\backslash F^{*}$
and $0010111\in L^{*}\backslash\Lambda$.

We recall some terminology and facts from [4]. For ${\mathbf{S}}\in\Lambda$,
we call the interval
$I^{\mathbf{S}}:=[\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}}]$ given
implicitly by

$\alpha(\beta_{\ell}^{\mathbf{S}})=\mathbb{L}({\mathbf{S}})^{\infty}\qquad\mbox{and}\qquad\alpha(\beta_{*}^{\mathbf{S}})=\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{-}\mathbb{L}({\mathbf{S}})^{\infty},$

a basic interval generated by the Lyndon word ${\mathbf{S}}$. We also define
an interval
$J^{\mathbf{S}}:=[\beta_{\ell}^{\mathbf{S}},\beta_{r}^{\mathbf{S}}]$, where
$\beta_{r}^{\mathbf{S}}$ is given by

$\alpha(\beta_{r}^{\mathbf{S}})=\mathbb{L}({\mathbf{S}})^{+}{\mathbf{S}}^{\infty}.$

The interval $J^{\mathbf{S}}$ is called a Lyndon interval generated by
${\mathbf{S}}$. For any ${\mathbf{S}}\in\Lambda$, the intervals
$J^{{\mathbf{S}}\bullet{\mathbf{r}}},{\mathbf{r}}\in F^{*}$ are pairwise
disjoint and contained in $J^{\mathbf{S}}\setminus I^{\mathbf{S}}$; let

(2.10) $E^{\mathbf{S}}:=(J^{\mathbf{S}}\setminus
I^{\mathbf{S}})\setminus\bigcup_{{\mathbf{r}}\in
F^{*}}J^{{\mathbf{S}}\bullet{\mathbf{r}}}.$

It is a consequence of [4, Proposition 5.4] that $\beta\in E^{\mathbf{S}}$ if
and only if $\alpha(\beta)=\Phi_{\mathbf{S}}(\alpha(\hat{\beta}))$ for some
$\hat{\beta}\in E$. In this case we say that $\beta$ is renormalizable222Our
definition differs slightly from that of Hubbard and Sparrow [21], who call
the above notion properly renormalizable and consider $\beta\in E$ to be
trivially renormalizable. by the word ${\mathbf{S}}$.

We further define the set

(2.11)
$E_{\infty}:=\bigcap_{k=1}^{\infty}\bigcup_{{\mathbf{S}}\in\Lambda_{k}}J^{{\mathbf{S}}}.$

These are the values of $\beta$ that lie in infinitely many of the intervals
$J^{{\mathbf{S}}}$, and are hence infinitely renormalizable. (In [4],
$E_{\infty}$ is called the infinitely Farey set, because its elements arise
from substitutions of an infinite sequence of Farey words.) The set
$E_{\infty}$ is uncountable but of zero Hausdorff dimension. By [4, Theorem
3], we have the decomposition in disjoint sets,

(2.12) $(1,2]=E\cup
E_{\infty}\cup\bigcup_{{\mathbf{S}}\in\Lambda}E^{{\mathbf{S}}}\cup\bigcup_{{\mathbf{S}}\in\Lambda}I^{{\mathbf{S}}}.$

In this paper, different parts of this decomposition require different
analysis, with the case of $\beta$ in a basic interval being the most
involved.

######  Remark 2.14.

Our presentation of the substitution map $\Phi_{\mathbf{s}}$ and its
properties differs somewhat from the substitutions used by most authors, but
can be seen to be equivalent. Precisely, let $\mathbf{w}_{+}$ (resp.
$\mathbf{w}_{-}$) be the word ${\mathbf{s}}$ (resp.
$\mathbb{L}({\mathbf{s}})$) shifted cyclically one place to the right.
Alternatively, $\mathbf{w}_{-}$ is the largest cyclic permutation of
${\mathbf{s}}$ beginning with $0$, and $\mathbf{w}_{+}$ is the smallest cyclic
permutation of ${\mathbf{s}}$ beginning with $1$; see [19]. For example, if
${\mathbf{s}}=01011$, then $\mathbf{w}_{+}=10110$ and $\mathbf{w}_{-}=01101$.
If $\mathbf{y}=y_{1}y_{2}\dots$ is a sequence beginning with $1$, then
$0\Phi_{\mathbf{s}}(\mathbf{y})$ is a concatenation of words from
$\\{\mathbf{w}_{+},\mathbf{w}_{-}\\}$, where the first block is
$\mathbf{w}_{-}$ and for each $i\geq 2$, the $i$th block is $\mathbf{w}_{-}$
if $y_{i-1}=0$, and $\mathbf{w}_{+}$ if $y_{i-1}=1$. Similarly, if
$\mathbf{y}$ is a sequence beginning with $0$, then
$1\Phi_{\mathbf{s}}(\mathbf{y})$ is also such a concatenation, but beginning
instead with $\mathbf{w}_{+}$.

Although the approach using the words $\mathbf{w}_{+}$ and $\mathbf{w}_{-}$
has been the traditional way to express renormalizations (see [21]), we prefer
here the map $\Phi_{\mathbf{s}}$ because (i) it emphasizes that the
renormalization is parametrized by a single word rather than two different
ones; (ii) it emphasizes the importance of Lyndon and Farey words; and (iii)
it greatly facilitates operations on finite words, which are used frequently
in future sections. On the other hand, our map $\Phi_{\mathbf{s}}$ is
essentially equivalent (up to the above-mentioned cyclical shift) to the map
$\rho_{r}$ of Glendinning and Sidorov [19], where
$r=|{\mathbf{s}}|_{1}/|{\mathbf{s}}|$. The maps $\rho_{r}$ are useful for
studying the sets $\Omega_{\mathbf{a},{\mathbf{b}}}$, but for dealing with the
sets $\Sigma_{\mathbf{a},{\mathbf{b}}}$ we find the maps $\Phi_{\mathbf{s}}$
more practical, despite their somewhat clunkier definition.

We point out also that it is possible to write the substitution maps
$\Phi_{\mathbf{S}}$, ${\mathbf{S}}\in\Lambda$ completely in terms of just
three basic substitutions, namely

$L:\begin{cases}0&\mapsto 0\\\ 1&\mapsto 10\end{cases},\qquad
M:\begin{cases}0&\mapsto 01\\\ 1&\mapsto 10\end{cases},\qquad
R:\begin{cases}0&\mapsto 01\\\ 1&\mapsto 1\end{cases},$

as was done by Komornik, Steiner and Zou [24] and before that (with different
notation) by Labarca and Moreira [25].

##  3\. Characterization of the bifurcation set $\mathscr{E}_{\beta}$

In the introduction we have followed the notation and definitions of Baker and
Kong [7]. However, we point out that $K_{\beta}(t)$ is slightly different from
the set denoted $K_{\beta}(t)$ in [22], and as a result, the bifurcation set
$\mathscr{E}_{\beta}$ defined in (1.2) is slightly different from the set
called $E_{\beta}$ in [22]. In fact, setting
$\check{K}_{\beta}(t):=\\{x\in[0,1):T_{\beta}^{n}(x)\not\in(0,t)\ \forall
n\geq 0\\}$, we have

(3.1)
$E_{\beta}=\left\\{t\in[0,1):\check{K}_{\beta}(t^{\prime})\neq\check{K}_{\beta}(t)~{}\forall
t^{\prime}>t\right\\}.$

It is also convenient to define the set

$\mathscr{E}_{\beta}^{+}:=\\{t\in[0,1):T_{\beta}^{n}(t)\geq t\ \forall n\geq
0\\}=\\{t\in[0,1):t\in K_{\beta}(t)\\}.$

(This is the set denoted $E_{\beta}^{+}$ in [22].) The following
characterization will play an important role throughout this article.

######  Lemma 3.1.

For each $\beta\in(1,2]$, we have

$\mathscr{E}_{\beta}=\left\\{t\in[0,1):K_{\beta}(t-\varepsilon)\neq
K_{\beta}(t+\varepsilon)~{}\forall\varepsilon>0\right\\}=\mathscr{E}_{\beta}^{+}.$

###### Proof.

If $t\in\mathscr{E}_{\beta}^{+}$, the definition of $K_{\beta}(t)$ implies
that $t\in K_{\beta}(t)\setminus K_{\beta}(t+\varepsilon)$ for any
$\varepsilon>0$, and so $t\in\mathscr{E}_{\beta}$.

Next, if $t\in\mathscr{E}_{\beta}$, then $K_{\beta}(t+\varepsilon)\neq
K_{\beta}(t)$ for all $\varepsilon>0$, so certainly
$K_{\beta}(t-\varepsilon)\neq K_{\beta}(t+\varepsilon)$ for all
$\varepsilon>0$.

Finally, suppose $t\not\in\mathscr{E}_{\beta}^{+}$. Then there is an
$n_{0}\in\mathbb{N}$ such that $T_{\beta}^{n_{0}}(t)<t$. By the continuity of
$T_{\beta}^{n_{0}}$ there is a sufficiently small $\varepsilon>0$ such that

(3.2) $T_{\beta}^{n_{0}}(x)<t-\varepsilon\quad\forall
x\in[t-\varepsilon,t+\varepsilon].$

Suppose there exists $x_{0}\in K_{\beta}(t-\varepsilon)\setminus
K_{\beta}(t+\varepsilon)$. Then there is an integer $n_{1}$ such that
$t-\varepsilon\leq T_{\beta}^{n_{1}}(x_{0})<t+\varepsilon$. Thus,
$T_{\beta}^{n_{0}+n_{1}}(x_{0})<t-\varepsilon$ by (3.2), contradicting that
$x_{0}\in K_{\beta}(t-\varepsilon)$. Hence,
$K_{\beta}(t-\beta)=K_{\beta}(t+\varepsilon)$. ∎

By a similar argument it can be shown that
$E_{\beta}=\left\\{t\in[0,1):T_{\beta}^{n}(t)\notin(0,t)~{}\forall n\geq
0\right\\}$. Thus, $\mathscr{E}_{\beta}\subseteq E_{\beta}$, and
$E_{\beta}\backslash\mathscr{E}_{\beta}$ is at most countable. As a result,
when considering questions of Hausdorff dimension the two sets may be freely
interchanged. Because of this, (1.3) is indeed equivalent to the conjecture of
Kalle et al.

##  4\. The key ideas

In this section we outline the main idea of the proof of Theorem 1.1. The key
is to prove the more technical Theorem 4.4 below; its proof takes up much of
the rest of the paper. First, we define the concepts appearing in the theorem.

The metric $d_{2}$ on $\\{0,1\\}^{\mathbb{N}}$ defined by

$d_{2}((x_{i}),(y_{i})):=2^{-\inf\\{i:x_{i}\neq
y_{i}\\}},\qquad(c_{i}),(d_{i})\in\\{0,1\\}^{\mathbb{N}},$

induces a Hausdorff dimension on $\\{0,1\\}^{\mathbb{N}}$, which we also
denote by $\dim_{H}$.

Next, we recall the definition of the $\beta$-shift from (1.5). It was shown
recently by Li [26, Theorem 1.1] that

(4.1) $\dim_{H}\pi_{\beta}(A)=\frac{\log
2}{\log\beta}\dim_{H}A\qquad\forall\,A\subseteq\Sigma_{\beta}.$

######  Definition 4.1.

Let $\mathcal{T}\subseteq(0,1)$ be an index set. A collection
$\\{\mathcal{X}(t):t\in\mathcal{T}\\}$ is a strictly descending collection of
subshifts if for any $t,t^{\prime}\in\mathcal{T}$ with $t<t^{\prime}$,
$\mathcal{X}(t^{\prime})$ is a proper subshift of $\mathcal{X}(t)$.

######  Definition 4.2.

A (finite or infinite) sequence $\mathcal{I}=(I_{1},I_{2},\dots)$ of intervals
is left-progressing if for each $k\geq 1$, $\sup I_{k+1}<\inf I_{k}$.

######  Definition 4.3.

Let $\mathcal{Y}$ be a subshift of $\\{0,1\\}^{\mathbb{N}}$, and $\mathcal{X}$
a subshift of $\mathcal{Y}$. We say $\mathcal{X}$ has full entropy in
$\mathcal{Y}$ if $h(\mathcal{X})=h(\mathcal{Y})$. Likewise, we say
$\mathcal{X}$ has full Hausdorff dimension in $\mathcal{Y}$ if
$\dim_{H}\mathcal{X}=\dim_{H}\mathcal{Y}$.

Recall from the Introduction that a $\beta$-Lyndon interval is an interval
$[t_{L},t_{R}]$ determined by

$b(t_{L},\beta)=\mathbf{w}0^{\infty},\qquad
b(t_{R},\beta)=\mathbf{w}^{\infty},$

where $\mathbf{w}$ is a $\beta$-Lyndon word. (Recall this means $\mathbf{w}$
is Lyndon and $\sigma^{n}(\mathbf{w}^{\infty})\prec\alpha(\beta)$ for all
$n\geq 0$.) We denote the set of all $\beta$-Lyndon words by $L^{*}(\beta)$.
We denote by $\mathcal{T}_{R}(\beta)$ the set of all right endpoints of
$\beta$-Lyndon intervals in $[0,\tau(\beta)]$, where $\tau(\beta)$ was defined
in (1.1).

######  Theorem 4.4.

For each $\beta\in(1,2]$ there is a (finite or infinite, possibly empty) left-
progressing sequence $\mathcal{I}$ of intervals such that

  1. (i)

For any $t_{R}\in\mathcal{T}_{R}(\beta)\backslash\bigcup_{I\in\mathcal{I}}I$,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ has a transitive subshift
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ of full entropy and full Hausdorff
dimension that contains the sequence $b(t_{R},\beta)$;

  2. (ii)

The collection

$\\{\mathcal{K}_{\beta}^{\prime}(t_{R}):t_{R}\in\mathcal{T}_{R}(\beta)\backslash\textstyle{\bigcup_{I\in\mathcal{I}}I}\\}$

is a strictly descending collection of subshifts;

  3. (iii)

The entropy function $t\mapsto h(\widetilde{\mathcal{K}}_{\beta}(t))$ is
constant throughout each interval $I\in\mathcal{I}$.

Theorem 4.4 describes the most general case; for many values of $\beta$ the
situation is simpler. In the following sections, we will prove Theorem 4.4 for
different cases of $\beta$. But first, we show how it implies Theorem 1.1.

######  Lemma 4.5.

Let $t\in\mathscr{E}_{\beta}\cap(0,1)$. Then $t$ is either the right endpoint
of a $\beta$-Lyndon interval, or $t$ is the limit from the left of a sequence
of right endpoints of $\beta$-Lyndon intervals.

###### Proof.

By Lemma 3.1, $b(t,\beta)\preccurlyeq\sigma^{n}(b(t,\beta))\prec\alpha(\beta)$
for all $n\geq 0$. If $b(t,\beta)$ is periodic, say $b(t,\beta)=(b_{1}\dots
b_{k})^{\infty}$ where $k$ is the minimal period of $b(t,\beta)$, then
$b_{1}\dots b_{k}\in L^{*}(\beta)$, and hence $t$ is the right endpoint of a
$\beta$-Lyndon interval.

Assume now that $b(t,\beta)=b_{1}b_{2}\dots$ is not periodic. Then by [22,
Lemma 3.5], there are infinitely many integers $k$ such that $b_{1}\dots
b_{k}$ is Lyndon. But then $b_{1}\dots b_{k}$ is in fact $\beta$-Lyndon, since
for each $n\geq 0$ there is an integer $j<k$ such that

$\sigma^{n}\big{(}(b_{1}\dots b_{k})^{\infty}\big{)}=b_{j+1}\dots
b_{k}(b_{1}\dots
b_{k})^{\infty}\preccurlyeq\sigma^{j}(b(t,\beta))\prec\alpha(\beta).$

Since $(b_{1}\dots b_{k})^{\infty}\to b(t,\beta)$ in the order topology as
$k\to\infty$ and $(b_{1}\dots b_{k})^{\infty}\prec b(t,\beta)$ for all $k$ by
the assumption that $t\in\mathscr{E}_{\beta}$, it follows that $t$ is the
limit from the left of a sequence of right endpoints of $\beta$-Lyndon
intervals. ∎

###### Proof of Theorem 1.1, assuming Theorem 4.4.

The proof is based roughly on the argument in the proof of [7, Lemma 3.8], but
requires several new ideas. Observe that we only need to prove the “$\geq$”
inequality, since $\mathscr{E}_{\beta}\cap[t,1)\subseteq K_{\beta}(t)$ by
Lemma 3.1. Furthermore, it suffices to establish the inequality for
$t\in\mathscr{B}_{\beta}$: For arbitrary $t\in(0,1)$, let
$t^{\prime}:=\inf(\mathscr{B}_{\beta}\cap[t,1])$; we then have

$\displaystyle\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1])$
$\displaystyle\geq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t^{\prime},1])=\dim_{H}K_{\beta}(t^{\prime})$
$\displaystyle=\dim_{H}K_{\beta}(t+)=\dim_{H}K_{\beta}(t),$

where the last equality follows from the continuity of
$t\mapsto\dim_{H}K_{\beta}(t)$.

Observe that each $t\in\mathscr{B}_{\beta}$ can be approximated from the right
by a sequence of right endpoints of $\beta$-Lyndon intervals
$[t_{L}^{(n)},t_{R}^{(n)}],n\in\mathbb{N}$. (If this were not the case, then
some right neighborhood $(t,t+\varepsilon)$ would not contain any points of
$\mathscr{E}_{\beta}$ by Lemma 4.5, hence $(t,t+\varepsilon)$ would not
contain any points of $\mathscr{B}_{\beta}$. But every point of
$\mathscr{B}_{\beta}$ is an accumulation point from the right of
$\mathscr{B}_{\beta}$ by the continuity of $t\mapsto\dim_{H}K_{\beta}(t)$.) We
can moreover choose the sequence $(t_{R}^{(n)})$ so that
$\dim_{H}K_{\beta}(t_{R}^{(n)})$ is strictly increasing in $n$. Thus, we have

$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R}^{(n)})\big{)}<h(\widetilde{\mathcal{K}}_{\beta}(t))\quad\forall\,n\qquad\mbox{and}\qquad
h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R}^{(n)})\big{)}\nearrow
h(\widetilde{\mathcal{K}}_{\beta}(t))\quad\mbox{as $n\to\infty$}.$

Since the sequence $\mathcal{I}$ of intervals in Theorem 4.4 is left-
progressing, there is at most one point all of whose neighborhoods intersect
infinitely many of the intervals in $\mathcal{I}$; if such a point exists, it
lies to the left of all the intervals in $\mathcal{I}$ and we denote it by
$t_{0}$.

We assume first that $t\neq t_{0}$. Then, except for finitely many values of
$n$, the points $t_{R}^{(n)}$ do not lie in any interval from the collection
$\mathcal{I}$, because $t\in\mathscr{B}_{\beta}$ and entropy is constant on
each interval of $\mathcal{I}$ by Theorem 4.4 (iii). Hence, by Theorem 4.4 (i)
it follows that for all large enough $n$,
$\widetilde{\mathcal{K}}_{\beta}(t_{R}^{(n)})$ has a transitive subshift
$\mathcal{K}_{\beta}^{\prime}(t_{R}^{(n)})$ of full Hausdorff dimension
containing the sequence $b(t_{R}^{(n)},\beta)$.

Write $\alpha(\beta)=\alpha_{1}\alpha_{2}\dots$. First assume
$t\in\mathscr{B}_{\beta}$ such that $t=t_{R}$ for a $\beta$-Lyndon interval
$[t_{L},t_{R}]$ generated by the $\beta$-Lyndon word $\mathbf{w}=w_{1}\dots
w_{p}$, so $b(t_{R},\beta)=\mathbf{w}^{\infty}$. Since the greedy expansion
$b(t,\beta)$ is right-continuous in $t$, we may assume that $t_{R}^{(n)}$ is
close enough to $t_{R}$ so that $b(t_{R}^{(n)},\beta)$ begins with
$\mathbf{w}$. Furthermore, since $b(t_{R}^{(n)},\beta)\succ
b(t_{R},\beta)=\mathbf{w}^{\infty}$, there is for each $n$ a sufficiently
large integer $k_{n}$ such that

(4.2) $b(t_{R}^{(n)},\beta)\succ\mathbf{w}^{k_{n}}1^{\infty}.$

Let

$\mathcal{F}_{n}:=\left\\{\mathbf{w}^{k_{n}}\mathbf{x}:\mathbf{x}=x_{1}x_{2}\dots\in\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)},x_{1}\dots
x_{p}=\mathbf{w},\ \mbox{and}\ \sigma^{j}(\mathbf{x})\prec\alpha(\beta)\
\forall\,j\geq 0\right\\},$

and

$F_{n}:=\pi_{\beta}(\mathcal{F}_{n}).$

Take $\mathbf{w}^{k_{n}}\mathbf{x}\in\mathcal{F}_{n}$. First we show that

(4.3) $\sigma^{j}(\mathbf{w}^{k_{n}}\mathbf{x})\prec\alpha(\beta)\quad\forall
j\geq 0.$

Since $\sigma^{l}(\mathbf{x})\prec\alpha(\beta)$ for all $l\geq 0$, (4.3)
holds for $j\geq k_{n}|\mathbf{w}|$. Observing that $\mathbf{x}$ begins with
$\mathbf{w}$, the inequality will follow for all $j<k_{n}|\mathbf{w}|$ once we
prove that

(4.4) $\mathbb{L}(\mathbf{w})=\mathbb{L}(w_{1}\ldots
w_{p})\prec\alpha_{1}\ldots\alpha_{p}.$

Since $\sigma^{n}(\mathbf{w}^{\infty})\prec\alpha(\beta)$ for all $n\geq 0$,
$\mathbb{L}(\mathbf{w})\preccurlyeq\alpha_{1}\ldots\alpha_{p}$. If
$\mathbb{L}(\mathbf{w})=\alpha_{1}\ldots\alpha_{p}$, then for some $i\leq p$,

$(\alpha_{1}\ldots\alpha_{p})^{\infty}=\sigma^{i}(\mathbf{w}^{\infty})\prec\alpha(\beta).$

But then $\alpha_{1}\ldots\alpha_{p}\prec\alpha_{\ell+1}\ldots\alpha_{\ell+p}$
for some $\ell\geq 0$, contradicting Lemma 2.1. Therefore, (4.4) holds, and
thus we have (4.3).

Since $\mathbf{x}\in\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$, we
have $\mathbf{x}\succcurlyeq b(t_{R}^{(n)},\beta)\succ\mathbf{w}^{\infty}$.
Furthermore, by (4.3), $\mathbf{w}^{k_{n}}\mathbf{x}$ is a greedy
$\beta$-expansion. Since $\sigma^{j}(\mathbf{w}^{\infty})\prec\alpha(\beta)$
for each $j\geq 0$, $\mathbf{w}^{\infty}$ is also a greedy $\beta$-expansion.
Thus,

$\pi_{\beta}(\mathbf{w}^{k_{n}}\mathbf{x})\geq\pi_{\beta}(\mathbf{w}^{\infty})=t_{R}.$

Furthermore, using (4.2),

$\sigma^{j}(\mathbf{x})\succcurlyeq
b(t_{R}^{(n)},\beta)\succ\mathbf{w}^{k_{n}}1^{\infty}\succ\mathbf{w}^{k_{n}}\mathbf{x}\qquad\forall
j\geq 0,$

and $w_{j+1}\dots w_{p}\succ w_{1}\dots w_{p-j}$ for all $1\leq j<p$ since
$\mathbf{w}$ is Lyndon. Hence

$\sigma^{j}(\mathbf{w}^{k_{n}}\mathbf{x})\succcurlyeq\mathbf{w}^{k_{n}}\mathbf{x}\qquad\forall
j\geq 0.$

We conclude that

(4.5) $F_{n}\subseteq\mathscr{E}_{\beta}\cap[t_{R},1].$

Next, observe that the condition $\sigma^{j}(\mathbf{x})\prec\alpha(\beta)\
\forall\,j\geq 0$ in the definition of $\mathcal{F}_{n}$ removes at most a
countable number of sequences from
$\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$. Since
$\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$ is transitive and
$\mathbf{w}$ is allowed in
$\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$, there is for each
sequence $\mathbf{z}\in\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$
a word $\mathbf{v}=\mathbf{v}(\mathbf{z})$ such that
$\mathbf{w}\mathbf{v}\mathbf{z}\in\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$.
For each finite word $\mathbf{v}\in\\{0,1\\}^{*}$, denote by
$\mathcal{X}_{\mathbf{v}}$ the set of all sequences
$\mathbf{z}\in\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$ such that
$\mathbf{v}(\mathbf{z})=\mathbf{v}$. Then

$\bigcup_{\mathbf{v}}\mathcal{X}_{\mathbf{v}}=\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}\qquad\mbox{and}\qquad\mathcal{F}_{n}\supseteq\left(\bigcup_{\mathbf{v}}\mathbf{w}^{k_{n}+1}\mathbf{v}\mathcal{X}_{\mathbf{v}}\right)\backslash\mathcal{C},$

where $\mathcal{C}$ is a countable set. (Here our stronger definition of
transitive is particularly convenient.) Hence,

$\dim_{H}\mathcal{F}_{n}\geq\sup_{\mathbf{v}}\dim_{H}\mathcal{X}_{\mathbf{v}}=\dim_{H}\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}=\dim_{H}\widetilde{\mathcal{K}}_{\beta}\big{(}t_{R}^{(n)}\big{)},$

where the last equality follows since
$\mathcal{K}_{\beta}^{\prime}\big{(}t_{R}^{(n)}\big{)}$ has full Hausdorff
dimension in $\widetilde{\mathcal{K}}_{\beta}\big{(}t_{R}^{(n)}\big{)}$. Since
all of the sets involved are subsets of $\Sigma_{\beta}$, it follows from
(4.1) that

$\dim_{H}F_{n}=\dim_{H}\pi_{\beta}(\mathcal{F}_{n})\geq\dim_{H}\pi_{\beta}\left(\widetilde{\mathcal{K}}_{\beta}\big{(}t_{R}^{(n)}\big{)}\right)=\dim_{H}K_{\beta}\big{(}t_{R}^{(n)}\big{)}.$

Thus, together with (4.5), we obtain

$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t_{R},1]\big{)}\geq\dim_{H}K_{\beta}\big{(}t_{R}^{(n)}\big{)}.$

Letting $t_{R}^{(n)}\searrow t_{R}$ gives
$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t_{R},1]\big{)}\geq\dim_{H}K_{\beta}(t_{R})$,
in view of the continuity of $t\mapsto\dim_{H}K_{\beta}(t)$.

Finally, for arbitrary $t\in\mathscr{B}_{\beta}$ (including the accumulation
point $t=t_{0}$ if it exists), $t$ can be approximated from the right by a
sequence $\big{(}t_{R}^{(n)}\big{)}$ of right endpoints of $\beta$-Lyndon
intervals, none of which are equal to $t_{0}$; so

$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}\geq\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t_{R}^{(n)},1]\big{)}\geq\dim_{H}K_{\beta}\big{(}t_{R}^{(n)}\big{)},$

and, again by the continuity of the map $t\mapsto\dim_{H}K_{\beta}(t)$, it
follows that
$\dim_{H}\big{(}\mathscr{E}_{\beta}\cap[t,1]\big{)}\geq\dim_{H}K_{\beta}(t)$.
∎

In this proof we did not use property (ii) of Theorem 4.4. However, it will be
needed later in the proof of Theorem 1.2. The following result illustrates the
main idea.

######  Proposition 4.6.

Let $\beta\in(1,2]$, and suppose that

  1. (i)

For any $t_{R}\in\mathcal{T}_{R}(\beta)$,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ has a transitive sofic subshift
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ of full entropy;

  2. (ii)

The collection
$\\{\mathcal{K}_{\beta}^{\prime}(t_{R}):t_{R}\in\mathcal{T}_{R}(\beta)\\}$ is
a strictly descending collection of subshifts; and

  3. (iii)

The $\beta$-Lyndon intervals are dense in $[0,\tau(\beta)]$.

Then

$\mathscr{E}_{\beta}\cap[0,\tau(\beta))=\mathscr{B}_{\beta}\cap[0,\tau(\beta)).$

###### Proof.

We show that the plateaus of $\delta_{\beta}:t\mapsto\dim_{H}K_{\beta}(t)$ in
$[0,\tau(\beta)]$ are precisely the $\beta$-Lyndon intervals. This gives the
desired conclusion, since $\mathscr{E}_{\beta}$ intersects the $\beta$-Lyndon
intervals only in their right endpoints, and the $\beta$-Lyndon intervals are
dense in $[0,\tau(\beta)]$.

It is enough to show that if $[t_{L},t_{R}]$ and $[u_{L},u_{R}]$ are two
distinct $\beta$-Lyndon intervals with $t_{R}<u_{R}$, then
$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R})\big{)}>h\big{(}\widetilde{\mathcal{K}}_{\beta}(u_{R})\big{)}$.
By hypotheses (i) and (ii), $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is sofic and
transitive, and $\mathcal{K}_{\beta}^{\prime}(u_{R})$ is a proper subshift of
$\mathcal{K}_{\beta}^{\prime}(t_{R})$. Hence, by [27, Corollary 4.4.9],

$h\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R})\big{)}=h\big{(}\mathcal{K}_{\beta}^{\prime}(t_{R})\big{)}>h\big{(}\mathcal{K}_{\beta}^{\prime}(u_{R})\big{)}=h\big{(}\widetilde{\mathcal{K}}_{\beta}(u_{R})\big{)}.$

This completes the proof. ∎

As we will see later, the hypotheses of Proposition 4.6 are satisfied when
$\beta\in\\{\beta_{\ell}^{\mathbf{S}},\beta_{*}^{\mathbf{S}},\beta_{r}^{\mathbf{S}}\\}$,
for any ${\mathbf{S}}\in\Lambda$. Thus, for these values of $\beta$, any
points of $\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ must lie to the
right of $\tau(\beta)$; see Theorem 8.1.

##  5\. The case $\beta\in E_{L}$

We begin with the case when $\beta\in E_{L}$. Our main result in this section
is:

######  Theorem 5.1.

Let $\beta\in E_{L}$. Then $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is
transitive for any right endpoint $t_{R}$ of a $\beta$-Lyndon interval
$[t_{L},t_{R}]$ with $t_{R}<\tau(\beta)=1-(1/\beta)$. In other words, the
conclusion of Theorem 4.4 holds with $\mathcal{I}=\emptyset$ and
$\mathcal{K}_{\beta}^{\prime}(t_{R})=\widetilde{\mathcal{K}}_{\beta}(t_{R})$
for each $t_{R}\in\mathcal{T}_{R}$.

We will prove Theorem 5.1 after developing some lemmas.

######  Lemma 5.2.

Let $1<\beta<2$, and $0<t<1-1/\beta$. Then for every sequence
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t)$ there exist a word
$\mathbf{u}$ beginning with $0$ and a word $\mathbf{v}$ beginning with $1$
such that
$\mathbf{u}\mathbf{z},\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t)$.

###### Proof.

Let $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t)$. Suppose
$1\mathbf{z}\not\in\widetilde{\mathcal{K}}_{\beta}(t)$. Then
$1\mathbf{z}\succ\alpha(\beta)$, and hence $0\mathbf{z}\succ
0\sigma(\alpha(\beta))=b(1-1/\beta,\beta)\succ b(t,\beta)$. Therefore,
$0\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t)$. So $\mathbf{z}$ can always
be extended to the left.

Let $N$ be the number of consecutive $1$’s at the beginning of
$\alpha(\beta)$, and $M$ be the number of consecutive $0$’s at the beginning
of $b(t,\beta)$. Then $N,M\geq 1$ since $\beta>1$ and $t<1-1/\beta<1/\beta$.
Clearly, no sequence in $\widetilde{\mathcal{K}}_{\beta}(t)$ can have more
than $N$ consecutive $1$’s or $M$ consecutive $0$’s anywhere. Hence, the lemma
follows. ∎

Recall that $\mathcal{L}(\mathcal{X})$ denotes the language of the subshift
$\mathcal{X}$, and $\mathcal{X}$ is transitive if for any word
$\mathbf{u}\in\mathcal{L}(\mathcal{X})$ and sequence
$\mathbf{z}\in\mathcal{X}$, there is a word $\mathbf{v}$ such that
$\mathbf{u}\mathbf{v}\mathbf{z}\in\mathcal{X}$. The next definition and lemma
will be used to simplify transitivity proofs in this section, by removing the
word $\mathbf{u}$ from consideration.

If $b(t,\beta)=\mathbf{w}^{\infty}$ for a Lyndon word $\mathbf{w}$, we also
write $\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$ instead of
$\widetilde{\mathcal{K}}_{\beta}(t)$. So

$\widetilde{\mathcal{K}}_{\beta}(\mathbf{w}):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\
\forall n\geq 0\\}.$

######  Definition 5.3.

Let $\beta\in(1,2]$ and write $\alpha(\beta)=\alpha_{1}\alpha_{2}\ldots$. We
say a Lyndon word $\mathbf{w}$ has Property TL for $\beta$ if for each
sequence $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$ there is a
word $\mathbf{v}=v_{1}\dots v_{n}$ such that

  1. (i)

$\mathbf{v}\prec\alpha_{j+1}\dots\alpha_{j+n}$ for all $j\geq 0$; and

  2. (ii)

$\mathbf{w}^{\infty}\preccurlyeq\sigma^{i}(\mathbf{v}\mathbf{z})\preccurlyeq\alpha(\beta)$
for all $i\geq 0$, i.e.
$\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$.

######  Lemma 5.4.

Let $\beta\in\overline{E}$, let $[t_{L},t_{R}]$ be a $\beta$-Lyndon interval
with $t_{R}<1-(1/\beta)$, and let $\mathbf{w}=w_{1}\dots w_{m}\in
L^{*}(\beta)$ such that $b(t_{R},\beta)=\mathbf{w}^{\infty}$. If $\mathbf{w}$
satisfies Property TL, then $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is
transitive.

###### Proof.

We first show that

(5.1)
$0\sigma(\alpha(\beta))=0\alpha_{2}\alpha_{3}\dots\preccurlyeq\sigma^{j}(\alpha(\beta))\qquad\forall\,j\geq
0.$

This is trivial for $j=0$, so assume $j\geq 1$. Suppose the contrary; then
there is an index $k\geq 2$ such that
$0\alpha_{2}\dots\alpha_{k-1}=\alpha_{j+1}\dots\alpha_{j+k-1}$ and
$\alpha_{k}=1$, $\alpha_{j+k}=0$. But then the word
$\alpha_{1}\dots\alpha_{k}$ has two more $1$’s than the word
$\alpha_{j+1}\dots\alpha_{j+k}$ of the same length, contradicting that
$\alpha(\beta)$ is balanced. Therefore, we have (5.1).

Let $\mathbf{u}=u_{1}\dots
u_{n}\in\mathcal{L}\big{(}\widetilde{\mathcal{K}}_{\beta}(t_{R})\big{)}$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given. Let $j$ be the
smallest index such that $u_{j+1}\dots u_{n}$ is a prefix of
$\mathbf{w}^{\infty}$, say $u_{j+1}\dots u_{n}=(w_{1}\dots
w_{m})^{k-1}w_{1}\dots w_{r}$ with $k\in\mathbb{N}$ and $0\leq r<m$. If no
such $j$ exists, set $j:=n$, $r:=0$ and $k:=1$. Let

$\mathbf{u}^{\prime}:=u_{1}\dots u_{n}w_{r+1}\dots w_{m}=u_{1}\dots
u_{j}\mathbf{w}^{k}.$

Let $\mathbf{v}$ be the word satisfying (i) and (ii) in Definition 5.3 for the
given $\mathbf{z}$. We claim that
${\mathbf{c}}:=\mathbf{u}^{\prime}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$.

We check first that $\sigma^{i}({\mathbf{c}})\succcurlyeq\mathbf{w}^{\infty}$
for all $i\geq 0$. For $0\leq i<j$ this follows from the minimality of $j$.
For $j\leq i<j+km$ and $m\nmid(i-j)$ it follows since $\mathbf{w}$ is Lyndon,
so $w_{l+1}\dots w_{m}\succ w_{1}\dots w_{m-l}$ for all $1\leq l<m$. And for
all other $i\geq j$ the inequality follows from assumption (ii).

Next, we check that $\sigma^{i}({\mathbf{c}})\preccurlyeq\alpha(\beta)$ for
all $i\geq 0$. For $0\leq i<j$, this follows since

  1. (a)

$u_{i+1}\dots u_{j}\preccurlyeq\alpha_{1}\dots\alpha_{j-i}$;

  2. (b)

$\mathbf{w}^{\infty}\prec b(1-1/\beta,\beta)=0\sigma(\alpha(\beta))$, so
$\mathbf{w}^{k}\preccurlyeq\alpha_{j-i+1}\dots\alpha_{j-i+km}$ by (5.1); and

  3. (c)

$\mathbf{v}\prec\alpha_{j-i+km+1}\dots\alpha_{j-i+km+n}$ by (i).

For $j\leq i<j+km$, the inequality follows since
$\sigma^{i-j}(\mathbf{w}^{k})\preccurlyeq\alpha_{1}\dots\alpha_{km-(i-j)}$
because $\mathbf{w}\in L^{*}(\beta)$, and
$\mathbf{v}\prec\alpha_{km-(i-j)+1}\dots\alpha_{km-(i-j)+n}$ by (i). Finally,
for $i\geq j+km$, $\sigma^{i}({\mathbf{c}})\preccurlyeq\alpha(\beta)$ follows
immediately from (ii). ∎

######  Lemma 5.5.

Let $\beta$ and $\hat{\beta}$ be bases such that
$0\alpha(\beta)=U_{0}(\alpha(\hat{\beta}))$. Let $\mathbf{z}$ and
$\hat{\mathbf{z}}$ be sequences in $\\{0,1\\}^{\mathbb{N}}$ such that
$\mathbf{z}=U_{0}(\hat{\mathbf{z}})$. Then

$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\quad\forall\,n\geq
0\qquad\Longleftrightarrow\qquad\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})\quad\forall\,k\geq
0.$

###### Proof.

Suppose first $\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})$
for all $k\geq 0$. Write $\mathbf{z}=z_{1}z_{2}\dots$. Clearly it is enough to
show that $\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ whenever
$z_{n+1}=1$. This implies $n\geq 1$, because
$\mathbf{z}=U_{0}(\hat{\mathbf{z}})$ begins with $0$. Since furthermore
$\mathbf{z}$ does not contain the word $11$, we have
$\sigma^{n-1}(\mathbf{z})=U_{0}(\sigma^{k}(\hat{\mathbf{z}}))$ for some $k\geq
0$. Since $U_{0}$ is increasing, this implies

$0\sigma^{n}(\mathbf{z})=\sigma^{n-1}(\mathbf{z})=U_{0}(\sigma^{k}(\hat{\mathbf{z}}))\preccurlyeq
U_{0}(\alpha(\hat{\beta}))=0\alpha(\beta),$

and hence, $\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$.

Suppose next that $\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ for all
$n\geq 0$. For given $k\geq 0$, we can find $n\geq 0$ such that
$\sigma^{n}(\mathbf{z})=U_{0}(\sigma^{k}(\hat{\mathbf{z}}))$ and moreover,
$\sigma^{n}(\mathbf{z})$ begins with $0$. So

$U_{0}(\sigma^{k}(\hat{\mathbf{z}}))=\sigma^{n}(\mathbf{z})=0\sigma^{n+1}(\mathbf{z})\preccurlyeq
0\alpha(\beta)=U_{0}(\alpha(\hat{\beta})),$

and as a result,
$\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})$. ∎

######  Lemma 5.6.

Let $\beta$ and $\hat{\beta}$ be bases such that
$0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta}))$. Let $\mathbf{z}$ and
$\hat{\mathbf{z}}$ be sequences in $\\{0,1\\}^{\mathbb{N}}$ such that
$\mathbf{z}=U_{1}(\hat{\mathbf{z}})$. Then

$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\quad\forall\,n\geq
0\qquad\Longleftrightarrow\qquad\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})\quad\forall\,k\geq
1.$

###### Proof.

Suppose first that
$\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})$ for all $k\geq
1$. Write $\mathbf{z}=z_{1}z_{2}\dots$. Since $\mathbf{y}\preccurlyeq
1\mathbf{y}$ for any sequence $\mathbf{c}$, it suffices to show
$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ whenever $z_{n}=0$. In that
case, there is a $k\geq 0$ such that
$0\sigma^{n}(\mathbf{z})=U_{1}(\sigma^{k}(\hat{\mathbf{z}}))$. This implies
that $\sigma^{k}(\hat{\mathbf{z}})$ begins with $0$, so
$\sigma^{k}(\hat{\mathbf{z}})=0\sigma^{k+1}(\hat{\mathbf{z}})$. Since $U_{1}$
is increasing, it follows that
$0\sigma^{n}(\mathbf{z})=U_{1}(0\sigma^{k+1}(\hat{\mathbf{z}}))\preccurlyeq
U_{1}(0\alpha(\hat{\beta}))=0\alpha(\beta)$, and hence,
$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$.

Conversely, suppose $\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ for all
$n\geq 0$. Write $\hat{\mathbf{z}}=\hat{z}_{1}\hat{z}_{2}\dots$. Let $k\geq 1$
be given. We may assume as above that $\hat{z}_{k}=0$. Then we can find $n\geq
0$ such that $\sigma^{n}(\mathbf{z})=U_{1}(\sigma^{k-1}(\hat{\mathbf{z}}))$,
and in particular, $\sigma^{n}(\mathbf{z})$ begins with $0$. Thus,

$U_{1}(0\sigma^{k}(\mathbf{z}))=U_{1}(\sigma^{k-1}(\hat{\mathbf{z}}))=\sigma^{n}(\mathbf{z})=0\sigma^{n+1}(\mathbf{z})\preccurlyeq
0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta})),$

and since $U_{1}$ is strictly increasing, it follows that
$\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})$. ∎

######  Remark 5.7.

(a) Both Lemmas 5.5 and 5.6 continue to hold if we replace the inequalities
with strict inequalities, because the maps $U_{0}$ and $U_{1}$ are strictly
increasing.

(b) In Lemma 5.6 we had to exclude the value $k=0$. For example, let
$\alpha(\hat{\beta})$ begin with $10$, so
$0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta}))$ begins with $01101$ and consists
of blocks $01$ and $011$. Let $\hat{\mathbf{z}}=110^{\infty}$, so
$\mathbf{z}=U_{1}(\hat{\mathbf{z}})=11(01)^{\infty}$. Then
$\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)$ for all $n\geq 0$, yet
$\hat{\mathbf{z}}\succ\alpha(\hat{\beta})$.

###### Proof of Theorem 5.1.

By Lemma 5.4 it is enough verify Property TL for each $\mathbf{w}\in L^{*}$
with $\mathbf{w}^{\infty}\prec 0\sigma(\alpha(\beta))$. We use induction on
the length of the generating word $\mathbf{w}$ of $[t_{L},t_{R}]$.

The shortest Lyndon word that can generate a $\beta$-Lyndon interval is
$\mathbf{w}=01$. Since $\mathbf{w}\in L^{*}(\beta)$, we have
$\alpha(\beta)\succ\mathbb{L}(\mathbf{w})^{\infty}=\alpha(\beta_{\ell}^{\mathbf{w}})$.
Furthermore, $\mathbf{w}\in F^{*}$, and $\beta\in E_{L}$. So by (2.3) it
follows that

$\alpha(\beta)\succ\alpha(\beta_{r}^{\mathbf{w}})=\mathbb{L}(\mathbf{w})^{+}\mathbf{w}^{\infty}=11(01)^{\infty}.$

In particular, $\alpha(\beta)$ begins with $11$. By Lemma 2.7, $\alpha(\beta)$
does not end in $(01)^{\infty}$, so there is an $M\in\mathbb{N}$ such that
$\alpha(\beta)$ does not contain the word $(01)^{M}$. Assume $\mathbf{z}$
begins with $0$, and set $\mathbf{v}:=(01)^{M}$. Since $\alpha(\beta)$ is
balanced by Lemma 2.5, and $\alpha(\beta)$ begins with $11$, it does not
contain the word $00$. Conditions (i) and (ii) of Definition 5.3 are clearly
satisfied. Hence $\mathbf{w}$ satisfies Property TL.

Now let $\mathbf{w}$ be a Lyndon word of length $\geq 3$. By Lemma 2.8, there
is a base $\hat{\beta}\in E_{L}$ such that
$0\alpha(\beta)=U_{0}(\alpha(\hat{\beta}))$ or
$0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta}))$.

Case (A). Assume first that

(5.2) $0\alpha(\beta)=U_{0}(\alpha(\hat{\beta})).$

Then $\alpha(\beta)$ does not contain the word 11, so $\mathbf{w}$ does not
contain this word either, since $\mathbf{w}\in L^{*}(\beta)$. Thus, since
$|\mathbf{w}|\geq 3$ and $\mathbf{w}$ is Lyndon, $\mathbf{w}$ must begin with
$00$, which implies $\mathbf{w}=U_{0}(\hat{\mathbf{w}})$ for some word
$\hat{\mathbf{w}}$ beginning with $0$ and ending with $1$. Since $U_{0}$ is
strictly increasing, $\hat{\mathbf{w}}$ is again Lyndon. Lemma 5.5 (with
strict, instead of weak inequalities) implies that
$\sigma^{k}(\hat{\mathbf{w}}^{\infty})\prec\alpha(\hat{\beta})$ for all $k\geq
0$. Thus, $\hat{\mathbf{w}}^{\infty}=b(\hat{t}_{R},\hat{\beta})$ for some
$\hat{t}_{R}\in(0,1)$, and

(5.3)
$U_{0}(b(\hat{t}_{R},\hat{\beta}))=U_{0}(\hat{\mathbf{w}}^{\infty})=\mathbf{w}^{\infty}=b(t_{R},\beta).$

Furthermore, the assumption $\mathbf{w}^{\infty}\prec 0\sigma(\alpha(\beta))$
implies $\hat{\mathbf{w}}^{\infty}\prec 0\sigma(\alpha(\hat{\beta}))$, because
(5.2) implies $\alpha(\beta)=1U_{0}\big{(}\sigma(\alpha(\hat{\beta}))\big{)}$,
and so

(5.4)
$0\sigma(\alpha(\beta))=0U_{0}\big{(}\sigma(\alpha(\hat{\beta}))\big{)}=U_{0}\big{(}0\sigma(\alpha(\hat{\beta}))\big{)}.$

Hence, $\hat{t}_{R}<1-1/\hat{\beta}$.

Now let $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given, and
assume $\mathbf{z}$ begins with $0$. Observe that

(5.5)
$b(t_{R},\beta)\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\alpha(\beta)\quad\forall
n\geq 0.$

Since $\alpha(\beta)$ begins with $10$, $\mathbf{z}$ cannot contain the word
$11$, and so $\mathbf{z}=U_{0}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}$. By (5.3), (5.5) and Lemma 5.5 it follows that

$b(\hat{t}_{R},\hat{\beta})=\hat{\mathbf{w}}^{\infty}\preccurlyeq\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})\quad\textrm{for
all }k\geq 0.$

As a result,
$\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$ with
$\hat{t}_{R}<1-1/\hat{\beta}$.

Since $\hat{\mathbf{w}}$ is shorter than $\mathbf{w}$, the induction
hypothesis now implies that $\hat{\mathbf{w}}$ has property TL for
$\hat{\beta}$, so with
$\alpha(\hat{\beta})=:\hat{\alpha}_{1}\hat{\alpha}_{2}\dots$, there is a word
$\hat{\mathbf{v}}$ such that

  1. (i)

$\hat{\mathbf{v}}\prec\hat{\alpha}_{j+1}\dots\hat{\alpha}_{j+|\hat{\mathbf{v}}|}$
for all $j\geq 0$; and

  2. (ii)

$\hat{\mathbf{w}}^{\infty}\preccurlyeq\sigma^{i}(\hat{\mathbf{v}}\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})$
for all $i\geq 0$.

Set $\mathbf{v}:=U_{0}(\hat{\mathbf{v}})$. Then $\mathbf{v}$ satisfies
condition (i) of Definition 5.3 because $U_{0}$ is strictly increasing.
Furthermore, the lower bound
$\sigma^{i}(\mathbf{v}\mathbf{z})\succcurlyeq\mathbf{w}^{\infty}$ is satisfied
because $U_{0}$ is increasing and $\mathbf{w}$ is Lyndon. Finally, the upper
bound $\sigma^{i}(\mathbf{v}\mathbf{z})\preccurlyeq\alpha(\beta)$ follows from
(ii) above and Lemma 5.5. Hence, $\mathbf{w}$ has Property TL for $\beta$.

Case (B). Assume next that

(5.6) $0\alpha(\beta)=U_{1}(0\alpha(\hat{\beta})).$

Then $\alpha(\beta)\neq(10)^{\infty}$, for otherwise $\alpha(\hat{\beta})$
would equal $0^{\infty}$, which isn’t possible. So $\alpha(\beta)$ begins with
$11$, and by Lemma 2.7, $\alpha(\beta)$ does not end in $(01)^{\infty}$. Hence
there is a positive integer $M$ such that $\alpha(\beta)$ does not contain the
word $(01)^{M}$. Furthermore, $\alpha(\beta)$ does not contain the word $00$.

If $\mathbf{w}$ begins with $00$, take $\mathbf{v}=(01)^{M}$; then it is easy
to see that $\mathbf{v}$ satisfies the conditions of Definition 5.3. Hence,
$\mathbf{w}$ has Property TL for $\beta$.

Assume therefore, for the remainder of the proof, that $\mathbf{w}$ begins
with $01$. Then $\mathbf{w}$ does not contain the word $00$ because
$\mathbf{w}$ is Lyndon, and hence $\mathbf{w}=U_{1}(\hat{\mathbf{w}})$ for
some word $\hat{\mathbf{w}}$. As in Case (A), we can deduce from Lemma 5.6
(with strict inequality) that $\hat{\mathbf{w}}\in L^{*}(\hat{\beta})$, so
$\hat{\mathbf{w}}^{\infty}=b(\hat{t}_{R},\hat{\beta})$ for some
$\hat{t}_{R}\in(0,1)$. Furthermore, the assumption (5.6) implies that
$U_{1}\big{(}0\sigma(\alpha(\hat{\beta}))\big{)}=0\sigma(\alpha(\beta))$,
because $0\sigma(\alpha(\hat{\beta}))$ is simply $0\alpha(\hat{\beta})$
without the first ‘1’, and $U_{1}(1)=1$. Therefore, since
$\mathbf{w}^{\infty}\prec 0\sigma(\alpha(\beta))$, we also have
$\hat{\mathbf{w}}^{\infty}\prec 0\sigma(\alpha(\hat{\beta}))$, and hence
$\hat{t}_{R}<1-1/\hat{\beta}$.

Now let $\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ be given, and
assume that $\mathbf{z}$ begins with $0$. Since
$\sigma^{n}(\mathbf{z})\succcurlyeq\mathbf{w}^{\infty}$ for all $n\geq 0$ and
$\mathbf{w}$ begins with $01$, $\mathbf{z}$ does not contain the word $00$.
Hence, $\mathbf{z}=U_{1}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}$ which again begins with $0$. As in Case (A), we can deduce
from Lemma 5.6 that

$b(\hat{t}_{R},\hat{\beta})=\hat{\mathbf{w}}^{\infty}\preccurlyeq\sigma^{k}(\hat{\mathbf{z}})\preccurlyeq\alpha(\hat{\beta})\quad\textrm{for
all }k\geq 0,$

where for $k=0$ the last inequality holds since $\hat{\mathbf{z}}$ begins with
$0$. Hence,
$\hat{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\hat{\beta}}(\hat{t}_{R})$ with
$\hat{t}_{R}<1-1/\hat{\beta}$. The induction hypothesis now comes in to
complete the proof in the same way as in Case (A), using Lemma 5.6 instead of
Lemma 5.5. ∎

###  5.1. Density of $\beta$-Lyndon intervals

In this subsection we show that the $\beta$-Lyndon intervals are dense in
$[0,1-1/\beta]$ for $\beta\in\overline{E}$.

######  Lemma 5.8.

Let $\beta\in\overline{E}$, and let $\mathbf{w}$ be a Lyndon word such that

  1. (i)

$\mathbf{w}0^{\infty}$ is a greedy $\beta$-expansion; and

  2. (ii)

$\mathbf{w}0^{\infty}\prec 0\sigma(\alpha(\beta))$.

Then $\mathbf{w}\in L^{*}(\beta)$.

###### Proof.

We use induction on the length of $\mathbf{w}$. Precisely, we prove: For any
Lyndon word $\mathbf{w}$, if $\beta\in\overline{E}$ is such that (i) and (ii)
are satisfied, then $\mathbf{w}\in L^{*}(\beta)$.

If $\mathbf{w}=01$, then (ii) implies $\alpha(\beta)\succcurlyeq
110^{\infty}$, so
$\sigma^{n}(\mathbf{w}^{\infty})\preccurlyeq(10)^{\infty}\prec\alpha(\beta)$
for each $n\geq 0$. Hence $\mathbf{w}\in L^{*}(\beta)$.

Now let $k\geq 3$ and assume the statement is true for all Lyndon words of
length $<k$. Let $\mathbf{w}$ be a Lyndon word of length $k$, and let
$\beta\in\overline{E}$ be such that (i) and (ii) are satisfied. By Lemma 2.8
(in which we can clearly replace $E_{L}$ with $\overline{E}$) there is a base
$\hat{\beta}\in\overline{E}$ such that
$0\alpha(\beta)=U_{0}(\alpha(\hat{\beta}))$ or
$\alpha(\beta)=1U_{1}(\alpha(\hat{\beta}))$.

First assume the former, i.e. $0\alpha(\beta)=U_{0}(\alpha(\hat{\beta}))$.
Then $\alpha(\beta)$ does not contain the word $11$, so $\mathbf{w}$ does not
contain this word either by assumption (i). Since $\mathbf{w}\in L^{*}$, this
implies $\mathbf{w}=U_{0}(\hat{\mathbf{w}})$ for some $\hat{\mathbf{w}}\in
L^{*}$. Furthermore, $U_{0}(\hat{\mathbf{w}}0^{\infty})=\mathbf{w}0^{\infty}$.
Thus, by Lemma 5.5, $\hat{\mathbf{w}}0^{\infty}$ is a greedy
$\hat{\beta}$-expansion. It also follows from (ii) and (5.4) that
$\hat{\mathbf{w}}0^{\infty}\prec 0\sigma(\alpha(\hat{\beta}))$. Hence, by the
induction hypothesis, $\hat{\mathbf{w}}\in L^{*}(\hat{\beta})$. But then,
again by Lemma 5.5, $\mathbf{w}\in L^{*}(\beta)$.

Next, suppose $\alpha(\beta)=1U_{1}(\alpha(\hat{\beta}))$. Then
$\alpha(\beta)$ does not contain the word $00$, so if $\mathbf{w}$ begins with
$00$, then $\sigma^{n}(\mathbf{w}^{\infty})\prec\alpha(\beta)$ for all $n\geq
0$: setting $m:=n\\!\\!\mod|\mathbf{w}|$, we have

$\displaystyle\sigma^{n}(\mathbf{w}^{\infty})$ $\displaystyle=w_{m+1}\dots
w_{|\mathbf{w}|}\mathbf{w}^{\infty}\preccurlyeq\alpha_{1}\dots\alpha_{|\mathbf{w}|-m}\mathbf{w}^{\infty}$
$\displaystyle\prec\alpha_{1}\dots\alpha_{|\mathbf{w}|-m}\sigma^{|\mathbf{w}|-m}(\alpha(\beta))=\alpha(\beta).$

So assume $\mathbf{w}$ begins with $01$. Then
$\mathbf{w}=U_{1}(\hat{\mathbf{w}})$ for some $\hat{\mathbf{w}}\in L^{*}$
because $\mathbf{w}\in L^{*}$. Now observe that
$\alpha(\beta)=1U_{1}(\alpha(\hat{\beta}))$ implies
$\sigma^{n}(\alpha(\beta))\succ(01)^{\infty}$ for all $n\geq 0$, so
$\mathbf{w}(01)^{\infty}=U_{1}(\hat{\mathbf{w}}0^{\infty})$ is also a greedy
$\beta$-expansion. Thus, by Lemma 5.6, $\hat{\mathbf{w}}0^{\infty}$ is a
greedy $\hat{\beta}$-expansion. The rest of the proof goes as in the first
case above, using Lemma 5.6 instead of Lemma 5.5. ∎

######  Proposition 5.9.

For each $\beta\in\overline{E}\cap(1,2]$, the $\beta$-Lyndon intervals are
dense in $\big{(}0,1-\frac{1}{\beta}\big{)}$.

###### Proof.

Fix $t\in\big{(}0,1-\frac{1}{\beta}\big{)}$ and write
$b(t,\beta)=t_{1}t_{2}\dots$. We will show that $t$ either lies in a
$\beta$-Lyndon interval or else $t\in\mathscr{E}_{\beta}^{+}$, in which case
$t$ can be approximated arbitrarily closely by endpoints of $\beta$-Lyndon
intervals by Lemma 4.5.

Note that $t_{1}=0$ since $t<1-\frac{1}{\beta}\leq\frac{1}{\beta}$. Suppose
$t\not\in\mathscr{E}_{\beta}^{+}$. Then $\sigma^{k}((t_{i}))\prec(t_{i})$ for
some $k$; choose $k$ as small as possible with this property. Then
$\mathbf{w}:=t_{1}\dots t_{k}$ is either Lyndon or a power of a Lyndon word.
Assume the former; in the latter case we replace $k$ with $k^{\prime}<k$ so
that $t_{1}\dots t_{k^{\prime}}$ is Lyndon and $k^{\prime}\mid k$. By Lemma
5.8, $\mathbf{w}\in L^{*}(\beta)$. Since furthermore,
$\sigma^{k}((t_{i}))\prec(t_{1}\dots t_{k})^{\infty}$, it follows that $t$
lies in the $\beta$-Lyndon interval generated by $\mathbf{w}$. ∎

The conclusion of Proposition 5.9 may fail if $\beta\not\in\overline{E}$; in
fact the $\beta$-Lyndon intervals may not even be dense in $[0,\tau(\beta)]$,
as the following example shows.

######  Example 5.10.

Let $\beta$ be any base such that $\alpha(\beta)$ begins with
$111\,01\,011\,001$. Set ${\mathbf{s}}=011$. Then $\beta$ lies in the basic
interval $[\beta_{\ell}^{{\mathbf{s}}},\beta_{*}^{\mathbf{s}}]$, where
$\alpha(\beta_{\ell}^{{\mathbf{s}}})=\mathbb{L}({\mathbf{s}})^{\infty}=(110)^{\infty}$
and
$\alpha(\beta_{*}^{\mathbf{s}})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}=111\,010\,(110)^{\infty}$.
Hence, by [4, Theorem 2],

$\tau(\beta)=\pi_{\beta}\big{(}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}\big{)}=\pi_{\beta}\big{(}010(110)^{\infty}\big{)}.$

Let $\mathbf{w}=01010111$ and $\tilde{\mathbf{w}}=01011$. Note that
$\mathbf{w}$ and $\tilde{\mathbf{w}}$ are both $\beta$-Lyndon words, and
$\mathbf{w}^{\infty}\prec\tilde{\mathbf{w}}^{\infty}\prec 010(110)^{\infty}$.
So the (symbolic) $\beta$-Lyndon intervals
$[\mathbf{w}0^{\infty},\mathbf{w}^{\infty}]$ and
$[\tilde{\mathbf{w}}0^{\infty},\tilde{\mathbf{w}}^{\infty}]$ are disjoint and
their projections under $\pi_{\beta}$ both lie inside $[0,\tau(\beta)]$. But
it is not difficult to see that there is no other $\beta$-Lyndon interval in
between these intervals, because there is no $\beta$-Lyndon word that extends
$\mathbf{w}$. Hence, the $\beta$-Lyndon intervals are not dense in
$[0,\tau(\beta)]$. We will have more to say about this example in Example
10.1.

######  Corollary 5.11.

If $\beta=\beta_{\ell}^{\mathbf{s}}$ is the left endpoint of a Farey interval
$[\beta_{\ell}^{\mathbf{s}},\beta_{r}^{\mathbf{s}}]$, then
$\mathscr{B}_{\beta}=\mathscr{E}_{\beta}$.

###### Proof.

Note that $\beta=\beta_{\ell}^{\mathbf{s}}\in E_{L}$, so
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is transitive for every
$\beta$-Lyndon interval $[t_{L},t_{R}]$ by Theorem 5.1. Furthermore, the
$\beta$-Lyndon intervals are dense in $[0,1-1/\beta]$ by Proposition 5.9.
Since $\alpha(\beta)=\mathbb{L}({\mathbf{s}})^{\infty}$, we have

$\widetilde{\mathcal{K}}_{\beta}(t_{R})=\left\\{\mathbf{z}\in\left\\{0,1\right\\}^{\mathbb{N}}:\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{s}})^{\infty}~{}\forall
n\geq 0\right\\},$

so $\widetilde{\mathcal{K}}_{\beta}(t_{R})$ is a SFT, hence sofic. Proposition
4.6 now implies that there are no points of
$\mathscr{E}_{\beta}\backslash\mathscr{B}_{\beta}$ in $[0,1-1/\beta)$. On the
other hand, as shown in the proof of [22, Proposition 5.2],
$\mathscr{E}_{\beta}\cap[1-1/\beta,1]\subseteq\mathcal{K}_{\beta}(1-1/\beta)=\emptyset$.
Hence, $\mathscr{E}_{\beta}=\mathscr{B}_{\beta}$. ∎

##  6\. Right endpoints of first-order basic intervals

Our main result in this section is the following.

######  Theorem 6.1.

Let $\beta_{*}=\beta_{*}^{\mathbf{s}}$ be the right endpoint of a basic
interval generated by ${\mathbf{s}}\in F^{*}$. Then for any $\beta_{*}$-Lyndon
interval $[t_{L},t_{R}]$ in $[0,\tau(\beta_{*})]$,
$\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R})$ is a transitive sofic subshift.
In particular, the conclusion of Theorem 4.4 holds for $\beta=\beta_{*}$.

We prove the theorem after establishing a useful lemma, which uses the
following fact from [22, Proposition 4.4].

######  Lemma 6.2.

For ${\mathbf{s}}\in F^{*}$, define the set

$\mathcal{X}({\mathbf{s}}):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:{\mathbf{s}}0^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{s}})^{\infty}\
\forall n\geq 0\\}.$

Then $\mathcal{X}({\mathbf{s}})=\\{\sigma^{j}({\mathbf{s}}^{\infty}):0\leq
j<|{\mathbf{s}}|\\}$.

######  Lemma 6.3.

Let $\beta\in(\beta_{\ell},\beta_{*}]$, where $[\beta_{\ell},\beta_{*}]$ is a
basic interval generated by a Farey word ${\mathbf{s}}=s_{1}\dots s_{m}$, and
let $[t_{L},t_{R}]$ be a $\beta$-Lyndon interval in $(0,\tau(\beta))$. Let
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$. Then $\mathbf{z}$ can
be extended to the left to a sequence
$\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning with
${\mathbf{s}}$, and also to a sequence
$\mathbf{z}^{\prime\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning
with $\mathbb{L}({\mathbf{s}})$.

###### Proof.

Set $\mathbf{z}^{(0)}:=\mathbf{z}$, and, for $k=0,1,2,\dots$, proceed
inductively as follows. If $\mathbf{z}^{(k)}$ has been constructed and does
not begin with the prefix ${\mathbf{s}}$, set
$\mathbf{z}^{(k+1)}=1\mathbf{z}^{(k)}$ if $1\mathbf{z}^{(k)}$ is legal in
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$, or
$\mathbf{z}^{(k+1)}=0\mathbf{z}^{(k)}$ otherwise. We end the procedure if we
reach a sequence $\mathbf{z}^{(k)}$ beginning with ${\mathbf{s}}$.

Suppose the procedure never ends. Let $k\geq 1$, and write
$\mathbf{z}^{(k)}=z_{1}^{(k)}z_{2}^{(k)}\dots$. Since
$\beta\in(\beta_{\ell},\beta_{*}]$, $\alpha(\beta)$ begins with
$\mathbb{L}({\mathbf{s}})^{+}$. So, if $z_{1}^{(k)}=0$, this means
$1\sigma(\mathbf{z}^{(k)})\succ\alpha(\beta)$ and hence

$1z_{2}^{(k)}\dots z_{m}^{(k)}\succcurlyeq\mathbb{L}({\mathbf{s}})^{+},$

which implies

$z_{1}^{(k)}z_{2}^{(k)}\dots z_{m}^{(k)}=0\sigma(1z_{2}^{(k)}\dots
z_{m}^{(k)})\succcurlyeq 0\sigma(\mathbb{L}({\mathbf{s}})^{+})={\mathbf{s}},$

where the last equality follows from Lemma 2.4 (i). Since $\mathbf{z}^{(k)}$
does not begin with ${\mathbf{s}}$, this means that
$z_{1}^{(k)}z_{2}^{(k)}\dots z_{m}^{(k)}\succ{\mathbf{s}}$. Clearly, this
inequality holds also when $z_{1}^{(k)}=1$.

On the other hand, for $k>m$ we claim that $\mathbf{z}^{(k)}$ cannot begin
with $\mathbb{L}({\mathbf{s}})^{+}$. For, suppose it did. Then, since
$\alpha(\beta)\preccurlyeq\alpha(\beta_{*})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}$
and $\mathbf{z}^{(k)}$ is legal in $\widetilde{\mathcal{K}}_{\beta}(t_{R})$,
we would have

$z_{1}^{(k-m)}\dots z_{m}^{(k-m)}=z_{m+1}^{(k)}\dots
z_{2m}^{(k)}\preccurlyeq{\mathbf{s}}^{-}\prec{\mathbf{s}},$

contradicting what we just showed. Hence, we also have $z_{1}^{(k)}\dots
z_{m}^{(k)}\preccurlyeq\mathbb{L}({\mathbf{s}})$.

We have now constructed a sequence, extending infinitely to the left, all of
whose subwords of length $m$ are strictly greater than ${\mathbf{s}}$ but
smaller than or equal to $\mathbb{L}({\mathbf{s}})$. But such a sequence
cannot exist, in view of Lemma 6.2. This contradiction shows that we
eventually end up with an extension $\mathbf{z}^{\prime}$ of $\mathbf{z}$
beginning with ${\mathbf{s}}$. But then
${\mathbf{s}}\mathbf{z}^{\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ as
well, and since ${\mathbf{s}}\mathbf{z}^{\prime}$ begins with
${\mathbf{s}}^{2}$ and ${\mathbf{s}}^{2}$ contains the word
$\mathbb{L}({\mathbf{s}})$, we can extend $\mathbf{z}$ to a sequence
$\mathbf{z}^{\prime\prime}\in\widetilde{\mathcal{K}}_{\beta}(t_{R})$ beginning
with $\mathbb{L}({\mathbf{s}})$. ∎

###### Proof of Theorem 6.1.

Let ${\mathbf{s}}={\mathbf{s}}^{(1)}{\mathbf{s}}^{(2)}$ be the standard
factorization of ${\mathbf{s}}$ into Farey words ${\mathbf{s}}^{(1)}$ and
${\mathbf{s}}^{(2)}$. It is well known that such a factorization exists and is
unique (cf. [9]). We can write

$\alpha(\beta_{*})=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}=\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{(1)}{\mathbf{s}}^{\infty},$

where the second equality can be deduced from [9, Lemma 2.6] and the
palindrome property of Farey words (see Lemma 2.4 (ii)).

Recall from [4, Theorem 2] that

$\tau(\beta_{*})=\pi_{\beta}\big{(}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}\big{)}.$

Let $[t_{L},t_{R}]$ be a $\beta_{*}$-Lyndon interval and assume
$t_{R}<\tau(\beta_{*})$; then
$b(t_{R},\beta_{*})\prec{\mathbf{s}}^{-}\mathbb{L}({\mathbf{s}})^{\infty}={\mathbf{s}}^{(1)}{\mathbf{s}}^{\infty}$.
So, letting $\mathbf{w}$ be the word such that
$b(t_{R},\beta_{*})=\mathbf{w}^{\infty}$, we can choose an integer $N$ so
large that

(6.1) $\mathbf{w}^{\infty}\prec{\mathbf{s}}^{(1)}{\mathbf{s}}^{N}0^{\infty},$

and

(6.2)
$\sigma^{n}(\mathbf{w}^{\infty})\prec\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{(1)}{\mathbf{s}}^{N}0^{\infty}\quad\forall
n\geq 0.$

(The latter condition holds for large enough $N$ since $\mathbf{w}\in
L^{*}(\beta_{*})$ and $\mathbf{w}^{\infty}$ is periodic.)

Now let $\mathbf{u}=u_{1}\dots
u_{r}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R}))$ and
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R})$. By Lemma 6.3 we may
assume that $\mathbf{z}$ begins with the word ${\mathbf{s}}$. We also extend
$\mathbf{u}$ to the right to a word $\mathbf{u}^{\prime}$ as follows: Let $k$
be the largest integer such that $u_{r-k+1}\dots
u_{r}=\alpha_{1}\dots\alpha_{k}$, where
$\alpha_{1}\alpha_{2}\dots:=\alpha(\beta_{*})$. If no such $k$ exists, set
$k=0$. Then set

$\mathbf{u}^{\prime}:=u_{1}\dots
u_{r-k}\mathbb{L}({\mathbf{s}})^{+}{\mathbf{s}}^{(1)}{\mathbf{s}}^{N_{1}},$

where $N_{1}\geq N$ is chosen large enough so that $\mathbf{u}^{\prime}$
actually extends $\mathbf{u}$. It is not difficult to see using (6.1) and
(6.2) that
$\mathbf{u}^{\prime}\in\mathcal{L}(\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R}))$.

Now if in fact $\mathbf{z}\preccurlyeq{\mathbf{s}}^{\infty}$, we have
immediately that
$\mathbf{u}^{\prime}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R})$,
in view of the inequalities

$s_{i+1}\ldots s_{m}s_{1}\ldots
s_{i}\prec\mathbb{L}({\mathbf{s}})^{+}\quad\textrm{and}\quad s_{i+1}\ldots
s_{m}\succ s_{1}\ldots s_{m-i}\succcurlyeq w_{1}\ldots
w_{m-i}\quad\forall\,0\leq i<m.$

Otherwise, there is an integer $l\geq 1$ and a word
$\mathbf{v}\succ{\mathbf{s}}$ such that $\mathbf{z}$ begins with
${\mathbf{s}}^{l}\mathbf{v}$. Now observe that
${\mathbf{s}}^{(1)}\mathbf{z}\succ{\mathbf{s}}^{(1)}{\mathbf{s}}^{\infty}\succ\mathbf{w}^{\infty}$,
and also ${\mathbf{s}}^{(1)}\mathbf{z}\prec{\mathbf{s}}^{\infty}$, since
${\mathbf{s}}$ being Lyndon implies
${\mathbf{s}}={\mathbf{s}}^{(1)}{\mathbf{s}}^{(2)}\prec{\mathbf{s}}^{(2)}{\mathbf{s}}^{(1)}$
and hence ${\mathbf{s}}^{(1)}{\mathbf{s}}\prec{\mathbf{s}}{\mathbf{s}}^{(1)}$.
Using the definition of $\mathbf{u}^{\prime}$ and the assumptions (6.1) and
(6.2) it now follows readily that

$\mathbf{w}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{u}^{\prime}{\mathbf{s}}^{(1)}\mathbf{z})\preccurlyeq\alpha(\beta)\qquad\forall
n\geq 0,$

and therefore,
$\mathbf{u}^{\prime}{\mathbf{s}}^{(1)}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R})$.
As a result, $\widetilde{\mathcal{K}}_{\beta_{*}}(t_{R})$ is transitive. ∎

##  7\. The case of finitely renormalizable $\beta$

In this section we consider value of $\beta$ in the set
$\overline{E^{\mathbf{S}}}$ for ${\mathbf{S}}\in\Lambda$. Recall
$E^{\mathbf{S}}$ is defined in (2.10). Recall further that
$\mathcal{T}_{R}(\beta)$ is the set of all right endpoints of $\beta$-Lyndon
intervals in $[0,\tau(\beta)]$.

######  Theorem 7.1.

Let $\beta\in\overline{E^{\mathbf{S}}}$ for some ${\mathbf{S}}\in\Lambda$.
Then:

  1. (i)

For every $t_{R}\in\mathcal{T}_{R}(\beta)$,
$\widetilde{\mathcal{K}}_{\beta}(t_{R})$ has a transitive subshift
$\mathcal{K}_{\beta}^{\prime}(t_{R})$ of full entropy and full Hausdorff
dimension that contains the sequence $b(t_{R},\beta)$. Moreover, these
transitive subshifts can be chosen so that
$\\{\mathcal{K}_{\beta}^{\prime}(t_{R}):t_{R}\in\mathcal{T}_{R}(\beta)\\}$ is
a strictly descending collection of subshifts, and if $\alpha(\beta)$ is
eventually periodic, then $\mathcal{K}_{\beta}^{\prime}(t_{R})$ is sofic for
each $t_{R}\in\mathcal{T}_{R}(\beta)$.

  2. (ii)

The $\beta$-Lyndon intervals are dense in $[0,\tau(\beta)]$.

Note that (i) implies Theorem 4.4 for $\beta\in\overline{E^{\mathbf{S}}}$. We
first establish a series of lemmas.

######  Lemma 7.2.

Let ${\mathbf{r}}\in\Lambda$, and let $\mathbf{x}$ and $\mathbf{y}$ be
sequences in $\\{0,1\\}^{\mathbb{N}}$ beginning with $0$ and $1$ respectively.
Suppose a sequence $\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}$ begins with
${\mathbf{r}}^{-}$ or $\mathbb{L}({\mathbf{r}})^{+}$ and satisfies

(7.1)
$\Phi_{\mathbf{r}}(\mathbf{x})\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\Phi_{\mathbf{r}}(\mathbf{y})\qquad\forall\,n\geq
0.$

Then $\mathbf{z}=\Phi_{\mathbf{r}}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}$.

###### Proof.

From Definition 2.10, we see that $\Phi_{\mathbf{r}}(\mathbf{x})$ begins with
${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{j}\mathbb{L}({\mathbf{r}})^{+}$ for
some $j\geq 0$, and $\Phi_{\mathbf{r}}(\mathbf{y})$ begins with
$\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{k}{\mathbf{r}}^{-}$ for some $k\geq
0$, where $j$ and $k$ may be infinite. So by (7.1),

${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{j}\mathbb{L}({\mathbf{r}})^{+}0^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{k}{\mathbf{r}}^{-}1^{\infty}\qquad\forall\,n\geq
0,$

and any block ${\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{l}$ (with $0\leq
l\leq j$) in $\mathbf{z}$ must be followed by $\mathbb{L}({\mathbf{r}})$ or
$\mathbb{L}({\mathbf{r}})^{+}$, whereas any block
$\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{m}$ (with $0\leq m\leq k$) in
$\mathbf{z}$ must be followed by ${\mathbf{r}}$ or ${\mathbf{r}}^{-}$. Hence
$\mathbf{z}\in X({\mathbf{r}})$, and so
$\mathbf{z}=\Phi_{\mathbf{r}}(\hat{\mathbf{z}})$ for some sequence
$\hat{\mathbf{z}}$. ∎

For any ${\mathbf{S}}\in\Lambda$, we define the set

(7.2)
$\Gamma({\mathbf{S}}):=\\{\mathbf{z}\in\\{0,1\\}^{\mathbb{N}}:{\mathbf{S}}^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})\preccurlyeq\mathbb{L}({\mathbf{S}})^{\infty}\
\ \forall n\geq 0\\}.$

The next result, which extends Lemma 6.2 to arbitrary words in $\Lambda$, was
proved in [4, Proposition 4.1]. It also follows from [19, Lemma 2.12] via the
relationship (1.8), or can alternatively be deduced, with some effort, from
[24, Theorem 2.5]. We include a short proof here for the reader’s convenience.

######  Lemma 7.3.

Let ${\mathbf{S}}\in\Lambda$. Then the set $\Gamma({\mathbf{S}})$ is
countable.

###### Proof.

We proceed by induction on the degree $k$ of ${\mathbf{S}}$. The case $k=1$
follows from Lemma 6.2. Let $k\geq 2$, and suppose the statement holds for all
${\mathbf{S}}\in\Lambda_{k-1}$. Take ${\mathbf{S}}\in\Lambda_{k}$, and write
${\mathbf{S}}={\mathbf{r}}\bullet{\mathbf{s}}$ where ${\mathbf{r}}\in F^{*}$
and ${\mathbf{s}}\in\Lambda_{k-1}$. Let $\mathbf{z}\in\Gamma({\mathbf{S}})$.
Since ${\mathbf{r}}\bullet{\mathbf{s}}$ begins with ${\mathbf{r}}^{-}$ and
$\mathbb{L}({\mathbf{r}}\bullet{\mathbf{s}})$ begins with
$\mathbb{L}({\mathbf{r}})^{+}$, it follows that

$\displaystyle{\mathbf{r}}^{-}0^{\infty}$
$\displaystyle\prec\Phi_{\mathbf{r}}({\mathbf{s}}^{\infty})=({\mathbf{r}}\bullet{\mathbf{s}})^{\infty}\preccurlyeq\sigma^{n}(\mathbf{z})$
$\displaystyle\preccurlyeq\mathbb{L}({\mathbf{r}}\bullet{\mathbf{s}})^{\infty}=\Phi_{\mathbf{r}}(\mathbb{L}({\mathbf{s}})^{\infty})\prec\mathbb{L}({\mathbf{r}})^{+}1^{\infty}\qquad\forall
n\geq 0,$

so either $\mathbf{z}\in\Gamma({\mathbf{r}})$, or else $\mathbf{z}$ eventually
contains ${\mathbf{r}}^{-}$ or $\mathbb{L}({\mathbf{r}})^{+}$, which by Lemma
7.2 implies $\sigma^{k}(\mathbf{z})=\Phi_{\mathbf{r}}(\hat{\mathbf{z}})$ for
some $k\geq 0$ and sequence $\hat{\mathbf{z}}$. But
$\mathbf{z}\in\Gamma({\mathbf{r}}\bullet{\mathbf{s}})$ implies that
$\hat{\mathbf{z}}\in\Gamma({\mathbf{s}})$ by Lemma 2.12 (i). Therefore,

$\Gamma({\mathbf{r}}\bullet{\mathbf{s}})\subseteq\Gamma({\mathbf{r}})\cup\bigcup_{{\mathbf{d}}\in\left\\{0,1\right\\}^{*}}{\mathbf{d}}\;\Phi_{\mathbf{r}}(\Gamma({\mathbf{s}})).$

Since both $\Gamma({\mathbf{r}})$ and $\Gamma({\mathbf{s}})$ are countable by
the induction hypothesis, we conclude that
$\Gamma({\mathbf{S}})=\Gamma({\mathbf{r}}\bullet{\mathbf{s}})$ is countable as
well. ∎

The next lemma describes the relationship between the substitution map
$\Phi_{\mathbf{r}}$ and the maps $U_{0}$ and $U_{1}$.

######  Lemma 7.4.

Let ${\mathbf{r}}\in F^{*}$. Suppose
$\alpha(\beta)=\Phi_{\mathbf{r}}(\mathbf{y})$ for some sequence
$\mathbf{y}\in\\{0,1\\}^{\mathbb{N}}$, and either

  1. (i)

${\mathbf{r}}=U_{0}(\tilde{{\mathbf{r}}})$ for some $\tilde{{\mathbf{r}}}\in
F^{*}$, and $\tilde{\beta}$ is the base such that
$0\alpha(\beta)=U_{0}(\alpha(\tilde{\beta}))$; or

  2. (ii)

${\mathbf{r}}=U_{1}(\tilde{{\mathbf{r}}})$ for some $\tilde{{\mathbf{r}}}\in
F^{*}$, and $\tilde{\beta}$ is the base such that
$\alpha(\beta)=1U_{1}(0\alpha(\tilde{\beta}))$.

Then $\alpha(\tilde{\beta})=\Phi_{\tilde{{\mathbf{r}}}}(\mathbf{y})$.

###### Proof.

Suppose ${\mathbf{r}}=U_{0}(\tilde{{\mathbf{r}}})$ and
$0\alpha(\beta)=U_{0}(\alpha(\tilde{\beta}))$. Since
$\alpha(\beta)=\Phi_{\mathbf{r}}(\mathbf{y})$ begins with 1 and
$\Phi_{\mathbf{r}}(0)={\mathbf{r}}^{-}$ begins with 0, it follows that
$\mathbf{y}$ begins with 1. Thus, according to the definition (2.7) we can
write

$\alpha(\beta)=\Phi_{\mathbf{r}}(\mathbf{y})=\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{k_{1}}{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{l_{1}}\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{k_{2}}{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{l_{2}}\dots,$

where $k_{i},l_{i}$ are nonnegative integers, possibly one of them taking the
value $+\infty$. By repeatedly applying the identities of Lemma 2.6 (i), it
follows that

$U_{0}(\alpha(\tilde{\beta}))=0\alpha(\beta)=U_{0}(\Phi_{\tilde{{\mathbf{r}}}}(\mathbf{y})).$

Thus, $\alpha(\tilde{\beta})=\Phi_{\tilde{{\mathbf{r}}}}(\mathbf{y})$. The
proof in case (ii) is similar, using Lemma 2.6 (ii). ∎

######  Lemma 7.5.

Let ${\mathbf{r}}\in F^{*}$ and let $\beta\in(1,2)$ such that
$\alpha(\beta)=\Phi_{\mathbf{r}}(\mathbf{y})$ for some sequence
$\mathbf{y}\succ 10^{\infty}$. Let $\mathbf{w}$ be a $\beta$-Lyndon word such
that
$\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}$.
Then for any sequence
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$ there is a word
$\mathbf{v}$ such that
$\mathbf{w}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$.

###### Proof.

We use induction on the length of the Farey word ${\mathbf{r}}$. For the
purpose of this proof, we will consider $\mathbf{w}=0$ a $\beta$-Lyndon word,
and the degenerate interval $[t_{L},t_{R}]=\\{0\\}$ a $\beta$-Lyndon interval.
Note that for this case, the statement of the lemma is trivial.

Step 1. First we take ${\mathbf{r}}=01$. Without loss of generality we may
assume that $\mathbf{z}$ begins with $0$. Since
$\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}=00(10)^{\infty}$,
there is an integer $M$ such that

$\mathbf{w}^{\infty}\prec 00(10)^{M}0^{\infty}.$

Likewise, since $\alpha(\beta)=\Phi_{\mathbf{r}}(\mathbf{y})$, $\alpha(\beta)$
begins with $\mathbb{L}({\mathbf{r}})^{+}$. Furthermore,
$1^{\infty}\succcurlyeq\mathbf{y}\succ 10^{\infty}$ implies
$\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{\infty}\succcurlyeq\alpha(\beta)\succ\mathbb{L}({\mathbf{r}})^{+}{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}=1100(10)^{\infty}$,
so there is a further integer $N\geq M$ such that

$11(01)^{\infty}\succcurlyeq\alpha(\beta)\succ 1100(10)^{N}1^{\infty}.$

Now set

$\mathbf{v}:=00(10)^{N}1.$

Then it is not difficult to see that
$\mathbf{w}\mathbf{v}\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$.
This completes the basis for the induction.

Step 2. Now let $k\geq 3$ and assume the statement of the lemma holds for all
Farey words of length $<k$. Let ${\mathbf{r}}$ be a Farey word of length $k$.
So either ${\mathbf{r}}=U_{0}(\tilde{{\mathbf{r}}})$ or
${\mathbf{r}}=U_{1}(\tilde{{\mathbf{r}}})$ for some Farey word
$\tilde{{\mathbf{r}}}$ of length $<k$.

Case (A): Suppose ${\mathbf{r}}=U_{0}(\tilde{{\mathbf{r}}})$. Let
$\tilde{\beta}$ be the base such that
$0\alpha(\beta)=U_{0}(\alpha(\tilde{\beta}))$. Then
$\alpha(\tilde{\beta})=\Phi_{\tilde{{\mathbf{r}}}}(\mathbf{y})$ by Lemma 7.4,
and $\alpha(\beta)$ begins with $10$. Thus, $\mathbf{w}$ cannot contain the
word $11$, and so $\mathbf{w}=U_{0}(\tilde{\mathbf{w}})$ for some
$\tilde{\beta}$-Lyndon word $\tilde{\mathbf{w}}$. Furthermore, by similar
reasoning as in the proof of Lemma 7.4, we obtain

$U_{0}(\tilde{\mathbf{w}}^{\infty})=\mathbf{w}^{\infty}\prec{\mathbf{r}}^{-}\mathbb{L}({\mathbf{r}})^{\infty}=U_{0}\big{(}\tilde{{\mathbf{r}}}^{-}\mathbb{L}(\tilde{{\mathbf{r}}})^{\infty}\big{)}.$

Thus,
$\tilde{\mathbf{w}}^{\infty}\prec\tilde{{\mathbf{r}}}^{-}\mathbb{L}(\tilde{{\mathbf{r}}})^{\infty}$.

Assume without loss of generality that
$\mathbf{z}\in\widetilde{\mathcal{K}}_{\beta}(\mathbf{w})$ begins with $0$. It
then follows in the same way as before that
$\mathbf{z}=U_{0}(\tilde{\mathbf{z}})$ for some sequence $\tilde{\mathbf{z}}$,
and
$\tilde{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\tilde{\beta}}(\tilde{\mathbf{w}})$.
Observe that $\tilde{{\mathbf{r}}}$, $\tilde{\beta}$, $\tilde{\mathbf{w}}$ and
$\tilde{t}_{R}$ satisfy the hypotheses of the lemma. Therefore, by the
induction hypothesis, there is a word $\tilde{\mathbf{v}}$ such that
$\tilde{\mathbf{w}}\tilde{\mathbf{v}}\tilde{\mathbf{z}}\in\widetilde{\mathcal{K}}_{\tilde{\beta}}(\tilde{\mathbf{w}})$.
Set $\mathbf{v}:=U_{0}(\tilde{\mathbf{v}})$; then